Graph Theory Notes

June 11, 2018

H.A. Kierstead

CHAPTER 1

Introduction

This is an on going project. There will be changes to the material not yet covered,especially just before each lecture. There may also be changes immediately after a lecturebased on class questions, etc. Later corrections to previous material will be marked in red.For these reasons, you should continuously download the current version.

Use at your own risk. There are bound to be typos. Corrections are appreciated, andrewarded with small amounts of extra credit, especially when they indicate mathematicalunderstanding. But you are also encouraged to ask questions in emails about things you donot understand. I will try to answer questions quickly, but if they are complicated, or thereare many, I may need to wait until the next class.

These notes are meant to enhance, not replace, the lectures and class discussions. Theyare intended to be concise records of proofs, to free students from the need to take carefulnotes during class. However the motivation for these proofs is left to the lectures anddiscussions in class.

1.1. Graphs

Formally a graph is an ordered pair G = (V,E) where E is an irreflexive, symmetric,binary relation on V . Since E is symmetric there is no need to keep track of the order ofpairs (x, y) ∈ E; since it is irreflexive there are no ordered singletons (x, x) in E. This leadsto a more intuitive formulation. We take E to be a set of unordered pairs of elements from V .Elements of V are called vertices ; elements of E are called edges. If x, y ∈ V are vertices and{x, y} ∈ E is an edge we usually (but not always) denote {x, y} by the shorthand notationxy. So xy = {x, y} = {y, x} = yx. The vertices x and y are called ends, or endpoints, of theedge xy. The ends x and y of an edge xy are said to be adjacent and the end x is said to beincident to the edge xy. We also say that x and y are joined or linked (not connected) by theedge xy. Two edges are said to be adjacent if they have a common end. In this course, allgraphs have a finite number of vertices, unless it is explicitly stated that they have infinitelymany. Graphs are illustrated by drawing dots for vertices and joining adjacent vertices bylines or curves.

Our definition of graph is what West calls a simple graph. Most of the time we will only beinterested in simple graphs, and so we begin with the simplest definition. When necessary, wewill introduce the more complicated notions of directed graphs, multigraphs and hypergraphs,but here is a quick hint. A directed graph G = (V,E) is any binary relation (not necessarilyirreflexive or symmetric) on V . In other words E is any set of ordered pairs of vertices. Amultigraph is obtained by letting E be a multiset; then two vertices can have more thanone edge between them. Sometimes this is not enough, and we need to distinguish betweenedges with the same ends. Also, we may need loops—edges whose ends are the same vertex.A hypergraph is obtained by letting E be a set of subsets of V , where the elements of E

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can have any size. If they all have size k then we get a k-uniform hypergraph, also called ak-graph. So ordinary graphs are 2-uniform hypergraphs, or 2-graphs.

The study of graph theory involves a huge number of of parameters—see the front andback inside covers of West or the last two pages of Diestel. This can be quite daunting. Mystrategy is to introduce these parameters as they are needed. Please feel free to interruptlectures to be reminded of their meanings. Most of the time my notation will agree withWest, and I will try to emphasize differences. Next we introduce some very basic notation.

Given a graph G, V (G) denotes the set of vertices of G and E(G) denotes the set of edgesof G. Set |G| := |V (G)| and ‖G‖ := |E(G)|; this is not standard, and instead the book usesv(G) = |V (G)| and e(G) = |E(G)|. Suppose v ∈ V (G) is a vertex of G. Define

NG(v) = {w ∈ V (G) : vw ∈ E(G)}; EG(v) = {e ∈ E(G) : v is an end of e}.The set NG(v) is called the (open) neighborhood of v, and its elements are called neighborsof v. So a vertex w is a neighbor of v iff it is adjacent to v. When there is no confusionwith other graphs the subscript G is often dropped. The closed neighborhood of v is N [v] =N(v)∪{v}—we dropped the subscript. The set EG(v) is the set of edges incident to v; again,we may drop the subscript G. The text does not provide notation for this set. For simplegraphs |N(v)| = |E(v)|. However for multigraphs this may not hold, since two verticesmight be joined by several edges. With this in mind, define the degree of a vertex v to bedG(v) := |EG(v)|, but note that for simple graphs dG(v) = |NG(v)|. A graph is k-regular ifevery vertex has degree k.

We use the following set theoretic notation. The sets of natural numbers, integers andpositive integers are denoted, respectively, by N, Z and Z+. For integers a and b let {a, . . . , b}indicate the set {i ∈ N : a ≤ i ≤ b}. Then {0, . . . ,−1} = ∅. For n ∈ N set [n] := {1, . . . , n};in particular [0] = ∅. For a setX and an element y, setX+y := X∪{y} andX−y := Xr{y}.Finally,

(Xn

)is the set of all n-element subsets of X.

In this course all graphs G satisfy 1 ≤ |G| <∞, unless a specific exception is stated.

1.2. Proofs by Mathematical Induction

Most proofs in graph theory involve mathematical induction, or at least the Least ElementAxiom. Here we quickly review this technique. Also see the discussion in West on pages19–20, and especially the induction trap on page 42.

Here we present mathematical induction in terms of the Least Element Axion (LEA).First we give a careful definition of “least element”.

Definition 1.1. Let B ⊆ N be a set of natural numbers. A number l ∈ N is a leastelement of B if

(L1) {0, . . . , l − 1} ⊆ NrB, and(L2) l ∈ B.

The following axiom (LEA) is fundamental.

Axiom (LEA). Every nonempty set of natural numbers has a least element.

Consider a set of “good” natural numbers S, and let B = N r S be the set of “bad”numbers. We would like to prove that every natural number is “good”, i.e., S = N. Here isa way to organize the argument.

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Theorem 1.2 (Principle of Induction). Suppose S ⊆ N. Then S = N, if

(*) ∀n ∈ N ({0, . . . , n− 1} ⊆ S → n ∈ S)

Proof. Suppose S ⊆ N satisfies (*). As S ⊆ N we only need prove N ⊆ S. Arguing bycontradiction, assume N * S. Then B := N r S 6= ∅. By LEA, B has a least element l.By (L1) applied to l, {0, . . . , l − 1} ⊆ N r B = S. By (*), l ∈ S, so l /∈ N r S = B. Thiscontradicts (L2). �

By the Principle of Induction, when proving that a set S ⊆ N is equal to N, it sufficesto prove (*). For this we consider any natural number n. If {0, . . . , n− 1} * S then we aredone, so we assume {0, . . . , n− 1} ⊆ S. We use this “induction hypothesis” to prove n ∈ S.

When proving (*) the case n = 0 is special, since there is no natural number k < n, andso {0, . . . , n − 1} = ∅ ⊆ S; thus the induction hypothesis has given us no new information.More generally, we say the cases of the argument that do not use the existence in S of asmaller number than n form the base step. Thus the case n = 0 is always part of thebase step, but the base step may include more cases. The cases that use that somesmaller number than n is in S form the induction step. Here is an example. Notice that thestatement of the theorem is carefully phrased because 0 and 1 do not have prime factors.This must be reflected in the definition of the set S.

Proposition 1.3. Every natural number greater than 1 has a prime factor.

Proof. Let S = {n ∈ N : n ≤ 1 or n has a prime factor}. It suffices to show (*).Consider any n ∈ N such that {0, . . . , n−1} ⊆ S. We must show n ∈ S. If n ≤ 1 then n ∈ Sby definition. So suppose n ≥ 2. If n is prime then it is a prime factor of itself, and so it isin S. Otherwise, there exist integers a, b such that 1 < a, b < n and ab = n. Since a < n, wehave a ∈ S. Since 1 < a this means that a has a prime factor p. Since p is a factor of a anda is a factor of n, p is a (prime) factor of n. �

In the above argument, the cases n ≤ 1 and n prime form the base step. Notice that inthe induction step we never used that n− 1 was in S; we only used that the factor a was inS, and it is easily seen that a is always less than n− 1.

Here is an example from graph theory. First we need some additional notation. For agraph G, let δ(G) = min{d(v) : v ∈ V (G)}. We say that a graph contains a cycle if it has asequence of at least three distinct vertices v1, . . . , vs such that vivi+1 ∈ E for all i ∈ [s − 1]and vsv1 ∈ E; in this case we call v1 . . . vsv1 a cycle. The following (typical) statement isabout all graphs, not all natural numbers, so it may be surprising that we can prove it byinduction.

Proposition 1.4. Every graph G = (V,E) with δ(G) ≥ 2 contains a cycle.

Proof. Let S = {n ∈ N : all graphs G with |G| = n and δ(G) ≥ 2 contain a cycle}.Using Theorem 1.2, it suffices to prove (*): then S = N; any graph G with δ(G) ≥ 2 satisfies|G| ∈ N = S; and G has a cycle by the definition of S.

Consider any n ∈ N, and any graph G = (V,E) with |G| = n and δ(G) ≥ 2. Thenn = |G| ≥ 3. As n− 1 ∈ N, we may assume n− 1 ∈ S.Case 1: δ(G) ≥ 3. Let v ∈ V and G′ = G− v, i.e., form a graph G′ by deleting the vertex vtogether with all edges incident to it. Then |G′| = n− 1, and δ(G′) ≥ 2, since no vertex lost

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more than one neighbor. Since n− 1 ∈ S, there is a cycle contained in G′, and this cycle isalso contained in G.Case 2: δ(G) = 2. Pick y ∈ V with d(y) = 2. Then N(y) = {x, z} for some distinctvertices x and z. If xz ∈ E then G contains the cycle xyzx. Otherwise, let G′ be the graphformed by deleting the vertex y, and the edges yx and yz, and adding the edge xz. Then|G′| = n − 1 and δ(G′) ≥ 2, since the loss of y is made up for by the fact that x and z arenow new neighbors of each other. Since n − 1 ∈ S, there is a cycle C contained in G′. If xand z do not appear consecutively in C then C is a cycle contained in G. If they do appearconsecutively, we may assume that C = xz . . . vsx. Then xyz . . . vsx is a cycle in G. �

What were the base and induction steps of this proof? The proof only applies to finitegraphs, why? Can you construct an infinite graph with δ(G) ≥ 2 and no cycle?

The proof of Proposition 1.4 illustrates an important technique. Suppose we want toprove that all graphs have a property P , and f is a function on graphs (such as |G|) whosevalues are natural numbers. We can argue by induction on f as follows. First set

S = {n ∈ N : all graphs G with f(G) = n have property P}.

Then argue by induction that S = N. Finally consider any graph G. As f(G) ∈ N = S, thedefinition of S implies G has property P . Often the cleverness of a proof is in picking theright function f on which to do induction.

I have one last comment about a common mistake made by beginning students (seeinduction trap on page 42 of West). When you argue by induction on f to prove a statementP → Q about graphs, the induction step starts with a big graph G that satisfies P ; then G isconverted into a smaller graph G′, i.e., f(G) > f(G′), and you prove that G′ also satisfiesP ; by the induction hypothesis G′ satisfies Q; finally, you use that G′ satisfies Q to provethat G satisfies Q. You do not start with a small graph, and then make it bigger!

1.3. Ramsey’s Theorem for Graphs

Ramsey’s Theorem is an important generalization of the Pigeonhole Principle. Here weonly consider its simplest version applied to graphs; the general version is a statement aboutk-uniform hypergraphs. In the past it was presented as part of MAT 415, but it has beenmoved to MAT 416 because its presentation benefits from the language of graph theory.

Let G = (V,E) be a graph, and suppose X ⊆ V . The set X is a clique in G if xy ∈ Efor all distinct vertices x, y ∈ X. It is an independent set, or coclique, in G if xy /∈ Efor all vertices x, y ∈ X. A clique (coclique) X is a b-clique (b-coclique) if |X| = b. Letω(G) := max{|X| : X is a clique in G}, and α(G) := max{|X| : X is a coclique in G}.

A graph H is a subgraph of G, denoted H ⊆ G, if V (H) ⊆ V (G) and E(H) ⊆ E(G). Itis an induced subgraph of G if H ⊆ G and E(H) = {xy ∈ E(G) : x ∈ V (H) and y ∈ V (H)}.For X ⊆ V , G[X] is the induced subgraph of G that has vertex set X. The complement ofG is the graph, G := (V (G), E(G)), where E(G) :=

(V (G)

2

)r E(G).

Theorem 1.5 (Ramsey’s Theorem). For all graphs G and a, b ∈ Z+, if |G| ≥ 2a+b−2

then ω(G) ≥ a or α(G) ≥ b.

Proof. Argue by induction on n = a+ b. (That is, let S be the set of natural numbersn such that for all positive integers a, b if n = a+ b, and G is a graph with |G| ≥ 2a+b−2 then

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NG(v)

NG(v)

G: v

Figure 1.3.1. Ramsey’s Theorem

ω(G) ≥ a or α(G) ≥ b. Show that for all n ∈ N if {0, . . . , n− 1} ⊆ S then n ∈ S.) Considerany n = a+ b with a, b ∈ Z+, and any graph G with |G| ≥ 2a+b−2.Base step: min{a, b} = 1. Since |G| ≥ 1, G has a vertex v. Since {v} is both a clique and anindependent set, both ω(G) ≥ 1 and α(G) ≥ 1. So we are done regardless of whether a = 1or b = 1.Induction Step: min{a, b} ≥ 2 (so a− 1, b− 1 ∈ Z+). (We assume the induction hypothesis:the theorem holds for all a′, b′ ∈ Z+ with a′ + b′ < a+ b.) Let v ∈ V (G). Then

1 + dG(v) + dG(v) = |G| ≥ 2a+b−2 = 2a+b−3 + 2a+b−3.

By the pigeonhole principle, either dG(v) ≥ 2a+b−3 or dG(v) ≥ 2a+b−3.Case 1: dG(v) ≥ 2a+b−3. Set H := G[NG(v)]. Then |H| = dG(v) ≥ 2a−1+b−2. By theinduction hypothesis H contains an (a− 1)-clique X or a b-coclique Y. In the latter case Yis a b-coclique in G. In the former case X + v is an a-clique in G.Case 2: dG(v) ≥ 2a+b−3. Set H := G[NG(v)]. Then |H| ≥ 2a+b−1−2. By the inductionhypothesis H contains an a-clique X or a (b − 1)-coclique Y . In the former case, X is ana-clique in G. In the latter case Y + v is a b-coclique in G. �

For a, b ∈ Z+, define Ram(a, b) to be the least integer n such that every graph G with|G| ≥ n satisfies ω(G) ≥ a or α(G) ≥ b. By Theorem 1.5, Ram(a, b) exists and satisfiesRam(a, b) ≤ 2a+b−2.

HW 1. Prove that for a, b ∈ Z+ with a, b ≥ 2:(1) Ram(a, b) = Ram(b, a).(2) Ram(a, 1) = 1.(3) Ram(a, 2) = a.(4) Ram(3, 3) = 6.(5) Ram(a, b) ≤ Ram(a− 1, b) + Ram(a, b− 1).(6) (+) Ram(3, 4) = 9 (tricky, see Proposition 1.11).(7) Ram(4, 4) ≤ 18 (use 6, even if you did not do it).(8) (+) Ram(4, 4) = 18.

It is known that Ram(4, 5) = 25 and 43 ≤ Ram(5, 5) ≤ 49, and conjectured that 43 is theright answer. Proving it would make for a truly exceptional doctoral thesis.

Theorem 1.6 (General Ramsey Theorem [5]). For all r, c, a1, . . . , ac ∈ Z+ there existsn ∈ Z+ such that for all sets V with |V | ≥ n and functions f :

(Vr

)→ [c] there exist i ∈ [c]

and H ⊆ V with |H| = ai such that f(S) = i for all S ∈(Hr

).

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Theorem 1.6 can be proved in the style of Theorem 1.5, but requires a “double induction”on r and then

∑ci=1 ai. Strong doctoral students (with effort) can work this out from scratch.

Our discussion of Ramsey’s Theorem is just the tip of the iceberg. Whole books havebeen written on the subject.

1.4. Graph Isomorphism and the Reconstruction Conjecture

In order to study graph theory we need to know when two graphs are, for all practicalpurposes, the same.

Definition 1.7. Two graphs G and H are isomorphic if there exists a bijection

f : V (G)→ V (H) such that xy ∈ E(G) iff f(x)f(y) ∈ E(H) for all x, y ∈ V (G).

In this case we say that f is an isomorphism from G toH and write G ∼= H. The isomorphismrelation is an equivalence relation on the class of graphs. The equivalence classes of thisrelation are called isomorphism types. In graph theory we generally do not differentiatebetween two isomorphic graphs. We say that H is a copy of G to mean that G ∼= H.

HW 2. Let f : V (G)→ V (H) be an isomorphism between two graphs G and H. Provecarefully that dG(v) = dH(f(v)) for all v ∈ V (G).

HW 3. Here we do distinguish between isomorphic graphs. Let V = {v, w, x, y, z} be aset of five vertices, and G = {G : G is a graph with V (G) = V and ‖G‖ = 4}. Determine(with proof) |G| and the number of isomorphism types of G.

If x is a vertex of a graph G then G−x is the induced subgraph G[V (G)−x]. The graphG− x is called a vertex deleted subgraph of G.

Definition 1.8. A complete set of vertex deleted subgraphs of a graph G = (V,E) is aset G such that there exists a bijection ψ : V → G with ψ(x) ∼= G− x for all x ∈ V . Noticethat for distinct vertices x and y it may be that G− x ∼= G− y. In this case ψ(x) and ψ(y)are distinct copies of G − x. See Figure 1.4.1. A complete set of vertex deleted subgraphsG of G is also called a deck of G, and the elements of G are called cards. Two graphs arehypomorphic if they have the same deck. See Figure 1.4.1.

Notice that G has infinitely many decks, but each deck G of G satisfies |G| = |G|. Wecannot determine V (G) from its deck G, but the following famous conjecture asks whetherwe can determine the isomorphism type of G from G.

Conjecture 1.9 (Reconstruction Conjecture 1.3.12). Any two hypomorphic graphs withat least three vertices are isomorphic.

Now we take a little detour before proving Proposition 1.12

Definition 1.10. An edge of a multigraph is called a link if it has two distinct ends, anda loop if both ends are the same vertex. The degree d(v) of a vertex of a multigraph is thenumber of links incident to v plus twice the number of loops incident to v. In a multigraphthere may be cycles of length 1—one loop—and cycles of length 2—two links between the

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H = = J

Figure 1.4.1. The vertex deleted subgraphs of a graph G. What is |G|?What is ‖G‖? What is the isomorphism type of G?

same two vertices. This definition is designed so that each edge is counted twice when yousum the degrees of a graph. For a graph G = (V,E) and pair (v, e) ∈ V × E, set

ι(v, e) :=

2 if e ∈ E(v) is a loop1 if e ∈ E(v) is a link0 otherwise

Proposition 1.11 (Handshaking 1.3.3.). Every (multi-)graph G := (V,E) satisfies∑v∈V d(v) = 2 ‖G‖. In particular, G has an even number of vertices with odd degree.

Proof. ∑v∈V

d(v) =∑v∈V

∑e∈E

ι(v, e) =∑e∈E

∑v∈V

ι(v, e) =∑e∈E

2 = 2 ‖G‖ . �

The degree sequence of a graph G = (V,E) is a nondecreasing sequence of integersd1, . . . , d|G|, where V = {v1, . . . v|G|} and di = d(vi) for all i ∈ [|G|]. For example the degreesequence for the graph H in Figure 1.4.1 is 2, 2, 3, 3, 3. The next Proposition shows that wecan determine the number of edges of a graph and its degree sequence from its deck.

Proposition 1.12 (1.3.11). For all simple graphs G = (V,E) with |G| ≥ 3 and v ∈ V ,

‖G‖ =

∑v∈V ‖G− v‖|G| − 2

and dG(v) = ‖G‖ − ‖G− v‖ .

Proof. Every edge e ∈ E satisfies e ∈ E(G− v) if and only if ι(v, e) = 0. Thus∑v∈V

‖G− v‖ =∑v∈V

∑e∈E

(1− ι(v, e)) =∑e∈E

∑v∈V

(1− ι(v, e)) =∑e∈E

(|G| − 2) = ‖G‖ (|G| − 2).

So the first equality holds. The second equality follows from E = E(G− v) ∪ E(v). �

Example 1.13. Suppose G is the deck for some unknown graph such that G has twodistinct copies of H and four distinct copies of J as shown in Figure 1.4.1. Find (with proof)a graph G such that if the deck of G′ is G then G′ ∼= G.

Solution. Let G = (V,E) be an arbitrary graph with deck G. Then |G| = |G| = 6.Using Proposition 1.12, we have:

‖G‖ = (2 ‖H‖+ 4 ‖J‖)/(|G| − 2) = (2 · 6 + 4 · 5)/4 = 8,

and G has degree sequence 2, 2, 3, 3, 3, 3. Consider x ∈ V with G− x ∼= H. Say V (G− x) ={w1, w2, w3, w4, y} and E = {w1w2, w2w3, w3w4, w4w1, yw2, yw3}. Then dG(x) = 8−‖H‖ = 2.As ∆(G) = 3 and dG−x(w2) = 3 = dG−x(w3), we have w2, w3 /∈ NG(x). Thus N(x) ⊆{w1, w4, y}. If xy ∈ E then G − w2 does not contain a cycle with three vertices, so H, J �

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Figure 1.4.2. Discovered graph G

G =

Figure 1.4.3. HW 5

G − y, a contradiction. Thus N(x) = {w1, w4}, and E(G) = E(G − x) ∪ {xw1, xw4}. SeeFigure 1.4.2, and check that the deck of G is G.

Now suppose G′ is another graph with deck G. Then there is x′ ∈ V (G′) such thatG− x ∼= H ∼= G′ − x′. Thus there is an isomorphism f from G− x to G′ − x′. Extending fto f+ : V (G)→ V (G′) by f+(x) = x′ yields an isomorphism from G to G′: f+ is a bijectionand by the same argument as above, N(x′) = N(f+(x)) = {f+(w1), f+(w4)}. �

HW 4. (*) Give a (small) example (with proof) of two graphs that have the same degreesequence, but are not isomorphic.

HW 5. (*) Find a graph G such that the graphs shown in Figure 1.4.3 form a deck G ofG. Prove that if G is also the deck of H then G ∼= H.

HW 6. (*) A graph is regular if all its vertices have the same degree. Prove that if tworegular graphs with at least three vertices have the same deck then they are isomorphic.

1.5. Some Important Graphs and Graph Constructions

A path is a graph P = (V,E) such that V can be ordered as v1, . . . , v|P | so thatE = {vivi+1 : i ∈ [|P |−1]}. The length of the path P is ‖P‖. A path with only one ver-tex is possible; such paths are said to be trivial. Clearly, any two paths with the samelength are isomorphic (HW 10). We use the notation Pn to denote a fixed path of lengthn− 1. Then if P is a path of length n− 1, we say that P is a copy of Pn, or more carelesslyP = Pn. We write v1v2 . . . vn (without commas) to denote a copy of Pn whose edge set is{vivi+1 : i ∈ [n − 1]}. For a path P = v1 . . . vn set viP = vi . . . vn, viPvj = vi . . . vj, andPvj = v1 . . . vj. The vertices v1 and vn are called the ends of Pn. The other vertices areinternal vertices. Suppose G is a graph and X ⊆ V (G). A path P ⊆ G is an X-path if itsends, but not its internal vertices, are contained in X. An {x, y}-path is usually called anx, y-path. The distance d(x, y) between x and y is the length of the shortest x, y-path.

A cycle is a graph C formed by adding the additional edge v1vn to a path v1v2 . . . vnwith n ≥ 3. Again, the length of C is ‖C‖. Clearly any two cycles with the same length areisomorphic. We use the notation Cn for a fixed cycle of length n. We write v1v2 . . . vnv1 todenote a copy of Cn whose edge set is {vivi⊕1 : i ∈ [n]}, where ⊕ denotes addition modulon. The girth of a graph G is the length of its shortest cycle C with C ⊆ G, if there is one;otherwise the girth is infinity. The circumference of G is the length of the longest cycle Cwith C ⊆ G, if there is one. Otherwise the circumference is zero.

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A complete graph is a graph K = (V,E) such that xy ∈ E for all x, y ∈ V . We usethe notation Kn for a fixed complete graph with n vertices. Notice that the vertices of acomplete graph are a clique. Then Kn is a graph with n vertices and no edges, and thevertices of Kn are a coclique. We call Kn the empty graph on n vertices. Now we introducesome notation not in the text: K(A,B) denotes the graph (V,E) such that V = A ∪B andE = {ab : a 6= b ∧ (a, b) ∈ A × B}. Then K(A,A) denotes a complete graph whose vertexset is A; we abbreviate this by K(A). Finally, for a, b ∈ Z+, let Ka,b denote a graph of theform K(A,B), where |A| = a, |B| = b, and A ∩ B = ∅. Such a graph is called a completebipartite graph. (We will have more to say about bipartite graphs shortly.)

Let G = (V,E) and H = (W,F ) be graphs. Define the sum of G and H by

G+H := (V ∪W,E ∪ F );

this may also be denoted by G ∪H. Define the join of G and H by

G ∨H := (G+H) +K(V,W ).

The k-th power of G is the graph Gk = (V,Ek), where Ek = {uv : d(u, v) ≤ k}.The Petersen graph has the form (

([5]2

), {AB : A ∩B = ∅, A,B ∈

([5]2

)}).

HW 7. (*) Prove that every graph G with |G| < ‖G‖ contains a P4.

Definition 1.14. A decomposition of a graph G is a set of subgraphs such that eachedge of G appears in exactly one subgraph of the set.

Example 1.15. K4 can be decomposed into two P4’s; it can also be decomposed intothree P3’s, and into K3, K1,3.

HW 8. Prove that Kn decomposes into three isomorphic subgraphs if and only if n + 1is not divisible by 3.

1.6. Connection in graphs

Definition 1.16 (1.2.2). A walk W = (V,E, σ) is a graph (V,E) and a sequence σ =(v1, . . . , vn) of all the vertices in V with possible repetitions such that E = {vivi+1 : i ∈[n− 1]}. Let G(W ) = (V,E), V (W ) = V , E(W ) = W and σ(W ) = σ. We denote the walkW by v1 . . . vn. If the vertices of σ are distinct then this notation also denotes a path. Ifv1 = vn then W is closed ; otherwise it is open. If W is open then v1 and vn are its ends,and all other vertices are internal. If all the edges vivi+1 are distinct then W is a trail. Thelength of W is n − 1. The walk W is a u, v-walk (trail, path) if u = v1 and v = vn. Thetrivial walk v1 of length 0 is closed. We say that W contains a walk W ′, or W ′ is a subwalkof W , if W ′ is a walk with G(W ′) ⊆ G(W ). We use the notation Wvj, viWvj, and viW toindicate the subwalks v1 . . . vj, vi . . . vj, and vi . . . vn. Also W ∗ := vnvv−1 . . . v1.

Lemma 1.17 (1.2.5). Every u, v-walk W = v1 . . . vn contains a u, v-path.

Proof. We argue by induction on the length l of W . If W is a path then we are done.Otherwise, there exist i, j ∈ [n] with i < j such that vi = vj. Then vivj+1 = vjvj+1 andW ′ = v1 . . . vivj+1 . . . vn is a shorter u, v-walk contained in W . By induction W ′ contains au, v-path P , and P is also contained in W . �

11

Definition 1.18 (1.2.6). Let G = (V,E) be a graph. Vertices x, y ∈ V are connected(regardless of whether xy ∈ E) if there is a u, v-walk in G. The graph G is connected if allvertices x and y are connected. The connection relation is the set of ordered pairs (u, v) ofG such that u and v connected.

Proposition 1.19. The connection relation is an equivalence relation.

HW 9. (-) Let G be a graph with x, y, z ∈ V (G). Prove that if G contains an x, y-pathand a y, z-path then it contains an x, z-path. Be careful; it is not completely trivial.

Definition 1.20 (1.2.8). A component of G is a subgraph H = G[X] induced by anequivalence class X of the connection relation.

HW 10. (+) Prove that if two connected graphs G and H on n vertices both have degreesequence 1, 1, 2, . . . , 2 (two ones followed by n− 2 twos) then they are isomorphic.

HW 11. (*) Let P ⊆ G be an x, y-path. Prove that G[P ] contains an x, y-path Q withQ = G[V (Q) ].

HW 12. (*) Prove that any two paths P and Q with maximum length in a connectedgraph have a common vertex.

1.7. Bipartite graphs

Definition 1.21. Consider a graph G = (V,E) and sets A,B ⊆ V . Let E(A,B) denotethe set of edges with one end in A and one end in B, and abbreviate E({x}, B) as E(x,B).Let ‖x,B‖ = |E(x,B)| and ‖A,B‖ =

∑a∈A |E(a,B)|, that is the number of edges in E(A,B)

counting those edges with both ends in A twice. A graph G = (V,E) is bipartite if it has abipartition, that is a partition of V into one or two independent sets. This means that E = ∅or there exists a partition {A,B} of V (with V = A ∪ B, A ∩ B = ∅) such that both Aand B are independent, or equivalently, E = E(A,B). An A,B-bigraph is a bipartite graphwith bipartition {A,B}. Notice that a graph is bipartite if and only if it is a subgraph of acomplete bipartite graph.

Many theorems in graph theory assert the existence of some special structure in agraph—say a bipartition. To show that a particular graph has such a structure it is enoughto make a lucky guess, and check that your guess provides the structure. In general, it ismuch harder to show that a graph does not have the desired structure. Typically this wouldrequire an exhaustive search of exponentially many possibilities—say all 2|G| partitions ofthe vertices into at most two parts. However for some structures we can prove the existenceof obstructions with the property that every graph either has the structure or it has anobstruction, but not both. In this case, a lucky guess of an obstruction provides a proof thatthe structure does not exist. Theorem 1.24 is an example of this phenomenon.

Definition 1.22. A path, cycle, trail, walk W is even (odd) if its length is even (odd).

Lemma 1.23. Every odd closed walk W = v1 . . . vnv1 contains an odd cycle.

Proof. Argue by induction on the length of W . If W is a cycle we are done. Otherwise,as ‖W‖ ≥ 3, there exist integers 1 ≤ i < j ≤ n with vi = vj. Then vivj+1 = vjvj+1 ∈ E andW ′ := v1 . . . vivj+1 . . . vnv1 and W ′′ := vivi+1 . . . vj are shorter closed walks, whose lengths

12

sum to the length of W . So one of them must be odd. By the induction hypothesis, the oddone contains an odd cycle, which is also contained in W . �

Observe that if xy ∈ E then xyx is an even closed walk that does not contain any cycle.

Theorem 1.24 (1.2.18). A graph G = (V,E) is bipartite iff it contains no odd cycle.

Proof. First suppose G is bipartite with bipartition {A,B}. It suffices to show that ifC ⊆ G is a cycle then it is even. Since G is bipartite, E(C) ⊆ E ⊆ E(A,B). So each edgee ∈ E(C) has exactly one end in A. Thus the length of C is the even number

‖C‖ =∑

e∈E(C)

∑v∈A∩V (C)

ι(v, e) =∑

v∈A∩V (C)

∑e∈E(C)

ι(v, e) =∑

v∈A∩V (C)

dC(v) = 2|A ∩ V (C)|.

Now suppose G is not bipartite. Then some component H ⊆ G is not bipartite (why?).Let x ∈ V (H). Set

A := {v ∈ V (H) : there exists an odd x, v-walk in H} andB := {v ∈ V (H)r A : there exists an even x, v-walk in H}.

Since H is a component of G, it is connected, so A ∪ B = V (H); by definition A ∩ B = ∅.As H is not bipartite there is an edge uv ∈ E(H[A]) ∪E(H[B]). Thus there are walks xPuand xQv with the same parity (odd if u, v ∈ A, even if u, v ∈ B). So W = xPuvQ∗x is anodd closed walk. By Lemma 1.23 there is an odd cycle C ⊆ W , and C ⊆ H ⊆ G. �

HW 13. (*) Prove that a 3-regular graph G decomposes into K1,3’s if and only if G isbipartite.

HW 14. (*) Prove that a graph G is bipartite if and only if α(H) ≥ 12|H| for all H ⊆ G.

1.8. Dirac’s Theorem

Let G = (V,E) be a graph, and suppose A,B ⊆ V . An A,B-walk is a walk whose firstvertex is in A, whose last vertex is in B and whose interior vertices are in neither A nor B. IfA = {a} or B = {b}, we may shorten this notation to an a,B-walk or an A, b-walk. Similarly,if A = B, we may shorten it to a B-walk. Also, if H ⊆ G an H-walk is a V (H)-walk, etc.

A path in a graph G is maximal if it is not a subpath of a longer path in G. It is maximumif there is no longer path in G.

Recall that the minimum degree of G is δ(G) = min{d(v) : v ∈ V }. Similarly, themaximum degree of G is ∆(G) = max{d(v) : v ∈ V }. If δ(G) = r = ∆(G) then G isr-regular.

An embedding of H into G is an isomorphism from H to a subgraph of G. If there existsan embedding of H in G then we say that H can be embedded in G, or that H is embeddablein G. A subgraph H ⊆ G is said to be a spanning subgraph of G if V (H) = V (G). Aspanning cycle of G is called a hamiltonian cycle. If G contains a hamiltonian cycle, G issaid to be hamiltonian.

Many questions in graph theory have the following form: Given two graphs G and Hwith |H| = |G| what “local” conditions on G ensure that H is embeddable in G? If G iscomplete then trivially H is embeddable in G. This is guaranteed by the local condition

13

v1

vtvivi+1

Figure 1.8.1. Hamiltonian cycle

δ(G) = |G| − 1, but in many cases we can do much better. Corollaries 1.27 and 1.26 beloware examples.

We have seen that not only can the question of whether a graph is bipartite be answeredpositively with proof by a lucky guess—the bipartition, it can also be answered negativelywith proof by a lucky guess—an odd cycle. The question of whether a graph is hamiltoniancan also be answered positively with proof by a lucky guess—the hamiltonian cycle. Howeverthere is no “efficient” guessing method known for deciding with proof that a graph has nohamiltonian cycle, and it is strongly believed that there is no such method.

Intuitively, if a graph has enough edges—for instance if it is complete—then it is hamil-tonian. Here are some ways of quantifying what "enough" means.

Theorem 1.25. Every connected graph G = (V,E) with |G| ≥ 3 contains a path or cycleof length at least l = min{|G|, d(x) + d(y) : xy /∈ E}.

Proof. Let P = v1 . . . vt be a maximum path in G. If ‖P‖ ≥ l then we are done. Soassume ‖P‖ ≤ l − 1, i.e. t ≤ l. We first prove:

G contains a cycle with V (C) = V (P ).(1.8.1)

As |G| ≥ 3 and G is connected, ∆(G) ≥ 2, so t ≥ 3. If v1vt ∈ E then, as t ≥ 3,C := v1Pvtv1 is the desired cycle. Otherwise, v1vt /∈ E. So

(1.8.2) t ≤ l ≤ d(v1) + d(vt).

Since P is maximum, N(v1), N(vt) ⊆ P . Let

X = {i ∈ [t− 1] : v1vi+1 ∈ E} and Y = {i ∈ [t− 1] : vtvi ∈ E}.

Then |X| = d(v1) and |Y | = d(vt). By (1.8.2) and inclusion-exclusion:

t− 1 ≥ |X ∪ Y | = |X|+ |Y | − |X ∩ Y | = d(v1) + d(vt)− |X ∩ Y | ≥ t− |X ∩ Y ||X ∩ Y | ≥ 1.

Let i ∈ X ∩ Y (Figure 1.8.1). Then the cycle C = v1vi+1PvtviPv1 spans P , proving (1.8.1).Since l ≤ |G| and |C| = ‖C‖, it suffices to show |G| = |C|. If not, there is x ∈ V rV (C).

Since G is connected, there is an x,C-path Q = x . . . u1. Choose notation so that C =u1 . . . utu1. Then P ′ = xQu1u2 . . . ut is a longer path than P , a contradiction. �

Here is a condition that ensures a graph is hamiltonian.

Corollary 1.26 (Ore’s Theorem 1960 7.2.9). If G is a graph with |G| ≥ 3 and d(x) +d(y) ≥ |G| for all distinct nonadjacent vertices x and y then G is hamiltonian.

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Proof. By Theorem 1.25, it suffices to show that G is connected. For any x, y ∈ V (G),

|G| ≥ |N [x] ∪N [y]| = |N [x]|+ |N [y]| − |N [x] ∩N [y]| ≥ |G|+ 2− |N [x] ∩N [y]||N [x] ∩N [y]| ≥ 2.

So x is connected to y by a path of length at most 2. �

Corollary 1.27 (Dirac’s Theorem 1952 [2] 7.2.8). If δ(G) ≥ 12|G| > 1 then G is

hamiltonian.

Proof. The hypothesis ensures that the hypothesis of Corollary 1.26 holds. �

Pósa (when he was in high school) posed a conjecture extending Dirac’s Theorem:

Conjecture 1.28 (Pósa 1963). If δ(G) ≥ 23|G| then G contains the square (2-power) of

a hamiltonian cycle.

The next Theorem answers a related question.

Theorem 1.29 (Fan & Kierstead 1996 [3]). If δ(G) ≥ 2|G|−13

then G contains the squareof a hamiltonian path. The degree condition is best possible.

The next conjecture generalizes Corollary 1.27 and Conjecture 1.28.

Conjecture 1.30 (Seymour 1974). If δ(G) ≥ kk+1|G| then G contains the k-th power of

a hamiltonian cycle.

Seymour’s Conjecture was proved for sufficiently large graphs (even more vertices thanthe number of electrons in the known universe when k ≥ 2).

Theorem 1.31 (Komlós, Sárközy & Szemerédi 1998 [4]). For every integer k there existsan integer n such that Conjecture 1.30 is true for graphs G with |G| ≥ n.

The next theorem improves the bound on n when k = 2.

Theorem 1.32 (Châu, DeBiasio & Kierstead 2011 [1]). Conjecture 1.28 is true for graphsG with |G| ≥ 2× 108.

HW 15. Prove that Ka,a−1 is not hamiltonian for any integer a > 1. More generally,prove that if α(G) > 1

2|G| then G is not hamiltonian. Determine δ(Ka,a−1).

HW 16. For disjoint sets A,B, {v}, let G = K(A,B) +K(v,A∪B). Prove that G is nothamiltonian. Determine δ(G) when |A| = |B|.

HW 17. Let P = v1 . . . vt ⊆ G be a maximum path and x ∈ V (G − P ). Prove that‖{vt, x}, P‖ ≤ t− 1.

HW 18. Prove that if δ(G) ≥ |G|−12

then G has a hamiltonian path.

HW 19. (+) Let G be an X, Y -bigraph with |X| = |Y | = k and δ(G) ≥ k+12≥ 2. Prove

that G contains a hamiltonian cycle. [Hint: First prove that if G contains a maximal pathP = Pt then G[P ] contains a cycle of length at least 2b t

2c. Then show that G contains a

hamiltonian path.] For all k give an example to show that the bound on the minimum degreecannot be lowered.

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1.9. Even graphs and Euler’s Theorem

Sometimes when proving a statement by induction it is easier to prove a stronger state-ment. This phenomenon is called the inventors paradox. The reason this is possible is thatwhile more must be proved, the induction hypothesis provides more to base an argumenton. In this section we see an elementary example of this. It is easier to prove our result formultigraphs. Proposition 1.35 is a minor example of this; we will see more serious exampleslater.

Definition 1.33. A multigraph is eulerian if it has a closed trail containing all edges.(Note that T = v is closed, since its only vertex is its first and its last.) Such a trail is saidto be an eulerian trail. A multigraph is even if every vertex has even degree. A componentis trivial if it has no edges.

For H ⊆ G and v ∈ V (G) r V (H), set dH(v) := 0. If T is a trail in a graph G then letdT (v) = dG(T )(v).

Fact 1.34. If H and G are even graphs with H ⊆ G then H ′ := G− E(H) is even.

Proof. Since G and H are even, every v ∈ V (G) satisfies

dH′(v) = dG(v)− dH(v) ≡ 0 mod 2. �

Proposition 1.35. Let T = v1...vn be a trail in a multigraph G = (V,E). Then dT (v)is even for every vertex v, except that if T is open then dT (v1) and d(vn) are odd.

Proof. Let T ′ := T + vnv1 (vnv1 may be multiple edge). Then T ′ is closed. It sufficesto show that dT ′(vi) is even for all v ∈ V , since dT (vi) 6≡ dT ′(vi) mod 2 if and only ifvi ∈ {v1, vn} and v1 6= vn. Argue by induction on n.

If T ′ is a cycle then dT ′(vi) = 2 for every v ∈ V (G). Otherwise, there exist 1 < i < j ≤n+ 1 with vi = vj. Let T1 = v1T

′vivj+1T′v1 and T2 = viT

′vj(= vi). Then T1 and T2 are bothclosed trails, shorter than T ′, and every edge of T ′ is in exactly one of T1 and T2. By theinduction applied to T1 and T2,

dT ′(v) = dT1(v) + dT2(v) ≡ 0 + 0 ≡ 0 mod 2. �

Theorem 1.36 (Euler (1736) 1.2.26). A multigraph G is eulerian iff it has at most onenontrivial component and it is even.

Proof. First suppose G has a Eulerian trail T . Since T is connected it only containsedges from one component. Since T contains all edges, G has only one nontrivial component.Since T is closed and contains all edges of G, Proposition 1.35 implies every vertex of G haseven degree (possibly 0).

Now suppose G has at most one nontrivial component H, v ∈ V (H), and every vertexhas even degree. Let T = v1 . . . vt be a maximum length trail in G; it exists because (v) isa candidate. Then T is closed: Otherwise vt is incident to an odd number of edges of T byProposition 1.35. Since d(vt) is even, vt is incident to some edge vtv that is not in T . So wecan extend T to T+ = v1Tvtv, contradicting the maximality of T .

It remains to show that E(H) ⊆ E(T ). Otherwise there is an edge ab ∈ E(H) r E(T ).Since H is connected there is an {a, b}, V (T )-path P = ab . . . vi or P = ba . . . vi. Since T isclosed, T+ = PviTvi is a longer trail than T , a contradiction. �

16

Lemma 1.37. Every graph G with δ(G) ≥ 2 contains a cycle.

Proof. Let P = v1 . . . vt be a maximum path in G; it exists since 1 ≤ |G| ≤ ∞. ThenN(vt) ⊆ V (P ). So there exist i < t − 1 such that vtvi ∈ E(G). Thus viPvtvi is a cyclecontained in G. �

Corollary 1.38 (1.2.25). If G is an even graph with ‖G‖ > 0 then G contains a cycle.

Proof. Some component H of G contains an edge. Since H is connected, δ(H) ≥ 1.Since G is even δ(H) ≥ 2. So by Lemma 1.37, H ⊆ G contains a cycle. �

Second proof of Theorem 1.36 (Sufficiency). SupposeG is even and has at mostone nontrivial component G′. We argue by induction on ‖G′‖. If G′ is a cycle then the cyclein its natural order is the Eulerian trail; if ‖G′‖ = 0 then or any vertex is an Eulerian trail.

Otherwise, by Corollary 1.38, G′ contains a cycle C. Let H be a nontrivial componentof G′ − E(C) (maybe H = G′ − E(C)), and set H ′ = G′ − E(H). Both H and H ′ areeven. Also H ′ is connected, since all components of G′ − E(C) that are contained in H ′

are connected to each other in H ′ by edges of C. Moreover, ‖H‖ ≤ ‖G′‖ − ‖C‖ < ‖G′‖and ‖H ′‖ = ‖G′‖ − ‖H‖ < ‖G′‖. So H and H ′ are nonempty even connected graphs with‖G′‖ = ‖H‖ + ‖H ′‖. By the induction hypothesis H and H ′ contain Eulerian trails T andT ′. Moreover, T contains a vertex v1 ∈ C and T ′ contains all vertices of C. Choose notationso that T = v1 . . . vnv1 and T ′ = v1u2 . . . umv1. Then v1Tvnv1T

′umv1 is an Eulerian trail inG′, and G. �

Theorem 1.39 (1.2.33). A connected graph G with exactly q vertices of odd degree de-composes into max{1, q

2} trails.

Proof. By Lemma 1.11, q is even. Let G+ be the result of adding a new vertex v+

to G so that N(v+) is the set of vertices with odd degree in G. Since q is even, and everyv ∈ V (G) satisfies dG+(v) ≡ dG(v) + 1 mod 2 if and only if dG(v) is odd, G+ is even. ByTheorem 1.36, G+ has an Eulerian Trail T . Removing v+ partitions T into q

2trails that

decompose G. �

Alternatively, we could have proved Theorem 1.39 by adding q edges connecting disjointpairs of odd degree vertices.

17

CHAPTER 2

Cut-vertices, -edges and trees

Definition 2.1 (1.2.12.). A cut-vertex is a vertex in a graph G is a vertex such thatG − v has more components than G. Similarly, a cut-edge is an edge of G such that G − ehas more components than G.

Notice that a vertex v is a cut-vertex of G if and only if H − v is not connected, whereH is the component of G (maybe H = G) containing v. Similarly an edge e is a cut-edge ofG if and only if H − e is not connected, where H is the component of G containing e. ThusG is not a cut-edge if its ends are connected in G− e.

Proposition 2.2 (1.2.14.). An edge e = xy in G is not a cut-edge iff it belongs to acycle.

Proof. First suppose e is not a cut edge. Then there exists an x, y-path P in G − e.So xPyx is a cycle in G. Now suppose e is on a cycle C in G. Then x(C − e)y is a pathconnecting the ends of e, and so e is not a cut-edge. �

Proposition 2.3. The ends of a maximal path P = x . . . y ⊆ G are not cut-vertices ofG.

Proof. Let H be the component of G containing x. If |H| ≤ 2 then we are done; elsey 6= x. It suffices to show that there is a u, y-walk for all u ∈ V (H − x). As H is connected,there is a u, y-walk W in H. If x /∈ V (W ) then W ⊆ H − x. Else the predecessor v of x onW is in N(x) ⊆ P − x; then uWvPy ⊆ H − x. �

Definition 2.4. A graph is acyclic if it contains no cycle. Acyclic graphs are also calledforests. A connected acyclic graph is called a tree. A leaf is a vertex v with d(v) = 1. We saythat a graph G satisfies (A) if it is acyclic, (C) if it is connected, and (E), if |G| = ‖G‖+ 1.

Lemma 2.5. Suppose a graph G with |G| ≥ 2 satisfies at least two of (A), (C) and (E).Then G has at least two leaves.

Proof. If (E) holds then ‖G‖ ≥ |G| − 1 ≥ 1; else (C) holds and G has distinct verticesthat are connected. Anyway, G has an edge.

Suppose (A) holds. Let P = v1 . . . vt be a maximum path in G. As ‖G‖ ≥ 1, v1 6= vt. AsP is maximum and G is acyclic, N(v1) = {v2} and N(vt) = {vt−1}. So v1 and vt are leaves.

Otherwise G satisfies (C) and (E). Let L be the set of leaves in G. Since G is connectedand has an edge, δ(G) ≥ 1. Since G satisfies (E),

2|G| − |L| = 2(|G| − |L|) + |L| ≤∑

v∈V (G)

d(v) = 2‖G‖ =(E) 2|G| − 2

2 ≤ |L|. �

18

Lemma 2.6. Suppose G is a graph with a leaf l and G′ = G − l. Then each condition(A), (C), (E) is satisfied by G iff it is satisfied by G′.

Proof. Suppose G is acyclic. Since removing a vertex cannot create a cycle, G′ isacyclic. Now suppose G′ is acyclic. Since every vertex in a cycle has degree 2, adding a leafl cannot create a cycle, and so G is acyclic.

Suppose G′ is connected. Since l has a neighbor in V (G′), G is connected. Now supposeG is connected. Since d(l) = 1, there is a maximal path P with an end l. Thus l is not acut-vertex, and so G′ is connected.

Since |G| = |G′|+ 1 and ‖G‖ = ‖G′‖+ 1, G satisfies (E) iff G′ does. �

Theorem 2.7 (2.1.4). If a graph G satisfies at least two of the conditions (A), (C), and(E) then it satisfies all three.

Proof. Argue by induction on |G|. If |G| = 1 then G satisfies all of (A), (C) and (E)by inspection. Else |G| ≥ 2. By Lemma 2.5, G has a leaf. Let G′ = G− l. By Lemma 2.6,G′ satisfies the two conditions that G does. By induction G′ satisfies all three conditions.By Lemma 2.6, G does too. �

For emphasis, we state that Theorem 2.7 implies every tree T satisfies |T | = ‖T‖+ 1.

Theorem 2.8 (2.1.4). G is a tree iff there is exactly one path between any two vertices.

Proof. Both statements imply G is connected. Arguing by contraposition, it remainsto show that G has a cycle if and only if it has two vertices x, y and distinct x, y-paths.

If C is a cycle with edge xy then P := xy and Q := x(C − xy)y are distinct x, y-paths.If P and Q are distinct x, y-paths then there is an edge uv with uv ∈ E(P − Q). ThenW := uP ∗xQyP ∗v is a u, v-walk in G− uv, so there is a u, v-path R ⊆ W by Lemma 1.17.Now R + uv is a cycle. �

For graphs H and G, H is a spanning subgraph of G if H ⊆ G and V (H) = V (G). Aspanning tree of G is a spanning subgraph of G that is a tree.

Corollary 2.9 (2.1.5). Let T = (V,E) be a tree. Then(1) Removing an edge disconnects T .(2) Adding an edge xy /∈ E with x, y ∈ V to T creates a unique cycle.

Proof. (1) Since T is acyclic, every edge is a cut edge by Theorem 2.2.(2) By Theorem 2.8 there is a unique x, y-path P ⊆ T . So C := P + xy is a cycle in

T ′ := T +xy. Consider any cycle D ⊆ T ′. Since T is acyclic, xy ∈ E(D). By the uniquenessof P, D − xy = P ; so D = P + xy = C. �

Corollary 2.10. Let G = (V,E) be a connected graph.(1) Every minimally connected spanning subgraph F ⊆ G is acyclic.(2) Every maximally acyclic spanning subgraph F ⊆ G is connected.(3) Every maximal subtree F ⊆ G is spanning.

Proof. (1) Let F be a minimally connected, spanning subgraph of G. Then every edgeof F is a cut-evdge. By Theorem 2.2, no edge of F is on a cycle, so F is acyclic.

(2) Let F be a maximally acyclic spanning subgraph of G. Consider any two verticesx, y ∈ V . As G is connected, there is a path P = x . . . y ⊆ G. Since the connectivity relation

19

is an equivalence relation, it suffices to show that each uv, uv ∈ E(P ), is connected in F. Ifuv ∈ E(F ) then u and v are connected in F . Else by maximality, there is a cycle C ⊆ F +uvwith uv ∈ E(C), and so u and v are connected by the path u(C − uv)v ⊆ F . Thus F isconnected.

(3) Let F ⊆ G be a subtree with |F | maximum. Consider any x ∈ V (G). Since Gis connected, there is an x, F -path (x =)x1 . . . xt ⊆ G. As |F | is maximum, t = 1, andx ∈ V (F ). �

Proposition 2.11 (2.1.5). Suppose T and T ′ are spanning trees of a graph G. Then forevery e = ab ∈ E(T )r E(T ′)

(1) there exists e′ ∈ E(T ′)r E(T ) such that T − e+ e′ is a spanning tree of G; and(2) there exists e′ ∈ E(T ′)r E(T ) such that T ′ + e− e′ is a spanning tree of G.

Proof. (1) The graph H := (T − e) ∪ T ′ is a connected since T ′ is. Let T ∗ ⊆ H be aminimum subject to T ∗ is connected and T −e ⊆ T ∗. Then all edges in E0 := E(T ∗)∩E(T ′)are cut edges, and so do not appear in cycles of T ∗. As T − e is acyclic, T ∗ is acyclic, andso it is a tree. By Theorem 2.7,

‖T − e‖+ |E0| = ‖T ∗‖ = |T ∗| − 1 = |T | − 1 = ‖T − e‖+ 1.

So E0 = {e′} for some e′ ∈ E(T ′), and T ∗ = T − e+ e′.(2) By Corollary 2.9(2), T ′ + e contains a unique cycle C. Since T is acyclic, C * T ; let

e′ ∈ E(C − T ) ⊆ E(T ′). Then T ∗ := T ′ + e− e′ is acyclic, since e′ is an edge of the uniquecycle in T ′ + e. By Theorem 2.7, T ∗ is a tree, since |T ∗| = |T | = ‖T‖+ 1 = ‖T ∗‖+ 1. �

Definition 2.12. If G = (V,E) is a graph with a function w : E → N then w is calleda weight function. For H ⊆ G set

w(H) =∑

e∈E(H)

w(e).

A minimal spanning tree of G (with respect to w) is a spanning tree T ⊆ G such thatw(T ) ≤ w(T ′) for all spanning trees T ′ ⊆ G. Consider the following algorithm.

Minimal Spanning Tree Algorithm: (1) input a connected graph G = (V,E) with aweight function w;

(2) let e′1, . . . , e′|G| be an enumeration of E that is increasing with w;(3) for i from 1 to |G| − 1 do

(a) choose ei ∈ E r {e1, e2 . . . , ei−1} such that (V, {e1, . . . , ei}) is acyclic andsubject to this w(ei) is minimum

(b) end do;(4) output T = (V, {e1, . . . , e|G|−1});

Theorem 2.13. The Minimal Spanning Tree algorithm computes a minimal spanningtree of the input graph G.

Proof. It suffices to prove by induction on i that there is a minimal spanning tree T ∗such that e1, . . . , ei ∈ E(T ∗). The base step i = 0 is trivial, so suppose i ≥ 1. By inductionthere is a minimal spanning tree T ′ with {ej : 1 ≤ j ≤ i − 1} ⊆ E(T ′). If ei ∈ E(T ′)we are done, so suppose not. By Proposition 2.11, there is e′ ∈ E(T ′) r E(T ) such that

20

T ∗ := T ′ − e′ + ei is a spanning tree. As e1, . . . , ei−1 ∈ T ′, (V, {e1, . . . , ei−1, e′}) is acyclic.

Thus e′ was a candidate when ei was chosen. Since ei was preferred, w(ei) ≤ w(e′), andw(T ′) ≤ w(T ∗) = w(T ′)− w(e′) + w(ei) ≤ w(T ′).

Thus w(T ∗) = w(W ′) and T ∗ is a minimal spanning tree with e1, . . . ei ∈ E(T ∗). �

Proposition 2.14. If T is a tree with k edges and G is a nontrivial graph with δ(G) ≥ kthen G contains a copy of T , i.e., a subgraph isomorphic to T .

Proof. Argue by induction on k.Base Step: k = 0. Then T ∼= K1, so T ∼= G[{v}] for any vertex v.Induction Step: k > 1. Let l be a leaf of T . Then T ′ := T − l is a tree with ‖T ′‖ = k−1. Bythe induction hypothesis there exists H ′ ⊆ G with H ′ ∼= T ′. Let p be the unique neighbor ofl in T , and let x be the image of p in H ′. Since |H ′| = ‖H ′‖+ 1 = k and x is not adjacent toitself, x has at most k− 1 neighbors in H ′. Since δ(G) ≥ k, there exists y ∈ NG(x)rV (H ′).Set H = H ′ + y + xy. Then H ⊆ G and we can extend the isomorphism between T ′ and H ′to an isomorphism between T and H by mapping l to y. �

HW 20. (+) Let d1, . . . , dn be positive integers with n ≥ 2. Prove that there exists atree with vertex degrees d1, . . . , dn iff

∑di = 2n− 2.

HW 21. (*) Every tree is bipartite. Prove that every tree has a leaf in its larger partiteset, and in both partite sets if they have equal size.

HW 22. (*) Let T be a tree such that 2k of its vertices have odd degree. Prove that Tdecomposes into k paths. Is it easier to prove this for forests?

HW 23. (+) Let T be a tree with even order. Prove that T has exactly one spanningsubgraph such that every vertex has odd degree.

HW 24. (+) A root of a graph G is a special vertex r. A spanning tree T of a graphwith root r is normal if every edge xy ∈ E(G) satisfies either x ∈ rTy or y ∈ rTx. Provethat every connected graph with root r has a normal spanning tree [Hint: Prove the strongerstatement that every path P with end r is contained in a normal spanning tree.]

HW 25. (+) Let T be a set of subtrees of a tree G such that T ∩T ′ 6= ∅ for all T, T ′ ∈ T .Prove that

⋂T 6= ∅.

HW 26. (*) A directed graph is a binary relation G = (V,E). So the edges are orderedpairs. We still write xy for the directed edge (x, y). Let E+

G(x) := {xy ∈ E}, d+G(x) :=

|E+G(x)|, and δ+(G) := minv∈V d

+G(v)}. A directed cycle C = (V,E) is a digraph with the

form V = {v1, . . . , vs} and E = {vivi⊕1 : i ∈ [s]}. This directed cycle is also denoted byv1 . . . vsv1 when it is clear form the context that the edges are directed.

Prove that a directed graph G has a directed cycle if δ+(G) ≥ 1. [Hint: You might wantto define the notion of a directed path.]

21

CHAPTER 3

Matchings

In this Chapter we will need the following easy proposition several times.

Proposition 3.1. Every connected multigraph G with ∆(G) ≤ 2 is a path or cycle.

Proof. Let P = v1 . . . vt be a maximum path in G. As ∆(G) ≤ 2, no other edge isincident to the internal vertices of P . As P is maximum, no edge joins an end of P with avertex of G − P . Since G is connected, P spans G. So G is the path P or G is the cycleP + v1vt. �

Definition 3.2. Let G = (V,E) be a graph. A matching in G is a set of edges M ⊆ Ewith no common ends. Amaximal matching is a matching that cannot be enlarged by addingan edge. A maximum matching is matching with maximum size among all matchings in theG. A vertex of G is said to be M -saturated if and only if it is the end of some edge in M ;otherwise it is M -unsaturated . A set of vertices X ⊆ V is said to be M -saturated if everyx ∈ X is M -saturated. The matching M is perfect if every vertex of G is M -saturated.

Definition 3.3. Given a matching M in a graph G = (V,E), an M -alternating path isa path P such that each vertex v ∈ V (P ) is incident to at most one edge in E(P )rM . Analternating path P is M -augmenting if |P | ≥ 2, and neither end of P is M -saturated in G.When M is clear from the context we may drop the prefix M .

Theorem 3.4 (3.1.10 Berge). A matching M in a graph G = (V,E) is not maximum inG iff G has an M-augmenting path.

Proof. Suppose P is an M -augmenting path. Then

M ′ = M 4 E(P ) =def (M r E(P )) ∪ (E(P )rM)

is a larger matching.Now suppose M is not maximum. Choose a matching M ′ with |M ′| > |M |. Let H be

the spanning submultigraph with edge set M ∪M ′, where we take two copies of an edgethat appears in both M and M ′. Since each vertex is incident to at most one edge of eachmatching, ∆(H) ≤ 2, and the components of H are cycles and alternating paths. Each cyclehas the same number of edges from M as M ′. Since |M | < |M ′|, some component Q of Hhas more edges from M ′ than M . Such a component must be an M -augmenting path. �

HW 27. (*) Two players Alice and Bob play a game on a graph G. Alice begins thegame by choosing any vertex. All other plays consist of the player, whose turn it is, choosingan unchosen vertex that is joined to the last chosen vertex. The winner is the last player toplay legally. Prove that Alice has a winning strategy if G has no perfect matching, and Bobhas a winning strategy if it does.

22

3.1. Bipartite matching

A bipartite G with bipartition {X, Y } is called an X, Y -bigraph. For S ⊆ X set N(S) :=⋃v∈S N(v). For a function f : A → B and S ⊆ A, let f(S) := {y ∈ B : ∃x ∈ S(f(x) = y)}

be the range of f restricted to S.

Theorem 3.5 (3.1.11 Hall’s Theorem [1935]). An X, Y -bigraph G has a matching thatsaturates X iff

(3.1.1) |S| ≤ |N(S)| for all S ⊆ X.

Proof. IfM is a matching saturating X and S ⊆ X then |S| = |E(S, Y )∩M | ≤ |N(S)|;so (3.1.1) holds.

Suppose (3.1.1) holds for some X, Y -bigraph with no matching saturating X; amongsuch counterexamples choose G with |G| minimal, and subject to this ‖G‖ maximal. Byminimality, (1) N(X) = Y : if y ∈ Y rN(X) then G− y satisfies (3.1.1), and it is a smallercounterexample. Also (2) all a ∈ X satisfy N(a) 6= Y : else G − a satisfies 3.1.1, so G − ahas a matching M saturating X − a; by (3.1.1) applied to S = X, there is an unsaturatedvertex b ∈ Y , so M + ab is a matching in G saturating X, a contradiction.

Let a ∈ X; by (2) there is b ∈ Y with ab /∈ E. Since G+ ab satisfies (3.1.1), maximalityimplies G+ab has a matchingM+ saturating X. By (1) there is a′ ∈ X with a′b ∈ E; by (2)there is b′ ∈ Y with a′b′ /∈ E. Again by maximality, G + a′b′ has a matching L+ saturatingX. Set M := M+−ab and L := L+−a′b′. Then M saturates X−a and L saturates X−a′.So M is maximum—any bigger matching would saturate X.

Let H be the spanning subgraph of G with E(H) = M 4 L. (Figure 3.1.1). Then∆(H) ≤ 2, dH(b) ≤ 1 = dH(a) = dH(a′), and if xy ∈ M ∩ L then dH(x) = 0 = dH(y).So the component of H containing a is an alternating a, v-path P , and dP (x) = 2 for allx ∈ V (P ) ∩ X r {a, a′}. If v ∈ Y then P is an M -augmenting path. Else v = a′. SincedH(b) ≤ 1, b /∈ V (P ). Thus aPvb is an M -augmenting path in G. Anyway, G has anM -augmenting path, so M is not maximum, a contradiction. �

Here is an other formulation of the backward direction of Theorem 3.5. The proof givenabove reflects how graph theorists would typically think when trying to prove the theorem.For students, it may also be useful to think in terms of induction. But aside from theframework, the two proofs are essentially the same.

Proof. Argue by induction on |G|+∥∥G∥∥. By (3.1.1), |X| ≤ |N(X)| ≤ |Y |.

Case 0: X = ∅. Then the result holds trivally.Case 1: There is y ∈ Y rN(X). Since G− y satisfies (3.1.1), we are done by induction.Case 2: There is a ∈ X with N(a) = Y . As G− a satisfies 3.1.1, by induction G− a has amatching M saturating X − a; since |X − a| < |Y |, there is an unsaturated vertex b ∈ Y .So M + ab is a matching in G saturating X.Case 4: None of Cases 0–3 hold. Let a ∈ X; by not Case 2, there is b ∈ Y with ab /∈ E. LetG′ = G+ ab; then |G′|+

∥∥∥G′∥∥∥ < |G|+∥∥G∥∥. By induction G′ has a matching M+ saturatingX. By Case 1, there is a′ ∈ X with a′b ∈ E. By Case 2, there is b′ ∈ Y with a′b′ /∈ E.Again by induction, G + a′b′ has a matching L+ saturating X. Set M := M+ − ab andL := L+ − a′b′. Then M saturates X − a and L saturates X − a′. If a is M -saturated or a′is L-saturated we are done. So suppose not; then b is M -unsaturated.

23

a a′

b b′Y

X

M,L

Figure 3.1.1. X, Y -bigraph H ⊆ G

Let H be the spanning subgraph of G with E(H) = M 4 L. (Figure 3.1.1). Then∆(H) ≤ 2, dH(b) = 1 = dH(a) = dH(a′), and if xy ∈ M ∩ L then dH(x) = 0 = dH(y).So the component of H containing a is an alternating a, v-path P , and dP (x) = 2 for allx ∈ V (P ) ∩ X r {a, a′}. If v ∈ Y then Q := P is an M -augmenting path. Else v = a′.Since dH(b) = 1, b /∈ V (P ). Thus Q := aPvb is an M -augmenting path in G. Anyway, Q isM -augmenting. Thus M 4 E(Q) is a matching that saturates X. �

Recall that a graph (or multigraph) is k-regular if every vertex has degree k. A k-factoris a k-regular subgraph. So a subgraph whose edge set is a matching is a 1-factor.

Corollary 3.6 (3.1.13). For k ∈ Z+, every k-regular bipartite multigraph has a 1-factor.

Proof. Suppose G is an k-regular X, Y -bimultigraph. Then

k|X| = |E(X, Y )| = k|Y |.

It follows that |X| = |Y |. Thus it suffices to show that G has a matching that saturates X.By Hall’s Theorem, it suffices to check 3.1.1. Consider any subset S ⊆ X. Then

k|S| = |E(S, Y )|=|E(S,N(S))|≤|E(X,N(S)|=k|N(S)|.

So |S| ≤ |N(S)|. �

HW 28. (*) Let S = {Si : i ∈ I} be a family of sets. A system of distinct representatives(sdr) for S is a sequence (ai : i ∈ I) of distinct elements such that ai ∈ Si for all i ∈ I. Provethat if S is finite then S has an sdr if and only if |J | ≤ |

⋃j∈J Sj| for all J ⊆ I.

HW 29. (*) Prove that there is an injection f :(

[2k+1]k+1

)→(

[2k+1]k

)such that f(S) ⊆ S

for all S ∈(

[2k+1]k+1

).

HW 30. (*) Let P = {P1, . . . , Pt} and Q = {Q1, . . . , Qt} be two partitions of a set Sinto t subsets of size k. Prove that P and Q have a common sdr, that is there exist an sdr(ai : i ∈ [t]) of P and a permutation σ : [t]→ [t] such that (aσ(i) : i ∈ [t]) is an sdr of Q.

HW 31. (+) Suppose G is an X, Y -bigraph with δ(G) ≥ 1 such that every edge xy withx ∈ X, satisfies d(x) ≥ d(y). Prove that G has a matching that saturates every vertex in X.[Hint: Consider assigning each edge xy with x ∈ X the weight w(xy) = 1

d(x).]

HW 32. (+) For k, n ∈ N, let G be an A,B-bigraph with |A| = n = |B| such thatδ(G) ≥ k, and for all X ⊆ A, Y ⊆ B, if |X|, |Y | ≥ k then |E(X, Y )| 6= ∅. Prove: G has aperfect matching.

24

HW 33. (++) Prove that for all partitions P = {Pi : i ∈ [n]} and Q = {Qi : i ∈ [n]} ofa region S of area n into parts of area 1 there exists a permutation σ : [n] → [n] such thatarea (Pi ∩Qσ(i)) ≥ f(n) for all i ∈ [n], where

f(n) :=

{4

(n+1)2if n is odd

4n(n+2)

if n is even.

Also show that the function f is optimal.

Definition 3.7. A cover of a graph G is a subset Q ⊆ V (G) that contains at least oneend of every edge.

Let C be an odd cycle with ‖C‖ = 2k + 1. Since C is 2-regular, every m-set Q ⊆ V (G)covers at most 2|Q| edges. Thus every vertex cover of C has at least k + 1 vertices. On theother hand, every m-matching in C has 2m distinct ends; so m ≤ k.

Theorem 3.8 (König, Egerváry [1931] 3.1.16). If G is a graph with matching M andcover W then |M | ≤ |W |; if G is bipartite then M and W can be chosen with |M | = |W |.

Proof. Order V := V (G) as v1 ≺ · · · ≺ v|G|. Since W is a cover, every edge of M isincident to some vertex of W (possibly two). Define a function g : M → W by g(e) is theleast w ∈ e ∩W . Since M is a matching, no vertex of W can be incident two edges of M .So g is an injection. Thus |M | ≤ |W |.

Now supposeG is anX, Y -bigraph, and letM be a maximummatching. We will constructa cover W with |M | = |W |. Set H = (V,M). As M is a matching, |NH(S)| ≤ |S| for allS ⊆ X. Let U be the set of unsaturated vertices in X. If U = ∅ then X is a cover with|X| = |M |; so assume U 6= ∅. Let A ⊆ V (G) be the set of ends of alternating paths startingin U . Set S = A ∩ X, S = X r S, T = A ∩ Y , and T = Y r T . Then U ⊆ S (witnessedby trivial paths). Consider any alternating path P = v0 . . . vn with v0 ∈ U . If i is even thenvi ∈ S, and if also i 6= 0 then vi−1vi ∈M ; if i is odd then vi ∈ T . We first show:

(i) NG(S) ⊆ T and (ii) T ⊆ NH(S r U).(3.1.2)

(i) Let z ∈ S and w ∈ N(z). Then w ∈ Y . Since z ∈ S, there is an alternating pathQ = y0 . . . y2k with y0 ∈ U and y2k = z. Either w ∈ V (Q) or Qzw is an alternating pathstarting in U . Anyway, z ∈ T , proving (3.1.2.i).

(ii) Let w ∈ T . Then there is an alternating path P = y0 . . . y2k+1 with y0 ∈ U , y2k+1 = wand y2kw /∈ M . Since M is maximum, G has no augmenting path. So w is saturated; saywx ∈ M . Then Pwx is an alternating path; so x ∈ S r U , and z ∈ NH(S r U), proving(3.1.2.ii).1

Set W = S ∪ T . Then W is a cover: Suppose xy ∈ E with x ∈ X. If x ∈ S theny ∈ T ⊆ W by (i); else x ∈ S ⊆ W .

Lastly, we show |W | ≤ |M |, by proving g (defined for M and W ) is onto. Let w ∈ W . Ifw ∈ T then there is x ∈ S with xw ∈M by (ii). As x /∈ W , g(xw) = w. If w ∈ S then thereis y with wy ∈M since U ⊆ S. As M is a matching, (ii) implies y ∈ T . So g(wy) = w. �

HW 34. (*) Prove that every bipartite graph has a matching of size at least ‖G‖∆(G)

.

1By (i) and (ii), (3.1.1) implies G has a perfect matching: |N(S)| ≤ |S r U |, so (3.1.1) fails if U 6= ∅.

25

3.2. General matching

Notice that if H is a component of a graph G and |H| is odd then G does not have aperfect matching.

Definition 3.9. Let CG be the set of components of the graph G. A component with anodd number of vertices is said to be an odd component. Let OG be the set of odd componentsof G and o(G) = |OG|.

Theorem 3.10 (Tutte [1947] 3.3.3). A graph G = (V,E) has a perfect matching iff

(3.2.1) o(G− S) ≤ |S| for all S ⊆ V.

Proof. Suppose G has a perfect matching. For all S ⊆ V , all odd components of G−Smust have vertices matched to distinct vertices in S. So o(G− S) ≤ |S|.

Now suppose (3.2.1) holds for some graph G0 with no perfect matching. Among all suchcounterexamples, choose one G = (V,E) with ‖G‖ maximum subject to |G| = |G0|. Addingany edge cannot cause G to violate (3.2.1), so it must create a perfect matching. Applying(3.2.1) to S = ∅ shows that G has no odd component; so |G| is even. Let U := {v ∈ V :N [v] = V }. For a contradiction we show that G has a perfect matching.

First suppose G−U consists of disjoint complete subgraphs. By (3.2.1) |U | ≥ o(G−U);so one vertex in each odd component can be matched to a vertex in U . The remainingvertices of G− U are in disjoint even cliques, and the remaining vertices of U form an evenclique, since |G| is even. Each of the even cliques has a perfect matching. Combining thesematchings yields a perfect matching of G.

Otherwise some component of G − U contains two nonadjacent vertices, and so a pathaba′ with aa′ /∈ E. Since b /∈ U , there exists b′ ∈ V with bb′ /∈ E. By maximality, G + aa′

and G + bb′ have perfect matchings M+ and L+, respectively. Set M := M+ − aa′ andL := L+ − bb′. The component of M 4 L containing a is a path P = a . . . x. Either P isM -augmenting, or P is M -alternating and x is L-unsaturated. The latter case implies baPis L-augmenting if x 6= b, and Pxa′ is M -augmenting if x = b. Anyway, G has an M - orL-augmenting path, and so a matching M∗ with |M∗| = |M+| = |L+|. Thus M∗ is perfect.

Here is an alternative line through the last paragraph proposed by Theo Molla:Otherwise some component of G − U contains two nonadjacent vertices, and so a path

aba′ with aa′ /∈ E. Since b /∈ U , there exists b′ ∈ V with bb′ /∈ E. By maximality, G + aa′

and G+ bb′ both have perfect matchings. So G has matchings M and L with a, a′ the onlyM -unsaturated vertices and b, b′ the only L-unsaturated vertices. Let H ⊆ G be a subgraphwith E(H) = M 4L+ aba′. Since b and b′ are the only vertices of H with odd degree, theyare in the same component of H; say P = b′ . . . xyb ⊆ H. Then E(b′Py) ⊆M4L and eitheryb ∈ M or y ∈ {a, a′} and xy ∈ L. Anyway, P is an L-augmenting path in G. So G has aperfect matching L∗ := L4 E(P ). �

The next theorem is a generalization of Tutte’s Theoerm. We will give two proofs. Thefirst builds on the previous proof. The second starts from scratch.

Theorem 3.11 (Tutte [1947] 3.3.3+3.3.7). The number of unsaturated vertices of a max-imum matching in a graph G = (V,E) is equal to

(3.2.2) d := maxS⊆V

o(G− S)− |S|.

26

First Proof. Say d = o(G− S)− |S| and M is a matching. Every odd component ofG − S contains an M -unsaturated vertex or a vertex x that is matched to a vertex y ∈ S,i.e. xy ∈M . Thus the number of M -unsaturated vertices is at least d.

Now we show that G has a matching with at most d unsaturated vertices. Using S = ∅shows that, d ≥ 0. By Theorem 3.10, assume d ≥ 1. Let G+ := G ∨Kd with Q = V (Kd).

We claim that o(G+ − T ) − |T | ≤ 0 for all T ⊆ V (G+). If Q * T then it is connected,so it has only one component—itself. Moreover, if T = ∅ then o(G+ − T ) = 0, since

|G+ − T | = |G+| = |G|+ d ≡ o(G− S) + |S|+ d = 2d+ 2|S| ≡ 0 mod 2.

Anyway, o(G+ − T ) ≤ 1 and o(G+ − T )− |T | ≤ 0. Else Q ⊆ T . So every odd component ofG+ − T is an odd component of G− (T −Q). It follows that

o(G+ − T )− |T | = o(G− (T −Q))− |T rQ| − |Q| ≤ d− d ≤ 0.

By Theorem 3.10, G+ has a perfect matching M+. Then M := M+ r E(Q, V rQ) is amatching in G with at most |Q| = d unsaturated vertices. �

Here is the second proof of Theorem 3.11. First some definitions. A graph G is factorcritical if G − v has a perfect matching for every vertex v ∈ V (G). A set S is matchableinto OG−S if there exists a matching M that saturates S such that each edge e ∈M has oneend in S and one end in an odd component of G − S, and at most one vertex of each oddcomponent is saturated.

Second Proof. Again, there are at least d unsaturated vertices for any matching. Soarguing by induction on |G| it suffices to show that some matching has d unsaturated vertices.For the base step |G| = 1, let S = ∅. Then o(G − S) − |S| = 1 and the only vertex of G isunsaturated by any matching. Now consider the induction step.

Choose a set S ⊆ V so that o(G− S)− |S| is maximum, and subject to this, |S| is alsomaximum. We first prove the following three claims:

Claim (1). Every component of G− S is odd.

Proof. Suppose H ∈ CG−S with |H| even. Choose a non-cut vertex x (end of a maximalpath) of H, and set S ′ = S + x. Then

OG−S′ = OG−S + (H − x) and |S ′| = |S|+ 1.

Thus o(G− S)− |S| = o(G− S ′)− |S ′|, contradicting the choice of S, since |S| < |S ′|. �

Claim (2). Every odd component of G− S is factor critical.

Proof. Consider any H ∈ OG−S and any vertex x ∈ V (H). We must show thatH ′ = H − x has a perfect matching. By the induction hypothesis, it suffices to show thato(H ′ − T ) − |T | ≤ 0 for all T ⊆ V (H ′). So consider any such T , and set S ′ = S ∪ T + x.Then |S ′| = |S|+ |T |+ 1 > |S|, and so by the choice of S

o(G− S)− |S| > o(G− S ′)− |S ′|.

Since T + x ⊆ V (H),OG−S′ = (OG−S −H) ∪ OH′−T .

27

Soo(G− S)− |S| > o(G− S ′)− |S ′| = o(G− S)− 1 + o(H ′ − T )− |S| − |T | − 1

2 > o(H ′ − T )− |T |.Moreover, by Claim (1), H is an odd component, and so |H ′| is even. Thus

0 ≡ |H ′| ≡ |H ′ − T |+ |T | ≡ o(H ′ − T ) + |T | ≡ o(H ′ − T )− |T | mod 2.

Hence 1 666= o(H ′ − T )− |T |, and so 0 ≥ o(H ′ − T )− |T |. �

Claim (3). S is matchable into OG−S.Proof. Let H be the S,OG−S-bigraph with edge set

F := {xD : x ∈ S,D ∈ OG−S and N(x) ∩ V (D) 6= ∅}.It suffices to show that H has a matching that saturates S. For this we apply Hall’s Theorem.

Consider any set T ⊆ S. Let S ′ := S − T . By the choice of So(G− S)− |S| ≥ o(G− S ′)− |S ′| ≥ o(G− S)−NH(T )− |S|+ |T |

|NH(T )| ≥ |T |. �

Finally, we obtain a maximum matching M as follows. By Claim (3) there is a matchingM0 that saturates S and one vertex of |S| odd components. For each H ∈ OG−S choosea vertex vH , and if possible, choose vH so that it is M0-saturated. Next use Claim (2) toobtain matchings MH of H − vH for every odd component H ∈ OG−S. Then

M := M0 ∪⋃

H∈OG−S

MH

is matching of G. Using Claim (1), it saturates every vertex of G except those o(G−S)−|S|vertices vH that are not saturated by M0. �

We have actually proved a stronger statement.

Theorem 3.12. Let G = (V,E) be a graph and S∗ ⊆ V be a set of vertices such that forall S ⊆ V

(1) o(G− S)− |S| ≤ o(G− S∗)− |S∗|; and(2) if equality holds in (1) then |S| ≤ |S∗|.

Then every component of G − S∗ is odd and factor-critical; and every maximum matchingsaturates S∗, matches S∗ into OG−S∗, and leaves at most one vertex of each component ofG− S∗ unsaturated.

A graph G = (V,E) is transitive if for all x, y ∈ V there is an automorphism ϕ of G withϕ(x) = y. For example Cn, Kn,n and the Petersen graph, but not Pn, are transitive.

HW 35. (*) Prove that a transitive graph does not have a cut vertex.

HW 36. (+) Prove that if G is a connected, transitive graph with |G| even then G hasa perfect matching. [Hint: Use Theorem 3.11 to show that if G does not have a perfectmatching then some, but not all, of its vertices have the property that they are saturatedin every maximum matching. Where does your argument use that G is connected and |G|is even?] (Lovasz has conjectured that every connected transitive graph has a hamiltonianpath; there are only four known examples of such graphs that are not hamiltonian.)

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3.3. Applications of Matching Theorems

A graph is k-regular if every vertex has degree k. A cubic graph is a 3-regular graph. Ak-regular spanning subgraph of a graph G is called a k-factor of G. Thus the edge set of a1-factor of G is a perfect matching. A cut-edge is also called a bridge. A bridgeless graph isa graph without cut-edges. It need not be connected.

Theorem 3.13 (Petersen [1891] 3.3.8). Every bridgeless cubic graph G = (V,E) containsa 1-factor.

Proof. By Tutte’s Theorem, it suffices to show that o(G − S) ≤ |S| for every subsetS ⊆ V . Fix any such S and consider any H ∈ OG−S. Since G is cubic and |H| is odd,

3|H| =∑

v∈V (H)

d(v) = 2‖H‖+ |E(V (H), S)| ≡ 1 mod 2.

It follows that |E(V (H), S)| is odd, and since G is bridgeless, |E(V (H), S)| ≥ 3. Thus

3o(G− S) ≤ |E(S, V r S)| ≤ 3|S|,

and so o(Gr S) ≤ |S|. �

Theorem 3.14 (Petersen [1891] 3.3.9). Every 2k-regular graph with k ∈ Z+ has a 2-factor.

Proof. Suppose G = (V,E) is 2k-regular with k ∈ Z+. It suffices to show that eachcomponent of G has a 2-factor, so we may assume G is connected. By Euler’s Theorem 1.36,G has an Eulerian trail T = v1 . . . vnv1. Let V ′ = {v′ : v ∈ V } and V ′′ = {v′′ : v ∈ V } besets of new vertices, disjoint from V and each other, where v 7→ v′ and v 7→ v′′ are bijections.Let H be the V ′, V ′′-bigraph defined by E(H) = {v′iv′′i⊕1 : i ∈ [n]}. Since each vertex v of Gis incident to 2k edges, it appears k times in T . Say v = vi1 = · · · = vik . Then

E(v) = {vi1−1vi1 , vi1vi1+1, . . . , vik−1vik , vikvik+1}

andNH(v′) = {v′′i1+1, . . . , v

′′ik+1} and NH(v′′) = {v′i1−1, . . . , v

′ik−1}.

So H is k-regular. By the Corollary 3.6, H has a perfect matching M . Let F = {xy ∈ E :x′y′′ ∈M}. Then (V, F ) is a 2-factor of G: for each y ∈ V there exists a unique x such thatx′y′′ ∈ M and a unique z such that y′z′′ ∈ M . Since T is a trail, and xy, yz ∈ T , we havexy 6= yz. �

HW 37. (*) Prove that a 3-regular graph has a 1-factor iff it decomposes into copies ofP4.

HW 38. Suppose G is a graph on 2k vertices with k ≥ 3, whose complement G does nothave a 1-factor. Let S be the set whose existence is guaranteed by Tutte’s Theorem (appliedto G). Prove that

(1) If |S| = 0 then G contains Kc,2k−c for some odd c.(2) If |S| ≥ k − 1 then G contains Kk+1.(3) If 1 ≤ |S| ≤ k− 2 then ∆(G) ≥ k+ 1. [Hint: x+1

x+2(2k− x) > k when 1 ≤ x ≤ k− 2.]

29

HW 39. (+) Let M be a matching in a graph G with an M -unsaturated vertex u. Provethat if G has no M -augmenting path starting at u then G has a maximum matching L suchthat u is L-unsaturated.

HW 40. (+) A graph is claw-free if it does not contain an induced K1,3. Prove that aconnected claw-free graph of even order has a 1-factor. Find (easy) a small counter exampleif the graph is not connected.

HW 41. (*) Let G be a k-regular graph with |G| even that remains connected when anyk − 2 edges are removed. Prove that G has a 1-factor.

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CHAPTER 4

Connectivity

Definition 4.1. A separating set or vertex cut of a graph G is a set S ⊆ V (G) suchthat G− S has more than one component. The connectivity κ(G) is the minimum size of avertex set S such that G− S has more than one component or only one vertex. A graph Gis k-connected if k ≤ κ(G). Two vertices x and y are separated by S if they are in differentcomponents of G− S.

Note that it is not possible to disconnect a complete graph by removing vertices. Thedefinition sets the connectivity of a complete graph K equal to |K| − 1.

Definition 4.2. Let G be a graph with |G| > 1. A disconnecting set of edges in G isa set F ⊆ E(G) such that G − F has more than one component. The edge-connectivityκ′(G) of G is the minimum size of a set F ⊆ E(G) such that G − F is disconnected. Itis k-edge-connected if k ≤ κ′(G). Two vertices x and y are separated by F if they are indifferent components of G− F .

Following the text, we write [S, T ] for E(S, T ). An edge cut in G is a set of edges of theform [S, S], where ∅ 6= S 6= V (G) and S denotes V (G)r S.

4.1. Basics

Theorem 4.3 (Whitney [1932] 4.1.9). Every graph G = (V,E) satisfies

κ(G) ≤ κ′(G) ≤ δ(G).

Proof. Choose a vertex v ∈ V with d(v) = δ(G). Then E(v) is a disconnecting set ofedges of size δ(G), and so κ′(G) ≤ δ(G).

For the first inequality, let [S, S] be a minimum edge cut; so∥∥S, S∥∥ = κ′(G). Note that

κ(G) ≤ |G| − 1. If every vertex in S is adjacent to every vertex in S, then

κ′(G) =∥∥S, S∥∥ = |S||S| ≥ |G| − 1 ≥ κ(G).

Else there are x ∈ S and y ∈ S with xy /∈ E. Define f : [S, S] → V by f(e) = z if e = xz;else f(e) ∈ e ∩ (S − x). So f(e) ∈ er {x, y}. Every x, y-path P contains an edge e ∈ [S, S],and f(e) is an interior vertex of P . So range(f) separates x from y. Thus

κ′(G) =∥∥S, S∥∥ ≥ |range(f)| ≥ κ(G). �

Theorem 4.4. Every cubic graph G = (V,E) satisfies κ(G) = κ′(G).

Proof. Put κ := κ(G) and κ′ := κ′(G). If κ = 0 then G is disconnected, and so κ′ = 0also. If κ = 3 then 3 = κ ≤ κ′ ≤ δ(G) = 3, and again κ = κ′. So assume κ ∈ [2]. LetS be a separating set with |S| = κ, and let H1, H2 be two components of G − S. Since Sis minimum, every vertex v ∈ S has a neighbor in each Hi. Since G is cubic, there exists

31

i such that v has a unique neighbor wv in Hi. Choose such a wv, preferring wv ∈ H1, andset F := {vwv : v ∈ S}. Then |F | = |S|. Moreover F is a disconnecting set of edges: Fdisconnects H1 from S unless wv /∈ H1 for some v ∈ S. In this case v has two neighbors inH1, by the preference for H1, and so no neighbors in S. Thus F separates H1 + v from H2.So κ′ ≤ |F | = |S| = κ ≤ κ′. �

Lemma 4.5 (Expansion Lemma 4.2.3). If G is k-connected and G′ is obtained from G byadding a new vertex x with at least k neighbors in G then G′ is k-connected.

Proof. Since |G′| = |G|+1, it suffices to show that G′ does not have a (k−1)-separatingset. Consider any (k− 1)-set S. Then G− S is connected and x has a neighbor in G− S, soG′ − S is connected. �

HW 42. Prove that an r-connected graph on an even number of vertices with no inducedsubgraph isomorphic to K1,r+1 has a 1-factor.

4.2. Low Connectivity

Definition 4.6. Suppose H is a subgraph of G. Recall that a path P ⊆ G is an H-pathif its ends, but not it internal vertices are in H. Let P := P0, P1, . . . , Pt be a sequence ofsubgraphs of G, and set Gi := P0 ∪ · · · ∪ Pi−1. Then P is a 2-witness for G if P0 is a cycle,Gt+1 = G, and Pi is a Gi-path in G for all i ∈ [t].

Theorem 4.7 (Whitney [1932] 4.2.8). A graph G = (V,E) is 2-connected iff it has a2-witness set.

Proof. First suppose G is 2-connected. Then δ(G) ≥ 2, and so G contains a cycle C.Let H ⊆ G be a maximal subgraph such that H has a 2-witness P0, . . . , Pt. It exists becauseC is a candidate. It suffices to show that H = G.

Suppose v0 ∈ V (G − H). Since G is connected, there exists a v0, H-path Q = v0 . . . vs.Since G is 2-connected, G − vs is connected. So there exists a vs−1, H-path P in G − vs.Then Pt+1 := vsvs−1P is an H-path in H + Pt+1, contradicting the maximality of H. Weconclude that H is a spanning subgraph of G.

Now suppose xy ∈ E(G − H). Then xy is an H-path of H + xy, contradicting themaximality of H. So H is an induced, spanning subgraph of G. Thus G = H.

Now suppose G has a 2-witness P0, . . . , Pt. Argue by induction on t that G is 2-connected.The base step t = 0 is easy since the cycle P0 is 2-connected. So consider the induction stept ≥ 1. By the induction hypothesis, H :=

⋃t−1i=0 Pi is 2-connected. Put Pt := v1 . . . vs.

Consider any x ∈ V (G). We must show that G − x is connected. Since H is 2-connected,H−x is connected. Also every vertex of Pt−x is connected to a vertex v ∈ {v1, vs} in G−x,even if x ∈ V (Pt). It follows that G is 2-connected. �

Definition 4.8. Let e = xy be an edge in a graph G, and fix a new vertex ve. Thegraph G · e obtained by contracting e is defined by

G · e := (G ∪K(ve, NG({x, y})))− x− y.Note that if P ′ is a path in G · e then either P ′ is a path in G or ve ∈ V (P ′). In the

latter case we can obtain a path in G by replacing ve by one of x, y, xy, yx. If P is a path inG then either P is a path in G · e or one or both of x, y are in V (P ). In the latter case wecan obtain a path P ′ in G · e by replacing one of x, y, xPy, yPx by ve.

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Lemma 4.9. Let G = (V,E) be a graph and xy ∈ E. If S ′ is a separating set of G′ := G·xythen either S ′ ⊆ V and S ′ is a separating set of G or vxy ∈ S ′ and S := S ′ − vxy + x + y isa separating set of G.

Proof. Suppose vxy /∈ S ′. Then S ′ ⊆ V (G) and vxy is in a component H of G′ − S ′.Choose a vertex w in another component of G′−S ′; then w ∈ G. Consider any w, {x, y}-pathP = w . . . u′u in G. Then P ′ := wPu′vxy is a w, vxy path in G′. Since S ′ ⊆ V is a separatingset in G′, there is v ∈ S ′ ∩ V (P ) ⊆ V . Thus S ′ separates x from w in G.

Else vxy ∈ S. Let u and v be vertices in distinct components of G′−S ′. If P is a u, v-pathin G then there is a u, v-path P ′ in G′ such that V (P ′) ⊆ V (P ) + vxy, and vxy ∈ V (P ′) onlyif x or y is in V (P ). As S ′ separates u and v in G′, there is w ∈ V (P ′)∩ S ′. If w 6= vxy thenw ∈ S. If w = vxy then x or y is in V (P ) ∩ S. So S separates u and v in G. �

Lemma 4.10 (Thomassen [1980] 6.2.9). Every 3-connected graph G with |G| ≥ 5 has anedge e such that G · e is 3-connected.

Proof. Suppose not. Consider any edge xy. Since G′ := G · xy is not 3-connected and|G′| ≥ 4, G′ has a separating 2-set S. By Lemma 4.9, S has the form S = {vxy, z} and{x, y, z} is a separating 3-set in G.

So far the edge xy ∈ E and the 2-separating set S are arbitrary. Now choose xy andS = {vxy, z} as above so that G − {x, y, z} has a component H that is as large as possibleamong all possible choices of xy and S. Let H ′ be another component of G−{x, y, z}. Since{x, y, z} is a minimal separating set, each of x, y, z has a neighbor in each of H and H ′.Let u be a neighbor of z in H ′. Then G · uz has a separating set {vuz, v}, and {u, v, z} is aseparating set for G.

Put H+ := G[V (H) + x + y]. Then H+ is connected, and u, z /∈ V (H+). Thus H∗ :=H+ − v is disconnected, since otherwise H∗ is a component of G − {u, v, z} with |H∗| =|H + {x, y} − v| ≥ |H| + 1, contradicting the choice of xy, S,H. As x and y are adjacent,they are in the same component of H∗. Let U be a component of H∗ containing neither xnor y, and consider a ∈ V (U). Since a ∈ V (H), we have N(a) ⊆ V (H) ∪ S = V (H+ + z).As a ∈ V (U), we have NH+(a) ⊆ V (U) + v. So N(a)rU ⊆ {v, z}. Thus {v, z} separates Ufrom H ′, contradicting κ(G) ≥ 3. �

Definition 4.11. A sequence of graphs G0, . . . , Gs is a 3-witness for G iff

(1) G0 = K4 and Gs = G; and(2) for each i ∈ [s] there is an edge xy ∈ E(Gi) such that Gi−1 = G · xy and

dGi(x), dGi

(y) ≥ 3.

Theorem 4.12. A graph G is 3-connected iff it has a 3-witness.

Proof. First suppose that G is 3-connected. Then |G| ≥ 4. We show by induction on|G| that G has a 3-witness. Suppose |G| = 4. If xy /∈ E(G) then V (G) r {x, y} is a 2-setthat separates x from y, a contradiction. So G = K4, and G0 = K4 = G is a 3-witness forG. Otherwise, |G| ≥ 5. By Lemma 4.10, there exists an edge xy ∈ E(G) such that G · xyis 3-connected. Since G is 3-connected, d(x), d(y) ≥ 3. By induction, G · xy has a 3-witnessG0, . . . , Gs. So G0, . . . , Gs, G is a 3-witness for G.

Now suppose G0, . . . , Gs is a 3-witness for G. We show by induction on s that G is 3-connected. If s = 0 then K4 = G0 = G is 3-connected. Otherwise, for some edge xy ∈ E(G),

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both Gs−1 = G · xy and dG(x), dG(y) ≥ 3. By induction G · xy is 3-connected. Suppose fora contradiction that S is a 2-separator in G. If S = {x, y} then vxy is a cut vertex of G · xy,a contradiction. So there is a component H of G − S that contains at least one, say x, ofx and y, and another component H ′ that contains neither x nor y. Since dG(x) ≥ 3 andN(x) ⊆ V (H) ∪ S, x has a neighbor v in H. It is not y, since then S separates vxy fromH ′ in G · xy. If y /∈ S then S separates v from H ′ in G · xy; otherwise S ′ := S − y + vxyseparates v from H ′ in G · xy. Regardless, κ(G · xy) ≤ 2, a contradiction. �

The last paragraph of the above proof is subtle. If dG(x) < 3 then we could haveS = N(x), and V (H) = {x}. Then H − x is not a component of G · xy − S ′ because H − xhas no vertices.

Conjecture 4.13 (Lovasz). There exists a function f : Z+ → Z+ such that for allk ∈ Z+ and f(k)-connected graphs G and all vertices x, y ∈ V (G), there exists a partition{V1, V2} of V (G) such that G[V1] is an x, y-path and G[V2] is k-connected.

HW 43. Show that Conjecture 4.13 is true in the case k = 1 with f(1) := 3.

4.3. Menger’s Theorem

Definition 4.14. Let G = (V,E) be a graph and A and B be subsets of V . An A,B-pathis a path with exactly one end in A and exactly one end in B, and whose internal verticesare in neither A nor B. Let l(A,B) be the maximum size of a set of disjoint A,B-paths.An A,B-separating set (or A,B-separator) is a set of vertices S such that G − S has noA,B-paths. Let k(A,B) be the minimum cardinality of an A,B-separating set.

Here is another fundamental theorem. We give two proofs.

Theorem 4.15 (Menger 1927 4.2.17). Let G = (V,E) be a graph, and suppose A,B ⊆ V .Then the size l := l(A,B) of a maximum set of disjoint A,B-paths is equal to the sizek := k(A,B) of a minimum A,B-separating set.

Proof 1. (l ≤ k) If P is a set of disjoint A,B-paths and S is an A,B-separator then Smust contain at least one vertex of each path, and each vertex of S is on at most one pathof P . Thus the function f : P → S defined by setting f(P ) equal to the first x ∈ S ∩ V (P )is an injection; so |P| ≤ |S|. Choosing P maximum and S minimum yields the inequality.

(k ≤ l) For a set of A,B-paths P let end(P) denote the set of ends in B of paths in P .It suffices to show (*) if P ′ is a set of disjoint A,B-paths with |P ′| < k then there existsa set P of disjoint A,B-paths such that |P| = |P ′| + 1 and end(P ′) ⊆ end(P). Argue byinduction on |G| − |B|. If |G| − |B| = 0 then A ⊆ B. So the A,B-paths are exactly thepaths consisting of a single vertex of A, and k = |A|. Thus (*) holds.

Suppose |G| > 0, and fix a set P ′ of disjoint A,B-paths with |P ′| < k. Since |end(P ′)| =|P ′| < k, there is an A,B-path R = Ry′ in G − end(P ′). If R ∩

⋃P ′ = ∅ then put

P := P ′ + R. If not, let x be the last vertex of R that is in⋃P ′; say x ∈ P ∈ P ′

and y is the end of P in B. Note that x /∈ B, since y 6= y′ and V (P ) ∩ B = y. PutB′ := B ∪ V (xRy′ ∪ xPy) and Q′ := P ′ − P + Px. Then end(Q′) = end(P ′) − y + x.Since every A,B-path contains an A,B′-path, k = k(A,B) ≤ k(A,B′). Since B + x ⊆ B′,|G| − |B′| < |G| − |B|. By induction, there exists a set Q of A,B′-paths and y′′ ∈ B′ suchthat |Q| − 1 = |Q′| = |P ′| and end(Q) = end(Q′) + y′′. So x 6= y′′. Let x, y′′ be the ends of

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Q,Q′′ ∈ Q. Set P0 = Q − Q − Q′′. If y′′ ∈ xPy then set P := P0 + QxRy′ + Q′′y′′Py; ify′′ ∈ xRy′ then set P := P0 +QxPy +Q′′y′′Ry′; else set P := P0 +QxPy +Q′′. Evidently,P witnesses (*). �

Proof 2 (k ≤ l). So it suffices to show k ≤ l. Argue by induction on ‖G‖.Base Step: ‖G‖ = 0. Then every A,B-path is trivial. So A ∩ B is the maximum set ofdisjoint A,B-paths and the minimum A,B-separating set. Thus l = |A ∩B| = k.Induction Step: ‖G‖ ≥ 1. Let e = xy ∈ E(G), and put G′ = G · e. For any U ⊆ V , define

U ′ =

{U − {x, y}+ ve if U ∩ {x, y} 6= ∅U otherwise

,

and note that for every T ⊆ V (G′) there exists S ⊆ V with T = S ′. Using Lemma 4.9 andthe discussion before, every set P ′ of disjoint A′, B′-paths corresponds to a set P of disjointA,B-paths with |P| = |P ′| (but not vice versa). So

lG′(A′, B′) ≤ l.

Also, if S is an A,B-separator in G if and only if S ′ is an A′, B′-separator in G′. So

kG′(A′, B′) ≤ k ≤ kG′(A

′, B′) + 1.

Choose a minimum A′, B′-separator T in G′. If kG′(A′, B′) = k then by the inductionhypothesis applied to G′ we have:

k = kG′(A′, B′) ≤ lG′(A

′, B′) ≤ l,

and we are done. Otherwise, k = kG′(A′, B′) + 1. In this case vxy ∈ T , and T = S ′, where

S := T − vxy + x+ y. In particular xy ∈ G[S].Set G′′ = G− e. Since e ∈ G[S],

kG(A, S) = kG′′(A, S) and kG(B, S) = kG′′(B, S)(4.3.1)

Since S separates A from B in G, every A, S-separator in G separates A from B, and so hassize at least |S|, and a similar statement holds for B. So we have

(4.3.2) kG(A, S), kG(S,B) ≥ k.

Thus

|S| ≥ lG(A, S) ≥ lg′′(A, S) =i.h. kG′′(A, S) =(4.3.1) kG(A, S) ≥(4.3.2) k = |S| and|S| ≥ lG(B, S) ≥ lg′′(B, S) =i.h. kG′′(B, S) =(4.3.1) kG(B, S) ≥(4.3.2) k = |S|.

Let KA be a collection of |S| = k disjoint A, S-paths and KB be a collection of |S| disjointS,B-paths. Then for each z ∈ S there is a unique A, z-path Pz and a unique z,B-path Qz. Ifv ∈ V (Pw)∩V (Qz) then v ∈ S, since otherwise PvQ is an A,B-walk in G−S, contradictingthe fact that S is an A,B-separator. Thus w = v = z, and so {PzzQz : z ∈ S} is a collectionof |S| = k disjoint A,B-paths. �

Definition 4.16. Let a and b be distinct vertices in a graph G. Two a, b-paths areinternally disjoint if they have no internal vertices in common. Let λ(a, b) be the maximumsize of a set of internally disjoint a, b-paths and if ab /∈ E then let κ(a, b) be the minimumsize of a set separating a and b.

35

If ab ∈ E(G) then no S ⊆ V separates a and b. This is why κ(a, b) is only defined whenab /∈ E.

Corollary 4.17 (4.2.17). If a and b are nonadjacent vertices of a graph G then λ(a, b) =κ(a, b).

Proof. Let A = N(a) and B = N(b). Then any A,B-path P ′ ⊆ G′ = G − {a, b} canbe extended to an a, b-path P = aP ′b ⊆ G and any a, b-path Q ⊆ G contains an A,B-pathQ′ = Q− a− b ⊆ G′. Thus S separates a and b in G iff it is an A,B-separator of G′ and themaximum number of internally disjoint a, b-paths in G is equal to the maximum number ofdisjoint A,B-paths in G′. Applying Menger’s Theorem to G′ we see that the minimum sizeof an A,B-separating set is equal to the size of a maximum set of disjoint A,B-paths. Thecorollary follows. �

Theorem 4.18 (4.2.21). Every graph G = (V,E) satisfies κ(G) = t := mina,b∈V,a6=b λ(a, b).

Proof. Choose a, b with t = λ(a, b). If G is complete then

t = λ(a, b) = 1 + (|G| − 2) = |G| − 1 = κ(G),

since ab is an a, b-path, and acb is also an a, b-path for all c ∈ V − a− b.Otherwise G has a separating set S with |S| = κ(G). Let x, y be vertices in distinct

components of G− S. Then

t = λ(a, b) ≤ λ(x, y) ≤ |S| = κ(G).

So t ≤ κ(G).Now we show t ≥ κ(G). If ab /∈ E then by Corollary 4.17, there exists an a, b-separating

set U with t = |U | ≥ κ(G). Otherwise, ab ∈ E; set G′ = G−ab. Then λG(a, b) = λG′(a, b)+1,since ab is an a, b-path.

We first show that κ(G′) + 1 ≥ κ(G). Suppose k := κ(G′) < κ(G). Then G′ has aseparating set W with |W | = κ(G′), and W is not a separating set of G. It follows thatG′ − W has exactly two components X and Y with (say) a ∈ V (X) and b ∈ V (Y ). If|X| > 1 then W + a is a separator of G with size κ(G′) + 1 ≥ κ(G). A similar argumentholds, if |Y | > 1. Otherwise |G| = |W |+ 2, and so κ(G) ≤ |G| − 1 = k + 1 = κ(G′) + 1.

By Corollary 4.17, we have

κ(G) ≤ κ(G′) + 1 ≤ κG′(a, b) + 1 = λG′(a, b) + 1 = λG(a, b) = t. �

Definition 4.19 (4.2.18). The line graph H = L(G) of a graph G = (V,E) is definedby

V (H) = E and E(H) = {ee′ : e ∩ e′ 6= ∅}.

Definition 4.20. Let x and y be distinct vertices in a graph G. An x, y-edge-cut is a setof edges F such that there are no x, y-paths in G− F ; let κ′(x, y) be the size of a minimumx, y-edge-cut. Two x, y-paths are edge-disjoint if they have no common edges; let λ′(x, y) bethe maximum size of a set of edge-disjoint x, y-paths.

Theorem 4.21 (4.2.19). Let G = (V,E) be a graph with distinct vertices x, y ∈ V. Thenκ′(x, y) = λ′(x, y).

36

Proof. Set G′ = G + x′ + xx′ + y′ + yy′, where x′ and y′ are new vertices. Thenκ′G(x, y) = κ′G′(x, y) and λ′G(x, y) = λ′G′(x, y). A set of edges disconnects x from y inG′ iff the corresponding set of vertices separates xx′ from yy′ in L(G′). Moreover, edgedisjoint x, y-paths in G′ correspond to internally disjoint xx′, yy′-paths in L(G′). Thus, byCorollary 4.17,

κ′G(x, y) = κ′G′(x, y) = κL(G′)(xx′, yy′) = λL(G′)(xx

′, yy′) = λ′G′(x, y) = λ′G(x, y).

�

Definition 4.22 (4.2.22). Let G = (V,E) be a graph with x ∈ V and U ⊆ V . Anx, U -fan is a set F of x, U -paths such that |F| = |U | and F ∩ F ′ = {x} for all distinctF, F ′ ∈ F .

Theorem 4.23 (4.2.23). A graph G = (V,E) is k-connected if and only if |G| ≥ k + 1and G has an x, U-fan for all x ∈ V and all k-sets U ⊆ V − x .

Proof. (Sketch) If G is k-connected then there exist k disjoint N(x), U -paths in G− x.Adding x yields the desired fan.

Conversely, the hypothesis implies δ(G) ≥ k, and for all x and y, there exist k disjointN(x), N(y)-paths. Thus by Theorem 4.18

κ(G) = minx,y∈V

λ(x, y) ≥ k.

�

Theorem 4.24 (HW 4.2.24). Let G = (V,E) be a k-connected graph with k ≥ 2. Thenfor any k-set S ⊆ V there is a cycle C ⊆ G with S ⊆ V (C).

Proof. Let C ⊆ G be a cycle containing as many vertices of S as possible. It existsbecause δ(G) ≥ κ(G) ≥ 2. We claim that S ⊆ V (C). Otherwise, let v ∈ S r V (C). Then|S ∩ V (C)| < k. Arguing by contradiction, it suffices to find a cycle containing S∩V (C) + v.

Orient C cyclically as−→C . Let t = min{k, |C|}. By Menger’s Theorem there is a set P

of t disjoint C,N(v)-paths; together with v they for a fan F . Set F =⋃F , let x1, . . . , xt

be a sequence of the leaves of F in cyclic order around−→C , and set

−→C i = xi

−→Cxi+1. Then

there exists i ∈ [k] such that−→P i contains no internal vertices from S: If t = |C| this is true

for all i ∈ [k]; otherwise it follows by the pigeonhole principle, since t = k > |S ∩ V (C)|. SoD = xi+1

−→CxiFvFxi+1 is a cycle containing S ∩ V (C) + v. �

HW 44. (*) Every 2-connected graph G has a cycle of length at least min{|G|, 2κ(G)}.

HW 45. (*) If G is a 2-connected graph with α(G) ≤ κ(G) then G is hamiltonian.

HW 46. (+) Let G be a 2-connected graph that does not induce K1,3. Then G has acycle of length at least min{|G|, 4κ(G)}.

HW 47. (++) Conjecture 4.13 is true in the case k = 2 with f(2) := 5. Prove it forf(2) ≤ 10. [Hint: Choose P so that G − P contains the biggest 2-connected subgraphpossible.]

37

HW 48. (*) Let G = (V,E) is a graph with x ∈ V and Y, Z ⊆ V and 4 = |Y | = |Z| − 1.Suppose Q = {Qy : y ∈ Y } is an x, Y -fan in G, where each Qy is an x, y-path. Similarly,suppose R = {Rz : z ∈ Z} is an x, Z-fan in G, where each Rz is an x, z-path. Prove: Thereexists a x, (Y + z)-fan in G for some z ∈ Z. [Hint: Add a new vertex w whose neighborhoodis Z and apply a theorem.]

38

CHAPTER 5

Graph coloring

Definition 5.1. Let G = (V,E) be a graph and C be a set (of colors). A proper C-coloring of G is a function f : V → C such that for all vertices x, y ∈ V if xy ∈ E(G)then f(x) 6= f(y). If k is a positive integer, we say that f is a proper k-coloring if it is aproper [k]-coloring. The chromatic number χ(G) is the least k such that G has a properk-coloring. In this case G is said to be k-chromatic. If G has a k-coloring then it is said tobe k-colorable. In this chapter we will assume that all colorings are proper unless otherwisestated. For i ∈ C, f−1(i) is called a color class.

Proposition 5.2 (HW). Every graph G satisfies ω(G), |G|α(G)≤ χ(G) ≤ ∆(G) + 1.

HW 49. (*) Prove Proposition 5.2.

5.1. Examples

Example 5.3 (5.2.3 Mycielski [1955]). For every positive integer k there exists a graphGk with ω(Gk) ≤ 2 and χ(Gk) = k.

Proof. We argue by induction on k. For k ≤ 2 let Gk = Kk. Now suppose k ≥ 3 andwe have constructed Gk−1 = (Vk−1, Ek−1) as required. We first construct Gk = (Vk, Ek) asfollows: Let V ′k−1 = {v′ : v ∈ Vk−1} be a set of new vertices, xk be a new vertex, and put

Vk = Vk−1 ∪ V ′k−1 + xk and,

Ek = Ek−1 ∪ {uv′ : uv ∈ Ek−1} ∪ {xkv′ : v′ ∈ V ′k−1}.So N(v′) ∩ Vk−1 = N(v) ∩ Vk−1 for all v ∈ Vk−1.

Suppose ω(Gk) ≥ 3, and choose Q = K3 ⊆ Gk. Then k ≥ 3. Since N(xk) = V ′k−1

is independent, xk /∈ Q, and |V ′k−1 ∩ Q| ≤ 1. As ω(Gk−1) =i.h. 2, there is at least onev′ ∈ V ′k−1∩Q. Hence N(v′)∩Q = N(v)∩Q, and Q−v′+v is a K3 in Gk−1, a contradiction.

Now χ(Gk) ≤ k: If k ≤ 2 this is obvious; else Gk−1 has a (k−1)-coloring f ′ by induction.Extend f ′ to a k-coloring f of Gk by setting f(v′) = k (new color), and f(xk) = 1.

Finally, χ(Gk) ≥ k: If k ≤ 2 this is obvious. If k ≥ 3 then it suffices to show that every(k − 1)-coloring g of Gk − xk satisfies g(V ′k−1) = [k − 1], since xk will require a new color.Suppose not. After possibly renaming color classes, let k − 1 ∈ [k − 1] r g(V ′k−1). For acontradiction, we construct a (k − 2)-coloring h of Gk−1. Define h : Vk−1 → [k − 2] by:

h(v) =

{g(v) if g(v) 6= k − 1

g(v′) if g(v) = k − 1.

If uv ∈ Ek−1 then g(u) 6= g(v) since g is proper; if g(u) 6= k − 1 6= g(v) then

h(u) = g(u) 6= g(v) = h(v);

39

else exactly one of u, v is colored with k − 1 by g; say g(v) = k − 1. Since uv′ ∈ Ek,h(u) = g(u) 6= g(v′) = h(v). �

HW 50. (*) Let k ≥ 2 and Gk be the graph in Example 5.3. Prove that Gk is critical, i.e.,χ(Gk − e) < χ(Gk) (= k) for all e ∈ Ek. [Hint: There are three types of edges to consider.]

HW 51. (*) Let G be graph, and suppose any two odd cycles C,C ′ ⊆ G have a commonvertex. Prove that χ(G) ≤ 5. [Caution: C could have chords.]

HW 52. (*) Let P = {v1, . . . , vn} be a path, and suppose G = (V,E) is a graph suchthat V is a subset of the set of subpaths of P , and E = {RQ ∈ V : R ∩Q 6= ∅}. Prove thatχ(G) = ω(G). [Hint: Order V so that if Q,R ∈ V (subpaths of P ), the first vertex of Q isvi, the first vertex of R is vj, and i < j then Q precedes R, and color V in this order.]

HW 53. (*) Let G = (V,E) be a k-colorable graph, and let P be a set of verticessuch that the distance dG(x, y) between any two points in P is at least 4. Prove that any[k + 1]-coloring of G[P ] can be extended to a [k + 1]-coloring of G.

5.2. Brooks’ Theorem

Proposition 5.4. Let G = (V,E) be a graph with V = A ∪B. If A ∩B is a clique thatseparates G then ω(G) = max{ω(G[A]), ω(G[B])} and χ(G) = k := max{χ(G[A]), χ(G[B])}.

Proof. LetA∩B := {v1, . . . vl} where l = |A∩B|. AsA∩B is a clique, l ≤ ω(G). As V =A∪B and A∩B is a separating set, every clique of G is contained in A or B. Thus ω(G) ≤max{ω(G[A]), ω(G[B])}. As χ(G[A]), χ(G[B]) ≤ k, there are partitions, {A1, . . . , Ak} and{B1, . . . , Bk} (allowing empty sets) of A and B, respectively, into k cocliques. Each cocliquecontains at most one vertex of A ∩ B. So we may assume Ai ∩ Bj = ∅ if i 6= j. PutCi := Ai∪Bi, and note that Ci is a coclique, since |Ci∩A∩B| ≤ 1 and A∩B is a separatingset. Then {C1, . . . , Ck} is a partition of V into k cocliques. So χ(G) = k. �

Lemma 5.5. Let G = (V,E) be a connected graph with v ∈ V . There is an orderingv1, . . . , v|G| of V such that (*) v = v|G| and for all i ∈ [|G| − 1] there is j ∈ [|G|] r [i] withvivj ∈ E.

Proof. Argue by induction on |G|. The base case n = 1 is trivial; so assume n > 1. LetP be a maximum path; as G is connected and |G| ≥ 2, |P | ≥ 2. Pick an end u of P withu 6= v. As u is not a cut-vertex, G− u is connected. By induction, there exists an orderingv2, . . . , vn of V − u satisfying (*) for G− u. So v1 := u, v2, . . . , vn satisfies (*) for G. �

Define a b-obstruction to be Kb or, if b = 3, an odd cycle. Let (non-standard) ω∗(G) bethe maximum b such that G has a b-obstruction. Then ω∗(G) ≤ χ(G) ≤ ∆(G) + 1.

Theorem 5.6 (Brooks (1941)). Every graph satisfies ω∗(G) = ∆(G)+1 or χ(G) ≤ ∆(G).

Proof. Set ∆ = ∆(G), χ = χ(G) and ω∗=ω∗(G). We assume ω∗ ≤ ∆ and show byinduction on |G| that χ ≤ ∆. Then ∆ ≥ 2. If ∆ = 2 then and G has no odd cycle, so χ = 2by Theorem 1.24. Now assume ∆ ≥ 3. By Proposition 5.4, assume G is 2-connected.

Let S be a maximal coclique, and put G′ := G− S. Then ∆(G′) < ∆ since every vertexof G′ has a neighbor in S. If ω∗(G′) < ∆ then inductively (∆ − 1)-color G′, and use a new

40

color for S. Else, there is a ∆-obstruction Q ⊆ G′. Then (1) Q = K∆ or (2) ∆ = 3 < |Q|and Q is an odd cycle. Anyway Q is (∆− 1)-regular. Pick y ∈ S ∩N(Q).

It suffices to prove (*) there exists an edge xw ∈ E(Q) with y ∈ N(x) r N(w) and ay,Q-path P = y . . . z in G − x − w. Then Q − w and P − y are connected through z, soLemma 5.5 implies H := G[Q ∪ P ] has an ordering L := w, y, v1, . . . , vt, x, where each vihas a neighbor to its right. Using induction, there is a ∆-coloring f of H ′ := G − H. AsQ is (∆ − 1)-regular, ‖w,H ′‖ ≤ 1; as x ∈ N(y) and N(y) ∩ P 6= ∅, ‖y,H ′‖ ≤ ∆ − 2. Thussome color β is not used on N(w) ∪N(y). Extend f to G by setting f(w) = β = f(y) andcoloring the remaining vertices in the order L. This is possible, since each vi has at most∆− 1 colored neighbors when it is colored, and x has two neighbors w, y colored the same.

Now we prove (*). If N(y) ∩ Q = {x} then, as G is 2-connected, there is a y,Q-pathP := y . . . z in G − x. Let w ∈ N(x) ∩ Q − z. Then wy /∈ E. Else ‖y,Q‖ ≥ 2. NowQ * N(y): if (1) then Q+ y is not a clique; if (2) then |Q| > d(y). As Q is connected, thereis wx ∈ E(Q) with y ∈ N(x)rN(w). Let z ∈ N(y) ∩Q− x, and set P = yz. �

Conjecture 5.7 (Borodin & Kostochka 1977). If a graph G satisfies 8, ω(G) < ∆(G)then χ(G) < ∆(G).

Reed used sophisticated methods to prove the conjecture for ∆(G) > 1014.

HW 54. (*) For a graphG let θ(G) = maxuv∈E(G)(d(u)+d(v)). Prove that if θ(G) ≤ 2r+1then χ(G) ≤ r + 1. Also prove that if θ(G) ≤ 2r and ω∗(G) ≤ r then χ(G) ≤ r.

5.3. Turán’s Theorem

Let n, s ∈ Z+. In this section we determine the number of edges a graph on n verticesmust have to ensure it contains Ks. In other words, how many edges can we put into a graphon n vertices without getting Ks.

Definition 5.8. A graph is said to be r-partite if it is r-colorable. Saying r-partiteinstead of r-colorable tends to emphasize the partition into r independent sets provided bythe r-coloring. These independent sets are called parts. The complete r-partite Kn1,...nr

graph is the r-partite graph with r parts of sizes n1, . . . , nr such that any two vertices indifferent parts are adjacent. The Turán graph Tn,r is the complete r-partite graph on nvertices such that any two parts differ in size by at most one.

Lemma 5.9 (5.2.8). The Turán graph Tn,r has more edges than any other r-partite graphon n vertices.

Proof. Let G be an r-partite graph on n vertices with as many edges as possible; sayX is an r-partition of G. Clearly, G is a complete r-partite graph. So, if G 6= Tn,r then thereexist parts X, Y ∈ X with |X| − |Y | ≥ 2 and x ∈ X. Let G′ be the complete r-partite graphwith r-partition X ′ := X −X − Y + (X − x) + (Y + x). Then

E(G′) ⊇ E(G)− {xy : y ∈ Y }+ {xx′ : x′ ∈ X − x}.

Thus‖G′‖ ≥ ‖G‖ − |Y |+ |X| − 1 ≥ ‖G‖+ 1,

a contradiction. So G ∼= Tn,r. �

41

xG′′ = G[S]

N := N(x)

G′ = G[N ]

G H := K(S) ∨ T∆,r−1

...

12

...

r − 1

T∆,r−1

K(S)

Figure 5.3: Turán’s Theorem

Theorem 5.10 (5.2.9 Turán [1941]). The Turán graph Tn,r has more edges than anyother graph G with |G| ≤ n and ω(G) ≤ r.

Proof. We will show by induction on r that any graph G = (V,E) with |G| = n,ω(G) ≤ r, and ‖G‖ ≥ ‖Tn,r‖ satisfies G ∼= Tn,r. If r = 1 then G ∼= Kn

∼= Tn,1; so supposer > 1.

Choose v ∈ V with d(v) = ∆ := ∆(G). Set N := N(v), G′ := G[N ], S := V − N andG′′ := G[S]. Then |G′| = ∆, and ω(G′) ≤ r − 1, since K + v is a clique in G for everyclique K in G′. Set H := T∆,r−1 ∨K(S). Then H is an r-partite graph on n vertices, andω(H) ≤ r, since any clique in H has at most r− 1 vertices in T∆,r−1 and one vertex in S. So

‖G‖ = ‖G′‖+ ‖G′′‖+ ‖N,S)‖

= ‖G′‖+∑v∈S

dG(v)− ‖G′′‖

≤ ‖T∆,r−1‖+∑v∈S

dG(v)− ‖G′′‖ (induction)(5.3.1)

≤ ‖T∆,r−1‖+ ∆|S| (maximum degree)(5.3.2)= ‖H‖≤ ‖Tn,r‖ (Lemma 5.9)(5.3.3)

Inequality (5.3.1) is strict unless G′ ∼= T∆,r−1. Inequality (5.3.2) is strict unless G′′ = K(S)and G = G′ ∨G′′. Inequality (5.3.3) is strict unless H ∼= Tn,r. If ‖G‖ ≥ ‖Tn,r‖ then all threeinequalities are tight, and so

G ∼= T∆,r−1 ∨K(S) ∼= H ∼= Tn,r. �

HW 55. (*) Prove that if ω(G) ≤ r then ‖G‖ ≤ (1− 1/r)|G|2/2.

HW 56 (*). Let G be a graph with |G|+12≤ δ(G) ≤ ∆(G) ≤ |G| − 2. Prove: (1) Every

edge of G is contained in a K3; (2) G contains two disjoint K3’s; and (3) for all n there is agraph H with n ≤ |H|+1

2≤ δ(H) without two disjoint K3’s. [Hint: For (2) first do the case

∆(G) > |G|+12

, and then use Turán’s Theorem on G−K3 if G is |G|+12

-regular.]42

5.4. Edge Coloring

Definition 5.11. Let G = (V,E) be a graph. A proper k-edge-coloring of G is a functionf : E → [k] such that f(e) = f(e′) implies that e and e′ are not adjacent (e ∩ e′ = 0). Thechromatic index χ′(G) of G is the least k such that G has a proper k-edge-coloring. In thissection we will assume that all edge colorings are proper. Note that this is not the case whenwe consider Ramsey Theory.

HW 57. (*) Let P be the Petersen graph and v ∈ V (P ). Determine χ′(P − v).

Theorem 5.12 (7.1.17 König [1916]). Every bipartite graph G satisfies χ′(G) = ∆(G).

Proof. Argue by induction on ∆ = ∆(G). The base step ∆ = 1 is trivial since G has noadjacent edges, and so all edges can receive the same color. Now suppose ∆ > 1. It sufficesto find a ∆-regular bipartite multigraph H with G ⊆ H: Then H has a perfect matchingM by Theorem 3.6. Color all edges in M ∩ E(G) with color ∆, and set G′ =G−M . Then∆(G′) ≤ ∆(H −M) ≤ ∆− 1, and so by induction, G′ has a (∆− 1)-edge-coloring.

It remains to construct H. Suppose G is an A,B-bigraph with |A| ≤ |B|. Let A ⊆ A′,where A′ ∩ B = ∅ and |A′| = |B|. Choose an A′, B-bipartite multigraph H ⊇ G with ‖H‖maximal subject to ∆(H) ≤ ∆. It exists because G is a candidate. Now

‖H‖ =∑v∈A′

dG′(v) =∑v∈B

dG′(v) ≤ ∆|B|.

If ‖H‖ = ∆|B| then H is ∆-regular. Else there exist vertices a ∈ A′ and b ∈ B withdH(a), dH(b) < ∆. Then H ′ := H + e, where e ∈ E(a, b) is a new, possibly parallel edge,contradicts the maximality of H. �

Now we consider edge coloring of general graphs. The fundamental result is Theorem 5.14due to Vizing. The following lemma does most of the work in its proof.

Lemma 5.13. Suppose G = (V,E) is a simple graph with ∆(G) ≤ k ∈ N, v ∈ V andG′ = G− v. If χ′(G′) ≤ k and d(x) = k for at most one x ∈ N(v) then χ′(G) ≤ k.

Proof. Argue by induction on k. If k = 1 then E is a matching, and so χ′(G) ≤ 1.Now suppose k > 1. For any edge-coloring f : E(G′)→ [k] , x ∈ N(v) and α ∈ [k], set

f(x) := [k]r {f(e) : e ∈ EG′(x)} and fα := {u ∈ N(v) : α ∈ f(u)}.By adding edges and vertices to G, we may assume k−1 ≤ d(x) ≤ k = d(v) for all x ∈ V −v,and d(y) = k for exactly one y ∈ N(v). So |f(x)| = 2 for all x ∈ N(v) − y and |f(y)| = 1.Pick a k-edge-coloring f of G′ with T (f) := {β ∈ [k] : 1 ≤ |fβ| ≤ 2} maximum.

Suppose |fα| 6= 1 for all α ∈ [k]. Since∑

α∈[k] |fα| =∑

x∈N(v) |f(x)| = 2k − 1, thereare β, γ ∈ [k] with |fβ| = 0 and |fγ| ≥ 3; say w ∈ fγ. Set Gβ,γ = (V − v, Eβ,γ), whereEβ,γ = {e ∈ E(G′) : f(e) ∈ {β, γ}}. The component of Gβ,γ containing w is a path P withends w and (say) z with f(z) ∩ {β, γ} 6= ∅. Switching γ and β on the edges of P yields anedge-coloring f ′ of G′. Then f ′(u) = f(u) for u ∈ V (G′)r{w, z}, and f ′(w) = f(w)−γ+β.So w ∈ f ′β ⊆ {w, z} and T (f) $ T (f) + β ⊆ T (f ′), contradicting the choice of f .

So fα = {z} for some α ∈ [k] and z ∈ N(v); say α = k. Set M = f−1(k) +vz. As neitherz nor v are incident to any edges colored k, M is a matching. Put H = G−M . As fk = {z}and vz ∈ M , every vertex of N [v] is M -saturated. So dH(v) = k − 1 and dH(x) ≤ k − 1

43

for every x ∈ NH(v), and equality holds once. Since f−1(k) ⊆ M , f is a (k − 1)-coloring ofH−v, and ∆(H−v) ≤ k−1. So ∆(H) ≤ k−1. By induction, χ′(G) ≤ χ′(H)+1 ≤i.h. k. �

Theorem 5.14 (7.1.10 Vizing (1964)). Every graph G satisfies χ′(G) ≤ ∆(G) + 1.

Proof. Set k := ∆(G)+1 and argue by induction on |G|. If |G| = 1 then χ′(G) ≤ 1 = k.Otherwise choose v ∈ V . By induction, χ′(G−v) ≤ k, and so by Lemma 5.13, χ′(G) ≤ k. �

Theorem 5.15 (Full Vizing (1964)). Every multigraph M satisfies χ′(M) ≤ ∆(M) +µ(M).

HW 58 (*). Let G be a graph with ∆(G) = k. Put X = {v ∈ V (G) : d(v) = k}. Provethat if G[X] is acyclic then χ′(G) ≤ k. [Hint: Use Lemma 5.13.]

HW 59 (+). Prove Theorem 5.15.

Conjecture 5.16 (Goldberg (1973), Seymour (1979)). Every multigraph loopless Mwith χ′(M) ≥ ∆(M) + 2 satisfies χ′(M) = maxH⊆Md ‖H‖b|H|/2ce.

HW 60 (*). Prove: Every loopless multigraph M satisfies χ′(M) ≥ maxH⊆Md ‖H‖b|H|/2ce.

Definition 5.17 (4.2.18). The line graph H = L(G) of a graph G = (V,E) is definedby

V (H) = E and E(H) = {ee′ : e ∩ e′ 6= ∅}.

If H is the line graph of a simple graph G then H contains neither an induced copyof K1,3 nor an induced copy of K5 − e (a K5 missing one edge). Also, χ(H) = χ′(G) andω(H) = ∆(G), unless ∆(G) = 2 and ω(G) = 3. So the following theorem (with an extraobservation for the case ∆(G) = 2 < ω(G)) extends Vizing’s Theorem for simple graphs.

Theorem 5.18 (Kierstead & Schmerl 1983). Every graph H that contains neither aninduced copy of K1,3 nor an induced copy of K5 − e satisfies χ(H) ≤ ω(H) + 1.

5.5. List Coloring

Definition 5.19. Let G = (V,E) be a graph and C a set of colors. We write 2C forthe power set of C. A list assignment for G is a function f : V → 2C . One should think off(v) ⊆ C as the set of colors that are available for coloring the vertex v. A k-list assignmentis a list assignment f such that |f(v)| = k for all v ∈ V . Given a list assignment f , anf -coloring is a proper coloring g such that g(v) ∈ f(v) for all v ∈ V . In this case G isf -colorable. The graph G is k-list-colorable (also k-choosable) if for every k-list assignmentf it is f -colorable. The list-chromatic number (also choosability, also choice number) χl(G)of G is the least k such that it is k-list colorable.

Example 5.20. Let G = Kt,tt . Then χ(G) = 2, but χl(G) ≥ t+ 1.

Proof. Say G = K(X, Y ), where |X| = t. Let f be a t-list assignment for G such thatthe vertices of X have disjoint lists of size t, and for each σ ∈

∏x∈X f(x) there exists yσ ∈ Y

with f(y) = range(σ). Then for any f -coloring σ of G[X], the vertex vσ cannot be coloredfrom the list f(yσ) = σ. �

44

Definition 5.21. An edge-list assignment for G is a function f : E → 2C . One shouldthink of f(e) ⊆ C as the set of colors that are available for coloring the edge e. A k-edge-list assignment is a list assignment f such that |f(e)| = k for all e ∈ E. Given anedge-list assignment f , an f -coloring is a proper edge-coloring g such that g(e) ∈ f(e) forall e ∈ E. In this case, G is f -list-colorable. The graph G is k-edge-list-colorable (alsok-edge-choosable) if for every k-edge-list assignment f , it is f -colorable. The list-chromaticindex (also edge-choosability, edge-choice number) χ′l(G) of G is the least k such that it isk-edge-list colorable.

Conjecture 5.22. Every graph G satisfies χ′l(G) = χ′(G).

Definition 5.23. A kernel of a digraph D = (V,A) is an independent set S ⊆ V suchthat for every x ∈ V r S there exists y ∈ S with xy ∈ A.

The 5-cycle G = (V,E) = v1v2v3v4v5v1 has several orientations. LetE1 = { ~v1v2, ~v2v3, ~v3v4, ~v4v5, ~v5v1} and E2 = { ~v1v2, ~v3v2, ~v3v4, ~v5v4, ~v5v1}.

Then ~G1 := (V, ~E1) does not have a kernel since ∆−(~G1) = 1, α(G) = 2, and |~G| = 5 >

α(G) · (∆−(~G1) + 1), but {v2, v4} is a kernel of ~G2 := (V, ~E2).

Lemma 5.24 (8.4.29 Bondy & Boppana & Siegel). Let D = (V,A) be a digraph all ofwhose induced subgraphs have kernels. If f is a list assignment for D satisfying d+(v) <|f(v)| for all v ∈ V then D has an f -coloring.

Proof. Argue by induction on |D|. Let v0 ∈ V . Since |f(v0)| > d+(v0) ≥ 0, thereis α ∈ f(v0). Set W = {v ∈ V : α ∈ f(v)}. Then v0 ∈ W . By hypothesis D[W ] hasa (nonempty) kernel S. Color every vertex in S with α. This is possible because S isindependent and S ⊆ W .

Now it suffices to f -color D′ := D − S so that no vertex in N(S) is colored α. For thispurpose, let f ′ be the list assignment for D′ defined by f ′(v) = f(v)− α. Since |D′| < |D|,using induction, it suffices to show |f ′(v)| > d+

D′(v) for all v ∈ V r S.If v /∈ W then α /∈ f(v), and so

|f ′(v)| = |f(v)| > d+D(v) ≥ d+

D′(v).

Else v ∈ W . Since S is a kernel of D[W ], there exists w ∈ S = V r V (D′) with vw ∈ A. So|f ′(v)| = |f(v)− α| > d+

D(v)− 1 ≥ d+D′(v). �

Theorem 5.25 (8.4.30 Galvin 1995). Every X, Y -bigraph G satisfies χ′l(G) = ∆(G).

Proof. Let ∆ := ∆(G) and set H := L(G). Then χ′l(G) = χl(H) and χ′(G) = ∆(Theorem 5.12). Fix a ∆-edge coloring c : E(G)→ [∆].

Let L be a ∆-edge-list assignment for G; so L is a ∆-list assignment for H. Our plan is toapply Lemma 5.24 to H to show that it has an L-coloring f ; then f is an L-edge-coloring ofG. So it suffices to show H has an orientation D := ( ~E(G), A) such that (i) ∆+(D) ≤ ∆− 1and (ii) every induced subgraph D′ ⊆ D has a kernel.

Each ~ee′ ∈ E(H) satisfies e ∩ e′ ⊆ X or e ∩ e′ ⊆ Y . Define D by putting~ee′ ∈ A iff (e ∩ e′ ⊆ X ∧ c(e) > c(e′)) ∨ (e ∩ e′ ∈ Y ∧ c(e) < c(e′)).

Each e ∈ E(G) = V (H) satisfies d+H(e) ≤ ∆− 1, since it has at most c(e)− 1 out-neighbors

e′ with e ∩ e′ ⊆ X and ∆− c(e) out-neighbors e′ with e ∩ e′ ⊆ Y . So (i) holds for D.45

For (ii), let D′ = D[F ], where F ⊆ E(G) = V (D). Argue by induction on |D′|. LetX ′ := {x ∈ X : EG(x) ∩ F 6= ∅}. For each x ∈ X ′, choose ex ∈ EG(x) ∩ F with c(ex)minimum. Then ~eex ∈ A for every e ∈ EG(x) ∩ F − ex. If Q = {ex : x ∈ X ′} is independentthen it is a kernel of D′; else fix distinct x, x′ ∈ X ′ with ex ∩ ex′ 6= ∅. Then ex ∩ ex′ = y forsome y ∈ Y ; say c(ex) < c(ex′). Let D′′ = D′ − ex. By induction, D′′ has a kernel S. Ifex′ ∈ S then S is a kernel for D′, since ~exex′ ∈ A. Otherwise, ~ex′e

∗ ∈ A for some e∗ ∈ S.The choice of ex′ implies ex′ ∩ e∗ /∈ X. Thus ex, ex′ , e∗ ∈ EG(y), and c(e∗) > c(ex′) > c(ex);so ~exe

∗ ∈ A. Hence S is a kernel for D′. �

46

CHAPTER 6

Planar graphs

We have been informally drawing graphs in the Euclidean plane R2 since the start of thesemester. Now we formalize the definition of a drawing of a graph in R2.

6.1. Very Basic Topology of the Euclidean Plane

For x ∈ R2 the open ball around x with radius r is the set Br(x) := {y ∈ R2 : ‖x, y‖ < r}.A set U ⊆ R2 is open if for all points p ∈ U there exists r > 0 such that Br(p) ⊆ U . Inparticular, R2 and ∅ are open. The complement of an open set is a closed set. The frontierof a set X is the set of all points x ∈ R2 such that Br(x)∩X 6= ∅ and Br(x)rX 6= ∅ for allr > 0. Note that if X is open, then its frontier lies in R2 rX.

Let p, q ∈ R2. The p, q-line segment Lp,q is the subset of R2 defined by L(p, q) :=

{p+λ(q−p) : 0 ≤ λ ≤ 1} and L(p, q) := L(p, q)r{p, q}. For distinct points p0, . . . , pk ∈ R2,the union A(p0, . . . , pk) :=

⋃i∈[k] L(pi−1, pi) is a (polygonal) p0, pk-arc provided L(pi−1, pi) ∩

L(pj−1, pj) = ∅ for all distinct i, j ∈ [k]. We say that p0 and pk are linked by A(p0, . . . , pk). IfA(p0, . . . , pk) is an arc and L(p0, . . . , pk)∩L(pk, p0) = ∅ then P (p0, . . . , pk, p0) := L(p0, . . . , pk)∪L(pk, p0) is a polygon. Note that arcs and polygons are closed in R2.

Let U be an open set. Two points x, y ∈ U are linked in U if there exists an x, y-arccontained in U . The relation of being linked is an equivalence relation on U . Its equivalenceclasses are called regions. Regions are open: Suppose R ⊆ U is a region and x ∈ R. Thenthere exists a r > 0 such that Br(x) ⊆ U . Clearly every y ∈ Br(x) is linked to x in U , sinceL(x, y) ⊆ Br(x). So Br(x) ⊆ R. A closed set X separates a region R if R r X has morethan one region.

Theorem 6.1 (Jordan Curve Theorem for Polygons). For every polygon P ⊆ R2, the setR2 r P has exactly two regions. Each of these regions has the entire polygon as its frontier.

Proof. To be continued ... �

6.2. Graph Drawings

Recall that in this class all graphs are finite. Let G = (V,E) be a (multi)graph. Adrawing of G is a graph G := (V , E) such that G ∼= G , V ⊆ R2, and each edge e ∈ Eis an arc linking its ends. So edges are no longer just pairs of vertices, but have their ownidentity and structure (we need this anyway to formally deal with different edges linkingthe same two vertices). It should be clear that every finite graph has a drawing. Moreover,by moving vertices slightly and readjusting edges, we can (and do) require the followingadditional properties for drawings, without restricting the set of (finite) graphs that can bedrawn.

(1) No three edges have a common internal point.47

(2) The only vertices contained in an edge are its endpoints.(3) No two edges are tangent.(4) No two edges have more than one common internal point.

A crossing in a drawing of a graph is a point that is in the interior of two edges. A plane(multi)graph is a drawing with no crossing of a (multi)graph. A planar (multi)graph is a(multi)graph that has a plane drawing.

Let G := (V , E) be a plane (multi)graph. The faces of G are the regions of R2r(V ∪⋃E).

The frontier of a face f is called its boundary ; it is denoted by G[f ].The edges of a face f can be ordered to form its boundary walk(s); this is a closed walk(s)

that contains every edge once or twice. Let l(f) be the length of its boundary walk. LetF (G) be the set of faces of G.

Note that a plane cycle is a polygon. By the Jordan Curve Theorem we have:

Proposition 6.2. A plane cycle is the boundary of exactly two faces.

6.3. Basic facts

Fact 6.3. [Top.]The boundary of a face is the union of some edges and vertices of G—soit is a subgraph of G.

Fact 6.4. [Top.]Let G be a plane graph. If the boundary of some face does not contain acycle then G has only one face whose boundary is G, and so G is a forest (whose edges arecut-edges.

Fact 6.5. [Top.]Every maximal planar graph is connected.

Fact 6.6. [Top.]Let G be a plane graph with a face f whose boundary is a cycle C. Letx, y ∈ V (C) and z ∈ f . Then there is an x, y-arc A with A ⊆ f and z ∈ A. Moreover,G + A is a plane graph with F (G + A) = F (G) − f + f1 + f2, where f = f1 ∪ f2 ∪ A is apartition of f and B(f1) = xCyA∗x and B(f2) = yCxAy.

Fact 6.7. [Top.]Let G be a plane multigraph with e ∈ E. If e is a cut-edge then |F (G)−e)| = |F (G)|; if e is not a cut-edge then |F (G− e)| = |F (G)| − 1.

Fact 6.8. [Top.]Let G be a plane graph with link (non-loop-edge) e = xy. Let G · e be aplane graph formed from G by deleting y and adding edges xv in the face of G-y containingy for all v ∈ N(y) r N(x). Then |F (G · e)| = |F (G)|. Deleting a loop of G, reduces thenumber of faces by 1.

For an edge e and a face f , let ι(e, f) be the number of times e appears on the boundarywalk of f .

Fact 6.9. [Top.]Let G be a plane multigraph. Every non-cut-edge appears exactly once onthe bounding walk of exactly two faces. Every cut-edge appears exactly twice on the boundingwalk of exactly one face. In particular,∑

f∈F (G)

l(f) =∑

f∈F (G)

∑e∈E(G)

ι(e, f) =∑

e∈E(G)

∑f∈F (G)

ι(e, f) = 2∥∥∥G∥∥∥ .

48

Theorem 6.10 (6.1.21 Euler’s Formula (1758)). All connected plane multigraphs G sat-isfy

|G| − ‖G‖+ |F (G)| = 2.

Proof. Argue by induction on ‖G‖. As G is connected, |G| = 1. Then |F (G)| = 1, so

|G| − ‖G‖+ |F (G)| = 1− 0 + 1 = 2.

Otherwise, there is e ∈ E(G). If e is a link xy then set G′ = G · e. Then G′ is connected.By Fact 6.8, |F (G)| = |F (G′)|, so by induction

|G| − ‖G‖+ |F (G)| = |G′|+ 1− ‖G′‖ − 1 + |F (G′)| = |G′| − ‖G′‖+ |F (G′)| =ind 2.

Else e is a loop. Now set G′ = G− e. By Fact 6.8, |F (G)| = F (G′) + 1, so by induction

|G| − ‖G‖+ |F (G)| = |G′| − ‖G′‖ − 1 + |F (G′)|+ 1 = |G′| − ‖G′‖+ |F (G′)| =ind 2. �

Theorem 6.11. If G is a simple planar graph with |G| ≥ 3 then ‖G‖ ≤ 3|G| − 6.Moreover, if G has girth greater than 3 then ‖G‖ ≤ 2|G| − 4.

Proof. Let G be a drawing of G. We may assume G has as many edges as any otherplanar graph on n vertices. Then G is connected by Fact 6.5. Since G is connected andhas at least three vertices, ‖G‖ ≥ 2; since it is simple it has no parallel edges. By Fact 6.4,|F (G)| = 1 and every edge is a cut edge, or every face boundary contains a cycle. Thusl(f) ≥ 3 for every f ∈ F (G). By Fact 6.9,

2‖G‖ =∑

f∈F (G)

l(f) ≥ 3|F (G)|.

So |F (G)| ≤ 23‖G‖. By Theorem 6.10

2 = |G| − ‖G‖+ |F (G)| ≤ |G| − 1

3‖G‖,

and so 3|G| − 6 ≥ ‖G‖.Now suppose that the girth of G is greater than 3. Then every face boundary has length

at least 4. So2‖G‖ =

∑f∈F (G)

l(f) ≥ 4|F (G)|.

Thus |F (G)| ≤ 12‖G‖. By Theorem 6.10

2 = |G| − ‖G‖+ |F (G)| ≤ |G| − 1

2‖G‖,

and so ‖G‖ ≤ 2|G| − 4. �

Corollary 6.12. Neither K5 nor K3,3, is planar.

Proof. If K5 is planar then by Theorem 6.11

10 = ‖K5‖ ≤ 3|K5| − 6 = 9,

49

a contradiction. If K3,3 is planar, then since it is bipartite, and so has girth greater than 3,Theorem 6.11 yields the contradiction:

9 = ‖K3,3‖ ≤ 2|K3,3| − 4 = 8. �

A graph H is a subdivision of a graph G if H is formed by replacing some of the edgesxy ∈ E(G) by an x, y-path P whose internal vertices are not vertices of G and have degree 2in H. The vertices of G are called branch vertices and the new vertices are called subdivisionvertices. Notice G is not a subgraph of a proper subdivision of itself.

Theorem 6.13 (6.2.2 Kuratowski (1930)). A graph is planar iff it contains neither asubdivision of K5 nor a subdivision of K3,3.

We will break the proof of Kuratowski’s Theorem into smaller pieces. Call a subdivisionof K5 or K3,3 a K-graph (for Kuratowski).

Lemma 6.14. Let e = x1x2 be an edge of a graph G. If G contains no K-graph then G · econtains no K-graph.

Proof. We prove the contrapositive. Suppose G · e contains a K-graph Q. If Q ⊆ Gwe are done; else ve ∈ V (Q). Then 2 ≤ dQ(ve) ≤ 4 and NQ(ve) ⊆ NG(x1) ∪ NG(x2). IfdQ(ve) = 2 then ve is a subdivision vertex of Q. Replacing ve with x, y or xy results in aK-graph in G. Else veis a branch vertex of Q.

If there is i ∈ [2] with |NQ(ve) r NG(xi)| ≤ 1 then replacing ve by xi or the path x1x2

yields a copy or subdivision of Q. Otherwise, dQ(ve) = 4, |NQ(ve)rNG(x1)| = 2 = |NQ(ve)rNG(x2)|, and Q is a subdivision of K5({ve, a, b, c, d}). We claim G contains a subdivision ofK3,3! Let Q′ = Q−ve+x1 +x2 +e. Then Q′ is a subdivision of K3,3{{x1, a, b}, {x2, c, d}}. �

Lemma 6.15. If G is planar then G does not contain a K-graph.

Proof. Arguing by contraposition, assume there is a K-graph Q ⊆ G, and show byinduction on the number h of subdivision vertices in Q that G is not planar. If h = 0 thenthis is Corollary 6.12. If h > 0, consider a subdivision vertex x and one of its two neighborsy. Contracting xy yields a K-graph with one less subdivision vertex. Thus by inductionG · xy is nonplanar, and by Fact 6.8, G is nonplanar. �

Let G = (V,E) be a graph, and S ⊆ V . An S-lobe is a subgraph of the form G[S∪V (H)],where H is a component of G− S.

Lemma 6.16. Let G be a planar 2-connected planar graph with plane drawing G. ThenG[f ] is a cycle for every face f ∈ F (G) .

Proof. By Theorem 4.7, G has an a 2-witness P0, . . . , Ph. Argue by induction on h. Ifh = 0, then P0 is a cycle that bounds the only two faces of G (Jordan Curve Theorem). Elseh ≥ 1. Let Ph have ends x, y, and note that ˚

Ph is contained in some face f ′ of H := G− ˚Ph.

By induction, every face of H is bounded by a cycle. Let C = xv1 . . . vayva+2 . . . vbx be thecycle bounding f ′. By Fact 6.8, F (G) = F (H)− f + f1 + f2 and both f1 and f2 are boundedby cycles. �

Theorem 6.17. Every 3-connected graph G that contains no K-graph is planar.50

Proof. Argue by induction on |G|. As K4 is planar, assume |G| ≥ 5. By Theorem4.10, H := G · e is 3-connected for some edge e = xy; so 3 ≤ d(x) and (say) d(x) ≤ d(y).By Lemma 6.14, H contains no K-graph. By induction, H has a plane drawing H. PutX = NG(x)−y and Y = NG(y)−x; then NH(ve) = X∪Y . As H is 3-connected, |X∪Y | ≥ 3.

Set H ′ = H−ve, and let f be the face of H−ve that contains ve. Since H ′ is 2-connected,C := H ′[f ] is a cycle by Lemma 6.16, and X ∪ Y ⊆ V (C). Let C = P1P2 . . . Pn where eachPi is an X-path in C with ends xi, xi⊕1. Draw G′ := G− y as G′ by replacing ve with x inH and deleting the edges vez with z ∈ Y rX. The faces of G′ that are contained in f havethe form fi where G′[fi] = xxiPxi⊕1x, i ∈ [n]. It suffices to show that N(y) ⊆ G′[fi] forsome i ∈ [n]—then using Fact 6.6 we can draw y and all e with e ∈ E(y) in fi.

If Y ⊆ X then X = Y as d(x) ≤ d(y); as |X ∪ Y | ≥ 3 G contains TK5 with branchvertices in (X ∩ Y ) ∪ {x, y}, a contradiction. Else there is y′ ∈ Y r X; say y′ ∈ P i. IfY * G′[fi], let y′′ ∈ Y r G′[fi]. Then G contains a TK3,3 whose branch vertices havebipartition {{y, xi, xi+1}, {x, y′, y′′}}, a contradiction. So N(y) ⊆ G′[fi]. �

Lemma 6.18. Let G = (V,E) be a graph with a minimum separating set S, where |S| ≤ 2.Suppose V1, V2 6= S with V = V1 ∪ V2 and S = V1 ∩ V2, and E(V1 r S, V2 r S) = ∅. If Gis edge-maximal among graphs without K-graphs then so are G1 := G[V1] and G2 := G[V2],and G[S] = K2.

Proof. Let S = {ui : i ∈ [|S|]}. As S is a minimum separating set, there are vij ∈N(ui)∩ (Vj r S) for all ui ∈ S and j ∈ [2]. Let wi ∈ Vir S for i ∈ [2]. By maximality G+ econtains a K-graph H with e ∈ E(H) for all e ∈ E(G).

If S = ∅ then e := w1w2 is a cut-edge in G + e. As K-graphs contain no cut-edges, thisis a contradiction. Now suppose |S| = 1. Then G′ := G + v1

1v21 contains a K-graph H. As

any two branch vertices of H are linked by 3 disjoint paths in H, all the branch vertices arecontained in, say, G1. Thus there is a subdivided edge P = v1

1v12 . . . u

1 with H[V2 + v11] = P .

Replacing P with v11u

1 yields a K-graph in G1 ⊆ G a contradiction. So |S| = 2. Supposee := u1u2 /∈ E. Then G + e contains a K-graph H with e ∈ E(H). Again, the branchvertices of H are contained in (say) G1. Then H ⊆ G1 + e. Replacing e by a u1, u2-path inG2 yields a K-graph in G, a contradiction. So G[S] = K2.

Finally, suppose e ∈ E(Gi); say i = 1. Then G+e contains a K-graph H with e ∈ E(H).If the branch vertices of H are contained in G2, then G1 ∩H is a u1, u2-path P . ReplacingP with u1u2 yields a K-graph in G2 ⊆ G, a contradiction. So the branch vertices of H arecontained in G1. If H ∩ G2 − S 6= ∅ then H ∩ G2 is a u1, u2 path P . Replacing P by u1u2

yields a K-graph in G1 + e. Thus G1 is maximum. �

Theorem 6.19. Let G = (V,E) be a graph with |G| ≥ 4. If G is edge-maximal amonggraphs with no K-graph then G is 3-connected.

Proof. Argue by induction on |G|. Suppose S is a minimum separating set. If |S| ≥ 3we are done, so suppose not. Choose sets V1, V2 6= S with V = V1 ∪ V2, S = V1 ∩ V2, andE(V1 r S, V2 r S) = ∅. By Lemma 6.18, G[S] = K2, say S = {x, y}, and each Gi := G[Vi]is edge-maximal among graphs with no K-graph. By induction each Gi is either K3 or 3-connected, so Theorem 6.17 implies each Gi is planar. For each Gi let Gi be a plane drawingof Gi. For i ∈ [2], let fi ∈ F (Gi) be a face whose boundary contains xy and a third vertex

51

zi. Then Gi + xzi + yzi is planar, and so contains no K-graph. By the maximality of Gi,xzi, yzi ∈ Gi. Let vi be a new vertex and G+

i := Gi ∪K({vi}, {x, y, zi}). Drawing v and itsincident edges in fi shows that G+

i is planar.As e := z1z2 ∈ E(G), G′ := G + e contains a K-graph H with e ∈ H. If all the branch

vertices of H are contained in the same Gi then there is a path P = ziz3−i . . . u in G′,where u ∈ S. Replacing P by uzi yields a K-graph in Gi, a contradiction. Thus H hasbranch vertices in each Gi − S. In G′ there are at most three V1, V2-paths. Thus the onlypossibility is that H is a subdivision of K3,3 with exactly one branch vertex v in (say) G2−S.Then H ∩G′[V2 + z1] is a v, {x, y, z1}-fan F . So H with F replaced by K(v1, {x, y, z1}) is asubdivision of K3,3 in G+

i , contradicting Lemma 6.15. �

Proof of Kuratowski’s Theorem 6.17. First supposeG contains aK-graph. Thenby Lemma 6.15, G is not planar. Now suppose G contains no K-graph. Then G is aspanning subgraph of a graph G′ that is edge-maximal among graphs with no K-graph. ByTheorem 6.19, G′ is 3-connected. By Theorem 6.17, G′ is planar, so G is planar. �

Theorem 6.20 (8.4.32 Thomassen (1994)). Every planar graph G is 5-list colorable.

Proof. It suffices to prove the following more technical statement by induction on G.

Claim. Suppose G is a drawing of a simple planar graph G such that every inner facehas length three, and the boundary of the outer face is a cycle C = v1v2 . . . vsv1 with x = v1

and y = v2. If L is a list assignment for G such that(1) L(x) = {α}, L(y) = {β}, and α 6= β,(2) |L(v)| = 3 for all vertices v ∈ {v3, . . . , vs}, and(3) |L(v)| = 5 for all vertices v ∈ V (G− C),

then G has an L-coloring.

The claim implies the theorem since adding edges and vertices to G, and deleting colorsfrom some lists of L does not make it easier to L-color G. Moreover, every face of anedge-maximal planar graph is bounded by a C3. So it suffices to prove the claim.

Proof of Claim. Argue by induction on |G|. Note that |G| ≥ |C| ≥ 3. First considerthe base step |G| = 3. Color x with α and y with β. The last vertex z has three colors inits list, so it can be colored with a color distinct from α and β.

Now consider the induction step |G| > 3. See Figure 6.3.1.Case 1: C has a chord vivj with i > j. Let C1 = vivi⊕1 . . . vjvi and C2 = vjvj⊕1 . . . vivj bethe two nonspanning cycles contained in C + vivj. Let Gi be the plane graph formed byCi and its interior. Then G1 ∪ G2 = G, G1 ∩ G2 = G[{vivj}], and x, y ∈ V (C1). By theinduction hypothesis, there exists an L-coloring g1 of G1. Set x′ = vi, α′ = g1(x′), y′ = vj,β′ = g1(y′), L′(x′) = {α′}, L′(y′) = {β′} and L′(v) = L(v) for all vertices of G2 − x′ − y′.Then by the induction hypothesis there exists an L′-coloring g2 of G2. Since the chord vivjseparates G1 from G2, setting f = g1 ∪ g2 yields an L-coloring of G.Case 2: C does not have a chord. Since every interior face is bounded by a C3, G[N(vs)]contains a hamiltonian x, vs−1-path P . Moreover, since C has no chords, the outer face ofG′ = G − vs is bounded by the cycle C ′ = xPvs−1C

∗v1 (= x). Of course, the interior faces52

G1 G2 G′

vn−1vi−1 x = v1x = v1

y = v2 y = v2

vi = x′

G G

Case 1: C has a chord. Case 2: C has no chord.

vj = y′

vn

C C

Figure 6.3.1. Proof of Thomassen’s Theorem

of G′ have length three. Let γ, δ ∈ L(vs) be distinct colors not equal to α. Define a listassignment L′ for L by

L′(v) =

{L(v)− γ − δ if v ∈ V (P )− x− vs−1

L(v) else

(and shrinking oversized lists). By induction G′ has an L′-coloring f ′. Pick ε ∈ {γ, δ} withε 6= f ′(vs−1). Finally, extend f ′ to an L-coloring f of G by setting f(vs) = ε. �

This completes the proof of the claim and the Theorem. �

HW 61 (*). Let G be a simple planar graph with a shortest cycle of length k. Provethat ‖G‖ ≤ k

k−2(|G| − 2).

HW 62 (*). Prove that every simple planar graph G with |G| ≥ 4 has at least fourvertices with degree less than six.

HW 63 (+). Prove that every simple planar graph G with δ(G) = 5 has a matching withat most 1

5|G| unsaturated vertices.

HW 64 (*). Prove that the vertices of a simple planar graph can be ordered so thatevery vertex is preceded by at most five of its neighbors. Similarly, prove that the verticesof every simple planar bipartite graph can be ordered so that each vertex is preceded by atmost three of its neighbors.

HW 65 (*). Prove that every planar bipartite graph satisfies χl(G) ≤ 4. [Hint: Use theprevious problem.

HW 66 (+). Prove that every orientation of every X, Y -bigraph has a kernel. [Hint:When is X a kernel?]

HW 67 (*). Prove that every bipartite planar graph G satisfies χl(G) ≤ 3. [Hint: Usethe previous problem.]

53

CHAPTER 7

Extras

7.1. Lower Bounds on Ramsey’s Theorem

Theorem 7.1. For every integer k ≥ 2 there exists a graph G such that ω(G) < k,α(G) < k, and |G| ≥ b2k/2−1/2c. In other words, Ram(k, k) ≥ 2k/2−1/2.

Proof. Fix k ≥ 2, and set n = b2k/2−1/2c. Let V be a set of n vertices, and G be theset of all graphs G with V (G) = V . So G = (V,E) ∈ G if and only iff E ⊆

(V2

). Since there

are 2(n2) choices for E,

(7.1.1) |G| = N := 2(n2).

For X ⊆ V with |X| = k, let GX be the set of graphs in G such that X is a clique or co-clique. So if G := (V,E) ∈ G then G ∈ GX iff E ∩

(X2

)∈ {∅,

(X2

)} and E r

(X2

)⊆(V2

)r(X2

).

There are two possibilities for the first conjunct and 2(n2)−(k

2) possibilities for the second.Thus

(7.1.2) |GX | = 2 · 2(n2)−(k

2) = 2 · 2−(k2−k)/2N.

Any graph G inG r

⋃X∈(V

k)

GX ,

satisfies ω(G), α(G) < k and |G| = n. So it suffices to prove |⋃X∈(V

k) GX | < |G|. Since (a)

|⋃X∈(V

k) GX | ≤(nk

)|GX |, (b)

(nk

)< nk

k!, and (c) n

2k/2−1/2 ≤ 1, this follows from:

|G| − |⋃

X∈(Vk)

GX | ≥ N −(n

k

)· 2 · 2−(k2−k)/2N ((a), (7.1.1), (7.1.2))

> N(1− 2nk

k!2−(k−1)k/2) (b)

≥ N(1− (n

2k/2−1/2)k) ≥ 0. (c) �

7.2. Equitable Coloring

Definition 7.2. An equitable k-coloring of a graph G = (V,E) is a proper coloringf : V → [k] such that difference ||f−1(i)|− |f−1(j)|| in the sizes of the the i-th and j-th colorclasses is at most 1 for all i, j ∈ [k]. In particular, every color is used if |G| ≤ k.

Theorem 7.3 (Hajnal & Szemerédi Theorem (1976)). Every graph G with maximumdegree at most r has an equitable (r + 1)-coloring.

54

The proof was long and sophisticated, and does not provide a polynomial time algorithm.Kierstead and Kostochka found a much simpler and shorter proof. This better understandinghas led to many new results, several of which are stated below.

Let θ(G) = max{d(x) = d(y) : xy ∈ E(G)}.

Theorem 7.4 (Kiestead & Kostochka (2008)). For every r ≥ 3, each graph G withθ(G) ≤ 2r + 1 has an equitable (r + 1)-coloring.

Theorem 7.5 (Kierstead, Kostochka, Mydlarz & Szemerédi). There is an algorithm thatconstructs an equitable k-coloring of any graph G with ∆(G) + 1 ≤ k, using time O(r|G|2).

Problem 7.6. Find a polynomial time algorithm for constructing the coloring in Theo-rem (7.4).

One might hope to prove an equitable version of Brooks’ Theorem, but the followingexample shows that the statement would require special care: For r is odd, Kr,r satisfies∆(Kr,r) = r and ω(G) = 2, but has no r-equitable coloring. Chen, Lih and Wu [?] proposedthe following common strengthening of Theorem 7.3 and Brooks’ Theorem.

Conjecture 7.7. Let G be a connected graph with ∆(G) ≤ r. Then G has no equitabler-coloring if and only if either (a) G = Kr+1, or (b) r = 2 and G is an odd cycle, or (c) ris odd and G = Kr,r.

Kierstead and Kostochka have proved the conjecture for r ≤ 4, and also for r ≥ 14|G|.

Proof of Theorem 7.3. Let G be a graph with ∆(G) ≤ r. We may assume that |G| isdivisible by r + 1: If |G| = s(r + 1) − p, where p ∈ [r] then set G′ := G + Kp. Then |G′| isdivisible by r+ 1 and ∆(G′) ≤ r. Moreover, the restriction of any equitable (r+ 1)-coloringof G′ to G is an equitable (r + 1)-coloring of G. So we may assume |G| = (r + 1)s.

We argue by induction on ‖G‖. The base step ‖G‖ = 0 is trivial, so consider the inductionstep. Let u be a non-isolated vertex. By the induction hypothesis, there exists an equitable(r + 1)-coloring of G − E(u). We are done unless some color class V contains an edge uv.Since ∆(G) ≤ r, some color class W contains no neighbors of u. Moving u to W yields an(r+ 1)-coloring of G with all classes of size s, except for one small class V − := V − u of sizes−1 and one large class V + := W +u of size s+1. Such a coloring is called nearly equitable.

Given a nearly equitable (r + 1)-coloring, define an auxiliary digraph H, whose verticesare the color classes, so that UW is a directed edge if and only if some vertex y ∈ U hasno neighbors in W . In this case we say that y witnesses UW . Let A be the set of classesfrom which V − can be reached in H, B be the set of classes not in A and B′ be the set ofclasses reachable from V + in H[B]. Set a := |A|, b := |B|, b′ := |B′|, A :=

⋃A, B :=

⋃B

and B′ :=⋃B′. Then r + 1 = a+ b. Since every vertex y ∈ B has a neighbor in every class

of A and every vertex z ∈ B′ also has a neighbor in every class of B − B′,

(*) dA(y) ≥ a for all y ∈ B and dA∪B\B′(z) ≥ a+ b− b′ for all z ∈ B′.

Case 0: V + ∈ A. Then there exists a V +, V −-path P = V1, . . . , Vk in H. Moving eachwitness yj of VjVj+1 to Vj+1 yields an equitable (r + 1)-coloring of G. �

We now argue by a secondary induction on b, whose base step b = 0 holds by Case 0.Also |A| = as− 1 and |B| = bs+ 1. Now consider the secondary induction step.

55

A class W ∈ A is terminal, if every U ∈ A−W can reach V − in H−W . Let A′ be theset of terminal classes, a′ := |A′| and A′ :=

⋃A′. An edge wz is solo if w ∈ W ∈ A′, z ∈ B

and NW (z) = {w}. Ends of solo edges are solo vertices and solo neighbors of each other.Order A as V −, X1, . . . , Xa−1 so that each Xi has a previous out-neighbor.Case 1: For some a − b ≤ i ≤ a − 1, class Xi is not terminal. Then some Xj ∈ A′

cannot reach V − in H−Xi. So j > i and Xj has no out-neighbors before Xi. In particular,d+A(Xj) < b. Then for each w ∈ Xj, dA(w) ≥ a− b, and so dB(w) < 2b. Let S be the set of

solo vertices in Xj, and D := Xj \ S. If v ∈ B −NB(S) then v has no solo neighbor in Xj,and so has at least two neighbors in D. Thus 2b |D| > 2 |B −NB(S)|. Using |S| + |D| = sand r |S| ≥ |E(S,A)|+ |NB(S)|,bs+ (a− 1) |S| = b |D|+ r |S| > |B −NB(S)|+ |E(S,A)|+ |NB(S)| > bs+ |E(S,A)| .

Thus (a−1) |S| > |E(S,A)|, and so there exists w ∈ S with dA(w) ≤ a−2. Thus w witnessessome edge XjX ∈ E(H[A]). Since w ∈ S, it has a solo neighbor y ∈ B.

Move w to X and y to Xj. This yields nearly equitable colorings of G[A+y] and G[B−y].Since Xj is terminal, X + w can still reach V −. Thus by Case 0, G[A+ y] has an equitablea-coloring. By (*), ∆(G[B − y]) ≤ b− 1. So by the primary induction hypothesis G[B − y]has an equitable b-coloring. After combining these equitable colorings we are done.

Case 2: All the last b classes Xa−b, . . . , Xa−1 are terminal. Then a′ ≥ b. For y ∈ B′, letσ(y) be the number of solo neighbors of y. Similarly to (*),

r ≥ d(y) ≥ a+ b− b′ + dB′(y) + a′ − σ(y) ≥ r + 1 + dB′(y) + a′ − b′ − σ(y).

So σ(y) ≥ a′− b′+dB′(y) + 1. Let I be a maximal independent set with V + ⊆ I ⊆ B′. Then∑y∈I(dB′(y) + 1) ≥ |B′| = b′s+ 1. Since a′ ≥ b,∑

y∈I

σ(y) ≥∑y∈I

(a′ − b′ + dB′(y) + 1) ≥ s(a′ − b′) + b′s+ 1 > a′s = |A′| .

So some vertex w ∈ W ∈ A′ has two solo neighbors y1 and y2 in the independent set I.Since the class Y of y1 is reachable from V +, we can equitably b-color G[B − y1]. Let

Y ′ be the new class of y2. If w witnesses an edge WX of G[A] then we are done by Case 1;otherwise we can move w to some class U ⊆ B′ − y1. Replacing w with y1 in W to get W ∗

and moving w to U yields a new nearly equitable (r + 1)-coloring of G. If U ∈ A then weare done by Case 0; otherwise at least a+ 1 classes, W ∗, Z ′, and all X ∈ A′ −W , can reachV −. In this case we are done by the secondary induction hypothesis.

56

APPENDIX A

Exceptional notation

|G| = |V (G)| = n(G), number‖G‖ = |E(G)| = e(G), numberE(v) = {vu : vu ∈ E}, edge setK(A,B) = (A ∪B, {ab : a 6= b ∧ (a, b) ∈ A×B}), graphK(A) = K(A,A), graphEG(A,B) = {ab ∈ E(G) : a 6= b ∧ (a, b) ∈ A×B}, edge set

57

APPENDIX B

Standards

Midterm 1 in red, Midterm 2 in blue, final in red, blue, and black. The syllabus for thequalifier is the syllabus as the final together with all old qualifiers.

B.0.1. MAT 416–Level C.(1) Theorem 1.24 (Characterization of bipartite graphs).(2) Theorem 1.36 (Euler’s Theorem). You may use Lemma 1.35.(3) Theorem 2.7 (Tree Theorem). Provide proofs for required parts of Lemma 2.5 and

Lemma 2.6.(4) Theorem 3.4 (Berge’s theorem on maximum matchings).(5) Corollary 3.6. You may use Hall’s Theorem.(6) Theorem 3.13 Petersens’s Matching Theorem.(7) Theorem 4.7 (Whitney’s Theorem).(8) Lemma 5.24 (Kernel Lemma).(9) Euler’s Formula Theorem 6.10 and Theorem 6.11.

B.0.2. MAT 416–Level A. All MAT 416–Level C and HWA #1–3, HWA #4,5, HW6 except (+) and:

(1) Theorem 1.5 (Ramsey’s Theorem for graphs).(2) Corollary 1.27 (Dirac’s Theorem). Provide proofs for required parts of Theorem 1.25.(3) Theorem 3.8 (König–Egerváry Theorem).(4) Theorem 3.5 (Hall’s Theorem).(5) Theorem 3.11 (Tutte’s Theorem).(6) Theorem 3.14 (Petersen’s 2-Factor Theorem).(7) Theorem 4.10 (Thomassen’s Contraction Theorem).(8) Theorem 4.15 (Menger’s Theorem).(9) Example 5.3 Mycielski’s Construction.(10) Theorem 5.6 (Brooks’ Theorem). You may assume Proposition 5.4 and Lemma 5.5,

but be prepared to prove them if asked.(11) Theorem 5.10 (Turan’s Theorem). You may assume Lemma 5.9, but be prepared

to prove it if asked.(12) Theorem 5.14 (Vizing’s Theorem). This includes proving Lemma 5.13.(13) Theorem 5.25 (Galvin’s Theorem).(14) Kuratowski’s Theorem 6.13. You may assume: 6.14, 6.15 and 6.16. You should be

able to prove Theorem 6.17 and also Theorem 6.13, using Theorems 6.17 and 6.19(easy).

(15) Thomassen’s 5-Choosability Theorem 6.20.(16) Lower bound on Ram(k, k) Theorem 7.1.

58

B.0.3. MAT–513. All MAT 416–Level A, HWA #1–3, HWA #4,5, HWA#6. Note:HWA #1–6 consists of HW #1–66.

B.1. Sample MAT 416/513 Midterm 1Directions: Use one sheet per problem. Order the sheets by problem before submitting.

MAT 513 Students: Do the last 4 problems.MAT 416 Students: Do any 5 problems; one will be treated as extra credit. The first

four problems are intended to be somewhat easier. Only the five chosen problems should beturned in.

(1) Prove: If an acyclic graph G satisfies |G| = ‖G‖+ 1 then it is connected.(2) Prove: Every k-regular bigraph G has a perfect matching.(3) Let P ⊆ G be an x, y-path. Prove: G[P ] contains an x, y-path Q with Q = G[Q].(4) Let T be a forest (acyclic graph) such that 2k of its vertices have odd degree. Prove

that T decomposes into k paths.***************************************************************************

(5) Prove: A graph is bipartite if and only if it contains no odd cycle.(6) Prove: A graph G = (V,E) has a perfect matching if o(G− S) ≤ |S| for all S ⊆ V .(7) Let n ≥ 2 and d1, . . . , dn ∈ Z+. Prove: If

∑ni=1 di = 2n− 2 then there is a tree with

vertices v1, . . . , vn such that d(vi) = di for all i ∈ [n].(8) For k, n ∈ N, let G be an A,B-bigraph with |A| = n = |B| such that δ(G) ≥ k, and

for all X ⊆ A, Y ⊆ B, if |X|, |Y | ≥ k then |E(X, Y )| 6= ∅. Prove: G has a perfectmatching.

59

APPENDIX C

Matching card trick

Consider a deck of 2k + 1 cards numbered 1, . . . , 2k + 1, and denoted by [2k + 1]. Theclass chooses a hand H consisting of k + 1 of these cards, and gives them to Professor A.Professor A looks at them, puts one of them in his pocket, and then has a student spread theremaining k cards face-up on a table. Professor B, who has observed none of this transaction,now enters the room, looks at the cards on the table and identifies the one in Professor A’spocket. How is this done?

Solution. Our arithmetic is done modulo k + 1, and we use k + 1 instead of 0 for therepresentative of its equivalence class. Arrange the cards of H in order as c1 < · · · < ck+1.Let x =

∑c∈H c mod k + 1. Professor A hides card cx. When Professor B arrives, he

sees that the cards d1 < · · · < dk+1 in [2k + 1] r (H − cx) are missing, and he calculatesy :=

∑c∈H−x = x − cx. The class is holding cx − 1 − (x − 1) = −y cards less than cx and

Professor A is holding cx. It follows that cx = d1−y, and Professor B can calculate the rhs.Another way of saying this is that Professor B knows the missing cards d1 > · · · > dk+1.

Then cx = dk+2−(1−y) = dy.

60

APPENDIX D

Alternative proofs

D.1. Hall’s Theoerm

Proof of Theorem 3.5. If M is a matching saturating X and S ⊆ X then |S| =|E(S, Y )∩M | ≤ |N(S)|; so (??) holds.

Suppose (??) holds for some X, Y -bigraph with no matching saturating X; among suchcounterexamples choose G with |G| minimal, and subject to this ‖G‖ maximal. By mini-mality, (1) N(X) = Y : if y ∈ Y rN(X) then G− y is a smaller counterexample. Also (2)all a ∈ X satisfy N(a) 6= Y : else, since G− a has a matching saturating X − a and there isan unsaturated vertex b ∈ Y by (??), G has a matching saturating X.

Let a ∈ X; by (2) there is b ∈ Y with ab /∈ E. Since G + ab satisfies (??), maximalityimplies G+ab has a matchingM+ saturating X. By (1) there is a′ ∈ X with a′b ∈ E; by (2)there is b′ ∈ Y with a′b′ /∈ E. Again by maximality, G + a′b′ has a matching L+ saturatingX. Set M := M+ − ab and L := L+ − a′b′.

Let H be the spanning submultigraph of G with edge set M ∪ L + a′b, where edges inM∩L have multiplicity 2. (Figure D.1.1). Then ∆(H) ≤ 2. As G is bipartite, Proposition ??implies the components of H are paths and even cycles. Each cycle has a perfect matching.As dH(a) = 1 and dH(x) = 2 for all x ∈ X − a, each path P has an end y ∈ Y and amatching saturating V (P )− y. Combining these matchings yields a matching saturating X.

Here is an alternative way through the last paragraph.Let H be the spanning subgraph of G with E(H) = M4L. (Figure D.1.1). Then

∆(H) ≤ 2 and dH(b), dH(b′) ≤ 1 = dH(a) = dH(a′) < dH(x) for all x ∈ X r {a, a′}. Sothe component of H containing a is an alternating a, v- path P . If v ∈ Y then P is an

a a′

b b′Y

X

M,L

Figure D.1.1. X, Y -bigraph H ⊆ G

61

M -augmenting path. Else v = a′, and aPvb is an M -augmenting path. Anyway, G has amatching saturating X. �

D.2. König, Egerváry Theorem

Proof of Theorem ??. Order V as v1 ≺ · · · ≺ v|G|. Since W is a cover, every edge isincident to some vertex of W (possibly two). Define a function g : M → W by g(e) is theleast w ∈ e∩W . Since M is a matching, no vertex of W can be incident to two edges of M .So g is an injection. Thus |M | ≤ |W |.

Now suppose G is an X, Y -bigraph. Let U be the set of unsaturated vertices in X. Setm = {(x, y) ∈ X × Y : xy ∈ M}. Since the ends of M in X are distinct, m is a functionwith domain X r U . Since the ends of m in Y are distinct, m is an injection.

If U = ∅ then X is a cover with |W | ≤ |X| = |M | ≤ |W |; so suppose U 6= ∅. LettingA ⊆ V (G) be the set of ends of alternating paths starting in U , set S := A∩X, S := XrS,T := A ∩ Y , and T := Y r T . Then U ⊆ S (witnessed by trivial paths). Consider anyalternating path P = v0 . . . vn with v0 ∈ U . If i is even then vi ∈ S, and if also i 6= 0 thenvi−1vi ∈M ; if i is odd then vi ∈ T . We first show:

(i) N(S) ⊆ T and (ii) T ⊆ m(S r U).(D.2.1)

(i) Let z ∈ N(S); say wz∈ E(S, z). Then there is an alternating path Q = y0 . . . y2k(= w)with y0 ∈ U and y2k−1w ∈ M . Either z ∈ V (Q) or Qwz is an alternating path starting inU . Regardless, z ∈ T , proving (D.2.1.i).

(ii) Let z ∈ T . Then there is an alternating path P = y0 . . . y2k+1(= z) with y0 ∈ U andy2kz /∈M . SinceM is maximum, G has no augmenting path. So z is saturated; say zx ∈M .Then Pzx is an alternating path; so x ∈ S r U , and z ∈ m(S r U), proving (D.2.1.ii).1

The set W = S ∪ T is a vertex cover of G: Suppose xy ∈ E with x ∈ X. If x ∈ S theny ∈ T ⊆ W by (i); else x ∈ S ⊆ W .

It remains to show that |W | ≤ |M |. For this it suffices to prove that g is onto. Considerw ∈ W . If w ∈ T then there is x ∈ S with xw ∈ M by (ii). As x /∈ W , g(xw) = w. Ifw ∈ S then there is y with wy ∈ M , since U ⊆ S. As M is a matching, (ii) implies y ∈ T .So g(wy) = w. �

D.3. Menger’s Theorem

Theorem D.1 (Menger 1927 4.2.17). Let G = (V,E) be a graph, and suppose A,B ⊆ V .Then the size l := l(A,B) of a maximum set of disjoint A,B-paths is equal to the sizek := k(A,B) of a minimal A,B-separating set.

Proof 1. (l ≤ k) If P is a set of disjoint A,B-paths and S is an A,B-separator then Smust contain at least one vertex of each path, and each vertex of S is on at most one pathof P . Thus the function f : P → S defined by setting f(P ) equal to the first x ∈ S ∩ V (P )is an injection; so |P| ≤ |S|. Choosing P maximum and S minimum yields the inequality.

(k ≤ l) For a set of A,B-paths P let end(P) denote the set of ends in B of paths in P . Itsuffices to show (*) if P ′ is a set of disjoint A,B-paths with |P ′| < k then there exists a setP of disjoint A,B-paths such that |P| = |P ′|+1 and end(P ′) ⊆ end(P). Argue by induction

1(i) and (ii) imply the sufficiency of Hall’s Criteria (??): |N(S)| ≤ |S r U |, so (??) fails if U 6= ∅.

62

on |G′|, where G′ := G − B. If |G′| = 0 then A ⊆ B. So the A,B-paths are exactly thepaths consisting of a single vertex of A, and k = |A|. Thus (*) holds.

Suppose |G′| > 0, and fix a set P ′ of disjoint A,B-paths with |P ′| < k. Since |end(P ′)| =|P ′| < k, there is an A,B-path R = Ry′ in G − end(P ′). If R ∩

⋃P ′ = ∅ then put

P := P ′ + R. If not, let x be the last vertex of R that is in⋃P ′; say x ∈ P ∈ P ′

and y is the end of P in B. Note that x /∈ B, since y 6= y′ and V (P ) ∩ B = y. PutB′ := B ∪ V (xRy′ ∪ xPy) and Q′ := P ′ − P + Px. Then end(Q′) = end(P ′) − y + x.Since every A,B-path contains an A,B′-path, k = k(A,B) ≤ k(A,B′). Since B + x ⊆ B′,|G − B′| < |G − B|. So by induction, there exists a set Q of A,B′-paths and y′′ ∈ B′ suchthat |Q| − 1 = |Q′| = |P ′| and end(Q) = end(Q′) + y′′. So x 6= y′′. Let x, y′′ be the ends ofQ,Q′′ ∈ Q. Set P0 = Q − Q − Q′′. If y′′ ∈ xPy then set P := P0 + QxRy′ + Q′′y′′Py′; ify′′ ∈ xRy′ then set P := P0 +QxPy +Q′′y′′Ry′; else set P := P0 +QxPy +Q′′. Evidently,P witnesses (*). �

Proof 2 (k ≤ l). So it suffices to show k ≤ l. Argue by induction on ‖G‖.Base Step: ‖G‖ = 0. Then every A,B-path is trivial. So A ∩ B is the maximum set ofdisjoint A,B-paths and the minimum A,B-separating set. Thus l = |A ∩B| = k.Induction Step: ‖G‖ ≥ 1. Let e = xy ∈ E(G), and put G′ = G · e. For any U ⊆ V , define

U ′ =

{U − {x, y}+ ve if U ∩ {x, y} 6= ∅U otherwise

,

and note that for every T ⊆ V (G′) there exists S ⊆ V with T = S ′. Using Lemma ?? andthe discussion before, every set P ′ of disjoint A′, B′-paths corresponds to a set P of disjointA,B-paths with |P| = |P ′| (but not vice versa). So

lG′(A′, B′) ≤ l.

Also, if S is an A,B-separator in G if and only if S ′ is an A′, B′-separator in G′. So

kG′(A′, B′) ≤ k ≤ kG′(A

′, B′) + 1.

Choose a minimum A′, B′-separator T in G′. If kG′(A′, B′) = k then by the inductionhypothesis applied to G′ we have:

k = kG′(A′, B′) ≤ lG′(A

′, B′) ≤ l,

and we are done. Otherwise, k = kG′(A′, B′) + 1. In this case vxy ∈ T , and T = S ′, where

S := T − vxy + x+ y. In particular xy ∈ G[S].Set G′′ = G− e. Since e ∈ G[S],

kG(A, S) = kG′′(A, S) and kG(B, S) = kG′′(B, S)(D.3.1)

Since S separates A from B in G, every A, S-separator in G separates A from B, and so hassize at least |S|, and a similar statement holds for B. So we have

(D.3.2) kG(A, S), kG(S,B) ≥ k.

Thus

|S| ≥ lG(A, S) ≥ lg′′(A, S) =i.h. kG′′(A, S) =(D.3.1) kG(A, S) ≥(D.3.2) k = |S| and|S| ≥ lG(B, S) ≥ lg′′(B, S) =i.h. kG′′(B, S) =(D.3.1) kG(B, S) ≥(D.3.2) k = |S|.

63

Let KA be a collection of |S| = k disjoint A, S-paths and KB be a collection of |S| disjointS,B-paths. Then for each z ∈ S there is a unique A, z-path Pz and a unique z,B-path Qz. Ifv ∈ V (Pw)∩V (Qz) then v ∈ S, since otherwise PvQ is an A,B-walk in G−S, contradictingthe fact that S is an A,B-separator. Thus w = v = z, and so {PzzQz : z ∈ S} is a collectionof |S| = k disjoint A,B-paths. �

Definition D.2 (4.2.18). The line graph H = L(G) of a graph G = (V,E) is defined byV (H) = E and E(H) = {ee′ : e ∩ e′ 6= ∅}.

64

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