NOTICE This is the Summer 2005 version of the Instructor's Solution Manual for Introduction to Graph Theory, by Douglas B. West. A few solutions have been added or clarified since last year's version.
Also present is a (slightly edited) annotated syllabus for the onesemester course taught from this book at the University of Illinois. This version of the Solution Manual contains solutions for 99.4% of the problems in Chapters 1—7 and 93% of the problems in Chapter 8. The author believes that only Problems 4.2.10, 7.1.36, 7.1.37, 7.2.39, 7.2.47, and 7.3.3 1 in Chapters 1—7 are lacking solutions here. There problems are too long or difficult for this text or use concepts not covered in the text; they will be deleted in the third edition. The positions of solutions that have not yet been written into the files are occupied by the statements of the corresponding problems. These problems retain the (—), (!), (+), (*) indicators. Also (.) is added to introduce the statements of problems without other indicators. Thus every problem whose solution is not included is marked by one of the indicators, for ease of identification. The author hopes that the solutions contained herein will be useful to instnictors. The level of detail in solutions varies. Instructors should feel free to write up solutions with more or less detail according to the needs of the class. Please do not leave solutions posted on the web. Due to time limitations, the solutions have not been proofread or edited as carefully as the text, especially in Chapter 8. Please send corrections to [email protected]. The author thanks Fred Galvin in particular for contributing improvements or alternative solutions for many of the problems in the earlier chapters. This will be the last version of the Solution Manual for the second edition of the text, The third edition will have many new problems, such as those posted at http:/Iwww.math.uiuc.edul westligtlnewprob.html. The effort to include all solutions will resume for the third edition. Possibly other pedagogical features may also be added later. Inquiries may be sent to [email protected]. Meanwhile, the author apologizes for any inconvenience caused by the absence of some solutions. Douglas B. West
Mathematics Department - University of Illinois
SYLLABUS FOR INSTRUCTORS Text: West, Introduction to Graph Theory, second edition, Prentice Hall, 2001.
Many students in this course see graph algorithms repeatedly in courses in computer science. Hence this course aims primarily to improve students' writing of proofs in discrete mathematics while learning about the structure of graphs. Some algorithms are presented along the way, and many of the proofs are constructive. The aspect of algorithms emphasized in CS courses is running time; in a mathematics course in graph theory from this book the algorithmic focus is on proving that the algorithms work. Math 412 is intended as a rigorous course that challenges students to think. Homework and tests should require proofs, and most of the exercises in the text do so. The material is interesting, accessible, and applicable; most students who stick with the course will give it a fair amount of time and thought. An important aspect of the course is the clear presentation of solutions, which involves careful writing. Many of the problems in the text have hints, either where the problem is posed or in Appendix C (or both). Producing a solution involves two main steps: finding a proof and properly writing it. It is generally beneficial to the learning process to provide "collaborative study sessions" in which students can discuss homework problems in small groups and an instructor or teaching assistant is available to answer questions and provide direction. Students should then write up clear and complete solutions on their own.
This course works best when students have had prior exposure to writing proofs, as in a "transition" course. Some students may need further explicit discussions of the structure of proofs. Such discussion appear in many texts, such as D'Angelo snd West, Mathematical Thinking: Problem-Solving and Proofs; Eisenberg, The Mathematical Method: A Transition to Advanced Mathematics; Fletcher/Patty, Foundations of Higher Mathematics; Galovich, Introduction to Mathematical Structures; Galovich, Doing Mathematics: An Introduction to Proofs and Problem Solving; Solow, How to Read and Do Proofs.
Suggested Schedule The subject matter for the course is the first seven chapters of the text, skipping most optional materiaL Modifications to this are discussed below. The 22 sections are allotted an average of slightly under two lectures each. In the exercises, problems designated by (—) are easier or shorter than most, often good for tests or for "warmup" before doing homework problems. Problems designated by (-F-) are harder than most. Those designated by (!)
are particularly instructive, entertaining, or important. Those designated by (*) make use of optional material. The semester at the University of Illinois has 43 fifty-minute lectures. The final two lectures are for optional topics, usually chosen by the students from topics in Chapter 8. Chapter 1 Fundamental Concepts 8 Chapter 2 Trees and Distance 5.5 Chapter 3 Matchings and Factors 5.5 Chapter 4 Connectivity and Paths 6 Chapter 5 Graph Coloring 6 Chapter 6 Planar Graphs 5 Chapter 7 Edges and Cycles 5 *
Optional Material No later material requires material marked optional. The "optional" marking also suggests to students that the final examination will not cover
that material. The optional subsections on Disjoint Spanning Trees (Bridg-It) in Section 2.1 and Stable Matchings in Section 3.2 are a'ways quite popthar with the students, The planarity algorithm (without proof) in 6.2 is appealing
to students, as is the notion of embedding graphs on the torus through Example 6.3.21. Our course usually includes these four items. The discussion of f-factors in Section 3.3 is also very instructive and is covered when the class is proceeding on schedule. Other potential additions include the applications of Menger's Theorem at 4.2.24 or 4.2.25. Other items marked optional generally should not be covered in class.
Additional text items not marked optional that can be skipped when behind schedule: 1.1: 31, 35 1.2: 16, 21—23 2.1: 8, 14—16 4,1: 4—6 6.1: 18—20, 28
2.2: 13—19 4,2: 20—21 6.3: 9—10, 13—15
3.2: 4 5.3: 10—11, 16(proof) 7.2: 17
Comments There are several underlying themes in the course, and mentioning these at appropriate moments helps establish continuity. These include 1) TONCAS (The Obvious Necessary Condition(s) are Also Sufficient). 2) Weak duality in dual maximation and minimization problems. 3) Proof techniques such as the use of extremality, the paradigm for inductive proofs of conditional statements, and the technique of transforming a problem into a previously solved problem.
Students sometimes find it strange that so many exercises concern the Petersen graph. This is not so much because of the importance of the Petersen graph itself, but rather because it is a small graph and yet has complex enough structure to permit many interesting exercises to be asked. Chapter 1. In recent years, most students enter the course having been exposed to proof techniques, so reviewing these in the first five sections has become less necessary; remarks in class can emphasis techniques as reminders. To minimize confusion, digraphs should not be mentioned until section 1.4; students absorb the additional model more easily after becoming comfortable with the first. 1.1: p3-6 contaln motivational examples as an overview of the course;
this discussion should not extend past the first day no matter where it ends (the definitions are later repeated where needed). The material on the Petersen graph estabhshes its basic properties for use in later examples and exercises. 1.2: The definitions of path and cycle are intended to be intuitive; one shouldn't dwell on the heaviness of the notation for walks. 1.3: Although characterization of graphic sequences is a classical topic,
some reviewers have questioned its importance. Nevertheless, here is a computation that students enjoy and can perform. 1,4: The examples are presented to motivate the model; these can be skipped to save time. The de Bruijn graph is a classical apphcation. It is desirable to present it, but it takes a while to explaln. Chapter 2. 2,1: Characterization of trees is a good place to ask for input from the class, both in listing properties and in proving equivalence.
2,2: The inductive proof for the Prufer correspondence seems to be easier for most students to grasp than the full bijective it also illustrates the usual type of induction with trees. Most students in the class have seen determinants, but most have considerable difficulty understanding the proof of the Matrix Tree Theorem; given the time involved, it is best
just to state the result and give an example (the next edition will include a purely inductive proof that uses only determinant expansion, not the Cauchy-Binet Formula). Students find the material on graceful labelings enjoyable and illuminating; it doesn't take long, but also it isn't required. The material on branchings should certaily be skipped in this course. 2.3: Many students have seen rooted trees in computer science and find ordinary trees unnatural; Kruskal's algorithm should clarify the distinction. Many CS courses now cover the algorithms of Kruskal, Dijkstra, and Huffrnan; here cover Kruskal and perhaps Dijkstra (many students have seen the algorithm but not a proof of correctness), and skip Huffman, Chapter 3. 3.1: Skip "Dominating Sets", but present the rest of the section. 3.2: Students find the Hungarian algorithm difficult; explicit examples
need to be worked along with the theoretical discussion of the equality subgraph. "Stable Matchings" is very popular with students and should be presented unless far behind in schedule. Skip "Faster Bipartite Matching". 3.3: Present all of the subsection on Tutte's 1-factor Theorem. The material on f-factors is intellectually beautiful and leads to one proof of the Erdös-Gallai conditions, but it is not used again in the course and is an "aside". Skip everything on Edmonds' Blossom Algorithm: matching algorithms in general graphs are important algorithmically but would require too much time in this course; this is "recommended reading". Chapter 4. 4.1: Students have trouble distinguishing "k-connected" from "connec-
tivity k" and "bonds" from "edge cuts". Bonds are dual to cycles in the matroidal sense; there are hints of this in exercises and in Chapter 7, but the full duality cannot be explored before Chapter 8. 4.2: Students find this section a bit difficult. The proof of 4.2.10 is similar to that of 4.2.7, making it omittable, but the application in 4.2.14 is appealing. The details of 4.2.20-21 can be skipped or treated lightiy, since the main issue is the local version of Menger's theorem. 4.2.24-25 are appealing appllcations that can be added; 4.2.5 (CSDR) is a fundamental result but takes a fair amount of effort.
4.3: The material on network flow is quite easy but can take a long time to present due to the overhead of defining new concepts. The basic idea of 4.3.13-15 should be presented without belaboring the details too much, 4.3.16 is a more appealing application that perhaps makes the point more effectively. Skip "Supplies and Demands".
Chapter 5. 5,1: If time is short, the proof of 5.1.22 (Brooks' Theorem) can be merely sketched.
5.2: Be sure to cover Turán's Theorem. Presentation of Dirac's Theorem in 5 +2+20 is valuable as an application of the Fan Lemma (Menger's Theorem). The subsequent material has limited appeal to undergraduates. 5.3: The inclusion-exclusion formula for the chromatic polynomial is derived here (5+3+10) without using inclusion-exclusion, making it accessible to this class without prerequisite. However, this proof is difficult for students to follow in favor of the simple inclusion-exclusion proof, which will be optional since that formula is not prerequisite for the course. Hence this formula should be omitted unless students know inclusion-exclusion. Chordal graphs and perfect graphs are more important, but these can also be treated lightly if short of time, Skip Acyclic Orientations". Chapter 6. 6+1: The technical definitions of objects in the plane should be treated very lightly. It is better to be informal here, without writing out formal
definitions unless explicitly requested by students. Outerplanar graphs are useful as a much easier class on which to solve problems (exercises!) like those on planar graphs; 6.18-20 are fundamental observations about outerplanar graphs, but other items are more important if time is short, 6.1.28 (polyhedra) is an appeahng application but can be skipped. 6.2: The preparatory material 6.2.4-7 on Kuratowski's Theorem can be presented lightly, leaving the annoying details as reading; the subsequent material on convex embedding of 3-connected graphs is much more interesting. Presentation of the planarity algorithm is appealing but optional; skip the proof that it works. 6.3: The four color problem is popular; for undergraduates, show the flaw in Kempe's proof (p271), but don't present the discharging rule unless ahead of schedule, Students find the concept of crossing number easy
to grasp, but the results are fairly difficult; try to go as far as the recursive quartic lower bound for the complete graph+ The edge bound and its geometric application are impressive but take too much time for undergraduates. The idea of embeddings on surfaces can be conveyed through the examples in 6.3+2 1 on the torus. Interested students can be advised to read the rest of this section. Chapter 7. 7.1: The proof of Vizing's Theorem is one of the more difficult in the
course, but undergraduates can gain follow it when it is presented with sufficient colored chalk. The proof can be skipped if short of time. Skip
"Characterization of Line Graphs", although if time and interest is plentiftil the necessity of Krausz's condition can be explained. 7.2: Chvátal's theorem (7.2.13) is not as hard to present as it looks if the instructor has the statement and proof clearly in mind. Nevertheless, the proof is somewhat technical and can be skipped (the same can be said of 7.2.17). More appealing is the Chvátal—Erdôs Theorem (7.2.19), which certainly should be presented. Skip "Cycles in Directed Graphs". 7.3: The theorems of Tait and Grinberg make a nice culmination to the required material of the course. Skip "Snarks" and "Flows and Cycle Covers". Nevertheless, these are lively topics that can be recommended for advanced students.
Chapter 8. If time permits, material from the first part of sections of Chapter 8 can be presented to give the students a glimpse of other topics. The best choices for conveying some understanding in a brief treatment are Section 8.3 (Ramsey Theory or Sperner's Lemma) and Section 8.5 (Random Graphs). Also possible are the Gossip Problem (or other items) from Section 8.4 and some of the optional material from earlier chapters. The first part of Section 8.1 (Perfect Graphs) may also be usable for this purpose if perfect graphs have been discussed in Section 5.3. Sections 8.2 and 8.6 require more investment in preliminary material and thus are less suitable for giving a "glimpse".
Chapter 1: Fundamental Concepts
I.FUNDAMENTAL CONCEPTS 1.1. WHAT IS A GRAPH? 1.1.1. Complete bipartite graphs and complete graphs. The complete bipartite graph Km,n isacompletegraphifandonlyifm = n = lor = . 1.1.2. Adjacency matrices and incidence matrices for a 3-vertex path.
(: ~ ~) (1 ~ 1) (~ ~ l) (l ~) (~ l) (l :) (: l) (~ :) (: ~) Adjacency matrices for a path and a cycle with six vertices.
010100 001010 000101 000010
1.1.3. Adjacency matrix for
010100 001010 000101 100010
:[ffi] 1.1.4. G;:; H if and only if G ;:; H. If f is an isomorphism from G to H, then f is a vertex bijection preserving adjacency and nonadjacency, and hence f preserves non-adjacency and adjacency in G and is an isomorphism from G to H. The same argument applies for the converse, since the complement ofG is G.
Section 1.1: What Is a Graph?
1.1.5. If every vertex of a graph G has degree 2, then G is a cycle—FALSE. Such a graph can be a disconnected graph with each component a cycle. (If infinite graphs are allowed, then the graph can be an infinite path.) 1.1.6. The graph below decomposes into copies of P4.
N1 1.1.7. A graph with more than six vertices of odd degree cannot be decomposed into three paths. Every vertex of odd degree must be the endpoint of some path in a decomposition into paths. Three paths have only six endpoints. 1.1.8. Decompositions of a graph. The graph below decomposes into copies
of K13 with centers at the marked vertices. It decomposes into bold and solid copies of P4 as shown.
1.1.9. A graph and its complement. With vertices labeled as shown, two vertices are adjacent in the graph on the right if and only if they are not adjacent in the graph on the left. b
1.1.10. The complement ofa simple disconnected graph must be connected--
TRUE. A disconnected graph G has vertices x, y that do not belong to a path. Hencex and y are adjacent in G. Furthermore, x and y have no common neighbor in G, since that would yield a path connecting them. Hence
Chapter 1: Fundamental Concepts
every vertex not in is adjacent in toat least one of . Hence every vertex can reach every other vertex in G using paths through . 1.1.11. Maximum clique and maximum independent set. Since two vertices have degree 3 and there are only four other vertices, there is no clique of size 5. A complete subgraph with four vertices is shown in bold. Since two vertices are adjacent to all others, an independent set con-
taining either of them has only one vertex. Deleting them leaves P4, in which the maximum size of an independent set is two, as marked.
1.1.12. The Petersen graph. The Petersen graph contains odd cycles, so it is not bipartite; for example, the vertices 12, 34, 51, 23, 45 form a 5-cycle. The vertices 12, 13, 14, 15 form an independent set of size 4, since any two of these vertices have 1 as a common element and hence are nonadjacent. Visually, there is an independent set of size 4 marked on the drawing
of the Petersen graph on the cover of the book. There are many ways to show that the graph has no larger independent set. Proof 1. Two consecutive vertices on a cycle cannot both appear in an independent set, so every cycle contributes at most half its vertices. Since the vertex set is covered by two disjoint 5-cycles, every independent set has size at most 4. Proof 2. Let ab be a vertex in an independent set 8, where a, b a [5].
leaves a subboard that cannot be partitioned into rectangles consisting of two adjacent unit squares. 2-coloring the squares of a checkerboard so that adjacent squares have opposite colors shows that the graph having a vertex for each square and an edge for each pair of adjacent squares is bipartite. The squares at opposite corners have the same color; when these are deleted, there are 30 squares of that color and 32 of the other
Section 1.1: What Is a Graph?
color. Each pair of adjacent squares has one of each color, so the remaining squares cannot be partitioned into sets of this type.
Generalization: the two partite sets of a bipartite graph cannot be matched up using pairwise-disjoint edges if the two partite sets have unequal sizes. 1.1.15. Common graphs in four families: .
1.1.16. The graphs below are not isomorphic. The graph on the left has four cliques of size 4, and the graph on the right has only two. Alternatively, the complement of the graph on the left is disconnected (two 4-cycles), while the complement of the graph on the right is connected (one 8-cycle).
1.1.17. There are exactly two isomorphism classes of 4-regular simple graphs with 7 vertices. Simple graphs G and H are isomorphic if and only if their complements G and H are isomorphic, because an isomorV(G) —* V(H) is also an isomorphism from G to H, and vice phism versa. Hence it suffices to count the isomorphism classes of 2-regular simple graphs with 7 vertices. Every component of a finite 2-regular graph is a cycle. In a simple graph, each cycle has at least three vertices. Hence each class it determined by partitioning 7 into integers of size at least 3 to be the sizes of the cycles. The only two graphs that result are C7 and C3 + C4 a
single cycle or two cycles of lengths three and four.
1.1.18. Isomorphism. Using the correspondence indicated below, the first ±* v1 if and two graphs are isomorphic; the graphs are bipartite, with only if i j. The third graph contains odd cycles and hence is not isomorphic to the others.
Visually, the first two graphs are Qa and the graph obtained by deleting four disjoint edges from K4,4. In each vertex is adjacent to the vertices whose names have opposite parity of the number of is, except for the complementary vertex. Hence Q g also has the structure of K4,4 with four disjoint edges deleted; this proves isomorphism without an explicit bijection.
i.i.19. Isomorphism of graphs. The rightmost two graphs below are isomorphic. The outside iO-cycle in the rightmost graph corresponds to the intermediate ring in the second graph. Pulling one of the inner 5-cycles of the rightmost graph out to the outside transforms the graph into the same drawing as the second graph. The graph on the left is bipartite, as shown by marking one partite set. It cannot be isomorphic to the others, since they contain 5-cycles.
i.i.20. Among the graphs below, the first (F) and third (H) are isomorphic, and the middle graph (G) is not isomorphic to either of these. F and H are isomorphic. We exhibit an isomorphism (a bijection from V(F) to V(H) that preserves the adjacency relation). To do this, we name the vertices of F, write the name of each vertex of F on the corresponding vertex in H, and show that the names of the edges are the same in H and F. This proves that H is a way to redraw F. We have done this below using the first eight letters and the first eight natural numbers as names. To prove quickly that the adjacency relation is preserved, observe that i, o, 2, b, 3, c, 4, ci, 5, e, 6, f, 7, g, 8, h is a cycle in both drawings, and the additional edges ic, 2d, 3e, 4f, Sg, 6h, 7a, 8b are also the same in both drawings. Thus the two graphs have the same edges under this vertex correspondence. Equivalently, if we list the vertices in this specified order for
Section 1.1: What Is a Graph?
the two drawings, the two adjacency matrices are the same, but that is harder to verif&.
G is not isomorphic to F or to H. In F and in H, the numbers form an independent set, as do the letters. Thus F and H are bipartite. The graph G cannot be bipartite, since it contains an odd cycle. The vertices above the horizontal axis of the picture induce a cycle of length 7. It is also true that the mid&e graph contains a 4-cycle and the others do not, but it is harder to prove the absence of a 4-cycle than to prove the absence of an odd cycle.
1.1.21. Isomorphism. Both graphs are bipartite, as shown below by mark-
ing one partite set. In the graph on the right, every vertex appears in eight 4-cycles. In the graph on the left, every vertex appears in only six 4-cycles (it is enough just to check one). Thus they are not isomorphic. Alternatively, for every vertex in the right graph there are five vertices having common neighbors with it, while in the left graph there are six such vertices.
1.1.22. Isoinorphism of explicit graphs.
Among the graphs below,
are pairwise isomorphic. Also G3 G4, and these are not isomorphic to any of the others. Thus there are exactly two isomorphism classes represented among these graphs. To prove these statements, one can present explicit bijections between vertex sets and veri& that these preserve the adjacency relation (such as by displaying the adjacency matrix, for example). One can also make other stnictural arguments. For example, one can move the highest vertex in G3 down into the middle of the picture to obtain G4; from this one can list the desired bijection.
Chapter 1: Fundamental Concepts
One can also recall that two graphs are isomorphic if and only if their complements are isomorphic. The complements of G1, G2, andG5 are cycles of length 7, which are pairwise isomorphic. Each of G3 and G4 consists
of two components that are cycles of lengths 3 and 4; these graphs are isomorphic to each other but not to a 7-cycle.
1.1.23. Smallest pairs of nonisomorphie graphs with the same vertex degrees For multigraphs, loopless multigraphs, and simple graphs, the required numbers of vertices are 2,4,5; constructions for the upper bounds appear below. We must prove that these constructions are smallest.
a) general b) loopless c) simple a) With 1 vertex, every edge is a loop, and the isomorphism class is
determined by the number of edges, which is determined by the vertex degree. Hence nonisomorphic graphs with the same vertex degrees have at least two vertices. b) Every loopless graph is a graph, so the answer for loopless graphs is at least 2. The isomorphism class of a loopless graph with two vertices is determined by the number of copies of the edge, which is determined by the vertex degrees. The isomorphism class of a loopless graph with three vertices is determined by the edge multiplicities. Let the three vertex degrees be a, b, c, and let the multiphcities of the opposite edges be x, y, z, where Since a = y -f- z, 1' = x + z, and c = x + y, we can solve for the multiplicities in terms of the degrees by x = (b + c a)/2, y = (a + c and z = (a + b c)/2. Hence the multiplicities are determined by the
degrees, and all loopless graphs with vertex degrees a, b, c are pairwise isomorphic. Hence nonisomorphic loopless graphs with the same vertex degrees have at least four vertices. c) Since a simple graph is a loopless graph, the answer for simple graphs is at least 4. There are 11 isomorphism classes of simple graphs with four vertices. For each of 0,1,5, or 6 edges, there is only one isomorphism class. For 2 edges, there are two isomorphism classes, but they have
Section 1.1: What Is a Graph?
different lists of vertex degrees (similarly for 4 edges). For 3 edges, the three isomorphism classes have degree lists 3111, 2220, and 2211, all different. Hence nonisomorphic simple graphs with the same vertex degrees must have at least five vertices. 1.1.24. Isomorphisrns for the Petersen graph. Isomorphism is proved by giving an adjacency-preserving bijection between the vertex sets. For pictorial representations of graphs, this is equivalent to labeling the two graphs
with the same vertex labels so that the adjacency relation is the same in both pictures; the labels correspond to a permutation of the rows and columns of the adjacency matrices to make them identical. The various drawings of the Petersen graph below illustrate its symmetries; the labelings indicate that these are all "the same" (uniabeled) graph. The number of isomorphisms from one graph to another is the same as the number of isomorphisms from the graph to itself. 24
1.1.25. The Petersen graph has no cycle of length 7. Suppose that the Petersen graph has a cycle C of length 7. Since any two vertices of C are connected by a path of length at most 3 on C, any additional edge with endpoints on C would create a cycle of length at most 4. Hence the third neighbor of each vertex on C is not on C.
Chapter 1: Fundamental Concepts
Thus there are seven edges from V (C) to the remaining three vertices, By the pigeonhole principle, one of the remaining vertices receives at least
three of these edges. This vertex x not on C has three neighbors on C. For any three vertices on C, either two are adjacent or two have a common neighbor on C (again the pigeonhole principle applies). Using x, this completes a cycle of length at most 4. We have shown that the assumption of a 7-cycle leads to a contradiction. Alternative completion of proof Let u be a vertex on C, and let v, w be the two vertices farthest from u on C. As argued earlier, no edges join vertices of C that are not consecutive on C. Thus u is not adjacent to v or w. Hence u, v have a common neighbor off C, as do u, w. Since u has only one neighbor off C, these two common neighbors are the same. The resulting vertex x is adjacent to all of u, v, w, and now x, v, w is a 3-cycle. 1.1.26. A k-regular graph ofgirth four has at least 2k vertices, with equality only for Kk,k. Let G be k-regular of girth four, and chose xy c E(G). Girth 4 implies that G is simple and that x and y have no common neighbors. Thus the neighborhoods N(x) and N(y) are disjoint sets of size k, which forces at least 2k vertices into G. Possibly there are others.
Note also that N(x) and N(y) are independent sets, since G has no triangle. If G has no vertices other than these, then the vertices in N(x) can have neighbors only in N(y). Since G is k-regular, every vertex of N(x)
must be adjacent to every vertex of N(y). Thus G is isomorphic to with partite sets N(x) and N(y). In other words, there is only one such isomorphism class for each value of k.
Comment. One can also start with a vertex x, choose y from among the k vertices in N(x), and observe that N(y) must have k 1 more vertices not in N(x) U tx>. The proof then proceeds as above. (An alternative proof uses the methods of Section 1.3. A triangle-free simple graph with a vertices has at most n2/4 edges. Since G is k-regular, this yields n2/4 > nk/2, and hence a > 2k. Furthermore, equality holds in the edge bound only for so this is the only such graph with 2k vertices, (C. Pikscher))
1.1.27. A simple graph of girth 5 in which every vertex has degree at least k has at least k2 -F- 1 vertices, with equality achieveable when k c . Let
G be k-regular of girth five. Let S be the set consisting of a vertex x and
Section 1.1: What Is a Graph?
its neighbors. Since G has no cycle of length less than five, G is simple, and any two neighbors of x are nonadjacent and have no common neighbor other than x. Hence each y c S has at least k 1 neighbors that are not in S and not neighbors of any vertex in S. Hence G has at least k(k 1) vertices outside S and at least k + 1 vertices in S for at least k2 + 1 altogether. The 5-cycle achieves equality when k = 2. Fork = 3, growing the graph symmetrically from x permits completing the graph by adding edges among
the non-neighbors of x. The result is the Petersen graph. (Comment: For k > 3, it is known that girth S with minimum degree k and exactly k2 + 1 vertices is impossible, except for k = 7 and possibly for k = 57.)
1.1.28. The Odd Graph has girth 6. The Odd Graph °k is the disjointness graph of the set of k-element subsets of 112k + 1]. Vertices with a common neighbor correspond to k-sets with k
mon elements. Thus they have exactly one common neighbor, and °k has no 4-cycle. Two vertices at distance 2 from a single vertex have at least k 2 common neighbors. For k > 2, such vertices cannot be adjacent, and thus °k has no 5-cycle when k > 2. To form a 6-cycle when k > 2, let
The Odd Graph also is not bipartite.
The successive elements
1.1.29. Among any 6 people, there are 3 mutual acquaintances orB mutual strangers. Let G be the graph of the acquaintance relation, and let x be one of the people. Since x has 5 potential neighbors, x has at least 3 neighbors or at least 3 nonneighbors. By symmetry (if we complement G, we still
have to prove the same statement), we may assume that x has at least 3 neighbors. If any pair of these people are acquainted, then with x we have 3 mutual acquaintances, but if no pair of neighbors of x is acquainted, then the neighbors of x are three mutual strangers. 1.1.30. The number of edges incident to is the ith diagonal entry in MMT and in A2. In both MMT and A2 this is the sum of the squares of the entries
Chapter 1: Fundamental Concepts
in the ith row. For MMT, this follows immediately from the definition of matrix multiplication and transposition, but for A2 this uses the graphtheoretic fact that A = AT; i.e. column i is the same as row i. Because
G is simple, the entries of the matrix are all 0 or 1, so the sum of the squares in a row equals the number of is in the row In M, the is in a row denote incident edges; in A they denote vertex neighbors. In either case, the number of is is the degree of the vertex. If i j, then the entry in position (1, j) of A2 is the number of common neighbors of v, and The matrix multiplication puts into position (i, J) the "product" of row i and column j; that is When G is simple, entries in A are i or 0, depending on whether the corresponding vertices = i if Vk is a common neighbor of and v1; are adjacent. Hence otherwise, the contribution is 0. Thus the number of contributions of i to entry (i, j) is the number of common neighbos of v, and v1. If i j, then the entry in position (i, J) of MMT is the number of edges joining v, and v1 (0 or 1 when G has no multiple edges). The ith row of
M has is corresponding to the edges incident to v,. The jth column of MT is the same as the jth row of M, which has is corresponding to the edges incident to v. Summing the products of corresponding entries will contribute i for each edge incident to both v, and v1; 0 otherwise. Comment. For graphs without loops, both arguments for (1, j) in general apply when i = j to explain the diagonal entries. 1.1.31. K,, decomposes into two isomorphic ("self-complementary") subgraphs if and only if a or a i is divisible by 4. a) The number of vertices in a self complementary graph is congruent to 0 or i (mod 4). If G and G are isomorphic, then they have the same number of edges, but together they have edges (with none repeated), so the number of edges in each must be n(n i)/4. Since this is an integer and the numbers a and a i are not both even, one of must be
divisible by 4. b) Construction of self complementary graphs for all such a.
Proof 1 (explicit construction). We generalize the structure of the self-complementary graphs on 4 and 5 vertices, which are P4 and C5. For a = 4k, take four vertex sets of size k, say X1, X2, Xa, X4, and join all vertices of X, to those of fori = i, 2,3. To the rest of G, within these sets let X1 and X4 induce copies of a graph H withk vertices, and let X2 and X3 induce H. (For example, H may be K4.) In G, both X2 and X3 induce H, while X1 and X4 induce H, and the connections between sets are X2 ÷* X4 ÷* X1 ÷* Xa. Thus relabellng the subsets defines an isomorphism between G and G. (There are still other constructions for G.)
Section 1.1: What Is a Graph?
For a = 4k + 1, we add a vertex x to the graph constructed above. Join x to the 2k vertices in Xi and X4 to form G. The isomorphism showing that G x is self-complementary also works for G (with x mapped to itself'), since this isomorphism maps N0(x) = Xi U X4 to Nw(x) = X2 U X3. Proof 2 (inductive construction). If G is self-complementary, then let H1 be the graph obtained from G and P4 by joining the two ends of P4 to
all vertices of G. Let H2 be the graph obtained from G and P4 by joining the two center vertices of P4 to all vertices of G. Both H1 and 112 are self-complementary. Using this with G = K1 produces the two selfcomplementary graphs of order 5, namely C5 and the bull. Self-complementary graphs with order divisible by 4 arise from re-
peated use of the above using G =
as a starling point, and self-
complementary graphs of order congruent to 1 modulo 4 arise from repeated use of the above using G = K1 as a starling point. This construction produces many more self-complementary graphs than the explicit construction in Proof 1. 1.1.32. K,n,n decomposes into two isomorphic subgraphs if and only if m and a are not both odd. The condition is necessary because the number of edges must be even. It is sufficient because Km,n decomposes into two copies of Km,n/2 when a is even.
1.1.33. Decomposition of complete graphs into cycles through all vertices. View the vertex set of Kn as 4, the values 0, . a 1 in cyclic order. Since each vertex has degree a 1 and each cycle uses two edges at each vertex, the decomposition has (a 1)/2 cycles. For a = 5 and a = 7, it suffices to use cycles formed by traversing the vertices with constant difference: (0, 1, 2, 3, 4) and (0, 2, 4, 1, 3) for a = 5 and (0, 1, 2, 3, 4, 5, 6), (0, 2, 4, 6, 1, 3, 5), and (0, 3, 6, 2, 5, 1, 4) for a = 7. This construction fails for a = 9 since the edges with difference 3 form three 3-cycles. The cyclically symmetric construction below treats the vertex set as 4 together with one special vertex.
Chapter 1: Fundamental Concepts
1.1.34. Decomposition of the Petersen graph into copies of P4. Consider the drawing of the Petersen graph with an inner 5-cycle and outer 5-cycle. Each P4 consists of one edge from each cycle and one cross edge joining them. Extend each cross edge e to a copy of P4 by taking the edge on each of the two 5-cycles that goes in a clockwise direction from e. In this way, the edges on the outside 5-cycle are used in distinct copies of P4, and the same holds for the edges on the inside 5-cycle. Decomposition of the Petersen graph into three pairwise-isomorphic connected subgraphs. Three such decompositions are shown below. We restricted the search by seeking a decomposition that is unchanged by 120° rotations in a drawing of the Petersen graph with 3-fold rotational symme-
try The edges in this drawing form classes of size 3 that are unchanged under rotations of 1200; each subgraph in the decomposition uses exactly one edge from each class.
decomposes into three pairwise-isomorphic subgraphs if and
only if a -F- 1 is not divisible by .3. The number of edges is n(n
is divisible by 3, then a and a 1 are not divisible by 3. Thus decomposition into three subgraphs of equal size is impossible in this case. If a + 1 is not divisible by 3, then e(K0) is divisible by 3, since a or a 1 is divisible by 3. We construct a decomposition into three subgraphs that are pairwise isomorphic (there are many such decompositions). When a is a multiple of 3, we partition the vertex set into three subsets Vi, V2, V3 of equal size. Edges now have two types: within a set or joining two sets, Let the ith subgraph consist of all the edges within V, and all the edges joining the two other subsets. Each edge of K0 appears in exactly
Section 1.1: What Is a Graph?
one of these subgraphs, and each G, is isomorphic to the disjoint union of and When a 1 (mod 3), consider one vertex w. Since a 1 is a multiple of 3, we can form the subgraphs G, as above on the remaining a 1 vertices. G, to form H, by joining w to every vertex of Each edge involving w has been added to exactly one of the three subgraphs. Each H, is isomorphic to the disjoint union of and
1.1.36. If decomposes into triangles, then a 1 or a 3 is divisible by 6. Such a decomposition requires that the degree of each vertex is even and the number of edges is divisible by 3. To have even degree, a must be odd. Also a(a 1)/2 is a multiple of 3, so3 divides a or a 1. If 3 divides a and a is odd, then a 3 is divisible by 6. If 3 divides a 1 and a is odd, then a 1 is divisible by 6.
1.1.37. A graph in which every vertex has degree 3 has no decomposition into paths with at least five vertices each. Suppose that G has such a decomposition. Since every vertex has degree 3, each vertex is an endpoint
of at least one of the paths and is an internal vertex on at most one of them. Since every path in the decomposition has two endpoints and has at least three internal vertices, we conclude that the number of paths in the decomposition is at least a(G)/2 and is at most a(G)/3, which is impossible. Alternatively, let k be the number of paths. There are 2k endpoints of paths. On the other hand, since each internal vertex on a path in the decomposition must be an endpoint of some other path in the decomposition,
there must be at least 3k endpoints of paths. The contradiction implies that there cannot be such a decomposition. 1.1.38. A 3-regular graph G has a decomposition into claws if and only if G is bipartite. When G is bipartite, we produce a decomposition into claws. We use all claws obtained by taking the three edges incident with a single vertex in the first partite set. Each claw uses all the edges incident to its central vertex. Since each edge has exactly one endpoint in the first partite set, each edge appears in exactly one of these claws. When G has a decomposition into claws, we partition V(G) into two independent sets. Let X be the set of centers of the claws in the decomposition. Since every vertex has degree 3, each claw in the decomposition
Chapter 1: Fundamental Concepts
uses all edges incident to its center. Since each edge is in at most one claw, this makes X an independent set. The remaining vertices also form an independent set, because every edge is in some claw in the decomposition, which means that one of its endpoints must be the center of that claw. 1.1.39. Graphs that decompose K6.
Triangle—No. A graph decomposing into triangles must have even degree at each vertex. (This excludes all decompositions into cycles.) Paw, P5—No. K6 has 15 edges, but each paw or P5 has four edges. House, Bowtie, Dart—No. K6 has 15 edges, but each house, bowtie, or dart has six edges. Claw—Yes. Put five vertices 0, 1, 2, 3, 4 on a circle and the other vertex z in the center. For i c , use a claw with edges from i to i + 1, I + 2, and z. Each edge appears in exactly one of these claws. Kite—Yes. Put all six vertices on a circle. Each kite uses two opposite edges on the outside, one diagonal, and two opposite edges of "length" 2. Three rotations of the picture complete the decomposition. Bull—Yes. The bull has five edges, so we need three bulls. Each bull uses degrees 3, 3, 2, 1, 1, 0 at the six vertices. Each bull misses one vertex, and each vertex of K6 has five incident edges, so three of the vertices will receive degrees 3, 2, 0 from the three bulls, and the other three will receive degrees 3, 1, 1. Thus we use vertices of two types, which leads us to position them on the inside and outside as on the right below. The bold, solid, and dashed bulls obtained by rotation complete the decomposition.
1.1.40. Automorphisms of and A path can be left alone or ifipped, a cycle can be rotated or ifipped, and a complete graph can be permuted arbtrarily. The numbers of automorphisms are 2, 2n, a!, respectively. Correspondingly, the numbers of distinct labelings using vertex set En] are n!/2, (a 1)!/2, 1, respectively. For these formulas require a > 1. 1.1.4 1. Graphs with one and three automorphisms. The two graphs on the left have six vertices and only the identity automorphism. The two graphs on the right have three automorphisms.
Section 1.1: What Is a Graph?
1.1.42. The set of automorphisms of a graph G satisfies the following:
a) The composition of two automorphisms is an automorphism. b) The identity permutation is an automorphism. c) The inverse of an automorphism is also an automorphism. d) Composition of automorphisms satisfies the associative property. The first three properties are essentially the same as the transitive, reflexive, and symmetric properties for the isomorphism relation; see the discussion of these in the text. The fourth property holds because composition of functions always satisfies the associative property (see the discussion of composition in Appendix A). 1.1.43. Every automorphism of the Petersen graph maps the 5-cycle
(12,34,51,23,45) into a 5-cycle with vertices ab, cd, ea, bc, de by a permutation of [5] tahing 1,2,3,4,5 to a, b, c, d, e, respectively. Let a denote the automorphism, and let the vertex ab be the image of the vertex 12 under a. The image of 34 must be a pair disjoint from ab, so we may let cd = a(34). The third vertex must be disjoint from the second and share an element with the first. We may select a to be the common element in the first and third vertices. Similarly, we may select c to be the common element in the second and fourth vertices. Since nonadjacent vertices correspond to sets with a common element, the other element of the fourth vertex must be Ii, and the fifth vertex can't have a or b and must have d and e. Thus every 5-cycle must have this form and is the image of (12,34,51,23,45) under the specified permutation a,
The Peterson graph has 120 automorphisms. Every permutation of [5] preserves the disjointness relation on 2-element subsets and therefore defines an automorphism of the Petersen graph. Thus there are at least 120 automorphism. To show that there are no others, consider an arbitrary automorphism a. By the preceding paragraph, the 5-cycle C maps to some 5-cycle (ab, cd, ea, bc, de). This defines a permutation f mapping 1, 2, 3, 4, 5 to a, b, c, d, e, respectively. It suffices to show that the other vertices must also have images under a that are described by f. The remaining vertices are pairs consisting of two nonconsecutive values modulo 5. By symmetry, it suffices to consider just one of them, say 24. The only vertex of C that 24 is adjacent to (disjoint from) is 51. Since
Chapter 1: Fundamental Concepts
ea, and the only vertex not on (ab, cd, ea, bc, de) that is adjacent to ea is bd, we must have cr(24) = bd, as desired.
P = (uo, u1, U2, and Q = v1, v2, v3) in the Petersen graph, there is an automorphism of the Petersen graph that turns P into Q. In the disjointness representation of the Petersen graph, suppose the pairs corresponding to the vertices of P are ob, cd, ef gh, respectively. Since consecutive pairs are disjoint and the edges are unordered pairs, we may write the pairs so that a, b, c, d, e are distinct, f = a, g = b, and h = c, Putting the vertex names of Q in the same format AR, CD, EF, G o H, we chose the isomorphism generated by 1.1.44. For each pair of 3-edge paths (vo,
the permutation of 115] that turns a, b, c, d, e into A, B, C, D, F, respectively.
1.1.45. A graph with 12 vertices in which every vertex has degree Band the only automorphism is the identity. p
There are many ways to prove that an automorphism must fix all the vertices. The graph has only two triangles (abc and uvw). Now an automorphism must fix p, since is the only vertex having no neighbor on a triangle, and also e, since it is the only vertex with neighbors on both triangles. Now d is the unique common neighbor of p and e. The remaining vertices can be fixed iteratively in the same way, by finding each as the only unlabeled vertex with a specified neighborhood among the vertices already fixed. (This
construction was provided by Luis Dissett, and the argument forbidding nontrivial automorphisms was shortened by Fred Galvin, Another such graph with three triangles was found by a student of Fred Galvin.) 1.1.46. Vertex-transitivity and edge-transitivity. The graph on the left in Exercise 1.1.21 is isomorphic to the 4-dimensional hypercube (see Section
1.3), which is vertex-transitive and edge-transitive via the permutation of coordinates. For the graph on the right, rotation and inside-out exchange takes care of vertex-transitivity. One further generating operation is needed to get edge-transitivity; the two bottom outside vertices can be switched with the two bottom inside vertices. 1.1.47. Edge-transitive versus vertex-transitive, a) If G is obtained from with a >- 4 by replacing each edge of with a path of two edges through
Section 1.2: Paths, Cycles, and Trails
a new vertex of degree 2, then G is edge-transitive but not vertex-transitive. Every edge consists of an old vertex and a new vertex. The a! permutations of old vertices yield automorphism. Let x&y denote the new vertex on the path replacing the old edge xy; note that x&y = y&x. The edge joining x and x&y is mapped to the edge joining u and u&v by any automorphism that maps x to u and y to v. The graph is not vertex-transitive, since x&y has degree 2, while x has degree a 1. b) If G is edge-transitive but not vertex-transitive and has no isolated vertices, then G is bipartite. Let uv be an arbitrary edge of G. Let S be the set of vertices to which u is mapped by automorphisms of G, and let T be the set of vertices to which v is mapped. Since G is edge-transitive and has no isolated vertex, S U T = V(G). Since G is not vertex-transitive, S V(G). Together, these statements yield S P T = 0, since the composition of two automorphisms is an automorphism. By edge-transitivity, every edge of G contains one vertex of S and one vertex ofT. Since S P T = 0, this implies that G is bipartite with vertex bipartition 5, T. c) The graph below is vertex-transitive but not edge-transitive. A composition of left-right reflections and vertical rotations can take each vertex to any other. The graph has some edges on triangles and some edges not on triangles, so it cannot be edge-transitive.
1.2. PATHS, CYCLES, AND TRAILS 1.2.1. Statements about connection.
a) Every disconnected graph has an isolated vertex—FALSE. A simple 4-vertex graph in which every vertex has degree 1 is disconnected and has no isolated vertex. b) A graph is connected if and only if some vertex is connected to all other vertices—TRUE. A vertex is "connected to" another if they lie in a common path. If G is connected, then by definition each vertex is connected to every other. If some vertex x is connected to every other, then because a u, x-path and x, v-path together contain a u, v-path, every vertex is connected to every other, and G is connected.
Chapter 1: Fundamental Concepts
c) The edge set of every closed trail can be partitioned into edge sets of cycles—TRUE. The vertices and edges of a closed trail form an even graph, and Proposition 1.2.27 applies.
d) If a maximal trail in a graph is not closed, then its endpoints have odd degree. If an endpoint v is different from the other endpoint, then the trail uses an odd number of edges incident to v. If v has even degree, then there remains an incident edge at v on which to extend the trail. 1.2.2. Walks in K4, a) K4 has a walk that is not a trail; repeat an edge. b) K4 has a trail that is not closed and is not a path; traverse a triangle and then one additional edge. c) The closed trails in K4 that are not cycles are single vertices. A closed
trail has even vertex degrees; in K4 this requires degrees 2 or 0, which forbids connected nontrivial graphs that are not cycles. By convention, a single vertex forms a closed trail that is not a cycle. 1.2.3. The non-coprimality graph with vertex set < 1. 15>.
1,11,13 are isolated. The remainder induce a single component. It has a spanning path 7,14,10,5,15,3,9,12,8,6,4,2. Thus there are four components, and the maximal path length is 11. 1.2.4. Effect on the adjacency and incidence matrices of deleting a vertex or edge. Assume that the graph has no loops. Consider the vertex ordering v1 v. Deleting edge v1v1 simply deletes the corresponding column of the incidence matrix; in the adjacency matrix it reduces positions i, j and j, i by one. Deleting a vertex v, ehminates the ith row of the incidence matrix, and it also deletes the column for each edge incident to In the adjacency matrix, the ith row and ith column vanish, and there is no effect on the rest of the matrix. 1.2.5. If v is a vertex in a connected graph G, then v has a neighbor in every component of G v. Since G is connected, the vertices in one component of G v must have paths in G to every other component of G v, and a path can only leave a component of G v via v. Hence v has a neighbor in each component. No cut-vertex has degree 1. If G is connected and G v has k components, then having a neighbor in each such component yields dG(v) > k. If v is a cut-vertex, then k > 2, and hence d0(v) > 2.
1.2.6. The paw. Maximal paths: acb, abcd, bacd (two are maximum paths). Maximal cliques: abc, cd (one is a maximum clique). Maximal independent sets: c, bd, ad (two are maximum independent sets).
Section 1.2: Paths, Cycles, and Trails
1.2.7. A bipartite graph has a unique bipartition (except for interchanging the two partite sets) if and only if it is connected. Let G be a bipartite graph. If u and v are vertices in distinct components, then there is a bipartition in which u and v are in the same partite set and another in which they are in opposite partite sets. If G is connected, then from a fixed vertex u we can walk to all other vertices. A vertex v must be in the same partite set as u if there is a u, vwalk of even length, and it must be in the opposite set if there is a u, v-walk of odd length. 1.2.8. The biclique Km,n is Eulerian if and only if m and a are both even or one of them is 0. The graph is connected. It vertices have degrees m and a (if both are nonzero), which are all even if and only if in and a are both even. When in or a is 0, the graph has no edges and is Eulerian.
1.2.9. The minimum number of trails that decompose the Petersen graph is 5. The Petersen graph has exactly 10 vertices of odd degree, so it needs at least 5 trails, and Theorem 1.2.33 implies that five trails suffice. The Petersen graph does have a decomposition into five paths. Given the drawing of the Petersen graph consisting of two disjoint 5-cycles and edges between them, form paths consisting of one edge from each cycle and one edge joining them.
1.2.10. Statements about Eulerian graphs. a) Every Eulerian bipartite graph has an even number ofedges—TRUE. Proof 1. Every vertex has even degree. We can count the edges by summing the degrees of the vertices in one partite set; this counts every edge exactly once. Since the summands are all even, the total is also even. Proof 2. Since every walk alternates between the partite sets, following an Eulerian circuit and ending at the initial vertex requires taking an even number of steps. Proof 3. Every Eulerian graph has even vertex degrees and decomposes into cycles. In a bipartite graph, every cycle has even length. Hence the number of edges is a sum of even numbers. b) Every Eulerian simple graph with an even number of vertices has an even number of edges—FALSE. The union of an even cycle and an odd cycle that share one vertex is an Eulerian graph with an even number of vertices and an odd number of edges.
Chapter 1: Fundamental Concepts
1.2.11. If G is an Eulerian graph with edges e, f that share a vertex, then G need not have an Eulerian circuit in which e, f appear consecutively. If G consists of two edge-disjoint cycles sharing one common vertex v, then edges incident to v that belong to the same cycle cannot appear consecutively on an Eulerian circuit. 1.2.12. Algorithm for Eulerian circuit. We convert the proof by extremality to an iterative algorithm. Assume that G is a connected even graph. Initialize T to be a closed trail of length 0; a single vertex. If T is not all of G, we traverse T to reach a vertex v on T that is incident to an edge e not in T. Beginning from v along e, traversing an arbitrary trail T' not using edges of T; eventually the trail cannot be extended. Since G E(T) is an even graph, this can only happen upon a return to the original vertex v, completing a closed trail. Splice T' into T by traversing T up to v, then following T', then the rest ofT. If this new trail includes all of E(G), then it is an Eulerian circuit, and we stop. Otherwise, let this new trail be T and repeat the iterative step. Since each successive trail is longer and G has finitely many edges, the procedure must terminate. It can only terminate when an Eulerian circuit has been found.
1.2.13. Each u, v-walk contains a u, v-path. a) (induction). We use ordinary induction on the length 1 of the walk, proving the statement for all pairs of vertices. A u, v-walk of length 1 is a u, v-path of length 1; this provides the basis. For the induction step, suppose 1> 1, and let W be a u, v-walk of length I; the induction hypothesis is that walks of length less than 1 contain paths linking their endpoints. If u = v, the desired path has length 0. If u v, let wv be the last edge of W, and let W' be the u, w-walk obtained by deleting wv from W. Since W' has length 1 1, the induction hypothesis guarantees a u, w-path P in W'. If w = v, then P is the desired u, v-path. If w v and v is not on P, then we
extend P by the edge wv to obtain a u, v-path. If w v and v is on P, then P contains a u, v-path. In each case, the edges of the u, v-path we construct all belong toW.
b) (extremality) Given a u, v-walk W, consider a shortest u, v-walk W'
contained in W. If this is not a path, then it has a repeated vertex, and the portion between the instances of one vertex can be removed to obtain a shorter u, v-walk in W than W1.
Section 1.2: Paths, Cycles, and Trails
1.2.14. The union of the edge sets of distinct u, v-paths contains a cycle.
Proof 1 (extremality). Let P and Q be distinct u, v-paths. Since a path in a simple graph is determined by its set of edges, we may assume (by symmetry) that P has an edge e not belonging to Q. Within the portion of P before P traverses e, let y be the last vertex that belongs to Q. Within the portion of P after P traverses e, let z be the first vertex that belongs to Q. The vertices y, z exist, because u, v E V(Q). The y, z-subpath of P combines with the y, z- or z, y-subpath of Q to form a cycle, since this subpath of Q contains no vertex of P between y and z. Proof 2 (induction). We use induction on the sum l of the lengths of the two paths, for all vertex pairs simultaneously. If P and Q are u, vpaths, then l 2. If 1 = 2, then we have distinct edges consisting of u and v, and together they form a cycle of length 2. For the induction step, suppose l > 2. If P and Q have no common internal vertices, then their union is a cycle. If P and Q have a common internal vertex w, then let P', P" be the u, w-subpath of P and the w, v-subpath of P. Similarly define Q', Q". Then P', Q' are u, w-paths with total length less than I. Similarly, P", Q" are w, v-paths with total length less than 1. Since P, Q are distinct, we must have P', Q' distinct or P", Q" distinct. We can apply the induction hypothesis to the pair that is a pair of distinct paths joining the same endpoints. This pair contains the edges of a cycle, by the induction hypothesis, which in turn is contained in the union of P and Q. The union of distinct u, v-walks need not contain a cycle. Let G = P3, with vertices u, x, v in order. The distinct u, v-walks with vertex lists u, x, u, x, v and u, x, v, x, v do not contain a cycle in their union.
1.2.15. If W is a nontrivial closed walk that does not contain a cycle, then some edge of W occurs twice in succession (once in each direction). Proof 1 (induction on the length 1 of W). We are given 1> 1. A closed walk of length 1 is a loop, which is a cycle. Thus we may assume 1 > 2. Basis step: 1 = 2. Since it contains no cycle, the walk must take a step
and return immediately on the same edge. Induction step: 1 > 2. If there is no vertex repetition other than first vertex = last vertex, then W traverses a cycle, which is forbidden. Hence there is some other vertex repetition. Let W' be the portion of W between the instances of such a repetition. The walk W' is a shorter closed walk than W and contains no cycle, since W has none. By the induction hypothesis, W' has an edge repeating twice in succession, and this repetition also appears in W,
Proof 2. Let w be the first repetition of a vertex along W, arriving from v on edge e. From the first occurrence of w to the visit to v is a w, vwalk, which is a cycle if v = w or contains a nontrivial w, v-path P. This
completes a cycle with e unless in fact P is the path of length 1 with edge e, in which case e repeats immediately in opposite directions. 1.2.16. If edge e appears an odd number of times in a closed walk W, then W contains the edges of a cycle through e. Proof 1 (induction on the length of W, as in Lemma 1.2.7). The shortest closed walk has length 1. Basis step (1 = 1): The edge e in a closed walk of length 1 is a loop and thus a cycle. Induction step (1 > 1): If there is no vertex repetition, then W is a cycle. If there is a vertex repetition, choose two appearances of some vertex (other than the beginning and end of the walk). This splits the walk into two closed walks shorter than W. Since each step is in exactly one of these subwalks, one of them uses e an odd number of times, By the induction hypothesis, that subwalk contains the edges of a cycle through e, and this is contained in W. Proof 2 (parity first, plus Lemma 1.2.6). Let x and y be the endpoints of e. As we traverse the walk, every trip through e is x, e, y or y, e, x. Since the number of trips is odd, the two types cannot alternate. Hence some two successive trips through e have the same direction. By symmetry, we may
The portion of the walk between these two trips through e is a walk that does not contain e. By Lemma 1.2.6, it contains a y, x-path (that does not contain e. Adding e to this path completes a cycle with e consisting of edges in W.
Proof 3 (contrapositive). If edge e in walk W does not lie on a cycle consisting of edges in W, then by our characterization of cut-edges, e is a cut-edge of the subgraph H consisting of the vertices and edges in W. This means that the walk can only return to e at the endpoint from which it most recently left e. This requires the traversals of e to alternate directions along e. Since a closed walk ends where it starts (that is, in the same component of H e), the number of traversals of e by W must be even. 1.2.17. The "adjacent-transposition graph" on permutations of [a] is connected. Note that since every permutation of [a] has a 1 adjacent pairs that can be transposed, is (a 1)-regular. Therefore, showing that is connected shows that it is Eulerian if and only if a is odd. Proof 1 (path to fixed vertex). We show that every permutation has a path to the identity permutation I = 1, . a. By the transitivity of the connection relation, this yields for all u, v c V(G) a u, v-path in G. To create a v, I-path, move element 1 to the front by adjacent interchanges, .
then move 2 forward to position 2, and so on. This builds a walk to I, which contains a path to I. (Actually, this builds a path.)
Proof 2 (direct u, v-path). Each vertex is a permutation of [a]. Let and v = b1, . we construct at u, v-path. The element 01, .
Section 1.2: Paths, Cycles, and Trails
b1 appears in u as some a1; move it to the front by adjacent transpositions,
beginning a walk from u. Next find b2 among 02, . and move it to position 2. Iterating this procedure brings the elements of v toward the front, in order, while following a walk. It reaches v when all positions have
been "corrected". (Actually, the walk is a u, v-path.) Note that since we always bring the desired element forward, we never disturb the position of the elements that were already moved to their desired positions. Proof 3 (induction on a). If a = 1, then G is connected (we can also start with a = 2). For a > 1, assume that is connected. In the subgraph H induced by the vertices having a at the end is isomorphic to Every vertex of G is connected to a vertex of H by a path formed by moving element a to the end, one step at a time. For u, v a we thus have a path from u to a vertex u' a V(H), a path from v to a vertex
v'-path in H that exists by the induction hypothe-
sis. By the transitivity of the connection relation, there is a u, v-path in G. This completes the proof of the induction step. (The part of G4 used in the induction step appears below.) 4132k
3241/ 3421/ 43211
Proof 4 (induction on a). The basis is as in Proof 3, For a > 1, note that for each i a [a], the vertices with i at the end induce a copy H, of
Ga_i. By the induction hypothesis, each such subgraph is connected. Also, has vertices with i in position a 1 whenever 1 -c i -c a 1. We can interchange the last two positions to obtain a neighbor in H,. Hence there is an edge from each H, to and transitivity of the connection relation again completes the proof.
1.2.18. For k > 1, there are two components in the graph Gk whose vertex set is the set of binary k-tuples and whose edge set consists of the pairs that differ in exactly two places. Changing two coordinates changes the number
of is in the name of the vertex by zero or by +2. Thus the parity of the
Chapter 1: Fundamental Concepts
number of is remains the same along every edge. This implies that Gk has at least two components, because there is no edge from an k-tuple with an even number of is to an k-tuple with an odd number of is. To show that Gk has at most two components, there are several approaches. In each, we prove that any two vertices with the same parity lie on a path, where "parity" means parity of the number of is. Proof 1. If u and v are vertices with the same parity, then they differ in an even number of places. This is true because each change of a bit in obtaining one label from the other switches the parity. Since they differ in an even number of places, we can change two places at a time to travel from u to v along a path in Gk. Proof 2. We use induction on k. Basis step (k = i): G1 has two components, each an isolated vertex. Induction step (Ic > i): when Ic > i, Gk consists of two copies of Gk_1 plus additional edges. The two copies are obtained by appending 0 to all the vertex names in Gk_1 or appending i to them all. Within a copy, the edges don't change, since these vertices
all agree in the new place. By the induction hypothesis, each subgraph has two components. The even piece in the 0-copy has 0• 000, which is adjacent to Oii in the odd piece of the i-copy. The odd piece in the .
0-copy has 0• OiO, which is adjacent to 0 OOi in the even piece of the i-copy. Thus the four pieces reduce to (at most) two components in Gk.
i.2.i9. For n, r, s c N, the simple graph G with vertex set 4 and edge set > has gcd(n, r, s) components. Note: The text gives the vertex set incorrectly. When r = s = 2 and a is odd, it is necessary to go up to a 0 to switch from odd to even. Let Ic = gcd(n, r, s). Since Ic divides a, the congruence classes modulo a fall into congruence classes modulo Ic in a wei-defined way. All neighbors of vertex i differ from i by a multiple of Ic. Thus all vertices in a component lie
in the same congruence class modulo Ic, which makes at least Ic components.
To show that there are only Ic components, we show that all vertices with indices congruent to i (mod Ic) lie in one component (for each i). It suffices to build a path from i to i -f-k. Letl = gcd(r, s), and let a = r/l and b = s/l. Since there are integers (one positive and one negative) such that pa -f-qb = i, moving p edges with difference +r and q edges with difference +s achieves a change of +1. We thus have a path from i to i + 1, for each i. Now, Ic = gcd(l, a). As above, there exist integers p', q' such that p'(l/k) + q'(a/k) = i. Rewriting this as p'l = Ic q'a means that if we use p' of the paths that add 1, then we will have moved from I to I + Ic (mod a).
i.2.20. If v is a cut-vertex of a simple graph G, then v is not a cut-vertex of G. Let 1/1. Vk be the vertex sets of the components of G v; note
Section 1.2: Paths, Cycles, and Trails
contains the complete multipartite graph with partite sets 1/1. T/k. Since this includes all vertices of G v, the graph G v is connected, Hence v is not a cut-vertex of G.
1.2.2 1. A self-complementary graph has a cut-vertex if and only if it has a vertex of degree 1. If there is a vertex of degree 1, then its neighbor is a cut-vertex (K2 is not self-complementary). For the converse, let v be a cut-vertex in a self-complementary graph G. The graph G v has a spanning biclique, meaning a complete bipartite subgraph that contains all its vertices. Since G is self-complementary, also G must have a vertex u such that G u has a spanning biclique. Since each vertex of G v is nonadjacent to all vertices in the other components of G v, a vertex other than u must be in the same partite set of the spanning biclique of G u as the vertices not in the same component as u in G v. Hence only v can be in the other partite set, anile has degree at least a 2. We conclude that v has degree at most 1 in G, so G has a vertex of degree at most 1. Since a graph and its complement cannot both be disconnected, G has a vertex of degree 1. 1.2.22. A graph is connected if and only if for every partition of its vertices into two nonempty sets, there is an edge with endpoints in both sets. Necessity. Let G be a connected graph. Given a partition of V(G) into nonempty sets 8, T, choose u c S and v c T. Since G is connected, G has a u, v-path P. After its last vertex in s, p has an edge from S to T. Sufficiency.
Proof 1 (contrapositive). We show that if G is not connected, then for some partition there is no edge across. In particular, if G is disconnected,
then let H be a component of G. Since H is a maximal connected subgraph of G and the connection relation is transitive, there cannot be an edges with one endpoint in V(H) and the other endpoint outside. Thus for the partition of V(G) into V(H) and V(G) V(H) there is no edge with endpoints in both sets. Proof 2 (algorithmic approach). We grow a set of vertices that lie in the same equivalence class of the connection relation, eventually accumulating all vertices. Start with one vertex in S. While S does not include all vertices, there is an edge with endpoints x c S and y S. Adding y to S produces a larger set within the same equivalence class, using the transitivity of the connection relation. This procedure ends only when there are no more vertices outside 5, in which case all of G is in the same equivalence class, so G has only one component.
Proof 3 (extremality). Given a vertex x e V(G), let S be the set of all vertices that can be reached from x via paths. If S V (G), consider the partition into S and V (G)
By hypothesis, G has an edge with endpoints
Chapter 1: Fundamental Concepts
u c S and v S. Now there is an x, v-path formed by extending an x, upath along the edge uv. This contradicts the choice of 5, so in fact S is all of V(G). Since there are paths from x to all other vertices, the transitivity of the connection relation implies that G is connected. 1.2.23. a) If a connected simple graph G is not a complete graph, then every vertex of G belongs to some induced subgraph isomorphic to P3. Let v be a vertex of G. If the neighborhood of v is not a clique, then v has a pair x, y of nonadjacent neighbors; induces P3. If the neighborhood of v is a clique, then since G is not complete there is some vertex y outside the set S consisting of v and its neighbors. Since G is connected, there is some edge between a neighbor w of v and a vertex x that is not a neighbor of v. Now the set induces P3, since x is not a neighbor of v. One can also use cases according to whether v is adjacent to all other vertices or not. The two cases are similar to those above. b) When a connected simple graph G is not a complete graph, G may have edges that belong to no induced subgraph isomorphic to P3. In the graph below, e lies in no such sub graph.
1.2.24. If a simple graph with no isolated vertices has no induced subgraph with exactly two edges, then it is a complete graph. Let G be such a graph. If G is disconnected, then edges from two components yield four vertices that induce a subgraph with two edges. If G is connected and not complete, then G has nonadjacent vertices x and y. Let Q be a shortest x, y-path; it has length at least 2. Any three successive vertices on Q induce with two edges. Alternatively, one can use proof by contradiction. If G is not complete, then G has two nonadjacent vertices, Considering several cases (common neighbor or not, etc.) always yields an induced subgraph with two edges. 1.2.25. Inductive proof that every graph G with no odd cycles is bipartite. Proof 1 (induction on e(G)). Basis step (e(G) = 0): Every graph with no edges is bipartite, using any two sets covering V(G). Induction step (e(G) > 0): Discarding an edge e introduces no odd cycles. Thus the induction hypothesis implies that G e is bipartite.
If e is a cut-edge, then combining bipartitions of the components of e so that the endpoints of e are in opposite sets produces a bipartition of G, If e is not a cut-edge of G, then let u and v be its endpoints, and let X, V be a bipartition of G e. Adding e completes a cycle with a u, v-path G
Section 1.2: Paths, Cycles, and Trails
in G e; by hypothesis, this cycle has even length. This forces u and v to be in opposite sets in the bipartition X, V. Hence the bipartition X, V of G e
is also a bipartition of G. Proof 2 (induction on n(G)). Basis step (n(G) = 1): A graph with one vertex and no odd cycles has no loop and hence no edge and is bipartite. Induction step (n(G) > 1): When we discard a vertex v, we introduce no odd cycles. Thus the induction hypothesis implies that G v is bipartite. Let G1 Gk be the components of G v; each has a bipartition. If v has neighbors u, w in both parts of the bipartition of G,, then the edges uv and vw and a shortest u, w-path in form a cycle of odd length. Hence we can specify the bipartition of so that contains all neighbors of v in WenowhaveabipartitionofGbylettingX = andY = U(UY3.
1.2.26. A graph G is bipartite if and only if for every subgraph H of G, there is an independent set containing at least half of the vertices of H. Every bipartite graph has a vertex partition into two independent sets, one of which must contain at least half the vertices (though it need not be a maximum independent set). Since every subgraph of a bipartite graph is bipartite, the argument applies to all subgraphs of a bipartite graph, and the condition is necessary.
For the converse, suppose that G is not bipartite. By the characterization of bipartite graphs, G contains an odd cycle H. This subgraph H has no independent set containing at least half its vertices, because every set consisting of at least half the vertices in an odd cycle must have two consecutive vertices on the cycle.
1.2.27. The "transposition graph" on permutations of [a] is bipartite. The partite sets are determined by the parity of the number of pairs i, j such that i (these are called inversions). We claim that each transposition changes the parity of the number of inversions, and therefore each edge in the graph joins vertices with opposite parity. Thus the permutations with an even number of inversions form an independent set, as do those with an odd number of inversions. This is a bipartition, and thus the graph is bipartite. Consider the transposition that interchanges the elements in position r and position s, with r 0. The assumption of connectedness is necessary, because the conclusion is not true for G = H1 + when H1 has some odd vertices and H2 is Eulerian. Proof 1 (induction on k). When k = 1, we add an edge between the
two odd vertices, obtain an Eulerian circuit, and delete the added edge. When k> 1, let P be a path connecting two odd vertices. The graph G' =
G E(P) has 2k —2 odd vertices, since deleting E(P) changes degree parity only at the ends of P. The induction hypothesis applies to each component of G' that has odd vertices. Any component not having odd vertices has an Eulerian circuit that contains a vertex of F; we sphce it into F to avoid
having an additional trail. In total, we have used the desired number of trails to partition E(G). Proof 2 (induction on e(G)). If e(G) = 1, then G = K2, and we have one trail. If G has an even vertex x adjacent to an odd vertex y, then G' = G xy has the same number of odd vertices as G. The trail decomposition of G' guaranteed by the induction hypothesis has one trail ending at x and no trail ending at y. Add xy to the trail ending at x to obtain the desired decomposition of G. If G has no even vertex adjacent to an odd vertex, then G is Eulerian or every vertex of G is odd, In this case, deleting an edge xy reduces k, and we can add xy as a trail of length one to the decomposition of G xy guaranteed by the induction hypothesis.
1.2.34. The graph below has 6 equivalence classes of Eulerian circuits. If two Eulerian circuits follow the same circular arrangement of edges, differing only in the starting edges or the direction, then we consider them equivalent. An equivalence class of circuits is characterized by the pairing of edges at each vertex corresponding to visits through that vertex. A 2-valent vertex has exactiy one such pairing; a 4-valent vertex has three possible pairings. The only restriction is that the pairings must yield a single closed trail. Given a pairing at one 4-valent vertex below there is a forbidden pairing at the other, because it would produce two edge-disjoint 4-cycles instead of a single trail. The other two choices are okay. Thus the
Section 1.2: Paths, Cycles, and Trails
Alternatively, think of making choices while following a circuit. Because each circuit uses each edge, and because the reversal of a circuit C is in the same class as C, we may follow a canonical representative of the class from a along ax. We now count the choices made to determine the circuit. After x we can follow one of 3 choices. This leads us through another neighbor of x to y. Now we cannot use the edge ya or the edge just used, so two choices remain. This determines the rest of the circuit. For each of the three ways to make the initial choice, there was a choice of two later, so there are 3• 2 = 6 ways to specif5r distinct classes of circuits. (Distinct ways of making the choices yields a distinct pairing at some vertex.) 1.2.35. Algorithm for Eulerian circuits. Let G be a connected even graph. At each vertex partition the incident edges into pairs (each edge appears in a pair at each endpoint). Start along some edge. At each arrival at a vertex, there is an edge paired with the entering edge; use it to exit. This can end only by arriving at the initial vertex along the edge paired with the initial edge, and it must end since the graph is finite. At the point where the first edge would be repeated, stop; this completes a closed trail. Furthermore, there is no choice in assembling this trail, so every edge appears in exactly one such trail, Therefore, the pairing decomposes G into closed trails.
If there is more than one trail in the decomposition, then there are two trails with a common vertex, since G is connected. (A shortest path connecting vertices in two of the trails first leaves the first trail at some vertex v, and at v we have edges from two different trails.) Given edges from trails A and B at v, change the pairing by taking a pair in A and a pair in B and switching them to make two pairs that pair an edge of A with an edge of B. Now when A is followed from v, the return to A does not end
the trail, but rather the trail continues and follows B before returning to the original edge. Thus changing the pairing at v combines these two trails into one trail and leaves the other trails unchanged. We have shown that if the number of trails in the decomposition exceeds one, then we can obtain a decomposition with fewer trails be changing the pairing. Repeating the argument produces a decomposition using one closed trail. This trail is an Eulerian circuit.
1.2.36. Alternative characterization of Eulerian graphs.
a) If G is loopless and Eulerian and G' = G uv, then G' has an odd number of u, v-trails that visit v only at the end. Proof 1 (exhaustive counting and parity). Every extension of every trail from u in G' eventually reaches v, because a maximal trail ends only at a vertex of odd degree. We maintain a list of trails from u. The number of choices for the first edge is odd. For a trail T that has not yet reached v, there are an odd number of ways to extend T by one edge. We replace T in the list by these extensions, This changes the number of trails in the list by an even number. The process ends when all trails in the list end at v. Since the list always has odd size, the total number of these trails is odd. Proof 2 (induction and stronger result). We prove that the same conclusion holds whenever u and v are the only vertices of odd degree in a graph H, regardless of whether they are adjacent. This is immediate if H
has only the edge uv. For larger graphs, we show that there are an odd number of such trails starting with each edge e incident to u, so the sum is odd. If e = uv, then there is one such trail. Otherwise, when e = uw with w v, we apply the induction hypothesis to H e, in which w and v are the only vertices of odd degree. The number of non-paths in this list of trails is even. If T is such a trail
that is not a path, then let w be the first instance of a vertex repetition on T. By traversing the edges between the first two occurrences of w in the opposite order, we obtain another trail T' in the list. For T', the first instance of a vertex repetition is again w, and thus T" = T. This defines an involution under which the fixed points are the u, v-paths. The trails we wish to delete thus come in pairs, so there are an even number of them. b) If v is a vertex of odd degree in a graph G, then some edge incident to v lies in an even number of cycles. Let c(e) denote the number of cycles containing e. Summing c(e) over edges incident to v counts each cycle through v exactly twice, so the sum is even. Since there are an odd number of terms in the sum, c(e) must be even for some e incident to v. c) A nontrivial connected graph is Eulerian if and only if every edge belongs to an odd number of cycles. Necessity: By part (a), the number of u, v-paths in G uv is odd. The cycles through uv in G correspond to the u, v-paths in G uv, so the number of these cycles is odd. Sufficiency: We observe the contrapositive. If G is not Eulerian, then G has a vertex v of odd degree. By part (b), some edge incident to v lies in an even number of cycles.
1.2.37. The connection relation is transitive. It suffices to show that if P is a u, v-path and P1 is a v, w-path, then P and P' together contain a u, wpath. At least one vertex of P is in P', since both contain v. Let x be the
Section 1.2: Paths, Cycles, and Trails
first vertex of P that is in P'. Following P from u to x and then P' from x to w yields a u, w path, since no vertex of P before x belongs to P'. 1.2.38. Every n-vertex graph with at least n edges contains a cycle. Proof 1 (induction on a). A graph with one vertex that has an edge has a loop, which is a cycle. For the induction step, suppose that a > 1. If our graph G has a vertex v with degree at most 1, then G v has a 1 vertices
and at least a 1 edges. By the induction hypothesis, G v contains a cycle, and this cycle appears also in G. If G has no vertex of degree at most 1, then every vertex of G has degree at least 2. Now Lemma 1.2.25 guarantees that G contains a cycle. Proof 2 (use of cut-edges). If G has no cycle, thenby Theorem 1.2.14 every edge is a cut-edge, and this remains true as edges are deleted. Deleting all the edges thus produces at least a + 1 components, which is impossible. 1.2.39. If G is a loopless graph and 3(G) > 3, then G has a cycle of even length. An endpoint v of a maximal path P has at least three neighbors on P. Let x, y, z be three such neighbors of v in order on P. Consider three v, y-paths: the edge vy, the edge vx followed by the x, y-path in P, and the edge vz followed by the z, y-path in P. These paths share only their endpoints, so the union of any two is a cycle. By the pigeonhole principle, two of these paths have lengths with the same parity. The union of these two paths is an even cycle. V
1.2.40. If P and Q are two paths of maximum length in a connected graph
G, then P and Q have a common vertex. Let m be the common length of P and Q. Since G is connected, it has a shortest path R between V(P) and
V(Q). Let 1 be the length of R. Let the endpoints of R be r e V(P) and e V(Q). The portion P' of P from r to the farther endpoint has length at least m/2. The portion Q' of Q from r to the farther endpoint has length at least m/2. Since R is a shortest path, R has no internal vertices in P or Q. If P and Q are disjoint, then P' and Q' are disjoint, and the union of P', Q', and R is a path of length at least m/2 + tn/2 + 1 = in + 1. Since the maximum path length is in, we have 1 = 0. Thus r = r', and P and Q have a common vertex.
The graph consisting of two edge-disjoint paths of length 2k sharing their midpoint is connected and hence shows that P and Q need not have a common edge.
1.2.41. A connected graph with at least three vertices has two vertices x, y such that 1) G is connected and 2) x, y are adjacent or have a common neighbor. Let x be a endpoint of a longest path P in G, and let v be
Chapter 1: Fundamental Concepts
its neighbor on P. Note that P has at least three vertices. If G
is connected, let y = v. Otherwise, a component cut off from P x v in G x v has at most one vertex; call it w, The vertex w must be adjacent to v, since otherwise we could build a longer path. In this case, let y = w. 1.2.42. A connected simple graph having no 4-vertex induced subgraph that is a path or a cycle has a vertex adjacent to every other vertex. Consider a vertex x of maximum degree. If x has a nonneighbor y, let x, v, w be the begining of a shortest path to y (w may equal y). Since d(v) -c d(x), some neighbor z of x is not adjacent to v. If z -e- w, then induce C4; otherwise, (z, x, v, w> induce P4. Thus x must have no nonneighbor.
1.2.43. The edges of a connected simple graph with 2k edges can be partitioned into paths of length 2. The assumption of connectedness is necessary, since the conclusion does not hold for a graph having components with an odd number of edges. We use induction on e(G); there is a single such path when e(G) = 2.
For e(G) > 2, let P = (x, y, z) be an arbitrary path of length two in G, and let G' = G . If we can partition E(G) into smaller connected subgraphs of even size, then we can apply the induction hypothesis to each piece and combine the resulting decompositions. One way to do this is to partition E(G') into connected subgraphs of even size and use P. Hence we are finished unless G' has two components of odd size (G' cannot have more than three components, since an edge deletion increases the number of components by at most one). Each odd component contains at least one of . Hence it is possible to add one of xy to one odd component and yz to the other odd component to obtain a partition of G into smaller connected subgraphs.
1.3. VERTEX DEGREES & COUNTING 1.3.1. A graph having exactly two vertices of odd degree must contain a path from one to the other. The degree of a vertex in a component of G is the same as its degree in G. If the vertices of odd degree are in different components, then those components are graphs with odd degree sum.
1.3.2. In a class with nine students where each student sends valentine cards to three others, it is not possible that each student sends to and receives cards from the same people. The sending of a valentine can be represented as a directed edge from the sender to the receiver. If each student sends to and receives cards from the same people, then the graph has x y if and
Section 1.3: Vertex Degrees and Counting
only if y —* x. Modeling each opposed pair of edges by a single unoriented
edge yields a 3-regular graph with 9 vertices. This is impossible, since every graph has an even number of vertices of odd degree.
If d(u) + d(v) = a + k for an edge u-v in a simple graph on a vertices, then uv belongs to at least k triangles. This is the same as showing that u and v have at least k common neighbors. Let S be the neighbors of u = j. Every vertex of G and T the neighbors of v, and suppose 5 P appears in S or T or none or both, Common neighbors are counted twice, son > = a + k j. Hence j > k, (Almost every proof of this + using induction or contradiction does not need it, and is essentially just this counting argument.) 1.3.3.
1.3.4. The graph below is isomorphic to
with the names of the vertices in
It suffices to label the vertices
that vertices are adjacent if and
only if their labels differ in exactly one place.
1.3.5. The k-dimensional cube Qk has copies of P3. Proof 1. To specify a particular subgraph isomorphic to Pg, the 3vertex path, we can the middle vertex and its two neighbors. For each vertex of Qk, there are ways to choose two distinct neighbors, since Qk is a simple k-regular graph. Thus the total number of Pg's is Proof 2. We can alternatively choose the starting vertex and the next two. There are 2k ways to pick the first vertex. For each vertex, there are k ways to pick a neighbor. For each way to pick these vertices, there are k 1 ways to pick a third vertex completing Pg, since Qk has no multiple edges. The product of these factors counts each P3 twice, since we build it from each end, Thus the total number of them is 2kk(k 1 (direct counting). The vertices two apart on a 4-cycle must differ in two coordinates. Their two common neighbors each differ from each in exactly one of these coordinates. Hence the vertices of a 4-cycle
Chapter 1: Fundamental
must use all 2-tuples in two coordinates while keeping the remaining coordinates fixed. All such choices yield 4-cycles. There are ways to choose
the two coordinates that vary and
to set a fixed value in the
remaining coordinates. Proof 2 (prior result). Every 4-cycle contains four copies of Pg, and every P3 contains two vertices at distance 2 in the cube and hence extends to exactly one 4-cycle. Hence the number of 4-cycles is one-fourth the number of copies of P3.
1.3.6. Counting components. If G has Ic components and H has 1 components, then G + H has Ic +1 components. The maximum degree of G + H is max.
1.3.7. Largest bipartite sub graphs. is already bipartite. loses one edge if a is odd, none if a is even. The largest bipartite subgraph of is which has edges. 1.3.8.
The lists (5,5,4,3,2,2,2,1), (5,5,4,4,2,2,1,1), and (5,5,5,3,2,2,1,1) are
graphic, but (5,5,5,4,2,1,1,1) is not.
The answers can be obtained from
the Havel-Hakimi test; a list is graphic if and only if the list obtained by deleting the largest element and deleting that many next-largest elements is graphic. Below are graphs realizing the first three lists, found by the Havel-Hakimi algorithm.
Fromthelastlist,wetest(4, 4,3,1,0,1,1), reorderedto(4, 4,3, 1,1,1,0), then (3, 2, 0,0, 1, 0). This is not the degree hst of a simple graph, since a vertex of degree 3 requires three other vertices with nonzero degree. 1.3.9. In a league with two divisions of 13 teams each, no schedule has each team playing exactly nine games against teams in its own division and four games against teams in the other division, If this were possible, then we could fonu a graph with the teams as vertices, making two vertices adjacent if those teams play a game in the schedule. We are asking for the subgraph induced by the 13 teams in a single division to be 9-regular. However, there is no regular graph of odd degree with an odd number of vertices, since for every graph the sum of the degrees is even. 1.3.10. If!, m, a are nonnegative integers with 1 + in = a > 1, then there exists a connected simple n-vertex graph with 1 vertices of even degree and in
Section 1.3: Vertex Degrees and Counting
vertices of odd degree if and only if m is even, except for (1, m, a) = (2, 0, 2).
Since every graph has an even number of vertices of odd degree, and the only simple connected graph with two vertices has both degrees odd, the condition is necessary. To prove sufficiency, we construct such a graph G. If m = 0, let G = C1 (except G = K1 ill = 1). For in > 0, we can begin with Ki,m_i, which has in vertices of odd degree, and then add a path of length 1 beyond one of the leaves. (Illustration shows 1 = 3, in = 4.) Alternatively, start with a cycle of length 1, and add in vertices of degree one with a common neighbor on the cycle. That vertex of the cycle has even degree because in is even. Many other constructions also work. It is also possible to prove sufficiency by induction on a for a> 3, but this approach is longer and harder to get right than an explicit general construction,
1.3.11. If C is a closed walk in a simple graph G, then the subgraph consisting of the edges appearing an odd number of times in C is an even graph.
Consider an arbitrary vertex v c V(G). Let S be the set of edges incident to v, and let f(e) be the number of times an edge e is traversed by C. Each time C passes through v it enters and leaves. Therefore, f(e) must be even, since it equals twice the number of times that C visits v. Hence there must an even number of odd contributions to the sum, which means there are an even number of edges incident to v that appear an odd number of times in C. Since we can start a closed walk at any of its vertices, this argument holds for every v c V (G). 1.3.12. If every vertex of G has even degree, then G has no cut-edge.
Proof 1 (contradiction). If G has a cut-edge, deleting it leaves two induced subgraphs whose degree sum is odd. This is impossible, since the degree sum in every graph is even. Proof 2 (construction/extremality). For an edge uv, a maximal trail in G uv starting at u can only end at v, since whenever we reach a vertex we have use an odd number of edges there. Hence a maximal such trail is a (u, v)-trail. Every (u, v)-trail is a (u, v)-walk and contains a (u, v)-path. Hence there is still a (u, v)-path after deletion of uv, so uv is not a cut-edge. Proof 3 (prior results). Let G be an even graph. By Proposition 1.2.27, G decomposes into cycles. By the meaning of every edge
Chapter 1: Fundamental Concepts
of G is in a cycle. By Theorem 1.2.14, every edge in a cycle is not a cut-edge. Hence every edge of G is not a cut-edge. For k a N, some (2k + 1)-regular simple graph has a cut-edge.
Construction 1. Let H, H' be copies of K2k,2k with partite sets X, V for H and X', Y' for H'. Add an isolated edge ye' disjoint from these sets. To H + H' + cv', add edges from v to all of X and from v' to all of X', and add k disjoint edges within V and k disjoint edges within Y'. The resulting graph Gk is (2k + 1)-regular with 8k +2 vertices and has cv' as a cut-edge. Below we sketch G2; the graph G1 is the graph in Example 1.3.26.
Construction 2a (inductive). Let G1 be the graph at the end of Example 1.3.26 (or in Construction 1). This graph is 3-regular with 10 vertices and cut-edge xy; note that 10 = 4• 1 + 6. From a (2k 1)-regular graph Gk_1 with 4k + 2 vertices such that Gk_1 xy has two components of order 2k + 1, we form Gk. Add two vertices for each component of Gk_1 xy, adjacent to all the vertices of that component. This adds degree two to each old vertex, gives degree 2k + 1 to each new vertex, and leaves xy as a cut-edge. The result is a (2k + 1)-regular graph Gj 2. We prove that (r, s) and (1, 1) are the only vertices of odd degree in G. This suffices, because every graph has an even number of vertices of
Section 1.3: Vertex Degrees and Counting
odd degree, which implies that (r, s) and (1, 1) are in the same component, connected by a path. The possible neighbors of (i, J) are the pairs obtained by changing i
or j by 1. Let X and V be the intervals of heights attained by x, and yy, and let I = X fl V. If the high end of I is the high end of exactly one of X and Y, then exactly one neighboring vertex can be reached by moving past
the end of the corresponding segment. If it is the high end of both, then usually one or three neighboring vertices can be reached, the latter when both segments reach "peaks" at their high ends. However, if (1, j) = (1, 1), then the high end of both segments is P and there is no neighbor of this type. Similarly, the low end of I generates one or three neighbors, except that when (1, j) = (r, s) there is no neighbor of this type. No neighbor of (i, j) is generated from both the low end and the high end of I. Since the contributions from the high and low end of I to the degree of (i, J) are both odd, each degree is even, except for (r, s) and (1, 1), where exactly one of the contributions is odd.
1.3.14. Every simple graph with at least two vertices has two vertices of equal degree. The degree of a vertex in an n-vertex simple graph is in a 1). These are a distinct values, so if no two are equal then all appear. However, a graph cannot have both an isolated vertex and a vertex adjacent to all others. This does not hold for graphs allowing loops. In the 2-vertex graph with one loop edge and one non-loop edge, the vertex degrees are 1 and 3. This does not hold for loopless graphs. In the 3-vertex loopless graph with pairs having multipllcity 0, 1, 2, the vertex degrees are 1, 3, 2. (0
1.3.15. Smallest k-regular graphs. A simple k-regular graph has at least k -f- 1 vertices, so Kk+1 is the smallest. This is the only isomorphism class of
k-regular graphs with k + 1 vertices. With k +2 vertices, the complement of a k-regular graph must be 1-regular. There is one such class when k is even ((k + 2)/2 isolated edges), none when k is odd. (Two graphs are isomorphic if and only if their complements are isomorphic.) With k + 3 vertices, the complement is 2-regular. For k> 3, there are distinct choices for such a graph: a (k + 3)-cycle or the disjoint union of a 3-cycle and a k-cycle. Since these two 2-regular graphs are nonisomorphic, their complements are nonisomorphic k-regular graphs with k + 3 vertices.
1.3.16. For k> 2 and g > 2, there exists a k-regular graph with girth g. We use strong induction on g. For g = 2, take the graph consisting of two vertices and k edges joining them. For the induction step, consider g > 2. Here we use induction on k. For k = 2, a cycle of length g suffices. Fork > 2, the induction hypothesis
Chapter 1: Fundamental Concepts
provides a (k 1)-regular graph H with girth g. Since Ig/21 0. Hence deleting a vertex of maximum degree in nontrivial graph reduces the average degree and cannot increase it. 1.3.18. If k >- 2, then a k-regular bipartite graph has no cut-edge. Since components of k-regular graphs are k-regular, it suffices to consider a connected k-regular X, Y-bigraph. Let uc be a cut-edge, and let G and H be the components formed by deleting uv. Let in = V(G) n and a = V(G) P By symmetry, we may assume that u a V(G) P Y and v a V(H) P X. We count the edges of G. The degree of each vertex of G in Xis k, so G has ink edges. The degree of each vertex of G in Y is k except for da(u) = k 1, SO G has nk 1 edges. Hence ink = nk 1, which is impossible because
one side is divisible by k and the other is not. The proof doesn't work if 1, and the claim is false then. If vertex degrees k and k + 1 are allowed, then a cut-edge may exist.
Consider the example of 2Kk,k plus one edge joining the two components.
1.3.19. A claw-free simple graph with maximum degree at least 5 has a 4cycle. Consider five edges incident to a vertex v of maximum degree in such a graph G. Since G has no induced claw, the neighbors of v must induce at
least three edges. Since these three edges have six endpoints among the five neighbors of v, two of them must be incident, say xy and yz. Adding the edges xv and zv to these two completes a 4-cycle. There are arbitrarily large 4-regular claw-free graphs with no 4-cycles.
Section 1.3: Vertex Degrees and Counting
Consider a vertex v in such a graph G. Since v has degree 4 and is not the center of an induced claw and does not lie on a 4-cycle, the subgraph induced by v and its neighbors consists of two edge-disjoint triangles sharing v (a bowtie). Since this happens at each vertex, G consists of pairwise edge-disjoint triangles, with each vertex lying in two of them. Hence each triangle has three neighboring triangles. Furthermore, two triangles that neighbor a given triangle in this way cannot neighbor each other; that would create a 4-cycle in the graph. Define a graph H with one vertex for each triangle in G; let vertices be adjacent in H if the corresponding triangles share a vertex in G. Now H is a 3-regular graph with no 3-cycles; a 3-cycle in H would yield a 4-cycle in G using two edges from one of the corresponding triangles. Also H must have no 4-cycles, because a 4-cycle in G could be built using one edge from
each of the four triangles corresponding to the vertices of a 4-cycle in H. Note that e(G) = 2n(G) and n(H) = e(G)/3 = 2n(G)/3. On the other hand, given any 3-regular graph H with girth at least 5, reversing the construction yields G with the desired properties and 3n (H)/2 vertices. Hence it suffices to show that there are arbitrarily large 3-regular graphs with girth at least 5. Disconnected such examples can be formed by taking many copies of the Petersen graph as components. The graph G is connected if and only if H is connected. Connected instances of H can be obtained from multiple copies of the Petersen graph by applying 2-switches (Definition 1.3.32). Alternatively, arbitrarily large connected examples can be constructed
by taking two odd cycles (say length 2m + 1) and joining the ith vertex on the first cycle to the 2ith vertex (modulo 2m + 1) on the second cycle (this generalizes the Petersen graph). We have constructed a connected 3-regular graph. Since we add disjoint edges between the cycles, there is no triangle. A 4-cycle would have to alternate edges between the two odd cycles with one edge of each, but the neighbors of adjacent vertices on the first cycle are two apart on the second cycle. 1.3.20. has (a. 1)!/2 cycles of length a., and has ni(n. 1)!/2 cycles of length 2n. Each cycle in is a listing of the vertices. These can be listed
in a.! orders, but we obtain the same subgraph no matter where we start the cycle and no matter which direction we follow, so each cycle is listed 2n times. In K. we can list the vertices in order on a cycle (alternating between the partite sets), in 2(n!)2 ways, but by the same reasoning each cycle appears (2n) 2 times. 6-cycles. To extend an edge in K. to a 6-cycle, we choose two more vertices from each side to be visited in order as we follow the cycle. Hence each edge in K. appears in (in 1)(n 1)(m 2)(n 2) 1.3.2 1. K. has
Chapter 1: Fundamental Concepts
6-cycles. Since each 6-cycle contains 6 edges, we conclude that mn(m 1)(n 1)(tn 2)(n 2)/6 6-cycles.
Alternatively, each 6-cycle uses three vertices from each partite set, which we can choose in ways. Each such choice of vertices induces a copy of K3,3 with 9 edges. There are 3! = 6 ways to pick three disjoint edges to be omitted by a 6-cycle, so each K3,3 contains 6 6-cycles.
1.3.22. Odd girth and minimum degree in nonbipartite triangle-free avertex graphs. Let k = 6(G), and let 1 be the minimum length of an odd cycle in G. Let C be a cycle of length 1 in G.
a) Every vertex not in V(C) has at most two neighbors in V(C). It suffices to show that any two neighbors of such a vertex v on C must have distance 2 on C, since having three neighbors would then require 1 = 6. Since G is triangle-free, v does not have consecutive neighbors on C. If v has neighbors x and y on C separated by distance more than 2 on C, then the detour through v can replace the x, y-path of even length on C to form a shorter odd cycle. b) a > kl/2 (and thus 1 2, we prove that each vertex mapped by f to a vertex of Qj having weight j differs from v in j positions of 5, by induction on j. Let x be a vertex mapped to a vertex of weight j in For j -c i, we have already argued that x differs from v in j positions of 5. For j > 2, let y and z be two neighbors of x whose images under f have weight j i in Qi. By the induction hypothesis, y and z differ from v in j positions of S. Since f(y) and f(z) differ in two places, they have two common neighbors in which are x and another vertex w. Since w has weight j 2, the induction hypothesis yields that w differs from v in j i positions of S. Since the images of x, y, z, w induce a 4-cycle in Qi, also x, y, z, w induce a 4-cycle in H. The only 4-cycle in Qk that contains all of y, z, w adds the vertex that differs from v in the j 2 positions of S where w differs, plus the two positions where y and z differ from w. This completes the proof that x has the desired property. b) The k-dimensional cube Qk has exactly 2kk! automorphisms. (Part (a) is unnecessary.) Form automorphisms of Qk by choosing a subset of the k coordinates in which to complement 0 and i and, independently, a permutation of the k coordinates. There are 2kk! such automorphisms.
Chapter 1: Fundamental Concepts
We prove that every automorphism has this form. Let 0 be the all-O vertex. Let f be the inverse of an automorphism, and let v be the vertex mapped to 0 by f. The neighbors of v must be mapped to the neighbors of 0. If these choices completely determine f, then f complements the coordinates where v is nonzero, and the correspondence between the neighbors of 0 and the neighbors of v detenuines the permutation of the coordinates that expresses f as one of the maps listed above. Suppose that x differs from v in coordinates r1, . , ry. Let u1, . u1 be the neighbors of v differing from v in these coordinates. We prove that f(x) is the k-tuple of weight j having 1 in the coordinates where f(u ') f(uy) have 1. We use induction on j. For j -c 1, the claim follows by the definition of u1 u7. For j >- 2, let y and z be two neighbors of x that differ from v in j 1 coordinates. Let w be the common neighbor of y and z that differs from v in j 2 coordinates. By the induction hypothesis, f(y) and f(z) have weight j 1 (in the appropriate positions), and f(w) has weight j 1. Since f(x) must be the other common neighbor of f(y) and f(z), it has weight j, with is in the desired positions.
i.3.30. The Petersen graph has twelve 5-cycles. Let G be the Petersen graph. We show first that each edge of G appears in exactly four 5-cycles. For each edge e = xy in G, there are two other edges incident to x and two others incident to y. Since G has no 3-cycles, we can thus extend xy at both ends to form a 4-vertex path in four ways. Since G has no 4-cycle, the endpoints of each such path are nonadjacent. By Proposition 1.1.38, there is exactly one vertex to add to such a path to complete a 5-cycle. Thus e is in exactly four 5-cycles. When we sum this count over the 15 edges of G, we have counted 60 5-cycles. However, each 5-cycle has been counted five times—once for each of its edges. Thus the total number of 5-cycles in G is 60/5 = 12. .1.
Combinatorial proofs with graphs.
a) For 0 -c k -c a, () =
k) + Consider the complete If we partition the vertices of into a then we can count the edges as those within one edges.
Section 1.3: Vertex Degrees and Counting
block of the partition and those choosing a vertex from each, Hence the total number of edges is + k(n k). + and b) If a, = a, then -c (,), Again consider the edges of partition the vertices into sets with a, being the size of the ith set. The left side of the inequality counts the edges in having both ends in the same 8,, which is at most all of
1.3.32. For a > 1, there are 2(t) simple even graphs with a fixed vertex set of size a. Let A be the set of simple even graphs with vertex set v1. Since is the size of the set B of simple graphs with vertex
we establish a bijection from A to B. Given a graph in A, we obtain a graph in B by deleting To show that each graph in B arises exactly once, consider a graph G e B. We form a new graph G' by adding a vertex and making it adjacent to each vertex with odd degree in G, as illustrated below. The vertices with odd degree in G have even degree in G'. Also, itself has even degree because the number of vertices of odd degree in G is even. Thus G' e A. Furthermore, G is the graph obtained from G' by deleting v. and every simple even graph in which deleting yields G must have adjacent to the same vertices as in G'. Since there is a bijection from A to B, the two sets have the same size.
1.3.33. Triangle-free graphs in which every two nonadjacent vertices have exactly two common neighbors.
a(G) = 1 + where k is the ckgree of a vertex x in G. For every pair of neighbors of x, there is exactly one nonneighbor of x that they have as a common neighbor. Conversely, every nonneighbor of x has exactly one pair of neighbors of x in its neighborhood, because these are its common neighbors with x. This establishes a bijective correspondence between the pairs in N(x) and the nonneighbors of x. Counting x, N(x), and N(x), we have a(G) = 1 + k -t- Q) = 1 + Since this argument holds for every x c V(G), we conclude that G is k-regular. Comment: Such graphs exist only for isolated values of k. Unique graphs exist for k = 1, 2, 5. Viewing the vertices as x, N(x) = [ki, and N(x) = (v), we have i adjacent to the pair if and only if e . The lack of triangles guarantees that only disjoint pairs in can be adjacent,
Chapter 1: Fundamental Concepts
but each pair in mast have exactly k 2 neighbors in (v), For k = 5, this implies that N(x) induces the 3-regular disjointness graph of which is the Petersen graph. Since the Petersen graph has girth 5 and diameter 2, each intersecting pair has exactly one common neighbor in N(x) in addition to its one common neighbor in N(x), so this graph has the desired properties. Numerical conditions eliminate k 3 (mod 4), because G would be regular of odd degree with an odd number of vertices. There are stronger necessary conditions. After k = 5, the next possibility is k = 10, then 26, 37, 82, etc. A realization for k = 10 is known to exist, but in general the set of realizable values is not known.
1.3.34. If G is a kite-free simple n-vertex graph such that every pair of nonadjacent vertices has exactly two common neighbors, then G is regular. Since nonadjacent vertices have common neighbors, G is connected. Hence it suffices to prove that adjacent vertices x and y have the same degree. To prove this, we establish a bijection from A to B, where A = N(x) N(y) and B = N(y) N(x). Consider u E A. Since u e y, there exists v e N(u) P N(y) with v x. Since G is kite-free, v x, so v c B. Since x and v have common neighbors y and u, the vertex v cannot be generated in this way from another vertex of A. Hence we have defined an injection from A to B. Interchanging the roles of y and x yields an injection from B to A. Since these sets are finite, the injections are bijections, and d(x) = d(y). 1.3.35. If every induced k-vertex subgraph of a simple n-vert cx graph G has the same number of edges, where 1 k and G' is a graph on 1 vertices in which every induced kvertex subgraph has tn edges, then e(G') = ,nG)/G1). Counting the edges in nil the k-vertex subgraphs of G' yields in but each edge appears in Cl) of these subgraphs, once for each k-set of vertices containing it. (Both sides of Cl)e(G') = mQ) count the ways to pick an edge of G' and a k-set of vertices in G' containing that edge. On the right, we pick the set first; on the left, we pick the edge first.) — or G = K. Given vertices u b) Under the stated conditions, G = and v, let A and B be the sets of edges incident to u and v, respectively. The set of edges with endpoints u and v is A P B. We compute = e(G) =e(G) = e(G)
In this fonnula, 1 and
are the edge sets of induced sub graphs of order and A P B is the edge set of an induced subgraph of order n 2. By part (a), the sizes of these sets do not depend on the choice of u and v. n
Section 1.3: Vertex Degrees and Counting
1.3.36. The unique reconstruction of the graph with vertex-deleted subgraphs below is the kite.
Proof 1. A vertex added to the first triangle may be joined to 0,1,2, or 3 of its vertices. We eliminate 0 and 1 because no vertex-deleted subgraph has an isolated vertex. We eliminate 3 because every vertex-deleted subgraph of K4 is a triangle. Joining it to 2 yields the kite.
V Proof 2. The graph G must have four vertices, and by Proposition 1.3.11 it has five edges. The only such simple graph is the kite.
1.3.37. Retrieving a regular graph. Suppose that H is a graph formed by deleting a vertex from a regular graph G. We have H, so we know n(G) = n(H) -F 1, but we don't know the vertex degrees in G. If G is dregular, then G has dn(G)/2 edges, and H has dn(G)/2 d edges. Thus d = 2e(H)/(n(G) 2). Having determined d, we add one vertex w to H and add d
dH(v) edges from w to v for each v c V(H).
1.3.38. A graph with at least 3 vertices is connected if and only if at least two of the subgraphs obtained by deleting one vertex are connected. The endpoints of a maximal path are not cut-vertices. If G is connected, then the subgraphs obtained by deleted such vertices are connected, and there are at least of these. Conversely, suppose that at least two vertex-deleted subgraphs are connected. If G v is connected, then G is connected unless v is an isolated vertex. If v is an isolated vertex, then all the other sub graphs obtained by deleting one vertex are disconnected. Hence v cannot be isolated, and G is connected.
1.3.39. Disconnected graphs are reconstructible. First we show that G is connected if and only if it has at least two connected vertex-deleted subgraphs. Necessity holds, because the endpoints of a maximal path cannot be cut-vertices. If G is disconnected, then G v is disconnected unless v is an isolated vertex (degree 0) in G and G v is connected, This happens for at most one vertex in G. After determining that G is disconnected, we obtain which disconnected graph it is from its vertex-deleted subgraphs. We aim to identif& a connected graph M that is a component of G and a vds in the deck that arises by deleting a specified vertex u of M. Replacing M u by M in that subgraph will reconstruct G.
Chapter 1: Fundamental Concepts
Among all components of all graphs in the deck, let M be one with maximum order. Since every component H of a potential reconstruction G appears as a component of some G v, M cannot belong to any larger component of G. Hence M is a component of G. Let L be a fixed connected subgraph of M obtained by deleting a leaf u of some spanning tree of M. Then L is a component of G u. We want to reconstruct G by substituting M for L in G U; we must identilS' G u. There may be several isomorphic copies of G
As in the disconnected graph G shown above, M may appear as a component of every vds G v. However, since M cannot be created by a vertex deletion, a vds with the fewest copies of M must arise by deleting a vertex of M, Among these, we seek a subgraph with the most copies of L as components, because in addition to occurrences of L as a component of G, we
obtain an additional copy if and only if the deleted vertex of M can play the role of u. This identifies G u, and we obtain G by replacing one of its components isomorphic to L with a component isomorphic to M.
1.3.40. Largest graphs of specified types.
a) Largest n-vertex simple graph with an indEpendent set of size a.
Proof 1. Since there are no edges within the independent set, such a graph has at most () edges, which equals (7) + (n a)a. This bound is achieved by the graph consisting of a copy H of Kn_a, an independent set S of size a, and edges joining each vertex of H to each vertex of S. Proof 2. Each vertex of an independent set of size a has degree at most n a. Each other vertex has degree at most n 1, Thus d(v) -c a(n a) + (n a)(n 1). By the Degree-Sum Formula, e(G) -c (n a)(n 1 + a)/2. This formula equals those above and is achieved by the same graph, since this graph achieves the bound for each vertex degree. b) The maximum size of an n-vertex simple graph with k components is The graph consisting of plus k 1 isolated vertices has Ac components and edges. We prove that other n-vertex graphs with Ac components don't have maximum size. Let G be such a graph. If G has a component that is not complete, then adding edges to make
it complete does not change the number of components. Hence we may assume that every component is complete. If G has components with r and s vertices, where r > s > 1, then we move one vertex from the s-clique to the r-clique. This deletes s 1 edges
Section 1.3: Vertex Degrees and Counting
and creates r edges, all incident to the moved vertex. The other edges remain the same, so we gain r s + 1 edges, which is positive. Thus the number of edges is maximized only when every component is a complete graph and only one component has more than one vertex. c) The maximum number of edges in a disconnected simple n-vertex graph is (v), with equality only for K1 + Kn_i. Proof 1 (using part (b)). The maximum over graphs with Ic components is which decreases as Ic increases. For disconnected graphs, Ic > 2. We maximize the number of edges when Ic = 2, obtaining Proof 2 (direct argument). Given a disconnected simple graph G, let S be the vertex set of one component of G, and let t = Since no edges join S and 5, e(G) -c t(n t). This bound is weakest when t(n t) is smallest, which for 1 -c -c n 1 happens when t E . Thus always = 1(n 1) and equality holds when G = K1 + Kn_i. e(G) -c Proof 3 (induction on n). When n = 2, the only simple graph with = 1 is K2, which is connected. For n >- 2, suppose e(G) >e(G) >If A(G) = n 1, then G is connected. Otherwise, we may select v with d(v) -c n 2. Then e(G v) > (n;i) n -F 2 = (n;2). By the induction hypothesis, G v is connected. Since e(G) > and G is simple, we have d(v) > 0, so there is an edge from v to G v, and G is also connected. Proof 4 (complementation). If G is disconnected, thenG is connected, = (n_i) In fact, G must contain so e(G) > n 1 and e(G) -c (ii 1) a spanning complete bipartite subgraph, which is as small as n 1 edges only when G = Ki,n_i and G = K1 + Kn_i. 1.3.41. Every n-vertex simple graph with maximum degree In/21 and minimum degree [n/2j 1 is connected. Let x be a vertex of maximum degree. It suffices to show that every vertex not adjacent to x has a common neighbor = fn/21 and > with x. Choose y N(x). We have 1. x, we have N(x), N(y) C V(G) . Thus Since y =
1.3.42. Strongly independent sets. If S is an independent set with no common neighbors in a graph G, then the vertices of S have pairwise-disjoint closed neighborhoods of size at least 5(G) + 1. Thus there are at most In(G)/(8(G) + 1)j of them. Equality is achievable for the 3-dimensional cube using S = . Equality is not achievable when G = since with 16 vertices and minimum degree 4 it requires three parwise-disjoint closed neighborhoods of size 5, If v c 5, then no vertex differing from v in at most two places is in S. Also, at most one vertex differing from v in at least three places is in
Chapter 1: Fundamental Concepts
5', since such vertices differ from each other in at most two places. Thus only two disjoint closed neighborhoods can be found in 1.3.43. Every simple graph has a vertex whose neighbors have average degree as large as the overall average degree. Let t(w) be the average degree of t(w) = the neighbors of w. In the sum d(y)/d(w), we have the terms d(u)/d(v) and d(v)/d(u) for each edge xc. Since
x/y -f- y/x > 2 whenever x, y are positive real numbers (this is equivalent to (x > > 0), each such contribution is at least 2. Hence > 2e(G). Hence the average of the neighborhood avd(v) erage degrees is at least the average degree, and the pigeonhole principle yields the desired vertex. It is possible that every average neighborhood degree exceeds the average degree. Let G be the graph with 2n vertices formed by adding a matching between a complete graph and an independent set. Since G has G) -f- n edges and 2n vertices, G has average degree (n + 1)72. For each vertex of the n-clique, the neighborhood average degree is n 1 + 1/n. For each leaf, the neighborhood average degree is n. 1.3.44. Subgraphs with large minimum degree. Let G be a loopless graph with average degree a. a) If x E V(G), then G' = G x has average degree at least a if and only
if d(x) 1, this implies that a' > a if and only if d(x) d, even there is a multigraph with vertex degrees d1, . .
Proof 1 (induction on sd,). If d, 0, the n-vertex graph with no edges has degree list d. For the induction step, suppose ci, > 0. If only one d, is nonzero, then it must be even, and the
Chapter 1: Fundamental Concepts
graph consisting of ii 1 isolated vertices plus degree list d (multigraphs allow loops).
ioops at one vertex has
Otherwise, d has at least two nonzero entries, and d1. Replacing these with ci 1 and d1 1 yeilds a list ci' with smaller even sum, By the induction hypothesis, some graph G' with degree llst d'. Form G by adding an edge with endpoints u and v to G', where 4i(u) = ci, 1 and dGl(v) =
Although u and v may already be adjacent in G', the
resulting multigraph G has degree list ci. Proof 2 (induction on n). For n = 1, put d1/2 loops at v1. If is even, put loops at and apply the induction hypothesis. Otherwise, put an edge from to some other vertex corresponding to positive ci, (which exists since >d is even) and proceed as before. 1.3.57. An n-tuple of nonnegative integers with largest entry k is graphic if the sum is even, k k + 1, then ci' has ks and (k— 1)s. Ifq = k+ 1, then ci' has only (k 1)s. Ifq k. Since the neighborhood of x1 is 0. Since the degree
consists of k copies of a and a k copies of b, with a >
sum must be even, the quantity ka + (a
k)b must be even. In addition,
the inequality ka -c k(k 1) + (a k) min must hold, since each vertex with degree b has at most min(k, b> incident edges whose other endpoint has degree a. We construct graphs with the desired degree sequence when these conditions hold. Note that the inequality implies a -c a 1. Case 1: b k and a a k. Begin with having partite sets X of size k andY of size a k. If k(a a +k) and (a k)(b k) are even, then add an (a a -f- k)-regular graph on X and a (b k)-regular graph on V. To show that this is possible, note first that 0 -c a a + k -c k 1 and 0 -c b k -c a k -c a k 1. Also, when pq is even, a q-regular graph on p vertices in a circle can be constructed by making each vertex adjacent to the [q/2j nearest vertices in each direction and also to the opposite vertex if q is odd (since then p is even). Note that k(a a +k) and (a k)(b k) have the same parity, since their difference ak (a k)b differs from the given even number ka + (a k)b by an even amount. If they are both odd, then we delete one edge from -k' and now one vertex in the subgraph on X should have degree a a + k + 1 and one in the sub graph on V should have degree b k+l. When pq is odd, such a graph on vertices v0, . vp_i in a circle (q-regular except for one vertex of degree q + 1) can be constructed by making each vertex adjacent to the (q 1)/2 nearest vertices in each direction and then adding the edges 0 -c i -c (p 1)/2. Note that all vertices are incident to one of the added edges, except that V(p_1)/2 is incident to two of them. Case 2: k 1 -c a - b. Put a set T of a k additional vertices in a circle. For each vertex in 5, add a k + 1 consecutive neighbors
in T, starting the next set immediately after the previous set ends, Since a -c a 1, each vertex in S is assigned a k + 1 distinct neighbors in T. Since k(a k + 1) ii k, no vertex of S receives
since b
ci,). The basis step is )ci, = 0, realized by an independent set, Suppose that >ci, > 0; we consider two cases. If cii = L=2 ci,, then the desired graph consists of cii edges from vi to v2, . v. If ci,, then the difference is at least 2, because the total degree sum cii - 0. If cii > ci2, then we can subtract 1 from cii and from ci2 to obtain ci' with smaller sum. Still cii 1 is a largest value in ci' and is bounded by the sum of the other values. If cii = ci2, then we subtract 1 from each of the two smallest values to form ci'. If these are cii and ci2, then ci' has the desired properties, and otherwise ci, exceeds cii by at least 2, and agaln ci' has the desired properties. In each case, we can apply the induction hypothesis to ci' and complete the proof as in Proof 1. Proof 3 (local change). Every nonnegative integer sequence with even sum is reahzable when loops and multiple edges are allowed. Given such a realization with a loop, we change it to reduce the number of loops without changing vertex degrees. Eliminating them all produces the desired realization. If we have loops at distinct vertices u and v, then we replace two loops with two copies of the edge uv. If we have loops only at v and have an edge xy between two vertices other than v, then we replace one loop and one copy ofxy by edges vx and vy. Such an edge xy must exist because the sum of the degrees of the other vertices is as large as the degree of v. 1.3.64. A simple graph with degree sequence cii
ci,, is connected
ifciy > jforallj such thatj -c a— 1—ci. Let V(G) = ,with ci(v,) = ci,, and let H be the component of G contalning v,,; note that H has at least 1 + ci,, vertices. If G is not connected, then G has another component H'. Let j
the number of vertices in H'. Since H has at least
1 + ci,, vertices, we have j -c a 1 ci. By the hypothesis, > j. Since H' has j vertices, its maximum degree is at least ci1. Since ci1> j, there are at
Chapter 1: Fundamental Concepts
least j + 1 vertices in H', which contradicts the definition off. Hence G is in fact connected. 1.3.65. If D = is a set of distinct positive integers, with 0 2, take a chque Q with vertices and an independent set S with vertices. Each vertex of S has neighborhood Q, and each vertex of Q is adjacent to all other vertices. Other vertices have neighbors in Q and none in 5, so the degree set of G Q S should By the induction hypthesis, there is a simple be 1 and 0 d(y'), there exists x' E X sothat y ÷*x' andy' Switchingxy',x'yforxy,x'y'increases
Chapter 1: Fundamental Concepts
N(x) P with the same bipartition. Iterating this reaches N(x) = 8; let G' be the resulting graph. Doing the same in H yields graphs G' from G and H' from H such that = NH'(x). Deleting x and applying the induction hypothesis to the graphs G* = G' x and H* = H' x completes the construction of the desired sequence of 2-switches. Proof 2 (induction on number of discrepancies). Let F be the X, Ybigraph whose edges are those belonging to exactly one of G and H. Let d = e(F). Since G and H have identical vertex degrees, each vertex of F has the same number of incident edges from E(G) E(H) and E(H) E(G). When d > 0, F therefore has a cycle alternating between E(G) and E(H) (when we enter a vertex on an edge of one type, we can exit on the other type, we can't continue forever, and all cycles have even length). Let C be a shortest alternating cycle in F, with first xy a E(G) E(H) and then yx' a E(H) E(G) and x'y' a E(G) E(H), We consider a 2switch involving (x, y, x', y'>. If y'x a E(H) E(G), then the 2-switch in G reduces d by 4. If y'x a E(G) E(H), then we would have a shorter cycle in F. If y'x E(G) U E(H), then we perform the 2-switch in G; if y'x a E(G) U E(H), then we perform the 2-switch in H. Each of these last two cases yields a new pair of graphs with d reduced by 2, and the induction hypothesis applies to this pair to provide the rest of the exchanges.
1.4. DIRECTED GRAPHS 1.4.1. Digraphs in the real world. Many digraphs based on temporal order have no cycles. For example, given a set of football games, we can put an edge from game x to game y if game x ends before game y begins. The relation a parent of" also works. Asymmetric digraphs without cycles often arise from tournaments. Each team plays every other team, and there is an edge for each game from the winner to the loser. The result can be without cycles, but usually cycles exist. Another example is the relation sent a letter to".
1.4.2. If the first switch becomes disconnected from the wiring in the lightswitch system of Application 1.4.4, then the digraph for the resulting system is that below.
Section 1.4: Directed graphs
1.4.3. Every u, v-walk in a digraph contains a u, v-path. The shortest u, vwalk contained in a u, v-walk W is a u, v-path, since the shortest walk has no vertex repetition. 1.4.4. Every closed walk of odd length in a digraph contains the edges ofan odd cycle. The proof follows that of the corresponding statement for graphs in Lemma 1.2.15, given that the definitions of walk and cycle require the head of each edge to be the tail of the next edge. We use induction on the length 1 of a closed odd walk W. Basis step: 1 = 1. A closed walk of length 1 traverses a cycle of length 1. Induction step: 1 > 1. Assume the claim for closed odd walks shorter than W. If W has no repeated vertex (other than first = last), then W itself forms a cycle of odd length. If vertex v is repeated in W, then we view W as starting at v and break W into two v, v-walks, Since W has odd length, one
of these is odd and the other is even. The odd one is shorter than W. By the induction hypothesis, it contains an odd cycle, and this cycle appears in order in W. 1.4.5. A finite directed graph contains a (directed) cycle if every vertex is the tail of at least one edge (has positive outdegree). (The same conclusion
holds if every vertex is the head of at least one edge.) Let G be such a graph, let P be a maximal (directed) path in G, and let x be the final vertex
of P. Since x has at least one edge going out, there is an edge xy. Since P cannot be extended, y must belong to P. Now xy completes a cycle with the y, x-subpath of P. 1.4.6. The Dc Bruijn graphs D2 and D3.
Chapter 1: Fundamental Concepts
1.4.7. In an orientation of a simple graph with 10 vertices, the vertices can have distinct outdegrees. Take the orientation of the complete graph with vertices 0, . 9 by orienting the edge ij from i to j if i > j. In this digraph, the outdegree of vertex I is I.
1.4.8. There is an n-vertex tournament with d+(v) = t(v) for every vertex
v if and only if n is odd. If n is even, then cft(v) + t(v) = n 1 is odd, so the summands can't be equal integers. For odd n, we construct such a tournament. Proof 1 (explicit construction). Place the n vertices equally spaced around a circle, and direct the edges from v to the (n 1)72 vertices that follow v in the clockwise direction, After doing this for each vertex, the (n 1)12 nearest vertices in the counterclockwise direction from v have edges directed to v, and each edge has been oriented.
Proof 2 (inductive construction). When n = 1, the 1-vertex tournament satisfies the degree condition. For k > 1, suppose that T is a tournament with 2k 1 vertices that satisfies the condition. Partition V(T) into = k and = k 1. Add two vertices x andy. Add sets A and B with all edges from x to A, from A to y, from y to B, and from B to x. Each vertex in V(T) now has one predecessor and one successor in . We have d(x) = k k 1, d(y) = k. Complete the construction of V by adding the edge yx. Now V is a tournament with 2k + 1 vertices that satisfies the degree condition. Proof 3 (Eulerian graphs). When a is odd, is a connected even graph and hence is Eulerian. Orienting edges of K,, in the forward direction while following an Eulerian circuit yields the desired tournament. 1.4.9. For each a, there is an n-vertex digraph in which the vertices have distinct indegrees and distinct outdegrees. Using vertices v1, . v. let the edges be [vt,1: 1-cl cc j 0. With equal indegree and outdegree, each vertex in the nontrivial component of the underlying graph of our digraph G has outdegree at least 1 in G. By Lemma 1.2.25, G has a cycle C. Let G' be the digraph obtained from G by deleting E(C). Since C has 1 entering and 1 departing edge at each vertex, G' also has equal indegree and outdegree at each vertex. Each component of the underlying graph H' of G' is the underlying graph of some subgraph of G'. Since G' has fewer than nz edges, the induction hypothesis yields an Eulerian circuit of each such subgraph of G'. To form an Eulerian circuit of G, we traverse C, but when a component of H' is entered for the first time we detour along an Eulerian circuit of the corresponding subgraph of G', ending where the detour began. When we complete the traversal of C, we have an Eulerian circuit of G.
1.4.20. A digraph is Eulerian if and only if indegree equals outdegree at every vertex and the underlying graph has at most one nontrivial compo-
nent. The conditions are necessary, since each passage through a vertex uses one entering edge and one departing edge. For sufficiency, suppose that G is a digraph the conditions. We prove first that every non-extendible trail in G is closed. Let T be a non-
extendible trail starting at u. Each time T passes through a vertex v other than u, it uses one entering edge and one departing edge. Thus upon each arrival at v, T has used one more edge entering v than departing v. Since d+ (v) = (v), there remains an edge on which T can continue. Hence a non-extendible trail can only end at v and must be closed. We now show that a trail of maximal length in G must be an Eulerian circuit. Let T be a trail of maximum length; T must also be non-extendible, and hence T is closed. Suppose that T omits some edge e of G. Since the underlying graph of G has only one nontrivial component, it has a shortest path from e to the vertex set of T. Hence some edge e' not in T is incident to some vertex v ofT. It may enter or leave v. Since T is closed, there is a trail T' that starts and ends at v and uses the same edges as T. We now extend T' along e' (forward or backward depending on whether e leaves or enters v) to obtain a longer trail than T, This contradicts the choice ofT, and hence T traverses all edges of G.
1.4.21. A digraph has anEulerian trail if and only if the underlying graph has only one nontrivial component and d(v) = d+(v) for all vertices or for all but two vertices, in which case in-degree and out-degree differ by one for the other two vertices. Sufficiency: since the total number of heads equals
the total number of tails, the vertices out of balance consist of x with an extra head and y with an extra tall. Add the directed edge xy and apply the characterization above for Eulerian digraphs.
Section 1.4: Directed graphs
If D is a digraph with d(v) = for every vertex v, except that t(x) = = t(y) then D contains kpairwise edge-disjoint Ac
x, y-paths. Form a digraph D' by adding Ac edges from yto x. Since indegree equals outdegree for every vertex of D', the of D' containing x
and y is Eulerian. Deleting the added edges from an Eulerian circuit cuts it at Ac places; the resulting Ac directed trails are x, y-trails in the digraph
D. As proved in Chapter 1, the edge set of every x, y-trail contains an x, y-path; the proof in Chapter 1 applies to both graphs and digraphs.
1.4.23. Every graph G has an orientation such that
Proof 1 (Eulerian circuits). Add edges to pair up vertices of odd degree (if any exist). Each component of this supergraph G' is Eulerian. Orient G' by following an Eulerian circuit in each component, orienting each edge forward as the circuit is traversed. The circuit leaves each vertex the same number of times as it enters, so the resulting orientation has equal indegree and outdegree at each vertex. Deleting the edges of E(G') E(G) now yields the desired orientation of G, because at most one edge was added at each vertex to pair the vertices of odd degree. Deleting at most one incident edge at v produces difference at most one between d+(v) and d (v).
Proof 2 (induction on e(G)). If e(G) = e(G) >
then the claim holds. For
if G has a cycle H, then orient H consistently, with no imbalance
anywhere. If G has no cycle, then find a maximal path H and orient it consistently. This creates imbalance of 1 at the endpoints and 0 elsewhere.
The endpoints have degree 1, so no further imbalance occurs there. In both cases, delete E(H) and apply the induction hypothesis to complete the orientation. 1.4.24. Not every graph has an orientation such that for every vertex subset, the numbers of edges entering and leaving differ by at most one. Let G be a graph with at least four vertices such that every vertex degree is odd. Let
D be an orientation of G. In D, no vertex of G has the same number of vertices entering and leaving. Let S = d(v)>. Since each d+(v) and edge within S contributes the same amount to d+(v) L05 d (v) more edges leaving S than entering. The there are difference is at least Similarly, for the absolute difference is at least so always some set has difference at least n(G)/2. 1.4.25. Orientations and P3-decomposition. a) Every connected graph has an orientation having at most one vertex with odd outdegree. Proof 1 (local change). Given an orientation of G with vertices x and y having odd outdegree, find an x, y-path P in the underlying graph and ifip
Chapter 1: Fundamental Concepts
the orientation of every edge on P. This does not change the parity of the outdegree for any internal vertex of P, but it changes the parity of the outdegree for the endpoints, which previously had odd outdegree. Hence this operation reduces the number of vertices of odd outdegree by 2. We can apply this operation whenever at least two vertices have odd outdegree, so we can reduce the number of vertices with odd outdegree to 0 or 1. Proof 2 (application of Eulerian circuits). Suppose that G has 2k vertices of odd degree. Add edges that pair these vertices to form an Eulerian supergraph G'. Follow an Eulerian circuit of G', starting from u along uv a E(G), producing an orientation of G as follows. Orient uv out from U; now u has odd outdegree and all other vertices have even outdegree. Subsequently, when the circuit traverses an edge xy a E(G), orient it so that x has even outdegree among the edges oriented so far, At each stage, the only vertex that can have odd outdegree among edges of G is the current vertex. The orientation chosen for the edges not in E(G) is unimportant. b) A simple connected graph with an even number of edges can be decomposed into paths with two edges. Since the sum of the outdegrees is the number of edges, the parity of the number of vertices with odd outdegree is the same as the parity of the number of edges. Hence part (a) implies that a connected graph with an even number of edges has an orientation in which every vertex has even outdegree. At each vertex, pair up exiting edges arbitrarily. Since G is simple, this decomposes G into copies of
1.4.26. De Bru4jn cycle for binary words of length 4, avoiding 0101 and 1010. Make a vertex for each of the 8 sequences of length 3 from the alphabet S = (0, 1>. Put an edge from sequence a to sequence b, with label a a 5, if b is obtained from a by dropping the first letter of a and appending a to
the end. Travehng this edge from a corresponds to having a in sequence after a. We want our digraph to have 14 edges corresponding to the desired 14 words, and we want an Eulerian circuit through them to generate the cyclic arrangement of labels. The difference between this digraph and the De Bruijn digraph in Application 1.4.25 is omitting the two edges joining 010 and 101. The resulting digraph still has indegree = outdegree at every vertex, so it is Eulerian. One arrangement of labels generated by an Eulerian circuit is 00001001101111. 1.4.27. Dc Bruijn cycle for any alphabet and length. When A is an alphabet of size k, there exists a cychc arrangement of k' characters chosen from A such that the kt strings of length 1 in the sequence are all distinct. Icka: The indegree and outdegree is k at each vertex of the digraph constructed in the matter analogous to that for k = 2. Thus the digraph is Eulerian, and recording the edge labels along an Eulerian circuit yields the desired sequence. Below we repeat the details.
Section 1.4: Directed graphs
Define a digraph Dk,7 whose vertices are the (F 1)-tuples with elements in A. Place an edge from a to b if the last a 2 entries of a agree with the first a 2 entries of b, Label the edge with the last entry of Ii. For each vertex a, there are Ac ways to append a element of A to lengthen its name, and hence there are Ac edges leaving each vertex. Similarly, there are Ac choices for a character deleted from the front of a predecessor's name to obtain name b, so each vertex has indegree Ac. Also, we can reach b = (b1 from any vertex by successively following the edges labeled b1, . Since Dkj is strongly connected and has indegree equal to outdegree at every vertex, the characterization of Eulerian digraphs implies that Dk,j is Eulerian.
Let C be an Eulerian circuit of Dk,j. When we are at the vertex with name a = (a1, . an_i) while traversing C, the most recent edge had label because the label on an edge entering a vertex agrees with the last
digit of the sequence at the vertex. Since we delete the front and shift the rest to obtain the rest of the label at the head, the successive earlier labels (looking backward) must have been al in order. If C next traverses an edge with label then the subsequence consisting of the a most recent edge labels at that time is a1, . Since the k'S' vertex labels are distinct, and the edges leaving each vertex have distinct labels, and we traverse each edge from each vertex exactly once along C, the Ac' strings of length F in the circular arrangement given by the edge labels along C are distinct.
1.4.28. Dc Bruijn cycle for length 4 without the constant words. Make a vertex for each of the m3 sequences of length 3 from the alphabet S. Put an edge from sequence a to sequence b, with label a c 5, if Ii is obtained from a by dropping the first letter and appending a to the end. Since there are in ways to append a letter, the out-degree of each vertex is in. For each sequence, there are in possible letters that could have been deleted to reach it, so the in-degree of each vertex is in. Deleting the loops at the in constant vertices (aaa, bbb, etc.) reduces the indegree and outdegree at those vertices by 1, so the resulting digraph has equal indegree and outdegree at every vertex. Also the underlying graph is connected, since vertex abc can be reach from any other vertex by following the edge labeled a, then b, then c, Thus an Eulerian circuit exists. Recording the edge labels while following an Eulerian circuit yields the desired arrangement. The 4-digit strings obtained are those formed by the 3-digit name of a vertex plus the label on an exiting edge. These in4 in strings are distinct and avoid the constant words, since the loops were deleted from the digraph. Alternative proof If we know (from Exercise 1.4.27, for example) that
Chapter 1: Fundamental Concepts
there exists a De Bruijn cycle including the constant words, then we can simply delete one letter from each string of four consecutive identical letters, without using graph theory. 1.4.29. A strong orientation of a graph that has an odd cycle also has an odd (directed) cycle. Suppose that D is a strong orientation of a graph G that has an odd cycle v1, . V2k+1. Since D is strongly connected, for each i there is a v,, i every such path has even length, then the edge between v and points from to v,, since the other orientation would be a v,, of length 1 (odd). In this case, we have an odd cycle through v and v Otherwise, we have a path of odd length from each v, to Combining these gives a closed trail of odd length. In a digraph as well as in a graph (by the same proof), a closed odd trail contains the edges of an odd cycle.
1.4.30. The maximum length of a shortest spanning closed walk in a strongly-connected n-vertex digraph is Ln + 1)2/4j if n > 3. For the lower bound, let G consist of a u, v-path P of n vertices, plus 1 vertices with edges from v and to u. When leaving a vertex not on P, P must be reached and traversed before the next vertex off P. Hence G requires l(n 1 + 1) steps to walk through every vertex, maximized by setting 1 = + 1)/2j. The length of the walk is then L(n + 1)2/4j. For any strongly-connected n-vertex digraph G, we obtain a spanning closed walk of length at most L(n + 1)2/4j. Let in be the maximum length of a path in G; from each vertex to every other, there is a path of length at most in. Begin with a path P of length in; this visits in + 1 vertices. Next 1
use paths to reach each of the remaining vertices in turn, followed by a path returning to the beginning of P. In this closed walk, 1 + (n in 1) +1 paths have been followed, each of length at most in. The total length is at most m(n + 1 in), which is bounded by L(n + 1)2/4j. 1.4.31. The smallest nonisomorphic pair of tournaments with the same score sequences have five vertices. At least five vertices are needed. The score sequence (outdegrees) of an
n-vertex tournament can have only one 0 or n 1, Nonisomorphic tournaments with such a vertex must continue to be nonisomorphic when that vertex is deleted. Hence a smallest nonisomorphic pair has no vertex with score 0 or n 1. The only such score sequences with fewer than 5 vertices are 111 and 2211. The first is reahzed only by the 3-cycle. For 2211, name the low-degree vertices as vi and v2 such that vi +— V2, and name the highvertices as v3 and v4 such that vg ÷— v4. The only way to complete a N+(v2) = with this score sequence is now N+(vi) = N+(v3) = [vi, 1)21, and N'jv4) = . Five vertices suffice, by construction. On five vertices, the sequences to degree
Section 1.4: Directed graphs
consider are 33211, 32221, and 22222. There is only one isomorphic class with score sequence 22222, but there are more for the other two sequences. In fact, there are 3 nonisomorphic tournaments with score sequence 32221. They may be characterized as follows: (1) the bottom player beats the top player, and the three middle players induce a cyclic subtournament; (2) the top player beats the bottom player, and the three middle players induce a cyclic subtournament; (3) the top player beats the bottom player, and the three middle players induce a transitive subtournament.
Five vertices suffice, by counting Each score sequence sums to 10 and
has maximum outdegree at most 4; also there is at most one 4 and at most one 0. The possibilities are thus 43210, 43111, 42220, 42211, 33310, 33220, 33211, 32221, 22222. There are 210 tournaments on five vertices; we show that they cannot fit into nine isomorphism classes. The isomorphism class consisting of a 5-cycle plus edges from each vertex to the vertex two later along the cycle occurs 4! times; once for each cyclic ordering of the vertices. Each of the other isomorphism classes occurs at most 5! times. Hence the nine isomorphism classes contain at most 24 -F- 8• 120 of the 210
tournaments. Since 1024 >
there must be at least 10 isomorphism
classes among the nine score sequences.
1.4.32. Characterization of bigraphic sequences. With P = P1, . Pm and
q = qi, . qn, the pair (p, q) is bigraphic if there is a simple bipartite graph in which pi, . Pm are the degrees for one partite set and qj, . are the degrees for the other.
If p has positive sum, then (p, q) is bigraphic if and only if (p', q') is bigraphic, where (p', q') is obtained from (p, q) by deleting the largest element A from p and subtracting 1 from each of the A largest elements of q. We follow the method of Theorem 1.3.3 1. Sufficiency of the condition follows by adding one vertex to a realization of the smaller pair. For necessity, choose indices in a realization G so that P1 Pm, qn, qi = p,, and d(yy) = qy. We produce a realization in If yy xl for some j Since q1 > there exists x, with i > 1 such that E N(y1) N(yk). We perform the 2-switch to replace with
x, yk>. This reduces the number of missing neighbors, so we can obtain
the desired realization. (Comment: the statement also holds when in =
Chapter 1: Fundamental Concepts
1.4.33. Bipartite 2-switch and 0,1-matrices with fixed row and column sums. With a simple X, Y-bigraph G, we associate a 0,1-matrix B(G) with rows indexed by X and columns indexed by Y. The matrix has a 1 in position i, j if and only if x, Applying a 2-switch to G that exchanges xy, x'y' for xy', x'y (preserving the bipartition) affects B(G) by interchanging the 0's and l's in the 2 by 2 permutation submatrix induced by rows x, and columns y, y'. Hence there is a sequence of 2-switches transforming G to H without changing the bipartition if and only if there is a sequence of switches on 2 by 2 permutation submatrices that transforms 11(G) to B(H). Furthermore, G and H have the same bipartition and same vertex degrees if and only if B(G) and B(H) have the same row sums and the same column sums. Therefore, in the language of bipartite graphs the statement about matrices becomes bipartite graphs with the same bipartition and vertex degrees can be reached from each other using 2-switches preserving the bipartition." We prove either statement by induction. We use the phrasing of bipartite graphs. Proof 1 (induction on in). If in = 1, then already G = H. For in> 1, let G be an X, Y-bigraph. Let x be a vertex of maximum degree in X, with d(x) = k. Let S be a set of k vertices of highest degree in Y. Using bipartition-preserving 2-switches, we transform G so that N(x) = S. If N(x) 5, we choose y C S and y' C Y S so that x y and x ±* y'. Since d(y) > d(y'), we have x' eX sothaty -e- x' andy' t÷x'. Switchingxy',x'y for xy, x'y' increases N(x) 11 Iterating this reaches N(x) = S. We can do
the same thing in H to reach graphs G' from G and H' from H such that N0(x) = NH'(x). Now we can delete x and apply the induction hypothesis to the graphs G" = G' x and H" = H' x to complete the construction of the desired sequence of 2-switches. Proof 2 (induction on number of discrepancies). Let F be the bipartite graph with the same bipartition as G and H consisting of edges belonging to exactly one of G and H. Let d = e(F). Orient F by directing each edge of G E(H) from X to Y and each edge of H e(G) from Y to X. Since G, H have identical vertex degrees, in-degree equals outdegree at each vertex
0, this implies that F contains a cycle. There is a 2-switch in G that introduces two edges of E(G) E(H) and reduces d by 4 if and
only if F has a 4-cycle. Otherwise, Let C be a shortest cycle in F, and let x, y,x', y' be consecutive vertices on C. We have xy e E(G) x'y C E(H) E(G), and x'y' c E(G) E(H), We also have xy' E(G), else we could replace these three edges of C by xy' to obtain a shorter cycle in F. We can now perform the 2-switch in G that replaces xy, x'y' with xy', x'y. This reduces d by at least 2.
1.4.34. 1/' G and H are two tournaments on a vertex set V, then
Section 1.4: Directed graphs
4(v) for all v c V if and only if G can be turned into H by a sequence of direction-reversals on cycles of length 3. Reversal of a 3-cycle changes no outdegree, so the condition is sufficient. For necessity, let F be the subgraph of G consisting of edges oriented the opposite way in H. Since 4(v) = 4(v) and 4(v) = 4(v) for all v, every vertex has the same indegree and outdegree in F. Let x be a vertex of maximum degree in F, and let S = and T = An edge from S to T in G completes a 3-cycle with x whose reversal in G reduces the number of pairs on which G and H disagree. An edge from
T to S in H completes a 3-cycle with x whose reversal in H reduces the number of disagreements. If neither of these possibilities occurs, then G orients every edge of S x T from T to 5, and H orients every such edge from
S to T. Also F has edges from T to x. This gives every vertex ofT higher outdegree than x in F, contradicting the choice of x. the sequence of outdegrees of a tournament if and n = mn\ Necessity. A tournament has )j1p, edges in total, and any k vertices have out-degree-sum at least within the subtournament they induce. = Sufficiency. Given a sequence p satisfying the conditions, let We prove sufficiency by induction on The Pi and e1 = a only sequence p with >e1 = 0 is 0, 1 1; this is realized by the transitive tournament having v1 —÷ vy if and only if k> j. If > > 0, 1.4.35. P1 -c
ey
0, and let s be the least index above r with = 0, which exists since = 0. We have q1_j > q1 = (), and > s and Pr 2. Similarly, q1--i> fl1). This yields ifr = 1 we have P1> 1, and ifr > 1 we have Pr Pr-i > 2. Hence we can subtract one from Pr and add one to to obtain a new sequence p' that is non-decreasing, satisfies the conditions, and reduces e1 by s r, By the induction hypothesis, there is a tournament with score sequence p'. If v1 —± in this tournament, we can reverse this edge to obtain the score sequence p. If not, then the fact that > implies there is another vertex u such that v1 —÷ u and u —÷ vr; obtain the desired tournament by reversing these two edges. 1.4.36. Let T be a tournament having no vertex with indegree 0. a) Ifx is a king in T, then T has another king in N(x). The subdigraph induced by the vertices of N (x) is also a tournament; call it T'. Since every
tournament has a king, T' has a king. Let y be a king in P. Since x is a successor of y and every vertex of N+(x) is a successor of x, every vertex of V(T) V(T') is reachable from y by a path in T of length at most T. Hence y is also a king in the original tournament T,
b) T has at least three kings. Since T is a tournament, it has some
Chapter 1: Fundamental Concepts
king, x, By part (a), T has another king y in N(x). By part (a) again, T has another king z in N(y). Since y —÷ x, we have x N(y), and hence x. Thus x, y, z are three distinct kings in T. z c) For n > 3, an n-vertex tournament T with no source and only three kings. Let S = (x, y, z> be a set of three vertices in V(T). Let the subtournament on S be a 3-cycle. For all edges joining S and V(T) 5, let the endpoint in S be the tail. Place any tournament on V(T) S. Now x, y, z are kings, but no vertex outside S is a king, because no edge enters S.
1.4.37. Algorithm to find a king in a tournament T: Select x c V(T). If x has indegree 0, call it a king and stop. Otherwise, delete U from T to form T', and call the output from T1 a king in T. We prove the claims by induction on the number of vertices. The algorithm terminates, because
it either stops by selecting a source (indegree 0) or moves to a smaller tournament. By the induction hypothesis, it terminates on the smaller tournament. Thus in each case it terminates and declares a king. We prove by induction on the number of vertices that the vertex declared a king is a king. When there is only one vertex, it is a king. Suppose that n(T) > 1. If the initial vertex x is declared a king immediately, then
it has outdegree n 1 and is a king. Otherwise, the algorithm deletes x and its successors and runs on the tournament P induced by the set of predecessors (in-neighbors) of x. By the induction hypothesis, the vertex z that the algorithm selects as
king in P is a king in T', reaching each vertex of P in at most two steps. It suffices to show that z is also a king in the full tournament. Since T' contains only predecessors of x, z —÷ x. Also, z reaches all successors of x in
two steps through x. Thus z also reaches all discarded vertices in at most two steps and is a king in T. 1.4.38. Tournaments with all players kings. a) If n is odd, then there is an tournament with n vertices such that every player is a king. Proof 1 (explicit construction). Place the players around a circle. Let each player defeat the (n 1)12 players closest to it in the clockwise direction, and lose to the (n 1)/2 players closest to it in the counterclockwise direction. Since every pair of players is separated by fewer players around one side of the circle than the other, this gives a well-defined orientation to each edge. All players have exactly (n 1)/2 wins. Thus every outdegree is the maximum outdegree, and we have proved that every vertex of maximum outdegree in a tournament is a king. It is also easy to construct explicit paths. Each player beats the next (n 1)/2 players. The remaining (n 1)12 players all lose to the last of these first (n 1)72 players. The construction is illustrated below for five players.
Section 1.4: Directed graphs
Proof 2 (induction on a). For a = 3, every vertex in the 3-cycle is a king. For a 3, given a tournament on vertex set S of size a in which every vertex is a king, we add two new vertices x, y. We orient S —* x -÷ y -÷ S. Every vertex of S reaches x in one step and y in two; x reaches y in one step and each vertex of S in two. Every vertex is a king. (The resulting tournaments are not regular.) Note: Since there is no such tournament when a = 4, one must also give an explicit construction for a = 6 to include in the basis. The next proof avoids this necessity. Proof 3 (induction on a). For a = 3, we have the cyclic tournament. For a = 5, we have the cyclically symmetric tournament in which each vertex beats the two vertices that follow it on the circle. For a > 5, let T be an (a 1)-vertex tournament in which every vertex is a king, as guaranteed by the induction hypothesis. Add a new vertex .x. If a is odd, then partition V(T) into pairs. For each pair, let a and b be the tail and head of the edge joining them, and add the edges xc and lix. If a is even, then among any four vertices of V (T) we can find a triple u, v, w > that induces a non-cyclic tournament. Pick one such triple, and partition the remaining vertices of V (T) into pairs. Treat the edges joining x to these pairs as in the other case. Letting u be the vertex of the special triple with edges to the two other vertices, add edges xu, vx, and wx. b) There is no tournament with four players in which every player is a king. Suppose G is such a tournament. A player with no wins cannot be a king. If some vertex has no losses, then no other vertex can be a king. Hence every player of G has 1 or 2 wins. Since the total wins must equal the total losses, there must be two players with 1 win and two players with 2 wins. Suppose x, y are the players with 1 win; by symmetry, suppose x
beats y. Since x has no other win and y has exactly one win, the fourth player is not reached in two steps from x, and x is not a king. 1.4.39. Every loopless digraph D has a vertex subset S such that D[SI has no edges but every vertex is reachable from S by a path of length at most 2. Proof 1 (induction). The claim holds when a(D) = 1 and when there is a vertex with edges to all others. Otherwise, consider an arbitrary vertex x, and let D' = D x Let 5' be the subset of V(D') guaranteed by the induction hypothesis. Observe that 5' fl = 0. If yx a E(D) for
Chapter 1: Fundamental Concepts
some y c 8', then x U N+(x) is reachable from y within two steps, and 8' is the desired set 8. Otherwise, the set 8 = 8' U works. Proof 2 (construction). Index the vertices as v1, . Process the list in increasing order; when a vertex v, is reached that has not been deleted, delete all successors of v, with higher indices. Next process the list in decreasing order; when a vertex v1 is reached that has not been deleted (in either pass), delete all successors of v, with lower indices. The set 8 of vertices that are not deleted in either pass is independent. Every vertex deleted in the second pass has a predecessor in 8. Every
vertex deleted in the first pass can be reached from 8 directly or from a vertex deleted in the second pass, giving it a path of length at most two from 8. Hence 8 has the desired properties. Proof 3 (kernels). By looking at the reverse digraph, it suffices to show that every loopless digraph D has an independent set 8 that can be reached by a path of length at most 2 from each vertex outside 8. Given a vertex ordering v1, . decompose D into two acyclic spanning subgraphs G and H consisting of the edges that are forward and backwards in the ordering, respectively. All subgraphs of G and H are acycic, and hence by Theorem 1.4.16 they have kernels. Let 8 be a kernel of the subgraph of G induced by a kernel T of H. Every vertex not in T has a successor in T, and every vertex in T 8 has a successor in 8, so every vertex not in 8 has a path of length at most 2 to 8. (Comment: The set 8 produced in this way is the same set produced in the reverse digraph by Proof 2. This proof is attributed to S. Thomasse on p. 163 of J. A. Bondy, Short proofs of classical theorems, J Graph Theory 44 (2003), 159—165.)
1.4.40. The largest unipathic subgraphs of the transitive tournament have edges. If a subgraph of contains all three edges of any 3-vertex induced subtournament, then it contains two paths from the least-indexed of these vertices to the highest. Hence a unipathic subgraph must have as its underlying graph a triangle-free subgraph of K. By Mantel's Theorem, the maximum number of edges in such a subgraph is achieved only by the complete equibipartite graph. This leaves the problem of finding unipathic orientations of K [n/2j [n/21 in T. Suppose G is such a subgraph, with partite sets X, Y. If there are four vertices, say i 4 all the vertices of X must precede all the vertices of Y, or vice versa. To obtain K[,,/2j[,,/2], we will have all edges ij such that i -c lfi/2j and and j > ['2721. Hence for a 4 I > [n/2j, or all edges such that i there are two extremal sub graphs when a is odd and only one when a is
Section 1.4: Directed graphs
even. (There is only one when ii = 1, and there are three when ii = 1.4.41. Given any listing of the vertices of a tournament, every sequence of switchings of consecutive vertices that induce a reverse edge leads to a list with no reverse edges in at most steps. Under this algorithm, each switch changes the order of othy one pair. Furthermore, the order of two elements in the list can change only when they are consecutive and induce a reverse
edge. Hence each pair is interchanged at most once, and the algorithm terminates after at most () steps with a spanning path. 1.4.42. Every ordering of the vertices of a tournament that minimizes the sum of lengths of the feedback edges puts the vertices in nonincreasing order of outdegree. For the ordering the sum is the sum of j i over edges such that j > i. Consider the interchange of and If some
vertex is a successor of both or predecessor of both, then the contribution to the sum from the edges involving it remains unchanged. If x E N+ +i), then the switch increases the contribution from these edges by 1. N+(v3, then the switch decreases the contribution from If x c these edges by 1. If then the switch increases the cost by 1, otherwise it decreases. Hence the net change in the sum of the lengths of feedback edges is This implies that if the ordering has any vertex followed by a vertex with larger outdegree, then the sum can be decreased. Hence minimizing the sum puts the vertices in nonincreasing order of outdegree. Furthermore, permuting the vertices of a given outdegree among themselves does not change the sum of the lengths of feedback edges, so every ordering in nonincreasing order of outdegree minimizes the sum.
Chapter 2: Trees and Distance
2.TREES AND DISTANCE 2.1. BASIC PROPERTIES 2.1.1. Trees with at most 6 vertices having specified maximum degree or diameter. For maximum degree k, we start with the star K1,4 and append leaves to obtain the desired number of vertices without creating a vertex of larger degree. For diameter k, we start with the path P4k' and append leaves to obtain the desired number of vertices without creating a longer path. Below we list all the resulting isomorphism classes. Fork = 0, the only tree is K1, and for k = 1, the only tree is K2 (diameter or maximum degree k). For larger k, we list the trees in the tables. Let denote the tree with i + j vertices obtained by starting with one edge and appending i 1 leaves to one endpoint and j 1 leaves at the other endpoint (note that T1,4 = K1,4 for k > 1). Let Q be the 6-vertex tree with diameter 4 obtained by growing a leaf from a neighbor of a leaf in P5. Let a denote the number of vertices. maximum degree k k
2.1.2. Characterization of trees. a) A graph is tree if and only if it is connected and every edge is a cutedge. An edge e is a cut-edge if and only if e belongs to no cycle, so there are no cycles if and only if every edge is a cut-edge. (To review, edge e = uv is a cut edge if and only if G e has no u, v-path, which is true if and only if G has no cycle containlng e.) b) A graph is a tree if and only if for all x, y c V(G), adding a copy of xy as an edge creates exactly one cycle. The number of cycles in G + uv
Section 2.1: Basic Properties
containing the new (copy of) edge uv equals the number of u, v-paths in G,
and a graph is a tree if and only if for each pair u, v there is exactly one u, v-path. Note that the specified condition must also hold for addition of extra copies of edges already present; this excludes cliques. 2.1.3. A graph is a tree if and only if it is loopless and has exactly one spanning tree. If G is a tree, then G is loopless, since G is acyclic. Also, G is a spanning tree of G. If G contains another spanning tree, then G contains another edge not in G, which is impossible.
Let G be loopless and have exactly one spanning tree T. If G has a edge e not in T, then T + e contains exactly one cycle, because T is a tree. Let f be another edge in this cycle. Then T + e contains no cycle. Also T+e is connected, because deleting an edge of a cycle cannot disconnect a graph. Hence T + e is a tree different from T. Since G contains no such tree, G cannot contain an edge not in T, and G is the tree T.
2.1.4. Every graph with fewer edges than vertices has a component that is a Since the number of vertices or edges in a graph is the sum of the number in each component, a graph with fewer edges than vertices must have a component with fewer edges than vertices. By the properties of trees, such a component must be a tree. 2.1.5. A maximal acyclic subgraph of a graph G consists of a spanning tree from each component of G. We show that if H is a component of G and F is a maximal forest in G, then F 1) H is a spanning tree of H. We may assume that F contains all vertices of G; if not, throw the missing ones in as isolated points to enlarge the forest. Note that F fl H contains no cycles, since F contains no cycles and F 1) H is a subgraph of F. We need only show that F fl H is a connected subgraph of H. If not, then it has more than one component. Since F is spanning and H is con-
nected, H contains an edge between two of these components. Add this edge to F and F P H. It cannot create a cycle, since F previously did not contain a path between its endpoints. We have made F into a larger forest (more edges), which contradicts the assumption that it was maximal. (Note: the subgraph consisting of all vertices and no edges of G is a spanning subgraph of G; spanning means only that all the vertices appear, and says nothing about connectedness. 2.1.6. Every tree with average degree a has 21(2 a) vertices. Let the tree have a vertices and in edges. The average degree is the degree sum divided = by a, the degree sum is twice in, and in is a 1, Thus a = 2(n 1)/n. Solving for a yields a = 2/(2 a). 2.1.7. Every n-vertex graph with in edges has at least in a + 1 cycles. Let k be the number of components in such a graph G. Choosing a spanning tree
Chapter 2: Trees and Distance
from each component uses ii k edges. Each of the remaining in ii + k edges completes a cycle with edges in this spanning forest. Each such cycle has one edge not in the forest, so these cycles are distinct. Since k > 1, we have found at least m ii + 1 cycles.
2.1.8. Characterization of simple graphs that are forests. a) A simple graph is a forest if and only if every induced subgraph has a vertex of degree at most 1. If G is a forest and H is an induced subgraph of G, then H is also a forest, since cycles cannot be created by deleting edges. Every component of H is a tree, which is an isolated vertex or has a leaf (a vertex of degree 1). If G is not a forest, then G contains a cycle. A shortest cycle in G has no chord, since that would yield a shorter cycle, and hence a shortest cycle is an induced subgraph. This induced subgraph is 2-regular and has no vertex of degree at most 1. b) A simple graph is a forest if and only if every connected subgraph is an induced subgraph. If G has a connected subgraph H that is not an induced subgraph, then G has an edge xy not in H with endpoints in V(H). Since H contains an x, y-path, H +xy contains a cycle, and G is not a forest. Conversely, if G is not a forest, then G has a cycle C, and every subgraph of G obtained by deleting one edge from C is connected but not induced. c) The number of components is the number of vertices minus the number of edges. In a forest, each component is a tree and has one less edge than vertex. Hence a forest with n vertices and k components has n k edges. Conversely, every component with n1 vertices has at least 1 edges, since it is connected. Hence the number of edges in an n-vertex is n minus vertices has the number of components only if every component with a1 1 edges. Hence every component is a tree, and the graph is a forest.
2.1.9. For 2 -c k -c n 1, the n-vertex graph formed by adding one vertex adjacent to every vertex of has a spanning tree with diameter k. Let be the vertices of the path in order, and let x be the vertex adjacent to all of them. The spanning tree consisting of the path v1, . and the edges XVk_1, has diameter k.
2.1.10. If u and v are vertices in a connected n-vertex simple graph, and d(u, v) > 2, then cl(u) + d(v) -c a + 1 d(u, v). Since d(u, v) > 2, we have N(s) 1) N(v) = 0, and hence d(u) + d(v) = N(s) U Let Ic = d(u, v). Between u and v on a shortest u, v-path are vertices x1, . Since this is a shortest u, v-path, vertices u, v and x2, . are forbidden from the neighborhoods of both u and v. Hence N(s) U -c a + 1 The inequality fails when d(u, v) -c 2, because in this case u and v can have many common neighbors. When d(u, v) = 2, the sum d(u) + d(v) can be as high as 2n 4. ,
Section 2.1: Basic Properties
2.1.11. If x and y are adjacent vertices in a graph G, then always -c 1. Az, y-path can be extended (or trimmed) to reach d0(x, z) dg(y, x, and hence d(z, x) 2, which yields radius and diameter 2. For the radius is 1 and diameter is 2 if a > 1. The radius and diameter of are 1. The radius and diameter of K0. are infinite if a > 1, and both are 0 for K0,1. 2.1.13. Every graph with diameter d has an independent set of size at least f(1 + d)/21. Let x, y be vertices with d(x, y) = d. Vertices that are nonconsecutive on a shortest x, y-path P are nonadjacent. Taking x and every second vertex along P produces an independent set of size 1(1 + d)/21,
2.1.14. Starting a shortest path in the hypercube. The distance between vertices in a hypercube is the number of positions in which their names differ. From u, a shortest u, v-path starts along any edge to a neighbor whose name differ from u in a coordinate where v also differs from u.
2.1.15. The complement of a simple graph with diameter at least 4 has diameter at most 2. The contrapositive of the statement is that if G has diameter at least 3, then G has diameter at most 3. Since G = G, this statement has been proved in the text. 2.1.16. The "square" of a connected graph G has diameter fdiam (G)/21. The square is the simple graph G' with x -e- y in G' if and only if dG(x, y) -c 2. We prove the stronger result that d0i(x, y) = Idcj(x, y)/21 for every x, y E V(G). Given an x, y-path P of length k, we can skip the odd vertices along
P to obtain anx, y-path of length Ik/21 in G'.
On the other hand, every x, y-path of length 1 in G' arises from a path of length at most 21 in G. Hence the shortest x, y-path in G' comes from the shortest x, y-path in G by the method described, and dçp (x, y) = fd0(x, y)/21. Hence y) =
2.1.17. If an n-vertex graph G has a 1 edges and no cycles, then it is connected. Let k be the number of components of G. If Ac > 1, then we adding an edge with endpoints in two components creates no cycles and reduces the number of components by 1. Doing this Ac 1 times creates a graph with (a
edges that is connected and has no cycles. Such 1 edges. Therefore, Ac = 1, and the original
a graph is a tree and has a graph G was connected.
Chapter 2: Trees and Distance
2.1.18. If G is a tree, then G has at least A(G) leaves. Let k = A(G), Given a > k > 2, we cannot guarantee more leaves, as shown by growing a path oflengthn —k lfromaleafofKl,k. Proof la (maximal paths). Deleting a vertex x of degree k produces a forest of k subtrees, and x has one neighbor w1 in the ith subtree G1. Let P1 be a maximal path starting at x along the edge x w1. The other end of P1 must be a leaf of G and must belong to G1, so these k leaves are distinct. Proof lb (leaves in subtrees). Deleting a vertex x of degree k produces a forest of k subtrees. Each subtree is a single vertex, in which case the vertex is a leaf of G, or it has at least two leaves, of which at least one is not a neighbor of x. In either case we obtain a leaf of the original tree in each subtree. Proof 2 (counting two ways). Count the degree sum by edges and by vertices. By edges, it is 2n 2. Let k be the maximum degree and 1 the number of leaves. The remaining vertices must have degree at least two each, so the degree sum when counted by vertices is at least k + 2(n 1) + 1. The inequality 2n 2 k + 2(n 1 1) + 1 simplifies to 1 k. (Note: Similarly, degree 2(n 1) k remains for the vertices other than a vertex of maximum degree. Since all degrees are 1 or at least 2, there must be at least k vertices of degree 1.) Proof 3: Induction on the number of vertices. For a -c 3, this follows by inspecting the unique tree on a vertices. For a > 3, delete a leaf u. If A(T u) = A(T), then by the induction hypothesis T u has at least k leaves. Replacing u adds a leaf while losing at most one leaf from T u. Otherwise A(T u) = A(T) 1, which happens only if the neighbor of u is the only vertex of maximum degree in T. Now the induction hypothesis yields at least k 1 leaves in T u. Replacing u adds another, since the vertex of maximum degree in T cannot be a leaf in T u (this is the reason for putting a = 3 in the basis step). 1
2.1.19. If a1 denotes the number of vertices of degree i in a tree T, then
depends only on the number of vertices in T. Since each vertex of degree i contributes i to the sum, the sum is the degree-sum, which equals twice the number of edges: 2n(T) 2.
2.1.20. Hydrocarbon formulas The global method is the simplest one. With cycles forbidden, there are k + 1 1 "bonds" - i.e., edges. Twice this must equal the degree sum. Hence 2(k + 1 1) = 4k + 1, or I = 2k + 2. Alternatively, (sigh), proof by induction, Basis step (k = 1): The formula holds for the only example. Induction step (k > 1): In the graph of the molecule, each H has degree 1. Deleting these vertices destroys no cycles, so the subgraph induced by the C-vertices is also a tree. Pick a leaf x in this tree. In the molecule it neighbors one C and three Hs. Replac-
Section 2.1: Basic Properties
ing x and these three Hs by a single H yields a molecule with one less C that also satisfies the conditions. Applying the induction hypothesis yields 1 =[2(k—1)-j-2]— 1-f-3=2k-f-2. 2.1.2 1. If a simple n-vertex graph G has a decomposition into k spanning trees, and A(G) = 6(G) + 1, then 2k 2 and d1, . . are positive integers, then there exists a tree = 2(n 1). with these as its vertex degrees if and only if c/n = 1 and (Some graphs with such degree lists are not trees.) Necessity: Every avertex tree is connected and has a 1 edges, so every vertex has degree at least 1 (when a> 2) and the total degree sum is 2(a 1). Sufficiency: We give several proofs.
Proof 1 (induction on a). Basis step (a = 2): The only such list is (1, 1), which is the degree list of the only tree on two vertices. Induction step (a > 2): Consider d1, . . the conditions. Since > a, some element exceeds 1. Since >d1 2. Create a path x, Ui. Un_k, y. For i -c i -, we compute the number of leaves as
Proof 3 (extremality). Because >d1 = 2(a 1), which is even, there is a graph with a vertices and a 1 edges that realizes a, Among such graphs, let G (having k components) be one with the fewest components. If Ac = 1, then G is a connected graph with a 1 edges and is the desired tree. If /c> 1 and G is a forest, then G has a edges. Therefore, G has a cycle. Let H be a component of G having a cycle, and let Ut) be an edge of the cycle. Let H' be another component of G. Because each a1 is positive, H' has an edge, xy. Replace the edges Ut) and xy by UX and vy (either Ut) or xy could be a loop.) Because Ut) was in a cycle, the subgraph induced by V(H) is still connected. The deletion of vy might disconnect H', but each piece is now connected to V(H), so the new graph G' realizes d with fewer components than G, contradicting the choice of G. Ac
Chapter 2: Trees and Distance
2.1.28. The nonnegative integers d1 > > c4 are the degree sequence of some connected graph if and only if d1 is even, tin> 1, and d1 > 2n 2. This claim does not hold for simple graphs because the conditions >d1 even, > 1, and > d1 > 2n 2 do not prevent d1 > a, which is impossible for a simple graph. Hence we allow loops and multiple edges. Necessity follows because every graph has even degree sum and every connected graph has a spanning tree with a 1 edges. For sufficiency, we give several proofs. Proof 1 (extremality). Since > d1 is even, there is a graph with degrees d1, . Consider a realization G with the fewest components;
since >d1 > 2n 2, G has at least a 1 edges. If G has more than one component, then some component as many edges as vertices and thus has a cycle. A 2-switch involving an edge on this cycle and an edge in another component reduces the number of components without changing the degrees. The choice of G thus implies that G has only one component. Proof 2 (induction on a). For a = 1, we use loops. For a = 2, if d1 = d2, then we use d1 parallel edges. Otherwise, we have a> 2 or d1 > d2. Form a new list d,_1 by deleting and subtracting units from other = 1, we subtract values. Ifa>> 2a—2 imphes d1 > 1. If a> 3 and > 1, we make the subtractions from any two of the other numbers. In each case, the resulting sequence has even sum and all entries at least 1. Letting D = >d1, we have >d = D = 1, then D 2a—2—2 = 2(a—1)—2. > 1,thenD > 2a 4 = 2(a 1) 2. Hence the new values satisfS7 the condition stated
a set of a 1 values. By the induction hypothesis, there is a connected graph G' with vertex degrees To obtaln the desired graph G, add a vertex v,, with d1 dl edges to the vertex with degree d1, for 1 -c i -c a 1. This graph G is connected, because a path from v to a to v in G', Proof 3 (induction on and prior result). If >1d1 = 2a 2, then Exercise 2.1.27 apphes. Otherwise, > 2a. If a = 1, then we use loops. If a > 1, then we can delete 2 from d1 or delete 1 from d1 and d2 without introducing a 0. After applying the induction hypothesis, adding one loop at v1 or one edge from v1 to v2 restores the desired degrees.
2.1.29. Every tree has a leaf in its larger partite set (in both if they have equal size). Let X andY be the partite sets of a tree T, with >If there is no leaf in X, thene(T) >2 = > = a(T). This + + contradicts e(T) 4, let v be a leaf of T, and let w be its neighbor. If w has no other leaf as neighbor, but has degree at least 3, then T v is a smaller tree the hypotheses. By the induction hypothesis, T v has a pair of leaves with a common neighbor, and these form such a pair in T. 2.1.31. A simple connected graph G with exactly two non-cut-vertices is a path. Proof 1 (properties of trees). Every connected graph has a spanning
tree. Every leaf of a spanning tree is not a cut-vertex, since deleting it leaves a tree on the remaining vertices. Hence every spanning tree of G has only two leaves and is a path. Consider a spanning path with vertices with i 2, if G is 2-regular, then G is a cycle, since G is connected. If G is not 2-regular, then it has a vertex v of degree 1. Let G' = G v. The graph G1 is connected and has n 1 vertices and a 1 edges. By the induction hypothesis, G' has exactly one cycle. Since a vertex of degree 1 belongs to no cycle, G also has exactly one cycle. 2.1.34. A simple n-vertex graph G with n > k and e(G) > a(G)(k 1) contains a copy of each tree with k edges. We use induction on a. For the basis step, let G be a graph with Ic + 1 vertices. The minimum allowed number of edges is (Ic + 1)(k 1) + 1, which simplifies to (c). Hence G = Kk+1, and T C G. For the induction step, consider a > Ic + 1. If every vertex has degree at least Ic, then containment of T follows from Proposition 2.1.8. Otherwise, deleting a vertex of minimum degree (at most Ic 1) yields a subgraph G' on a 1 vertices with more than (a 1) (Ic 1) edges. By the induction hypothesis, G' contains T, and hence T C G. 2.1.35. The vertices of a tree T all have odd degree if and only if for all e a E(T), both components of T e have odd order. Necessity. If all vertices have odd degree, then deleting e creates two of even degree. By the Degree-sum Formula, each component ofT e has an even number of odd-degree vertices. Together with the vertex incident to e, which has even degree in T e, each component of T e has odd order, Sufficiency.
Proof 1 (parity). Given that both components ofT e have odd order, a(T) is even. Now consider v a V(T). Deleting an edge incident to v yields a component containing v and a component not containing v, each of odd order, Together, the components not containing v when we delete the various edges incident to v are d(v) pairwise disjoint subgraphs that together
Section 2.1: Basic Properties
contain all of V(T) . Under the given hypothesis, they all have odd order. Together with ii, they produce an even total, n(T). Hence the number of these subgraphs is odd, which means that the number of edges in T incident to v is odd. Proof 2 (contradiction). Suppose that such a tree To has a vertex vi of even degree. Let e1 be the last edge on a path from a leaf to x. Let T1 be the e1 containing v1. By hypothesis, T1 has odd order, and v1 is a vertex of odd degree in Ti, Since the number of odd-degree vertices in T1 must be even, there is a vertex v2 ofT1 (different from vi) having even degree (in both T1 and T). Repeating the argument, given of even degree in T1_1, let be the last edge on the v-path in and let be the component of
containing v1. Also T1 is the component ofT0
e1 that contains v1, so T1 has
odd order. Since has odd degree in there must be another vertex vi with even degree in T1. In this way we generate an infinite sequence v1, v2, . of distinct vertices in T0. This contradicts the finiteness of the vertex set, so the assumption that T0 has a vertex of even degree cannot hold.
2.1.36. Every tree T of even order has exactly one subgraph in which every vertex has odd degree. Proof 1 (Induction). For n(T) = 2, the only such subgraph is T itself.
Suppose n(T) > 2. Observe that every pendant edge must appear in the subgraph to give the leaves odd degree. Let x be an endpoint of a longest path P, with neighbor u. If u has another leaf neighbor y, add ux and uy to the unique such subgraph found in T . Otherwise, d(u) = 2, since P is a longest path. In this case, add the isolated edge ux to the unique such subgraph found in T . Proof 2 (Explicit construction). Every edge deletion breaks T into two components. Since the total number of vertices is even, the two components of T e both have odd order or both have even order. We claim that the desired subgraph G consists of all edges whose deletion leaves two components of odd order. First, every vertex has odd degree in this subgraph. Consider deleting
the edges incident to a vertex u. Since the total number of vertices in T is even, the number of resulting components other than u itself that have odd order must be odd. Hence u has odd order in G. Furthermore, G is the only such subgraph. If e is a cut-edge of G, then in G e the two pieces must each have even degree sum. Given that G is a subgraph of T with odd degree at each vertex, parity of the degree sum forces G to e if T e has components of odd order and omit e if T e has components of even order.
Chapter 2: Trees and Distance
Comment: Uniqueness also follows easily from symmetric difference. Given two such subgraphs G1, G2 , the degree of each vertex in the symmetric difference is even, since its degree is odd in each This yields a cycle in G1 U G2 c T, which is impossible.
2.1.37. If T and T' are two spanning trees of a connected graph G, and e c E(T) E(T'), then there is an edge e' c E(T') E(T) such that both T e + e' and T' e' + e are spanning trees of G. Deleting e from T leaves a graph having two components; let U, U' be their vertex sets. Let the endpoints of e be u c U and u' C U'. Being a tree, T' contains a unique u, u'-path. This path must have an edge from U to U'; choose such an edge to be e', and then T e + e' is a spanning tree. Since e is the oniy edge ofT between U and U', we have e' c E(T') E(T). Furthermore, since e' is on the u, u'-path in T', e' is on the unique cycle formed by adding e to T', and thus T' e' + e is a spanning tree. Hence e' has all the desired properties.
2.1.38. If T and T' are two trees on the same vertex set such that dr(v) = for each vertex v, then T' can be obtained from T' using 2-switches (Definition 1.3.32) with every intermediate graph being a tree. Using induction on the number a of vertices, it suffices to show when a> 4 that we can apply (at most) one 2-switch to T to make a given leaf x be adjacent to its
neighbor w in T'. We can then delete x from both trees and apply the induction hypothesis. Since the degrees speciQy the tree when a is at most 3, this argument also shows that at most a 3 2-switches are needed. Let y be the neighbor of x in T. Note that w is not a leaf in T, since dr'(w) = and xw c E(T) and a > 4. Hence we can choose a vertex z in T that is a neighbor of w not on the x, w-path in T. Cutting xy and wz creates three components: x alone, one containing z, and one containing w. Adding the edges zy and xw to complete the 2-switch gives x its desired neighbor and reconnects the graph to form a new tree. 2.1.39. If G is a nontrivial tree with 2k vertices of odd degree, then G decomposes into k paths.
Proof 1 (induction and stronger result). We prove the claim for every forest G, using induction on k, Basis step (k = 0): If k = 0, then G has no leaf and hence no edge. Induction step (k > 0): Suppose that each forest with 2k 2 vertices of odd degree has a decomposition into k 1 paths. Since k > 0, some component of G is a tree with at least two vertices. This component has at least two leaves; let P be a path connecting two leaves, Deleting E(P) changes the parity of the vertex degree oniy for the endpoints of P; it makes them even. Hence G E(P) is a forest with 2k 2 vertices of odd degree. By the induction hypothesis, G E(P) is the union of k 1 pairwise edgedisjoint paths; together with P, these paths partition E(G).
Section 2.1: Basic Properties
Proof 2 (extremality). Since there are 2k vertices of odd degree, at least k paths are needed. If two endpoints of paths occur at the same vertex of the tree, then those paths can be combined to reduce the number of paths. Hence a decomposition using the fewest paths has at most one
endpoint at each vertex. Under this condition, endpoints occur only at vertices of odd degree. There are 2k of these. Hence there are at most 2k endpoints of paths and at most k paths. Proof 3 (applying previous result). A nontrivial tree has leaves, so k > 0. By Theorem 1.2.33, G decomposes into k trails. Since G has no cycles, all these trails are paths. 2.1.40. If G is a tree with k leaves, then G is the union of Ik/21 pairwise intersecting paths We prove that we can express G in this way using paths that end at leaves. First consider any way of pairing the leaves as ends of I k/21 paths (one leaf used twice when k is odd). Suppose that two of the paths are disjoint; let these be a u, v-path P and an x, y-path Q. Let R be the path connecting P and Q in G. Replace P and Q by the u, x-path and the v, y-path in G. These paths contain the same edges as P and Q, plus they cover R twice (and intersect). Hence the total length of the new set of paths is larger than before. Continue this process; whenever two of the paths are disjoint, make a switch between them that increases the total length of the paths. This process cannot continue forever, since the total length of the paths is bounded by the number of paths (fk/21) times the maximum path length (at most a 1). The process terminates only when the set of paths is pairwise intersecting. (We have not proved that some vertex belongs to all the paths.) Finally, we show that a pairwise intersecting set of paths containing all the leaves must have union G. If any edge e of G is missing, then G e has two components H, H', each of which contains a leaf of G. Since e belongs to none of the paths, the paths using leaves in H do not intersect the paths using leaves in H'. This cannot happen, because the paths are pairwise intersecting. (Comment: We can phrase the proof using extremality. The pairing with maximum total length has the desired properties; otherwise, we make a switch as above to increase the total length.)
Chapter 2: Trees and Distance
2.1.4 1. For n > 4, a simple n-vertex graph with at least 2n 3 edges must have two cycles of equal length. For such a graph, some component must have size at least twice its order minus 3. Hence we may assume that G is connected. A spanning tree T has n 1 edges and diameter at most n 1. Each remaining edge completes a cycle with edges of T. The lengths of these cycles belong to . Since there are at least n 2 remaining edges, there are two cycles of the same length unless there are exactly n 2 remaining cycles and they create cycles of distinct lengths with the edge of T. This forces T to be a path. after adding the edge e between the endpoints of T that produces a cycle of length n, the other remaining edges each produce two additional shorter cycles when added. These 2n 6 additional cycles fall
pigeonhole principle yields two cycles of equal length.
2.1.42. Extendible vertices. In a nontrivial Eulerian graph G, a vertex is extendible if every trail beginning at v extends to an Eulerian circuit. a) v is extendible if and only if G v is a forest. Necessity. We prove the contrapositive. If G
v is not a forest, then has a cycle C. In G P2(C), every vertex has even degree, so the component of G P2(C) containing v has an Eulerian circuit. This circuit
starts and ends at v and exhausts all edges of G incident to v, so it cannot be extended to reach C and complete an Eulerian circuit of G. Sufficiency. If G
is a forest, then every cycle of G contains v. Given a
trail T starting at v, extend it arbitarily at the end until it can be extended no farther, Because every vertex has even degree, the process can end only at v. The resulting closed trail P must use every edge incident to v, else it could extend farther, Since P is closed, every vertex in G P2(T') has even degree. If G P2(T') has any edges, then minimum degree at least two in a component of G
yields a cycle in G
cycle avoids v, since
P exhausted the edges incident to v. Since we have assumed that G v has no cycles, we conclude that G P2(T') has no edges, so P is an Eulerian circuit that extends T. (Sufficiency can also be proved by contrapositive.) b) If v is extendible, then d(v) = A(G). An Eulerian graph decomposes into cycles. If this uses in cycles, then each vertex has degree at most
Section 2.1: Basic Properties
2m. By part (a) each cycle contains v, and thus d(v) >
Alternatively, since each cycle contains v, an Eulerian circuit must visit v between any two visits to another vertex u. Hence d(v) > d(u). c) For n(G) > 2, all vertices are extendible if and only if G is a cycle. If G is a cycle, then every trail from a vertex extends to become the complete cycle. Conversely, suppose that all vertices are extendible. By part (a), every vertex lies on every cycle. Let C be a cycle in G; it must contain all vertices. If G has any additional edge e, then following the shorter part of C between the endpoints of e completes a cycle with e that does not contain all the vertices. Hence there cannot be an additional edge and G = C. d) If G is not a cycle, then G has at most two extendible vertices. From part (c), we may assume that G is Eulerian but not a cycle. If v is extendible, then G v is a forest. This forest cannot be a path, since then G is a cycle or has a vertex of odd degree. Since G v is a forest and not a path, G v has more than A(G v) leaves unless G v is a tree with exactly one vertex of degree greater than two. If G v has more than A(G v) leaves, all in N(v), then no vertex of G v has degree as large as v in G,
and by part (b) no other vertex is extendible. In the latter case, the one other vertex of degree d(v) may also be extendible, but all vertices except those two have degree 2.
2.1.43. Given a vertex u in a connected graph G, there is a spanning tree of G that is the union of shortest paths from u to the other vertices. Proof 1 (induction on n(G)). When n(G) = 1, the vertex u is the entire tree. For n(G) > 1, let v be a vertex at maximum distance from u. Apply the induction hypothesis to G v to obtain a tree T in G v. Shortest paths in G from u to vertices other than v do not use v, since v is farthest from u. Therefore, T consists of shortest paths in G from u to the vertices other than v. A shortest u, v-path in G arrives at v from some vertex of T. Adding the final edge of that path to T completes the desired tree in G. Proof 2 (explicit construction). For each vertex other than u, choose an incident edge that starts a shortest path to u, No cycle is created, since as we follow any path of chosen edges, the distance from u strictly decreases. Also n(G) 1 edges are chosen, and an acyclic subgraph with n(G) 1 edges is a spanning tree. Since distance from u decreases with each step, the v, u-path in the chosen tree is a shortest v, u-path. Comment: The claim can also be proved using BPS to grow the tree. Proof 1 is a short inductive proof that the BPS algorithm works. Proof 2 is an explicit description of the edge set produced by Proof 1.
2.1.44. If a simple graph with diameter 2 has a cut-vertex, then its complement has an isolated vertex—TRUE. Let v be a cut-vertex of a simple
Chapter 2: Trees and Distance
graph G with diameter 2. In order to have distance at most 2 to each vertex in the other component(s) of G v, a vertex of G v must be adjacent to v. Hence v has degree n(G) 1 in G and is isolated in G. 2.1.45. If a graph G has spanning trees with diameters 2 and 1, then G has spanning trees with all diameters between 2 and 1. Proof 1 (local change). The only trees with diameter 2 are stars, so G has a vertex v adjacent to all others. Given a spanning tree T with leaf u, replacing the edge incident to u with uv yields another spanning tree T'. For every destroyed path, a path shorter by 1 remains. For every created path, a path shorter by 1 was already present. Hence diam P differs from diam T by at most 1. Continuing this procedure reaches a spanning tree of diameter 2 without skipping any values along the way, so all the desired values are obtained. Proof 2 (explicit construction). Since G has a tree with diameter 2, it has a vertex v adjacent to all others. Every path in G that does not contain v extends to v and to an additional vertex if it does not already contain all vertices. Hence fork 2, the number of isomorphism classes of n-vertex trees with diameter at most 3 is [n/2j. If n 4, every tree has diameter at least 2. There is one having diameter 2, the star. Every tree with diameter 3 has two centers, x, y, and every non-central vertex is adjacent to exactly one of x, y, so d(x) -Id(y) = a. By symmetry, we may assume d(x) -c d(y). The unlabeled tree is now completely specified by d(x), which can take any value from 2 through [n/2j. Together with the star, the number of trees is [n/2j. 2.1.47. Diameter and radius.
a) The distance function d(u, v) satisfies the triangle inequality: d(u, v) + d(v, w) d(u, w). A u, v-path of length d(u, v) and a v, w-path of length d(v, w) together form a u, w-walk of length 1 = d(u, v) + d(v, w). Every u, w-walk contains a u, w-path among its edges, so there is a u, wpath of length at most 1. Hence the shortest u, w-path has length at most 1.
b) d -c 2r, where d is the diameter of G and r is the radius of G. Let u, v be two vertices such that d(u, v) = d. Let w be a vertex in the center of G; it has eccentricity r. Thus d(u, w) -c r and d(w, v) -c r. By part (a), d = d(u, v) d(v, w) 1 = s(v) 1. Summing these inequalities yields s(x) + s(y) > s(v) + s(v). The smallest graph where this inequality can fail is the kite K4 a Let
v be a vertex of degree 2; it has eccentricity 2. Its neighbors x and y has degree 3 and hence eccentricity 1. 2.1.52. Eccentricity of vertices outside the center a) If G is a tree, then every vertex x outside the center of G has a neighbor with eccentricity s(x) 1. Let y be a vertex in the center, and let w be a
vertex with distance at least c(x) 1 from x. Let v be the vertex where the unique x, w- and y, w-paths meet; note that v is on the x, y-path in G. Since d(y, w) -c s(y) -c s(x) 1 -c d(x, w), we have d(y, v) -c d(x, v). This implles that v x, Hence x has a neighbor z on the x, v-path in G. This argument holds for every such w, and the x, v-path in G is always part of the x, y-path in G. Hence the same neighbor of x is always chosen asz. Wehaveprovedthatd(z, w) = d(x, w)—lwheneverd(x, w) > s(x)—1. On the other hand, since z is a neighbor ofx, we have d(z, w) -c d(x, w) + 1 -c s(x) 1 for every vertex w with d(x, w) 2, then a tree S with leaf distance set D has a shortest path P from Xk to ajunction w. Since P has no internal vertices on paths joining other leaves, deleting V(P) (w> leaves a subtree with leaf set realizing the distances not involving Xk. By the induction hypothesis, this distance set is uniquely realizable; call that tree 5'. It remains only to show that the vertex w in V(S') and ds(xt, w) are uniquely determined. Let t = w). The vertex w must belong to the path Q joining some leaves x1 and x1 in 5'. The paths from x1 and x1 to Xt in S together
use the edges of Q, and each uses the path P from w to Xk. Thus t = (d5(x1, xt)
Section 2.1: Basic Properties
For arbitrary x1 and x1, this formula gives the distance in S from Xk to the junction with the x1, xy-path. If w is not on the x1, xy-path, then the value of the formula exceeds t, since w is the closest vertex of 5' to Xk. Hence t = mm1 J g, every path from N(x) to N(y) has length at least 1. Also every path whose endpoints are within N(x) has length at least g 2; otherwise, G has a short cycle through x. Every cycle through a new g edge uses one new edge and a path from N(x) to N(y) or at least two new edges and at least two paths of length at least g 2. Hence every new cycle has length at least g.
2.1.62. Connectedness and diameter of the 2-switch graph on spanning trees of G. Let G be a connected graph with a vertices. The graph G' has one vertex for each spanning tree of G, with vertices adjacent in G' when the corresponding trees have exactly n(G) 2 common edges. a) G' is connected.
Proof 1 (construction of path). For distinct spanning trees T and P in G, choose e E E(T) E(T'). By Proposition 2.1.6, there exists e' E E(T') E(T) such that T e+e' is a spanning tree of G. Let T1T= e-f- e'. The trees T and T1 are adjacent in G'. The trees T1 and P share more edges than T and T' share. Repeating the argument produces a T, P-path in G' via vertices T, T1, T2, . Tk, T1. Formally, this uses induction on the number in of edges in E(T) E(T'). When m = 0, there is a T, P-path of length 0. When in > 0, we generate
Ti as above and apply the induction hypothesis to the pair T1, T'. K1.
Proof2(inductionone(G)). If e(G) = a— 1,thenGisatree, and G' = For the induction step, consider e(G) > a 1. A connected n-vertex
Section 2.1: Basic Properties
graph with at least n edges has a cycle C. Choose e c E(C), The graph G e e)' is connected. Every spanning tree of G e is a spanning tree of G, so (G e)' is the induced subgraph of T(G) whose vertices are the spanning trees of G that omit e. Since (G e)' is connected, it suffices to show that every spanning tree of G containing e is adjacent in G' to a spanning tree not containing e. If T contains e and P does not, then there exists e' E E(T') E(T) such that T e + e' is a spanning tree of G omitting e. Thus T e -F e' is the desired tree in G e adjacent to T in G'. b) The diameter of C is at most n 1, with equality when G has two is connected, and by the induction hypothesis (G
spanning trees that share no edges. It suffices to show that T') = E(T) Each edge on a path from T to P in C discards at most one edge ofT, so the distance is at least E(T) Since for each e E E(T) E(T') there exists e' E E(T') E(T) such that T e + e' E V(G'), the path built in Proof 1 of part (a) has precisely this length. Since trees in n-vertex graphs have at most n 1 edges, always E(T) -c n 1, sodiamG' -c n lwhenGhas nvertices. WhenG has two edge-disjoint spanning trees, the diameter of C equals n 1. 2.1.63. Every n-vertex graph with n + 1 edges has a cycle of length at most 1(2n + 2)/3j. The bound is best possible, as seen by the example of three paths with common endpoints that have total length n +1 and nearly-equal lengths. Note that R2n + 2)/3j = 2n/31. Proof 1. Since an n-vertex forest with k components has only n k edges, an n-vertex graph with n + 1 edges has at least two cycles. Let C be a shortest cycle. Suppose that e(C) > 12n/31. If G E(C) contains a path connecting two vertices of C, then it forms a cycle with the shorter path on C connecting these two vertices. The length of this cycle is at most F
ie(C) k> 2, then diamG 3. Let (vo, . be a path joining vertices at distance d. For a ver= NEc31] for 0 - 3, there are 1 + Ld/3i such sets, pairwise disjoint (since we have a shortest v0, cd-path), and each has at least k + 1 vertices. Furthermore, Vd_2 does not appear in any of these sets if j = 1, and both Vd_2 and Vd_3 do not appear if j = 2. Hence a. is as large as claimed. To obtainanupperbound ond interms of a., wewrite ]jl/3j as (d—j)/3. Solving for d in terms of a., we find in each case that d 3(a. 2)/(k + 1) 1— j[1 3/(k + 1)]. Since k > 2, the bound d -c 3(a. 2)/(k + 1) 1 is valid for every congruence class of d modulo 3. When a. 2 is a multiple ofk + 1, the bound is sharp. Ifa. 2 = k + 1, then deleting two edges incident to one vertex of K,, yields a graph with the desired diameter and minimum degree (also C,, suffices). For larger multiples, let in = (a. 2)/(k + 1); note that in > 2. Begin with cliques
Qi. Q. such that Qi and Q. have order k + 2 and the others have order k + 1. For 1 -c i -c in, choose x1, Yf e and delete the edge XIYI.
Section 2.1: Basic Properties
1 3, define the k 1 subtrees G1 P By Gk_1 P the case k = 3, these subtrees are pairwise intersecting. There are k 1 of them, so by the induction hypothesis they have a common vertex. This vertex belongs to all of the original k trees.
2.1.73. A simple graph G is a forest if and only if pairwise intersecting paths in G always have a common vertex. Sufficiency. We prove by contradiction that G is acydllc. If G has a cycle, then choosing any three vertices on the cycle cuts it into three paths that pairwise intersect at their endpoints. However, the three paths do not all have a common vertex. Hence G can have no cycle and is a tree. Necessity. Let G be a forest. Pairwise intersecting paths lie in a single component of G, so we may assume that G is a tree. We use induction on the number of paths. By definition, two intersecting paths have a common vertex. For k > 2, let P1. Pk be pairwise intersecting paths. Also P1, . Pk_1 are pairwise intersecting, as are P2, . Pk; each consists of k 1 paths. The induction hypothesis guarantees a vertex u belonging to all of P1. Pkl and a vertex v belonging to all of P2, . Pk. Since each of P2,
. Pkl contains both u and v and G has exactly one u, v-path Q,
this path Q belongs to all of P2, . Pkl. By hypothesis, P1 and also have a common vertex z. The unique z, path I? lies in Pi, and the unique z, v-path S lies in Pk. Starting from z, let w be the last common vertex of R and S. It suffices to show that w c V(Q),
Otherwise, consider the portion of R from w until it first reaches Q, the
Chapter 2: Trees and Distance
portion of S from w until it first reaches Q, and the portion of Q between these two points. Together, these form a closed trail and contain a cycle, but this cannot exist in the tree G. The contradiction implies that w belongs to Q and is the desired vertex. 2.1.74. Every simple n-vertex graph G with n complement. (We need e(G) 3, let G be an n-vertex graph with a 2 edges. Suppose first that G has an isolated vertex x, Since e(G) = a 2, the Degree-Sum Formula yields a vertex y of degree at least 2. Let G' = G ; this is a graph with a 2 vertices and at most a 4 edges. By the induction hypotheses, every graph with a 2 vertices and a 4 edges appears in its complement, so the same holds for smaller graphs (since they are contained in graphs withn —4 edges). A copy of G' contained in G extends to a copy of G in G by letting x represent y and letting y represent x.
Hence we may assume that G has no isolated vertices. Every nontree component of G has at least as many edges as vertices, and trees have one less. Hence at least two components of G are trees. We may therefore choose vertices x and y of degree 1 with distinct neighbors. Let N(x) = and N(y) = with x' y'. Let G' = G ; this graph has a 2 vertices and a —4 edges. By the induction hypothesis, G' C G' = G x y. Let H be a copy of G' in G x y. If x' or y' represents itself in H, then we let x and y switch identities to add their incident edges. Otherwise, we let x and y represent themselves to add their incident edges. 2.1.75. Every non-star tree is (isomorphic to) a subgraph of its complement. Proof 1 (loaded induction on a). We prove the stronger statement that,
given an n-vertex tree T other than the graph K,, with vertex set contains two edge-disjoint copies ofT in which the two copies of each non-leaf vertex of T appear at distinct vertices. The only non-star tree with at most 4 vertices is the path F4, which is self-complementary via a map that moves each vertex. Now consider a> 4, We show first that T has a leaf x such that T x
is not a star. If T is a path, let x be either leaf. Otherwise, T has at least three leaves; let P be a longest path in T, and let x be a leaf other than the endpoints of P. In either case, T x has a path of length at least 3. Let P = T x, and let y be the neighbor ofx in T. If y is not a leaf in T', then the induction hypothesis yields embeddings of V in K,,i in which y occurs at distinct vertices. We can extend both embeddings to K,, by placing x at v,, in each and adding the distinct edges to the images of y.
Section 2.1: Basic Properties
In this case the non-leaves of T are the same as the non-leaves of T', and the loaded claim holds for T.
If y is a leaf in T, we use the same argument unless f(y) = where f, g are the mappings from V(T') to
for the two embeddings
of T' guaranteed by the induction hypothesis. In this case, let z be the other neighbor of y; we have z as a non-leaf of T', and hence f(z) g(z). We cannot have both g(z) = f(w) for some w E N(z) and f(z) = g(u) for some u e N(z), because then the edge between f(z) and g(z) is used in both embeddings ofT'. By symmetry, we may assume f(z) g(w) for all to E N(z). For T, we define f', g' : V(T) —÷ for the edge-disjoint embeddings of T as follows: If w , let f'(w) = f(w) and g'(w) = g(w). For the other vertices, let f'(z) = f(z), f'(y) = f(y), f'(x) = g'(y) = g(z), g'(x) = g(y), as illustrated below, By construction g'(z) = the non-leaves of T have pairs of distinct images. The edges not involving x, y, z are mapped as before and hence become edge-disjoint subgraphs of f(y), f(z), g(z)>. The path x, y, z is explicitly given edge-disjoint images under f', g'. This leaves only the edges involving z. Those under f are the same as under f'. The shift of z from g(z) to g'(z) = does not produce a common edge because f'(z) = f(z) is not the image under g of any neighbor of z.
Proof 2. (induction on n(T) by deleting two leaves—proof due to Fred Galvin). To cover the basis step, we prove first that the claim is true when T has a path P of length at least 3 that includes a endpoint of every edge (see "caterpillars" in Section 2.2). First we embed P in its complement so that every vertex moves. If n(P) is even, say n(P) = 2k, then we apply the vertex permutation When n(P) = 2k 1, we use Now, since every vertex on P has moved, we can place the leaves at their original positions and add incident edges from T to make them adjacent to their desired neighbors. All non-star trees with at most six vertices have such a path For the induction step, consider a tree T with n(T) > 6. Let u and v be endpoints ofa longest path in T, so d(u, ii) = diamT, and let V = T u v. Let x and y be the neighbors of u and v, respectively. If T is not a star and V is a star, then T is embeddable in its complement using the construction above.
Chapter 2: Trees and Distance
If T' is not a star, then by the induction hypothesis V embeds in r.
If the embedding puts x or y at itself; then adding the edges xv and yu
yields a copy ofT in T. Otherwise, make u adjacent to the image of x and v adjacent to the image of y to complete the copy ofT in T. 2.1.76. If A1, . are distinct subsets of [ii], then there exists x c En] such that A1 U , . U are distinct. We need to find an element x such that no pair of sets differ by x. Consider the graph G with V(G) = and A1 -e- A1 if only if A1 and A1 differ by the addition or deletion of a single element. Color (label) an edge A1 A1 by the element in which the endpoints differ. Any color that appears in a cycle of G must appear an even number of times in that cycle, because as we traverse the cycle we return to the original set. Hence a subgraph F formed by selecting one edge having each edge-label that appears in G will contain no cycles and must be a forest, Since a forest has at most a 1 edges, there must be an element that does not appear on any edge and can serve as x.
2.2. SPANNING TREES & ENUMERATION 2.2.1. Description of trees by Prüfer codes. We use the fact that the degree of a vertex in the tree is one more than the number of times it appears in the corresponding code. a) The trees with constant Prufer codes are the stars. The a 1 labels that don't appear in the code have degree 1 in the tree; the label that appears a 2 times has degree a 1. b) The trees whose codes contain two values are the double-stars. Since a 2 labels don't appear in the code, there are a 2 leaves in the tree. c) The trees whose codes have no repeated entries are the paths. Since a 2 labels appear once and two are missing, a 2 vertices have degree 2, and two are leaves. All trees with this degree sequence are paths. 2.2.2. The graph K1 v C4 has 45 spanning trees. For each graph G in the computation below, we mean r(G).
Section 2.2: Spanning Trees and Enumeration
2.2.3. Application of the Matrix Tree Theorem. The matrix Q = D A for this graph appears on the right below. All rows and columns sum to 0. If we delete any row and column and take the determinant, the result is 106, which is the number of spanning trees. Alternatively, we could apply the recurrence. The number of trees not containing the diagonal edge is
Thenumberoftrees containing the diagonal edge is 5• 6, which is 30. V2
2.2.4. If a graph G with m edges has a graceful labeling, then K2m+i decomposes into copies of G. As in the proof of Theorem 2.2.16, view the vertices modulo 2m + 1. Let a1, . a,, be the vertex labels on in a graceful labeling of G. By definition, 0 -c a1 -c in for each j. For 0 -c i -c 2in, the ith copy of G uses vertices i + al I + a. Each copy uses one edge from each difference class, and the successive copies use distinct edges from a class, so each edge of K2m+1 appears in exactly one of these copies of G.
2.2.5. The graph below has 2000 spanning trees. The graph has 16 vertices
and 20 edges; we must delete five edges to form a spanning tree. The 5-cycles are pairwise edge-disjoint; we group the deleted edges by the 5cycles. Each 5-cycle must lose an edge; one 5-cycle will lose two. To avoid disconnecting the graph, one edge lost from the 5-cycle that loses two must be on the 4-cycle, and thus the 4-cycle is also broken. Every subgraph satisf5ring these rules is connected with 15 edges, since every vertex has a path to the central 4-cycle, and there is a path from one vertex to the next on the 4-cycle via the 5-cycles that lose just one edge). Hence these are the spanning trees. We can pick the 5-cycle that loses two edges in 4 ways, pick its second lost edge in 4 ways, and pick the edge lost from each remaining 5-cycle in five ways, yielding a total of 4- 4- 5- 5 - 5 spanning trees. The product is 2000,
Chapter 2: Trees and Distance
2.2.6. The 3-regular graph that is a ring of in kites (shown below for in = 6) has 2mW" spanning trees. Call the edges joining kites the "link edges". Deleting two link edges disconnects the graph, so each spanning tree omits at most one hnk edge. If a spanning tree uses in 1 link edges, then it also contains a spanning tree from each kite. By Example 2.2.6, each kite has eight spanning trees, (Each such spanning tree has three edges; each choice of three edges works except the two forming triangles, and 8 = 2.) To form a spanning tree of this type, we pick one of the in link edges to
delete and pick a spanning tree from each kite in 8' ways. Thus there are msk_t spanning trees of this sort. The other possibility is to use all in link edges. Now we must have exactly one kite where the vertices of degree 2 in the kite are not connected
by a path within the kite. Since we avoid cycles and spanning trees but must connect the two 3-valent vertices of the kite out to the rest of the graph, we retain exactly two edge from the kite that is cut. Each way of choosing two edges to retain works exept the two that form a path between the 2-valent vertex through one 3-valent vertex: 8 = 2. Since we pick one kite to cut in in ways, pick one of 8 ways to cut it, and pick one of 8 spanning trees in each other kite, there are m8m spanning trees of this type, for 2,n8m spannning trees altogether.
2.2.7. e has (a 2)n"3 spanning trees. Proof 1 (symmetry and Cayley's Formula—easiest!). By Cayley's Formula, there are spanning trees in Since each has a 1 edges, there are (a 1)n"2 pairs (e, T) such that T is a spanning tree in and e a E(T). When we group these pairs according to the edges in we divide by to obtain as the number of trees containing any given edge, since by symmetry each edge of appears in the same number of
spanning trees. To count the spanning trees in e, we subtract from the total number of spanning trees in K,, the number that contain the particular edge e. Subtracting t = 2n"3 from leaves (a spanning trees in K,, that do not contain e.
Section 2.2: Spanning Trees and Enumeration
Proof 2 (Prfifer correspondence). Given vertex set [n], we count the trees not containing the edge between a 1 and a. In the algorithm to generating the Prufer code of a tree with vertex set En], we never delete vertex a. Also, we do not delete vertex a 1 unless a 1 and a are the only leaves, in which case the remaining tree at that stage is a path (because it is a tree with only two leaves). If the tree contains the edge (a
1, a), then (a 1, a) will be the final edge, and the label last written down is a 1 or a. If not, then the path between a 1 and a has at least two edges, and we will peel off vertices
from one end until only the edge containing a remains, The label a is never
recorded during this process, and neither is a
Thus a Prufer code
corresponds to a tree not containing (a 1, a) if and only if the last term of the list is not a 1 or a, and there are (a such lists. Proof 3 (Matrix Tree Theorem). For e, the matrix D A has
diagonala—1. a—1,a—2,a—2,withpositionsa—1,aanda,a—1 equal to 0 and all else —1, Delete the last row and column and take the determinant to obtain the number of spanning trees. To compute the determinant, apply row and column operations as follows: 1) add the a 2 other columns to the first so the first column becomes 1, . 1, 0. 2) subtract the first row from all but the last, so the first row is 1, —1, . —1, the last is 0, —1. —1, a 2, and the others are 0 except for a on the diagonal. The interior rows can then be used to reduce this to a diagonal matrix with entries 1, a, . a, a 2, whose determinant is (a 2.2.8. With vertex set [a], there are ()(2n_2 2) trees with a 2 leaves and a!/2 trees with 2 leaves. Every tree with two leaves is a path (paths along distinct edges incident to a vertex of degree k leads to k distinct leaves, so having only two leaves in a tree implies maximum degree 2). Every tree with a 2 leaves has exactly two non-leaves. Each leaf is adjacent to one of these two vertices, with at least one leaf neighbor for each of the two vertices. These trees are the "double-stars". To count paths directiy, the vertices of a path in order form a permu-
tation of the vertex set. Following the path from the other end produces another permutation. On the other hand, every permutation arises in this way. Hence there are two permutations for every path, and the number of paths is a!/2. To count double-stars directly, we pick the two central vertices in one of ways and then pick the set of leaves adjacent to the lower of the two central vertices. This set is a subset of the a 2 remaining vertex labels, and it can be any subset other than the full set and the empty set. The number of ways to do this is the same no matter how the central vertices is chosen, so the number of double-stars is 2).
Chapter 2: Trees and Distance
To solve this using the Prilfer correspondence, we coimt Prfifer codes for paths and for double-stars. In the PrUfer code corresponding to a tree, the labels of the leaves are the labels that do not appear. For paths (two leaves), the other a 2 labels must each appear in the Priifer code, so they must appear once each. Having chosen the leaf labels in ways, there are (a 2)! ways to form a Prüfer code in which all the other labels appear. The product is n!/2. For double-stars (a —2 leaves), exactly two labels appear in the Prufer code. We can choose these two labels in
ways. To form a Prafer code (and
thus a tree) with these two labels as non-leaves, we choose an arbitrary nonempty proper subset of the positions 1, .. ., a 2 for the appearances of the first label, There are 2n-2 2 ways to do this step. Hence there are ways to form the Prilfer code. 2.2.9. There are (n!/k!)S(n 2, a k) trees on a fixed vertex set of size a that have exactly k leaves. Consider the Prilfer sequences of trees. The leaves of a tree are the labels that do not appear in the sequence. We can choose the labels of the leaves in Q) ways. Given a fixed set of leaves, we must count the sequences of length a 2 in which the remaining a k labels all appear. Each label occupies some set of positions in the sequence. We partition the set of positions into a k nonempty parts, and then we can assign these parts to the labels in (a k)! ways to complete the sequence. The number of ways to perform the partition, by definition, is S(a —2, a—k). Since these operations are independent, the total number of legal Prftfer sequences is Q)(a—k)!S(n—2,a—k). 2)
2.2.10. K2,m has spanning trees. Let X, Y be the partite sets, with = 2. Each spanning tree has one vertex of Y as a common neighbor
of the vertices in X; it can be chosen in in ways. The remaining vertices are leaves; for each, we choose its neighbor in X in one of two ways. Every spanning tree is formed this way, so there are m2m_l trees.
Alternatively, note that K2. is obtained from the two-vertex multigraph H with in edges by replacing each edge with a path of 2 edges. Since H itself has in spanning trees, Exercise 2.2.12 allows the spanning trees of = 2m_1, to be counted by multiplying in by a factor of K2,m has [On + 1)/2j isomorphism classes of spanning trees. The vertices in X have one common neighbor, and the isomorphism class is determined by splitting the remaining in 1 vertices between them as leaves.
We attach k leaves to one neighbor and in o
Also, every tree arises in this way. A tree T in H uses at most one edge between each pair of vertices. Since T is connected and acyclic, the edges in G whose copies are used in
T form a spanning tree of G that generates T. Hence r(H) -c Proof 2 (induction on in using the recurrence for r). If in = 0, then r(G) = r(H) = 0, unless a = 1, in which case 1 = k°- 1. If in > 0, choose e c E(G). Let H' be the graph obtained from H by contracting all k copies of e. Let H" be the graph obtained from H by deleting all k copies of e. The spanning trees of H can be grouped by whether they use a copy of e (they cannot use more than one copy). There are k x r(H') of these trees that use a copy of e and r(H") that do not. We can apply the induction hypothesis to H' and H", since each arises from a graph with fewer than in edges by having k copies of each edge: H' from G- e and H" from G e. Thus
r(H) = k x r(H') + r(H") = k =
Proof 3 (matrix tree theorem). Let Q, Q' be the matrices obtained from G, G', from which we delete one row and column before taking the determinant. By construction, Q' = k Q. When we take the determinant of a submatrix of order a 1, we thus obtain r(G') = k"1r(G). b) If H is obtained from G by replacing each e e E(G) with a path P(e) of k edges, then -r(H) =
Proof 1 (combinatorial argument). A spanning tree T of G yields spanning trees of H as follows. If e c E(T), include all of P(e). If E(T), use all but one edge of P (e). Choosing one of the k edges of P (e) to omit for each e c E(G) E(T) yields distinct trees (connected and acycic) in H. Again we must show that all spanning trees have been gene
erated. A tree T' in H omits at most one edge from each path P(e), else some vertex in P (e) would be separated from the remainder of H. Let T be the spanning subgraph of G with E(T) = . If T'
Chapter 2: Trees and Distance
is connected and has no cycles, then the same is true ofT, and V is one of the trees generated from T as described above. Proof 2 (induction on m). The basis step m = 0 is as in (a). Form >
an edge e a E(G). The spanning trees of H use k or k 1 edges of P(e). These two types are counted by r(H') and r(H"), where H' is the select
graph obtained from H by contracting all edges in P(e), and H" is the graph obtained from H by deleting P(e) (except for its end-vertices). Since these graphs arise from G e and G e (each with m 1 edges) by replacing each edge with a path of length k, applying the induction hypothesis yields
e) + kEk(m_l)_n+hr(G km_n+lEr(G e) +
2.2.13. Spanning trees in For each spanning tree T of a list f(T) of pairs of integers (written vertically) is formed as follows: Let u, v be the least-indexed leaves of the remaining subtree that occur in X and Y. Add the pair to the sequence, where a is the index of the neighbor of u and Ii is the index of the neighbor of v. Delete and iterate until ii 2 pairs are generated and one edge remains. a) Every spanning tree of has a leaf in each partite set, and hence f is well-defined. If each vertex of one partite set has degree at least 2, then at least 2n edges are incident to this partite set, which are too many to have in a spanning tree of a graph with 2n vertices. b) f is a bijection from the set of spanning trees of to the set of n 1-element lists of pairs of elements from [nil, and hence Ku. has spanning trees. We use an analogue of Prufer codes. Consider with partite sets X =
vertex unmarked. At the ith step, when the ith ordered pair is let u be the least index of an unmarked vertex in Y that does not appear in first coordinates of L at or after position i, and let v be the least index of an unmarked vertex in X that does not appear in second coordinates of L at or after position i. We add the edges Xa(i)Yu and Yb(i)Xv, and then we mark
Section 2.2: Spanning Trees and Enumeration
and to eliminate them from further consideration. After a 1 pairs, we add one edge joining the two remaining unmarked vertices. After the ith step, we have 2n 2i components, each containing one
unmarked vertex. This follows by induction on i; it holds when i =
Since indices cannot be marked until after they no longer appear in the list, the two edges created in the ith step join pairs of unmarked vertices. By the induction hypothesis, these come from four different components, and the two added edges combine these into two, each keeping one unmarked vertex. Thus adding the last edge completes the construction of a tree. In computing f(T), a label no longer appears in the sequence after it is deleted as a leaf. Hence the vertices marked at the ith step in computing g(L) are precisely the leaves deleted at the ith step in computing f(g(L)), which also records Thus L = f(g(L)). Similarly, the leaves deleting
at the ith step in computing f(T) are the vertices marked at the ith step in computing g(f(T)), which yields T = g(f(T)). Hence each maps inverts the other, and both are bijections. 2.2.14. The number of trees with vertices 1, . r + s that have partite sets of sizes r and s is ( S)sr_1f_1 if r s. It suffices to count the Prufer codes for such trees. The factor counts the assignments of labels to the two partite sets (half that amount if r = s). When deleting a vertex in computing the Prufer code, we record a vertex of the other partite set. Since an edge remains at the end of the construction, the final code has s 1 entries from the r-set and r 1 entries from the s-set. It suffices to show that the sublists formed from each partite set detennine the full list, because there are ?_1T1_l such pairs of sublists. In reconstructing the code and tree from the pair of lists, the next leaf to be "finished" by receiving its last edge is the least label that is unfinished and doesn't appear in the remainder of the list. The remainder of the list is the remainder of the two sublists. We know which set contains the next leaf to be finished. Its neighbor comes from the other set. This tells us which sublist contributes the next element of the full list. Iterating this merges the two sublists into the full Prufer code. When r = s, the given fonnula counts the lists twice.
2.2.15. For a> 1, the number of spanning trees in the graph vertices and 3n 2 edges pictured below satisfies the recurrence for a > 3, with t1 = 1 and t2 = 4.
Chapter 2: Trees and Distance
(Comment: The solution to the recurrence is t,, = (2 + Using the recurrence, this follows by induction on ii.) We derive the recurrence. Let t,, = r(G,j.
Proof 1 (direct argument for recurrence). Each spanning tree in G,, uses two or three of the three rightmost edges. Those with two of the rightmost edges are obtained by adding any two of those edges to any spanning
tree of G,,_1. Thus there are 3t,,_1 such trees. To prove the recurrence = 4tn_i tn_2, it suffices to show that there are t,,_1 t,,_2 spanning trees that contain the three rightmost edges. Such trees cannot contain the second-to-last vertical edge e. Therefore,
deleting the three rightmost edges and adding e yields a spanning tree of G,,_1. Furthermore, each spanning tree of G,,_1 using e arises exactly once in this way, because we can invert this operation. Hence the number of spanning trees of G,, containing the three rightmost edges equals the number of spanning trees of G,,_1 containing e. The number of spanning trees of G,,_1 that don't contain e is t,,_2, so the number of spanning trees of G,,_1 that do contain e is Proof 2 (deletion/contraction recurrence). Applying the recurrence introduces graphs of other types. Let H,, be the graph obtained by contracting the rightmost edge of G. and let F,,_1 be the graph obtained by contracting one of the rightmost edges of H. Below we show G4, and Fg.
By using r(G) = r(G e) + r(G e) on a rightmost edge e and observing that a pendant edge appears in all spanning trees while a loop appears in
r(G,,) = r(G,,_1) + v(H,,) r(H,,) = r(G,,_1) + v(F,,_1) r(F,,) = r(G,,) + r(H,,_1) Substituting in for r(H,,) and then for v(F,,_i) and then for r(H,,_1) yields the desired recurrence:
r(G,,) = r(G,,_1) + r(G,,_1) + r(F,,_1) = 2r(G,,_1) + v(G,,_1) + r(H,,_2) = 3r(G,,_1) + r(G,,_1) r(G,,_2) = 4r(G,,_1) 2.2.16. Spanning trees in K1 v P. The number a,, of spanning trees sat> 1,witha1= 1. Letx1. x,,bethe
vertices of the path in order, and let z be the vertex off the path. There are a,,_1 spanning trees not using the edge zx,,; they combine the edge x,,_1x,,
Section 2.2: Spanning Trees and Enumeration
with a spanning tree of K1 v Among trees containing zx,1, let i be the highest index such that all of the path .. ., appears in the tree. For each i, there are such trees, since the specified edges are combined with a spanning tree of K1 v P1. The term 1 corresponds to I = 0; here the entire tree is U This exhausts all possible spanning trees.
2.2.17. Cayley's formula from the Matrix Tree Theorem. The number of labeled n-vertex trees is the number of spanning trees in Using the Matrix Tree Theorem, we compute this by subtracting the adjacency matrix from the diagonal matrix of degrees, deleting one row and column, and taking the determinant. All degrees are n 1, so the initial matrix is n 1 on the diagonal and —1 elsewhere. Delete the last row and column. We compute the determinant of the resulting matrix. Proof 1 (row operations). Add every row to the first row does not change the determinant but makes every entry in the first row 1. Now add the first row to every other row. The determinant remains unchanged, but every row below the first is now 0 everywhere except on the diagonal, where
the value is n. The matrix is now upper triangular, so the determinant is the product of the diagonal entries, which are one 1 and n 2 copies of n. Hence the determinant is as desired. Proof 2 (eigenvalues). The determinant of a matrix is the product of its eigenvalues. The eigenvalues of a matrix are shifted by X when XI is added to the matrix. The matrix in question is nIh_i where 'ni is the n 1-by-n 1 identity matrix and is the n 1-by-n 1 matrix with every entry 1. The eigenvalues of are —(n—i) with multiplicity land O with multiplicity n 2. Hence the eigenvalues of the desired matrix are 1 with multiplicity 1 and n with multiplicity n 2. Hence the determinant is as desired. =
2.2.18. Proof that adjacency matrix of
using the Matrix Tree Theorem. The where 0 and 1 denote matrices of all Os and
all is, and both the row partition and the column partition consist of r in the first block and s in the second block. The diagonal matrix of degrees is (sjr where 4 is the identity matrix of order n. Hence we may delete the first row and column to obtain Q * = (u1r4 ;fl. We apply row and column operations that do not change the determinant. We subtract column r 1 (last of the first block) from the earlier columns and subtract column r (first of the second block) from the later columns. This yields the matrix on the left below, where the values outside the matrix indicate the number of rows or columns in the blocks. Now we add to row r 1 the earlier rows and add to row r the later rows, yielding the matrix on the right below.
Chapter 2: Trees and Distance
i in position r (on the diagonal). Adding row r to the first r 2 rows (and r i limes row r to row r i) now leaves the i in row r as the only nonzero entry in column r. Also, the s in column r i of row r i is now the only nonzero entry in row r i. Hence we can add i/s times row r i to each of the last s 1 rows to eliminate the other nonzero entries in column r i. The resulting matrix is diagonal, with diagonal entries consisting of r i copies of s, one copy of i, and s i copies of r, Since adding a multiple of a row or column to another does not change the determinant, the determinant of our original matrix equals the determinant of this diagonal matrix. The determinant of a diagonal matrix is the product of its diagonal entries, so the determinant is 2.2.19. The number of labeled trees on n vertices satisfies the recurrence = For an arbitrary labeled tree on n vertices, delete the edge incident to v2 on the path from v2 to v1. This yields labeled trees on Ic and a Ic vertices for some Ic, where v1 belongs to the tree on Ic vertices and to the tree on a Ic vertices. Each such pair arises from exactly labeled trees on a vertices. To see this, reverse the process. First choose the Ic i other vertices to be in the subtree contalning vi. Next, choose a tree on Ic labeled vertices and a tree on a Ic labeled vertices (any such choice could arise by deleting the specified edge of a tree on a vertices). Finally, reconnect the tree by adding an edge from v2 to any one of the Ic vertices in the tree containing v1. This counts the trees such that the v2
subtree contalning v1 has Ic vertices, and summing this over Ic yields 2.2.20. A d-regular graph G has a decomposition into copies of if and only if G is bipartite. If G has bipartition X, Y, then for each x e X we include the copy of K1,d obtained by taking all d edges incident to x. Since every edge has exactly one endpoint in X, and every vertex in X has degree d, this puts every edge of G into exactly one star in our list. If G has a Ki,d-decomposition, then we let X be the set of centers of the copies of Ki,d in the decomposition. Since G is d-regular, each copy of K1,d uses all edges incident to its center. Since the list is a decomposition, each
edge is in exactly one such star, so X is an independent set. Since every edge belongs to some Ki,d centered in X, there is no edge with both endpoints outside X. Thus the remaining vertices also form an independent set, and G has bipartition X, X.
Section 2.2: Spanning Trees and Enumeration
Alternative proof of sufficiency. If G is not bipartite, then G contains an odd cycle. When decomposing a d-regular graph into copies of K1,d, each
subgraph used consists of all d edges incident to a single vertex, Hence each vertex occurs only as a center or only as a leaf in these subgraphs. Also, every edge joins the center and the leaf in the star containing it. These statements require that centers and leaves alternate along a cycle, but this cannot be done in an odd cycle. 2.2.2 1. Decomposition of K2m_i,2m into m spanning paths. We add a vertex to the smaller partite set and decomposition K2m,2m into in spanning cycles.
Deleting the added vertex from each cycle yields pairwise edge-disjoint spanning paths of K2m_1,2m. Let the partite sets Of K2m,2m be x1, . and . Y2m. Let the kth cycle consist of the edges of the forms x1 Yi+2k- 1 and x1 yi+2k, where subscripts ,
above 2m are reduced by 2m. These sets are pairwise disjoint and form spanning cycles. 2.2.22. If G is an n-vertex simple graph having a decomposition into k spanning trees, and A(G) = 6(G) f 1, then G has a 2k vertices of degree 2k and 2k vertices of degree 2k 1. Each spanning tree has a 1 edges, so e(G) = k(n 1). Note that k 2, and hence x is on the spinal path. Growing a leaf at x yields obtain a larger caterpillar G with degree list ci. This completes the induction step.
2.2.29. Every tree transforms to a caterpillar with the same degree list by operations that delete an edge and add another rejoining the two compo-
nents. Let P be a longest path in the current tree T. If P is incident to every edge, then T is a caterpillar. Otherwise a path P' of length at least two leaves P at some vertex x, Let uv be an edge of P', with u between x and v, and let y be a neighbor of x on P. Cut xy and add yu. Now cut uv and add cx. Each operation has the specified type, and together they form a 2-switch preserving the vertex degrees. Also, the new tree has a path whose length is that of P plus dT(x, u). Since the length of a path cannot exceed the number of vertices, this process terminates. It can only terminate when the longest path is incident to all edges and the tree is a caterpillar. 2.2.30. A connected graph is a caterpillar if and only if it can be drawn on a channel without edge crossings.
Necessity. If G is a caterpillar, let P be the spine of G. Draw P on a channel by alternating between the two sides of the channel. The remaining edges of G consist of a leaf and a vertex of P. If u, v, w are three consecutive vertices on P, then v has an "unobstructed view" of the other side of the channel between the edges vu and vw. Each leaf x adjacent to v can be placed in that portion of the other bank, and the edge cx can then be drawn straight across the channel without crossing another edge. Sufficiency. Suppose that G is drawn on a channel. The endpoints of an edge e cannot both have neighbors in the same direction along the channel, since that would create a crossing. Hence G has no cycle, since a cycle would leave an edge and return to it via the same direction along the channeL We conclude that G is a tree. If G contains the 7-vertex tree that is not a caterpillar, then let v be its central vertex. The three neighbors of v occur on the other side of the channel in some order; let u be the middle neighbor. The other edge incident to u must lle in one direction or the other from uv, contradicting the preceding paragraph. Hence G avoids the forbidden subtree and is a caterpillar. (Alternatively, we can prove this directly by moving along the channel to extract the spine, observing that the remainder of the tree must be leaves attached to the spine.)
Section 2.2: Spanning Trees and Enumeration
2.2.31. Every caterpillar has an up/down labeling. Constructive proof. Let P = vo, . be a longest path in a caterpillar G with in edges; by the argument above P is the spine of G, We iteratively construct a graceftil labeling f for G. Define two parameters 1, u that denote the biggest low label and smallest high label used; after each stage the unused labels are . Let r denote the lowest edge difference achieved; after each stage r in have been achieved. Begin by setting f(vo) = 0 and f(vi) = in; hence 1 = 0, u = in, r = in. Before stage i, we will have ; this is true by construction before stage 1, Suppose this is true before stage 1, along with the other claims made for 1, u, d. Let d = d the d 1 numbers nearest f(v1_i) that have = not been used, ending with vf+1. Since we start with
u—l=r,thenewdifferencesarer—1. r—d-f-1,whichhavenotyet = been achieved. To finish stage i, reset r to r d + 1; also, if = reset 1 to 1 + d 1, but if u reset u to u d + 1. Now stage i is complete, and the claims about I, u, r are satisfied as we are ready to start 1
stage i + 1: (f(v1+i), f(v1)> = , r = u—i, and the edge differences so far are r, . in. After stage k 1, we have assigned distinct labels in
to all in f 1 vertices, and the differences of labels of adjacent vertices are all distinct, so we have constructed a graceftil labeling. The 7-vertex tree that is not a caterpillar has no up/down labeling. In an up/down labeling of a connected bipartite graph, one partite set must have all labels above the threshold and the other have all labels below the threshold. Also, we can interchange the high side and the low side by subtracting all labels from a 1. Hence for this 7-vertex tree we may assume the labels on the vertices of degree 2 are the high labels 4,5,6. Since 0,6 must be adjacent, this leaves two cases: 0 on the center or 0 on the leaf next to 6. In the first case, putting 1 or 2 next to 6 gives a difference already present, but with 3 next to 6 we can no longer obtain a difference of 1 on any edge. In the second case, we can only obtain a difference of 5 by
putting 1 on the center, but now putting 2 next to 5 gives two edges with difference 3, while putting 2 next to 4 and 3 next to 5 give two edges with difference 2. Hence there is no way to complete an up/down labeling. 2.2.32. There are isomorphism classes of n-vertex caterpil+ lars. We describe caterpillars by binary lists. Each 1 represents an edge on the spine. Each 0 represents a pendant edge at the spine vertex between the edges corresponding to the nearest is on each side. Thus n-vertex caterpillars correspond to binary lists of length a 1 with both end bits being 1. We can generate the lists for caterpillars from either end of the spine; reversing the list yields a caterpillar in the same isomorphism class. Hence
Chapter 2: Trees and Distance
we count the lists, add the symmetric lists, and divide by 2. There are lists of the specified type. To make a symmetric list, we specify [(a 3)121 + 2[(n_3)121)/2, bits. Thus the result is
2.2.33. If T is an orientation of a tree such that the heads of the edges are all distinct, then T is a union of paths from the root (the one vertex that is not a head), and each each vertex is reached by one path from the root. We use induction on a, the number of vertices. For a = 1, the tree with one vertex satisfies ali the conditions. Consider a > 1. Since there are a 1 edges, some vertex is not a tail. This vertex v is not the root, since the root is the tail of all its incident edges. Since the heads are distinct, v is incident to only one edge and is its head. Hence T v is an orientation of a smaller tree where the heads of the edges are distinct, By the induction hypothesis, it is a tree of paths from the root (one to each vertex), and replacing the edge to v preserves this desired conclusion for the full tree. 2.2.34. An explicit de Bruijn cycle of length is generated by starting with a 0's and subsequently appending a 1 when doing so does not repeat a previous string of length a (otherwise append a 0). A de Bruijn cycle is formed by recording the successive edge labels along an Eulerian circuit in the de Bruijn digraph. The vertices of the de Bruijn digraph are the binary strings of length a 1. From each vertex two edges depart, labeled o and 1. The edge 0 leaving v goes to the vertex obtained by dropping the first bit of v and appending 0 at the end. The edge 1 leaving v goes to the vertex obtained by dropping the first bit of v and appending 1 at the end. Let v0 denote the all-zero vertex, and let e be the loop at v0 labeled 0. The 2"' 1 edges labeled 0 other than e form a tree of paths in to v0. (Since a path along these edges never reintroduces a 1, it cannot return to a vertex with a 1 after leaving it.) Starting at v0 along edge e means starting with a 0's. Algorithm 2.4.7 now tells us to follow the edge labeled 1 at every subsequent step unless it has already been used; that is, unless appending
a 1 to the current list creates a previous string of length a. Theorem 2.4.9 guarantees that the result is an Eulerian circuit. 2.2.35. Tarry's Algorithm (The Robot in the Castle). The rules of motion are: 1) After entering a corridor, traverse it and enter the room at the other end. 2) After entering a room whose doors are all unmarked, mark I on the door of entry. 3) When in a room having an unmarked door, mark 0 on some unmarked door and exit through it. 4) When in a room having all doors marked, find one not marked 0 (if one exists), and exit through it. 5) When in a room having all doors marked 0, STOP. When in a room other than the original room u, the number of entering edges that have been used exceeds the number of exiting edges. Thus an
Section 2.3: Optimization and Trees
exiting door has not yet been marked 0. This implies that the robot can only terminate in the original room u. The edges marked I grow from u a tree of paths that can be followed back to u. The rules for motion establish an ordering of the edges leaving each room so that the edge labeled I (for a room other than u) is last. In order to terminate in u or to leave a room v by the door marked I, every edge entering the room must have been used to enter it, including all edges marked I at the other end. Therefore, for every room actually entered, the robot follows all its incident corridors in both directions. Thus it suffices to show that every room is reached. Let V be the set of all rooms, and let S be the set of rooms reached in a particular robot tour. If S V, then since the castle is connected there is a corridor joining rooms s c S and r S (the shortest path between S and Since every reached vertex has its incidence corridors followed in both directions, the corridor sr is followed, and r is also reached. The contradiction yields S = V. Comment. Consider a digraph in which each corridor becomes a pair of oppositely-directed edges. Thus indegree equals outdegree at each vertex. The digraph is Eulerian, and the edges marked I form an intree to the
initial vertex. The rules for the robot produce an Eulerian circuit by the method in Algorithm 2.4.7.
The portion of the original tour after the initial edge e = tour formed according to the rules for a tour in G
because in the original
tour no door of u is ever marked I. If e is not a cut-edge, then tours that follow e, follow G e from v, and return along e do not include tours that do not start and end with e. There may be such tours, as illustrated below, so such a proof falls into the induction trap.
2.3.1. In an edge-weighting of the total weight on every cycle is even if and only if the total weight on every triangle is even. Necessity is trivial, since triangles are cycles. For sufficiency, suppose that every triangle has even weight. We use induction on the length to prove that every cycle C has even weight. The basis step, length 3, is given by hypothesis. For the
Chapter 2: Trees and Distance
induction step, consider a cycle C and a chord e. The chord creates two shorter cycles C1, C2 with C. By the induction hypothesis, C1 and C2 have even weight. The weight of C is the sum of their weights minus twice the weight of e, so it is still even.
2.3.2. If T is a minimum-weight spanning tree of a weighted graph G, then the u, v-path in T need not be a minimum-weight u, v-path in G. If G is a cycle of length of length at least 3 with all edge weights 1, then the cheapest path between the endpoints of the edge omitted by T has cost 1, but the cheapest path between them in T costs n(G) 1.
2.3.3. Computation of minimum spanning tree. The matrix on the left below corresponds to the weighted graph on the right. Using Kruskal's algorithm, we iteratively select the cheapest edge not creating a cycle. Starting with the two edges of weight 3, the edge of weight 5 is forbidden, but the edge of weight 7 is available. The edge of weight 8 completes the minimum spanning tree, total weight 21. Note that if the edge of weight 8 had weight 10, then either of the edges of weight 9 could be chosen to complete the tree; in this case there would be two spanning trees with the minimum value. 1
2.3.4. Weighted trees in K1 v C4. On the left, the spanning tree is unique, using all edges of weights 1 and 2. On the right it can use either edge of weight 2 and either edge of weight 3 plus the edges of weight 1.
2.3.5. Shortest paths in a digraph. The direct i to j travel time is the entry
in the first matrix below. The second matrix recordes the least i to j travel time for each pair i, j. These numbers were determined for each i
by iteratively updating candidate distances from i and then selecting the closest of the unreached set (Dijkstra's Algorithm). To do this by hand,
Section 2.3: Optimization and Trees
make an extra copy of the matrix and use crossouts to update candidate distances in each row, using the original numbers when updating candidate distances. The answer can be presented with more information by drawing the tree of shortest paths that grows from each vertex. 0
2.3.6. In an integer weighting of the edges of the total weight is even on every cycle if and only if the subgraph consisting of the edges with odd weight is a spanning complete bipartite subgraph. Sufficiency. Every cycle contains an even number of edges from a spanning complete bipartite subgraph. Necessity. Suppose that the total weight on every cycle is even. We claim that every component of the spanning subgraph consisting of edges with even weight is a complete graph. Otherwise, it has two vertices x, y at distance 2, which induce P3 with their common neighbor z. Since xy has odd weight, x, y, z would form a cycle with odd total weight. If the spanning subgraph of edges with even weight has at least three components, then selecting one vertex from each of three components yields a triangle with odd weight. Hence there are at most two components. This implies that the complement (the graph of edges with odd weight) is a spanning complete bipartite subgraph of G.
2.3.7. A weighted graph with distinct edge weights has a unique minimumweight spanning tree (MST). Proof 1 (properties of spanning trees). If G has two minimum-weight spanning trees, then let e be the lightest edge of the symmetric difference. Since the edge weights are distinct, this weight appears in only one of the two trees. Let T be this tree, and let T' be the other. Since e E E(T) there exists e' c E(T') E(T) such that P -F- e e' is a spanning tree. By the choice of e, w(e') > w(e). Now w(T' + e e') 0, then G is not a tree; let e
be the heaviest edge of G that appears in a cycle, and let C be the cycle containing e. We claim that e appears in no MST of G. If T is a spanning tree containing e, then T omits some edge e' of C, and T e + e' is a cheaper spanning tree than T, Since e appears in no MST of G, every MST of G is an MST of G e. By the induction hypothesis, there is only one such tree.
Chapter 2: Trees and Distance
Proof 3 (Kruskal's Algorithm). In Kruskal's Algorithm, there is no choice if there are no ties between edge weights. Thus the algorithm can produce only one tree. We also need to show that Kruskal's Algorithm can produce every MST. The proof in the text can be modified to show this; if e is the first edge of the algorithm's tree that is not in an MST T', then we obtain an edge e' with the same weight as e such that e' c E(T') E(T) and e' is available when e is chosen. The algorithm can choose e' instead. Continuing to the choices in this way turns T into P. 2.3.8. No matter how ties are broken in choosing the next edge for Kruskal's Algorithm, the list of weights of a minimum spanning tree (in nondecreasing order) is unique. We consider edges in nondecreasing order of cost. We prove that after considering all edges of a particular cost, the vertex sets of the components of the forest built so far is the same independent of the order of consideration of the edges of that cost. We prove this by induction on the number of different cost values that have been considered. At the start, none have been considered and the forest consists of isolated vertices. Before considering the edges of cost x, the induction hypothesis tells us that the vertex sets of the components of the forest are fixed. Let H be a graph with a vertex for each such component, and put two vertices adjacent in H if G has an edge of cost x joining the corresponding two components. Suppose that H has k vertices and 1 components. Independent of the order
in which the algorithm consider the edges of cost x, it must select some l edges of cost x in G, and it cannot select more, since this would create a cycle among the chosen edges.
2.3.9. Among the cheapest spanning trees containing a spanning forest F is one containing the cheapest edge joining components ofF. Let T be a cheapest spanning tree containing F. If e E(T), then T +e contains exactly one cycle, since T has exactly one u, v-path. Since u, v belong to distinct components of F, the u, v-path in T contains another edge e' between distinct components of F. If e' costs more than e, then P = T e' + e is a cheaper
spanning tree containing F, which contradicts the choice of T. Hence e' costs the same as e, and P contains e and is a cheapest spanning tree containing F. Applying this statement at every step of Kruskal's algorithm proves that Kruskal's algorithm finds a minimum weight spanning tree. 2.3.10. Prim's algorithm produces a minimum-weight spanning tree, Let v1 be the initial vertex, let T be the tree produced, and let P be an optimal tree that agrees with T for the most steps. Let e be the first edge chosen for T that does not appear in T*, and let U be the set of vertices in the subtree ofT that has been grown before e is added. Adding e to T* creates a cycle C; since e links U to U, C must contain another edge e' from U to U. Since P + e e' is another spanning tree, the optimality of P yields
Section 2.3: Optimization and Trees
w(e') -c w(e). Since e' is incident to U, e' is available for consideration when e is chosen by the algorithm; since the algorithm chose e, we have w(e) -c w(e'). Hence w(e) = w(e'), and Tt -F- e e' is a spanning tree with the same weight as T*. It is thus an optimal spanning tree that agrees with T longer than Tt, which contradicts the choice of P. e
2.3.11. Every minimum-weight spanning tree achieves the minimum of the maximum weight edge over all spanning trees. Let T be a minimum-weight spanning tree, and let P be one that minimizes the maximum weight edge. If T does not, then T has an edge e whose weight is greater than the weight of every edge in T*. If we delete e from T, Then we can find an edge e* c
E(Tt) that joins the components of T e, since T* is connected. Since w(e) > w(e*), the weight of T e + e' is less than the weight of T, which contradicts the miimality of T. Thus T has the desired property. 2.3.12. The greedy algorithm cannot guarantee minimum weight spanning paths. This fails even on four vertices with only three distinct vertex weights. If two incident edges have the minimum weight a, such as a = 1, the algorithm begins by choosing them. If the two edges completing a 4cycle with them have maximum weight c, such as c = 10, then one of those must be chosen to complete a path of weight 2a + c. However, if the other two edges have intermediate weight b, such as b = 2, there is a path of weight 2b + a, which will be cheaper whenever b 4, the construction generalizes in many possible ways using three weights a Owhenqo+qi = 1. Because f(x) = x lgx is convex for 0 2, if M is a perfect matching of Qk, then there are an even number of edges in M whose endpoints differ in coordinate i. Let V0 and Vi be the sets of vertices having 0 and 1 in coordinate i, respectively. Each has even size. Since the vertices of not matched to Vt_p must be matched within the number of vertices matched by edges to must be even. has nine perfect matchings. There are four edges in each such b) matching, with an even number distributed to each coordinate. The possible distributions are (4, 0, 0) and (2, 2, 0). There are three matchings of the first type. For the second type, we pick a direction to avoid crossing, pick one of the two matchings in one of the 4-cycles, and then the choice of the matching in the other 4-cycle is forced to avoid making all four edges change the same coordinate. Hence there are 3 . 2• 1 perfect matchings of the second type.
Section 3.1: Matchings and Covers
the k-dimensional hypercube Qk has at least
perfect matchings. Proof 1 (induction on k). Let mk denote the number of perfect matchings. Note that mQ = 2, which satisfies the inequality. When k > 2, we can choose matchings independently in each of two disjoint subcubes of dimension k 1. The number of such matchings is By the induction hypothesis, this is at least (22k_3)2, which equals 221c_2.
Comment: Since we could choose the two disjoint subcubes in k ways, we can recursively form perfect matchings in this way, some of which are counted more than once.
Proof 2 (direct construction). Pick two coordinates. There are 2k2 of Q2 in which those two coordinates var4', and two choices of a perfect matching in each copy of Q2. This yields perfect matchings. (Since we can choose the two coordinates in ways, we can generate percopies
fect matchings, but there is some repetition.)
3.1.17. In every perfect matching in the hypercube Qk, there are exactly (k—i) edges that match vertices with weight i to vertices with weight i + 1, where the weight of a vertex is the number of is in its binary k-tuple name. Proof 1 (induction on i). Since the vertex of weight 0 must match to a vertex of weight 1, the claim holds when i = 0. For the induction step, the induction hypothesis yields (tii) vertices of weight i 1 matched to vertices of weight i. The remaining vertices of weight i must match to vertices of = (k71), the claim follows. weight i + 1. Since Proof 2 (canonical forms). Let M* be the matching where every edge matches vertices with 0 and 1 in the last coordinate. The number of edges matching weight i to weight i + 1 is the number of choices of i ones from the first k 1 positions, which is (k71). It now suffices to prove that every perfect matching M has the same weight distribution as M*. The symmetric difference of M and M* is a union of even cycles alternating between M and plus isolated vertices saturated by the same edge in both matchings. It suffices to show that the weight distribution on each cycle is the same for both matchings. The edges joining vertices of weights i and i -F- 1 along a cycle C alternate appearing with increasing weight and with decreasing weight, since weight changes by 1 along each edge. For the same reason, the number of edges along C from a vertex to the next appearance of a vertex with the same weight is even. Since C alternates between M and Mt, this means that the edges joining vertices of weights i and i + alternate between M and M*. Hence there is the same number of each type, as desired. 3.1.18. The game of choosing adjacent vertices, where the last move wins,
Suppose that G has a perfect matching M. Whenever the first player
Chapter 3: Matchings and Factors
chooses a vertex, the second player takes its mate in M. This vertex is available, because after each move of the second player the set of vertices visited forms a set of full edges in M, and the first player cannot take two vertices at a time. Thus with this strategy the second player can always make a move after any move of the first player and never loses. If G has no perfect matching, then let M be a maximum matching in G. The first player starts by choosing a vertex not covered by M. Thereafter, whenever the second player chooses a vertex x, the first player chooses the mate of x in M. The vertex x must be covered by M, else x completes an M-augmenting path using all the vertices chosen thus far, Thus the first player always has a move available and does not lose. 3.1.19. A family A1 Am of subsets of Y has a system of distinct represenS C [in]. Form an X, Y-bigraph tatives ifand only if U105 A1 G with X = (1 m> and Y = . Include the edge iyy if and only
E A1. A set of edges in G is a matching if and only if its endpoints in Y form a system of distinct representatives for the sets indexed by its endpoints in X. The family has a system of distinct representatives if and only if G has a matching that saturates X. It thus suffices to show that the given condition is equivalent to Hall's
conditionforsaturatingX. JfSc X,thenN0(S) =
3.1.20. An extension of Halt's Theorem using stars with more than two vertices. We form an X, Y-bigraph G with partite sets X = x1, . x0 for the trips and Y = . y0, for the people, and edge set . To fill each trip to its capacity c1, we seek a subgraph whose .
components are stars, with degree c1 at x1. Form an X', Y-bigraph G' by making a1 copies of each vertex x1. Now
G has the desired stars if and only if G' has a matching that saturates X'. Thus the desired condition for G should become Hall's Condition for G'. In G', the neighborhoods of the copies of a vertex of x are the same. Hence Hall's Condition will hold if and only if it holds whenever S C X' consists of all copies of each vertex of X for which it includes any copies. That is, Hall's Condition reduces to requiring c1 for all T C > X. This condition is necessary, since the trips in T need this many distinct people. It is sufficient, because it implies Hafl's Condition for G'. 3.1.21. If G isanX, Y-bigraph such that > wheneverc Sc X, then every edge of G belongs to some matching that saturates X. Let xy be an edge of G, withx e X and ye 1', and let G' = G—x —y. Each set S C X [x > loses at most one neighbor when y is deleted. Combining this with the hypothesis yields 1 > Thus G' satisfies >
Section 3.1: Matchings and Covers
Hall's Condition and has a matching that saturates X . With the edge xy, this completes a matching in G that contains xy and saturates X.
3.1.22. A bipartite graph G has a perfect matching if and only
> S C V (G). This conclusion does not hold for non-bipartite graphs.
In an odd cycle, we obtain neighbors for a set of vertices by taking the vertices immediately following them on the cycle. Thus S
C V, but the graph has no perfect matching. Complete graphs of odd
order also form counterexamples. For bipartite graphs, we give two proofs. Proof 1 (graph transformation). Let G' be a bipartite graph consisting of two disjoint copies of G, where each partite set in G' consists of one copy of X and one copy of Y; call these X' and Y'. Then G' has a perfect matching if and only if G has a perfect matching. Since X' = 1", G' has a perfect matching if and only if it has a matching that completely saturates X'.
By Hall's Theorem, G' has a matching saturating X' if and only if > for all 5' C X'. Given S'C X', let T1 = S'nX and T2 = 5'— T1. Let S C V(G) be the set of vertices consisting of Ti in X plus the vertices of V having copies in T2. This establlshes a bijection between subsets 5' = = ofX' and subsets S of V(G), with Also by the construction of G'. Hence Hall's condition is satisfied for G' if and only if the condition of this problem holds in G. In summary, we have shown G' has a 1-factor > for all SC V(G).
[G has a 1-factor]
for all S'C X' Proof 2 (by Hall's Theorem). Necessity: Let M be a perfect matching, >
and let S be a subset of V(G). Vertices of S are matched to distinct vertices Sufficiency: If of N(S) by M, so > > S for all C the graph thus has a matching M that saturates X. Thus > XL and the condition > = yields CY. Thus and Mis aperfectmatching.
3.1.23. Alternative proof of Hall's Theorem. Given an X, Y-bigraph G, we prove that Hall's Condition suffices for a matching that saturates X. Let m= For m = 1, the statement is immediate.
Induction step: m > S
X, select any neighbor y of any vertex x c X. Deleting y reduces
the size of the neighborhood of each subset of X by at most 1. Hence Hall's Condition holds in G' = G x y. By the induction hypothesis, G' has a matching that saturates X , which combines with xy to form a
matching that saturates X. Otherwise, N(S) = = G[S U N(S)], and let G2 =
S c X. Let V(Gt). Because the neighbors of
Chapter 3: Matchings and Factors
vertices in S are confined to N(S), Hall's Condition for G implies Hall's Condition for G1. For G2, consider T c X we obtain NG2(T) =
Thus Hall's Condition holds for both G1 and G2. By the induction hypoth-
esis, G1 has a matching that saturates 5, and G2 has a matching that saturates X 5. Together they form a matching that saturates X. 3.1.24. A square matrix of nonnegative integers is a sum of k permutation matrices if and only if each row and column sums to Ic. If A is the sum of
permutation matrices, then each matrix adds one to the sum in each row and column, and each row or column of A has sum Ic.
For the converse, let A be a square matrix with rows and columns summing to Ic. We use induction on Ic to express A as a sum of k permutation matrices. For Ic = 1, A is a permutation matrix. 1, form a bipartite graph G with vertices xl. x,, and so that the number of edges joining x1 and yy is oj,j. The graph G is bipartite and regular, so by the Marriage Theorem it has a perfect matching. Let b,1 = 1 if x1y1 belongs to this matching and b,1 = 0 otherwise; the
resulting matrix B is a permutation matrix. Each row and column of B has exactly one 1. Thus A' = B A is a nonnegative integer matrix whose rows and columns sum to Ic 1. Applying the induction hypothesis to A' yields Ic 1 additional permutation matrices that with B sum to A. 3.1.25. A nonnegative doubly stochastic matrix can be expressed as a convex combination of permutation matricea For simplicity, we allow multiples of
doubly stochastic matrices and prove a superficially more general statement. We use induction on the number of nonzero entries to prove that if Q is a matrix of nonnegative entries in which every row and every column sums to t, then Q can be expressed as a linear combination of permutation matrices with nonnegative coefficients summing to t. If Q has exactly a nonzero entries, then Q is z' times a permutation matrix, because Q must have at least one nonzero entry in every row and column, If Q has more nonzero entries, begin by defining a bipartite graph G with x1 ±* yy if and only if > 0. If G has a perfect matching, then the edges XIYa(i) of the matching correspond to a permutation a with permuta-
tion matrix P. Let e be the minimum (positive) value in the positions of Q corresponding to the l's in P. The matrix Q' = Q eP is a nonnegative matrix with fewer nonzero entries than Q, and row and column sums t s. By the induction hypothesis, we can express Q' as a nonnegative combination
Section 3.1: Matchings and Covers
= t —s. Hence Q = >1 c1 P1 -f-sP. With Cm+i = sand L_ic1P1, with P, we have expressed Q in the desired form. Pm+i = It remains to prove that G has a perfect matching; we show that it satisfies Hall's condition. IfS is a subset of X corresponding to a particular set of rows in Q, we need only show that these rows have nonzero entries in at least columns altogether. This follows because the total nonzero amount in the rows S is t Since each column contains only a total oft, it is not possible to contain a total oft in fewer than columns. Comment, When the entries of Q are rational, the result follows directly from the Marriage Theorem. Multiplying Q by the least common denominator d of its positive entries converts it to an integer matrix in which all rows and columns sum to d. The entry in position 1, j now is the number of edges joining x1 and yy in a d-regular bipartite graph (multiple edges allowed). The Marriage Theorem implies that the graph has a perfect matching. By induction on d, it can be decomposed into perfect matchings. These matchings correspond to permutation matrices. In the expression of Q as a convex combination of these matrices, we give weight
a/d to a permutation matrix arising a times in the list of matchings.
3.1.26. Achieving columns with all suits. The cards in an n by in array have in values and n suits, with each value on one card in each suit. a) It is always possible to find a set of in cards, one in each column, having the in different values. Form a X, Y-bigraph in which X represents the columns and Y represents the values, with r edges from x c X to y c Y if value y appears r times in column x. Since each column contains n cards and each value appears in n positions (once in each suit), the multigraph is n-regular. By the Marriage Corollary to Hall's Theorem, every nontrivial regular bipartite graph has a perfect matching. (This applies also when
multiple edges are present, which can occur here.) A perfect matching selects in distinct values occurring in the in columns. (Using Hall's Theorem directly, a set S of k columns contains nk cards. Since there are n cards of each value, S contains cards of at least k values. Hence the graph satisfies Hall's condition and has a perfect matching.) b) By a sequence of exchanges of cards of the same value, the cards can be rearranged so that each column consists of n cards of distinct suits. Making each column consist of n cards of different suits is equivalent to spreading each suit across all columns. The frill result follows by induction on n, with n = 1 as a trivial basis step. For the induction step, when n > 1, use part (a) to find cards of distinct values representing the in columns. Then perform at most one exchange for each value to bring the values in a single suit to those positions. Positions
within a column are unimportant, so we can treat the other suits as an
Chapter 3: Matchings and Factors
instance of the problem with n to fix up the remaining suits.
suits, We apply the induction hypothesis
The problem can always be solved using at most ma >k
where a is the minimum size of a winning set and b is the maximum number of winning sets containing a particular position. Let P be the set of positions, and let W1. Wm be the winning sets of positions. With = a, let G be the bipartite graph on n + 2,n vertices with partite sets 2b,
and W = w1 and P1 w9 for each incidence
by creating two edges of a position in a winning set.
If G has a matching M that saturates W, then Player 2 can use M to force a draw. When the position taken by Player 1 on a given move is matched to one of 2b by Hail's condition. IfS C W, then S has representatives (w or w') of at least S /2 winning sets. Since each position appears in at most b winning sets, the number of positions in the union of these winning sets is at least /2)/b> for every SC W. Thus > 3.1.28. A graph with no perfect matching Proof 1 (vertex cover). The graph has 42 vertices, so a perfect matching would have 21 edges. The marked vertices form a vertex cover of size 20. The edges of a matching must be covered by distinct vertices in a vertex cover, so there is no matching with more than 20 edges.
Proof 2 (Hail's condition). Using two labels X and Y, we obtain a bipartition of the graph. Partite set X consists of the marked vertices in the left half of the picture and the unmarked vertices in the right half. This is an independent set of size 21, and the remaining vertices also form an independent set Y of size 21.
Hall's Condition is a necessary condition for a perfect matching; we show that Hail's Condition does not hold. Let S be the subset of X consisting of the 11 unmarked vertices in the right half of the graph. The neighbors of vertices in S are the 10 marked vertices in the right half of the graph. Thus
Section 3.1: Matchings and Covers
Proof 3 (other dual problems). In every graph a'(G) + fl'(G) = suffices to show that at least 22 edges are needed to cover V(G). Also fl'(G) > a(G) always, since distinct edges are needed to cover the vertices of an independent set. Thus it suffices to show that G has an independent set of size at least 22. Such a set is given by the unmarked vertices above so it
(the complement of a vertex cover).
Proof 4 (augmenting paths). Having found a matching M of size 20, one can prove that there is no perfect matching by following all possible M-aiternating paths from one M-unsaturated vertex to show that none reaches the other unsaturated vertex. In this particular example, this method is not too difficult. Proof 5 (symmetry and case analysis). The graph has two edges whose deletion leaves two isomorphic components of order 21. Since 21 is odd, a perfect matching must use exactly one of the two connecting edges. By symmetry, we may assume it is the one in bold above. This forces a neighboring
vertex of degree 2 to be matched to its other neighbor, introducing other bold edge. Repealing this argument yields a path of bold edges forced into the matching. As soon as this leaves a vertex with no available neighbor, we have proved that a perfect matching cannot be completed. 3.1.29. Every bipartite graph G has a matching of size at least e(G)/A(G), Each vertex of G covers at most A(G) edges. Since all edges must be covered
in a vertex cover, this yields /3(G) > e(G)/A(G). By the Kbnig—Egerváry Theorem, a'(G) = /3(G) when G is bipartite. Thus a'(G)> e(G)/A(G), Every subgraph of Ks. with more than (k 1)n edges has a matching of size at least Such a graph G is a simple bipartite graph with partite sets of size a. Thus A(G) -c a, and we compute a'(G) > e(G)/A(G) > (Ic 1)ii/n = Ic 1. Thus G has a matching of size Ic. 3.1.30. The maximum number of edges in a simple bipartite graph that has no matching with Ic edges and no star with 1 edges is (Ic 1)(l 1). If G is a bipartite graph having no matching with Ic edges, then G has a vertex cover using at most Ic 1 vertices. If G is a simple graph having no star
Chapter 3: Matchings and Factors
with 1 edges, then each vertex covers at most 1 1 edges. Hence the vertex cover covers at most (Ic 1)(l 1) edges, which must be all the edges of G. The bound is achieved by (Ic 1)K1,1_1. 3.1.3 1. Hall's Theorem from the Konig—Egervdry Theorem. By the Konig—
Egerváry Theorem, an X, Y-bigraph G fails to have a matching that saturates X if and only if G has a vertex cover of size less than X . Let Q be such a cover, with R = Q fl X and T = Q fl Y. Because Q is a vertex cover, there is no edge from X R to Y T, which means that N(X R) c T. This yields N(X
We have used the Konig—Egerváry Theorem to show that absence of a matching that saturates X yields a violation of Hall's Condition. Thus Hall's Condition is sufficient for such a matching. Similarly, for some S C X, then N(S) U X
cover of size less than X , and there is no matching of size X
Hall's Condition also is necessary.
3.1.32. If G is a bipartite graph with partite sets X, Y, then a'(G) = Let d = The case S = 0 implies = d. Because saturated that d > 0. Choose T C X such that vertices of T must have distinct neighbors in any matching and only T d neighbors are avallable, every matching leaves at least d vertices (of T) unsaturated. Thus a'(G) -c d. To prove that G has a matching as large as d, we form a new graph G' by adding d vertices to the partite set Y and making all of them adjacent to all of X. This adds d vertices to N(S) for each S ci X, which yields >S ci X. By Hall's Theorem, G' has a matching saturating all of X. When we delete the new vertices of G', we lose at most d edges of the matching. Hence what remains is a matching of size at least —din G, as desired. 3.1.33. Konig—Egerudry from Exercise 3.1.32. Always a'(G) -c fl(G), so it suffices to show that a bipartite graph G has a matching and a vertex cover of the same size. Consider an X, Y-bigraph G in which S is a subset of X with maximum deficiency. By part (a), a'(G) = + Let R = (X 5) U(N(S)). By the definition of N(S), there are no edges joining Sand Y N(S). Therefore, R is a vertex cover of G, The size of R is which equals a'(G). Thus G has a matching and a vertex + cover of the same size, as desired.
3.1.34. When G is an X, Y-bigraph with no isolated vertices and the defiN(S), the graph G has a matching that saturates X
ciency of a set S is
Section 3.1: Matchings and Covers
if and only if each subset of Y has deficiency at most Theorem, it suffices to show that > follows from Thus G2 has a matching saturating X S.
The subgraph G1 also satisfies Hall's condition, since it retains all neighbors of each vertex of S. By the induction hypothesis, S has a vertex x such that every edge incident to x belongs to a matching in G1 that saturates S. These matchings can be combined with a single matching in G2 that saturates X S to obtain matchings in G that saturate X. Hence the vertex x serves as the desired vertex in G. It appears that part (a) is not needed to solve part (b). 3.1.38. Pairing up farms and hunting ranges. Suppose the unit of area is the size of one range. Let G be the bipartite graph between hunting ranges and farms formed by placing an edge between a hunting range and a farm if the area of their intersection is at least s, where s = 4/(n + 1)2 if a is odd and s = 4/[n(n -F 2)] if a is even. We prove that this graph has a perfect matching, which yields the desired assignments. Let H be the union of some set of k hunting ranges. Let fi be the areas of intersection with H of the farms, and let F be the set of k farms having largest intersection with H. If = a, then the area of H is 1-a(n+ 1 —k)-ck 1—k). It also equals k, so we have a > 1/(n + 1 k). Since we have k farms meeting H with area at least 1/(n + 1 k), we find for each farm in F a hunting range contained in H that intersects the farm with area at least 1/[k(n + 1 k)] > e. Hence any set of k hunting ranges has at least k neighbors in G, which guarantees the matching. Note that s is the largest possible guaranteed minimum intersection. Let k = fn/21. With hunting ranges in equal strips, we can arrange that some set of k 1 farms intersects each of the first k hunting ranges with
area 1/k, and the remaining farms intersect each of the first k hunting ranges with area s, since (k 1)/k -F (a -F 1 k)s = k hunting ranges must be matched with area s.
Now one of the first
3.1.39. a(G) -ctz(G) —e(G)/A(G). Let Sbe anindependent set ofsizea(G). Since V(G) S is a vertex cover, summing the vertex degrees in V(G) S provides an upper bound on e(G). Thus e(G) -c (n(G) a(G))A(G), which is equivalent to the desired inequality. If G is then a(G) -c a(G)/2. In the previous inequality, set e(G) = n(G)A(G)/2.
3.1.40. If G is a bipartite graph, then a(G) = n(G)/2 if and only if G has a perfect matching Since a(G) = a(G) fl(G) = a(G) a'(G) by Lemma 3.1.21 and the König—Egerváry Theorem, we have a(G) = n(G)/2 if and only if a'(G) = a(G)/2.
Chapter 3: Matchings and Factors
3.1.4 1. (corrected statement) If G is a nonbipartite n-vertex graph with exactly one cycle C, then a(G) > (n 1)/2, with equality if and only if G V(C) has a perfect matching. The cycle C must have odd length, say
k. Let e be an edge of C, and let G' = G e. The graph G' is bipartite, so a(G e) > n/2. An independent set 5' in G e is also independent in G unless it contains both endpoints of e. If > n/2, then we can afford to = n/2, then we can take the other partite drop one of these vertices. If set instead to avoid the endpoints of e. In each case, a(G) > (a 1)/2. If G V(C) has a perfect matching, then an independent set is limited to(k— 1)/2vertices of C and (n—Jc)/2 vertices outside C, soa(G) -c (a— 1)/2 and equality holds. For the converse, observe that deleting E(C) leaves a forest F in which each component has a vertex of G. Let H be a component ofF, with x being
its vertex on C, and let r be its order. If H
x has no perfect matching, then a'(H —x) -c r/2 1 (that is, it cannot equal (r 1)/2). Now /3(H x) -c r/2 1, by Konig-Egervdry, and a(H x) > r/2, since the complement of a vertex cover is an independent set. Since this independent set does not use x, we can combine it with an independent set of size at least (a r)/2 in the bipartite graph G V(H) to obtain a(G) > n/2. Since this holds for each component of F, a(G) = (n 1)/2 requires a perfect matching in G V(C). (This direction can also be proved by induction on n ic.) 3.1.42. The greedy algorithm produces an independEnt set of size at least
in a graph G. The algorithm iteratively selects a vertex of minimum degree in the remaining graph and deletes it and its neighbors. We prove the desired bound by induction on the number of vertices. Basis step: a = 0. When there are no vertices, there is no contribution to the independent set, and the empty sum is also 0. Induction step a > o. Let x be a vertex of minimum degree, let S = U N(x), and let G' = G S. The algorithm selects x and then seeks an independent set in G'. We apply the induction hypothesis to G' to obtain a lower bound on the contribution that the algorithm obtains from G'. Thus the size of the independent set found in G is at least 1 + Note that U N(x) is a set of size d0(x) + 1, and the choice of x as a vertex of minimum degree implies that each vertex in this set contributes at most dG(x) + 1 to the desired sum. Thus 1+ veV(G')
Thus the algorithm finds an independent set at least as large as desired.
Section 3.1: Matchings and Covers
3.1.43. Consequences of Gallai's Theorem (G has no isolated vertices).
a) A maximal matching M is a maximum matching if and only if it is contained in a minimum edge cover. If M is a maximal matching, then the smallest edge cover L containing M adds one edge to cover each Munsaturated vertex, since no edge covers two M-unsaturated vertices. We = have = a + (a ML By Gallai's Theorem (Theorem = a'(G) if and only if = fl'(G). 3.1.22), b) A minimal edge cover L is a minimum edge cover if and only if it contains a maximum matching As observed in proving Theorem 3.1.22, every minimal edge cover consists of disjoint stars. The largest matching contained in a disjoint union of stars consists of one edge from each component. The size of this matching is a Hence a minimal edge cover L has size a a'(G) if and only if L contains a matching of size a'(G).
3.1.44. If G is a simple graph in which the sum of the degrees of any k vertices is less than a k, then every maximal independent set in G has more than k vertices. Let S be an independent set. If -c k, then the sum of the degrees of the vertices in S is less than a k. This means that some vertex x outside S is not a neighbor of any vertex in 5, and hence x can be added to form an independent set containing S. Thus maximal independent sets must have more than k vertices. 3.1.45. Ifxy and xz are a-critical edges in G and y z, then G contains an induced odd cycle (through xy and xz). Let 1', Z be maximum stable sets in G xy and G xz, respectively. Since 1', Z are not independent in G, we havex, y c Y andx, z c Z. Proof 1. Let H = Since x e V fl Z, H is a bipartite graph with bipartition V Z, Z V. If some component of H has partite sets of different sizes, then we can substitute the larger for the smaller in V or Z to obtain a stable set in G of size exceeding a(G). If y and z belong to different components and of H, then let S be the union of P V, one partite set of each other component fl Z, ofH, and VPZ. Sincex c VPZ andy, z 5, the set Sis independentinG. = = Also, > a(G). Hence y and z belong to the same component of H. A shortest y, z-path in H is a chordless path of odd length in G, and it completes a chordless odd cycle with zx and xy. Proof 2. Let H' = G[(Vz1Z)] U . Note that is even, since = Let 2k = If H' is bipartite, then it has an independent set of size at least k +1, which combines with VP Z to form an independent set of size a(G) + 1 in G. Hence H' has an odd cycle. Since H is bipartite, this cycle passes through x. Since NH'(x) = , the odd cycle contains the desired edges.
Chapter 3: Matchings and Factors
3.1.46. A graph has domination number 1 if and only if some vertex neighbors all others. This is immediate from the definition of dominating set.
3.1.47. The smallest tree where the vertex cover number exceeds the domination number is P6. In a graph with no isolated vertices, every vertex cover is a dominating set, since every vertex is incident to an edge, and at least one endpoint of that edge is in the set. Hence y(G) -c fl(G). We want a tree where the inequahty is strict, If y(G) = 1, then a single vertex is adjacent to all others, and since G is = 1. Hence we need y(G) > 2 and a tree there are no other edges, so fl(G) > 3. A tree is bipartite, so al'(G) = fl(G) > 3. A matching of size 3 requires at least 6 vertices. There are two isomorphism classes of 6-vertex trees with perfect matchings, and P6 is the only one having a dominating set of size 2. (Smaller trees can also be excluded by case analysis instead of using the Konig—Egervdry Theorem.)
-Di = = [n/31. With maximum degree 2, vertices can dominate only two vertices besides themselves. Therefore, fn/31 is a lower bound. Picking every third vertex starting with the second (and using the last when n is not divisible by 3) yields a dominating set of size ln/31. 3.1.48.
3.1.49. In a graph G without isolated vertices, the complement of a minimal
dominating set is a dominating set, and hence y(G) -c n(G)/2. Let S be a minimal dominating set. Every vertex of S has a neighbor in 5, and by miimality this fails when a vertex is omitted from 5. For each x c 8, there is thus a vertex of S whose only neighbor inS is x. In particular, every x a S has a neighbor in 5, which means that S is a dominating set. Since S and S are disjoint dominating sets, one of them has size at most n(G)/2. 3.1.50. If G is a n-vertex graph without isolated vertices, then y(G) n/2. As discussed in the proof of Theorem 3.1.22, the components of a minimum edge cover are stars, and the number of stars is n ,8'(G). Since
the union of these stars is a spanning subgraph, choosing the centers of these stars yields a dominating set. Proof 2. Since a'(G) = n fl'(G) by Theorem 3.1.22, it suffices to show that y(G) -c a'(G). In a maximum matching M, the two endpoints of an edge in M cannot have distinct unsaturated neighbors. Also the unsaturated neighbors all have saturated neighbors. Therefore, picking from
Section 3.1: Matchings and Covers
each edge of M the endpoint having unsaturated neighbor(s) (or either if neither has such a neighbor) yields a dominating set of size a'(G). Construction of n-vertex graphs with domination number k, for 1 -c k -cC n/2. Form G from a matching of size k by selecting one vertex from each
edge and adding edges to make these vertices pairwise adjacent. 3.1.51. Domination in an n-vertex simple graph G with no isolated vertices. a) In/(1 -F A(G)1 4. One vertex and three of these, since the graph is 3-regular. Hence its neighbors form a total dominating set of size four.
3.1.55. Dominating sets in the hypercube Since is 4-regular, each vertex dominates itself and four others, Now n(Q4) = 16 yields > 116/51 = 4. Since is an independent dominating set and is a total dominating set, the domination, independent domination, and total domination numbers all equal 4. Adding two vertices to this total dominating set of size 4 completes a connected dominating set of size 6. We show there is no smaller connected dominating set. A connected 5-vertex subgraph contains two incident edges. Let S be the set of three vertices in two such edges. The set T of vertices undominated by S has size 6. Each neighbor of a vertex of S dominates at most two vertices in T. Each vertex ofT dominates at most three vertices in T, except for one vertex that dominates itself and four others (For example, ifS = , then the high-degree vertex of Q4[T] is 1111.) To dominate T with only two additional vertices, we must therefore use the high-degree vertex of T. However, its distance to S is 3, so it cannot be used to complete a connected set of size 5.
3.1.56. Five pairwise non-attacking queens can control an 8-by-8 chessboard. As shown below, they can also control a 9-by-9 chessboard. Five queens still suffice for an 11-by-li chessboard, but this configuration does not exist on the 8-by-8 board.
Section 3.1: Matchings and Covers
3.1.57. An n-vertex tree with domination number 2 in which the minimum size of an independent dominating set is Consider the tree of diam-
eter 3 with two central vertices u and v in which one central vertex has 2)/2j leaf neighbors and the other has [(a 2)121 leaf neighbors. The set is a dominating set, but these cannot both appear in an independent dominating set. If u does not appear in a dominating set, then all its leaf neighbors must appear. We also must include at least one vertex from the set consisting of v and its leaf neighbors, since these are not dominated by the other leaves. Hence the independent dominating set must have at least [(a 2)/2j + 1 vertices. 3.1.58. Every Ki,r-free graph G has an independent dominating set of size at most (r 2)y(G) (r 3). Let 8 be a minimum dominating set in G. Let 8' be a maximal independent subset of 8. Let T = V(G) R, where R is the set N(S') U 8' of vertices dominated by 8'. Let T' be a maximal independent subset of T, Since T' contains no neighbor of 8', 8' U T' is independent. Since 8' is a maximal independent subset of 8, every vertex of 8 8' has a neighbor in 8'. Similarly, F dominates T T'. Hence 8' U T' is a dominating set. It remains to show that 8' U -c (r 1)y(G) (r 3). Each vertex of 8 8' has at most r 2 neighbors in T', since it has a neighbor in 8', and 8'UT' is independent, and G is K1,1-free. Since 8 is dominating, each vertex -c (r 2) 8 ofT' has at least one neighbor inS 8'. Hence which = y(G) and -c (r (r —3) yields 8' U Since 8' U a, a : v > u, b : u > v. Ifbothmengettheirfirst choices, then they prefer no one to their assigned partner, so the matching is stable. The same argument applies when the women get their first choices. However,
Section 3.2: Algorithms & Applications
the matchings with men getting their first choices and women getting their first choices are different. 3.2.4. Stable matchings under proposal algorithm. Consider the preference orders hsted below. Men
u:a>b>d>c>f>e a:z>x>y>u>v>w v: a>b>c>f>e>d b: y>z>w>x>v>u w:c>b>d>a>f>e c:v>x>w>y>u>z x: c>a>d>b>e>f d: w>y>u>x>z>v y: c>d>a>b>f>e e: u>v>x>w>y>z z:d>e>f>c>b>a f:u>w>x>v>z>y When men propose, the steps of the algorithm are as below. For each round, we list the proposals by u, v, w, x, y, z in order, followed by the resulting rejections. Round 1: a, a, c, c, c, d; a x v, c x w, c x y. Round 2: a, b, b, c, d, d; b x v, d x z. Round 3: a, c, b, c, d, e; c x x. Round 4: a, c, b, a, d, e; a x u. Round 5: b, c, b, a, d, e; b x u. Round 6: d, c, b, a, d, e; dxu. Round 7: c, c, b, a, d, e; cxu. RoundS: f, c, b, a, d, e; stablematching. When women propose, the steps of the algorithm are as below. For each round, we llst the proposals by a, b, c, d, e, f in order, followed by the resulting rejections. Round 1: z, y, v, w, u, U; U X e. Round 2: z, y, v, w, v, u; v x e. Round 3: z, y, v, w, x, u; stable matching.
Note that the pairs uf and tic occur in both results, and in all other cases the women are happier when the women propose and the men are happier when the men propose. 3.2.5. Maximum weight transversal. For each matrix below, we underscore a maximum weight transversal, and the labels on the rows and columns form a cover whose total cost equals the weight of the transversal. For every position (i, J)' the label on row i plus the label on column j is at least the entry in position (i, f) in the matrix. Hence the labeling is feasible for the dual problem. Equality between the sum of the labels in a feasible labeling and the sum of the entries of a transversal implies that the transversal is one of maximum weight and feasible labeling is one of minimum weight, because every feasible labeling has sum as large as the weight of every matching (since the positions in the matching must be covered disjointly by the labels).
00102 444436 311434 414535 756479 853683
778987 687676 696546 585764 576555
312345 767872 313445 736287 441354
Chapter 3: Matchings and Factors
Review of the "Hungarian Algorithm" for maximum weighted matching in the assignment problem. Find a feasible vertex labeling for the dual, i.e. weights 1(v) such that 1(x1) + > w(ij). (This can be done by using the maximum in each row as the row label, with 0's for the columns.) Subtract out to find the value" matrix 1(x1) + l(yy) w(ij). Find a maximum matching and minimum cover in the equality subgraph (0's in the excess matrix). If this is a perfect matching, its value equals the dual value being minimized, hence is optimal. If not, let S be the set of rows not in the cover, T the set of columns in the cover, and s the minimum excess value in the uncovered positions. Subtract e from the row labels in 5, add e to
the row labels in T, readjust the excess matrix, and iterate. (Note: any minimum cover can be used, and we know from the proof of the Konig— Egerváry Theorem that we can obtain a minimum cover by using T U (X
where T and S are the subsets of Y and X reachable by alternating paths from the unsaturated vertices in the row-set X.) If the matching was not complete, then > T and 1(v) decreases, which guarantees the finiteness of the algorithm. The positions are of four types, corresponding to edges from S toT, X Sto T, Sto Y —T, and X Sto 5),
T. The change to the excess in the four cases is 0, +8, —8, 0, respectively.
Note that s was defined to be the minimum excess corresponding to edges from S to V T, so every excess remains positive. For the first matrix above, the successive excess matrices computed in the algorithm could look like those below. These are not unique, because different matchings could be chosen in the equality subgraphs. The entries in the matching and the rows and columns in the cover (X S and T) are indicated with underscores.
622230 433010 943520 835205 5
00101 511220 322000 -÷4 832510 835306 3
Finding a transversal of minimum weight. Let the rows correspond . and let the weight
to vertices x1, . .. , x5, the columns to vertices of edge x1
be the value in position ij. Optimality of the answer can be
proved by exhibiting an optimal matching and exhibiting a feasible labeling for the dual problem that has the same total value. Alternatively, finding a minimum transversal is the same as finding a minimum weight perfect matching in the corresponding graph, which corresponds to a maximum weight matching in the weighted graph obtained by subtracting all the weights from a fixed constant. In the example given,
Section 3.2: Algorithms & Applications
we could subtract the weights from 13, and then the answer would be 13 minus the maximum weight of a transversal in the resulting matrix. In the direct approach, the dual problem is to maximize 1(v) subject to 1(x1) + l(yy) -c w (x1 yy). Subtracting the labels from the weights yields the
"reduced cost" matrix. At each iteration, we detennine the equality subgraph and a as before, but this time add a to the labels of vertices in S (rows not in the cover) and subtract a from the labels of vertices in T (columns in the cover). Since S T , 1(v) increases. Every matching has weight at least >1(v). When G1 contains a complete matching, mm w and max
are attained and equal. In the matrix below, the underscored positions form a minimum-weight
transversal; the weight is 30. In the dual problem, the indicated labeling has total value 30, and the labels 1(x1) and sum to at most the matrix entry w1,1. Hence these solutions are optimal.
3.2.7. The Bus Driver Problem. Bus drivers are paid overtime for the time by which their routes in a day exceed r. There are a bus drivers, a morning routes with durations x1, . . x. and a afternoon routes with durations Assign to the edge a1b1 the weight w11 = max
problem is then to find the perfect matching of minimum total weight. In> dex the morning runs so that x1 > Index the afternoon runs so > A feasible solution matches that Yl a of En]. If there exists i cc j with a(i) > a(j), then we have
It suffices to prove that a > fi, because then there exists a minimizing permutation with no inversion. The nonzero terms in the maximizations have the same sum for each pair. Also, X1 + Ya(i)
Chapter 3: Matchings and Factors
If the central terms in the inequahties are both positive, then a is at least their sum, which equals fi. If both are nonpositive, then a > 0 = fi. If the first is positive and the second nonpositive, then
If instead the second is positive, then a=
3.2.8. When the weights in a matrix are the products of nonnegative numbers associated with the rows and columns, a maximum weight transversal is obtained by pairing the row having the kth largest row weight with the column having the kth largest column weight, for each 1 a1 but the weight b matched with a1 is less than the weight b' matched with a1. To show that switching these assignments increases the total weight, we compute a1b' f ayb = a1b f a1(l/ b) + a1(b b') + a1b' = a1b + a1!1 + (a1 a1)(b' b) > a1b + a1b' When the weights in a matrix are the sums of nonnegative numbers asevery transversal has the same weight. sociated with the rows and Since a transversal uses one element in each row and each column, w1,1 = a1 + b1 means that every transversal has total weight b1. a1 + 3.2.9. One-sided preferences. There are k seminars and n students, each
student to take one seminar. The ith seminar will have k1 students, where = a. Each student ranks the seminars; we seek a stable assignment where no two students can both improve by switching. Form an X, Y-bigraph where X is the set of students and Y has k1 vertices for each vertex i. For each edge from student x1 to a vertex rep-
resenting the ith seminar, let the weight be k minus the rank of the ith seminar in the preference of x1. A maximum weight matching in this weighting is a stable assigmuent,
since if two students can both improve by trading assignments, then the result would be a matching of larger weight. 3.2.10. Weighted preferences need not be stable. Consider men . Each assigns 3 her preference list. Hence we indicate a preference order by a triple whose entries are 0, 1, 2 in some order, with the position of the integer i being the index of the person of the opposite sex to whom this person assigns i points. In the matrix below, the preference vectors of the men and women label the rows and coloumns, respectively. An entry in the matrix is the sum
Section 3.2: Algorithms & Applications
of the points assigned to that potential edge by the two people. The underlined diagonal is the only matching that uses the maximum entry in each row and column, so it is the only maximum-weight matching. However, it is not a stable matching, because man x1 and woman Yi prefer each other to their assigned mates. 120
120(2 0211 1 120\2
The example can be extended for all larger numbers of men and women by adding pairs who are each other's first choice and are rated last by the six people in this example.
3.2.11. In the result of the Gale-Shapley Proposal Algorithm with men proposing, every man receives a mate at least as high on his list as in any other stable matching. We prove that under the G-S Algorithm with men proposing, no man is ever rejected by any woman who is matched to him in any stable matching. This yields the result, since each man's sequence of proposals proceeds downward from the top of his list, and he can only wind up with a woman less desirable than his most desirable match over all stable matchings if he is rejected by some women who matches him in some stable matching. Consider the first time when some man x is rejected by a woman a to whom he is matched in some stable matching M. The rejection occurs because a has a proposal from a man y higher than x on her list. In M, man y is matched to some woman Ii. Since M is stable, y cannot prefer a to b. Thus b appears above a on the list for y. But now the decreasing property of proposals from men implies that y has proposed to b in the GS Algorithm before proposing to a. If y is now proposing to a, then y was previously rejected by b. Since y is matched to b in M, this contradicts our hypothesis that a rejecting x was the first rejection involving a pair that occurs in some stable matching. 3.2.12. The Stable Roommates Problem defined by the preference orderings below has no stable matching. There are only three matchings to consider: and In each, two non-paired people prefer each other to their current roommates. The problematic pairs are , , in the three matchings, respectively.
a:b>c>d b:c > a c:a>b>d
Chapter 3: Matchings and Factors
3.2.13. In the stable roommates problem, suppose that each individual declares a top portion of the preference hst as "acceptable". Define the acceptability graph to be the graph whose vertices are the people and whose edges are the pairs of people who rank each other as acceptable. Prove that all sets of rankings with acceptability graph G lead to a stable matching if and only if G is bipartite. (Abeledo—Isaak [19911). In the stable roommates problem with each individual declaring a top portion of the preference list as "acceptable", and the acceptability graph being the graph on the people whose edges are the mutually acceptable pairs,
all sets of ranhings with acceptability graph G allow stable matchings if and only if G is bipartite. If G is bipartite, then we view the two partite sets as the two groups in the classical stable matching problem (isolated vertices may be added to make the partite sets have equal size). The unacceptable choices for an individual x may be put in any order, since they are all (equally) unacceptable, so we can ensure that all choices for x that are in the same partite set appear at the bottom of the preference order for x. In the outcome of the Gale-Shapley Proposal Algorithm, there is no pair (x, o) from opposite partite sets such that o and x prefer each other to their assigned mates. Also no x prefers an individual in its own partite set to the person assigned to x, since all individuals in its own partite set are unacceptable. Hence the stable matching produced for the bipartite version is also stable in the original problem. If G is not bipartite, then G has an odd cycle [xi, .. ., Define a set of x1 prefers rankings such that x1 prefers x1_1 to all others. The preferences of people not on the cycle are irrelevant. Since the cycle has odd length, the people on the cycle cannot be paired up using edges of the cycle. Given a candidate matching M, we may assume by symmetry that x1 is not matched to x2 or to Xk in M. Now x1 prefers x2 to M-mate of x1, and x2 prefers x1 to the M-mate of x2 (which might be xe), Hence the matching M is not stable. Thus there is no stable matching for these preferences, which means that this acceptabillty graph does not always permit a stable matching.
3.2.14. In the Proposal Algorithm with men proposing, no man is every rejected by all the women.
Proof 1, By Theorem 3.2.18, the Proposal Algorithm succeeds, so it ends with each men being accepted before being rejected by all women.
Proof 2. Once a woman has received a proposal, she thereafter receives a proposal on each round, since the key observation is her sequence of"maybe"s is nondecreasing inherlist. If a round has j rejections and a j "maybe"s, then the a j unrejected men are distinct, since men propose to exactly one woman on each round.
Section 3.3: Matchings in General Graphs
When a man has been rejected by k women, those k women have received proposals, and thereafter by the remarks above they always receive proposals from k distinct men. In particular, when a man has been rejected by a 1 women, on the next round they receive proposals from a 1 distinct men other than him, and he proposes to the remaining women, so the algorithm ends successfully on that step.
3.3. MATCHINGS IN GENERAL GRAPHS 3.3.1. The graph G below has no 1-factor Deleting the four vertices with degree 3 leaves six isolated vertices; thus o(G 8) > for this set S.
3.3.2. The maximum size of a matching in the graph G below is 8. A matching of size 8 is shown. Since n(G) = 18, it suffices to show that G has no perfect matching. For this we present a set S such that o(G 5) > tolating Tutte's condition. Such a set S is marked. (Note: the smallest vertex cover has size 9, so duality using vertex cover is not adequate.)
3.3.3. k-factors in the 4-regular graph below, The full graph is a 4-factor, and the spanning subgraph with no edges is a 0-factor. There is a 2-factor consisting of the outer 4-cycle and the 6-cycle on the remaining vertices. Since these cycles have even length, taking alternating edges from both cycles yields a 1-factor. Deleting the edges of the 1-factor leaves a 3-factor.
Chapter 3: Matchings and Factors
3.3.4. A k-regular bipartite graph is r-factorable if and only if r divides k. The edges incident to a single vertex demonstrate necessity. For sufficiency, a k-regular bipartite graph has a perfect matching, and hence by induction on k is 1-factorable; take unions of the 1-factors in groups of r. 3.3.5. Join of graphs G and H. As long as G and H have at least one vertex each, G v H is connected (it has as a spanning subgraph).
In forming G v H, every vertex of G gains n(H) neighbors in H, and every vertex in H gains n(G) neighbors in G. Hence A(G V H) = max.
3.3.6. A tree T has a perfect matching if and only if o(T v) = 1 for every v c V(T). Necessity. Let M be a perfect matching in T in which u is the vertex matched to v. Each component of T v not containing u must have a perfect matching and hence even order. The component containing u is matched by M except for u, so it has odd order. Sufficiency. Proof 1 (construction of matching). Suppose that o(T v) = 1 for all v c V(T). Each vertex has a neighbor in one component of odd order. We claim that pairing each w to its neighbor in the odd component of T w yields a matching. It suffices to prove that if u is the neighbor of v in the unique odd component T1 of T v, then v is the neighbor of u in the unique odd component l'2 of T u. Since o(T v) = 1, the components of T v other than T1 have even order, The subtree T2 consists of these components and edges from these to v. Hence T2 includes some even vertex sets and , and T2 thus has odd order. Proof 2 (induction on n(T)). The claim is immediate for n(T) = 2. If n(T) > 2 and o(T v) = 1 for all v, then the neighbor w of any leaf u has only one leaf neighbor. Let V = T . The components of V v are the same as the components of T v, except that one of them in T v includes and the corresponding component of V v omits them. Hence the parities are the same, and o(T' v) = 1 for all v c V(T'). By the induction hypothesis, V has a perfect matching, and adding the edge uw to this completes a perfect matching in T. (Comment: It is also possible to do the induction step by deleting an
arbitrary vertex, but it is then a bit more involved to prove that every
Section 3.3: Matchings in General Graphs
component T' of the forest left by matching v to its neighbor in the odd component of T v satisfies the condition o(T' x) = 1 for all x. Proof 2a (induction and extremality). The basis again is n(T) = 2.
For n(T) > 2, let P be a longest path. Let x be an endpoint of P, with y) = 1 and P is a longest path, dy(y) = 2. Deleting x and y yields a tree P such that o(T' v) = o(T v) = 1 for all v E V(T'), since x and y lie in the same component of T' v. Hence the induction hypothesis yields a perfect matching in T', which combines with xy to form a perfect matching in T. Proof 3 (Tutte's Condition). By Tutte's Theorem, it suffices to prove
neighbor y. Since o(T
for all S C V(T) that o(T —5) -c We prove this by induction on Since = 1, o(T v) = 1, we have n(T) even, and hence o(T 0) = 0. When the hypothesis o(T v) = 1 yields the desired inequality for S = .
For the induction step, suppose that > 1. Let T' be the smallest subtree of T that contains all of S. Note that all leaves of V are elements of S. Let v be a leaf of T', and let 5' = S (v>. By the induction hypothesis, = o(T 5') -c 1. It suffices to show that when we delete v from T 5', the number of odd components increases by at most 1. Let T" be the component ofT 5' containing v. Deleting v from T 5' replaces T" with the components of T" v. We worry only if T" v has at least two odd components. Since v is a leaf of T', all of 5' lies in one component of T v. Hence the components of T" v are the same as the components ofT v except for the one component ofT v containing 5'. Since o(T v) = 1, we can have two odd components in T" v only if the one odd component ofT v is a component ofT" v and the component of T" v that is not a component ofT v is also odd. Since the remaining components ofT" v are even, this means that T" itself has odd order (it includes v and the two odd components ofT" v). Therefore, the replacement of T" with the components of T" v increases the number of odd componentsonlybyone. Weconcludethato(T—S) -c o(T—S')+l -c = which completes the induction step. 3.3.7. There exist k-regular simple graphs with no perfect matching When k is even, Kk+1 is a k-regular graph with no perfect matching, since it has an odd number of vertices. When k is odd, there are two usual types of constructions. Construction 1. Begin with k disjoint copies of Kk+1. Delete (k 1)/2 disjoint edges from each copy, which drops the degree of k 1 vertices in each copy to k 1. Add a new vertex v1 to the ith copy, joining it to each of these vertices of degree k 1. Add one final vertex x joined to Uk. The graph has been constructed to be k-regular, Deleting x leaves k components of order k + 2 (odd); hence the graph fails Tutte's
Chapter 3: Matchings and Factors
condition and has no perfect matching. A slight variation is to start with k copies of Kk_1,k, add a matching of size (k 1)/2 to the larger side in each copy, and join the leftover vertices from each larger side to a final vertex x. Construction 2. Begin with Ic disjoint copies of Kk+1. Subdivide one edge in each copy, which introduces Ic new vertices of degree 2. To raise their degree to Ic, add an independent set of Ic 2 additional vertices in the center joined to each of these Ic vertices, Deleting the Ic 2 vertices in the center violates Tutte's condition.
3.3.8. No graph with a cut-vertex is 1-factorable. Suppose v is a cut-vertex of G. If G is 1-factorable, then G has even order, and G v has a component H of odd order. For any 1-factor using an edge incident to v whose other endpoint is not in H, the vertices of H cannot all be matched. The contradiction implies there is no 1-factorization. A 3-regular simple graph having a 1-factor and connectivity 1.
3.3.9. Every graph G with no isolated vertices has a matching of size at least n(G)/(1 + A(G)). We use induction on the number of edges. In the induction step, we will delete an edge whose endpoints have degree at least 2 (other edge deletions would isolate a vertex). This tells us what we need to cover in the basis step. Basis step: every edge of G is incident to a vertex of degree 1. In such a graph, every component has at most one vertex of degree exceeding 1,
and thus each component is a star. We form a matching using one edge from each component. Since the number of vertices in each component is 1
Section 3.3: Matchings in General Graphs
plus the degree of the central vertex, the number of components is at least n(G)/(1 + A(G)). Induction step: G has an edge e whose endpoints have degree at least 2. Since G' = G e has no isolated vertex, we can apply the induction hypothesis to obtaina'(G) > a'(G') > n(G')/(l+A(G')) > n(G)/(1+A(G)). 3.3.10. The maximum possible value of /3(G) in terms of a'(G) is 2a'(G).
If G has a maximal matching of size k, then the 2k endpoints of these edges form a set of vertices covering the edges, because any uncovered edge could be added to the matching. Hence /3(G) - k and n(G) >- 2k, then a'(G) >- k. Let a = a(G). By the Berge—Tutte Formula, it suffices to show that the deficiency o(G 5) is at most a 2k for every S C V(G). We prove this by contradiction; suppose that o(G 5) > a 2k.
Section 3.3: Matchings in General Graphs
8) > a 2k-Es. Thus there are more than 8L We have o(G + s vertices outside 8. Together with 8, we have a > a 2k + 2s. Thus s m > 3o(G 8) 4. Thus > o(G 8) 4/3. Since a(G) is even, and o(G 8) have the same parity, which means that o(G 8) exceeds only if is greater by at least 2. This contradicts o(G 8) -c + 4/3. Hence Tutte's condition holds, and Tutte's Theorem implies that G has a 1-factor. 3.3.16. If G is k-regular and remains connected when any k 2 edges are deleted, then G has a 1-factor By Tutte's Theorem, it suffices to show that o(G 8) -c 8 for 8 = 0 from the assumption that a(G) is even; hence we may assume that 8 Let H
be an odd component of G
and let m be the number of edges joining H
to 8. In the subgraph H, the sum of the degrees is ka(H) m. Since this must be even and a(H) is odd, k and m must have the same parity. By the hypothesis, there are at least k 1 edges between H and 8. The requirement of equal parity thus yields m > k. Summing over all odd components of G 8 yields at least k - o(G 8) edges between 8 and V(G) 8. Since the degree sum of the vertices in 8 is exactly k we 3.3.17. Under the conditions of Exercise 3.3.16, each edge belongs to some 1-factor in G. We want to show that G' = G x y has a 1-factor, By Tutte's Theorem, since G has even order, it suffices to show that o(G' 8') -c 8' + 1 8 C V(G) that for all 8' c V(G'). Equivalently, o(G 8) k. Summing over all odd components of G 8, we have m > k- o(G 8), where m is the number of edges between 8 and the rest of the graph. Since G is k-regular and G[8] contains the edge xy, we have m —2. Thus o(G —8) 1, let H be a 2factor of G, and let i be a leaf of T, with neighbor j. Note that the distance in T from j to any other vertex of T is at most diam (T) 1. The induction hypothesis guarantees a labeled T i-decomposition of G E(H). For each vertex w in each cycle in H, we add to the copy of T i with j at w by adding the edge to the next vertex in the cycle, which will then receive label i. This vertex does not already appear in this copy of T i, because the girth of G exceeds the diameter of T.
3.3.22. Hall's Theorem follows from Tutte's Theorem. Given an X, Ybigraph G, let H be the graph obtained from G by adding one vertex to V if n(G) is odd and then adding edges to turn V into a clique. a) G has a matching of size if and only if H has a 1-factor Each edge of a matching in G has one vertex of X and one vertex of V. Since H[Y] is a clique, we can pair the remaining vertices arbitrarily to obtain a 1-factor in H from a matching of size in G. Conversely, if H has a 1factor, it must use edges to saturate X, since HEX] is an independent set. These edges from the desired matching in G. S C X), then H b) If G satisfies Hall's Condition satisfies Tutte's Condition (o(H T) -c T a clique, the odd components obtained by deleting T are the vertices of X whose neighbors all lie in T and perhaps the one large remaining component. Let S = . Since G satisfies Hall's Condition, -c Tn Thus o(H T) 2): Let T be a Breadth-First Search tree from r. Let u be a vertex at maximum distance from r, and let s be the parent of v. If s has no other child, then 1" = T is connected. if s has another child v, then let V = T . In each case, V is connected, so G'[V(T')J is a smaller connected claw-
free graph of even order. The induction hypothesis guarantees a perfect matching in G'. To this matching we add the edge between the two vertices we deleted to obtain G' (in the first case, the edge us exists because s is the parent of U; in the second case, the edge ut' exists by part (a).)
3.3.24. Maximum number of edges with no 1-factor A maximal n-vertex graph with no 1-factor consists of in vertices of degree n 1, with the remaining n in vertices inducing a union of in +2 cliques of odd order Since adding an edge cannot reduce a' or increase it by more than one, we may = 2; the assume that a'(G) = n/2 1. Hence 8) maximum matching omits 2 vertices, Let S be a set achieving equality, so o(G 5) = S must induce a clique, G S has no component of even order (else add edges from even to odd components), and vertices of S have degree n 1, all because adding the edges that would be missing if any of these failed would not reduce the deficiency of S. This completely describes the maximal graphs.
The maximum number of edges in a graph with minimum degree (n_2k_1) k 0, which requires only
Section 3.3: Matchings in General Graphs
k 8, we can build counterexamples when Ac Ac. By direct computation, f(t + 1) f(t) = 3t a + 4. If t > n/2, then there is no graph with minimum degree at least t that has no 1-factor. Since f(t) is a parabola centered at 1 = (a 4)/S and 8(G) q > Ac in + 3, then we gain 2p and lose 2q 4 edges by moving two vertices
from the smaller to the larger clique in G 5, still maintaining the same minimum degree. Hence for fixed in, e(G) is maximized by using cliques of size Ac in + 1 for all but one component of G S. Now the degree sum in G is in (n—i) + (Ac in +1) (in + i)Ac + [a (Ac in f 1) (in f 1)] [a (Ac in +1) (in +1)]. If in Ac/2. If Ac > in 1, then we can increase in by moving two vertices from a small clique to 5, since 2(in + 1) > 2(Ac in 1) guarantees that we can then take vertices from the other small components to make new components of size Ac in 1. This increases the number of edges (computation omitted), so we may assume in = Ac. Now e(G) = f(Ac).
3.3.25. A graph G is factor-critical ifand only if a(G) is odd and o(G 5) -c S c V(G). Necessity. Factor-critical graphs are those where every subgraph obtained by deleting one vertex has a 1-factor. Thus factor-critical graphs have odd order. Given a nonempty subset S of V (G), let v be a member of 5, and let G' = G v. Since G' has a 1-factor and = G S = G' 5', we have o(G —5) = o(G' 5') d1 even and d1 > = then G has an f-factor if and only if d1
for all k, s with k + s
k, then G is k-connected. Suppose S is a separating f-set of H, where j k and X has edges to at least k distinct vertices of H X. If Sfl = 2, then H S = G —(5— XUx), and 5— XUx is aseparatingf 1-set ofG. If = 1, then SUX separates H, since X S is not a component of H S. Hence S X U x is a separating f-set of G, which requires j > k. Finally, suppose S n X = 0. Now must belong to the same component of H S. Contracting an edge of a component in a disconnected graph leaves a disconnected graph, so in this case S separates G. = K1 v C,,1 by a sequence b) Every graph obtained from a "wheel"
of edge additions and vertex 3-splits on vertices of degree at least 4 is 3connected. Since wheels are 3-connected, part (a) implies that every graph arising from wheels by 3-splits and edge additions is also 3-connected. The Petersen graph arises by successively splitting off vertices from the central vertex of the wheel K1 v 06. Each newly-split vertex acquires two neighbors on the outside and remains adjacent to the central vertex.
4.2.28. If X and Y are disjoint vertex sets in a k-connected graph G and u(x) = w(y) = are assigned nonnegative integer weights with then G has k pairwise internally disjoint X, Y-paths from X to Y such that u(x) of them start at x and w(y) of them end at y. We may assume that all weights are positive, since otherwise we delete vertices of weight 0 from X and Y and apply the argument to the sets that remain, We construct a related G' and apply Menger's Theorem. Add copies of vertices in X and Y, with each new vertex having the same neighborhood as the vertex it copies. Since G is k-connected, these neighborhoods have size at least k, and by the Expansion Lemma the new graph is k-connected. We do this until there are u(x) copies of each x and w(y) copies of each y. Next add two additional vertices s and tjoined to the copies ofallx e X and the copies of all y E Y, respectively. Note that s and t each have degree k in this final graph G'. By the Expansion Lemma, G' is k-connected, By Menger's Theorem, there are k pairwise internally disjoint s, t-paths in G'.
Section 4.2: k connected Graphs
These must depart s via its k distinct neighbors and reach t via its k distinct neighbors, so each path connects a copy of some x a X to a copy of some
Y, and no x or y appears in one of these paths except at endpoints. Collapsing G' to G by the copies of each original vertex turns these into the desired paths, since there are u(x) copies of each x and w(y) copies of each y and one path at the original vertex arising from each copy ofit in G'. y a
4.2.29. Graph connectivity from connectivity in the corresponding symmetric digraph. From a graph G, we form D be by replacing each edge with two oppositely-directed edges. Given two vertices a, b on a path P, let P(a, b) denote the a, b-path along P. y) = Xb(x, y). It suffices to prove If 4)(x, y) = X'0(x, y), then that y) -c 4,(x, y), since the weak duality y) > X'D(x, y) and X0(x, y) -c y) holds always. Let F be a family of X'D(x, y) pairwise edge-disjoint x, y-paths in D. If
there is some vertex palr u, v such that uv appears in a path P in F and vu appears in another path Q in F, then we modi& F. Let P' be path consisting of P(x, u) followed by Q(u, y), and let Q' be the path consisting of Q(x, v) followed by P(v, y). Replacing P and Q with P' and Q' in F reduces the number of edges that used in both directions. Repeating this replacement yields a family F' with no such doubly-used palr. Now F' becomes a family of X'D(x, y) pairwise edge-disjoint x, y-paths in G using the same succesion of vertices, and hence y) > )JD(x, y). Let R be a set of y) edges in D whose removal makes y unreachable from x. By the construction of D from G, every x, y-path in G must use an edge having a copy in R. Hence the corresponding edges in G form an x, y-disconnecting set, and x y) y y) = Xa(x, y). It suffices to prove that Xa(x, y) ?: XD(x, y) and .'ca(x, y) -c .'CD(X, y), since the weak duality Xg(x, y) -c icg(x, y) holds always. Let F be a family of XD(x, y) pairwise internally-disjoint x, y-paths in D, Since these pairs pairwise share no vertices other than their endpoints,
there is no pair u, v such that the edges uv and vu are both used. In particular, the paths (listed by vertices) in F also form a family of XD(x, y) pairwise internally-disjoint x, y-paths in G, and Xa(x, y) XD(x, y). Let R be a set ofKD(x, y) vertices in D whose removal makes y unreach-
able from x. By the construction of D from G, every x, y-path in G uses a vertex of R, Hence R is an x, y-separating set in G, and Ka(x, y) 1(C) + 1(D), because then C and D are not longest cycles in G. Consider k = 2. Let e be an edge of C, and e' an edge of D, chosen to share the vertex of 5' if 5) = 1. Since G is 2-connected, there is a cycle R
containing both e and e'. The two portions of R between e and e' contain paths P, Q that travel from V(C) to V(D) with no vertices of V(C) U V(D) along the way. (If 5) = 1, then one of these paths is a single vertex and has length 0.) Note that since R is a cycle, P and Q are disjoint. The vertices where P and Q intersect C and D partition C and D into paths C1, C2 and Di, D2, respectively. Let C' = C1 U P U D1 U Q and D' = C2 U P U U Q; we have 1(C') + 1(D') = 1(C) + 1(D) + 21(P) + 21(Q) > 1(C) + 1(D).
Since G is also 2-connected, we may assume by the
Section 4.2: k connected Graphs
= 2. Now G S is connected and has a shortest argument above that path P between C S and D S. The vertices where P meets C and D, together with the vertices S = , partition C and D into three paths C3 and D1, D2, D3, where D1 are y, x-paths, C2, D2 are x, V(P)paths, and C3, D3 are y, V(P)-paths. Let C' = U C2 U P U D3 and D' = C1, C2,
D1 U D2 U P U C3.
Now 1(C') -F 1(D') = 1(C) -F 1(D) + 21(P) > 1(C) -F 1(D).
b) For k > 2, one cannot guarantee more than k common vertices. The graph Kk,2k is k-connected and has two cycles sharing only the smaller partite set.
2, let G1 and G2 be disjoint k-connected graphs, with c V(G1) and v2 c V(G2), If B is a bipartite graph with parts N01(v1) and
(v2) that has no isolated vertex and has a matching of size at least k, then (Gi vt)U(G2 —v2)UB is k-connected. Let G = (G1 —vt)U(G2 —v2)UB. It suffices to show that for distinct vertices x, y c V(G), there is a family of k independent x, y-paths. If x, y c V(Gjj, then there are k such paths from G1, except that one of them may pass through v. If x' and y' are the neighbors of v along this path, then we replace (x', v, y') with a path through G2, using edges in B incident to x' and y'. The argument is symmetric when x, y E V(G2).
Ifx e V(G1) andy e V(G2), then let X c NG1(vi) andY c be the partite sets of a matching M of size k in B. Deleting v1 from k independent x, vi-paths in G1 leaves an x, X-fan. Similarly, deleting v2 from k independent y, v2-paths in G2 leaves an y, Y-fan. Combining M with these two fans yields the desired x, y-paths. The claim fails fork = 1. If G1 and G2 are stars, with centers v1 and 1)2, then the resulting graph G is simply the bipartite graph B. The only requirement on B is that it have no isolated vertices. In particular, it need not be connected. 4.2.33. Ford-Fulkerson CSDR Theorem implies Hall's Theorem. Given an X, Y-bigraph G with X =
SDR, and thus G has a matching saturating X. Thus it suffices to show that Hall's Condition on G implies the Ford-Fuilkerson condition for these systems. Let I, J c [m] be sets of indices. Since B1 = A1, we have A1 P
By Hall's Condition, Uieznj > 1 Fulkerson condition holds in G, as desired.
Chapter 4: Connectivity and Paths
If Ai. Am and B1, . are partitions of a set K into sets of size s, then the two systems have a CSDR. It suffices to show that the systems satisfy the Ford-Fulkerson condition. By the defining condition, A1 = sand B4 = s. Thus A1 U 1€!
-in. 4.2.34. Every minimally 2-connected graph has a vertex of degree 2. Consider an ear decomposition of a minimally 2-connected graph G. If the last ear adds just one edge e, then G e also has an ear decomposition and is 2-connected, Hence the last ear added contains a vertex of degree 2. A minimally 2-connected graph G with at least 4 vertices has at most 2n(G) —4 edges, with equality only for The graph is minimally 2-connected and has 2n 4 vertices. For the upper bound, we use induction on n(G). When n(G) = 4, K2,2 is the only minimally 2-connected graph. When n(G) > 4, consider an ear decomposition of G. If G is only a cycle,
then the bound holds, with strict inequality. Otherwise, delete the last added ear from G to obtain G'. This deletes k vertices and k + 1 edges, where k >- 1 as observed above.
The graph G' is also minimally 2-connected, since if G' e is 2connected, then also G e is 2-connected. Hence e(G') 6(H'), and thus f(a) > 1 + 8(H') = 1
When greedy coloring is run with respect to a', each v, is a vertex of minimum degree in G. Thus f(a') = 1+max,8(G,) -c 1+maxHCG8(H). By the first paragraph, equality holds.
5.1.37. The vertices of a simple graph G can be partitioned into 1 + maxncG 8(H)/i classes such that the subgraph induced by the each class has a vertex of degree less than r. Let v,, be a vertex of minimum degree in G, and for i e(G') >
If G is the union of in cliques of order in,
then x(G) n(G). If a proper coloring partitions n vertices into k color classes, there must be at least n/k vertices in some class, by the pigeonhole principle. These vertices form a clique in the complement, which forces x(G) > n/k. Hence x(G) > n/x(G), or x(G)- x(G) > n. x(G) + x(G) >
Two numbers with a fixed product x have
smallest sum when they are equal; then their sum is inequality implies this bound. For n
is achieved by G = rnKm,
Hence the first
Since the complement is a complete
both graphs have chromatic number
5.1.41. x(G) + x(G) -c n(G) + lfor every graph G.
Proof 1 (induction on n(G)). The inequality holds (with equality) if n = 1. For n> 1, choosev c V(G), and let G' = G v. By the induction hypothesis, x(G') + x(G') -c n. When we replace v to obtain G and G, each chromatic number increases by at most 1. We have the desired bound unless they both increase. If both increase, then v must have at least x(G') neighbors in G (else we could augment a proper coloring of G' to include v) and similarly at least x(G') neighbors in G. Since v has altogether n 1 neighbors in G and we conclude that in this case x(G') + x(G') -c n 1, and adding 2 again yields the desired bound + x(G) k and d, + 1 > k for i k. This becomes dl a k + 1 and > a k for j cc a k + 1. Therefore
maximin(dl + 1,j> = a —kf 1, so x(G)+ x(G) -c k+(a —k+ 1) = a+ 1. Proof 3 (degeneracy). By the Szekeres—Wilf Theorem, it suffices to show that maxaco8(H) + maxHCG6(H) -c a 1. Let H1 and H2 be subgraphs of G and achieving the maximums. Let k, = 8(H3, Note that a(H) > + 1. Jfk1 -f-k2 > a, then H1 and have a common vertex v. Now v must have at least k, neighbors in 8, for each i, but only a 1 neighbors are available in total. 5.1.42. Analysis of the ratio of x(G) to a(G)/a(G). a) x(G) x) a(G). Hence x(G)a(G)/(a + 1) x(G) x(G)/(a -F 1) f(v1). Obtainfrom D two digraphs F and H defined as follows. Given the edge zy in D, put zy in F if f(x) -c f(y), and put zy in H if f(x) > f(y). If D has no nondecreasing path of length r and no decreasing path of length s, then F has no path of length r and H has no path of length s. By the Gallai—Roy Theorem, this
implies x(F) -c r and x(H) -c s. By part (a), we have x(G) -c is, where G = F U H, but this contradicts the hypothesis on G. Hence one of the specified long paths exists.
c) Every sequence of rs + 1 distinct real numbers has an increasing subsequence of size r + 1 or a decreasing subsequence of size s + 1. Let D be the tournament with vertices v1 and —÷ vy if i > j, and let f (v,) be the ith value in the sequence a. Every path in D corresponds to a subsequence of a, where the vertex labels are the values in a. Because x(D) = is + 1, part (b) guarantees an increasing path with r -f- 1 vertices or a decreasing path with s + 1 vertices. 5.1.44. Minty's Theorem. Given an acydic orientation D of a connected graph G, let r(D) = maxc fa/bl, where a counts the edges of C that are forward in D and b counts those that are backward in D. Fix a vertex x e V(G), andlet Wbeawalkfromx i. Letg(W) = a —b-r(D),where a counts the steps along W followed forward in D and Ii counts those followed
backward in D. For y a V(G), let g(y) = max. a) g(y) is finite and thus well-defined, and G is 1 + r(D)-colorable.
By the definition of r, every cycle with a forward edges has at least ra backward edges. Hence traversing a cycle makes no positive contribution to g(W), and g(y) = g(W) for some x, y-path W. Thus there are only finitely many paths to consider, and g(y) is well-defined. To obtain a proper coloring of G, let the color of y be the congruence class of g(y) modulo 1 + r(D). If U —÷ v in D, then g(v) > g(U) + 1, since wu can be appended to an x, U-walk. On the other hand g(U) g(v) r(D), since VU can be appended to anx, v-walk. Thus g(U)+ 1 -c g(v) -c g(U)+r(D) when U and v are adjacent in G, which means that g(U) and g(v) do not lie in the same congruence class modulo r(D) + 1. b) x(G) = minDOD 1 + r(D), where D is the set of acyclic orientations of G. The upper bound follows immediately from part (a). For the lower
bound, we present an acydic orientation D such that r (D) -c x (G) 1. Given an optimal coloring f with colors 1, . x(G), orient each edge zy in the direction of the vertex with the larger color. Since colors increase
ChapterS: Coloring of Graphs
strictly along every path, the orientation is acyclic and has maximum path length at most x(G) 1.
5.1.45. Gallai-Roy Theorem from Minty's Theorem. We first prove that 1+1(D) is minimized by an acychc orientation, to which we can then apply Minty's Theorem. If D is an arbitrary orientation, let D' be a maximal acydic subgraph of D. Let xy be an edge of D D', Since adding xy to D' creates a cycle, D' contains a y, x-path. Let Dt be the orientation of G obtained from D by reversing the orientation on each edge of D D'. If contains a cycle C, then for each reversed edge yx on C corresponding to an edge xy of D D', we replace yx with a y, x-path that exists in D1. The result is a closed (directed) walk in D', This yields a cycle in D', because a shortest closed walk in a digraph that has a closed walk is a cycle. Since by construction D' is acyclic, we conclude that Dt is acyclic. We also claim that -c 1(D'). Let P be a u, v-path in Dt; some edges of P may have opposite orientation in D and Dt. For such an edge yx c E(P), there is a y, x-path in D'. When we replace all such edges of Dt D in P by paths in D', we obtain a u, v-walk in D'. This must in fact be a u, v-path in D', because D' is acyclic. Finally, the path we have found in D' is at least as long as P, because we replaced each edge of P not in D' with a nontrivial path in D'. Since D' C D, also 1(D') -c 1(D), so r(Dt), and we obtain the desired inequahty x(G) k unless G is a complete graph or an odd cycle. Let H be ak-critical subgraph of G. Since H is k-critical, 3(H) > k 1. If A(G) 2k and G has no odd cycle longer than 2k 1, then G has a cycle of length at least 4k. Let P = x1, . . .x, be a maximal path in G, so N(x1) C V(P). Let Xr be the neighbor ofx1 farthest along P; d(x1) 2k implies r >- 2k + 1. By the odd cycle condition, r is even, and neither nor can belong to N(xi) ifi k. If and 2 k> are disjoint, then together with N(x1) we have at least r 2k + 2k = r
vertices with indices from 2 to r. This is impossible, so we must have 2k—f-14k, c) If G is a 2-connected graph having no odd cycle longer than 2k 1, then x(G) -c 2k. We use induction on n(G). For n(G) = 2, the claim holds using k = 1. For the induction step, suppose n(G) = a > 2 and the claim holds for graphs with fewer than a vertices. Since x(G) is the maximum chromatic number of its blocks, we may assume G is 2-connected. Suppose the longest odd cycle in G has length 2k 1, but x(G) > 2k. For any x E V(G), the induction hypothesis implies x(G x) -c 2k. Hence G is vertex-(2k + 1)-critical, which imphes 3(G) > 2k. By part (b), G has a cycle C of length at least 4k. By part (a), G has a path P joining two vertices x, y
Section 5.2: Structure of k chromatic Graphs
of C such that P together with either x, y-path along C forms an odd cycle. The sum of the lengths of these two odd cycles is at least 4k + 2. Hence one of them has length at least 2k + 1, contradicting the hypothesis. The contradiction yields x(G) -c 2k.
5.2. STRUCTURE OF k CHROMATIC GRAPHS = x (G) 2 for all pairs x, y of distinct vertices, then G is a complete graph. If x y, then a proper coloring of G x y can be augmented with one new color on x and y to obtain a proper coloring of G, This yields x(G) -c x(G x y) + 1, so the given condition forces x -e- y
for allx, ye V(G). 5.2.2. A simple graph is a complete multipartite graph if and only if it has no induced three-vertex subgraph with one edge. If a connected graph is not
a clique, then the shortest of all paths between nonadjacent pairs of vertices has length two, and the three vertices of this path induce a subgraph with exactly two edges. Hence each successive pair of the following statements are equivalent: (1) G has no induced 3-vertex subgraph with one edge. (2) G has no induced 3-vertex subgraph with two edges. (3) Every component of G is a clique. (4) G is a complete multipartite graph. 5.2.3. The smallest k-critical graphs. a) If x, y are vertices in a color-critical graph G, then N(x) C N(y) is impossible, and hence there is no k-critical graph with k -F 1 vertices. If G is k-critical, then G x is (k 1)-colorable, but N(x) C N(y) would allow us to return x with the same color as y to obtain a (k 1)-coloring of G. If n(G) = k + 1, then we have 6(G) k 1 by the properties of k-critical graphs. Hence 6(G) = k 1, which implies that nonadjacent vertices x, y have the same set of neighbors (the remaining k 1 vertices), which contradicts the statement just proved. Hence there is no k-critical graph with k -F 1 vertices. b) x(G v H) = x(G) + x(H), and G v H is color-critical if and only
if both G and H are color-critical, and hence there is a k-critical graph with k -F 2 vertices, Coloring G and H optimally from disjoint sets yields a proper coloring of G v H, so x(G v H) -c x(G) + x(H). The colors used on the subgraph of G v H arising from G must be disjoint from the colors on the copy of H, since each vertex of the former is adjacent to each of the latter; hence x(G v H) > x(G) + x(H).
For criticality, consider an arbitrary edge xy c E(G v H). If xy c E(G), then (Gv H) xy = (G xy) v H, and hence x(G v H) xy =
ChapterS: Coloring of Graphs
being color-critical implies that G and H are color-critical. For the converse, assume that G and H are color-critical. We have already considered G v H xy for xy a E(G) U E(H); we must also consider xy a E(G v H) with x a V(G) and y a V(H). By the properties of color-critical Hence G v H
graphs, we know that G and H have optimal colorings in which x and y, respectively, are the only vertices in their color classes. In G v H xy, we use these colorings but change the color of y to agree with x. This uses x(G) + x(H) 1 colors. Since C5 is 3-critical and Kk_3 is (k 3)-critical, we conclude that C5 v Kk_3 is a k-critical graph with k + 2 vertices.
5.2.4. Blocks and coloring in a special graph. Let G be the graph with -c 2 and I + I vertex set (v0 defined by v, v1 if and only if i is not divisible by 6.
a) The blocks of G. Because consecutive integers sum to a number that is odd and hence not divisible by 6, the vertices vo form a path in order. Edges of the form modulo 6, but not when i is congrent to 2 or 5 modulo 6. Thus G is the graph below, and there are a blocks. The blocks are the subgraphs induced by
b) Adding the edge to G creates a 4-critical graph. In a proper 3coloring of G, the induced kites force successive vertices whose indices are multiples of 3 to have the same color. When the edge 1)O1)3n is added to form G', the graph is no longer 3-colorable. If an edge in the ith kite is deleted, then giving its endpoints the same color permits properly 3-coloring the remainder of the subgraph induced by
5.2.5. A subdivision of K4 in the Grötzsch graph. The subgraph in bold below is a subdivision of K4.
Section 5.2: Strueture of k chromatic Graphs
5.2.6. The minimum number of edges in a connected n-vertex graph with chromatic number k is + ii k. Equality holds for the graph obtained by a vertex of Kk with an endpoint of Pn_k+1. The desired lower bound on e(G) when k = 2 is ii 1 and holds trivially for connected graphs, so we may assume that k > 3. Proof 1 (critical subgraph). Let G be a connected k-chromatic n-vertex graph. Let H be a k-critical subgraph of G. If H has t vertices, then e(H)> (k 1)t/2, since 8(H) > k 1. With H and the remaining n z' vertices of G as n t -F 1 components, we add at least n t additionall edges to reduce the number of components to 1. Hence e(G) > (k 1)t/2 + n t = (k 3)t/2 -F n. Ic, this is minimized when t = k, yielding the desired value. Proof 2 (induction on ii). For ii = k, the bound again is trivial. For n > k, let G be a minimal connected k-chromatic n-vertex graph. By the
choice of G, deletion of any edge disconnects G or reduces k. If G e is disconnected for some e e E(G), then it has two components. At least one of these must be k-chromatic, else we can recolor G with fewer than k colors. Letting 1 be the number of vertices in a k-chromatic component of G e, the induction hypothesis yields 1) = -F n e(G) > [C) -F 1 k] -F 1 -F (ii where the additional terms count e itself and the edges of a spanning tree of the other component. 1
In the remaining case, x(G
k 1. Hence e(G) > n(k 1)/2 = n-F n(k 3)/2 > n + k(k
5.2.7. In an optimal coloring of a graph, for each color there is a vertex of
that color that is adjacent to vertices of all other colors. Let C be the set of vertices of color i, and consider v e C. If v has no neighbor of color j, then we can switch the color of v to j. Since we are changing colors othy for vertices in C, moving several of them to color j in this way creates no
ChapterS: Coloring of Graphs
conflicts, since C is an independent set, After relabeling all vertices of C, we have obtained a proper coloring without using color i. Hence C must have some "unmovable" vertex, adjacent to vertices of every other color. 5.2.8. Critical subgraph approach to x(G) k 1. Hence x(G) = k = min 1, let x be vertex of maximum degree, with d(x) = k. Since G has no (r + 1)-clique, the subgraph G' induced by N(x) has no i-clique. Hence G' has at most edges, by the induction hypothesis. The remaining edges are incident to the remaining n k
vertices; since each such vertex has degree at most k, there are at most k(n k) such edges. Summing the two types of contributions, we have e(G) -c k(n oh), where a = r/(2r —2). The functionk(n ak) is maximized by settingk = where it equals Hence e(G) -c = (1— Proof 2 (by Turán's Theorem). By Turán's Theorem, the maximum number of edges in a graph with no (r + 1)-clique is achieved by the complete r -partite graph with no two part-sizes differing by more than one. If the part-sizes are >, the degree-sum is (n nj = n2 By the convexity of the squaring function, the sum of the squares of numbers summing to n is minimized when they all equal n/i. Hence if G has no (r + 1)-clique, we have 2e(G) (). If these two statements are true, then we can discard edges from Tnr to obtain a graph G with n vertices and in edges such that x(G) = co(G) = r. If i is an integer that divides n, then = (1 1/i)n2/2 = in and the desired properties hold. However, when n = 12 and in = 63, there are
ChapterS: Coloring of Graphs
three edges in We have rn2'(n2 2,n)1 = 8, but every 12-vertex simple graph with only three edges in the complement has clique number 9. 2mfl vertices, and this is sometimes sharp. We can b) a(G) transfonu this question into an instance of part (a) by taking complements. Every clique in G becomes an independent set in G, and vice versa. Let
H = G. Let m' = in be the number of edges in H. If the largest independent set in G has s vertices, then the largest chque in H has s vertices. From part (b), we have s rn2'(n2 2/n')l. Substituting nil =
() m yields s rn2/(n + 2,n)1. Since this lower bound for s is achieved for some in, n by letting H be the appropriate Turán graph, it is also achieved by letting G be the complement of that graph. 5.2.18. Counting edges in the Turan graph. Let denote the r-partite Turán graph on n vertices, and let a = Ln/rj and b = n To. = (1 1/T)n2/2 bfr b)/(2T). By the degree-sum fornuila, a) we need only show that the vertex degrees sum to (1 1/r)n2 b(r b)/T. Every vertex has degree a a or a a 1, with (T b)a of the former and b(a + 1) of the latter. Hence the degree sum is n(n a) b(a + 1). Substituting a = (a b)/T yields a2 n(n b)/T— b(n b + T)/T, which equals the desired formula. b) The least T where can differ from L(1 1/r)n2/2j is T = 8, and
1. For fixed T, the left side is maximized by b = r/2, where it equals i78. Hence the condition occurs if and only if T > 8,
andwhenT =Sit occurs if andonlyifbz= 4. 5.2.19. Comparison of the Turan graph with the graph L + Kn_a yields e(Tn,r) = Here a = [n/TJ. The initial graph + (T + Kn_a has edges. We transform it into Ta,,- and study the change in the number of edges. Let A be the independent set of size a. We create Tnr by iteratively removing the edges within a set of size a or a + 1 to make it one of the desired partite sets, replacing these edges by edges to A.
Section 5.2: Strueture of k chromatic Graphs
The number of edges from A to a new partite set B is a B. Whether is a or a + 1, this numerically equals Thus replacing the + edges of the clique on B with these edges gains edges. Repeating this r 1 times to create the other partite sets gains (r 1) edges, and thus = 5.2.20.
For positive integers n and Ic, if q =
andi = n—s(k+l), then (jk+rq >
Turán graph has partite sets of sizes q and q + 1, with r of the latter. Hence its complement has + rq edges. Similarly, has G) (Ic + 1) + ts edges. To prove the desired inequality, it thus suffices to show that or This follows from Turán's Theorem. Since is the maximum number of edges in an n-vertex graph not containing Kk+2, and -c a graph, we have 5.2.2 1.
is the unique n-vertex K1+i-free graph of maximum size. We use
induction on r. The statement is immediate for r = 1. For the induction step, suppose r> 1. Let G be an n-vertex graph, and let x be a vertex of maximum degree in x. Let G' = GIIN(x)]. Let H' = and let H = I e(G) e(G'). Hence e(H) > e(G), with equality only if equality occurs in both transformations. We have seen (by the induction hypothesis) that equality in the first transformation requires G' = H'. Equality in the second transformation requires each edge of E(G) E(G') to have exactly one endpoint in S and e(G)
requires each vertex of S to have degree d(x). Thus every vertex of S is adjacent to every vertex of N(x) and to no other vertex of 5, which means that G is the join of G' with an independent set. Since G' is a complete (r 1)-partite graph, this makes G a complete r-partite graph. Finally, we know by shifting vertices between partite sets that is the only n-vertex complete r-partite graph that has the maximum number of edges. 5.2.22. Vertices of high degree. We have 18 vertices in a region of diameter 4, with E(G) consisting of the pairs at most 3 units apart. Since 3 > Application 5.2.11 (in particular the absence of independent 4-sets) guar-
antees that G lacks at most 108 edges of its 153 possible edges and thus has at least 45 edges. If at most one vertex has degree at least five, then the degree-sum is at most (17)4+ (1)17 = 85, which only permits 42 edges.
ChapterS: Coloring of Graphs
The result can be strengthened by a more detailed argument (communicated by Fred Galvin). Let S be the set of vertices with degree less than 5. Because there cannot be four vertices that are pairwise separated by at least 3 units, the subgraph induced by S has no independent set of size 4. Thus -c 15, since the edges incident to the vertices of a maximal independent set in S must cover all the vertices in S. This shows among any 17 vertices there must be two with degree at least 5. Furthermore, consider a set with 16 vertices. If r. Let G be a largest simple n-vertex graph avoiding Kr+i. By part (a), G contains Kr; let 5 be an r-vertex clique in G. Since G avoids Kr+i, every vertex not in 5 has at most r 1 neighbors in S. Therefore, deleting 5 loses at most () + (a r)(r 1) edges. The remaining graph G' avoids Kr+i. By the induction hypothesis, e(G') 0. Let G be a graph with exactly one (r + 1)-clique Q; we first use Turán's Theorem to bound e(G). Note that e(G Q) -c tr(n —r 1), and fiirthennore each v c V(G) Q has at most r 1 neighbors in Q. Thus r
To express this in terms Of tr(n), we compute t1(n) t1(n r 1). First, deleting one vertex from each partite set in loses the edges among them plus an edge from each remaining vertex to r 1 deleted vertices. Hence t1(n) t1(n r) = 1)(n r). Also, Tn_r,r becomes Tn_r_i,r when -F- (r
we delete a vertex from a largest partite set, which has degree n Rn
e(G) 1 copies of By iteratively deleting an edge that does not belong to every (r + 1)-clique, we can delete fewer than s edges from G to obtain a graph G' with exactly one (r + 1)-clique. By the preceding argument, e(G') -c fr(n). Since e(G) e(G') (m 1)(), then G contains K2,m. If a) If G is simple and any pair of vertices has m common neighbors, then G contains K2,m. Since there are Q) pairs of vertices , this means by the pigeonhole principle that a graph with no K2,m has at most (m 1)() selections (v, ) such that v is a common neighbor of x and y. Counting such selections by v shows that there are exactly of them, which completes the proof. b) If G has e edges, then (t) > e(2e/n 1). Because is a > convex function of x, Hence + (t) is minimized S
ChapterS: Coloring of Graphs
over fixed degree sum (number of edges) by setting all d(v) = d(v)/n = 2e/n, in which case the sum is e(2e/n 1). c) A graph with more than edges contains K2,m. 1)"2n312 + Since this edge bound implies 2e/n 1 > (m 1)'12n"2 we conclude e
By (b), this implies the hypothesis of (a) (if in > 2), and then (a) implies
that G contains K2,m. d) Among n points in the plane, there are at most + pairs with distance exactly one. Let V(G) be the n points, with edges corresponding to
the pairs at distance 1. If G has more than the specified number of edges, then (c) with in = 3 implies that G contains K2,3. However, no two points in the plane have three points at distance exactly 1 from each of them. 5.2.26. Every n-vertex graph G with more than 1 edges has girth at most 4. The sum counts the triples u, v, w such that v is a common neighbor of u and w. If G has no 3-cycle and no 4-cycle, then we can bound the common neighbors of pairs u, w. If u -e- w in G, then they have no common neighbor. If u w in G, then they have at most one common neighbor. Thus (t) 6, the maximum number of edges in a simple n-vertex graph G not having two disjoint cycles is 3n 6. To construct such a graph, form
ChapterS: Coloring of Graphs
a triangle on a set S of three vertices, and let S be the neighborhood of each
remaining vertex. Each cycle uses at least two vertices from 5, so there cannot be two disjoint cycles. The graph has 3 + 3(n 3) = 3n 6 edges. For the upper bound, we use induction on a. Basis step (a = 6): G has at most two missing edges. We find one triangle incident to all the missing edges, and then the remaining three vertices also form a triangle. Induction step (a > 6): If G has a vertex v of degree at most 3, then the induction hypothesis apphed to G v yields the claim. Thus we may assume that 6(G) > 4. Since e(G) > a, there is a cycle in G. Let C be a shortest cycle in G, and let H = G V(C). We may assume that H is a forest, since otherwise we have a cycle disjoint from C. Since 6(G) > 4, every leaf or isolated vertex in H has at least three
neighbors on C. This yields a shorter cycle than C unless C is a triangle, Hence we may assume that C is a triangle, and now 6(G) > 4 implies that H has no isolated vertices. Since every leaf of H is adjacent to all of V(C), two leafs in a single component of H plus one additional leaf yield two disjoint cycles. Hence we may assume that H is a single path. Thus every internal vertex of H has at least two neighbors in C, and there is at least one such vertex since a > 6. We now have two disjoint triangles: the first two vertices of the path plus one vertex of C, and the last vertex of the path plus the other two vertices of C.
5.2.29. Let G be a claw-free graph (no induced K1,3).
a) The subgraph induced by the union of any two color classes in a proper coloring of G consists of paths and even cycles. Let H be such a
subgraph. Since H is 2-colorable, it is triangle-free. Hence a vertex of degree 3 in H is the center of a claw. Since G is claw-free, every induced subgraph of G is claw-free. Hence A(H) + 2, there must be a component of H that is a path P with one more vertex from A than from B. Switching the colors on P brings the two color classes closer together in size. Iterating this procedure leads to all pairs of classes differing in size by at most 1.
Section 5.2: Strueture of k chromatic Graphs
If G has a proper coloring in which each color class has at least two vertices, then G has a x (G)-coloring in which each color class has at least two vertices. (Note that C5 doesn't have either type of coloring.) Proof 1 (induction on x(G); S. Rajagopalan). The statement is immediate if x(G) = 1. If x(G) > 1, let f be an optimal coloring of G, and let g: V(G) —* N be a coloring in which each class has at least two vertices. If f has a singleton color set , let S = , and let = G S. Since f restricts to a (x(G) 1)-coloring of G' (because x is omitted) and g restricts to a coloring of G' in which every color is used at least twice (because only vertices with a single color under G were omitted), the induction hypothesis implies that G' has a (x(G) 1)-coloring in which every color is used at least twice. Replacing S as a single color class yields such a coloring for G. Proof 2 (algorithmic version). Define f and g as above, if x is a singleton color in the current x (G)-coloring f, change all vertices in (v: g(v) = g(x)> to color f(x). The new coloring is proper, since f(x) appeared only on x and since the set of vertices with color g(x) in g is independent. No new colors are introduced, so the new coloring is optimal. Vertices that have been recolored are never recolored again, so the procedure terminates after at most x (G) steps. It can only terminate with an optimal coloring in which each color is used at least twice. 5.2.30.
5.2.31. If G is a connected graph that is not a complete graph or a cycle whose length is an odd multiple of3, then in every minimum proper coloring of G there are two vertices of the same color with a common neighbor. For odd cycles, if every two vertices having the same color are at least three apart, then the coloring must be 1, 2, 3, 1, 2, 3, - --, cyclically, so the length is an odd multiple of 3. For other graphs, Brooks' Theorem yields x(G) -c A(G). Since only A(G) 1 colors are available for the neighborhood of a vertex of maximum degree, the pigeonhole principle implies that a vertex of maximum degree has two neighbors of the same color in any optimal coloring.
5.2.32. The Hajos construction. Applied to graphs G and H sharing only vertex v, with vu E E(G) and vw e E(H), the Hajós construction produces the graph F = (G —vu)U(H vw)Uuw,
a) If G and H are k-critical, then F is k-critical. A proper (k
coloring ofF contains proper (k 1)-colorings of G vu and H vw. Since G and H are k-critical, every (k 1)-coloring of F gives the same color to v and u and gives the same color to v and w. Since this gives the same color to u and w, there is no such coloring of F. Thus x(F) > k, and equality holds because we can combine proper (k 1)-colorings of G vu and H vw and change w to a new color.
ChapterS: Coloring of Graphs
Finally, for e a E(F) we show that F e is (k 1)-colorable. For F uw, the coloring described above is proper. Let xy be another edge of F; by symmetry, we may assume that xy a E(G). Since G is k-critical, we have a proper (k 1)-coloring f of G xy. Since uv is an edge in G xy, this coloring gives distinct colors to u and v. In a proper (k 1)-coloring of H vw that gives v and w the same color, we can permute labels so this color is f (v). Combining these colorings now yields a proper (k 1)-coloring ofF —xy. b) For k > 3, a k-critical graph other than K4. Apply the Hajós construction to the graph consisting of two edge-disjoint k-cliques sharing one vertex v. This deletes one edge incident to v from each block and then adds an edge joining the two other vertices that lost an incident edge. The resulting graph is (k 1) -regular except that v has degree 2k 4. c) Construction of 4-critical graphs with n vertices for all a > 6. Since the join of color-critical graphs is color-critical, we can use v K 1, which yields 4-critical graphs for all even a. In particular, this works for a a , which has a member of each congruence class modulo 3. If we apply the Hajós construction to a 4-critical graph G with 2! ver-
tices and the 4-critical graph H = K4, we obtain a 4-critical graph F with 2! -F 3 vertices. Thus we obtain a 4-critical n-vertex graph whenever a exceeds one of by a multiple of 3. This yields all a> 4 except a =
5.2.33. a) If a k-critical graph G has a 2-cut S = , then 1) x y, 2) G has exactly two S-lobes, and .3) we may index them as G1 and G2 such
that G1 + xy and G2 xy are k-critical. Since no vertex cut of a k-critical graph induces a dllque, we have x y. By k-criticality, every S-lobe of G 1)-colorable. If each S-lobe has a proper (k 1)-coloring where x, y have the same color, then colors can be permuted within S-lobes so they
agree on , so G is (k 1)-colorable. The same can be done if each S-lobe has a proper (k
1)-coloring where x, y have different colors, Hence there must be an S-lobe G1 such that u, v receive the same color in every proper (k 1)-coloring and an S-lobe G2 such that u, v receive the different colors in every proper (k 1)-coloring. Deletion of any other S-lobe would therefore leave a graph that is not (k 1)-colorable, so criticallty implles that there is no other S-lobe. Since every proper (k 1)-coloring of G1 gives x and y the same color, G1 + xy is not (k 1)-colorable. Since every proper (k 1)-coloring of G2 gives x and y different colors, G2 xy is not (k 1)-colorable. To see that G1 + xy is k-critical, let G' = G1 + xy and consider edge deletions. First G' xy = G1, which is (k 1)-colorable. For any other edge e of G', G e has a proper (k 1)-coloring that contains a proper (k 1)-coloring of G2,
hence it gives distinct colors to x and y. Therefore the colors it uses on
Section 5.2: Strueture of k chromatic Graphs
the vertices of G1 form a proper (k 1)-coloring of G' e. The analogous argument holds for G2 . zy. b) Every 4-chromatic graph contains a K4-subdivision. Part (a) can be used to shorten the proof of this. We use induction on n(G), with the basis n(G) = 4 and K4 itself. Given n(G) > 4, let G' be a 4-critical subgraph of G. We know G' has no cutvertex. If G' is not 3-connected, then we have a 2-cut S. Part (a) guarantees an S-lobe G1 such that G1 + zy is 4critical. By the induction hypothesis, G1 -F- uv contains a subdivision of K4; if this subdivision uses the edge uv, then this edge can be replaced by a path through G2 to obtain a subdivision of K4 in G. If C is 3-connected, the proof is as in the text. 5.2.34. In a 4-criti cal graph G with a separating set Sof size 4, e(G[S]) 3, then for any e, f c E(G) there is a (k 1)-critical subgraph of G containing e but not f. Any (k 1)-coloring q5 of G e assigns the same color to both endpoints of e. The endpoints of f get distinct colors under by renumbering colors, we may assume one ofthemgetscolork— 1. Let S = ; notethat G—e— S However, G S is (k 1)-chromatic, since S is an is (k 2)-colored by independent set, so any (k 1)-critical subgraph of G S must contain e and be the desired graph. (Toft [1974]) b) If G is k-critical, with k > 3, then G is (k 1)-edge-connected. Since the 3-critical graphs are the odd cycles, this is true for k = 3, and we pro-
ceed by induction. For k >
consider an edge cut with edge set F. If
= 1, we permute colors in one component of G F to obtain a (k 1)coloring of G from a (k 1)-coloring of G F, so we may assume > 2. Choose e, f c F. By part (a), there is a (k 1)-critical subgraph C con-
taining e but not f. Deleting F from C separates it, since it separates the endpoints of e. By the induction hypothesis, F k 2, and thus >k
5.2.36. If G is k-critical and every (k 1)-critical subgraph of G is isomor1 4). Since K4 is k-critical, a k-critical graph cannot properly contain K4, so if we can find K4 in G, then G = K4. Let G have the specified properties; since k > 4, G has a triangle x, y, z. Toft's critphic to K4_1, then G =
ChapterS: Coloring of Graphs
ical graph lemma says that for any edges e, f, G contains a (k 1)-critical subgraph that contains e and avoids f. Let G1 be such a graph that contains xy but omits yz. Since every (k 1)-critical subgraph is a chque, by hypothesis, G1 cannot contain z at all. Similarly, let G2 be a (k 1)-critical graph that contains yz but omits x. Both G1 and G2 are (k 1)-cliques, so a proper (k 2)-coloring of G1 xy must give x and y the same color, and a proper (k 2)-coloring of G2 yz must give y and z the same color, This means that the graph H = (Gi xy) U (G2 yz) U xz is not (k 2)-colorable, so it contains some (k 1)-critical subgraph H', which by hypothesis is a (k 1)-clique. Furthermore, the set of vertices common to G1 and G2 induce a clique, which means that the (k 2)-colorings of G1 and G2 can be made to agree on their intersection. This means that H xz is (k 2)-colorable, which implies that xz e H'. By construction, NH(x) = V(Gi) y and NH(z) = V(G2) y. Since H' is a clique containing x, z, this forces G1, G2 to have k 3 common vertices other than y. We add x, y, z to these to obtain a k-clique in G, which as noted earlier implies that G = Kk. 5.2.37. Vertex-color-critical graphs. a) Every color-critical graph is vertex-color-critical. Every proper sub-
graph of a color-critical graph has smaller chromatic number, including those obtained by deleting a vertex, which is all that is needed for vertexcolor-critical graphs.
b) Every 3-chromatic vertex-color-critical graph G is color-critical. Since it needs 3 colors, G is not bipartite, but G v is bipartite for every v c V(G). Hence every vertex of G belongs to every odd cycle of G; let C be a spanning cycle of G. If G has any edge e not on C, then e creates a shorter odd cycle with a portion of C, leaving out some vertices. Since G is vertex-color-critical, this cannot happen, and G is precisely an odd cycle. c) the graph below is vertex-color-critical but not color-critical. This graph G is obtained from the Grotzsch graph by adding an edge, so x (G)> 4. An exphcit coloring shows that x(G) = 4. Hence G is not color-critical. Explicit 3-colorings of the graphs obtained by deleting one vertex show that G is vertex-color-critical.
Section 5.2: Structure of k chromatic Graphs
5.2.38. Every nontrivial simple graph with at most one vertex of degree less than 3 contains a K4-subdivision. Call a vertex with degree less than 3 a deficient vertex. By considering the larger class of graphs that may have one deficient vertex, we obtain a stronger result than 8(G) > 3 forcing a 1C4-subdivision, but one that is easier to prove inductively.
We use induction on n(G); the only graph with at most four vertices that satisfies the hypothesis is K4 itself. For the induction step, we seek a graph G' having at most one deficient vertex and having n(G') 3, the maximum number of edges in a simple n-vertex graph G having no K4-subdivision is 2n 3. If G has at least 2n 2 edges, then a> 4; we prove by induction on a that G has a K4-subdivision. For a = 4, G has (at least) 6 edges and must be K4. For a > 4, if 8(G) > 3, then Dirac's Theorem guarantees that G has a K4-subdivision. When 8(G) n(G7)/2 = 6.5 and x(G8) >- n(G8)/2 = 7.5. Since x(G) is always an integer, we have x(Gk) k. If Gk has a Kk-subdivision H, then H must have two vertices u, v of
degree k 1 in nonadjacent groups, since adjacent groups together have size at most k 2. Since there must be k 1 pairwise internally disjoint a, v-paths in H, this is impossible when Gk has a u, v-separating set of size k 2. In all cases except one, has such a a, v-separating set consisting of two groups. The exception is a, v chosen from the groups of size 2 in G7. In this exceptional case, we have forbidden the high-degree vertices of
H from the consecutive groups of size 3, since that would yield the case already discussed. Thus the seven high-degree vertices must consist of the two groups of size 2 and the triangle between them. Now the four needed paths connecting the two groups of size 2 must use the two consecutive groups of size 3, but only three paths can do this.
5.2.41. If in = k(k + 1)/2, then Km,m_i contains no subdivision of K2k. In there is such a subdivision: place k branch vertices in each partite set,
and then there remain unused vertices in each partite set to subdivide edges joining the branch vertices in the other partite set. We prove that if an X, Y-bigraph G contains a subdivision of K2k, then n(G) >
Proof 1 (counting argument). The paths representing edges of K2k are pairwise internally-disjoint. When a partite set has o "branch vertices" (degree more than two in the subdivision), the other partite set has at least vertices that are not branch vertices. If the subdivision of (2k_a) has a branch vertices in X, we thus need at least + 2k a + vertices. Using the identity + a (ii a) + (fl formula becomes (), the = (2) + 2k o(2k a). Since a(2k a) in and n(G) > n(F), then F C G. We may assume that F has no isolated vertices, since those could be added at the end.
Let F' be a subgraph of F obtained by deleting one leaf from each nontrivial component of F. Let R be the set of neighbors of the deleted vertices. Map R onto an rn-set X C V(G) that minimizes e(G[X]). Since 6(G) > in and n(F') = in, we can extend X to a copy of F' in G (each vertex has at least in neighbors, but fewer than in of its neighbors are used already in F' when we need to add a neighbor to it). To extend this copy of F' to become a copy of F, we show that G contains a matching from X into the set Y of vertices not in this copy of F'. Let H be the maximal bipartite subgraph of G with bipartition X, Y. By Hall's Theorem, the desired matching exists unless there is a set S C X such that in, we have N0(t) fl > On the other hand, since u NH(S), we have 1,letfbethecoloring,andletC, = . Suppose that x is a leaf of T with neighbor y and that we seek label p for x and q for y. Let S c Cq be the vertices in Cq adjacent to no vertex of Cj. We have S Cq, else we can combine color classes in f. The vertices of S cannot be used in the desired embedding of T, so we will discard them, Let G' = G (SU Ce). We have x(G') -c k 1 because we have discarded all vertices of color p in f, and we have x(G') > k 1 because SU is an independent set in G. By the induction hypothesis, G' has T x as a labeled subgraph H, and the image of y in H belongs to Cq. We have retained in Cq S only vertices having a neighbor with color p in
Chapter 5: Coloring of Graphs
f (by part (a), this set is non-empty). Hence G has a vertex in
can use as the image of x to obtain T as a labeled subgraph. 5.2.44. Every k-chromatic graph with girth at least 5 contains every k-vertex tree as an induced subgraph. If x(G) = k and d(x)
Suppose T is a k-vertex tree, x is a leaf of T with neighbor y, and T
x. By the induction hypothesis, G has T as an induced subgraph
f(T') contains a vertex adjacent to no vertex of f(T') except u. Each vertex in f(N(y)) has no neighbor in 5, because G has no triangles. Each vertex in f(T N[yI) has at most one neighbor in 5, else it would complete a 4-cycle in G with two such vertices and u. Hence S has at most n(T') 1 d(y) unavailable vertices. Since > k 1 d(y), there remains an available vertex in S to assign to x. f(T'); let u = f(y). It suffices to show that S = N(u)
5.3. ENUMERATIVE ASPECTS 5.3.1. The chromatic polynomial of the graph below is k(k 1)2(k 2). The graph is chordal, and the polynomial follows immediately from a simphcial elimination ordering. It can also be obtained from the recurrence, from the inclusion-exclusion formula, etc.
5.3.2. The chromatic polynomial of an n-vertex tree is k(k by the chromatic recurrence. We use induction on a. For a = 1, the polynomial is k, as desired. Contracting an edge of an n-vertex tree leaves a tree with a 1 vertices. Deleting the edge leaves a forest of two trees, with orders in and a m for some in between 1 and a 1. The polynomial for a disconnected graph is the porduct of the polynomials for the components. We use the induction hypothesis and the chromatic recurrence and extract the factors k and (k to obtain the polynomial k(k
Section 5.3: Enumerative Aspects
+ 3k2 is not a chromatic polynomial. In x (G; k), the degree
is n(G), and the second coefficient is —e(G). Hence we need a 4-vertex graph with four edges. The only such graphs are C4 and the paw, which have chromatic polynomials k(k
each with nonzero linear term. (Note: The linear term of the chromatic polynomial of a connected graph is nonzero; see Exercise 5.3.12.) Alternatively, observe that the value at 2 is negative, so it cannot count the proper 2-colorings in any graph.
5.3.4. a) The chromatic polynomial of the n-cycle is (k 1)" -F 1 (induction on n). The chromatic polynomial of the loop (C1) is 0, which the formula reduces to when n = 1. Those considering only simple graphs can start with x(Cs; k) = k(k 1)(k 2) = (k (k 1). For larger n, the chromatic recurrence yields x(Cn; k) = x(Pn; k) x (Cn_i; k). By
the induction hypothesis and the formula for trees, this
Proof 2 (Whitney's formula). We use x(G; k) =
For every set S of size j, the number of components of G(S) is n
that for S = E(G) the number of components is 1, not 0. Since there are sets with j edges, we obtain x(Cn; k) = + (—1f'k. By the binomial theorem, (k if' = ( + (—if'. Thus we obtain x(Cn; k) from (k if' by adding (—1f'k and subtracting (—if'. b) If H = G v K1, then x(H; Ic) = kx(G; k 1). Let x be the vertex added to G to obtain H. In every proper coloring, the color used on x is forbidden from the rest of H. Each of the k ways to color x combines with each of the x (G; k 1) ways to properly color the rest of H to form a proper coloring of H, Hence x(H; k) = kx(G; k 1);in particular, x(Cn v K1; k) = k(k 2f' + (—1f'k(k 1). If Gn = K2 o then x(Gn; k) = (k2 3k + 3f''k(k 1). Proof 1 (induction on n). Since G1 is a 2-vertex tree, x(Gi; k) = k(k 1). For n > 1, let u,, be the two rightmost vertices of Gn. The proper colorings of G,, are obtained from proper colorings of G,,1 by assigning colors also to u, and Each proper coloring f of Gn_m satisfies f(un_i) Thus each such f extends to the same number of colorings of Gn. f 1)2 There are (k ways to specify f(un) and f(vn) so that f(un) f(un_i) and f(vn) f(vn_m). Of these extensions, k 2 give and vn the same color, and we delete them. Since (k 1)2 (k 2) = k2 3k + 3, the induction hypothesis yields 5.3.5.
ChapterS: Coloring of Graphs
Proof 2 (induction plus chromatic recurrence). Again x(Gi; k) = k(k Let e = unvn. For a > 1, observe that x(Gn e; k) = x(Gn_i; k)(k 1)2 and x(Gn e; k) = x(Gn_i; k)(k 2), by counting the ways to extend each coloring of Gn_i to the last column. Thus 1).
x(Gn; k) = unvn; k) = x(Gn_i; k)[(k
5.3.6. Non-inductive proof that the coefficient of
Let a be the number of vertices in G. By Proposition 5.3,4, x(G; k) = p1k(1), where Pr is the number of partitions of G into exactly r nonempty independent sets. Since k(1) is a polynomial in k of degree r, contributiong to the coefficient of in x(G; k) can arise only from the terms for r = a and r = a 1. The only partition of V(G) into a independent sets is the one with each = 1. When partitioning V(G) into a 1 vertex in a set by itself, so independent sets, there must be one set of size 2 and a 2 sets of size 1. Thus each such partition is determined by choosing two nonadjacent = () e(G). vertices. There are e(G) such pairs (G is simple), so The term involving in k(n_i) arises only by choosing the term k from each factor when expanding the product. Thus the coefficient of in k(ni) is 1. Contributions to the coefficient of k"' in k(n) arise by choosing the term k from a 1 factors and the constant from the remaining term. Thus the contributions are —1, —2. —(a 1), and the coefficient 15
i, which equals Combining these computations yields the coefficient of .
5.3.7. Roots of chromatic polynomials. a) The chromatic polynomial x (G; k) of an arbitrary graph G is a non-
negative linear combination of chromatic polynomials of cliques with at most a(G) vertices. This holds trivially when G itself is a chque, which is the situation where e(G) = 0. This is the basis step for a proof by induction on e(G). For e(G) > 0, let G' be the graph obtained by adding the edge e = uv and contracting it; we have x(G; k) = x(G + uv; k) + x(G'; k) by the chromatic recurrence. To apply the induction hypothesis, note that e(G + uv) = e(G) 1 and e(G') = e(G) 1 N(u) fl where e = uv. Hence we can express x(G'; k) and x(G' e; k) as nonnegative linear combinations of the polynomials x(K7; k) for j -c a. b) The chromatic polynomial of a graph on a vertices has no real root larger than a 1. The combinatorial definition of the chromatic polynomial as the function of k that counts the proper colorings of G using at
Section 5.3: Enumerative Aspects
most k colors guarantees that the value cannot be 0 for k > it, because we can arbitrarily assign the vertices distinct colors to obtain at least k(k 1) > 0 proper colorings. However, this argument applies only to integers. To forbid all real roots exceeding it, we use part (a). Observe that x(K1; x) is the product of positive real numbers whenever x > j 1; hence these polynomials have no real roots larger than 1. Since any chromatic polynomial is a nonnegative linear combination of these for j n 1 is the sum of at most n positive numbers and therefore is also positive.
5.3.8. The number of proper k-colorings of a connected graph G is less than if k > 3 and G is not a tree. If G is connected but not a tree, let T be a spanning tree contained in G, and choose e E E(G) E(T). Every proper coloring of G must be a proper coloring of the subgraph T, and there are exactly k(k proper k-colorings ofT. It suffices to show that at least one of these is not a proper k-coloring of G. Since T is bipartite and k > 3, we can construct such a coloring by using a 2-coloring ofT and then changing the endpoints of e to a third color. This is still a proper k-coloring ofT, but it is not a proper k-coloring of G. If k = 2, then T has exactly two proper k-colorings, and these are both proper colorings of G if G is bipartite. Thus the statement fails when k = 2 if G is bipartite (if G is not bipartite, then it still holds when k = 2). k(k
5.3.9. x(G; x +y) = x(G[U]; x)x(G[U]; y). Polynomials of degree a that agree at a + 1 are equal everywhere. Hence it suffices to prove the claim when x and y are nonnegative integers. We show that then each side counts the proper (x + y)-colorings of G. In each proper (x + y)-coloring, the first x colors are used on some subset U c V(G), and U receives colors among the remaining y colors. Since there is no interference between the colors, we can put an arbitrary x-coloring on G[U] and an arbitrary y-coloring on G[U] and form such a coloring in x(G[UI; x)x(G[UI; y) ways. Furthermore, the set U that receives colors among the first x colors is uniquely determined by the coloring. Hence summing over U counts each coloring exactly once. The left side by definition is the total number of colorings. 5.3.10. If G isa connected n-vertex graphwith x(G; k) = then a, > for 1 -c i -c a. In order to prove this inductively using the
chromatic recurrence, we must guarantee that the graphs in the recurrence are connected and appear "earlier". We use induction on a, and to prove the induction step we use induction on e(G) ft + 1. The statement holds for the only 1-vertex graph, so consider a >
and G is connected, then G is a tree and has chromatic
ChapterS: Coloring of Graphs
polynomial k(k 1)"'. The term involving k' is magnitude of the coefficient is as desired.
Now consider e(G) > a 1. If G is connected and has more than 1 edges, then G has a cycle, and deleting any edge of the cycle leaves a connected graph. Let e be such an edge, and define , in the chromatic polynomials by x(G e;k) = n—i bn_,k n—i and x(G . e;k) = a
The recurrence x(G; k) = x(G e; Ic) x(G e; Ic) implies that (—1)"'a, is the sum of the coefficients of Ic' in the other two polynomials. Since G e and G e are connected, the induction hypothesis implies a, = b, (— 1)c, = b, + for 1 > the bound for any of these coefficients only if G is a tree.
5.3.11. The coefficients of x(G; Ic) sum to 0 unless G has no edges The sum of the coefficients of a polynomial in Ic is its value at Ic = 1. The value of x(G; 1) is the number of proper 1-colorings of G. This is 0 unless G has no edges. (The inductive proof from the chromatic recurrence is longer.)
5.3.12. The exponent in the last nonzero term in the chromatic polynomial of G is the number of components of G. We use induction on e(G). When e(G) = 0, we have x(G; Ic) = and G has n(G) components. Let c(G) count the components in G. Both G e and e have fewer edges than G. Also G e has the same number of components as G, and G e has the same number or perhaps one more. Since n(G e) = n(G e) 1 and coefficients alternate signs, the coefficients of have opposite signs in x(G e; Ic) and x(G e; Ic). Thus we have positive a and noimegative a' such that x(G
a + a' > 0, the last coefficient of x (G; Ic) is as claimed. Alternatively, one can reduce to the case of connected graphs by observing that the chromatic polynomial of a graph is the product of the chromatic polynomials of its components. Since an n-vertex tree has chromatic polynomial Ic(k its last nonzero term is the linear term. For a connected graph that is not a tree, the chromatic recurrence can be applied as above to obtain the result inductively. If p(Ic) = Ic" a a chromatic polynomial. If p is a chromatic polynomial of a (simple) graph G, then G has a vertices, a edges, and r components. The maximum number of edges in a simple graph with a vertices and r components is achieved by r 1 isolated vertices and one clique of order a r + 1, This has edges (Exercise 1.3.40), which is less than a.
Section 5.3: Enumerative Aspects
5.3.13. Chromatic polynomials and clique cutsets. Let F = G U H, with S = V(G) P V(H) being a clique. Every proper k-coloring ofF yields proper k-colorings of G and H, and proper k-colorings of G and H together yield a proper k-coloring of F if they agree on S. Since S induces a clique, in every proper k-coloring of G or H the vertices of S have distinct colors. Therefore, given a proper k-coloring of G P H, the number of ways to extend it to a proper k-coloring of H [or G, or F] is independent of which proper k-coloring of G PH is used. For each k > 0, the value of the chromatic polynomial simply counts proper colorings. We have partitioned the proper k-colorings of these
graphs into equal-sized classes that agree on S. For a fixed coloring f of G P H, the number of ways to extend it to a coloring of G, H, or F is thus x(G; k)/x(G PH; k), x(H; k)/x(G PH; k), or x(F; k)/x(G P H; k), respectively. Since every extension of f to G is compatible with every extension of f to H to yield an extension of f to F, the product of the first two of these equals the third, and x(G U H; k) = x(G; k)x(H; k)/x(G P H; k). (Comment: 1) When G and H intersect in a clique, it need not be true that
x(G;k)= x(G—GPH;k)x(GPH;k);forexample,letGandHbe4-cycles sharing a single vertex.) When G P H is not a clique, this argument breaks down. For example, consider G = H = P3, F = G U H = C4, G P H = 2K1. We have
x(F; k)x(G P H; k) = k3(k
5.3.14. Minimum vertex partitions of the Petersen graph into independent sets.
Let P be the Petersen graph. The Petersen graph P has odd cy-
cles, so it requires 3 colors, and it is easy to partition the vertices into 3 independent sets using color classes of size 4,3,3, as described below. a) IfS is an independent 4-set, then P S = 3K2. The three neighbors
of a vertex have among them an edge to every other vertex, so S cannot contain all the neighbors of a vertex. Hence P S has no isolated vertex. Deleting S deletes 12 edges, so P S has 3 edges and 6 vertices. With no isolated vertices, this yields P S = 3K2. b) P has 20 partitions into three independent sets. Since P has 10 vertices, every such partition has an independent set of size at least four. There is no independent 5-set, because we have seen that every independent 4-set has two edges to each remaining vertex. For each independent 4-set 5, there are 4 ways to partition the vertices of the remaining 3K2 into two independent 3-sets. Hence it suffices to count the independent 4-sets and multiply by 4. The number of independent 4-sets containing a specified
vertex is 2, since deleting that vertex and its neighbors leaves C5, which has two independent 3-sets. Summing this over all vertices counts each
ChapterS: Coloring of Graphs
independent 4-set four times. Hence there are 2 10/4 = 5 independent 4-sets and 20 partitions of the vertices. c) If r = x(G), then V(G) has x(G; r)/r! partitions into r independent sets. Each such partition can be converted into a coloring in exactly r! ways.
5.3.15. A graph with chromatic number k has at most
vertex partitions into k independent sets, with equality achieved only by Kk -1- (a k)K1 (complete graph plus isolated vertices. For Kk + (a k)K1, the sets of the
partition are identified by the vertex of the clique that they contain, and the isolated vertices can be assigned to these sets arbitrarily, so this is the correct number of vertex partitions for this graph. If G has only k vertices, then G has be a k-chque, and there is only one partition. If a > k, choose a vertex v 0 V(G). We consider two cases; x(G v) = k and x(G v) = k 1. If x(G v) = k, then partitions of G v can be extended to partitions of G by putting v in any part to which it has no edges. Thus it extends in at most k ways, with equality only if v is an isolated vertex. If x(G v) = k 1, then G has a k-partition in which v is by itself and is adjacent to vertices X = xk_1> of the other parts. Let R be the independent set containing v in an arbitrary k-partition, and suppose R = 1 f r. Note that x(G R) = k 1. By the induction hypothesis, G R has at most (k partitions into k 1 independent sets. Allowing R to be an arbitary subset of G (X U ), we obtain at most partitions of G into k independent sets, which equals kn_k by the binomial theorem. For equality, we must have N(v) = X and G (X U) = (a k) K1 for each such choice of v, which again yields G = K1 + (a k)Ki.
If G is a simple graph with a vertices and in edges, then G has at
triangles. Each triangle has three pairs of incident edges, and each edge pair of incident edges appears in at most one triangle. Hence the number of triangles is at most 1/3 of the number of pairs of edges. The coefficient of k"2 in x(G; k) is positive unless G has at most one edge. In the expression for the chromatic polynomial in Theorem 5.3.10, contributions to the coefficient of arise from spanning subgraphs with a 2 components. These include all ways to choose two edges (weighted
positively) and all ways to choose three edges forming a triangle (weighted negatively). With in edges and t triangles, the coefficient is t. Since -c the coefficient is positive unless G has at most one edge. z'
5.3.17. Chromatic polynomial via the inclusion-exclusion principle. In the universe of all k-colorings of G, let A, be the set of colorings that assign the same color to the endpoints of edge e,. The proper k-colorings of G are the k-colorings outside all the sets A,. By the inclusion-exclusion formula,
Section 5.3: Enumerative Aspects
the number of these is
where g(S) is the number of
k-colorings in These are the colorings in which every edge in S has its endpoints given the same color, To count these, we can choose a color independently for each component of the spanning subgraph of G with edge set S. Hence g(S) = where c(G(S)) is the number of these components. We have obtained the formula of Theorem 5.3.10. 5.3.18. Two chromatic polynomials.
a) The graphs G, H above have the same chromatic polynomial. Applying the chromatic recurrence using the edge labeled e shows that each of these graphs has a chromatic polynomial that is the difference of the chromatic polynomials of the two graphs below.
b) The chromatic polynomial of G. The first graph G is G'
G' is the graph on the left below and e' is the indicated edge. The graph
e' appears on the right. Each of these graphs is chordal, as shown by exhibiting a simplicial eliminationordering. For each, the chromatic polynomial is a product of linear factors arising from the reverse of a simplicial elimination ordering. Thus
k) = x(G'; k) -f1)(k —2)2(k— 3)2+k(k
ChapterS: Coloring of Graphs
5.3.19. The chromatic polynomial of the graph G obtained from K6 by subdividing one edge is a product of linear factors, although G is not a chordal graph. Let v be the vertex of degree 2 in G, and let e be an edges incident to e. The cycle consisting of v, its incident edges, and the edges from its neighbors to one other vertex form a chordless 4-cycle, so G is not chordal. To compute x(G; k), observe that G e consists of a 5-clique Q, an
additional vertex w adjacent to four vertices of Q, and v adjacent to w. Hence G e is a chordal graph, with x(G e; k) = k(k 1)(k 2)(k 3)(k 4)(k 4)(Jc 1). Let f(k) = i). The graph G e is K6, with
x(CK)s)=f(k)(k—5).Thus e; k) x(G . e; Ic) = f(k)[(k 4)(k 1) x(G; k) = 5)1 —f(k)[k2—Gk-j-9]——k(k—1)(k—2)(k—3)(k—4)(k—3)2
Properties of a chordal graph G with a vertices. G has at most a maximal cliques, with equality if and only if G has no edge.s As each vertex v is added in the reverse of a simplicial elimination ordering, it creates one new maximal chque (containing v) if N(v) is not already a maximal clique. If N(v) is already a maximal clique, then the clique grows. No other maximal clique appears or changes. Thus there is at most one new maximal clique for each vertex, The first time an edge is added, a maximal clique is enlarged, not created, so there is a new clique at most a 1 times if G has an edge. (Comment: A more formal version of this argument uses the language of induction on a.) b) Every maximal clique of G that contains no simplicial vertex of G is a separating set of G. b) Every maximal clique of G that contains no simplicial vertex of G is 5.3.20. a)
separating set of G.
Proof 1 (construction ordering, following part (a).) When a maximal clique Q of G acquires its last vertex v in the construction ordering, v is then simplicial. If all vertices of Q that are simplicial when Q is created are not simplicial in G, then the rest of the construction gives them additional neighbors that are separated by Q from each other and from the vertices of G Q that are earlier than v. If there are no such earlier vertices, then Q has at least two simplicial vertices at the time it is formed; each of these acquires a later neighbor, so Q separates those later neighbors. Proof 2 (induction on a.) When G = there is no separating set, but all the vertices are simplicial, so the statement holds. When G let Q be a maximal clique containing no simplicial vertex of G. Every chordal graph that is not a complete graph has two nonadjacent simplicial vertices (this follows, for example, from Lemma 5.3.16). Let u and v be
Section 5.3: Enumerative Aspects
such vertices. Note that Q cannot contain both u and v; we may assume that v Q. Hence Q is a maximal clique in G v. If Q contains no simplicial vertex of G v, then the induction hypothesis implies that Q separates G v. All neighbors of v in G lie in one component of G v Q, since N(v) is a clique in G v. Hence Q is also a separating set in G. If Q contains at least one simplicial vertex of G v, then all such vertices lie in N (v), since they are not simplicial in G. Therefore u and Q separates v from u. 5.3.21. A graph G is chordal if and only if s(H) = w(H) for every induced subgraph H of G, where s(H) is the Szekeres—Wilf number of H, defined to be 1 + maxHcG 3(H). Sufficiency. We prove the contrapositive. If G is not chordal, then G
has a chordless cycle with length at least 4. Such a cycle is an induced subgraph. Its clique number is 2, and its Szekeres—Wilf number is 3. Necessity. Since every induced subgraph of a chordal graph is chordal, it suffices to show that s(G) = co(G) (the argument for G also applies to each induced subgraph). Since always s(G) > cc (G), it suffices to show that s(G) f(x). The maximum path length in this orientation is less than x(G), and hence it is smaller by at least two than the length of any cycle.
An edge in an acyclic orientation is dependent if and only if there is another path from its tail to its head, The length of such a path would be one less than the length of the resulting cycle, but we have shown that our orientation has no paths this long. 5.3.31. Comparison between acyclie orientations and spanning trees. The number r(G) satisfies the recurrence r(G) = r(G e) + r(G e). This is the recurrence satisfied by a(G), but the initial conditions are different. A graph with no edges has one acyclic orientation, but it has no spanning tree unless it has only one vertex. A connected graph containing a loop has spanning trees but no acyclic orientation. A tree of order a has one spanning tree and acyclic orientations. A clique of order a has spanning trees and a! acychc orientations; > a! if a 6.
5.3.32. Compatible pairs: ij(G; Ac) = —Ac). Suppose D is an acychc orientation of G and f is a coloring of V(G) from the set [hi. We say that (D, f) is a compatible pair if u —f v in D implies f(u) -c f(v). Let ij(G; Ac) be the number of compatible pairs. If f(u) f(v) for every adjacent pair u, v, then only one orientation is compatible with f. Therefore, x (G; Ac) counts the pairs (D, f) under a slightly different condition: D is acycic and u v in D implies f(u) 3, we can form a simple connected 4-regular plane graph with 2n vertices by using an inner n-cycle, an outer n-cycle, and a cycle in the region between them that uses all 2n vertices. Below we show this for n = 8.
6.1.14. For each a E
6.1.15. A 3-regular planar graph of diameter 3 with 12 vertices. By inspec-
tion, the graphs below are 3-regular and planar. To show that that they
Chapter 6: Planar Graphs
have diameter 3, we conduct a breadth-first search (Dijkstra's Algorithm) to compute distances from each vertex. By symmetry, this need only by done for one vertex each of "type" (orbit under automorphisms). In this sense, the rightmost graph is the best answer, since it is vertex transitive, and the distances need only be checked from one vertex. The graph on the left has five types of vertices, and the graph in the middle has two.
6.1.16. An Eulerian plane graph has an Eulerian circuit with no crossings. As the graph on the left below illustrates, it is not always possible to do this by splitting the edges at each vertex into pairs that are consecutive around
the vertex. The figure on the right illustrates a non-consecutive planar splitting. We give several inductive proofs. The result does not require the graph to be simple.
Proof 1 (induction on e(G) n(G)), A connected Eulerian plane graph G has at least as many edges as vertices; if e(G) = n(G), it is a single cycle and has no crossings. If e(G) > n(G), then G has a vertex x on the outer face with degree at least 4. Form G' by splitting x into two vertices: a vertex y incident to two consecutive edges of the outer face that were incident to x, and a vertex z incident to the remaining edges that were incident to x. Locating y in the unbounded face of G yields a planar embedding of G'. Since G' is an even connected plane graph and e(G') n(G') = e(G) n(G) 1, the induction hypothesis applies to G'. Since the edges incident to y are consecutive at x, the resulting Eulerian circuit of G' translates back into an Eulerian circuit without crossings in G. This proof converts to an algorithm for obtaining the desired circuit.
Section 6.1: Embeddings and Euler's Formula
Proof 2 (induction on e(G)), The basis is a single cycle. If G has more than one cycle, find a cycle C with an empty interior. Delete its edges and apply induction to the resulting components G1, . For i = 1, .. ., absorb
G, into C as follows. Find a vertex x where G, intersects C; all
edges incident to x but not in C belong to Consider a visit el, x, e2 that the tour on G, makes to x using an edge e1 next to an edge fi of C in the embedding. Let f2 be the other edge of C at x. Then replace the visits e1, e2 and fi, fQ by e1, fi and e2, This absorbs G, into C while maintaining planarity of the circuit. Each component of G C is inserted at a different vertex, so no conflicts arise. Proof 3 (local change). Given an Eulerian circuit that has a crossing in a plane graph G, we modify it to reduce the number of crossings formed by pairs of visits to vertices. By symmetry, it suffices to consider four edges a, b, c, d incident to v in counterclockwise order in the embedding such that one visit to v enters on a and leaves on c, and a later visit enters on b and leaves on d. We eliminate this crossing by traversing the portion c, . b of the circuit in reverse order, as b c. Crossings at other vertices are unchanged by this operation. At v itself, if a passage e, f through v now crosses a, b, then e, f crossed a, c or b, d before, and if it now crosses a, b and c, d, then it crossed both a, c and b, d before. Thus there is no increase in other crossings, and we obtain a net decrease by un-crossing a, b, c, d. 6.1.17. The dual of a 2-connected simple plane graph with minimum degree 3 need not be simple. For the 2-connected 3-regular plane graph G below, G* has a double edge joining the vertices of degree 6.
6.1.18. Duals of connected plane graphs. a)IfG isa plane graph, then G* isconnected, One vertex of G* is placed in each face of G. If u, v a V (G*), then any curve in the plane between u and
v (avoiding vertices) crosses face boundaries of G in its passage from the face of G containing u to the face of G containing v. This yields a u, v-walk in Gt, which contains a u, v-path in Gt. b) If G is connected, and G* is drawn by placing one vertex inside each face of G and placing each dual edge in G* so that it intersects only the corres ponding edge in G, then each face of Gt contains exactly one vertex of G. The edges incident to a vertex ii a V(G) appear in some order around v. Their duals form a cycle in Gt in this order. This cycle is a face of Gt. If w is another vertex of G, then there is a v, w-path because G is connected, and this path crosses the boundary of this face exactly once. Hence every
Chapter 6: Planar Graphs
face of G* contains at most one vertex of G. Equality holds because the number of faces of G* equals the number of vertices of G: since both G and Gt are connected, Euler's formula yields a e + f = 2 and at et + f* = 2. We have e = et and jjt = f by construction, which yields ft = n,. c) For a plane graph G, Gtt G ifand only if G is connected. Since Gtt
is the dual of the plane graph Gt, part (a) implies that Gtt is connected. Hence if Gtt is isomorphic to G, then G is connected. Conversely, suppose that G is connected. By part (b), the usual drawing of Gt over the picture of G has exactly one vertex of G inside each face of Gt. Associate each vertex x c V(G) with the vertex x' of Gtt contained in the face of G that contains x; by part (b), this is a bijection. Consider .xy e E(G). Because the only edge of Gt crossing xy is the edge of Gt dual to it, we conclude that the faces of Gt that contain x and
y have this edge as a common boundary edge. When we take the dual of Gt, we thus obtain x'y' as an edge. Hence the vertex bijection from G to Gtt that takes x to x' preserves edges. Since the number of edges doesn't change when we take the dual, Gtt has no other edges and thus is isomorphic to G. 6.1.19. Fora plane graph G, a set D C E(G) forms a cycle in G if and only if the corresponding set Dt C E(Gt) forms a bond in Gt, by induction on e(G). We prove also that if D forms a cycle, then the two sides of the edge
cut that is the bond in Gt corresponding to D are the sets of dual vertices corresponding to the sets of faces inside and outside D. Basis step: e(G) = 1. When G and Gt have one edge, in one it is a loop (a cycle), and in the other it is a cut-edge (a bond). Induction step: e(G) > 1. If D is a loop or a cut-edge, then the statement holds. Otherwise, D has more than one edge. If D forms a cycle, then let e be an edge of the cycle, and let G' be the graph obtained from G by contracting e. In G', the contracted set D' forms a cycle. Also, the set of faces in G' is the same as the set of faces in G; the only change is that the lengths of the faces bordering e (there are two of them since e is not a cut-edge) have shrunk by 1. Since e(G') = e(G) 1, the induction hypothesis implies that in the dual (G')t, the edges dual to D' form a bond, and the sets of vertices separated by the bond are those corresponding to the faces inside and outside D. By Remark 6.1.15, the effect of contracting e in G was to delete et from Gt. Since joins vertices for faces that are inside and outside D, replacing it would reconnect Gt. Hence Dt forms a bond as claimed, and the sets of vertices on the two sides are as claimed. Now consider the induction step for the converse. We assume that Dt forms a bond, so —? forms a bond in Gt et separating the same
Section 6.1: Embeddings and Euler's Formula
two vertex sets that D* separates in G*, By Remark 6.1.15, G* e* is the dual of G', and the edges of D* e* are the duals to D'. By the induction hypothesis, D' forms a cycle in G', and the two sides of the bond Dt et in e* correspond to the faces inside and outside D'. Since e* joins vertices from these two sets, e (when we re-expand it in G) must bound faces from these two sets. With D being the boundary between two sets of faces, we can argue that D is a cycle. 6.1.20. A plane graph is bipartite if and only if every face length is even. A face of G is a closed walk, and an odd closed walk contains an odd cycle, so a bipartite plane graph has no face of odd length. Conversely, suppose that every face length is even; we prove by induction on the number of faces that G is bipartite. If G has only one face, then by the Jordan Curve Theorem G is a forest and is bipartite. If G has more than one face, then G has an edge e on a cycle. This edge belongs to two faces F1, F2 of even length; these faces are distinct because the cycle embeds as a closed curve, and by the Jordan Curve Theorem the
regions on the inside and outside are distinct. Thus deleting e yields a combined face F whose length is the sum of the lengths of F1 and F2, less two for the absence of e from each. Hence F has even length. Lengths of other faces remain the same. Thus every face of G e has even length, and we apply the induction hypothesis to conclude that G e is bipartite. To show that G also is bipartite, we replace e. Since Fi has even length, there is an odd walk in G e connecting the endpoints of e, so they lie in opposite parts of the bipartition of G e. Hence when we add e to return to G, the graph is still bipartite. (Comment: Since we deleted one edge to obtain G e, we could phrase this as induction on e(G). Then we must either put all forests into the basis step or consider the case of a cut-edge in the induction step.) 6.1.2 1. A set of edges in a connected plane graph G forms a spanning tree of G if and only if the duals of the remaining edges form a spanning tree of G*. Since (Gt)t = G when G is connected, it suffices to prove one direction of the equivalence; the other direction is the same statement applied to Gt. Let T be a spanning tree of G, where G has a vertices and f faces. Let P be the spanning subgraph of G* consisting of the duals of the remaining edges; T* has f vertices. Proofs 1,2,3. (Properties of trees). It suffices to prove any two of(1) T* has f 1 edges, (2) P is connected, (3) P is acydic. (1) By Euler's Formula, e(G) = a + f 2; hence if T has a 1 edges there are f 1 edges remaining. (2) Since T has no cycles, the edges dual to T contain no bond of Gt (by Theorem 6.1.14). Hence P is connected.
Chapter 6: Planar Graphs
(3) Since T is spanning and connected, the remaining edges contain no bond of G. Thus P contains no cycle in G* (by Theorem 6.1.14 for G*). Proof 4 (extremahty and duality). A spanning tree of a graph is a minimal connected spanning subgraph. "Connected" is equivalent to "omits no bond" (see Exercise 4.1.29). Hence the remaining edges fonu a maximal subgraph containing no bond. By Theorem 6.1.14, the duals of the remaining edges form a maximal subgraph of Gt containing no cycle. A maximal subgraph of C containing no cycle is a spanning tree of G*. Proof 5 (induction on the number of faces). If G has one face, then G is a tree, G* = K1, and Tt is empty and forms a spanning tree of C. If G has more than one face, then G is not a tree, and hence G has an edge e not in the given tree T, Since e lies on a cycle (in T + e) and is not a cut-edge, G e is a connected plane graph with one less face. Let G' = G e. The induction hypothesis implies that the duals of E(G') E(T) form a spanning tree in (G5*. Note that (G e)* = C e*; we obtain the dual
of G' by contracting the edge dual to e in C. Returning to G keeps et in E(G) E(T), so what happens to the duals of the edges outside T is that the vertex of (G e)* representing the two faces that merged when e was deleted sphts into two vertices joined by et. This operation turns a tree into a tree with one more vertex, and it has all the vertices of G*, so it is a spanning tree. 6.1.22. The weak dual of an outerplane graph is a forest. A cycle in the dual graph Gt passes through faces that surround a vertex of G. When every vertex of G lies on the unbounded face, every cycle of G* therefore passes through the vertex of C that represents the unbounded face in G. Hence G* vt is a forest when G is an outerplane graph. 6.1.23. Directed plane graphs. In foliowing an edge from tail to head, the dual edge is oriented so that it crosses the original edge from right to left. a) If D is strongly connected, then
has no directed cycle. Such a cycle
C encloses some region R of the plane. Let S be the set of vertices of D corresponding to the faces of Dt contained in R. Since has a consistent orientation, the construction implies that all the edges of I) corresponding to C are oriented in the same direction across C (entering R or leaving R). This contradicts the hypothesis that D is strongly connected. b) If D is strongly connected, then 8 (Di) = = 0. A finite acycic directed graph has 3 = = 0, because the initial vertex of a maximal directed path can have no entering edge, and the terminal vertex of such a path can have no exiting edge. c) If D is strongly connected, then D has a face on which the edges form
a clockwise directed cycle and a face on which the edges form a counterclockwise directed cycle. A vertex of D* with indegree 0 corresponds to a
Section 6.1: Embeddings and Euler's Formula
face of D on which the bounding edges must form a clockwise directed cycle,
and a vertex of D* with outdegree 0 corresponds to a face of D on which the edges must form a counter-clockwise directed cycle. 6.1.24. Alternative proof of Euler's Formula. a) Faces of trees. Given a planar embedding of a tree, let x, y be two points of the plane not in the embedding. Ifthe segment between them does
not intersect the tree, then x and y are in the same face. If the segment does intersect the tree, then we create a detour for it closely following the embedding. Induction on the number of vertices yields a precise proof that this is possible. Using the detour yields a polygonal x, y-path that does not cross the embedding, so again x and y are in the same face. b) Euler by edge-deletion. Ether's formula states that for a connected n-vertex plane graph with in edges and f faces, a in + f = 2. If every edge of such a graph is a cut-edge, then the graph is a tree. This implies in = a 1 and f = 1, in which case the formula holds. For an induction on e, we need only consider graphs that are not trees in the induction step. Such a graph G has an edge that is not a cut-edge. If e lies on a cycle, then
both the interior and the exterior of the cycle have e on their boundary, and hence e is on the boundary of two faces. Therefore, deleting e reduces the number of faces by one but does not disconnect G. By the induction hypothesis, a (in 1) + (f 1) = 2, and hence also a in f f = 2. 6.1.25. Every self-dual plane graph with a vertices has 2n 2 edges. If G is isomorphic to then G must have the same number of vertices as faces. Ether's formula then gives a e + a = 2 (and hence e = 2n 2) if G is connected. Every self-dual graph is connected, because the dual of any graph contains a path to the vertex for the outside face of the original. For every a > 4, the n-vertex "wheel" is self-dual. This is a cycle on a 1 vertices, plus an nth vertex joined to all others. The triangular faces becomes a cycle, and each is adjacent to the remaining face; this is the same description as the original graph.
6.1.26. The maximum number of edges in a simple outerplanar graph of order a is 2n
For the lower bound, we provide a construction. A simple
Chapter 6: Planar Graphs
cycle on a vertices together with the chords from one vertex to the a 3 vertices not adjacent to it on the cycle forms an outerplanar graph with 2n 3 edges. For the upper bound, we give three proofs. a) (induction on a). When a = 2, such a graph has at most 1 edge, so the bound of 2n 3 holds. When a > 2, recall from the text that every simple outerplanar graph G with a vertices has a vertex v of degree at most two. The graph G' = G v is an outerplanar graph with a 1 vertices; by the induction hypothesis, e(G') -c 2(a 1) 3. Replacing v restores at most two edges, so e(G) k(2 a + e) and thus e - 13, the Petersen graph is not planar, and at least two edges must be deleted to obtain a planar subgraph. The figure below shows that deleting two edges suffices,
6.1.31. The simple graph G with vertex set
j 3. The graph G' obtained by deleting vertex a from G is the previous graph. By the induction hypothesis, it has 3(a 1) 6 edges and has an embedding with on one face. We add Vn_2, edges from to these vertices to obtain G. Thus e(G) = 3a 6. To embed G we place inside the face of the embedding of G' having Vn_2, on its boundary. When we add the edges from a to those vertices to complete
the embedding, we form a face with on the boundary. The resulting embedding is illustrated below, with the bold path bein order. The special face remains the outside face as the ing Vt, . induction proceeds.
Section 6.1: Embeddings and Euler's Formula
6.1.32. If G is a maximal plane graph, and S is a separating 3-set of G*, then G* S has two components. A maximal plane graph is a triangulation and has no ioops or multiple edges. Hence its dual is 3-regular and 3-edge-connected. The connectivity of a 3-regular graph equals its edgeconnectivity (Theorem 4.1.11). If G* has a separating 3-set 5, then it is a minimal separating set, and each vertex of S has a neighbor in each component of G* 5. Extract a portion of a spanning tree in each component of Gt S that links the chosen neighbors of S. Combine these with the edges from S to the chosen neighbors. If G* S has at least three components, then we obtain a subdivision of Since G* is planar, we conclude that S has at most two components. 6.1.33. If G is a triangulation, and a, is the number of vertices of degree i in G, then >11(6 i )n, = 12. A triangulation with a vertices has 3n 6 edges and hence degree-sum Ga 12. The sum i a, also equals the degree-sum. 12 = sin,, as desired. Hence 6.1.34. An infinite family of planar graphs with exactly twelve vertices of degree 5. Begin with (at least two) concentric 5-cycles; call these "rungs". For each consecutive pair of rungs, add the edges of a 10-cycle in the region between the two 5-cycles. Inside the innermost rung, place a single vertex adjacent to the 5 vertices of the rung. Outside the outermost rung, place a single vertex adjacent to the S vertices of the rung. The vertices of degree 5 are the innermost vertex, the outermost vertex, and the vertices of the
innermost and outermost rungs. The other vertices have degree 6. The case with exactly two 5-cycles is the icosahedron.
6.1.35. Every simple planar graph with at least four vertices has at least four vertices with degree less than 6. It suffices to prove the result for maximal planar graphs, since deleting an edge from a graph cannot make the statement become false. Let G be a maximal planar graph with a vertices.
Chapter 6: Planar Graphs
In a maximal planar graph with at least four vertices, every vertex has degree at least 3. The sum of the vertex degrees is 6n 12. Therefore, the sum of 6— d(v) over the vertices with degree less than 6 is at least 12. Since 3(G) > 3, each term contributes at most 3, so we must have at least four such vertices. For each even value of a with a >- 8, there is an n-vertex simple planar graph G that has exactly four vertices with degree less than 6. By the analysis above, such a graph must be a triangulation with four vertices of degree 3 and the rest of degree 6. The graph sketched below has eight vertices. If we extend the two halfedges at the left and right to become a single edge, then we have the desired
8-vertex graph. To enlarge the graph, we could instead place vertices at the ends of the two half-edges, make them adjacent also to the top and bottom vertices, and extend half-edges from the top and bottom. If those half-edges become a single edge, then we have the desired 10-vertex graph. Otherwise, we can continue adding pairs of vertices to obtain the sequence of examples.
6.1.36. IfS is a set of a points in the plane such that the distance in the plane between any pair of points in S is at least 1, then there are at most 3n 6 pairs for which the distance is exactly 1, If two unit-distances cross, then one of the other distances among these four points is less than 1. Hence the condition implies that the graph of unit distances is a planar graph with a vertices. A planar graph with a vertices has at most 3a 6 edges.
6.1.37. Given integers Ic > 2, 1 > 2, and Il even, there is a planar graph with exactly Ic faces in which every face has length 1. (For 1 = 1 and Ic > 2, this does not work.) When 1 > 1 and Ic is even, use two vertices with degree Ic joined by Ic paths of lengths 11/21 and [l/2J (alternating) through vertices of degree 2. Each face is formed by a path of length 11/21 and a path of length [l/2J. When Ic is odd, 1 is even and 11/21 = [l/2J, so Ic paths of this length suffice.
Section 6.2: Characterization of Planar Graphs
6.2. CHAR'ZN OF PLANAR GRAPHS 6.2.1. The complement of the 3-dimensional cube is nonplanar. The vertices of are the binary triples. Those with an odd number of is form an independent set, as do those with an even number of is. Each vertex is adjacent to three in the other independent set. Hence consists of two 4cliques with a matching between them. This graph contains a subdivision of K5 in which four branch vertices lie in one of the 4-cliques.
6.2.2. Nonplanarity of the Petersen graph. a) via Kuratowski's Theorem. Since the Petersen graph has no vertices of degree at least 4, it contains no K5-subdivision. Below we show a K33subdivision.
b) via Euler's Formula. To apply Euler's formula, assume a planar embedding. Since the Petersen graph has no cycle of length less than 5, each face has at least 5 edges on its boundary. Each edge contributes twice to boundaries of faces. Counting the appearances of edges in faces grouped by edges or by faces yields 2e> Sf. Since the graph is connected, Euler's formula yields a e + f = 2. Substituting for f in the inequality yields 2e> 5(2 a + e), or e -c (5/3)(n 2). For the Petersen graph, iS -c (6/3)8 is a contradiction.
c) via the planarity-testing algorithm. We may start with any cycle. When we start with a 9-cycle C as illustrated, every C-fragment can go inside or outside, so we can pick one of the chords and put it inside. Now the other two chords can only go outside, but after embedding one of them, the remaining chord cannot go on any face, This occurs because this cycle has three pairwise crossing chords.
Chapter 6: Planar Graphs
6.2.3. A convex embedding. This is the graph of the icosahedron. It is 3-connected and has a convex embedding in the plane.
6.2.4. Planarity of graphs.
The first graph is planar; a straight-line embedding with every face convex appears below. The second graph is nonplanar. It has many subgraphs that are subdivisions of K3,3; one is shown below.
6.2.5. The minimum number of edges that must be deleted from the Pete rsen
graph to obtain a planar subgraph is 2. The drawing on the left below illustrates a subdivision of K3,3 in the Petersen graph. Since this does not use every edge of the Petersen graph, the graph obtained by deleting one edge from the Petersen graph is still nonplanar (all edges are equivalent, by the disjointness description of the Petersen graph). Deleting two edges from the Petersen graph yields a planar subgraph as shown on the right below.
Section 6.2: Characterization of Planar Graphs
6.2.6. Fary's Theorem.
a) Every simple polygon with at most 5 vertices contains a point that sees every point in the polygon. In a convex polygon, by definition, the segment joining any pair of points lies entirely in the polygon. Hence every point in a convex polygon sees the entire polygon. Proof 1. If a 4-gon is not convex, then the vertex opposite the interior reflex angle (exceeding 180 degrees) sees the entire polygon. A non-convex pentagon has one or two reflex angles, and if two they may be consecutive or not. The cases are illustrated below.
Proof 2. Triangulate the polygon by adding chords between corners that can see each other. This can be done by adding one chord to a 4-gon and by adding two to a 5-gon, with cases as illustrated above. The resulting triangles have one common vertex. Since a corner of a triangle sees the entire triangle, the common corner sees the entire region. b) Every planar graph has a straight-line embedding. By induction on we prove the stronger statement that the edges of any plane graph G can be "straightened" to yield a straight-line embedding of G without changing the order of incident edges at any vertex. The statement is true by inspection for n(G) -c 3.
For n(G) > 4, we may assume that G is a triangulation, since any plane graph can be augmented to a maximal plane graph, and deleting extra edges in a straight-line embedding of the maximal planar supergraph yields a straight-line embedding of the original graph. Every planar graph has a vertex of degree at most 5; let x be such a vertex in G. Since G is a triangulation, the neighborhood of G is a cycle C, and G x has C as a face boundary. By the induction hypothesis, G x has a straight-line embedding with C as a polygonal face boundary. By part
Chapter 6: Planar Graphs
(a), we can place x at a point inside C and draw straight lines from x to all vertices of C without crossings.
6.2.7. G is outerplanar if and only if G contains no subdivision of K4 or K2,3. Let G' = G v K1 denote the graph obtained by from G by adding a single vertex x joined to all vertices of G. Then G has an embedding with all vertices on a single face G' is planar G' has no subdivision of K5 or K3,3
G has no subdivision of K4 or K2,3.
Additional details for these statements of equivalence: 1) If G is outerplanar, then we place x in the unbounded face of an outerplanar embedding of G and join it to all vertices on the face to obtain a planar embedding of G'. Conversely, if G' is planar, then it has a planar embedding in which x lies on the unbounded face. Deleting x from this embedding yields an outerplanar embedding of G, because it has an unobstructed curve from each vertex to the point that had been occupied by x and is now in the unbounded face. 2) Kuratowski's Theorem. 3) If G has a subdivision of K4 or K2,3, then adding x as a additional branch vertex yields a subdivision of K5 or K3,3 in G'. Conversely, if G' has a subdivision F of K5 or K3,3, then deleting x destroys at most one branch vertex or one path of F, which leaves a subdivision of K4 or K2,3 in G. 6.2.8. Every 8-connected graph with at least 6 vertices that contains a subdivision of K5 also contains a subdivision of K3,3. Let H be a K5-subdivision in G, with branch vertices x, y, t, u, v. If H itself has only five vertices, then G has another vertex p, and G has a p, V(H)-fan of size 3. By symmetry,
we may assume that the paths of the fan arrive at x, y, t. Then G has a subdivision of K3,3 with branch vertices x, y, t in one partite set and p, u, v
in the other partite set. If H has more than five vertices, then by symmetry we may assume that the x, y-path P in H has length at least two, Since G is 3-connected, G has a shortest path Q from V(P) to V(H) V(P). Let the endpoints of Q be p on P and q in H' = H V(P). Ifq is on the cycle in H' through t, u, v, then by symmetry we may assume q is on the branch path between u and v and not equal to u. In H U Q we now have a subdivision of K3,3 with branch vertices x, y, q in one partite set and p, t, u in the other partite set. On the other hand, if q is not on the cycle through z', u, v, then by symmetry we may assume q is on the x, t-path in H' (and not equal to 9, Now H U Q has a subdivision of K3,3 with branch vertices x, y, tin one partite set and p, u, v in the other partite set.
Section 6.2: Characterization of Planar Graphs
6.2.9. For a > 5, the maximum number of edges in a simple planar n-vertex graph not having two disjoint cycles is 2n 1. For the construction, begin with a copy of P3 and a —5 isolated vertices. Add two vertices x, y adjacent to all of these and each other. In a set of pairwise-disjoint cycles, at most one cycle can avoiding using both x and y, so no two cycles are disjoint. The number of edges is 2 + 2(n —2) + 1 = 2n 1. For the upper bound, we use induction on a. Basis step (a = 5): There is no 5-vertex planar graph with 10 edges, so the bound holds. Induction step (a >- 6): We need only consider a planar graph G with exactly 2a edges and no disjoint cycles. If any vertex has degree at most 2, then we delete it and apply the induction hypothesis to the smaller graph. Hence 8(G)> 3. Since G is planar, e(G) >- 2a 4 forces a triangle on some set S C V(G). Since G does not have disjoint cycles, G S is a forest H. If H has three isolated vertices, then 8(G) > 3 yields a copy of K3,3 with S as a partite set. Hence H has a nontrivial component. Main case. If x, y are vertices in a nontrivial component of H, and z is a vertex of H not on the unique x, y-path, and z has two neighbors in S other than a vertex of N(x) P N(y), then we form one cycle using the x, ypath in H and a vertex of N(x) P N(y), and we form a second cycle using z and the rest of S. Any two vertices of degree 1 in H have a common neighbor in S. If
H has an isolated vertex z, then using two leaves x, y from a nontrivial component of H yields the main case. Hence H has no isolated vertex. Suppose that H has a component with at least three leaves x, y, z. If x and y both have a neighbor in S outside N(z), then the main case occurs. Otherwise, symmetry yields N(y) P 5 = N(z) P 5, and the main case occurs
unless x, y, z all have the same two neighbors in 5. Now G contains a subdivision of K3,3 with one partite set being and the other consist
of their two common neighbors in S and the vertex that is the common vertex of the x, y-, y, z-, and z, x-paths in H. Hence every component of H
is a nontrivial path. If any component of H has endpoints with a common neighbor in 5 distinct from a common neighbor of the endpoints of another component, then we obtain two disjoint cycles. Hence there is a single vertex t e 5 that
Chapter 6: Planar Graphs
is adjacent to all endpoints of components in H. In each component the two ends have distinct second neighbors in 5; otherwise n(G) > 6 yields the main case. If H has at least two components, we now form one cycle using one component of H plus t and another cycle using another component of H plus the rest of S. Hence H is a single path. If any internal vertex u of H has a neighbor w in S other than t, then let v be the leaf of H that also neighbors w. We now obtain one cycle using
w and the u, v-path in H, and we obtain another cycle using the other endpoint of that component of H plus S in H is adjacent only to t in S.
, Hence every internal vertex
We now have determined H exactly. Every cycle in H contains t or avoids only t. In fact, G is the wheel v K1, where K1 is the vertex t. However, this graph has only 2n 2 edges. This final contradiction completes the proof. 6.2.10. Simple n-vertex graphs containing no K3,3-subdivision. Let f(n) be the maximum number of edges in such a graph. a) If n 2 is divisible by 3, then f(n) > 3n 5. We form G by pasting together (n 2)/3 copies of K3 as shown. Since K3,3 is 3-connected, a subdivision of K3,3 cannot have branch vertices in different S-lobes when = 2. This confines the branch vertices to a single S-lobe and yields an inductive proof that this graph has no K3,g-subdivision.
2 is divisible by 3, and otherwise f(n) = 3n 5 when n (for n > 2). Note that f(n) > 3n 6 for all n by using maximal planar graphs. For the upper bound, we use induction on n, checking the
small cases (2 -c a -c 5) by inspection. If e(G) > Sn 5, then G is nonplanar. By Kuratowski's Theorem, G contains a subdivision of K5 or K3,g. If G is 3-connected, then a subdivision of K5 also yields a subdivision of K3,3, by Exercise 6.2.8. Hence we may assume that G has a separating 2-set S. We avoid a K3,3-subdivision in G if and only if each S-lobe with the addition of the edge joining the vertices of S has no such subdivision. Since we are maximizing e(G), we may assume
that this edge is present in G. Now the number of edges and existence of K33-subdivisions is unaffected by how we add the S-lobes. If there are more than two, then we can paste one onto an edge in one of the other S-lobes and maintain the same properties. Hence we may assume that there are only two S-lobes.
Section 6.2: Characterization of Planar Graphs
Let the two S-lobes have iti and it2 vertices, repectively. The induction hypothesis yields e(G) -c f(iti) + f(it2) 1. Since it1 + it2 = it + 2 and we count the shared edge only once, this total is 3it c, where c depends on the congruence classes of it and it1 modulo 3. If it1 and it2 are congruent to 2 modulo 3, then the sum is 3it1 5 + 3it2 —5 1 = 3it 5. In other cases, at least one of the contributions is smaller by one. Hence 3it 5 is achievable only when it 2 (mod 3), and otherwise 3it 6 is an upper bound. Comment. When it 2 is divisible by 3, the only way to achieve the bound is by pasting together copies of K3 at edges.
6.2.11. If A(H) - 3, at most three edges of H depart from Let be the union of the paths in that connect the vertices of from which edges of H depart. In particular, if x, y, z are the vertices of departure for the paths leaving we can let Tj be the x, y-path P in plus the path in from z to P. Discard from H' all edges except those of each Ti,' and those that in the paths that contract to edges of H. The remaining graph is a subdivision of H in H'. Proof 2. An alternative proof follows the process from H (that is, K3, itself back to G, undoing the sequence of deletions and contractions (in the reverse order), keeping only a graph that is a subdivision of H and at the end is H', a subdivision of H contained in G. Deletions are undone by doing nothing (don't add the edge back). Undoing a contraction is splitting a vertex v. At most three edges incident to v have been kept in the current subdivision of H, If u and w are the adjacent vertices resulting from the split, then at least one of them, say w, inherits at most one of these important edges. Keeping that edge and the edge uw allows u to become the vertex playing the role of v in the subdivision, with the same number of paths entering as entered v, going to the same places. If a path went off along an edge now incident to w, then that path is one edge longer. Comment. The claim fails for graphs with maximum degree 4. Consider the operation of vertex split, which replaces a vertex x with two new adjacent vertices x1 and x2 such that each former neighbor of x is adjacent to x1 or x2. Contracting the edge x1x2 in the new graph produces the original graph. In applying a split to a vertex x of degree 4, the two new vertices may each inherit edges to two of the neighbors of x and thus wind up with
Chapter 6: Planar Graphs
degree 3. If H has maximum degree 4, then applying vertex splits to vertices of maximum degree can produce a graph G in which each new vertex has degree at most 3. This graph G has H as a minor, but G contain no H-subdivision since G has no vertex of degree 4. 6.2.12. Wagner's characterization of planar graphs. The condition is that neither K5 nor Kg,3 can be obtained from G by performing deletions and contractions of edges. a) Deletion and contraction of edges preserve planarity. Given an embedding of G, deleting an edge cannot introduce a crossing. Also, there is a dual graph G*, Contracting an edge e in G has the effect of deleting the dual edge e* in Gt. In other words, Gt et is planar, and G e is its planar
dual, so G e is also planar. Alternatively, one can follow the transformation that shrinks one endpoint of e continuously into the other and argue that at no point is a crossing introduced. Since deletion and contraction preserve planarity and K5 and K3,3 are not planar, we cannot obtain these graphs from a planar graph by deletions and contractions. Hence the condition is necessary. b) Kuratowski's Theorem implies Wagner's Theorem. We prove sufficiency by proving the contrapositive: if G is nonplanar, then K5 or K3,3 can be obtained by deletions and contractions.
If G is nonplanar, then by Kuratowski's Theorem, G contains a subdivision of K5 or K3,3. Every graph containing a subdivision of a graph F can be turned into F by deleting and contracting edges (delete the edges not in the subdivision, then contract edges incident to vertices of degree 2). Hence K5 or Kg,g can be obtained by deleting and contracting edges.
6.2.13. G is planar if and only if every cycle in G has a bipartite conflict graph. The condition is necessary because in any planar embedding a cycle C separates the plane into two regions, and the C-bridges embedded in each of the regions must form an independent set in the conflict graph. Conversely, if G is non-planar, then by the preceding theorem it is K5 (with conflict graph C5), or it has a cycle C with three crossing chords that produce a triangle in the conflict graph of C. 6.2.14. If x and y are vertices of a planar graph G, then there is a planar embedding with x and y on the same face if and only if G has no cycle C avoiding (x, y> such that x and y belong to conflicting C-fragments. If there is a cycle C such that x and y belong to conflicting C-fragments, then in every planar embedding of G, one of these fragments goes inside
G and the other goes outside it. Hence in every embedding, x and y are separated by C. (This argument applies when C does not contain x or y, but it suffices to consider such cycles.)
Section 6.2: Characterization of Planar Graphs
Conversely, suppose there is no such cycle; we show that G + xy is planar. If not, then G + xy contains a Kuratowski subgraph using xy. If this is a K3,3-subdivision H with xy on the path between branch vertices u and v, then x and y belong to fragments with alternating vertices of attachment on the cycle in H through the other four branch vertices. If this is a K5-subdivision H with xy on the path between branch vertices u and v if and only if x and y belong to fragments with three common vertices
of attachment on the cycle in H through the other three branch vertices. In either case, x and y belong to conflicting C-fragments for some cycle C.
6.2.15. A cycle C in a 3-connected plane graph G is the boundary of a face in G if and only if G has exactly one C-fragment. If G has exactly one Cfragment, then it must be embedded inside C or outside C, and the other of these regions is a face with boundary C. If C is a face boundary, then all C-fragments are embedded on one side of C, say the inside. This prevents two C-fragments H1, H2 from having
alternating vertices of attachment along C. This means that there is a path P along C that contains all vertices of attachment of H1 and none of H2. Now the endpoints of P separate G, which contradicts the assumption of 3-connectedness.
6.2.16. If G is an n-vertex outerplanar graph and P is a set of n points in the plane with no three on a line, then G has a straight-line embedding with its vertices mapped onto P. It suffices to consider maximal outerplanar
graphs. We prove the stronger statement that if vi, V2 are two consecutive vertices of the unbounded face of a maximal outerplanar graph G, and p1, P2 are consecutive vertices of the convex hull of P, then G has a straight-llne embedding f on P such that f(v1) = P1 and f(v2) = p2.) The statement is trivial for n = 3; assume n > 3. Let vi, V2, . denote the counterclockwise ordering of the vertices of G on the outside face in a particular embedding. Let be the third vertex on the triangle containing vi, v2. Claim: there is a point p c P with the two properties (a) no point of P is inside P1P2P, and (b) there is a line 1 through p that separates pi from P2, meets P only at p, and has exactly i 2 points of P on the side of 1 containing To obtain p, we rotate the line P1P2 about P2 until we
reach a line 1' = P2P' with p' e P such that exactly i 3 points of P are separated from P1 by 1'. Among the points of P in the closed halfplane determined by 1' that contains p be the point minimizing the angle P2P1P. By this choice, p satisfies (a), and there are at most i 2 points of P on the side of pip containing P2. If we rotate this line about p, then before it becomes parallel to 1' it reaches a position 1 satisfying (b). Let H1 and H2 denote the closed halfplanes determined by 1 contain-
Chapter 6: Planar Graphs
ing P1 and P2, respectively. By the induction hypothesis, the subgraphs of G induced by (v1, v2, . v and v1, . v1> can be straight-line embedded on H1 fl P and P P so that vi, v2, v, are mapped to Combining these embeddings yields a straight-line embedding of G with the desired properties. >
6.3. PARAMETERS OF PLANARITY 6.3.1. A polynomial-time algorithm to properly color a planar graph G.
First find an embedding in the plane and augment to a maximal plane graph G'. Now delete a vertex v of degree at most 5. Recursively find a proper 5-coloring of G' v. To extend the coloring to v, use Kempe chains if necessary to remove a color from the neighborhood of v. 6.3.2. If every subgraph of G has a vertex of degree at most k, then G is k+ 1-
colorable. We use induction on n(G). For the basis, K1 is k + 1-colorable whenever k > 0. For the induction step, let v be a vertex of degree at most k in a graph G with at least two vertices. By the definition of k-degenerate, every subgraph of a k-degenerate graph is k-degenerate. He the induction hypothesis yields a proper k + 1-coloring of G v. Extend the coloring to v by giving v a color that does not appear on its neighbors. 6.3.3. Every outerplanar graph G is .3-colorable, by the Four Color Theorem.
Adding a vertex v adjacent to all of G yields a planar graph G', which is 4-colorable. A proper 4-coloring of G' restricts to a proper 3-coloring of G, because the colors used on the vertices of G must all be different from the color used on v. 6.3.4. Crossing number of K2,2,2,2, 1(4,4, and the Petersen graph. Let k = [(a 2)g/(g 2)j. The maximum number of edges in a planar n-vertex
graph with girth g is k, so v(G) >
if G has girth g. This yields
and v(G) > 2 when G is the Petersen graph. The drawings below achieve these lower bounds. >
Section 6.3: Parameters of Planarity
6.3.5. Every planar graph G decomposes into two bipartite graphs. By the Four Color Theorem, G is 4-colorable. Let the four colors be 0, 1, 2, 3. Let H consist of all edges of G joining a vertex of odd color with a vertex of even
color. Let H' consist of all edges joining two vertices whose color has the same parity. Now H and H' are bipartite and have union G.
6.3.6. Small planar graphs. We use induction on n(G); every graph with at most four vertices is planar. A planar graph G with at most 12 vertices has degree-sum at most 6- 12 12, with equality only for triangulations. The bound is 60. Hence 6(G) 1,
is obtained from Gk_1 by adding the three vertices Xk, Yk, Zk and the five edges Xk_lXk, xkYk, YkZk, ZkYk_1, ZkXk_2. The graph G3 is shown below. Moving the edges x,_2z, outside yields a planar embedding.
We prove by induction on Ic that c4Gk) = Ic + 1 = (n(G) + 1)/3 and that, flirthermore, every maximum stable set in Gk contains Xk or Yk or .
In G1 the maximum stable sets are the nonadjacent pairs of vertices; the only one not containing x1 or is , so the stronger statement holds. Suppose that the claim holds for Gk_1. A maximum stable set S in Gk
uses at least one vertex not in Gk_1. If it uses two, then they are Xk and Zk. Since Yk-1, Xk_1, Xk_2 a N(
satisfies the stronger statement. To complete the proof of the stronger statement, we must show that a stable set S of size Ic + 1 in Gk that contains Zk but not Xk also contains S must contain a stable set of size kin Gk_1, which Xk_1. Since contains Xk_1 or Yk—1 or , by the induction hypothesis. Since Zk
is adjacent to Xk_2 and Yk-1, the only possibility here is Xk_1 a 5, which completes the proof of the statement.
6.3.11. For the graph defined below, when a is even, every proper 4coloring of uses each color on exactly a. vertices, Let G1 be C4. For a > 1, obtain from by adding a new 4-cycle surrounding making each vertex of the new cycle also adjacent to two consecutive vertices of the previous outside face. Each two consecutive "rungs" of form a subgraph isomorphic to G2, shown below on the left. This graph is 4-chromatic but not 4-critical, since it contains the 4-chromatic graph shown on the right. Since the remaining graph after deleting any one vertex still needs four colors, every color must appear at least twice (and hence exactly twice) in each copy of G2.
Chapter 6: Planar Graphs
6.3.12. Every oute rplanar graph is 3-colorable. The fact that every induced
subgraph of an outerplanar graph is outerplanar yields inductive proofs. Proof 1 (induction on n(G)). If every edge of G is on the outside face, then every block of G is an edge or a cycle, and G is 3-colorable. Otherwise, suppose xy is an internal edge. Then S = is a separating set. The S-lobes of G are outerplanar; by the induction hypothesis, they are 3colorable. Since S induces a clique, we can make these colorings agree on 5, which yields a 3-coloring of G.
Proof 2 (induction on n(G)). Every simple outerplanar graph has a vertex of degree at most 2 (proved in the text); we can delete such a vertex x, 3-color G x by the induction hypothesis, and extend the coloring to x. Proof 3 (prior results), Every graph with chromatic number at least 4 contains a subdivision of K4 (Dirac's Theorem), but a graph containing a subdivision of K4 cannot be outerplanar. Every art gallery laid out as a polygon with n segments can be guarded by [n/3j guards so that every point of the interior is visible to some guard. The art gallery is a drawing of an n-cycle in the plane. We add straightline segments to obtain a maximal outerplanar graph with a vertices. To do this, observe that 3-gons are already triangulated without adding segments. For a > 3, some corner can see some other corner across the interior of the polygon. We add this segment and proceed inductively on the two resulting polygons with fewer corners.
Consider a proper 3-coloring of the resulting maximal outerplanar graph (outerplanar graphs are 3-colorable). Since each bounded region is a triangle, its vertices are pairwise adjacent and receive distinct colors. Thus each color class contains a vertex of each triangule. Any point in a trian-
gle, such as a corner, sees all points in the triangle. Thus guards at the vertices of a color class can see the entire gallery. Since the three classes partition the set of vertices, the smallest class has at most ]n/3j elements.
The bound of [n/3J guards is best possible. The alcoves in the polygon
below require their own guards; no guard can see into more than one of them. There are [n/3J alcoves. When a is not divisible by 3, we can add the extra vertex (or two) anywhere.
Section 6.3: Parameters of Planarity
6.3.13. Every art gallery with walls whose outer boundary is an n-gon can be guarded by ](2n 3)/3j guards, where a > 3, and this is sharp. Adding walls cannot make it easier to guard the gallery, so we may assume that the polygon is triangulated by nonintersecting chords. A guard in a doorway can see the two neighboring triangles; we use such guards and guards on the outside walls. The bound is achieved by an art gallery of the type below.
Proof 1. The embedded n-gon plus the interior walls form a planar embedding of an outerplanar graph whose vertices are the corners; it has a + (a 3) = 2n 3 edges. Every outerplanar graph is 3-colorable (this can be proved inductively by cutting along chords formed by walls, as in Thomassen's proof of 5-choosability, or by using the existence of a vertex of degree at most 2, which can be proved inductively or by Euler's Formula.)
From a proper 3-coloring of the vertices of the outerplanar graph, 3color the edges of the graph by assigning to each edge the color not used on its endpoints. Now each triangle has each color appearing on its incident edges. If we put guards at the edges occupied by the least frequent color, then each room is guarded, and we have used at most K2n 3)/3j guards.
Proof 2. Again triangulate the region to obtain an outerplanar graph In the dual graph G*, let v denote the vertex corresponding to the unbounded face of G. The graph Gt v is a tree with a 2 vertices and G.
maximum degree at most 3. Each edge corresponds to a guard in a doorway, so an edge cover (a set of edges covering the vertices) corresponds to a set of guards in doorways that together can see all the rooms.
It suffices to show that a tree T with a
vertices and maximum
degree at most 3 has an edge cover with at most (2a 3)/3 edges, for a 4, We study a E expllcitly. For larger a, consider the endpoint x of
a longest path in T. By the choice of x, its neighbor y has one non-leaf neighbor and at most two leaf neighbors. We use the pendant edges at y in the edge cover and delete y and its leaf neighbors to obtain a smaller tree
Chapter 6: Planar Graphs
T'. We have placed k edges in the cover and deleted k + 1 vertices, where k a . Using the induction hypothesis and Ic -c 2, we obtain an edge cover of size at most 2(n-k-1)-3 3
4. Since K5 is nonplanar, the next smallest Eulerian triangu-
lation is the octahedron, with six vertices of degree 4; this is 3-colorable, as illustrated below.
For the induction step, suppose that n(G) > 6. Since every planar graph has a vertex of degree less than 6, G has a 4-valent vertex. If G has a triangle T of 4-valent vertices, then G the neighbors ofT induce a 3-cycle containing T, as in the octahedron. Deleting T reduces the degrees of the neighboring vertices by 2 each, so we can apply the induction hypothesis to the resulting subgraph G'. The coloring assigns distinct colors to the neighbors of T, and this proper coloring extends also to T.
Hence we may assume that when G has two adjacent 4-valent vertices x, y, their two common neighbors a, b have degree greater than 4. Suppose G has adjacent 4-valent vertices , with u and v being the fourth neighbors of x andy, respectively. Form G' by deleting and adding the edge uv. Because d(o), d(b) > 4, u and v are not already adjacent. All vertices
Section 6.3: Parameters of Planarity
still have even degree; hence G' is an Eulerian triangulation, We apply the induction hypothesis and extend the resulting coloring to a coloring of G, as indicated above. Finally, suppose that G has no adjacent 4-valent vertices. Choose a 4-valent vertex x with neighbor y, and define a, b, u as before. Form G' by deleting and adding edges from u to all of N(y) N(x). Because d(a), d(b) > 4, z is not already adjacent to any vertex of N(y) N(x). All vertices still have even degree; hence G' is an Eulerian triangulation. We apply the induction hypothesis and extend the resulting coloring to a coloring of G, as indicated below.
Proof 2. All faces are triangles; we start with an arbitrary 3-coloring on some face F. The color of the remaining vertex on any neighboring face is forced. We claim that iterating this yields a proper 3-coloring f. Otherwise, a contradiction is reached at some vertex v. This means there are two paths of faces from F, distinct after some face F' (we start at F' to obtain disjoint dual paths), that reach v but assigning different colors to v. Let C be the cycle enclosing the faces on these two paths and the regions inside them. Choose an example in which C encloses the smallest possible number of vertices. The contradiction cannot arise when C encloses only one vertex x. In this case, the faces causing the conflict are only those incident to x, and C is the cycle through the neighbors of x. Since d(x) is even, the colors alternate on C when following the path of faces, and there is no conflict. We obtain a contradiction by finding such a cycle enclosing fewer vertices. Since the initial face starts two distinct paths of faces, one of its vertices (x below) must be enclosed by C and not on C. Together, the two paths contain a portion of the faces containing x, say from J to J' around x. We replace these by the other faces involving x, but keeping J, P. Because d(x) is even, the coloring forced on J by f(V(F)) forces onto P the same coloring that f(V(F)) forced onto P directly. From J, P outward, the paths of faces lead to the same conflict as before. Hence we can start with one of the inner faces involving x and obtain a conflict using paths that enclose fewer vertices than before (x is no longer inside).
Chapter 6: Planar Graphs
6.3.15. The vertices of a simple outerplanar graph can be partitioned into two sets so that the subgraph induced by each set is a disjoint union of pat ha
Let one set be the set of vertices with even distance from a fixed vertex u, and let the other set be the remainder; call these acolor classes". Since no adjacent vertices can have distance from x differing by more than 1, each component of the graph induced by one color class consists of vertices with the same distance from u. Let H be such a component. To show that H is a path, it suffices to show that H has no cycle and has no vertex of degree at least 3. Given three vertices x1, x2, x3 in H, let be a shortest x,, u-path in G. Since x1, x2, x3 have the same distance from u, each P, has only x in H. Also, since the paths eventual merge, Pi U P2 U P3 contains a subdivision of a claw; call this F (note that F need not contain u, as the paths may meet before reaching u). If H contains a cycle C, let x1, x2, x3 be three vertices on C. Now F UC is a subdivision of K4. If H contains a vertex w of degree 3, let x1, x2, x3
be neighbors of w. Now F together with the claw having center w and leaves x1, x2, x3 is a subdivision of K2,3. Since an outerplanar graph has no subdivision of K5 or K2,3, H is a path. 6.3.16. The 4-dimensional cube is nonplanar and has thickness 2. The graph is isomorphic to C4o C4. On the left below, we show a subdivision of K3,3 in bold. The graph is also isomorphic to 0 K2, consisting of two 3cubes with corresponding vertices adjacent. Taking one of the 3-cubes and the edges to the other 3-cube from one of its 4-cycles yields a planar graph that is isomorphic to the subgraph consisting of the remaining edges.
Section 6.3: Parameters of Planarity
6.3.17. Thickness. a) The thickness of is at least Rn -F 1)/6j. Each planar graph used to form G has at most 3n(G) —6 edges, so the thickness of G is at least e(G)/[3n(G) —61. For G = this yields [n(n 1)/[6(n 2)11, since thickness must be an integer. We compute n(n 1)/(n 2) = n(1 + 1/(n—2)) = n+n/(n—2) = n+1+2/(n—2). Since k/ri = Rx +r 1)/rj, the thickness is at least 1 + 2/(n 2)1/61 = [En + 6 + 2/(n 2)]/6j = Rn + 7)/6j. The last equality holds because there is no integer between these two arguments to the floor function. (Comment: this lower is the exact answer except for a = 9, 10,)
b) A self-complementary planar graph with 8 vertices. See the solution to Exercise 6.1.29 for examples of self-complementary planar graphs with 8 vertices. To show that the thickness of K8 is 2, it suffices to present any 8-vertex planar graph with a planar complement. Many examples are
possible. A natural approach is to use a triangulation to eliminate the most possible edges from the complement. An example is C6 v 2K1, putting
one vertex inside and one vertex outside a 6-cycle. The complement is (C3o K2) + K2, which is planar as shown below.
Since K8 is nonplanar, these examples show that K8 has thickness 2, and that in fact K5, K8, K7 also have thickness 2. The bound in (a) implies that the thickness of is at least 3 when a > 11, which is the same as saying there is no planar graph with more than 10 vertices having a planar complement. In fact, there is also no planar graph on 9 or 10 vertices having a planar complement, but the counting argument in (a) is not strong enough to show that. 6.3.18. Decomposition of K9 into three pairwise-isomorphic planar graphs. View the vertices as the congruence classes of integers modulo 9. Group them into triples by their congruence class modulo 3. The graph below consists of a triangle on one triple, a 6-cycle between that and a second triple, and a matching from the second triple to the third. Rotating the picture on the left yields three pairwise isomorphic graphs decomposing K9. The drawing on the right shows that the graph is planar.
Chapter 6: Planar Graphs
6.3.19. The chromatic number of the union of two planar graphs is at most 12. Let G be a graph with ii vertices that is the union of two planar graphs H1 and 1-12. For coloring problems, we may assume that G, H1, 1-12 are sim-
ple. We claim that G has a vertex of degree at most 11. Since each 11 has at most 3n 6 edges, G has at most 6n 12 edges. The degree-sum in G is at most 12n 24, and by the pigeonhole principle G has a vertex of degree at most 11. It now follows by induction on n(G) that x(G) 12, we delete a vertex x of degree at most 11 to obtain G'. Since G' = x) U (112 x), we can apply the induction hypothesis to G' to obtain a proper 12-coloring. Since d(x) 11, we can replace x and give it one of these 12 colors to obtain x (G) -c 12. The chromatic number of the union of two planar graphs may be as large as 9. The graph C5 v K6 has chromatic number 9, since the three colors on the 5-cycle must be distinct from the six colors on the 6-clique, and such a coloring is proper. It thus suffices to show that C5 v K8 is the union of two planar graphs. Since C5 v K6 contains K8 and K9 e (for some edge e) as induced subgraphs, it is reasonable to start with one of these and then try to add the missing vertices with their desired neighbors to the two graphs. Let the vertices of the C5 be a, b, c, d, e in order, and let the vertices of the K6 be 1, 2, 3, 4, 5, 6. Exercise 6.1.29 requests a self-complementary graph with S vertices; in other words, an expression of K8 as the union of two planar graphs. 6.3.20. Thickness of K,-1. Let X, Y be the partite sets of K,-,1, with = r. The graph is planar. Taking all of Y and two vertices from X yields a
copy of K2,-. Taking two vertices at a time from X thus yields r/2 planar subgraphs decomposing K,-,1.
Since K,-,1 is triangle-free, a planar subgraph of K,-,1 has at most 2(r + s) 4 edges. Thus the number of planar subgraphs needed in a decomposition is at least 2r+2s—4 = 2+(2r—4)/1 As s increases the denominator decreases and the quotient increases. Thus when s > (r 2)2/2, the value of the lower bound is larger than the result of setting s = (r 2)2/2 in the fonnula. Since (2r 4)/s = 4/(r —2) when s has this value,our lower bound is bigger than 2+4Rr2) = 2r4+4 = r/2 1, Since the crossing number is an integer bigger than r/2 1 when r is even and s > (r 2)2/2, it is at least r/2. Hence our construction is optimal. 6.3.2 1. Crossing number of K1,2,2,2. This simple graph has 7 vertices and 18 edges. The maximum number of edges in a simple planar graph with 7 vertices is 3 - 7 6 = 15. Hence in any drawing of this graph, a maximal .
plane subgraph has at most 15 edges, and the remaining edges each yield at
Section 6.3: Parameters of Planarity
least one crossing with the maximal plane subgraph. Hence v(Ki,2,2,2) > 3, and the drawing of this graph on the left below shows that equality holds. Crossing number of K2,2,2,2. Deleting any vertex in a drawing of K2,2,2,2 yields a drawing of K1,2,2,2, which must have at least 3 crossings. Doing this
for each vertex yields a total of at least 24 crossings. Since each crossing is formed by two edges involving 4 vertices, we have counted each crossing 8 4 = 4 times. Thus the drawing of K2,2,2,2 has at least 6 crossings. We have proved that v(K1,2,2,2) > 6, and the drawing of this graph on the right below shows that equality holds.
6.3.22. K3,2,2 has no planar subgraph with 15 edges, and thus v(K3,2,2)> 2. The graph has 16 edges, so it suffices to show that deleting one edge leaves a nonplanar graph. Let X be the partite set of size 3. Every 6-vertex induced subgraph containing X contains a copy of K3,3, which is nonplanar. Since every edge e is incident to a vertex not in X, the 6-vertex induced subgraph avoiding such an endpoint remains when e is deleted.
6.3.23. The crossing number of the graph obtained from the cycle by adding chords between vertices that are opposite (if n is even) or nearly opposite (if n is odd) is 0 if n -c 4 and 1 otherwise. For n 6, the cycle with vertices graphs are planar. For n = 5, plus the chords vltJl+[n/2j, V2V2+[n/2J, VSVS+Ln12J is a subgraph of that is a subdivision of K3,g, so the crossing number is at least 1. The drawings below, by avoiding crossings among the chords and allowing a crossing within the drawing of the cycle, show that one crossing is enough.
is a maximal planar graph. The graph is K4, with six 6.3.24. a) edges. Note that 6 = 3 4 6. Each successive vertex in is adjacent to the last three of the earlier vertices, so = 3n 6. Together with having 3n —6 edges, showing that is planar implies that it is a maximal -
Chapter 6: Planar Graphs
planar graph. An embedding is obtained by drawing the path in a spiral as suggested below. Alternatively, we can prove planarity by proving inductively that there is a planar embedding with all of on the same face. This holds for an embedding of (we could also start with a = 3 as the basis). For the induction step (a > 4), take such an embedding of Since all of lie on a single face, we can place a in that face and
draw edges to all three. This yields a planar embedding of
a> on a single face. = a 4. The graph
b) is K5, with 10 edges. Each additional = 4a 10. In any drawing, a vertex provides four more edges, so maximal plane subgraph H has at most 3a 6 edges and thus leaves at least a —4 edges that each cross an edge of H. That bound is achieved with equality by adding the second diagonals of the trapezoids in the picture below, making each vertex adjacent to the vertex four earlier on the path.
(Alternatively, the earlier induction proof can be strengthened to guarantee an embedding with all of on a single face and a —4 on an adjacent face across the edge joining a 1 and a 3. That enables the a 4 additional edges to be added so that each crosses only the specified edge of and no added edges cross each other.) 6.3.25. There are toroiclal graphs with arbitrarily large crossing number in the plane. The cartesian product of two cycles, Cm o embeds naturally
on the torus; each face is a 4-cycle. The graph has inn vertices and 2ma edges. View the copies of Cm as vertical slices (columns) and the copies of as horizontal slices (rows). A sub graph of Cm D consisting of three full columns and three full rows is a subdivision of C3 Since C3 contains a subdivision of K3,3, it is nonplanar. Therefore, a planar subgraph of Cm D cannot contain
three full columns and three full rows. This means it must omit at least 2> edges. By Proposition 6.3.13, v(Cm DCX) > min. By making in and a at least k +2, we make the crossing number
k while having a toroidal graph.
Section 6.3: Parameters of Planarity
6.3.26. Lower bounds on crossing numbers. As stated correctly in Example 6.3.15 (not stated correctly in this exercise), the crossing number of is 6 1)/2j. a) v(Km,n) mfr1 LU Consider copies of in a drawing of Kmn, with the partite set of size 6 in the subgraph selected from the partite set of size m in the fill graph. There are such copies, and each has at least 6 ljn 1)/2j crossings. Each crossing appears in of the subgraphs. Hence v(Km,n)
Cancellation of common factors
in the numerator and denominator yields the bound claimed, b) + 0(p3). Consider copies of in a drawing of There are of these copies, and each has at least 6 1(p 6)/2j 1(p 7)/2j crossings. Each crossing appears in of these subgraphs, since the four vertices involved in the crossing can contribute to the smaller partite set in four ways (assuming that a > 12), and then four vertices not involved in the crossing must be chosen to fill that partite set. /[4(P4)] The numerator has four Hence >
more linear factors than the denominator, so the growth is quartic. The leading coefficient is
which simplifies to 1/80.
6.3.27. If the conjecture that v(Kmn) =
holds for LU Km,n and m is odd, then the conjecture holds also for Km+in. In a drawing of Km+i,n, there are m + 1 copies of Km,n obtained by deleting a vertex of the partite set of size m. Each crossing in the drawing of Km+i,n appears in m 1 of these copies. Hence (m 1)v(Km+i,n) (m + 1)V(Km,n). Sincemis odd, 1)/2j = (m—1)2/4,and (m + 1)(m 1)/4. Therefore,
'K m,n)\_m+1m—h1f11n—h1_1m+h11m11f11n—1 rr LU LTT] LTT] LU
6.3.28. If m and a are odd, then in all drawings of Kmn, the parity of the
number of pairs of edges that cross is the same. (We consider only drawings where edges cross at most once and edges sharing an endpoint do not cross.) Any drawing of Km. can be obtained from any other by moving vertices and edges. The pairs of crossing edges change only when an edge e is moved
through a vertex v not incident to it (or vice versa). Let S be the set of edges incident to v other than those also incident to endpoints of e. When e is moved through v (or vice versa), the set of edges incident to v that e crosses is exchanged for its complement in S.
Chapter 6: Planar Graphs
Since the degree of each vertex is odd and v is adjacent to exactly one endpoint of e, the parity of these two sets is the same. Hence the parity of the number of crossings never changes.
Ifm and n are odd, then v(Km,n) is odd when in —3 and a —3 are divisible
by 4 and even otherwise. In the naive drawing of Km,n with the vertices on
opposite sides of a channel and the edges drawn as segments across the channel, the number of crossings is It suffices to determine the parity of this, since the parity is the same for all other drawings including those with fewest crossings. The binomial coefficient r(r 1)/2 is odd if and only if r is congruent to 2 or 3 modulo 4. Since we require in and a odd, the additional requirement for () being odd is thus in and a being congruent to 3 modulo 4.
6.3.29. If n is odd, then in all drawings of the parity of the number of pairs of edges that cross is the same. (We consider only drawings where edges cross at most once and edges sharing an endpoint do not cross.) Any drawing of Km,n can be obtained from any other by moving vertices and edges. The crossing pairs change only when an edge a moves through a vertex v not incident to it (or vice versa). Let S be the set of edges incident to v that are not incident to endpoints of e. When a moves through v (or vice versa), the set of edges incident to v that a crosses is exchanged for its complement in S. Since d(v) is even and v is adjacent to both endpoints of e, we have even, so the sizes of complementary subsets of S have the same parity. Hence the parity of the number of crossings does not change.
is even when a is congruent to 1 or B modulo 8 and is odd when a is congruent to 5 or 7 modulo 8. Since the parity is the same in all drawings of we need only look at one, such as the straight-line drawing with the vertices on a circle. Its number of crossings is (fl, which equals n(n 1)(n 2)(n
When the congruence class is 1 or 3, the numerator has a
multiple of 8 and an odd multiple of 2, so it has four factors of 2, and only
Section 6.3: Parameters of Planarity
is even. When the class is 5 or 7, the numerator has an odd multiple of 4, an odd multiple of 2, and two odd factors, so the factors of 2 are canceled out, and is odd. three are canceled by the denominator. Hence
6.3.30. V(Cm n -c (m 2)n and v(K4 n C,,) -c 3n. For Cm o C. we draw the copies of C. along concentric circles. The vertices arising from a single copy of C,, are laid out along a spoke. The "long" edge in each copy of C,, generates m 2 crossings, as shown below on the left. For K4 o C. we make the cycles concentric again, almost: the two outside cycles weave in and out of each other, as shown on the right below. We draw each copy of K4 around a spoke, but each copy is the reflection of its neighbors. For each copy of K. the two outer cycles cross, and the central cycle crosses two edges in that copy of K.
The weaving in and out requires a to be even. When a is odd, a special construction is needed for the case a = 3 (shown below); there is one crossing in a copy of K4, and the inner two triangles and outer two triangle each provide four crossings. Copies of K4 can then be added in pairs by breaking the four edges joining two "neighboring" copies of K4 and inserting two copies of K4 with six crossings as in the middle of the figure above.
6.3.31. Crossing number of complete tripartite graphs. Let f(s) = prove that n3(n 1)16 -c -c (9/16)n4 -F- 0(n3). ,,2 a) 3v(K. ) -c f(n) 3C) . The lower bound follows from the existence of three pairwise edge-disjoint copies of K. in K. The upper bound
Chapter 6: Planar Graphs
follows from a straight-line drawing with the vertices of each part placed on a ray leaving the origin. b) v(K3,2,2) = 2+ Lower bounds: Since a triangle-free 7-vertex planar graph has at most 2n 4 = 10 edges, v(K3,4) 2, and K3,4 is a subgraph of K3,2,2. Alternatively, a counting argument for the crossings shows both 2 and u(K3,3,1) > 3. Consider each vertex-deleted subgraph for some embedding; if it contains K3,3, its includes a crossing, and each crossing is counted n 4 = 3 times. Hence v(Kg,2,2) 14/31 = 2 and 17/31 = 3. Extending this approach yields v(K3,3,2) 118/41 = and v(K3,3,3) 145/51 = 9. Constructions for v(Kg,2,2)
(5/18)n4 + 0(n3), which is approximately a factor of 2 from the upper bound below.
d) Improving the upper bound (3/4)n4 + 0(n3) of (a) to f(n)
(9/16)n4 + 0(n3). The layout on the tetrahedron splits the n vertices of each part into two sets of size n/2 laid out along opposite edges. For the points on a given edge, the four neighboring edges of the tetrahedron contain all points of the other two parts, to which these points have edges laid out directly on the surface of the tetrahedron. Crossings on a face of the tetrahedron are formed by pairs of vertices from two incident edges or by a pair from one edge with one vertex each from the other two edges. If the parts have sizes 1, in, n and 1' = (i!2) ,& = (m/2) W = (ri/2) then the total number of crossings on a single face of the tetrahedron is [l'rn' + l'n' + m'n'I + [l'Øn/2)(n/2) + m'(l/2)(n/2) + n'(l/2)Øn/2)]. Over the four identical faces,
Section 6.3: Parameters of Planarity
-F 12n2 -F m2n2 + 2!2 ma + 2m21n -F 2n21m), plus lower
order terms. When 1 = m = a, this becomes (9/16)n4. For the other construction, begin with an optimal drawing of Turn each vertex into an independent set consisting of one vertex from each part.
When there are three parts, each edge of the original drawing has now become a bundle consisting of 6 edges. For each crossing in the drawing of we get 36 crossings between the two bundles. For each edge in the drawing of we get at most 15 crossings within the bundle. Near a vertex of the original drawing, we get at most 36 crossings (actually slightly less)
between the bundles corresponding to incident edges. There are () edges in and pairs of incident edges, but always (a4) crossings, so the other contributions are of smaller order, Therefore, we have only + 0(n3) crossings. With the best known bound of -c n4/64 + 0(n3), we
get the same constant 9/16. This generalizes easily to complete multipartite graph with r parts of size a.
6.3.32. An embedding ofa 3-regular nonbipartite simple graph on the torus such that every face has even length. It suffices to use K4 as shown below. Larger examples can be obtained from this.
6.3.33. If a is at least 9 and a is not a prime or twice a prime, then there is a 6-regular toroidal graph with a vertices. Given these conditions, express a as rs with r and s both at least 3, Now form Cr D this 4-regular graph
embeds naturally on the torus with each face having length 4. On the combinatorial description of the tons as a rectangle, the embedding looks like the interior of Pr+i o but the top and bottom rows of vertices are the same, and the left and right columns of vertices are the same. Now add a chord in each face from its lower left corner in this picture
to its upper right corner. The resulting graph is 6-regular, toroidal, and has a vertices. 6.3.34. Regular embeddings of K4,4, K3,6, and K3,3 on the torus, The num-
ber of faces times the face-length is twice the number of edges, and the number of faces is the number of edges minus the number of vertices. For K4,4, we need eight 4-faces. For K3,6, we need nine 4-faces, For K3,3, we need three 6-faces. Such embeddings appear below.
Chapter 6: Planar Graphs
6.3.35. Euler's Formula for genus y: For every 2-cell embedding of a graph on a surface with genus y, the numbers of vertices, edges, and faces satisfy a e + f = 2 2y. We use induction on e(G) via contraction of edges.
For the basis step, we need the number of edges required to cut S> into
2-cells. Each face in an embedding is a 2-cell; lay it flat. Combining neighboring faces along shared edges yields a large 2-cell R. Identifying shared edges reassembles the surface. The number of edge-pairs needed on the boundary of R is at least the number of cuts required to lay the surface flat, because that is what these boundary edges do. It takes two cuts to lay a handle flat. If every cut is one edge, then every cut is a loop and there is only one vertex. So, the only 2-cell embeddable graphs on that have at most 2y edges are those with 1 vertex and 2y edges, and the resulting embeddings have 1 face. The polygonal representation of the surface is itself such an embedding, if we view the vertices of the polygonal as copies of the single vertex in the graph, and the edges of the polygon as paired loops. Since 1 2y + 1 = 2 2y, all is well.
Given a 2-cell embedding with more than 2y edges, contract an edge e that is not a loop surrounding another loop of the embedding. If e is not a loop, then following the boundaries of the face(s) bounded by e shows they are still 2-cells, and we now have a 2-cell embedding of G e on the same surface. The induction hypothesis provides Euler's Fonuula for G has one more vertex and edge than G e but the same number of faces, Euler's Formula holds also for G. On the other hand, if e is a loop, then G has one more edge and face than G• e but the same number of vertices; again the formula holds.
Section 6.3: Parameters of Planarity
An n-vertex simple graph embeddable on has at most 3(n 2 -F 2y) edges. If G embeds on S,, then G has a 2-cell embedding on S, for some y' with y' 3f. Substituting in Euler's Formula n e -F f = 2 2y yields e -c 3(n 2 + 2y). 6.3.36. Genus of Since has n + 6 vertices and 6n -F 9 edges, Euler's formula yields 1 + (6n + 9 3n 18)/6 = (n 1)/2. This can be improved by apply Euler's Formula to the bipartite subgraph K6. Here the genus is at least 1 + (6n 2n 12)/4, which simplifies to n 2. For 0 k. Let H = G o K2. The definition of cartesian product yields n(H) = 2n(G) and e(H) = 2e(G) + n(G). Since an n-vertex graph embeddable on has at most 3(n 2 + 2y) edges, we have y(H) > 1 + (e(H) 3n(H))/6 = 1 + (2e(G) 5n(G))/6, If G is a triangulation with n vertices, then e(G) = 3n 6, and we obtain y(H)> —1 + n/12. It suffices to choose n> 12k + 12,
Chapter 7: Edges and Cycles
7.EDGES AND CYCLES 7.1. LINE GRAPHS & EDGE COLORING 7.1.1. Edge-chromatic number and line graph for the two graphs below. The labelings are proper edge colorings, the number of colors used is the maximum degree, so the colorings are optimal.
In the cube Qk, the edges
7.1.2. x'(Qk) = A(Qk), by explicit coloring.
between vertices differing in coordinate j form a complete matching. Over the k choices of j, these partition the edges. 7.1.3.
The lower bound is given by the maximum degree.
For the upper bound, when n is even colors 0 and 1 can alternate along the two cycles, with color 2 appearing on the edges between the two copies of the factor When a is odd, colors 0 and 1 can alternate in this way except for the use of one 2. Color 2 appears on all cross edges except those incident to edges on the cycles with color 2, as shown below.
Section 7.1: Line Graphs and Edge Coloring
7.1.4. For every graph G, x'(G)> e(G)/a'(G). In a proper edge-coloring, each color class has at most a'(G) edges. The lower bound follows because all e(G) edges must be colored. 7.1.5. The Petersen graph is the complement ofL(K5). The vertices of L(K5) are the edges in K5, which can be named as the 2-element subsets of [51.
Two such pairs are adjacent in the Petersen graph if and only if they are disjoint, which is the condition for them being nonadjacent in L(K5). 7.1.6. The line graph of the Petersen graph has 10 triangles. For a simple graph G, there is a triangle in L(G) for every set of three edges in G that share one common endpoint and for every set of three edges that form
a triangle in G. The Petersen graph has no triangles, so the latter type does not arise. However, the Petersen graph has 10 triples of edges with a common endpoint, one for each of its vertices. 7.1.7. T5 is a line graph. The complement of P5 is a 5-cycle with a chord. It is the line graph of a 4-cycle with a pendant edge.
7.1.8. The line graph of Kmn is the cartesian product of Km and For each edge in Kmn, we have a vertex (i, f) in L(Kmn); these are also the vertices of Km u Pairs (i, J) and (Ic, 1) in V(Km o are adjacent in if and only if i = Ic or j = 1. This is the same as the condition for Km o adjacency in L(Km,n), because x1 yy and xkyl share an endpoint in Km,n if and
only ifi = Ic or j = 1. 7.1.9. A set of vertices in the line graph of a simple graph G form a clique if and only if the corresponding edges in G have one common endpoint or form a triangle. Let S be the corresponding set of edges in G, and choose e c S. If all other elements of S intersect e at the same endpoint of e, we have one common endpoint. Otherwise, we have edges f and g such that
f shares endpoint x with e and g shares endpoint y with e, Since f and g must share an endpoint, they share their other endpoint z and complete a triangle. Since no single vertex lies in all of e, f, g, no additional edge of the simple graph G can share a vertex with all of these.
Chapter 7: Edges and Cycles
7.1.10. If L(G) is connected and regular, then either G is regular or G is a bipartite graph in which vertices of the same partite set have the same degree. If L(G) is connected, then G is connected (except for isolated vertices, which we ignore). For e = uv e E(G), the degree of e in L(G) is d(u) + d(v) 2. If the edges incident to v in G have the same degree in L(G), then they must join v to vertices of the same degree in G. If G is not regular, then G has adjacent vertices u, v with different degrees, since G is connected. By the observation above about maintaining constant degree, every walk from v in G must alternate between vertices of degrees d(v) and d(u). Thus G has no odd walk and is bipartite. Furthermore, the vertices of one partite set have degree d(v), and those of the other partite set have degree d(u). 7.1.11. Line graphs of simple graphs. (d(v)) a) e(L(G)) =
Proof 1 (bijective argument). The edges of L(G) correspond to the incident pairs of edges in G. Such pairs share exactly one vertex, and each vertex v a V(G) contributes exactly such incident pairs. Proof 2 (degree-sum formula). The degree in L(G) of the vertex corresponding to uv a E(G) is da(u) + d0(v) 2, the number of edges of G
sharing an endpoint with it. When this is summed over all edges of G, the term da(u) appears da(u) times. Hence the degree sum in L(G) is 2e(G), and L(G) has e(G) edges. Replacing e(G) by yields (Comment: The formula holds also for graphs with multiple edges un-
der the convention that when edges share both endpoints we have two edges between the corresponding vertices of L(G).) b) G is isomorphic to L(G) if and only if G is 2-regular Sufficiency. A 2-regular graph is a disjoint union of cycles. The line graph of any cycle is a cycle of the same length (successive edges on a cycle in G turn into successive vertices on a cycle in L(G)). Necessity.
Proof 1 (numerical argument). If G is isomorphic to L(G), then L(G) has the same number of vertices and edges as G. Thus n(G) = n(L(G)) = e(G) = e(L(G)), By (a), this becomes n(G) = Using the degree-sum formula, = 2e(G) = 2n(G). We have shown that the average degree is 2. When the degrees all equal 2, the sum equals n(G), as desired, It suffices to show that when the average degree is 2 but the individual degrees do not all equal 2, y is larger than n(G). In this case, there is at least one number bigger than 2 (the average) and one smaller than 2. Since + (1 > (r_1) + when r > s + 1, we can
Section 7.1: Line Graphs and Edge Coloring
iteratively bring the values toward the average while always decreasing (d(2v)). Hence the equality n(G) = is achieved only when every vertex degree is 2.
(Comment: This is the discrete version of a calculus argument. Because is quadratic in x with positive leading coefficient, it is convex. For a convex function, the sum of values at a set of n arguments with fixed sum s is minimized by setting each argument to s/n.) Proof 2 (graph structure). As above, n(G) = (v). If all degrees are at least 2, then equality holds only when all equal 2. Hence it suffices to forbid vertices of degree less than 2. For a graph H, observe that L(H) is a path if and only if H is a nontrivial path. If G has any component that is a path, then let k be the maximum
number of vertices in such a component. In L(G) there is no component isomorphic to Pk. Hence G does not have a component that is a path. In particular, G has no isolated vertex. Suppose that G has a path (vo, . v1) such that d(vo) > 3, d(v1) = 1,
and internal vertices have degree 2. Let ei, . e1 be the edges of P. In L(G), the vertices e1, . e1 form a path such that d(et) >- 3, d(e1) = 1, and internal vertices have degree 2. This path is shorter than P. Also, a pendant path in L(G) can only arise in this way. Let m be the maximum length of a path from a vertex of degree at least 3 through vertices of degree 2 to a vertex of degree 1. By the reasoning above, L(G) has no such path of length in. Hence L(G) cannot be isomorphic to G if G has a vertex of degree 1,
7.1.12. If G is a connected simple n-vertex graph, then e(L(G)) - k, we have 3(L(G)) >2k 2, since each edge is incident to at least k 1 others at each endpoint. Let [T, T'] be a minimum edge cut of L(G), with ic' edges. Because a minimum edge cut yields only two components, T, T' corresponds to a partition of E(G) into two connected subgraphs, which we call F, F', respectively. There is an edge of L(G) in [T, T'] each lime an edge of F is incident to an
edge ofF'. These incidences take place at vertices of G. At a vertex x E V(G), there are 4(x) edges of F (corresponding to vertices ofT) and 4' (x) edges of F (corresponding to vertices ofT'). Since each such edge of F is incident to each such vertex of F', this vertex x in G yields (x) edges in [T, T']. Since 4(x) -F = dG(x) > k, this product is at least k 1 whenever x is incident to edges of both F and F'. Hence it suffices to show that there are at least two vertices of G that are incident to edges from both F and F'. If F and F' are incident at only one vertex x, then this must be a cut-vertex of G, because any path from F to F' that avoids x would yield another vertex where F and F' are incident. Deleting the edges of F incident to x or the edges of F' incident to x disconnects G. Since G is k-edge-connected, we conclude that 4(x), 4'(x) H > k > k2 > 2k —2. and [T, 7.1.15. Every connected line graph of even order has a perfect matching. Note that a graph without isolated vertices has the same number of components as its line graph. Let 5' be the set of edges in G corresponding to a set S c V(L(G)). Deleting S from L(G) corresponds to deleting 5' from G,
Chapter 7: Edges and Cycles
but each edge deletion increases the number of components by at most one. Thus G 8' (and L(G) 5) have at most 1 + components of any sort, odd or otherwise. For a graph of even order, o(L(G) 8) -c 1 + 5) implies Tutte's condition o(L(G) 5) -c 5), since the order is even. The edges of a connected simple graph of even size can be partitioned into paths of length two. The paired vertices of a perfect matching in L(G) correspond in G to paired edges forming paths of length 2. Since the matching saturates V(L(G)), the corresponding paths partition E(G).
7.1.16. If G is a simple graph, then y(L(G)) > y(G), where y(G) denotes the genus of G (Definition 6.3.20). Consider an embedding of L(G) on a surface 8; it suffices to obtain an embedding of G on the same surface. For each x E V(G), the edges of G with endpoint x form a clique in L(G). For the embedding of G, locate x at one vertex xx' of in the embedding of L(G). For each edge xy, embed it along the path in the embedding of L(G) from xx' to xy to yy'. Since xy is used in only one such path, the edges of the new embedding of G on this surface have no crossings.
7.1.17. The number of proper 6-edge-colorings of the graph below (from a specified set of six colors) is 900 512.
It suffices to count the ways to assign pairs of colors to the double edges so that the pairs at two double edges with a common endpoint are disjoint, because we can then multiply by to assign the colors within the pairs. We can view such an assignment as a 3-by-3 matrix in which the entry in position (1, j) is the pair assigned to the two edges joining the ith top vertex and the jth bottom vertex. Each color must appear exactly once in some pair in each row and each column. We can choose entry (1, 1) in
and for each such way there are choices for entry (1, 2). Thus we can choose the first row in 90 ways, and for each way the number of ways,
completions will be the same. Let the pairs in the first row be , , and , in order. If entry (2, 1) is one of the pairs in the first row, then we have two such pairs to choose from, By symmetry, let it be . Now entry (2, 2) must be , and entry (2, 3) is , and the bottom row is determined. If entry (2, 1) is not one of the pairs in the first row, then we fill it using one element from entry (1, 2) and one element from entry (1, 3); these can be chosen in 4 ways. For example, suppose that entry (2, 1) is . Now
Section 7.1: Line Graphs and Edge Coloring
f must appear in entry (2, 2) and d in entry (2, 3), and the second row is
completed by deciding which of goes into entry (2, 2) and which goes
into entry (2, 3). There are two ways to make this choice, and again the bottom row is determined. Thus after choosing the first row; there are two ways to complete the matrix with entry (2, 1) not being a pair from the first row. Since there are two ways when entry (2, 1) is a pair from the first row; the total number of colorings is 10- 90-
7.1.18. x'(Km,n) = A(Km,n), by explicit coloring We may assume that m -c a, so the maximum degree is a. If the vertices are X U Y with X = x1, . . and V = we give the edge x,y1 the color i + j (moda). Since incident edges differ in the index of the vertex in X or the vertex in Y, they receive different colors.
7.1.19. Every simple bipartite graph G has a A(G)-regnlar simple bipartite supergraph. Let k = A(G), and let X and V be the partite sets of G.
Construction 1. A huge simple k-regular supergraph of G can be constructed iteratively as follows: If G is not regular, add a vertex to X for each vertex of V and a vertex to V for each vertex of X. On the new vertices, construct another copy of G. For each vertex in G with degree less than k, join its two copies in the new graph to get G'. Now k is the same as before, the minimum degree has increased by one, and G' is a supergraph of G. Iterating this k 5(G) times yields the desired simple supergraph H. It is connected if G was connected. Construction 2. We may assume that = by adding vertices d(x,); this is the total to the smaller side, if necessary. Let M = ak "missing degree". Add M vertices to both X and V, and place a (k 1)regular graph H on these, which may be constructed using successively tilted matchings as in the natural 1-factorization of KMM. Now add edges joining deficient vertices of X and V to vertices of H on the opposite side.
Each vertex of H receives one such edge, which remedies the M deficiencies
ineachofX andY. 7.1.20. Edge-coloring of digraphs. Given a digraph D with indegrees and outdegrees at most d, form a bipartite graph H as follows. The partite sets are A = and B = x c V(D)>. For each edge xy in D, place an edge x inherits the edges exiting x and the vertex inherits the edges entering x. The resulting bipartite graph H is the "spllt" of D (Section 1.4). Since the maximum number of edges entering or exiting a vertex of D is d, A(H) = d, Since H is bipartite, f(H) = d, The d-edge-coloring on the edges of H is the desired coloring of the corresponding edges in D.
Chapter 7: Edges and Cycles
7.1.21. Algorithmic proof of x1(G) = A(G) for bipartite graphs. Let G be a bipartite graph with maximum degree k. Let f be a proper k-edge-coloring of a subgraph H of G. Let uv be an edge of G not in H. We produce a proper A(G)-edge-coloring of the subgraph consisting of H plus the edge uv. Since uv is uncolored, among the A(G) available colors there is a color a not used at u. Similarly, some color fi is not used at v. If a is missing at v or to at u, then we can extend the coloring to uv using a or .8. Otherwise,
follow the path P from u that alternates in colors a and fi. The path is well-defined, since each color appears at most once at each vertex. Since a does not appear at u, the path P ends somewhere and does not complete a cycle. The path reaches the partite set of v along edges of color fi, and it reaches the partite set of u along edges of color a. Hence P cannot reach v, where to is missing. We can now interchange colors a and ,8 on the edges of P to make to available for the edge uv.
7.1.22. If G isasimple graph with maximum degreeS, then f(G) -c 4. Let H = L(G); since x'(G) = x(L(G)), we seek abound on x(H). By making the same argument for each component, we may assume that G and H are connected. Since A(G) = 3, an edge of G intersects at most two other edges at each end, and hence A(H) -c 4. If H is 4-regular, then G must be 3-regular. The smallest 3-regular simple graph has 6 edges, so H K5. By Brooks' Theorem, x(H) -c A(H) if H is not a clique or odd cycle. If H is an odd cycle, then x(H) A(GoK2) = A(G) -f-i. b)IfGi, G2 are edge-disjoint graphs with the same vertex set and H1, H2 are edge-disjoint graphs with the same vertex set, then (G 1UG2)o(H1U H2) =
We view G1 and G2 as a red/blue edge-coloring of G1 U G2, and we view H1 and H2 as a yellow/green edge-coloring of H. Since every edge of G o H is a copy of an edge of G or H, this induces a red/blue/yellow/green edge-coloring of the product. The spanning subgraph containing the red and green edges is G1o H2, and the spanning subgraph containing the blue and yellow edges is G2 o H1. c) Go H is 1-factorable if G and H each have a 1-factor Let G1 be a 1(G1 0 H2) U (G2 0
factor of G, G2 = G E(G1), H1 a 1-factor of H, and H2 = H E(H1). Since H1 = mK2,wehave G2DH1 = G2onzK2 = tn(G2oK2). Bypart(a),thereis a proper edge-coloring of G2 o H1 with A(G2) + 1 = A(G) colors. Similarly,
there is a proper edge-coloring of G1 o H2 with A(H) colors. By part (b), these together yield a proper edge-coloring of G 0 H with A(G) + A(H) = A(GDH) colors. (This result is Kotzig's Theorem, usually stated for regular graphs; the proof is from the thesis of J. George.) 7.1.26. If G is a regular graph with a cut-vertex x, then x'(G) > A(G). Proof 1. Because G is regular, x'(G) = A(G) requires that each color class be a 1-factor. Hence n(G) is even. Since n(G) 1 is odd, G x has a component H of odd order. Let y be a neighbor of x not in H. A 1-factor of G that contains xy must contain a 1-factor of H, which is impossible since H has odd order. Proof 2. Again each color class must be a 1-factor. Let M1 and M2 be color classes containing edges incident to x whose other endpoints are in different components of G x. Since these are perfect matchings, their symmetric difference consists of isolated vertices and even cycles. In particular, it contains a cycle through x that visits different components of G x, but there is no such cycle. 7.1.27. Density conditions for x'(G)> A(G), a) If G is regular and has 2tn + 1 vertices, then x'(G) > A(G). For a regular graph, being A(G)-edge-colorable means being 1-factorable, which is impossible with odd order since such graphs have no 1-factor. b)IfG has 2m + 1 vertices and more than in A(G) edges, then f(G)> A(G). Each color class is a matching, and each matching has size at most m, so A matchings cover at most mA edges. Since G has more edges than that, every proper edge-coloring of G requires more than A colors. c) If G arises from a k-regular graph with 2m + 1 vertices by deleting fewer than k/2 edges, then x'(G) > A(G). Since fewer than k vertices have
Section 7.1: Line Graphs and Edge Coloring
lost an edge and k -c 2m, some vertex of degree k remains; hence A(G) = k. Also e(G) > (2m + 1)k/2 k/2 = mA(G), so (b) implies x'(G) > A(G).
7.1.28. The Petersen graph has no overfull subgraph. A subgraph H is overfull if and only if it has an odd number of vertices and has more than (n(H) 1)A(G)/2 edges. Subgraphs of order 3, 5, 7, 9 would need more than 3, 6, 9, 12 edges, respectively. Since the Petersen graph has no cycle of length less than 5, the smaller cases are excluded. For the last case, deleting a single vertex leaves a subgraph with 9 vertices and 12 edges, but 12 is not more than 12. 7.1.29. A non-i-factorable regular graph with high degree. Let G be the 1)-regular connected graph formed from 2Km by deleting an edge from each component and adding two edges between the components to restore regularity. If m is odd and greater than 3, then G is not 1-factorable. To see this, observe that the central edge cut of size 2 leaves an odd number of vertices on both sides. Hence every 1-factor in G includes an edge of this cut. If G is 1-factorable, this forces the degree to be at most 2, and hence m -c 3. (m
i-factorization Conjecture. Let G be a 7.1.30. Over full Conjecture regular simple graph of order 2m. An induced subgraph of G is overfull if and only if the subgraph induced by the remaining vertices is overfull. Let H be the subgraph induced by vertex set 8. We have 2e(H) = kn(H) (Proposition 4.1.12). [8, Overfullness for H is thus the inequality kn(H) > k(n(H) 1) (and n(H) odd), since A(G) = k, This inequality simplifies to 3, suppose that e(G) > (n_i) + 1. Thus e(G) a- 2. Since 2) = the induction hypothesis provides a Hamiltonian (a ), cycle C in G v. Since v has at most one nonneighbor in V(G) and a 1 > 3, vertex v has two consecutive neighbors on C. Hence we can enlarge C to include v and obtain a spanning cycle in G. 7.2.28. Generalization of the edge bound. a)Iff(i) = 2i2—i-f-(a—i)(a—i—1)anda >6k, then onthe intervalk - f(a/2) and complete the proof, it suffices to show that k is farther from the axis a/3 than a/2 is. This is the inequality a/3 k > a/2 a/3, which is equivalent to the hypothesis a > 6k. b) If 3(G) = k and G has at least 6k vertices and has more than +k2 edges, then G is Hamiltonian. By Chvátal's Condition, it suffices to show >-a—iforeveryi vertex degrees of G. If this condition fails for some i, we have i; this
requires i > k, since every vertex has degree at least k. Hence we may assume k a/2 k whenever k -c a/4, then G is Hamiltonian. b) If c/k > k or By part (a), it suffices to show that this condition on G implies that G' satisfies Chvátal's Condition. In G', the vertices of V are the a/2 vertices of largest degree (otherwise, G has a vertex in V with degree 0 and a vertex in X with degree a/2, which is impossible). If there is a value k (n_i) -t- 2, then G is Hamiltonian; if e(G) > ("f) + 3, then G is Hamiltonian-connected. We prove the two statements simultaneously by induction on a. The statements are vacuous for very small graphs. For a = 4, both conditions can hold; K4 e is Hamiltonian and K4 is Hamiltonian-connected. For the induction step, suppose that a > 4. For clarity, we write the conditions as e(G) 3a/2; since the number of edges is an integer, this means e(G)> F 3a/21. If a vertex has degree 0 or 1, there is no Hamiltonian path or no Hamiltonian path without it as an endpoint. If x has degree 2, then since there is no Hamiltonian path that has the neighbors of x as the endpoints (when a > 4), since the two neighbors of x appear immediately next to x in any Hamiltonian path where x is not the endpoint. b) If in is odd, then G = Cm o K2 is Hamiltonian-connected, We phrase the cases for general odd in but illustrate with C7 o K2. Express V(G) as U U W, where U = and W = ; thus G[U] = 1). We G[WI = Cm, and the remaining edges are
Chapter 7: Edges and Cycles
Case 1: Au0, w21-path. Beginthe path by zig-zagging: uc, wQ, Wi, ul
The step is from U to W on even indices and from W to U on odd indices,
thus finishing at u21_i after W2j_i. Now finish the path by traversing U
from to urn_i and W from Wrn_1 to W2J. Case 2: A u0, Begin in the same way, stopping the zig-zag at WQj. Now finish the path by traversing W from w27 to Wrn_i and U from urn_i to u21+i. •
7.2.35. Hamiltonian-connected graphs - sufficient condition.
a) A simple n-vertex graph G is Hamiltonian-connected if 8(G) > n/2. We must guarantee a Hamiltonian path from each vertex to every other; let u, v be an arbitrary pair of vertices in G, Let G' be the graph obtained from G by adding a vertex w and adding the edges wu, wv. Then G has a Hamiltonian u, v-path if and only if G' has a Hamiltonian cycle. We prove that G' has a Hamiltonian cycle. A graph is Hamiltonian if and only if its closure is Hamiltonian. The closure of G' contains a clique induced by the vertices of G, because dG(x) +
d0(y) > n(G) + 1 = n(G') when x and y are nonadjacent vertices of G. After adding all the edges on V(G), the degrees are high enough that the edges to w will also be added. Thus the closure of G' is a clique and G' is Hamiltonian, which yields the spanning u, v-path in G. b)An n-vertex graph with minimum degree n/2 that is not Hamiltonianconnected. Let consist of two cliques of order n/2 + 1 sharing an edge xy. The minimum degree is n/2, and because is a separating 2-set, there is no Hamiltonian path with endpoints x, y. Another example is Since a spanning path must alternate between the partite sets and the total number of vertices is even, there is no spanning x, y-path when x and y lie in the same partite set. 7.2.36. Las Vergnas' Condition. The condition, which implies that the nclosure is complete, is the existence of a vertex ordering vi, .. .,
there is no nonadjacent pair v1, v1 such that i n(G) > n(P) forces a neighbor of u following a neighbor of v by the usual switch argument; again we have a cycle C through V(P). By following P, we find C in linear time. The condition d(u) + d(v) > n(G) also forces diameter at most 2. If V(C) V(G), we select a vertex not on C. Either it has a neighbor on C, or it has a neighbor with a neighbor on C. Thus we find an edge from V(C) to V(G) V(C) in hnear time. This gives us a path longer than P, which we extend greedily through the new vertex. We repeat the process. Each iteration takes only linear time, and the length of P increases fewer than a times, so in quadratic time we find a spanning cycle of G. -c 7.2.39. (.) Prove that if a simple graph G has degree sequenced1 and d1 + d2 3. (Ore [1967b])
7.2.40. Every 2k-regular simple graph G on 4k + 1 vertices is Hamiltonian (using Dirac's theorem that a 2-connected simple graph has a cycle of length at least 23). To apply Dirac's theorem, we first must show that G is 2connected. Suppose G has a vertex x whose removal leaves a disconnected
graph (this includes the case where G is not connected). Let H1 be the smallest component of G x, and let H2 be another component. H1 has at most 2k vertices. If any vertex in H1 was not joined to x in G, then it still has degree 2k in G x. This is impossible, since H1 has at most 2k vertices and G is simple. So, H1 must have exactly 2k vertices; all joined to x. This means that H2 also has at most 2k vertices. The same argument requires that every vertex of H2 have x as a neighbor in G, but this assigns 4k neighbors to x. This contradiction means there could not have been such
Section 7.2: Hamiltonian Cycles
an x, and G is 2-connected.
Dirac's theorem now implies that G has a cycle C of length at least If G is not Hamiltonian, let x be the vertex not included in C, and let X and V denote the neighbors and non-neighbors of x. If X has two adjacent vertices on C, then we can visit x between them and augment C 4k.
to a Hamiltonian cycle (see first figure below). Since X has half the vertices on C, C therefore alternates between X and V. Now, if any two vertices of V are adjacent, then it is possible to form a Hamiltonian cycle as indicated in the second figure below. On the other hand, if the only neighbors of vertices in V U are the vertices in X, they must each neighbor every vertex in
X (since there are only 2k of them), and thus every vertex in V has 2k + 1 neighbors. Since this contradicts 2k-regularity, one of the possibilities mentioned above, in which G is Hamiltonian, must occur. (Note: the fact that an (a 1)12-regular graph is Hamiltonian if a 1 (mod 4) is just a slight improvement over minimum degree n/2. It has in fact been proved that an n/3-regular graph is Hamiltonian).
7.2.41. Scott Smith's Conjecture (for k 2, then G' has a spanning cycle through any specified edge t1t2, where t1 and t2 are adjacent tours. We have verified this for 1 = 2. For the induction
step, consider 1 > 3. Let v be the vertex where the reversal occurs to obtain from t1. Since A(G) = 4, we have d(v) = 4, with incident edges e0, e1, e2, e3 c E(G).
Let V' be the subset of V(G') consisting of tours in which the visit through v that uses e0 also traverses e,, for i e . Each vertex of G' lies in exactly one of these sets; call this the v-partition of G'. Let is isomorphic to the Eulerian cir= GIV]. The induced subgraph emit graph of the graph G, obtained from G by splitting v into two 2-valent vertices x, x', where the edges incident to x are , and those incident to x' are the other two edges at v. For any tour, the tour adjacent to it by
the reversal at v lies in a different set in the v-partition. Reversal at any other vertex does not change the pairing at v and thus reaches another tour in the same block of the v-partition. Therefore, the edges of G' that
Section 7.2: Hamiltonian Cycles
join two sets in form a perfect matching of G' and correspond to
reversals at v. Call these the cross-edges. If v is a cut-vertex, then because all vertex degrees are even, v has two edges to each component of G
and e3 to one component and e1 and
e2 to the other. In this case, G3 is empty and Gi G2, with corresponding vertices joined by an edge. That is G' = Gin K2. Since Gi is Hamiltonian or is a single edge, G' has a spanning cycle through any cross-edge (see Exercise 7,2.17), If v is not a cut-vertex, then each set in the v-partition is nonempty. The reason is that G v is connected, and hence the graph obtained from
by adding vertices x and x' whose neighbors are the endpoints of (eo, ej and the other neighbors of v, respectively, is connected. This is precisely the graph G,; being even and connected, it is Eulerian. (The sets V need not have the same size, as shown by letting G be the 4-regular graph consisting of K4 with an extra copy of two disjoint edges, where the G
sizes of the are 16,16,6.) By the induction hypothesis, each G, has a Hamiltonian cycle through any specified edge (or a path through the single edge, if it has two vertices).
Thus it suffices to find a cycle C in G' that contains nt2 and alternates between cross-edges and non-cross-edges, using exactly one edge within each G, (consecutive cross-edges are acceptable if G, = Ki). Using the cross-edges on C plus a Hamiltonian path of each G, joining its vertices on C yields a Hamiltonian cycle of G containing 1112
the specified edge, using a reversal at v. We may assume E G2. The vertex v cuts into two segments. Since v is not a cut-vertex, the two segments share another vertex u, which therefore has degree (at least) 4. The desired cycle C is now obtained by alternating reversals at v and u. To list the tours of C explicitly, break into four successive trails with endpoints v and U; that is, express 11 as Lv, Q, u, R, v, 8, u, TI, in the sense that A starts at v and ends at u, etc. We may further assume that Q starts with el, R ends with e2, and T ends with so that c Gi and 12 c G2. Let 7k, T denote the reversals of these trails. For the six successive tours Let 1112 be
ti=[v,Q,u,R,v,8,u,T]eGi t4=[v,8,u,R,v,Q,u,T]eG3 =[v,
8, v, Q, ii, TI e G2
= Ri, Q, u, 8, v, R, u, TI e Gi
With this approach, the construction of the desired Hamiltonian cycle is easy. The approach also works for the general case without limits on A(G). For the general problem, Zhang and Guo [19861 use three cases like this when d(v) = 6 and two cases when d(v) = 2t > 6.
Chapter 7: Edges and Cycles
7.2.43. For a graph G, the Eulerian circuit graph G' of Exercise 7.2.42 is
2))-regular, which is not enough to apply general results on Hami ltoni city of regular graphs. The formula for the degree is obtained in the first paragraph of the solution to Exercise 7.2.42. For a given Eulerian orientation, Theorem 2.2.28 computes the number of Eulerian circuits as c (d (v) /2 1)!, where c is the number of in-trees or out-trees from any
vertex. Already this number is very much bigger than the degree, and in addition there are many Eulerian orientations. Summing over all the orientations and dividing by 2 counts the vertices in G'. Hence n(G') is hugely bigger than the degree, not bounded by a factor of 2 or 3 times the degree, which would be needed to apply general sufficiency conditions for Hamiltonian cycle. This explains why a specialized structural argument is needed in Exercise 7.2.42.
7.2.44. Every tournament has a Hamiltonian path. Proof 1. If a directed path P of maximum length omits x, then u —> x v, where u and v are the origin and terminus of P. Considering the vertices of P in order, there must therefore be a consecutive pair y, z on P such that y —÷ x z. This detour absorbs x to form a longer path. Hence a path of maximum length in a tournament omits no vertex. Proof 2. The result follows immediately from the Gallai-Roy Theorem, since x (K,,) = n and every tournament is an orientation of K. 7.2.45. Strong tournaments are Hamiltonian. We prove first that a vertex on a k-cycle is also on a (k + 1)-cycle, if k n/2. Take two sets A and B of size n/2. Add edges A o A and B riB (hence there is a loop at each vertex and opposed edges joining each pair in one set), and add a matching from A to B. Each vertex has indegree and outdegree n/2 within its own set, but the fill digraph is not strongly connected.
7.2.49. Ghouild-Houri's Theorem implies Dirac's Theorem for Hamiltonian cycles. Suppose that a simple graph G satisfies Dirac's Condition 3(G) > n(G)/2. From G we form a digraph D be replacing each edge with a pair of oppositely directed edges having the same endpoints. Thus
Chapter 7: Edges and Cycles
= d5(x) = for all x c V(G). Since n(D) = n(G), we obtain min <6tD), ?r(D)>= 6(G) > n(G)/2 = n(D)/2. Hence Ghouilà-Houri's The-
orem implies that D is Hamiltonian. Since a Hamiltonian cycle C in D does not use two oppositely directed edges from G, the edges of G giving rise to the edges in C also form a Hamiltoruian cycle in G.
7.3. PLANARITY, COLORING, & CYCLES 7.3.1. Every Hamiltonian 3-regular graph has a Tait coloring. A 3-regular graph has even order, so two colors can alternate along a Hainiltonian cycle C. Deleting E(C) leaves a 1-factor to receive the third color.
7.3.2. Examples of 3-regular simple graphs: a) planar but not 3-edgecolorable. b) 2-connected but not 3-edge-colorable. c) planar with connectivity 2, but not Hamiltonian. For part (b), the Petersen graph is an example. For (a) and (c), suitable graphs appear below. Regular graphs with cut-vertices are not 1-factorable, and graphs having 2-cuts that leave 3 components are not Hamiltonian.
7.3.3. Every maximal plane graph other than K4 is 3-face-colorable. With fewer than four vertices, the maximal plane graphs have fewer than three faces. For larger graphs, every face is a triangle, so the dual is 3-regular. Since the dual is planar, it does not contain K5, Hence the dual is not a complete graph, and by Brooks' Theorem it is 3-colorable. This becomes a proper 3-coloring of the original graph.
7.3.4. Every Hamiltonian plane graph G is 4-face-colorable. It suffices to show that the faces inside C can be properly 2-colored, since the same argument apphes to the faces outside C using two other colors. View the union of C and the edges embedded inside C as an outerplane graph H; all the vertices are on the outer face. Proof 1. In the dual H*, the bounded faces in H become vertices. We
clalm that the subgraph of H* induced by these vertices is a tree P. If
Section 7.3: Planarity, Coloring, and Cycles
they induce a cycle, then that cycle lies inside C in the embedding of G and encloses a face of H*, which in turn contains a vertex of H. This is a vertex of G that does not lie on the outer face of H, which contradicts C being a spanning cycle.
Proof 2. We properly 2-color the faces inside C using induction on the number of edges inside. With no such edges, H has one bounded face and is 1-colorable. Otherwise, let e be an inside edge whose endpoints are as close together as possible on C. By the choice of e, there is a face whose boundary consists of e and edges of C. This face F is adjacent to only one other, F'. Deleting e merges F into F' in a smaller graph H'. By the induction hypothesis, H' has a proper 2-face-coloring f'. To obtain the proper 2-face-coloring f of H, let f give the same color as f' for each face other than F and give F the opposite color from f(F'). 7.3.5. A 2-edge-connected plane graph is 2-face-colorable if and only if it is
Eulerian. Let G be a 2-edge-connected plane graph; note that (G*)* = We have G 2-face-colorable Wand only WG* is bipartite, which by (G*)* =
and Theorem 6.1.16 is equivalent to G being Eulerian. 7.3.6. The graph below is 3-edge-colorable. By Tait's Theorem, it suffices to show that the graph is 4-face-colorable.
7.3.7. Let G be a plane triangulation. a) The dual G" has a 2-factor The dual of a plane triangulation is 3regular and has no cut-edge (since G has no loop). Hence the dual has a 1-factor, by Petersen's Theorem (Corollary 3.3.8). Deleting the 1-factor leaves a 2-factor. b) The vertices of G can be 2-colored so that every face has vertices of both colors. Given the 2-factor F of G* resulting from part (a), we can 2-
color the faces of the dual by given each face the parity of the number of cycles in F that contain it. This assigns colors to the vertices of G, which correspond to the faces of Gt. Each face of G correspond to a vertex v of G*, with degree 3. The 2factor F uses two edges at v, lying on one cycle of F. Hence each face of G is entered by one cycle of F. This cycle cuts one of the vertices of F from the other two, and hence the face has vertices of both colors.
Chapter 7: Edges and Cycles
7.3.8. The icosahedron is Class 1, The graph is 5-regular; we describe a proper 5-edge-coloring. Show in bold is a 2-factor consisting of even cycles;
on this we use two colors. For the remaining three colors, we color by the angle in the picture. Color 0 goes on the six edges that are vertical or horizontal. Colors 1 and 2 go on the edges obtained by rotating this 1-factor by 120 or 240 degrees in the picture.
7.3.9. Every proper 4-coloring of the icosahedron uses each color exactly 3 times. The icosahedron has 12 vertices; it suffices to show that it has no independent set of size 4. In the figure above, an independent set takes at most one vertex from the inner triangle, one vertex from the outer triangle, and at most three from the 6-cycle C between them. If it takes three vertices from C, then they alternate on C and include neighbors of all other vertices Two opposite vertices on C also kill off the rest. Two vertices at distance 2 along C kill off one triangle but leave one vertex on the other triangle that can be added. 7.3.10. By Whitney's result that every 4-connected planar triangulation is Hamiltonian, the Four Color Problem reduces to showing that every Hamil-
tonian planar graph is 4-colorable. The Four Color Problem reduces to showing that triangulations are 4-colorable. Let S be a minimal separating set in a triangulation G; we show first that S of G S. Since there is no edge joining two components of G S and every face is a triangle, in the embedding of G edges must emerge from x between edges to different components of
G S. These edges go to other vertices of S. Hence G[S] has minimum degree at least two, and 3. If = 3, then 6(G[S]) > 2 implies that S is a clique. Hence a proper 4-coloring of each S-lobe of G uses distinct colors on 5, and we can permute the names of the colors to agree on S. This yields a proper 4-coloring of G. Hence a minimal planar triangulation that is not 4-colorable must be 4connected. Since every such graph is Hamiltonian, it suffices to show that Hamiltonian planar graphs are 4-colorable.
Section 7.3: Planarity, Coloring, and Cycles
7.3.11. Highly connected planar graphs. The icosahedron is 5-connected. The symmetry of the solid icosahedron is such that two vertices at distance d in the graph can be mapped into any other pair at distance d by rotating the solid. Hence it suffices to consider one pair at distance d, for each d, and show that they are connected by five pairwise internally disjoint paths. This can be done on any drawing of the graph. Since every planar graph has a vertex of degree at most 5, there is no 6-connected planar graph. 7.3.12. A plane triangulation has a vertex partition into two sets inducing forests if and only if the dual is Hamiltonian. Every plane triangulation F is connected, so (F*)* = F. Let G = F*, Let G be a Hamiltonian plane graph. A spanning cycle C is embedded as a closed curve, and the subgraph H of G consisting of C and all edges drawn inside C is outerplanar. In the dual of an outerplane graph H, every cycle contains the vertex for the outer face, since every cycle in H* encloses a vertex, and thus a cycle in H* not including the vertex for the outer face in H would yield a vertex of H not on the outer face. We conclude that in Gt, the vertices for faces of H induce a forest. The same argument applies to the graph consisting of C and the edges of G drawn outside G. Conversely, let F be a plane triangulation with such a vertex partition. Since F is connected, there exist edges joining components in the union of these two forests. We add edges joining components, possibly changing the vertex partition while doing this, until we obtain a vertex partition into two sets 5', 5' inducing trees. Adding any edge from 5' to 5' yields a spanning tree of G, 50 [8, 5'] is a bond. Hence the duals of the edges in [5', 5'] form a cycle. We claim that = f. this is a spanning cycle in the dual. It suffices to show that [8, Since F has 3n 6 edges and we use a 2 edges in the two trees, we have 2n 4 edges from S to 8. By Euler's Formula, this is indeed the number of faces in a triangulation. 7.3.13. Grinberg's Theorem. Neither of the graphs below is Hamiltonian. Grinberg's Theorem requires )ji = 0, where and are the number of i-faces inside and outside the Hamiltonian cycle. The plane graph on the left has six 4-faces and one 8-face. Since must be a multiple of 4 and 6(1 1) cannot be a multiple of 4, there is no way these can sum to 0. Similarly, redrawing the graph on the right yields a plane graph with three 4-faces and six 6-faces, This lime cannot be a multiple of 4, but must be; again they cannot sum to 0.
Chapter 7: Edges and Cycles
7.3.14. A non-Hamiltonian graph. In any spanning cycle of the graph
below, both edges incident to a vertex of degree 2 must appear. Applying this to the vertices of degree 2 on the outside face generates a non-spanning cycle that must appear. Irrelevance and relevance of Grinberg's Theorem. This plane graph has four 5-faces, three 6-faces, and one 14-face. It is possible to choose = 3, f4 + nonnegative integers f' and 111 such that + f5" = 4, + = 1, and = = = = 3, 2, 2)07 f/') 0. This is achieved by f5" = 1, and f6" = = 0. Hence the graph does not violate the numerical conditions of Grinberg's Theorem. On the other hand, since the four long horizontal edges in the drawing are incident to vertices of degree 2 and therefore must appear in any Hamiltonian cycle, subdividing them once each does not affect whether the graph is Hamiltonian. The new plane graph has seven 6-faces and one 18face. Since the difference of two numbers summing to 7 is odd, Grinberg's Condition now requires an odd multiple of 4 to equal an even multiple of 4, which is impossible. Hence the graph is not Hamiltonian.
:;i;:;: 7.3.15. Proof of Grinberg's Theorem from Euler's Formula, Let C be a Hamiltonian cycle in a plane graph G, and let J7 be the number of faces of length i inside C. It suffices to prove that (i 2) /7 = a 2, since the same argument applies to the regions outside the cycle. We apply Euler's Formula to the outerplanar graph G' formed by C and the chords inside it.
Section 7.3: Planarity, Coloring, and Cycles
We can rewrite the desired formula as 2 = a Xi if,' + 2 Xi f,'. Note if,' counts every internal edge of G' twice and every edge on the cycle once. Thus Xi if,' = 2e a. Also, 1, the total number of f,' = bounded faces in G'. Thus we want to prove that 4 = 2n 2e + 2f, which follows immediately from Euler's Formula. 7.3.16. The Grinberg graph is not Hamiltonian. In the plane graph below, all faces have length 5, except for three of length S and the one unbounded face of length 9. If it is Hamiltonian and f,', f," denote the number of faces of length i inside and outside the cycle, respectively, then Grinberg's Condition requires that + + = 0. This can happen only when is divisible by 3, which is impossible since there is exactly one face of length 9.
7.3.17. The smallest known 3-regular p1 anar non-Hamiltonian graph. The triangular portion on both ends is the subgraph of the Tutte graph called H. Since it has three entrance points here, it must be traversed by a spanning path connecting the entrance points. Example 7.3.6 in the text shows that no such path exists joining the top and bottom entrances. Hence edges a'b' and oh must be used. By symmetry, we may assume that t'c is used. If cli' is used, then completion of a cycle will miss d or the portion on the top. Hence ccl is used. Since each copy of H can be visited only once, de must be used. Now the cycle must traverse the left copy of H,
emerge at Ii', and turn up to f. On the other end, the cycle exits the right copy of H at g. Now the cycle cannot be completed without missing one of the common neighbors of f and g.
Chapter 7: Edges and Cycles
7.3.18. A Hamiltonian path between opposite corners of a grid splits the squares of the grid into two sets of equal size. Suppose Q is a Hamiltonian path from the upper-leftmost vertex to the lower-rightmost vertex of PmDPn. Adding an edge through the unbounded face from the upper-leftmost vertex
to the lower-rightmost vertex completes a Hamiltonian cycle. Each face containing the added edge has length m + a 1, and they are on opposite sides of the cycle. By Grinberg's Theorem, then, the number of 4-faces inside the cycle must equal the number of 4-faces outside the cycle. One of these measures the area of the regions escaping to the top and right, and the other measures the area of the regions escaping to the bottom and left.
7.3.19. The generalized Petersen graph P(n, 1 3. Thus we may assume that G is simple. It remains only to consider a nontrivial 2-edge cut (we have eliminated the case where e, e' share a vertex of degree 2). The bridgeless graphs are those where every two vertices lie in a common circuit, and contract-
ing an edge of such a graph with at least three vertices does not destroy this property. Thus we may assume that G e' has a nowhere-zero k-flow
Section 7.3: Planarity, Coloring, and Cycles
(D, f). Let 5, T be the vertex sets of the components of G , and let w be the vertex of G - e' obtained by contracting e'. We may assume that e is oriented from S to T in D, Let m = f(e). Because f*(5 U w) = 0, the edges between w and T contribute —m to 7' (w). Similarly, the edges between S and w contribute m to 7(w). Thus we let f(e') = m, oriented from T to 5, to obtain a nowherezero k-flow on G.
7.3.27. Every Hamiltonian graph G has a nowhere-zero 4-flow. Since G has a nowhere-zero 4-flow if and only if it is the uthon of 2 even subgraphs (Theorem 7.3.25), we express G in this way. The Hamiltonian cycle C is
one such subgraph. Let P be a spanning path obtained by omitting one edge of C. For each e c G E(C), let C(e) be the cycle created by adding e to P. Each edge outside C appears in exactly one of these cycles. Let C' be the spanning subgraph whose edge set consists of all edges appearing in an odd number of the cycles . Since C' is a binary sum of even graphs, it is an even graph. It also contains E(G) E(C). 7.3.28. Every bridgeless graph G with a Hamiltonian path has a nowherezero 5-flow. If G is Hamiltonian, then G is 4-flowable (Exercise 7.3.27). Otherwise, let G' be the graph obtained from G by adding the edge e joining the endpoints of a spanning path in G. We claim that G' has a nowhere-zero 4-flow with weight 1 on e. Let
C be a spanning cycle in G' through e, with vertices vi, . in order starting and ending at the endpoints of e. The remaining edges are chords of C; let there be m of them. Let u1, . U2m be the endpoings of chords of C, in order on C, listed with multiplicity (a vertex may be an endpoint of many chords. Let C' be the subgraph of G consisting of the chords of C together with the u2,_i, u2,-path on C not containing e, for 1 -c is the degree sequence of a simple graph G, and m is the largest value of k such that dk Ic 1, then G is a split graph if and only if ti = m(m 1) + L=m+i d,. The Erdôs-Gallai condition
says that d1 > --- >- ci,, are the vertex degrees of such a graph if and only if k. Since by (2) G has no k + 2-clique, we have -c k. Hence d(v) = k. To complete the proof, we must show that deleting a simplicial vertex of degree k does not destroy the conditions, so we can complete a "k-valent" simplicial
elimination sequence by applying induction. Deleting a simplicial vertex
Chapter 8: Additional Topics
does not disconnect a graph or create a k + 2-clique, and if G is not a clique, then (3) implies that G v retains ak-clique. To prove that (3) is preserved, if a minimal vertex separator of G v is a minimal vertex separator of G, then it induces a k-clique. We claim that every minimal x, y-separator of an induced subgraph of
a graph is contained in a minimal x, y-separator of the fill graph. If so, then a minimal x, y-separator of G v that is not a k-clique must be part of a minimal x, y-separator of G that contains v, which is impossible since no simplicial vertex belongs to a minimal vertex separator. To prove the claim, suppose S is a minimal x, y-separator in an induced subgraph H of G, so S U (V(G) V(H)) separates x and y in G. Hence this set contains a minimal x, y-separator of G, but such a separator must include all of 5, else we retain an x, y-path from H. 8.1.2 1. An n-vertex chordal graph with no (k + 2)-clique has at most kn (k-I-i) edges, with equality if and only if it is a k-tree. This is the special case of Exercise 8.1.23 obtained by setting r = k + 1. 8.1.22. The number of k-trees with vertex set En] is ()[k(n k) + 1]n—k—2 We show that the number of rooted k-trees with vertex set En] that have a fixed set of k vertices as a root clique is Ek(n k) + To obtain the formula from this, note that every k tree has 1 + k(n k) k-cliques, beginning with a root and adding k each time a new vertex is grown from an old k-clique. On the other hand, there are () ways to pick a set of k vertices to form a root clique; hence we multiply by () and divide by [k(n k) + 1] to obtain the final formula. Note that when a = k there is oniy one k-tree, which agrees with the formula, so henceforth we may assume a > k. To count the k-trees with label set En] and a fixed root R C [n], we put them in 1-1 correspondence with lists of length a k 1 chosen from a fixed alphabet of size 1 + k(n k). The alphabet consists of 0, which refers to the root, together with pairs (v, i) such that v c (En] R) and i c [k]. Since
a > k, every vertex belongs to a k + 1-clique; when we deal with rooted k-trees, the leaves are the vertices not in the root that belong to only one k + 1-clique. Given a k-tree with root R, we form a list by iteratively deleting
the leaf u with the least label and recording an appropriate member of the alphabet. If N(u) = R, we record 0. If N(u) R, we want to record some other code in the alphabet that will enable us to recover the k-clique to which u is joined in growing the current tree from the root.
In growing the current tree from the root, there is a unique list of vertex additions that leads from the root to u (ignoring other additions not needed to reach u). When N(u) R, there is a last non-root vertex v before
u on this list; let this be the vertex part of the code recorded. When we add u, N(u) consists of v together with all but one vertex of the k-clique to
Section 8.1: Perfect Graphs
which v was connected when added. Let the index part of the code recorded be the position among this list of k of the vertex not in N(u). After a k 1 iterations, there remains one non-root vertex joined to the root.
This defines a unique list from each k-tree. To reconstruct from any list on these labels the unique k-tree that generates it, and thereby show that the map is a bijection, there are two phases. In the first phase, at each iteration select the least non-root label u that has not yet been marked finished. If the current code is 0, create edges from u to R. If the code is a vertex-index pair, create an edge from u to the vertex v that is the vertex part of the code, Mark u finished. After a k 1 iterations, there remains one unfinished non-root vertex; join it to R. The first phase produces a "skeleton" describing possible ways to grow the k-tree from the root. If we shrink the root to a single node, this is in fact a rooted tree that, for each non-root vertex, describes by its path to the root the list of vertices that must be added before it is added. The second phase fleshes out this skeleton. Moving outward from the root as the construction
procedure would, we iteratively "expand" a non-root vertex u such that every previous vertex on the path to the root has already been expanded; this expansion creates the other edges formed when the k-tree is grown to u. Let u be a vertex whose deletion generated a non-root code (v, 1). When we expand u, the vertex v is the last vertex on the path to it from the root and has already been expanded, which means that we know the set of vertices S to which v was joined when the tree grew to it. The code i tells us which element of S should not be joined to u. This two-phase procedure generates a unique k-tree from every list, and the k-tree generated from a list r yields r under the first procedure, so this is a bijection.
8.1.23. An n-vertex chordal graph G with clique number r has at most (a r) cliques of order j, with equality (for all j simultaneously) () + if and only if G is an r
We use induction on a. The formula holds
for a = r. For a > r, let v be the first vertex to be deleted in a simplicial elimination order. Since v has at most r 1 neighbors, it is involved in at most cliques of order j. The f-cliques not containing v are bounded by the induction hypothesis. Furthermore, equality holds if and only if it holds for G v and adding v adds cliques of order j, which by the inductive hypothesis implies that G is an r 1-tree. 8.1.24. Pairwise intersecting real intervals have a common point. Let a be the rightmost left endpoint among these intervals, and let b be the leftmost right endpoint. If some right endpoint occurs before some left endpoint, then those two intervals do not intersect. Hence a -c Ii. For every interval,
its left endpoint is at most a, and its right endpoint is at least b. Hence
Chapter 8: Additional Topics
every interval in the family contains the interval [a, hi, which we have shown is nonempty.
8.1.25. A tree is an interval graph if and only if it is a caterpillar We prove the following equivalent for a tree G. A) G is an interval graph. B) G is a caterpillar. C) G does not contain the tree Y formed by subdividing each edge of a claw. B A. Create an interval for each vertex on the spine of the caterpillar, such that each interval intersects its the intervals for its neighbors on the spine and no others. This leaves part of each interval intersecting no other. Place small intervals for the leaf neighbors of each vertex x of the spine in the "displayed" area of the interval for x. C B. A longest path P contains an endpoint of every edge. If some edge is missed, then there is an edge with neither endpoint on P but having a neighbor x on P (since the tree is connected). Since P is a longest path, P continues at least two edges in each direction from x. Now the tree contains Y, consisting of these six edges within distince 2 of x. A C. If G contains Y but is an interval graph, then in an interval representation of G the intervals for the leaves of Y are pairwise disjoint. Name the leaves x, y, z in the order of the corresponding intervals, from left to right. The union of the intervals for the x, z-path in G must cover the gap between the intervals for x and z in the representation. Since this gap contains the interval for y, we obtain a contradiction, because y has no neighbor on this path. 8.1.26. Every interval graph is a chordal graph and is the complement of a comparability graph. If it is not a chordal graph, then it has a chordless cycle. A chordless cycle has no interval representation, because the two paths along the cycle between the vertices corresponding to the leftmost and rightmost intervals among these vertices must occupy all the space between them on the line, which produces chords between the two paths
when the intersections are taken. Hence the full graph has no interval representation. — Given an interval representation of a graph G, orienting G by x —f y if the interval for x is completely to the right of the interval for y expresses G as a comparability graph.
8.1.27. A graph G has an interval representation if and only if the cliquevertex incidence matrix of G has the consecutive is property. Necessity. From an interval representation, we obtain a natural ordering of the maximal cliques. By the Helly property (Exercise 8.1.24) the intervals corresponding to the vertices of a maximal clique have a common point. These points are different for distinct maximal cliques, because the
Section 8.1: Perfect Graphs
interval for a vertex nonadjacent to some vertex of a maximal clique must be disjoint from the interval for that vertex. Therefore, we can place the cliques in a linear order by the order of the chosen points. Using this ordering on the clique-vertex incidence matrix exhibits the consecutive is property, because the interval for a vertex extends from the first chosen point for a clique containing it to the last. The vertex belongs to all maximal cliques whose chosen cliques are between these, and it belongs to no other maximal cliques, since intervals have no gaps. Sufficiency. Let M be the clique-vertex incidence matrix of G, and suppose that M has the consecutive is property. We construct an interval
representation. Permute the rows of M so the is are consecutive in the columns. On a line, select points in order left to right corresponding to the rows of M. For each column of M (vertex of G), an interval from
the point for the first i in it to the point for the last i in it. This defines one interval for each vertex because the is are consecutive. It yields an interval representation of G because vertices are adjacent if and only if there is a maximal clique that contains both of them. 8.1.28. A graph is an interval graph if and only if it has a vertex ordering such that the neighborhood of each Vk among the lower-indexed vertices is a terminal segment v. Vk_1. Given an interval representation f, index the vertices in order of the right endpoints of the corresponding intervals. If Uk -e- v, with i p, then q p is even, else wi. Uk, Wq and the wi,, we-portion of P2 contradict minimahty. We have now partitioned P2 into three subpaths, of which the middle path has even length and the two extreme paths have the same parity; this is impossible and implies that Uk belongs to no chords. Now consider Ic = 1. As before, the y-paths yield t + 1 q even, P2
and when q > 1 the x, we-paths yield q even. This is impossible, since -f- 1 is odd. We conclude that is the only possible chord involving Vt. Similarly, V1Wt is the only possible chord involving We have proved that v1w1 and are the only possible chords of C, and they do not cross; this
contradicts the hypothesis.
8.1.37. Every perfectly orderable graph is strongly perfect. Let G be a perfectly orderable graph and L an admissible ordering of G. I.e., G has no induced P4 such that in L each endpoint appears before its neighbor. Let 5' be the greedy stable set with respect to L, i.e., place the first vertex of L in 5', delete its neighbors, and iterate this step with the remaining vertices. Note that 5' is the set receiving color 1 under the greedy coloring for L. We show that 5' meets every maximal chque. If 5' misses a maximal
clique Q, then each vertex of Q must be deleted from the ordering due to having a prior neighbor that is in 5'. If all vertices of Q share a prior neighbor in 5', then Q is not maximal. Hence we can choose x, y c Q and U, V 0 5' such that u -c-* X, V -c-* y, but u
x. Since x -c-* y and u
these vertices induce P4; since u comes before x and V before y, they induce an obstruction, contradicting the assumption that L is admissible.
8.1.38. The graphs below are strongly perfect. In each case, the marked stable set intersects all maximal cliques, but strong perfection also requires this for all induced subgraphs.
Section 8.1: Perfect Graphs
For G1, an induced subgraph that omits a vertex of the central triangle is bipartite. Every bipartite graph is strongly perfect, because we can form a stable set intersecting all maximal cliques by taking one partite set from each nontrivial component plus all isolated vertices, and the family is hereditary This takes care of all induced subgraphs of G1 except those that retain the central triangle. For such a subgraph H, deleting an edge of the triangle yields a bipartite graph H' in which the three central vertices are in the same component. From this component of H', we choose the partite set containing only one vertex of the triangle in H; from others we take either partite set. The resulting set is stable in H' and intersects all maximal cliques in H', and it has the same properties in H. For G2, suppose that some induced subgraph H has a maximal clique Q avoiding the marked stable set S. This requires H to omit a vertex of S on a triangle. We may assume that Q is the lower horizontal edge. Now a "rotation" of S around the triangles intersects all maximal cliques in H unless H omits both of the top vertices. Now H C P4 + P2, but every disjoint union of paths has the desired property. The graphs above are not perfectly orderable. A perfectly orderable graph has an orientation (associated with a perfect ordering) such that no induced P4 has its pendant edges oriented outward. We show that these graphs have no such orientation; suppose that one exists. For G1, two of the cut-edges must be oriented in toward the triangle. Let yz be the oriented edge joining them, with xy being the entering cut-edge at its tail. The edges in a matching of size 3 on the 6-cycle containing z must be consistently oriented along the cycle, but one choice of this orientation conflicts with xy, and the other choice conflicts with yz. For G2, in the top half of the drawing, two of the three vertical edges must be oriented upward to avoid completing an obstruction with the top triangle. By symmetry, we may assume that these are the left and right vertical edges, but now either orientation of the horizonal edge on the bottom completes an obstruction with one of them.
Chapter 8: Additional Topics
8.1.39. The graphs in Exercise 8.1.38 are a Meyniel graph but are not perfectly orderable. The graphs have no chordless odd cycle (of length at least 5), so they vacuously satisQy the definition of a Meyniel graph. The task of showing they are not perfectly orderable is done in Exercise 8.1.38. The graph is perfectly orderable but is not a Meyniel graph. The graph P5 is the "house", a 5-cycle with one chord, so the cycle does not have the requisite two chords. There are two induces 4-vertex paths (each containing one endpoint of the cycle. If the cycle is numbered vi, v2, v4, 1)5 in order so that the chord is then both copies of P4 have one endpoint at so the associated orientation directs that pendant edge in toward the center, and there is no obstruction. Hence this is a perfect ordering.
8.1.40. Every chordal graph is weakly chordal. If a graph has no chordless cycle, then it has no chordless cycle of length at least five, Suppose in order induce in G a chordless cycle, meaning that G contains an antihole on these vertices. If k = 5, then is a chordless 5-cycle in G. If k > 6, then is a chordless 4-cycle in G. The graph H below is weakly chordal. Any cycle with more than four vertices has at least three in the central clique Q and hence has a chord. In H, we need only forbid induced for k > 6, since C5 = C5. Note that H has 16 edges (too many for C8), of which 3 are incident to each vertex of Q and 5 to each of the other vertices. Hence every 7-vertex subgraph has at least 11 edges. The 6-vertex induced sub graphs of H with only 6 edges are those where the deleted vertices are neighboring vertices of degree 2 in H (deleting 10 edges from H), but such a subgraph of H is a 4-cycle with two pendant edges. H is not strongly perfect. Proof 1. Since V(H) is covered by three disjoint cliques, a(H) -c 3.
However, each vertex appears in two maximal cliques, so three vertices cannot meet ali 7 maximal cliques. Proof 2. There are 7 maximal cliques in H: one 4-clique and 6 edges. In a chorditess path of three edges, a stable set meeting every maximal clique must contain at least two vertices, including at least one endpoint. Hence if a stable set S meets every maximal clique, the paths on the left and right force S to contain two vertices of the central clique.
8.1.41. SPOC Skew Partition Conjecture Star-C utset Lemma. The Skew Partition Conjecture states that no p-critical graph has a skew parti-
Section 8.1: Perfect Graphs
tion (a skew partition of G is a partition of V(G) into nonempty sets X and V such that G[X] is disconnected and G[Y] is disconnected).
The SPGC states that every p-critical graph is an odd cycle or the complement of an odd cycle. Since a skew partition of G is also a skew partition of G, we obtain the Skew Partition Conjecture from the SPGC by showing that an odd cycle has no skew partition. A skew partition requires X to use more than one segment along the cycle, but then the subgraph of the complement induced by the remaining vertices is connected. To prove that the Skew Partition Conjecture implies the Star-Cutset Lemma, which states that no p-critical graph has a star-cutset, it suffices to show that a graph with a star-cutset has a skew partition. If C is a starcutset in G, let X = V(G) C and V = C. Now G[X] and G[V] are both disconnected, since the dominating vertex in C becomes an isolated] vertex in G[Y].
8.1.42. The graph below is 3, 3-partitionable. Due to the horizontal symmetry through the vertical axis, we need only check six classes of vertices to show that each V(G x) partitions into three 3-cliques and into three stable 3-sets. This is easy but tedious and seems to require a picture for each vertex. Alternatively, by Theorem 8.1.39, since a(G) = co(G) = 3, it suffices to show that (1) each vertex belongs to three 3-cliques and to 3 stable 3sets, and (2) G has 10 3-cliques and 10 stable 3-sets, paired so that each intersects every set of the other type except its mate. We show this giving a matrix that lists the 3-cliques and stable 3-sets in the rows and columns and has the elements of 1) '1y as the entries. Each vertex appears in three of the row labels and three of the column labels. However, the matrix does not contain a proof that there are no other cliques or stable sets of size 3. Curiously, the maximum cliques and stable sets are the same as in except for a switch of membership in two cliques and two stable sets, underlined below.
672 789 890 901
Chapter 8: Additional Topics
8.1.43. If x and v are nonadjacent vertices in apartitionable graph G, then every maximum clique containing x consists of one vertex from each stable set that is the mate of a clique containing v. (The complementary assertion
is that if x and v are adjacent vertices, then every maximum stable set containing x consists of one vertex from each clique that is the mate of a stable set containing v.) By Theorem 8.1.41, the unique minimum coloring of G v consists of the co(G) stable sets that are mates of the maximum cliques containing v. Since x and v are nonadjacent, a maximum clique containing x omits v and hence must contain exactly one vertex from each stable set in this coloring.
8.1.44. No p-critical graph has antitwins. Antitwins are a pair of vertices such that every vertex outside them is adjacent to exactly one of them. Consider a p-critical graph G, and let co = co(G) and a = a(G). We first prove that a p-critical graph with antitwins has a clique of size co 1 in N(x) that doesn't extend into N(y). Recall that co(G 5) = co(G) for any stable set S in a p-critical G (reminder of proof - since G S is perfect, smaller clique-number would give a smaller coloring, extending to an co-coloring of G by replacing 5). Since G is partitionable, G x has a unique coloring by co stable sets of size a; let S be the stable set containing y in this coloring, and let Q be an co-clique in G S. Since G x S is co 1-colorable (S is a color in the co-coloring of G 9, Q must contain x. Since G x S has no co-clique, Q' = Q x c N(x) is the desired clique. Reversing the roles of x and y yields a similar co 1-clique in N(y). Since the complement of a p-critical graph is p-critical, we also can apply the same argument to obtain a 1-cliques in and that trans-
lates into the desired a
sets in N(y) and N(x). Let 5' be the
resulting stable set of size a 1 in N(y) that doesn't extend in N(x) U N(y). Choose u to be the vertex of Q' with the minimum number of neighbors in 5'; u must have at least one neighbor v in 5', else 5' extends to u. Similarly, v must have a non-neighbor z in Q'. Since v e N(u) N(z) and z has at least as many neighbors as u in 5', z must have a neighbor w in 5' that is not adjacent to u. Now y, v, u, z, w induce a chordless 5-cycle in G. This misses x, so G is not p-critical. (Note: For the special non-circulant partitionable graph pictured in the text, which is not p-critical, the top and bottom vertices are antitwins.)
8.1.45. Stable sets and even pairs in partitionable graphs. a) If Si, S2 are maximum stable sets in a partitionable graph G, then is connected. Let S = S1zS2. Let R1 be the vertex set of a component of G[S], and let 1?2 = S R1. The sets T1 = (Si R1) U (82 R2) and = (Si I?2) U (S2 R1) are stable sets with the same union and intersection as Si and S2 (see figure). Hence = 2a(G), which implies +
Section 8.1: Perfect Graphs
= = a(G) since each has size at most a(G). Since the rows of the incidence matrix between maximum stable sets and vertices are linearly independent, we cannot have two pairs of stable sets with the same union and intersection. Either Ti = Si and T2 = S2, which yields the contradiction R1 = 0, or T1 = S2 and T2 = Si, in which case R1 = S and G[S] is connected.
b) No partitionable graph (and hence no p-critical graph) has an even pair. Let x, y be any two vertices in a partitionable graph G. Let S be a maximum stable set containing x in G y, and let T be a maximum stable set containing y in G x. Let H = GIISAT]. By part (a), H is connected. Since S and T are stable sets, H is bipartite, with partite sets S T and T — S. By construction, x E S — T and y E T — 5, so every x, y-path in H has odd length. Since H is an induced subgraph of G, a shortest x, y-path in H is a chordless x, y-path in G. Hence x, y is not an even pair in G. 8.1.46. If G is partitionable, and Si, 52 are stable sets in the optimal coloring of G x, then G[51 U U ] is 2-connected. Since Si, S2 are maximal, x is adjacent to a vertex of each. Since Si, S2 are disjoint, Si e S2 = U S2. Thus part (a) of the preceding problem implies that H = is con-
nected and that x cannot be a cut-vertex of H. If H has a cut-vertex, we may assume it is s c Let be a component of H s not containing x, and let G2 be the rest of H s, with V2 = V(G3, Recall (*): whenever v, x are nonadjacent vertices of a partitionable graph G, any maximum clique containing v omits x and therefore consists of one vertex from each stable set in the unique minimum coloring of G x. Since x has no neighbor in Vi, we can apply (21 to any v E Vi. I.e., each clique in x) that contains a vertex of Gi must contain exactly one vertex of each of Si, Both these vertices must be in G i, else we introduce an edge between G i and G2. Thus Vi has an equal number of vertices from Si and S2, both equal to the number of cliques in 8(G x) that meet Gi, Choose u E fl To any clique Q of ®(G u), we can apply (*) again to guarantee that Q contains one vertex each of Si, 52. In particular, for each v c Si U Vi, there is a clique of u) containing it, and this yields
a vertex v' e 52 P Vi
adjacent to it. Since Si is stable, these cliques
Chapter 8: Additional Topics
are disjoint, and so the vertices are distinct. This implies V1 contradicting the result of the previous n
paragraph. 8.1.47. The graph G below is a circular-arc graph but not a circle graph. To represent G as a circular-arc graph, we let the arcs for the inner cycle and the outer cycle in the drawing each cover the circle, More precisely, consider a circle of circumference 9, with points on the circle described by numbers modulo 9. Assign arcs as in the middle table below to form a circular-arc representation. a U
To show that G is not a circle graph, suppose that G has an intersection representation by chords in a circle. The chords for are pairwise intersecting, so their endpoints occur in the order a, b, c, a, b, c on the circle.
The chord for v cannot cross the chord for c, so to intersect the chords for a and b the endpoints for v must precede an a and follow the subsequent b, yielding a, b, c, v, a, b, v, c. We make the analogous argument for x and for z. However, tv, x, z> is independent, so the endpoints of chords for any two of them cannot alternate. This means that when we add the endpoints for x and z to the constraints, we must obtain a, z, x, b, c, x, v, a, b, v, z, c, as shown above. Now we cannot add the chord for u to cross the chords for without crossing the chord for b or c.
The graph H below is a circle graph but not a circular-arc graph. A circle representation is shown in the middle below. b y
Section 8.1: Perfect Graphs
To show that H has no circular-arc representation, note that the arcs for must be pairwise disjoint. Since the arc for v must intersect all three, it must contain one of them completely; by symmetry; we may let it be a. Now the arc for x cannot intersect the arc for a without intersecting the arc for v. 8.1.48. Paw-free graphs satisfy the SPOC. The "paw" is the graph obtained from the claw K1,3 by adding an edge joining two leaves. We must prove that every paw-free graph having no odd hole and no odd antihole is perfect. It suffices to prove that every paw-free graph G having no odd hole is a Meyniel graph, meaning that odd cycles of length at least 5 have at least two chords. Let C be an odd cycle of length at least 5 in G. Since G has no odd hole, C has a chord xy. This forms two cycles with the x, y-paths on C; one is odd. If the odd one has length at least 5, we obtain another chord of C. Otherwise, it has length 3, Since the subgraph induced by these three vertices and the next vertex on C must not be a paw, it contains an additional chord of C.
8.1.49. Sets Sand T of sizes a +2 and w + 2 that intersect every maximum clique and every maximum stable set, respectively, in the cycle-power C;11. (This completes the proof of Theorem 8.1.51.) 2 -c i -c a 1). The maximum Let S = U cliques in are the sets of w vertices with consecutive indices. The first four indices listed for S are separated successively by 2, w 1, and 2, respectively. The next step is w 1, and the subsequent gaps are w until the final step of w 1 that returns to the beginning. Since the set never skips as many as w consecutive indices, it intersects all maximum cliques. Let T = U . The maximum
stable sets in are the sets of a vertices whose indices increase successively by w (cyclically) starting from some point. In particular, a set of w successive vertices intersects all but one maximum stable set. The set T has w 1 of w successive indices from w through 2w 1. The stable set skipping this interval starts at V2w and contains Vaw, so it intersects T. The remaining stable sets are those containing These all contain v1 except the one that starts at w + 1, but this stable set intersects T at (a
8.1.50. SPGC for circle graphs. a) If x is a vertex in a partitionable graph G, then G NEz] is connected. If G NEz] is disconnected, then NEz] is a star-cutset. It thus suffices to show that partitionable graphs have no star-cutsets. Since x(G z) = w for each z c V(G), every proper induced subgraph of G is co(G)-colorable.
Chapter 8: Additional Topics
x has a partition into a(G) disjoint maximum cliques, a stable set intersecting ali maximum cliques must be a maximum stable set. However, every maximum stable set misses its mate, so no stable set intersects every maximum clique. These are the hypotheses of the Star-Cutset Lemma Lemma, so G has no star-cutset. b) Partitionable circle graphs are claw-free. Three pairwise-disjoint
chords Y, X, Z of a circle can be intersected by a single chord W only if the endpoints occur as shown below. Suppose that a circle graph G has a claw induced by central vertex w and stable set . If x is the vertex corresponding to the middle chord among in the circle representation of G, we have G N
tain a vertex whose chord intersect the chord for x in the representation. By part (a), this cannot occur in a partitionable circle graph. c) Circle graphs satisfy the SPOC. By part (b), partitionable circle graphs are claw-free. By Corollary 8.1.53, claw-free graphs satisfy the SPGC. Thus every p-critical circle graph is an odd hole or an odd antihole, and circle graphs satisfSi the SPGC. w
8.2. MATROIDS 8.2.1. The family of independent vertex sets of a graph need not be the family of independent sets of a matroid, In the star let the leaves have weight 1 and the remaining vertex have weight 2. The resulting maximum weighted stable set has weight a, but the greedy algorithm stops with a stable set of weight 2.
8.2.2. The family of stable sets of a graph G is the family of independent sets of a matroid on its vertex set if and only if every component of G is a complete graph. If some component of G is not complete, then G has P3 as an induced subgraph. The stable set of size 1 consisting of the middle of this path cannot be augmented from the stable set of size 2 consisting of its endpoints, so the augmentation inequality fails.
Section 8.2: Matroids
Conversely, if every component is complete, then the hereditary system is a partition matroid, with the stable sets being those sets of vertices having at most one vertex in each component. 8.2.3. Every partition matroid isa transversal matroid A partition matroid on E is defined by sets E1, . partitioning E such that a subset of E is independent if and only if it contains at most one element of each E1. This is the same as the transversal matroid on E arising from the E, [k]-bigraph whose ith component is the star with center i and leaf set E1, for 1 -c i e(H). By construction, the sizes of the partite sets are k(X) and k(Y), Components ofGg and that share a vertex lie within a single component in Hence k(X U Y) is the number of components of H. Every edge of H has the form CxCy, where Cg and Cy are components of Gx and Gy, respectively. Let S = V(Cy) P V(Cy); since is an edge, S 0. Every vertex outside S is outside V(Cg) or outside V(Cy). Hence X Y has no edge leaving 5, and is a nonempty union of components of Thus k(X P Y) > e(H), since we generate at least one component of for each edge of H (maybe more than one, such as when G is a 4-cycle and X and Y decompose G into two copies of P3).
Section 8.2: Matroids
b) For the cycle matroid M(G), the submodularity property r(X P Y) -F r(X U Y) -c r(X) + r(Y) holds. In the cycle matroid, r(X) = n(G) k(X), so it suffices to show that Ic is supermodular.
A graph with ii vertices and c components has at least ii
Since H has k(X) + k(Y) vertices and k(X U Y) components, we conclude that e(H) > k(X) -F k(Y) k(X U Y). By part (a), also k(X P Y) > e(H). Hence k(X P Y) + k(X U Y) > k(X) + k(Y), as desired.
8.2.11. Submodularity of rank functions of transversal matroids, using matching theory. A transversal matroid on a set E is induced by a family A1 Am of subsets of E by letting the independent sets be the systems of distinct representatives of subfamilies. Equivalently, the independent sets are the subsets of E that can be saturated by matchings in the E, [in]bigraph G. that is the incidence bigraph of the family. By definition, then, the rank of a set X C E is the maximum size of a matching in the subgraph GEX U [m]], which we denote by Gg. By the Konig—Egerváry Theorem, a'(Gy) equals the minimum size of a vertex
cover in Gx. For S c X, the smallest vertex cover Q such that S = X Q is (X 5) U N(S). Hence the minimum size of a vertex cover of Gg is The quantity is the deficiency of 5, = denoted def(S), and the fact that maxscg def(S) is due to Ore (Exercise 3.1.32).
Now consider subsets X, Y C E. For the submodularity inequality, we must bound r(X U Y) + r(X P 1') by r(X) + r(Y). For this we begin by studying the neighborhoods of the union and intersection of two sets
S C X and T C Y. The key to the inequality is that for 5, T C E, we have N(S P T) c N(S) P N(T) (equality need not hold!). Also N(S U T) = N(S) U N(T). Thus =
def(S) + def(T). Furthermore, the deficiency of a set S is the same in each Gg such that X D S. Therefore, if we let S and T be subsets of X and Y with maximum deficiency in Gx and Gy, we obtain r(X)
using in the last step that S U T and S P T are particular subsets of X U Y and X P Y, respectively. Thus the submodularity inequality holds.
8.2.12. For a digraph D with distinguished source s and sink t, andr(X) defined for X c V(D) to be the number of edges from s U X to X Ut,
Chapter 8: Additional Topics
the function r is submodular. When we view D as a network by giving each edge capacity 1, the statement of submodularity for r is precisely the statement of part (a) of Exercise 4.3.12.
8.2.13. For an element x in a hereditary system, the following properties are equivalent and characterize loops. The definition of a loop (an element comprising a circuit of size 1) is statement C. A) r(x) = 0. D) x belongs to no base. B) x E a(ø). E) Every set containing x is dependent. C) x is a circuit. F) x belongs to the span of every X C E.
Ifx e a(X)forallX ci E,thenx e
Ifx c a(ø), then x completes a circuit with 0;
A. The rank ofa circuit C is 1. D. Every subset of every base is independent. If x belongs to a base, then r() = 1. D E. If x belongs to an independent set, then it can be augmented to a maximal independent set (a base) containing x. E F. Let Y be a maximal independent subset of X. If every set containing x is dependent, then Y U x contains a circuit C, which must contain x since Y is independent. Hence x completes a circuit with a subset C A
ofX,sox c a(X). 8.2.14. The following characterizations of parallel elements in a hereditary system are equivalent, assuming that x y and neither is a loop. Property B is the definition of parallel elements, given that neither is a loop. A) r() = 1. B) e C. C) x E a(y), y e a(x), r(x) = r(y) =
The rank of a circuit C is 1, so B A. Conversely, if r() r(Y) always, so equality holds. Proof 2 (span function and absorption). The hypothesis implies X c a(X P Y), which in turn is contained in a(Y) since a is order-preserving. Now X c a(Y) and the absorption property yield r(X U Y) = r(Y). 8.2.16. If M is a hereditary system that satisfies the base exchange property (B), then the greedy algorithm generates a maximum-weighted base whenever the elements have nonnegative weights. This is actually more direct
using the dual version of the base exchange property (Lemma 8.2.33): if B1, B2 e Bande e Bi—B2,thenthereexistsf e B2—BjsuchthatB2-f-e—f is a base. This follows from the induced circuit property in Lemma 8.2.33, and the induced circuit property follows directly from the base exchange property in Exercise 8.2.17. Since the weights are nonnegative, the greedy algorithm generates a base. Let B be a base generated by the greedy algorithm. Among the bases of maximum weight, let B* be one having largest intersection with B. If B, then there exists an element e E B fi*, since the bases form an antichain. Let e be a heaviest element of B B*. By the dual base exchange property, there exists f c B such that B* f e is a base. Since B* is optimal, w(f)> w(e). Since the greedy algorithm chose e after choosing the heavier elements of B, even though f was also avalable, w(e) > w(f). Hence w(e) = w(f), and Bt + e is an optimal base having larger intersection with B than Bt does. Hence in fact B = Bt.
8.2.17. Exercises in axiomatic& a) In a hereditary system, the submodularity property implies the weak absorption property. Applying submodularity to X -I-- e and X + f yields r(X + e + f) + r(X) 0 we may select e c X Y (by symmetry) such that r((X P Y) + e) = r(X P Y) + 1. Let Y' = V + e. By the induction hypothesis, r(XPY')+r(XUYO -c r(X)+r(Y'). The left side equals r(XPY)+1+r(XUY) and the right side is bounded by r(X) +r(Y) + 1, so subtracting 1 from both sides yields the desired inequality. c) The base exchange property (B) implies the induced circuit property
(J). Proof 1 (contradiction). For I c I, if I + e contains distinct circuits C1, C2, then each consists of e plus a subset of I. Since C1
Chapter 8: Additional Topics
Both C1— a and (C1 U e are independent; augment them to bases B1 and B2, respectively. Since C1 e c B2, every element of B1 B2 except e is outside C1 a. Using (B), delete such elements from B1, replacing them with elements of
This transforms B1 to a base B such that the only element of
is e, and still C1
= 1. is in B. However, e a B,
Since (B) implies that bases have the same size, also B2 Since a a B2 B, the rest of B2, including C2 so C2 a B, contradicting that B is a base.
Proof 2 (extremality). Since every independent set lies in a base, it suffices to prove for B a B that B + e contains exactly one circuit. Let A be a minimal subset of B containing an element of each circuit in B + e. Thus (B A) + e a I, but (B A) + e + a I for all a a A. LetB'beabasecontaining(B —A)+e; notethat B—B' = A. IfB'—B has an element b other than e, then (B) yields an element a a B B' such that B' b + a a B, but this contradicts the dependence of(B A) + e + a. = 1. Since every minimal transversal Hence B' B = , and therefore of the circuits in B -F- e has one element, there is only one such circuit. d) The uniqueness of induced circuits (J) implies the weak elimination property (C). Suppose that C1, C2 a C and e a (C1capC2). If(C1 U C2) e is independent, then adding e creates a unique circuit, which contradicts the distinctness of C1 and C2. e) In a hereditary system, uniqueness of induced circuits (J) implies the augmentation property (I). Choose it, '2 a I with > IlL We obtain the augmentation by induction on lit '21 = k. If it c 12, any element of 12 '1 works; this is the basis step k = 0. For k > 0, select e a ft '2. If '2 + e a I, then the induction hypothesis allows us to augment 11 from '2 + e. Hence we may assume that '2 -F- e contains a unique circuit C. Choose f a C fl '2, and let I' = '2 + e f; we = = k—i, so theinductionhypothesis have I' El. Now and guarantees an augmentation of 11 from 1'. Any such element is also in
8.2.18. A hereditary system is a matroid if and only if it satisfies the fol= 1, then 1i+e a Ifor lowing: If 11,12 a Iwith > 'il and some e a 12 ft. This is a weaker form of the augmentation property, so it suffices to show that this implies the augmentation property. The stated property provides the basis for induction on k = it 12. If k > 1, select x a ft 12, and let I = ft x. The induction hypothesis yields el a 12 I such that I + a1 a I. Also 1 + eil = and 1 + a1 = k 1,so Proof 2: We seek a transversal of , where 1(e) U e is the unique circuit in B2 -Fe ife e B1 —112, and 1(e) = ife e Bin B2. For X C Bi, let Y = Ue€x 1(e). Since e c a(1(e)), we have X C a(Y). Since = r(Y) = r(c(Y)) > r(X) = X, V e I, the incorporation property yields S
Hence Hairs Condition holds. (Comment: we can similarly establish a bijection it:
+ 21(e) e B for all e e 111,)
There is a bijection it:
such that for each e c B1, the set
B2— x(e)-I-e is a base of M. Such a bijection is given by the perfect matching obtained in part (a). Elements of B1 n B2 yield isolated edges in G.
8.2.24. For any e E
there exists f e B2 such that B1
B2—f-FeEB. Ifee B1nB2,thenletf=e. Hencewemayassume
Proof 1 (transitivity of dependence). Let 1(e) + e be the unique circuit
1(e), then 1(e) C c(B1
Since 1(e) + e is a circuit, this implies
which is impossible since B1 is independent. Proof 2 (cocircuits). In B2 -F e there is a unique circuit C containing e. Since if1 is a cobase, + e contains a unique cocircuit containing e. Since C n C = 1 is forbidden, there exists another element f c C n C. Hence B2 + e is independent, has size B2, and therefore is a base. Similarly B1 + e is independent in the dual, has size B1, and is a cobase. Therefore B1 e + f is a base and f is the desired element. b) There may be no bijection it: B1 -÷ B2 such that e and f = x(e) e e cr(1(e)) C
satisfy part (a) for all e c B1, Consider the cycle matroid M(K4). Let B1 and B2 be the edge sets of two complementary 4-vertex paths. If e is a pendant edge of B1, then e can only be matched with the central edge of B2, since one pendant edge of B2 completes a triangle with B1 e, and the other
is not in the triangle of B2 + e. This argument applies for both pendant edges, but they cannot both be paired with the one central edge of B2. 8.2.25. Every matroid has a fundamental set of circuits (a collection of r(E) circuits such that C, contains but no higher-indexed element). If the elements e1, . form a base B, then addition of any other e e B B creates a unique circuit in B + e. The set of these generated by the elements of B B form a fundamental set of circuits.
. Ck are distinct circuits in a matroid, with none contained
Section 8.2: Matroids
in the union of the others, and X is a set with X contains a circuit. We use induction on k. For k = 1 the statement is trivial, and for k = 2 it is the statement of the weak elimination property C1 fl C2, then C1 or C2 itself is the desired circuit). For k > 2, choose X . By the induction hypothesis, X' contains a circuit C'. The case k = 2 yields a circuit in (C' U Ck) x; this (if x
x e X, and let X' =
circuit has the desired properties. 8.2.27. (+) For a hereditary system, prove that the weak elimination property implies the strong elimination property, by induction on C1 U C2.
8.2.28. Mm-max formula for maximum weighted independent set. Given weight w(e) e N U (O> for each element e, we prove maxi€r L61 w(e) = C mm r(X3, where the minimum is taken over all chains Xi C of sets in E such that each element e E E appears in at least w(e) sets in the chain (sets may repeat). Max -c mm. This inequality holds for every I and every acceptable chain . Independence of I implies r (X,) > Ifl X, . Now the appearance
ofeachec Iinatleastw(e) setsof(X,>yields
establish equality, let I be a maximum weighted independent set, and define a chain by X, = W + 1 i>, where W is the maximum weight and 1 6. For the upper bound, consider a 2coloring of E(K6). Since R(P4, C4) = S by Claim 2, we may assume that the coloring has a red P4, with vertices u, v, w, x in order. If both of the remaining vertices, y and z, have both edges to blue, then [x, y, u, z] is a blue C4. So, we may assume that one of the edges is red and extend the path to a red PS, such as u, v, w, x, y. The chords uy, vy, ux must all be blue, else we have a red 4-cycle or 5-cycle, which suffices, by Claim 1. Now consider z. The edges zv and zx cannot be both red or both blue, else liz, v, w, x] is a red 4-cycle or liz, w, y, u, xl is a blue 5-cycle. Hence we may assume by symmetry that to is blue and zx is red. Now we cannot color zu. 8.3.26. R(2K3, 2K3) = 10. The lower bound is provided by the construction in Theorem 8.3.15: the red graph is K1,3 -f-Ks. For the upper bound, consider a 2-coloring of E(K15). Any six vertices contain a monochromatic triangle,
and the seven vertices outside that triangle yield another monochromatic triangle. We are finished if they have the same color; if not, then the edges joining them are used to collapse the configuration to a "bow tie" as in the proof of Theorem 8.3.15. Thus we have vertices such that lix, v, w] is a bold triangle and lix, y, z] is a solid triangle. To avoid a monochromatic triangle on the remaining five vertices, we must have a bold 5-cycle [q, r, s, t, u] and a solid 5-cycle [q, s, u, r, 1], as
Section 8.3: Ramsey Theory
Vertex x must have three neighbors of the same color in , of which two must be adjacent on the cycle in that set having that color. By symmetry, we may thus assume that [x, q, r] is a bold triangle. If any edge e with endpoints in and is solid, then the edges joining the endpoint of e in (v, w> to the solid neighbors in of the other endpoint
of e (for example, vt and vu when e = vi) must be bold to avoid disjoint solid triangles, but this makes disjoint bold triangles (Li, q, r] and Eu, t, u] in the example e = yr. Hence the edges joining (5, 6> to (1, 2> are all bold. Now consider the edges vs and vu. If both are solid, then Li, y, z] and lu, s, u] are disjoint solid triangles. Hence one is bold; we still have symmetry and may assume that it is vu. Now we have disjoint solid triangles [v, q, u] and Li, w, r]. Hence a monochromatic 2K2 is forced.
8.3.27. R(mK2, mK2) = 3m 1. For the lower bound, let the red graph be K2m_i + Km_i. Every red edge has both endpoints in the (2m 1)-clique, so there cannot be m independent red edges. The complementary blue graph is the join of the complete graph on m 1 vertices with an independent set on 2m 1 vertices (K2m_i V Km_i. Every edge has at least one endpoint in the (m 1)-clique, so again there cannot be m independent edges. For the upper bound, we use induction. Note that R(K2, K2) = 2. For m > 1, consider an arbitrary 2-coloring of the edges of K3m_ i. There must be incident edges of differing colors, else the entire clique gets one color and
has enough points to contain mK2. Remove the three points hit by these two incident edges of different color, and apply the induction hypothesis. To the resulting monochromatic (m 1) K2, add the edge of the appropriate color from the deleted three vertices. 8.3.28. If G, is a graph of order p,, for 1 -c i -c/c, then R(miGi, . l)p, + R(Gi, . Each m, is the disjoint union of m, copies of G,. Given a 2-coloring with the specified number of vertices, we iteratively extract disjoint monochromatic copies of these graphs in the specified col-
ors. As long as R(Gi, . Gk) vertices remain that have not been touched by the extracted graphs, we can find another monochromatic G, in color i for some i. If ever we obtain m, copies of G,, then we are done, 0therwise, we have obtained at most 1 copies of each so we have
Chapter 8: Additional Topics
eliminated at most
l)p, vertices from consideration. In this case vertices remain, and we can continue. Hence the
process terminates only by finding ,n,G, in color i, for some 8.3.29. Graphs_with n vertices having no clique or independent set with size as large as 2c71&gnloglogn yield a lower bound for R(p, p) in terms of p that grows faster than every polynomial in p but slower than every exponential
in p. The existence of such a graph imphes that R(p, p) > n, where p = n log logn
To find the behavior of the lower bound, we need to solve this
equation for n in terms of p, but we do not need the complete solution to answer the question. Taking logs and squaring both sides yields c'(logp)2 = lognloglogn, where c' = 1/(clog2)2. To study the form of the function, we express n in terms of p and a parameter t. First suppose that n In this case c'(logp)2 > (plogt)(logp + loglogi). Again, this is impossible when t is a constant. Hence n cannot grow faster than any exponential function of p.
8.3.30. If G is an n-vertex graph such that a'(G) =
k, then R(Pg, G) = We seek the minimum r such that recL/blue-colorings of E(Kr) yield a red P3 or a blue G. If r r, so n is a lower bound. If r = 2(n —k 1), then we color E(Kr) with a perfect matching in red and the rest in blue. The red matching avoids and every set of
n vertices contains at least k + 1 pairs from the red matching (if it has s vertices whose mates are omitted and t matched pairs, then s + 2t = n and s + z' -c n k 1, 50 ii + t -c ft k 1). There is no blue G on such a set of vertices, since G has no matching of size k + 1.
For the upper bound, consider a 2-coloring of E(Kr) with r = 2k 1>. If there is no red P3, then the red graph is restricted to a matching. Thus all edges are blue except for 1) at most n k 1 pairwise disjoint edges and an isolated vertex if k (nw), then G contains a 4-cycle.
counts the triples u, v, w such that v is a common neighbor of u and w. If G has no 4-cycle, then every pair of vertices has at most one common neighbor. 3), then G contains a 4-cycle. Since b) If e(G) > + is a convex function of x, the minimum of for fixed
d(v) are equal (even though this = 2e(G), we conclude that may not be realized by a graph). Since > If then the condi>
tion of part (a) holds. This inequality reduces to the stated condition. Hence some color class is as large as
and the result of part
= max, except possibly if in is even and does 8.3.33. R(Cm, not exceed 2n. For the lower bound, first consider in > 2n + 1. Form a red clique on in 1 vertices; it has no red Cm and no blue edge, hence no vertex
with blue degree a. If in 2n + 1 and any 2-coloring of E(Km). If no vertex has blue degree at least a, then the red degree of every vertex is at least in a, which exceeds tn/2 since a 3, consider the
coloring on v). If this has a monochromatic cycle, then we can replace an arbitrary edge of the cycle by the edges from its endpoints to v.
Section 8.3: Ramsey Theory
If it has two monochromatic paths whose union is a cycle, then let x, y, z be three consecutive vertices on the cycle with xy red and yz blue. We may assume that yv is red. Now the cycle obtained by replacing yz with (y, v, z) has the desired property. 8.3.35. Ramsey numbers for cycles. a) A 2-coloring of that contains a monochromatic C2k+1 for some k > 3 also contains a monochromatic C2k. Let C be a red (2k + 1)-cycle, with vertices vo, . V2k in order. If there is no monochromatic 2k-cycle,
then each is blue, which yields a blue 2k + 1-cycle C' and implies that each is red, where indices are mod 2k. For each i, consider the cycle obtained from C by replacing the path (x,, . with
from C to obtain a blue 2k-cycle.
Note that this requires k > 3. b) A 2-coloring of that contains a monochromatic C2k for some k > 3 also contains a monochromatic C2k1 or 2Kk. Let C be a red 2k-cycle, with vertices v0, . in order. We prove that if there is no monochromatic (2k 1)-cycle, then all edges of the form x,x,+21 are blue. This suffices, since it implies that the odd-indexed vertices and even-indexed vertices along C both induce blue copies of Kk. If there is no monochromatic (2k 1)-cycle, then each
we may assume that 2 -c j
k 2. Replacing X,+21X1+2j+1 and (x,x1+1x1+2) on C with x,x,+21 and x,+2x,+21+1 yields a cycle of length 2k 1 in which -c
every edge except the two new edges is red. Ifx,x,+21 is red,
must therefore be blue to avoid a red (2k 1)-cycle. (In the figure below, bold means blue and solid means red.) Similarly, replacing and (x,x,_1x,_2) on C with X,X,+2j and X,_2X,+2j_1 forces to be blue. Now replacing (x,_2x,, x,+21_ix,+21+i> with in the set of edges of the form yields a blue (2k
x,. Hence if there is no monochromatic (2k
Chapter 8: Additional Topics
5, then R(Cm, Cm) 1)(n(G) 1) forces a cycle of length at least in in G. Since > 1)[2rn —21, we conclude that the coloring has a red cycle of length at least in. By parts (a) and (b), there is also a red rn-cycle or two disjoint blue complete graphs of equal order exceeding rn/2; we may assume the latter. Let Qi and Q2 be disjoint sets inducing blue complete graphs, each of order exceeding rn/2, chosen to maximize Qi U If two nonincident blue edges join and Q2, then we can take PFm/21 from Qi and P[m/2J from Q2 to form a blue rn-cycle with these edges. Hence all blue edges joining Q i and Q2 are incident to a single vertex x, which we may assume is in Q If in is even, then we can now take rn/2 vertices from each Q,, avoiding x, and form a red rn-cycle using the edges between them. We may therefore assume that in is odd. Let T = Qi U Q2 and S = . If there is no blue in-cycle within or Q2, then Q' U 3, there are at least two blue edges from x to R. Hence
there is a blue cycle spanning Q2 U , and we may assume that Q2 U in 1. Therefore 5 U > (2in 1) (in 1) = in.
We now have a blue in-cycle in the graph induced by S U T if >= (in 1)72 and Iin/21, so we may assume that > (in + 1)12> 3. If an edge within T is blue, then we complete a blue cycle by using it and otherwise alternating between S and T. Hence we may assume that all edges induced by T are red. If there is a blue edge in [T, RI, then we can form a blue cycle by following it with any portion of R, then x, then any portion of S. The length is any value from 4 to at least Q 1 U which exceeds in. Hence we may assume that all of [T, RI is red. Now we can form a red in-cycle by using a path alternating between S and R using (in 3)72 vertices of 5, (in 1)72 vertices of R, and two vertices of T. ,
Section 8.3: Ramsey Theory
8.3.36. The Ramsey multiplicity of K3 is 2, where the Ramsey multiplicity of G is the minimum number of monochromatic copies of G in a 2-coloring of E(KR(G,G)). To color E(K6) with only two monochromatic triangles, let the red graph be K33, which is triangle-free. The complementary graph is 2K3, with two triangles. Now we show that every coloring has at least two monochromatic triangles. Since R(3, 3) = 6, there is at least one monochromatic triangle T, say in red. If we delete one vertex of T, then there remains a monochro-
matic triangle on the remaining five vertices unless the color classes on that subgraph are complementary 5-cycles. Let C be the red 5-cycle, and let z be the deleted vertex. To form T, we have edges from z to consecutive vertices on C, which we call x and y. Let u, x, y, v be the consecutive vertices on C including the edge xy. A red edge from z to u or v completes another red triangle, but if uz and vz are both blue they complete a blue triangle with uv.
8.3.37. Each point in a triangular region has a unique expression as a convex combination of the vertices of the triangle. We observe first that each point on a segment has a unique expression as a convex combination of the endpoints. Now, given a point x inside the triangle with corners u, v, w, let y be the point at which the ray from u through x reaches the opposite side. Nowy = Xv-1-(1—X)w), forauniqueX, andx = for a unique ,tt. Hence x = (1 ,tt)u + (Xp4v + .k)w. The coefficients are uniquely determined in terms of X and ,a, and these constants also are uniquely determined by x and the corners. 8.3.38. Sperner's Lemma in higher dimensions. In a proper labeling of a simplicial subdivision of a k-dimensional simplex, there is a cell receiving all k + 1 labels, where "proper labeling" is a labeling such that label i does not appear at any vertex on the ith outer face. We prove the stronger result, by induction on k, that there are an odd number of completely labeled cells, When k = 1, we have a 0, 1-labeling of a path segment with 0 and 1 on the ends, and there must be an odd number of switches between 0 and 1 along the path. For k > 1, define a graph G with a vertex for each cell plus one vertex v for the outside region. Two vertices of G are adjacent if the corresponding regions share a (k 1)-dimensional face with corners having labels 0, . .. , k 1. If the vertex for a cell is nonisolated, then the cell has all these k labels among its k + 1 corners. If it repeats one of the labels, then it has two incident edges in G. Otherwise, it is a completely labeled cell and has degree 1. Hence the only cells with odd degree are the completely labeled cells. To prove that there are an odd number of them, it suffices to prove that the
Chapter 8: Additional Topics
vertex v also has odd degree. A cell having a (k 1) dimensional face on the ith outside face cannot have label i on it. Therefore, having an edge to v happens only through the (k + 1)th face, where label k + 1 is forbidden. This face is a simplicial subdivision of a (k 1)-dimensional simplex. The labeling is proper on this face, as it inherits the needed properties from the full labeling (think of the edges of a triangle in the 2-dimensional case). By the induction hypothesis, this lower-dimensional labeling has an odd number of completely labeled cells. Hence the frill-dimensional labeling has an odd number of cells with edges to v. 8.3.39. The badwidths of P. and are 1, a 1, and 2, respectively. A nontrivial graph has bandwidth 1 if and only if its vertices can be ordered so that no nonconsecutive vertices are adjacent, which means that its components are paths. In ordering, the vertices of the first and last vertices are adjacent. Since is not a path, its bandwidth is at least 2. To achieve this, number the vertices around the cycle . 5, 3, 1, 2, 4, .. . in order, reaching to a 1 in one direction and to a in the other direction. 8.3.40. The bandwidth of is a 1 Ln'/2j, where a = a, and a' = max, a,. Consider an optimal numbering. If the vertices given labels 1 and a come from different partite sets, then the bandwidth is a 1. If they come from the same partite set, with c vertices of this partite set at the beginning and c' at the end of the lab ellng, then the bandwidth is at least maxa c, a c'. To minimize this lower bound, we split the largest partite set between the front and back. The lower bound becomes a 1 La'/2i. Also, splitting the largest partite set in this way achieves equality for any ordering of the remaining vertices in the remaining middle positions. 8.3.4 1. The bandwidth ofa tree with Ic leaves is at most fk/21. Let m Ik/21.
We use the fact that every tree with Ic leaves is the union of in palrwise intersecting paths (Exercise 2.1.40). We repeat the proof: Let T be a tree with Ic leaves. By pairing leaves arbitrarily, we form a set of in paths that together cover the leaves, Among all such sets of paths, choose one with maximum total length; we claim it has the desired properties. If some pair of paths is disjoint, say an x, y-path P and a u, v-path Q, consider the path R in T from V(P) to V(Q). Replace P and Q with the x, u-path and the y, v-path in T. The new paths still cover the leaves, and the total length has increased by twice the length of R. If some edge e of T is omitted by the longest covering set of paths, then consider the two components of T e.
Each contains a leaf of T, so each contains at least one path in the set. Again making the switch increases the total length. To prove the upper bound, we provide an injective integer embedding in which the difference along every edge is at most in; the set of labels need not be consecutive. Let Pc. be a set of pairwise-intersecting paths
Section 8.3: Ramsey Theory
with union T, and let 1 = U[=0 P,. Because the paths are pairwise intersecting, each T1 is connected. Because Ti-i is connected and contains no cycle, a traversal of cannot leave T1_4 and then return to it, First assign successive multiples of in to the vertices along Po. For have received labels congruent to j > 0, suppose that the vertices of 1, . j 1 modulo in so that edges have dilation at most in. We use labels congruent to j on vertices of V(T_1). Let u, v be the vertices of V(P1) fl V(I>_1) that are closest to the two ends of (these may be equal). By symmetry, we may assume that f(u) -c f(v). Let a be the largest integer less than f(u) congruent to j (mod in), and let b be the smallest integer greater than f(v) congruent to j (mod in). From the neighbor of u [or vi out to the corresponding leaf of assign the label a (1 1)in [or b + (1 1)m] to the ith vertex of V(l>_1) encountered. (If k is odd, then for one value off, one of these subpaths is empty). The new labels are in the new congruence class, and the newly-included edges have difference at most in.
8.3.42. If G is a caterpillar with -cm for all H c G, then 11(G) -c in. (Note that the least such in is a lower bound, so equality with hold.) Let P be the spine of G, having vertices (v0, . v,) in order, where v0 and are leaves. Assign the number urn to v, for 0 -c -c p. It suffices to show that this allows us to assign numbers to the remaining vertices so that all leaf neighbors of v, receive numbers between (1 1)rn and (1 + 1)rn. We can then compress the numbering to eliminate gaps without increasing any edge difference. u
Let L, = N(v,) V(P), and let 1, = U. For 1 c, give the remaining vertices 1, c labels starting with urn + 1. We show that this works by proving that 4 c -c rn 1 at step i. For k + 1. o When k is larger than (n 1)/2, the full set we have described does not fit inside the n-by-n gnd. We must subtract 4 k—(n—1)/2 (2z 1) vertices
after putting w in the center of the grid. The largest subgraph now has 1)2 + 1 vertices. After subtracting 1 and dividing 2k(2n 1 k) (n by the diameter 2k, we have a lower bound of 2n 1 Ek + (n 1)2/2k]. This is maximized by setting k to be about (n where the resulting lower bound is (n 1)[2 This is about .59n, which is still short of the desired lower bound of n. b) Sliding the elements of a vertex subset of o to the extreme left within their rows does not increase the size of the boundary. Choose S c
with a vertices in the ith row, for each i. Let T consist of the ci first a in the ith row, for each i. We show that -c If a1 = n, then each set has the same number of boundary elements in row j. Furthermore, no vertex outside row j becomes a boundary element due to an edge to row j. Therefore, we may assume that a, --- > and hence Harper's lower bound for B(P,, o is a, By part (b), sliding vertices to the left within their rows does not increase the boundary, and it produces a set whose column populations are in nonincreasing order. By symmetry, sliding vertices to the top within their columns also does not increase the boundary This leaves the column populations unchanged and produces a set whose row populations also are in nondecreasing order. To show that the boundary bound is at least a, it suffices to prove that in o there is some k such that every k-set of vertices has boundary at View least a. We choose k such that C) a(G) for any graph without isolated vertices, it suffices to obtain a stable set consisting of one vertex from each member of in a minimum clique cover 8 of E(G). We may assume that every member of 8 is a maximal clique. Since a simplicial vertex belongs to only one maximal clique, this implies that each clique of S contains a vertex belonging only to that clique. The 6'(G) vertices from distinct cliques thus selected must be independent, because no edge among them is covered by S.
8.4.3. If b(G) is the minimum number of bipartite graphs needed to partition the edges of G, and a(G) is the minimum number of classes needed to partition E(G) such that every cycle of G contains a non-zero even number of edges in some class, then b(G) = c(G) = [lgy(G)1. We prove
and [lgy(G)1 differbylessthan 1,
the integers in this string of inequalities must be the same. Let E1 U-- U Eb(c) be a minimum partition of E(G) into bipartite subgraphs; we may assume these are spanning subgraphs. We can define a proper f by giving each v e V(G) a binary b(G)-sequence f(v) in which f(v) indicates which partite set in E, contains v. Since each edge belongs to some E, the endpoints of each edge receive different labels. This proves x(G) -c 2b(0), i.e. -c b(G). Let E1 U- - U Ea(o) be a minimum partition having the cycle intersection -
property defined above. If E contains an odd cycle, then this cycle in G does not contain a non-zero even number of edges of any color. Hence each
E, is bipartite, and b(G) (r—l)!
Allowing r + 1 coordinates, we can represent this graph using (i, . i) for the ith vertex of the clique and (1 r, j) for the jth vertex of the stable set, with the extra coordinate included to ensure that the vertices of the stable set get distinct encodings. Since Kr + K1 is an induced subgraph, the answer thus is always r or r + 1. In every coordinate, the vertices of the clique must have distinct val-
ues. By permuting the labels used within a coordinate, we may assume that the code for the ith vertex of the clique is (1. 1). If pdim G = r, then each vertex of the stable set must be encoded by a permutation of Fr] in order to establish all non-adjacencies to clique vertices. These permutations must be distinct, and each pair of them must agree in some coordinate to
avoid edges in the stable set. Hence no pair of the permutations can be cyclic permutations of each other.
This partitions the r! permutations into (r 1)! classes of size r, from each of which we can take at most 1. Therefore, pdim (Kr -F mK1) = r requires in -c (r
r coordinates do suffice; give each
vertex of the stable set value 1 in coordinate r to prevent edges, and use the (r 1)! distinct permutations of Er 1] in coordinates 1, . , r 1. 8.4.7. The product dimension of the three-dimensional cube is 2. Since Qg is not a complete graph, we need at least two coordinates, and we can encode it with two coordinates by using the binary triples. Each triple x is a vertex that is adjacent to every vertex of opposite parity except the complement of x. We use coordinate 1 to destroy edges to vertices of the same parity and coordinate 2 to destroy edges between complements. In coordinate 1, assign 0 to each sequence of even weight and 1 to each sequence of odd weight. In coordinate 2, assign 0 to 000 and 111, and assign i to each sequence in which the ith coordinate has a value that appears only once in the sequence. The resulting vectors are <(i, j): 0 ]jz/2j and fn/21,yieldingf(n) >(n2 1)/4. = flg(n 8.4.10. If n > 4, then 1 + [lg(n 1)1 and 1 + [lgnl
3, then [lg(m 2)1. Thus pdim C2,,> [lg(2n 2)1 = 1 + [lg(n 1)1 and pdim [lg(2n 1)1 = [lg(2n)1 = 1 + [lgnl. We complete the proof for paths by embedding in the weak product of k triangles, beginning with Ac = 2. Let xk(i) be the encoding of the ith vertex on the path, for 1 2, we obtain
Section 8.4: More Extremal Problems
xkQ) from the previous codes, by appending a suitable value in the new coordinate. Here i runs from 0 to index i parity of i in first k 1 coords in kth coord even > >
