Graph Colorings

Abstract

Graph theory would not be what it is today if there had been no coloring problems. In fact, a major portion of the 20th-century research in graph theory has its origin in the four-color problem. (See Chap. 8 for details.)

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7.1 Introduction

Graph theory would not be what it is today if there had been no coloring problems. In fact, a major portion of the 20th-century research in graph theory has its origin in the four-color problem. (See Chap. 8 for details.)

In this chapter, we present the basic results concerning vertex colorings and edge colorings of graphs. We present two important theorems on graph colorings, namely, Brooks’ theorem and Vizing’s theorem. We also present a brief discussion on “snarks” and Kirkman’s schoolgirl problem. In addition, we also present a detailed description of the Mycielskian of a graph.

7.2 Vertex Colorings

7.2.1 Applications of Graph Coloring

We begin with a practical application of graph coloring known as the storage problem. Suppose a university’s Department of Chemistry wants to store its chemicals. It is quite probable that some chemicals cause violent reactions when brought together. Such chemicals are incompatible chemicals. For safe storage, incompatible chemicals should be kept in distinct rooms. The easiest way to accomplish this is, of course, to store one chemical in each room. But this is certainly not the best way of doing it since we will be using more rooms than are really needed (unless, of course, all the chemicals are mutually incompatible!). So we ask: What is the minimum number of rooms required to store all the chemicals so that in each room only compatible chemicals are stored?

We convert the above storage problem into a problem in graphs. Form a graph G = (V, E) by making V correspond bijectively to the set of available chemicals and making u adjacent to v if and only if the chemicals corresponding to u and v are incompatible. Then, any set of compatible chemicals correspond to a set of independent vertices of G. Thus, a safe storing of chemicals corresponds to a partition of V into independent subsets of G. The cardinality of such a minimum partition of V is then the required number of rooms. The minimum cardinality is called the chromatic number of the graph G.

Definition 7.2.1.

The chromatic number $\chi (G)$ of a graph G is the minimum number of independent subsets that partition the vertex set of G. Any such minimum partition is called a chromatic partition of V (G).

The storage problem just described is actually a vertex coloring problem of G. A vertex coloring of G is a map $f\, :\, V \rightarrow S,$ where S is a set of distinct colors; it is proper if adjacent vertices of G receive distinct colors of S. This means that if uv ∈ E(G), then f(u)≠f(v). Thus, $\chi (G)$ is the minimum cardinality of S for which there exists a proper vertex coloring of G by colors of S. Clearly, in any proper vertex coloring of G, the vertices that receive the same color are independent. The vertices that receive a particular color make up a color class. This allows an equivalent way of defining the chromatic number.

Definition 7.2.2.

The chromatic number of a graph G is the minimum number of colors needed for a proper vertex coloring of G. G is k-chromatic if χ(G) = k.

Definition 7.2.3.

A k-coloring of a graph G is a vertex coloring of G that uses at most k colors.

Definition 7.2.4.

A graph G is said to be k-colorable if G admits a proper vertex coloring using at most k colors.

In considering the chromatic number of a graph, only the adjacency of vertices is taken into account. Hence, multiple edges and loops may be discarded while considering chromatic numbers, unless needed otherwise. As a consequence, we may restrict ourselves to simple graphs when dealing with (vertex) chromatic numbers.

It is clear that $\chi ({K}_{n}) = n.$ Further, $\chi (G) = 2$ if and only if G is bipartite having at least one edge. In particular, $\chi (T) = 2$ for any tree T with at least one edge (since any tree is bipartite). Further (see Fig. 7.1),

$$\chi ({C}_{n}) = \left \{\begin{array}{@{}l@{\quad }l@{}} 2\quad &\text{ if}\ n\ \text{ is even}\\ 3\quad &\text{ if} \ n\ \text{ is odd.} \end{array} \right.$$

(7.1)

Fig. 7.1

Illustration of proper vertex coloring

Exercise 2.1.

Prove $\chi (G) = 2$ if and only if G is a bipartite graph with at least one edge.

Exercise 2.2.

Determine the chromatic number of

• (i) The Petersen graph

• (ii) Wheel W _n (see Sect. 1.7, Chap. 1)

• (iii) The Herschel graph (see Fig. 5.4)

• (iv) The Grötzsch graph (see Fig. 7.6)

We next consider another application of graph coloring. Let S be a set of students. Each student of S is to take a certain number of examinations for which he or she has registered. Undoubtedly, the examination schedule must be such that all students who have registered for a particular examination will take it at the same time.

Let $\mathbb{P}$ be the set of examinations and for $p \in \mathbb{P},$ let S(p) be the set of students who have to take the examination p. Our aim is to draw up an examination schedule involving only the minimum number of days on the supposition that papers a and b can be given on the same day provided they have no common candidate and that no candidate shall have more than one examination on any day.

Form a graph $G = G(\mathbb{P},E),$ where $a,b \in \mathbb{P}$ are adjacent if and only if $S(a)\, \cap \, S(b) \neq {\o}.$ Then each proper vertex coloring of G yields an examination schedule with the vertices in any color class representing the schedule on a particular day. Thus, $\chi (G)$ gives the minimum number of days required for the examination schedule.

Exercise 2.3.

Draw up an examination schedule involving the minimum number of days for the following problem:

Exercise 2.4.

If G is k-regular, prove that $\chi (G) \geq \frac{n} {n-k}.$

Theorem 7.2.5 gives upper and lower bounds for the chromatic number of a graph G in terms of its independence number and order.

Theorem 7.2.5.

For any graph G with n vertices and independence number $\alpha,$

Table 1

$\frac{n} {\alpha } \leq \chi \leq n - \alpha + 1.$

Proof.

There exists a chromatic partition $\{{V }_{1},{V }_{2},\ldots,{V }_{\chi }\}$ of V. Since each V _i is independent, $\vert {V }_{i}\vert \leq \alpha,$ $1 \leq i \leq \chi.$ Hence, $n ={ \sum \nolimits }_{i=1}^{\chi }\,\vert {V }_{i}\vert \leq \alpha \,\chi,$ and this gives the inequality on the left.

To prove the inequality on the right, consider a maximum independent set S of $\alpha$ vertices. Then the subsets of $V \setminus S$ of cardinality 1 together with S yield a partition of V into $(n - \alpha ) + 1$ independent subsets.

Remark 7.2.6.

Unfortunately, none of the above bounds is a good one. For example, if G is the graph obtained by connecting C _2r with a disjoint K _2r (r ≥ 2), by an edge, we have n = 4r, $\alpha = r + 1,$ and $\chi = 2r,$ and the above inequalities become $\frac{4r} {r+1} \leq 2r \leq 3r.$ For a simple graph G, the number ${\chi }^{c} = {\chi }^{c}(G) = \chi ({G}^{c}),$ the chromatic number of G ^c is the minimum number of subsets in a partition of V (G) into subsets each inducing a complete subgraph of G. Bounds on the sum and product of $\chi (G)$ and ${\chi }^{c}(G)$ were obtained by Nordhaus and Gaddum [148] (see also reference [93]), as given in Theorem 7.2.7.

Theorem 7.2.7 (Nordhaus and Gaddum [148]).

For any simple graph G

$$2\sqrt{n} \leq \chi + {\chi }^{c} \leq n + 1,\mathrm{and}n \leq \chi \,{\chi }^{c} \leq {\left (\frac{n + 1} {2} \right )}^{2}.$$

Proof.

Let $\chi (G) = k$ and let ${V }_{1},{V }_{2},\ldots,{V }_{k}$ be k color classes in a chromatic partition of G. Then ${\sum \nolimits }_{i=1}^{k}\,\vert {V }_{i}\vert = n,$ and so $ma{x}_{1\leq i\leq k}\,\vert {V }_{i}\vert \geq \frac{n} {k}.$ Since each V _i is an independent set of G, it induces a complete subgraph in G ^c. Hence, ${\chi }^{c} {\geq \max }_{1\leq i\leq k}\,\vert {V }_{i}\vert,$ and so $\chi \,{\chi }^{c} = k\,{\chi }^{c} \geq k {\circ \max }_{1\leq i\leq k}\,\vert {V }_{i}\vert \geq k \circ \frac{n} {k} = n.$ Further, since the arithmetic mean of $\chi$ and ${\chi }^{c}$ is greater than or equal to their geometric mean, $\frac{\chi +{\chi }^{c}} {2} \geq \sqrt{\chi \,{\chi }^{c}} \geq \sqrt{n}.$ Hence, $\chi + {\chi }^{c} \geq 2\,\sqrt{n}.$ This establishes both lower bounds.

To show that $\chi + {\chi }^{c} \leq n + 1,$ we use induction on n. When n = 1, $\chi = {\chi }^{c} = 1,$ and so we have equality in this case. So assume that $\chi + {\chi }^{c} \leq (n - 1) + 1 = n$ for all graphs G having n − 1 vertices, $n \geq 2.$ Let H be any graph with n vertices, and let v be any vertex of H. Then G = H − v is a graph with n − 1 vertices and ${G}^{c} = {(H - v)}^{c} = {H}^{c} - v.$ By the induction assumption, $\chi (G) + \chi ({G}^{c}) \leq n.$

Now $\chi (H) \leq \chi (G) + 1$ and $\chi ({H}^{c}) \leq \chi ({G}^{c}) + 1.$ If either $\chi (H) \leq \chi (G)$ or $\chi ({H}^{c}) \leq \chi ({G}^{c}),$ then $\chi (H) + \chi ({H}^{c}) \leq \chi (G) + \chi ({G}^{c}) + 1 \leq n + 1.$ Suppose then $\chi (H) = \chi (G) + 1$ and $\chi ({H}^{c}) = \chi ({G}^{c}) + 1.$ $\chi (H) = \chi (G) + 1$ implies that removal of v from H decreases the chromatic number, and hence ${d}_{H}(v) \geq \chi (G).$ [If ${d}_{H}(v) < \chi (G),$ then in any proper coloring of G with $\chi (G)$ colors at most $\chi (G) - 1$ colors would have been used to color the neighbors of v in G, and hence v can be given one of the left-out colors, and therefore we have a coloring of H with $\chi (G)$ colors. Hence, $\chi (H) = \chi (G),$ a contradiction.] For a similar reason, $\chi ({H}^{c}) = \chi ({G}^{c}) + 1$ implies that $n - 1 - {d}_{H}(v) = {d}_{{H}^{c}}(v) \geq \chi ({G}^{c});$ thus, $\chi (G) + \chi ({G}^{c}) \leq {d}_{H}(v) + n - 1 - {d}_{H}(v) = n - 1.$ This implies, however, that $\chi (H) + \chi ({H}^{c}) = \chi (G) + \chi ({G}^{c}) + 2 \leq n + 1.$

Finally, applying the inequality $\sqrt{\chi \,{\chi }^{c }} \leq \frac{\chi +{\chi }^{c}} {2},$ we get $\chi \,{\chi }^{c} \leq {(\frac{\chi +{\chi }^{c}} {2} )}^{2} \leq {(\frac{n+1} {2} )}^{2}.$

Note 7.2.8.

Since the publication of Theorem 7.2.7, there had been similar results for other graph parameters (see, for instance, [115] for the domination number γ). All these results have now come to be known as Nordhaus–Gaddum inequalities, with reference to the parameters in question.

Exercise 2.5.

For a simple graph G, prove that $\chi ({G}^{c}) \geq \alpha (G).$

Exercise 2.6.

Prove $\chi (G) \leq \mathcal{l} + 1,$ where ℓ is the length of a longest path in G. For each positive integer ℓ, give a graph G with chromatic number $\mathcal{l} + 1$ and in which any longest path has length ℓ.

Exercise 2.7.

Which numbers can be chromatic numbers of unicyclic graphs? Draw a unicyclic graph on 15 vertices with $\Delta = 3$ and having each of these numbers as its chromatic number.

Exercise 2.8.

If G is connected and $m \leq n,$ show that $\chi (G) \leq 3.$

Exercise 2.9.

Let G _n be the graph defined by $V ({G}_{n}) =\{ (i,j)\, :\, 1 \leq i < j \leq n\},$ and $E({G}_{n}) =\{ \big{(}(i,j)(k,l)\big{)}\, :\, i < j = k < l\}.$ Prove

• (i) $\omega ({G}_{n}) = 2$.

• (ii) $\chi ({G}_{n}) = {\lceil \log }_{2}n\rceil.$ [Note that $\chi ({G}_{n}) \rightarrow \infty$ as $n \rightarrow \infty$.]

Exercise 2.10.

Prove that $\chi (G\square H) =\max (\chi (G),\chi (H)).$

Exercise 2.11.

Prove $\chi (G \times H) \leq \min (\chi (G),\,\chi (H)))$ (A celebrated conjecture of Hedetniemi [104] states that $\chi (G \times H) =\min (\chi (G),\,\chi (H))).$

7.3 Critical Graphs

Definition 7.3.1.

A graph G is called critical if for every proper subgraph H of G, $\chi (H) < \chi (G).$ Equivalently, $\chi (G - e) < \chi (G)$ for each edge e of G. Also, G is k-critical if it is k-chromatic and critical.

Remarks 7.3.1.2.

If $\chi (G) = 1,$ then G is either trivial or totally disconnected. Hence, G is 1-critical if and only if G is K ₁. Again, $\chi (G) = 2$ implies that G is bipartite and has at least one edge. Hence, G is 2-critical if and only if G is K ₂. For an odd cycle C, $\chi (C) = 3,$ and if G contains an odd cycle C properly, G cannot be 3-critical.

Exercise 3.1.

Prove that any critical graph is connected.

Exercise 3.2.

Prove that for any graph G, $\chi (G - v) = \chi (G)$ or $\chi (G) - 1$ for any $v \in \, V,$ and $\chi (G - e) = \chi (G)$ or $\chi (G) - 1$ for any e ∈ E.

Exercise 3.3.

Show that if G is k-critical, $\chi (G - v) = \chi (G - e) = k - 1$ for any v ∈ V and e ∈ E.

Exercise 3.4.

[If $\chi (G - e) < \chi (G)$ for any e of G, G is sometimes called edge-critical, and if $\chi (G - v) < \chi (G)$ for any vertex v of G, G is called vertex-critical.] Show that a nontrivial connected graph is vertex-critical if it is edge-critical. Disprove the converse by a counterexample.

Exercise 3.5.

Show that a graph is 3-critical if and only if it is an odd cycle. It is clear that any k-chromatic graph contains a k-critical subgraph. (This is seen by removing vertices and edges in succession, whenever possible, without diminishing the chromatic number.)

Theorem 7.3.3.

If G is k-critical, then $\delta (G) \geq k - 1.$

Proof.

Suppose $\delta (G) \leq k - 2.$ Let v be a vertex of minimum degree in G. Since G is k-critical, $\chi (G - v) = \chi (G) - 1 = k - 1$ (see Exercise 3.3). Hence, in any proper (k − 1)-coloring of G − v, at most (k − 2) colors would have been used to color the neighbors of v in G. Thus, there is at least one color, say c, that is left out of these k − 1 colors. If v is given the color c, a proper (k − 1)-coloring of G is obtained. This is impossible since G is k-chromatic. Hence, $\delta (G) \geq (k - 1).$

Corollary 7.3.4.

For any graph G $\chi (G) \leq 1 + \Delta (G).$

Proof.

Let G be a k-chromatic graph, and let H be a k-critical subgraph of G. Then $\chi (H) = \chi (G) = k.$ By Theorem 7.3.3, $\delta (H) \geq k - 1,$ and hence $k \leq 1 + \delta (H) \leq 1 + \Delta (H) \leq 1 + \Delta (G).$

Exercise 3.6.

Give another proof of Corollary 7.3.4 by using induction on n = | V (G) | .

Exercise 3.7.

If $\chi (G) = k,$ show that G contains at least k vertices each of degree at least k − 1.

Exercise 3.8.

Prove or disprove: If G is k-chromatic, then G contains a K _k .

Exercise 3.9.

Prove: Any k ( ≥ 2)-critical graph contains a (k − 1)-critical subgraph.

Exercise 3.10.

For each of the graphs G of Exercise 2.2, find a critical subgraph H of G with $\chi (H) = \chi (G).$

Exercise 3.11.

Prove that the wheel ${W}_{2n-1} = {C}_{2n-1}\, \vee \, {K}_{1}$ is a 4-critical graph for each n ≥ 2. Does a similar statement apply to W _2n ?

Theorem 7.3.5.

In a critical graph G, no vertex cut is a clique.

Proof.

Suppose G is a k-critical graph and S is a vertex cut of G that is a clique of G (i.e., a complete subgraph of G). Let H _i , $1 \leq i \leq r,$ be the components of $G\setminus S,$ and let ${G}_{i} = G[V ({H}_{i})\, \cup \, S].$ Then each G _i is a proper subgraph of G and hence admits a proper (k − 1)-coloring. Since S is a clique, its vertices must receive distinct colors in any proper (k − 1)-coloring of G _i . Hence, by fixing the colors for the vertices of S, and coloring for each i the remaining vertices of G _i so as to give a proper (k − 1)-coloring of G _i , we obtain a proper (k − 1)-coloring of G. This contradicts the fact that G is k-chromatic (see Fig. 7.2).

Fig. 7.2

$G[<CitationRef CitationID="CR104">104</CitationRef>]\, \simeq \, {K}_{4}$ (r = 2)

Corollary 7.3.6.

Every critical graph is a block.

Exercise 3.12.

^∗Prove that every k-critical graph is (k − 1)-edge connected (Dirac [53]).

Exercise 3.13.

Show by means of an example that criticality is essential in Exercise 3.12; that is, a k-chromatic graph need not be (k − 1)-edge connected.

7.3.1 Brooks’ Theorem

We next consider Brooks’ [31] theorem. Recall Corollary 7.3.4, which states that $\chi (G) \leq 1 + \Delta (G).$ If G is an odd cycle, $\chi (G) = 3 = 1 + 2 = 1 + \Delta (G),$ and if G is a complete graph, say K _k , $\chi (G) = k = 1 + (k - 1) = 1 + \Delta (G).$ That these are the only extremal families of graphs for which $\chi (G) = 1 + \Delta (G)$ is the assertion of Brooks’ theorem.

Theorem 7.3.7 (Brooks’ theorem).

If a connected a graph G is neither an odd cycle nor a complete graph, then $\chi (G) \leq \Delta (G).$

Proof.

If $\Delta (G) \leq 2,$ then G is either a path or a cycle. For a path G (other than K ₁ and K ₂), and for an even cycle G, $\chi (G) = 2 = \Delta (G).$ According to our assumption, G is not an odd cycle. So let $\Delta (G) \geq 3.$

The proof is by contradiction. Suppose the result is not true. Then there exists a minimal graph G of maximum degree $\Delta (G) = \Delta \geq 3$ such that G is not $\Delta$-colorable, but for any vertex v of G, G − v is $\Delta$-colorable.

Claim 1.

Let v be any vertex of G. Then in any proper Δ-coloring of G − v, all the Δ colors must be used for coloring the neighbors v in G. Otherwise, if some color i is not represented in N _G (v), then v could be colored using i, and this would give a Δ-coloring of G, a contradiction to the choice of G. Thus, G is a Δ-regular graph satisfying Claim 1.

For $v \in V (G),$ let $N(v) =\{ {v}_{1},{v}_{2},\ldots,{v}_{\Delta }\}.$ In a proper $\Delta$-coloring of G − v = H, let v _i receive color i, $1 \leq i \leq \Delta.$ For i≠j, let H _ij be the subgraph of H induced by the vertices receiving the ith and jth colors.

Claim 2.

v _i and v _j belong to the same component of H _ij. Otherwise, the colors i and j can be interchanged in the component of H _ij that contains the vertex v _j . Such an interchange of colors once again yields a proper Δ-coloring of H. In this new coloring, both v _i and v _j receive the same color, namely, i, a contradiction to Claim 1. This proves Claim 2.

Claim 3.

If C _ij is the component of H _ij containing v _i and v _j , then C _ij is a path in H _ij. As before, N _H (v _i ) contains exactly one vertex of color j. Further, C _ij cannot contain a vertex, say y, of degree at least 3; for, if y is the first such vertex on a ${v}_{i} - {v}_{j}$ path in C _ij that has been colored, say, with i, then at least three neighbors of y in C _ij have the color j. Hence, we can recolor y in H with a color different from both i and j, and in this new coloring of H, v _i and v _j would belong to distinct components of H _ij (see Fig. 7.3a). (Note that by our choice of y, any v _i − v _j path in H _ij must contain y. ) But this contradicts Claim 3.

Claim 4.

${C}_{ij}\, \cap \, {C}_{ik} =\{ {v}_{i}\}$ for j≠k. Indeed, if $w \in {C}_{ij}\, \cap \, {C}_{ik},$ $w\neq {v}_{i},$ then w is adjacent to two vertices of color j on C _ij and two vertices of color k on C _ik (see Fig. 7.3b). Again, we can recolor w in H by giving a color different from the colors of the neighbors of w in H. In this new coloring of H, v _i and v _j belong to distinct components of H _ij , a contradiction to Claim 2. This completes the proof of Claim 4.

We are now in a position to complete the proof of the theorem. By hypothesis, G is not complete. Hence, G has a vertex v, and a pair of nonadjacent vertices v ₁ and v ₂ in N _G (v) (see Exercise 5.11, Chap. 1). Then the ${v}_{1} - {v}_{2}$ path C ₁₂ in H ₁₂ of H = G − v contains a vertex y ($\neq \,{v}_{2}$) adjacent to v ₁. Naturally, y would receive color 2. Since $\Delta \geq 3,$ by Claim 1, there exists a vertex ${v}_{3} \in {N}_{G}(v).$ Now interchange colors 1 and 3 in the path C ₁₃ of H ₁₃. This would result in a new coloring of H = G − v. Denote the v _i -v _j path in H under this new coloring by C _ij ′ (see Fig. 7.3c). Then $y \in {C}_{23}\prime$ since v ₁ receives color 3 in the new coloring (whereas y retains color 2). Also, $y \in {C}_{12} - {v}_{1} \subset {C}_{12}\prime .$ Thus, $y \in {C}_{23}\prime \, \cap \, {C}_{12}\prime .$ This contradicts Claim 4 (since $y\neq {v}_{2}$), and the proof is complete.

Fig. 7.3

Graphs for proof of Theorem 7.3.7 (The numbers inside the parentheses denote the vertex colors)

7.3.2 Other Coloring Parameters

There are several other vertex coloring parameters of a graph G. We now mention three of them. Let f be a k-coloring (not necessarily proper) of G, and let $({V }_{1},{V }_{2},\ldots,{V }_{k})$ be the color classes of G induced by f. Coloring f is pseudocomplete if between any two distinct color classes, there is at least one edge of G. f is complete if it is pseudocomplete and each V _i , $1 \leq i \leq k,$ is an independent set of G. Thus, $\chi (G)$ is the minimum k for which G has a complete k-coloring f.

Definition 7.3.8.

The achromatic number a(G) of a graph G is the maximum k for which G has a complete k-coloring.

Definition 7.3.9.

The pseudoachromatic number $\psi (G)$ of G is the maximum k for which G has a pseudocomplete k-coloring.

Example 7.3.10.

Figure 7.4 gives (a) a chromatic, (b) an achromatic, and (c) a pseudoachromatic coloring of ${K}_{3,3} - e.$

Fig. 7.4

Different colorings of ${K}_{3,3} - e$

It is clear that for any graph G, $\chi (G) \leq \text{ a}(G) \leq \psi (G).$

Exercise 3.14.

Let G be a graph and H a subgraph of G. Prove that $\chi (H) \leq \chi (G)$ and $\psi (H) \leq \psi (G).$ Show by means of an example that { a}(H) ≤ { a}(G) need not always be true.

Exercise 3.15.

Prove

• (i) $\psi (\psi - 1) \leq 2m.$

• (ii) $\psi ({K}_{a}\, \vee \, {K}_{b}^{c}) = a + 1.$

From (ii) deduce that for any graph, $\psi \leq n - \alpha + 1.$

Exercise 3.16.

If G has a complete coloring using k colors, prove that $k\, \leq \, \tfrac{1+\sqrt{1+8m}} {2}.$ (m = size of G).

Exercise 3.17.

Prove that for a complete bipartite graph G, { a}(G) = 2.

Exercise 3.18.

What is the minimum number of edges that a connected graph with pseudoachromatic number ψ can have? Construct one such tree.

Exercise 3.19.

If G is a subgraph of H, prove that $\psi (G) \leq \psi (H).$

Exercise 3.20.

Prove: $\psi ({K}_{n,n}) = n + 1.$

7.3.3 b-Colorings

Definition 7.3.11. A b-coloring of a graph G is a proper coloring with the additional property that each color class contains a color-dominating vertex (c.d.v.), that is, a vertex that has a neighbor in all the other color classes. The b-chromatic number of G is the largest k such that G has a b-coloring using k colors; it is denoted by b(G).

The concept of b-coloring was introduced by Irving and Manlove [111].

Exercise 3.21 guarantees the existence of the b-chromatic number for any graph G and shows that $\chi (G) \leq \text{ b}(G).$ Note that { b}(K _n ) = n while { b}(K _{m, n} ) = 2.

Exercise 3.21.

Show that the chromatic coloring of a graph G is a b-coloring of G.

Exercise 3.22.

Prove that K _{n, n} − F, n ≥ 2, where F is a 1-factor of K _{n, n} , has a b-coloring using 2 colors and n colors but none with k colors for any k in 2 < k < n.

Exercise 3.23.

Prove $\text{ b}(G)\, \leq \, 1 + \Delta (G).$ A better upper bound for { b}(G) is given in the next exercise.

Exercise 3.24.

Let ${d}_{1} \leq {d}_{2}\, \leq \ldots \leq \, {d}_{n}$ be the degree sequence of the graph G with vertex set $V =\{ {v}_{1},\ldots,{v}_{n}\},$ and ${d}_{i} = d({v}_{i}),$ $1 \leq i\, \leq \, n.$ Let $M(G) =\max \{ i\, :\, {d}_{i} \geq i - 1,\,1 \leq i \leq n\}.$ Prove that $\text{ b}(G)\, \leq \, M(G).$ Show further that the number of vertices of degree at least M(G) in G is at most M(G).

Exercise 3.25.

Let Q _p be the hypercube of dimension p. Prove $\text{ b}({Q}_{1}) = \text{ b}({Q}_{2})=2,$ and $\text{ b}({Q}_{3}) = 4.$ [A result of Kouider and Mahéo [125] states that for $p\, \geq \, 3,$ $\text{ b}({Q}_{p}) = p + 1$.]

We complete this section by presenting a result of Kratochvíl, Tuza, and Voigt [126] that characterizes graphs with b-chromatic number 2. Let G be a bipartite graph with bipartition (X, Y ). A vertex x ∈ X (respectively, y ∈ Y ) is called a full vertex (or a charismatic vertex) of X (respectively, Y ) if it is adjacent to all the vertices of Y (respectively, X).

Theorem 7.3.12 ([126]).

Let G be a nontrivial connected graph. Then { b}(G) = 2 if and only if G is bipartite and has a full vertex in each part of the bipartition.

Proof.

Suppose G is bipartite and has a full vertex in each part, say x ∈ X and y ∈ Y. Naturally, in any b-coloring, the color class containing x, say W ₁, is a subset of X and that containing y, say W ₂, is a subset of Y. If G has a third color class W ₃ disjoint from W ₁ and W ₂, then W ₃ must have a c.d.v. adjacent to a vertex of W ₁ and a vertex of W ₂. This is impossible, as G is bipartite. Therefore, { b}(G) = 2.

Conversely, let { b}(G) = 2. Then $\chi (G) = 2$ and therefore G is bipartite. Let (X, Y ) be the bipartition of G. Assume that G does not have a full vertex in at least one part, say, X. Let x ₁ ∈ X. As x ₁ is not a full vertex, there exists a vertex y ₁ ∈ Y to which it is not adjacent. Let X ₁ be the maximal subset of X such that ${V }_{1} = {X}_{1} \cup \{ {y}_{1}\}$ is independent in G. Now choose a new vertex ${x}_{2} \in X\setminus {X}_{1}$. Again, as X has no full vertex, we can find a ${y}_{2} \in Y \setminus \{{y}_{1}\}$ to which x ₂ is not adjacent. Let X ₂ be the maximal subset of $X\setminus {X}_{1}$ such that ${V }_{2} = {X}_{2} \cup \{ {y}_{2}\}$ is independent in G. In this way, all the vertices of X would be exhausted and let ${V }_{1},{V }_{2},\ldots,{V }_{k}$ be the independent sets thus formed. Also, let Y ₀ denote the set of uncovered vertices of Y, if any. Since G is connected, $G\neq \langle {V }_{i} \cup {V }_{j}\rangle,$ and $G\neq \langle {V }_{l} \cup {Y }_{0}\rangle,$ $i,j,l \in \{ 1,2,\ldots,k\}.$ Hence, $k\, \geq \, 2$ when ${Y }_{0}\neq {\o}$ and $k\, \geq \, 3$ when ${Y }_{0} = {\o}.$ Thus, the partition $V = {V }_{1} \cup {V }_{2} \cup \ldots \cup {V }_{k} \cup \{ {V }_{k+1} = {Y }_{0}\}$ has at least 3 parts. If each of these parts has a c.d.v., we get a contradiction to the fact that { b}(G) = 2. If not, assume that the class V _l has no c.d.v. Then for each vertex x of V _l , there exists a color class V _j , $j\neq l,$ having no neighbor of x. Then x could be moved to the class V _j . In this way, the vertices in V _l can be moved to the other V _i ’s without disturbing independence. Let us call the new classes ${V }_{1}\prime ,{V }_{2}\prime ,\ldots,{V }_{l-1}\prime ,{V }_{l+1}\prime ,\ldots,{V }_{k+1}\prime .$ If each of these color classes contains a c.d.v., we get a contradiction as $k\, \geq \, 3.$ Otherwise, argue as before and reduce the number of color classes. As G is connected, successive reductions should end up in at least three classes, contradicting the hypothesis that { b}(G) = 2.

A description of several other coloring parameters can be found in Jensen and Toft [116].

7.4 Homomorphisms and Colorings

Homomorphisms of graphs generalize the concept of graph colorings.

Definition 7.4.1.

Let G and H be simple graphs. A homomorphism from G to H is a map $f\, :\, V (G) \rightarrow V (H)$ such that $f(x)f(y) \in E(H)$ whenever xy ∈ E(G). The map f is an isomorphism if f is bijective and $xy \in E(G)$ if and only if $f(x)f(y) \in E(H).$

We write $f\, :\, G \rightarrow H$ to denote the fact that f is a homomorphism from G to H and write $G \simeq H$ to denote that G is isomorphic to H. If $f\, :\, G \rightarrow H$ is a graph homomorphism, then $\langle \{f(x) : x \in V (G)\}\rangle,$ the subgraph induced in H by the image set f(V (G)) is the image of f. If f(V (G)) = V (H), f is an onto-homomorphism. If $f\, :\, G \rightarrow H,$ then for any vertex v of H, ${f}^{-1}(v)$ is an independent set of G. (If ${f}^{-1}(v) = {\o},$ then f ^{− 1}(v) is an independent subset of V (G), while if f ^{− 1}(v) contains an edge uw, then f(u) = v = f(w), and hence H has a loop at v, a contradiction to the fact that H is a simple graph).

Lemma 7.4.2.

Let ${G}_{1},\,{G}_{2}$ , and G ₃ be graphs and let ${f}_{1}\, :\, {G}_{1} \rightarrow {G}_{2},$ and ${f}_{2}\, :\, {G}_{2} \rightarrow {G}_{3}$ be homomorphisms. Then ${f}_{2} \circ {f}_{1}\, :\, {G}_{1} \rightarrow {G}_{3}$ is also a homomorphism. [Here $({f}_{2} \circ {f}_{1})(g) = {f}_{2}({f}_{1}(g))$ .]

Proof.

Follows by direct verification.

A graph homomorphism is a generalization of graph coloring. Suppose G ia a given graph and there exists a homomorphism $f\, :\, G \rightarrow {K}_{k},$ where k is the least positive integer with this property. Then f is onto and the sets ${S}_{i} =\{ {f}^{-1}({v}_{i})\, :\, {v}_{i} \in {K}_{k}\},$ $1 \leq i\, \leq \, k,$ form a partition of V (G). Moreover, between any two sets S _i and S _j , i≠j, there must be an edge of G. Otherwise, $A = {S}_{i} \cup {S}_{j}$ is an independent set of G, and we can define a homomorphism from G to K _{k − 1} by mapping A to the same vertex of K _{k − 1}. Thus, $\chi (G) = k.$ We state this result as a theorem.

Theorem 7.4.3.

Let G be a simple graph. Suppose there exists a homomorphism $f\, :\, G \rightarrow {K}_{k},$ and let k be the least positive integer with this property. Then $\chi (G) = k.$

Corollary 7.4.4.

If there exists a homomorphism $f\, :\, G \rightarrow {K}_{p},$ then $\chi (G)\, \leq \, p.$

Corollary 7.4.5.

Let $f\, :\, G \rightarrow H$ be a graph homomorphism. Then $\chi (G)\, \leq \, \chi (H).$

Proof.

Let $\chi (H)\, =\, k.$ Then there exists a homomorphism $g\, :\, H \rightarrow {K}_{k}.$ By Lemma 7.4.2, $g \circ f\, :\, G \rightarrow {K}_{k}$ is a homomorphism. Now apply Corollary 7.4.4.

Example 7.4.6.

Let ${V }_{1} =\{ {u}_{1},\ldots,{u}_{7}\}$ and ${V }_{2} =\{ {v}_{1},\ldots,{v}_{5}\}$ be the vertex sets of the cycles C ₇ and C ₅, respectively. Then the map $f({u}_{1}) = {v}_{1},$ $f({u}_{2}) = {v}_{2},$ $f({u}_{3}) = {v}_{3},$ $f({u}_{4}) = {v}_{2},$ $f({u}_{5}) = {v}_{3},$ $f({u}_{6}) = {v}_{4},$ and $f({u}_{7}) = {v}_{5}$ is a homomorphism of C ₇ to C ₅.

7.4.1 Quotient Graphs

Let $f\, :\, G \rightarrow H$ be a graph homomorphism from G onto H. Let $V (H) =\{ {v}_{1},\ldots,{v}_{k}\},$ and ${S}_{i} = {f}^{-1}({v}_{i}),$ $1 \leq i \leq k.$ Then no S _i is empty. The quotient graph G ∕ f is defined to be the graph with the sets S _i as its vertices and in which two vertices S _i and S _j are adjacent if ${v}_{i}{v}_{j} \in E(H).$ This defines a natural isomorphism $\tilde{f}\, :\, G/f\, \simeq \, H.$

A consequence of the above remarks is the fact that a complete k-coloring of G is just a homomorphism of G onto K _k . Recall that both the chromatic and achromatic colorings are complete colorings. We now establish the coloring interpolation theorem for the complete coloring.

Theorem 7.4.7 (Interpolation theorem for complete coloring).

If a graph G admits a complete k-coloring and a complete l-coloring, then it admits a complete i-coloring for every i between k and l.

Proof.

Let ${A}_{1},{A}_{2},\ldots,{A}_{k}$ and ${B}_{1},{B}_{2},\ldots,{B}_{l}$ be the color partitions in the two complete colorings. We assume without loss of generality that k < l. Clearly, it suffices to construct a complete (k + 1)-coloring of G. For each $i = 0,1,2,\ldots,l,$ let ${C}_{i} ={ \bigcup \nolimits }_{1\leq j\leq i}{B}_{j}.$ Let ${\Theta }_{i}$ denote the partition of V (G) by the nonempty sets of the sequence ${B}_{1},{B}_{2},\ldots,{B}_{i};\,{A}_{1} - {C}_{i},{A}_{2} - {C}_{i},\ldots,{A}_{k} - {C}_{i}.$ The partition Θ ₀ has parts ${A}_{1},{A}_{2},\ldots,{A}_{k};$ the partition Θ _l has parts ${B}_{1},{B}_{2},\ldots,{B}_{l}$ (since C _l = V (G), ${A}_{i} - {C}_{l} = {\o}$ for each j). Hence, $G/{\Theta }_{0}\, \simeq \, {K}_{k}$ and $G/{\Theta }_{l}\, \simeq \, {K}_{l}.$ Hence, there must exist a first suffix j, 0 < j ≤ l, such that $G/{\Theta }_{j}$ is not k-colorable. By the choice of j, this implies that $G/{\Theta }_{j}$ is (k + 1)-colorable, and hence by Lemma 7.4.2, G is (k + 1)-colorable. (Just compose the two onto homomorphisms $G \rightarrow G/{\Theta }_{j} \rightarrow {K}_{k+1}$.)

Exercise 3.22 shows that an interpolation theorem similar to that of complete coloring does not hold well for the b-coloring.

Exercise 4.26.

Let $f\, :\, G \rightarrow H$ be a graph homomorphism and let $x,y \in V (G).$ Prove ${d}_{H}(x,y)\, \leq \, {d}_{G}(x,y).$

Exercise 4.27.

Assume that there exists a homomorphism from G onto C _k , where k is odd. Show that G must contain an odd cycle. Show by means of an example that a similar statement need not hold good if k is even.

Exercise 4.28.

Prove that there exists a homomorphism from C _{2l + 1} to C _{2k + 1} if and only if l ≤ k.

7.5 Triangle-Free Graphs

Definition 7.5.1.

A graph G is triangle-free if G contains no K ₃.

Remark 7.5.2.

Triangle-free graphs cannot contain a K _k , k ≥ 3, either. It is obvious that if a graph G contains a clique of size k, then $\chi (G) \geq k.$ However, the converse is not true. That is, if the chromatic number of G is large, then G need not contain a clique of large size. The construction of triangle-free k-chromatic graphs, for k ≥ 3, was raised in the middle of the 20th century. In answer to this question, Mycielski [144] developed an interesting graph transformation known as the Mycielskian of a graph.

Definition 7.5.3.

Let G be a finite simple connected graph with vertex set V = V (G) and edge set E = E(G). The Mycielskian μ(G) of G is defined as follows: The vertex set V (μ(G)) of μ(G) is the disjoint union $V \cup V \prime\cup \{ u\},$ where $V \prime=\{ x\prime \, :\, x \in V \}$ and the edge set of μ(G) is $E(\mu (G)) = E \cup \{ x\prime y\, :\, xy \in E\} \cup \{ x\prime u\, :\, x\prime\in V \prime \}.$

We denote V (μ(G)) by the triad {V, V′, u}. For x ∈ V, we call x′ ∈ V′, the twin of x in μ(G), and vice versa, and u, the root of μ(G). Figure 7.5 displays the Mycielskian $\mu ({K}_{1,3} + e).$

Fig. 7.5

$\mu ({K}_{1,3} + e)$

Remark 7.5.4.

The following facts about μ(G), where G is of order n and size m, are obvious:

• (i) | V (μ(G)) | = 2n + 1.

• (ii) For each v ∈ V, ${d}_{\mu (G)}(v) = 2{d}_{G}(v).$

• (iii) For each v′ ∈ V′, ${d}_{\mu (G)}(v\prime ) = {d}_{G}(v) + 1.$

• (iv) ${d}_{\mu (G)}(u) = n.$

We now establish some basic results concerning the Mycielskian.

Theorem 7.5.5.

$\chi (\mu (G)) = \chi (G) + 1.$

Proof.

Assume that $\chi (G) = k.$ Consider a proper (vertex) k-coloring c of G using the colors, say, $1,2,\ldots,k.$ We now give a proper (k + 1)-coloring c′ for μ(G). For v ∈ V, set c′(v) = c(v). For the twin v′ ∈ V′, set c′(v′) = c(v). For the root u of μ(G), set c′(u) = k + 1. Then c′ is a proper coloring for $\mu (G)$ using k + 1 colors and therefore $\chi (\mu (G))\, \leq \, k + 1.$ [c′ is proper because for any edge xy′, $c\prime (x) = c(x)\neq c(y) = c\prime (y\prime ).$] We now show that it is actually k + 1.

Suppose $\mu (G)$ has a proper k-coloring c′ using the colors $1,2,\ldots,k.$ Assume, without loss of generality, that c′(u) = 1. Then for any $v\prime\in V \prime ,$ $c\prime \prime (v\prime )\neq 1.$ Recolor each vertex of V that has been colored by 1 in c′ by the color of its twin under c′. Then this gives a proper coloring of V using the k − 1 colors $2,3,\ldots,k.$ This is impossible as $\chi (G) = k.$ This proves that $\chi (\mu (G)) = k + 1 = \chi (G) + 1.$

Theorem 7.5.6.

If G triangle-free, then μ(G) is also triangle-free.

Proof.

Assume that G is triangle-free. If μ(G) contains a triangle, it can only be of the form vwz′, where v ∈ V, w ∈ V, and z′ ∈ V′, so that vz′ and wz′ are edges of μ(G). This means, by the definition of μ(G), that vz and wz are edges of G and hence vwz is a triangle in G, a contradiction.

Theorem 7.5.7 (Mycielski [144]).

For any positive integer p, there exists a triangle-free graph with chromatic number p.

Proof.

For p = 1, 2, the result is trivial. [For p = 1, take G = K ₁, and for p = 2, take G = K ₂. For p = 3, take $G = \mu ({K}_{2}).$ $\mu ({K}_{2}) = {C}_{5}$ is triangle-free and $\chi ({C}_{5}) = 3$.] For $p\, \geq \, 3,$ by Theorems 7.5.5 and 7.5.6, the iterated Mycielskian ${\mu }^{p-2}({K}_{2}) = \mu ({\mu }^{p-3}({K}_{2}))$ is triangle-free and has chromatic number p.

Remark 7.5.8.

The graph ${\mu }^{2}({K}_{2}) = \mu ({C}_{5})$ is the Grötzsch graph of Fig. 7.6.

Fig. 7.6

The Grötzsch graph, μ(C ₅)

Theorem 7.5.9.

If G is critical, then so is μ(G).

Proof.

Assume that G is k-critical. Since by Lemma 7.4.7, $\chi (\mu (G)) = k + 1,$ we have to show that μ(G) is (k + 1)-critical.

Start with a (k + 1)-coloring c with colors $1,2,\ldots,k + 1$ of μ(G) with vertex set {V, V′, u}.

We first show that $\chi (\mu (G) - u) = k.$ Without loss of generality, assume that c(u) = 1. Then 1 is not represented in V′. Let S be the set of vertices receiving the color 1 in V under c. Recolor each vertex v of S by the color of its twin v′ ∈ V′. This gives a proper coloring of $\mu (G) - u$ using k colors and hence $\chi (\mu (G) - u) = k.$ [Recall that adjacency of v and w in G implies adjacency of vw′ and v′w in μ(G).]

Next remove a vertex v′ of V′ from μ(G). Without loss of generality, assume that c(u) = 1 and c(v′) = 2. Now recolor the vertices of G − v by the k − 1 colors $3,\ldots,k,k + 1$ (this is possible as G is k-critical) and recolor the vertices of V′ − v′, if necessary, by the colors of their twins in V − v. Also, give color 1 to v. This coloring of μ(G) − v′ misses the color 2 and gives a proper k-coloring to μ(G) − v′.

Lastly, we give a k-coloring to μ(G) − v, v ∈ V. Color the vertices of G − v by $1,2,\ldots,k - 1$ so that the resulting coloring of G − v is proper. Let A be the subset of G − v whose vertices have received color 1 in this new coloring and $A\prime\subset V \prime$ denote the set of twins of the vertices in A. Now color the vertices of $(V \prime \setminus A\prime ) - v\prime$ by the colors of their twins in G, the vertices of $A\prime\cup \{ v\prime \}$ by color k, and u by color 1. This coloring is a proper coloring of $\mu (G) - v,$ which misses the color k + 1 in the list $\{1,2,\ldots,k + 1\}.$ Thus, μ(G) is (k + 1)-critical.

Remark 7.5.10.

Apply Theorem 7.5.12 to observe that for each k ≥ 1, there exists a k-critical triangle-free graph. Not every k-critical graph is triangle-free; for example, the complete graph K _k (k ≥ 3) is k − critical but is not triangle-free.

Lemma 7.5.11.

Let $f\, :\, G \rightarrow H$ be a graph isomorphism of G onto H. Then $f({N}_{G}(x)) = {N}_{H}(f(x)).$ Further $G - x\, \simeq \, H - f(x),$ and $G - {N}_{G}[<CitationRef CitationID="CR144">144</CitationRef>]\, \simeq \, H - {N}_{H}[f(x)]$ under the restriction maps of f to the respective domains.

Proof.

The proof follows from the definition of graph isomorphism.

Theorem 7.5.12 ([13]).

For connected graphs G and H $\mu (G)\, \simeq \, \mu (H)$ if and only if G ≃ H.

Proof.

If $G\, \simeq \, H,$ then trivially $\mu (G)\, \simeq \, \mu (H).$ So assume that G and H are connected and that $\mu (G)\, \simeq \, \mu (H).$ When n = 2 or 3, the result is trivial. So assume that $n\, \geq \, 4.$ If G is of order n, then μ(G) and μ(H) are both of order 2n + 1, and so H is also of order n. Let $f\, :\, \mu (G) \rightarrow \mu (H)$ be the given isomorphism, where V (μ(G)) and V (μ(H)) are given by the triads $({V }_{1},{V }_{1}\prime ,{u}_{1})$ and $({V }_{2},{V }_{2}\prime ,{u}_{2}),$ respectively.

We look at the possible images of the root u ₁ of μ(G) under f. Both u ₁ and u ₂ are vertices of degree n. If $f({u}_{1}) = {u}_{2},$ then by Lemma 7.5.11, $G = \mu (G) - N[{u}_{1}]\, \simeq \, \mu (H) - N[{u}_{2}] = H.$

Next we claim that $f({u}_{1})\notin {V }_{2}.$ Suppose $f({u}_{1}) \in {V }_{2}.$ Since ${d}_{\mu (H)}(f({u}_{1})) = {d}_{\mu (G)}({u}_{1}) = n,$ it follows from the definition of the Mycielskian that in μ(H), $\frac{n} {2}$ neighbors of f(u ₁) belong to V ₂ while another $\frac{n} {2}$ neighbors (the twins) belong to V ₂ ′. (This forces n to be even.) These n neighbors of f(u ₁) form an independent subset of μ(H). Then $H\prime= \mu (H) - {N}_{\mu (H)}[f({u}_{1})]\, \simeq \, \mu (G) - {N}_{\mu (G)}[{u}_{1}] = G.$ Now if x ∈ V ₂ is adjacent to f(u ₁) in μ(H), then x is adjacent to f(u ₁)′, the twin of f(u ₁) belonging to V ₂ ′ in μ(H). Further, ${d}_{H\prime }(f({u}_{1})\prime ) = 1 = {d}_{G}(v),$ where v ∈ V ₁ (the vertex set of G) corresponds to f(u ₁)′ in μ(H). But then ${d}_{\mu (G)}(v) = 2,$ while ${d}_{\mu (H)}(f(u)\prime ) = \frac{n} {2} + 1 > 2,$ as $n\, \geq \, 4.$ Hence, this case cannot arise.

Finally, suppose that $f({u}_{1}) \in {V }_{2}\prime .$ Set f(u ₁) = y′. Then y, the twin of y′ in μ(H), belongs to V ₂. As ${d}_{\mu (G)}({u}_{1}) = n,$ ${d}_{\mu (H)}(y\prime )\, =\, n.$ The vertex y′ has n − 1 neighbors in V ₂, say, ${x}_{1},{x}_{2},\ldots,{x}_{n-1}.$ Then ${N}_{H}(y) =\{ {x}_{1},{x}_{2},\ldots,{x}_{n-1}\},$ and hence y is also adjacent to ${x\prime }_{1},{x\prime }_{2},\ldots,{x\prime }_{n-1}$ in V ₂ ′. Further, as ${N}_{\mu (G)}({u}_{1})$ is independent, N _μ(H)(y′) is also independent. Therefore, H = star ${K}_{1,n-1}$ consisting of the edges $y{x}_{1},y{x}_{2},\ldots,y{x}_{n-1}.$ Moreover, $G = \mu (G) - N[{u}_{1}]\, \simeq \, \mu (H) - N[y\prime ]\, =$ star K _{1, n − 1} consisting of the edges $y{x\prime }_{1},y{x\prime }_{2},\ldots,y{x\prime }_{n-1}.$ Thus, $G\, \simeq \, {K}_{1,n-1}\, \simeq \, H.$

7.6 Edge Colorings of Graphs

7.6.1 The Timetable Problem

Suppose in a school there are r teachers, ${T}_{1},{T}_{2},\ldots,{T}_{r},$ and s classes, ${C}_{1},{C}_{2},\ldots,{C}_{s}.$ Each teacher T _i is expected to teach the class C _j for p _ij periods. It is clear that during any particular period, no more than one teacher can handle a particular class and no more than one class can be engaged by any teacher. Our aim is to draw up a timetable for the day that requires only the minimum number of periods. This problem is known as the “timetable problem.”

To convert this problem into a graph-theoretic one, we form the bipartite graph G = G(T, C) with bipartition (T, C), where T represents the set of teachers T _i and C represents the set of classes C _j . Further, T _i is made adjacent to C _j in G with p _ij parallel edges if and only if teacher T _i is to handle class C _j for p _ij periods. Now color the edges of G so that no two adjacent edges receive the same color. Then the edges in a particular color class, that is, the edges in that color, form a matching in G and correspond to a schedule of work for a particular period. Hence, the minimum number of periods required is the minimum number of colors in an edge coloring of G in which adjacent edges receive distinct colors; in other words, it is the edge-chromatic number of G. We now present these notions as formal definitions.

Definition 7.6.1.

An edge coloring of a loopless graph G is a function $\pi \, :\, E(G) \rightarrow S,$ where S is a set of distinct colors; it is proper if no two adjacent edges receive the same color. Thus, a proper edge coloring π of G is a function $\pi \, :\, E(G) \rightarrow S$ such that $\pi (e)\neq \pi ({e}^{{\prime }})$ whenever edges e and ${e}^{{\prime }}$ are adjacent in G, and it is a proper k-edge coloring of G if | S | = k.

Definition 7.6.2.

The minimum k for which a loopless graph G has a proper k-edge coloring is called the edge-chromatic number or chromatic index of G. It is denoted by ${\chi }^{{\prime }}(G).$ G is k-edge-chromatic if ${\chi }^{{\prime }}(G) = k.$

Further, if an edge uv is colored by color c, we say that c is represented at both u and v. If G has a proper k-edge coloring, E(G) is partitioned into k edge-disjoint matchings.

It is clear that for any (loopless) graph G, ${\chi }^{{\prime }}(G) \geq \Delta (G)$ since the $\Delta (G)$ edges incident at a vertex v of maximum degree $\Delta (G)$ must all receive distinct colors. For bipartite graphs, however, equality holds.

Theorem 7.6.3 (K$\ddot{\mbox{ o}}$nig).

If G is a bipartite graph ${\chi }^{{\prime }}(G) = \Delta (G).$

Proof.

The proof is by induction on the size (i.e., number of edges) m of G. The result is true for m = 1. Assume the result for bipartite graphs of size at most m − 1. Let G have m edges. Let $e = uv \in E(G).$ Then G − e has [since $\Delta (G - e) \leq \Delta (G)$] a proper $\Delta$-edge coloring, say c. Out of these $\Delta$ colors, suppose that one particular color is not represented at both u and v. Then in this coloring the edge uv can be colored with this color, and a proper $\Delta$-edge coloring of G is obtained.

In the other case (that is, in the case in which each of the $\Delta$ colors is represented either at u or at v in G − e), since the degrees of u and v in G − e are at most $\Delta - 1,$ there exists a color out of the $\Delta$ colors that is not represented in G − e at u, and similarly there exists a color not represented at v. Thus, if color j is not represented at u in c, then j is represented at v in c, and if color i is not represented at v in c, then i is represented at u in c. Since G is bipartite and u and v are not in the same parts of the bipartition, there can exist no u-v path in G in which the colors alternate between i and j.

Let P be a maximal path in G − e starting from u in which the colors of the edges alternate between i and j. Interchange the colors i and j in P. This would still yield a proper edge coloring of G − e using the $\Delta$ colors in which color i is not represented at both u and v. Now color the edge uv by the color i. This results in a proper $\Delta$-edge coloring of G.

Exercise 6.1.

Disprove the converse of Theorem 7.6.3 by a counterexample.

Next, we determine the chromatic index of the complete graphs.

Theorem 7.6.4.

${\chi }^{{\prime }}({K}_{n}) = \left \{\begin{array}{@{}l@{\quad }l@{}} n - 1\ \ \ \mathrm{if}\ n\ is\ even,\quad \\ n\ \ \ \ \ \ \ \ \mathrm{if } \ n\ is odd.\quad \end{array} \right.$

Proof.

(Berge) Since K _n is regular of degree n − 1, ${\chi }^{{\prime }}({K}_{n}) \geq n - 1.$

Case 1.

n is even. We show that ${\chi }^{{\prime }}({K}_{n}) \leq n - 1$ by exhibiting a proper (n − 1)-edge coloring of K _n . Label the n vertices of K _n as $0,1,\ldots,n - 1.$ Draw a circle with center at 0 and place the remaining n − 1 numbers on the circumference of the circle so that they form a regular (n − 1)-gon (Fig. 7.7). Then the $\frac{n} {2}$ edges (0, 1), (2, n − 1), (3, n − 2), $\ldots,$ $(\frac{n} {2}, \frac{n} {2} + 1)$ form a 1-factor of K _n . These $\frac{n} {2}$ edges are the thick edges of Fig.7.7. Rotation of these edges through the angle $\frac{2\pi } {n-1}$ in succession gives (n − 1) edge-disjoint 1-factors of K _n . This would account for $\frac{n} {2} (n - 1)$ edges and hence all the edges of K _n . (Actually, the above construction displays a 1-factorization of K _n when n is even.) Each 1-factor can be assigned a distinct color. Thus, ${\chi }^{{\prime }}({K}_{n}) \leq n - 1.$ This proves the result in Case 1.

Fig. 7.7

Graph for proof of Theorem 7.6.4

Case 2.

n is odd. Take a new vertex and make it adjacent to all the n vertices of K _n . This gives K _{n + 1}. By Case 1, ${\chi }^{{\prime }}({K}_{n+1}) = n.$ The restriction of this edge coloring to K _n yields a proper n-edge coloring of K _n . Hence, ${\chi }^{{\prime }}({K}_{n}) \leq n.$ However, K _n cannot be edge colored properly with n − 1 colors. This is because the size of any matching of K _n can contain no more than $\frac{n-1} {2}$ edges, and hence n − 1 matchings of K _n can contain no more than $\frac{{(n-1)}^{2}} {2}$ edges. But K _n has $\frac{n(n-1)} {2}$ edges. Thus, ${\chi }^{{\prime }}({K}_{n}) \geq n,$ and hence ${\chi }^{{\prime }}({K}_{n}) = n.$

Exercise 6.2.

Show that a Hamiltonian cubic graph is 3-edge-chromatic.

Exercise 6.3.

Show that the Petersen graph is 4-edge-chromatic.

Exercise 6.4.

Show that the Herschel graph (see Fig. 5.4) is 4-edge-chromatic.

Exercise 6.5.

Determine the edge-chromatic number of the Grötzsch graph (Fig. 7.6).

Exercise 6.6.

Show that a simple cubic graph with a cut edge is 4-edge-chromatic.

Exercise 6.7.

Describe a proper k-edge coloring of a k-regular bipartite graph.

Exercise 6.8.

Show that any bipartite graph G of maximum degree $\Delta$ is a subgraph of a $\Delta$-regular bipartite graph. Hence, furnish an alternative proof of Theorem 7.6.3, using Exercise 6.7.

7.6.2 Vizing’s Theorem

Although it is true that for any loopless graph G, ${\chi }^{{\prime }}(G) \geq \Delta (G),$ it turns out that for any simple graph G, ${\chi }^{{\prime }}(G) \leq 1 + \Delta (G).$ This major result in edge coloring of graphs was established by Vizing [183] and independently by Gupta [81].

Theorem 7.6.5 (Vizing-Gupta).

For any simple graph G $\Delta (G) \leq {\chi }^{{\prime }}(G) \leq 1 + \Delta (G).$

Proof.

In a proper edge coloring of G, $\Delta (G),$ colors are to be used for the edges incident at a vertex of maximum degree in G. Hence, ${\chi }^{{\prime }}(G) \geq \Delta (G).$

We now prove that ${\chi }^{{\prime }}(G) \leq 1 + \Delta,$ where $\Delta = \Delta (G).$

If G is not $(1 + \Delta )$-edge-colorable, choose a subgraph H of G with a maximum possible number of edges such that H is (1 + Δ)-edge-colorable. We derive a contradiction by showing that there exists a subgraph H ₀ of G that is $(1 + \Delta )$-edge-colorable and has one edge more than H.

By our assumption, G has an edge $u{v}_{1}\notin E(H).$ Since $d(u) \leq \Delta,$ and $1 + \Delta$ colors are being used in H, three is a color c that is not represented at u (i.e., not used for any edge of H incident at u). For the same reason, there is a color c ₁ not represented at v ₁. (See Fig. 7.8, where the color not represented at a particular vertex is enclosed in a circle and marked near the vertex.)

Fig. 7.8

Graph for proof of Theorem 7.6.5

There must be an edge, say uv ₂ of H, colored c ₁; otherwise, uv ₁ can be assigned the color c ₁, and $H\, \cup \, (u{v}_{1}),$ which has one edge more than H, would have a proper $(1 + \Delta )$-edge coloring. Again, there is a color, say c ₂, not represented at v ₂. Then as above, there is an edge uv ₃ colored c ₂ and there is a color, say c ₃, not represented at v ₃.

In this way, we construct a sequence of edges $\{u{v}_{1},u{v}_{2},\ldots,u{v}_{k}\}$ such that color c _i is not represented at vertex v _i , $1 \leq i \leq k,$ and the edge uv _{j + 1} receives the color c _j , $1 \leq j \leq k - 1$ (see Fig. 7.8).

Suppose at some stage, say the rth stage, where $1 \leq r \leq k,$ c (the missing color at u) is not represented at v _r . We then “cascade” (i.e., shift in order) the colors ${c}_{1},\ldots,{c}_{r-1}$ from $u{v}_{2},u{v}_{3},\ldots,u{v}_{r}$ to $u{v}_{1},u{v}_{2},\ldots,u{v}_{r-1}.$ Under this new coloring, c is not represented both at u and at v _r , and therefore we can color uv _r with c. This yields a proper $(1 + \Delta )$-edge coloring to $H\, \cup \, (u{v}_{1}),$ contradicting the choice of H. Hence, we may assume that c is represented at each of the vertices ${v}_{1},{v}_{2},\ldots,{v}_{k}.$

Now we need to know why the sequence of edges uv _i , $1 \leq i \leq k,$ had stopped. There are two possible reasons. Either there is no edge incident to u that is colored c _k , or the color ${c}_{k} = {c}_{j}$ for some j < k − 1 and so has already been represented at u. Note that the sequence must stop at some finite stage since d(u) is finite; however, it may as well stop before all the edges incident to u are exhausted.

If c _k is not represented at u in H, then we can cascade as before so that uv _i gets color c _i , $1 \leq i \leq k - 1,$ and then color uv _k with color c _k . Once again, we have a contradiction to our assumption on H.

Thus, we must have ${c}_{k} = {c}_{j}$ for come j < k − 1. In this case, cascade the colors ${c}_{1},{c}_{2},\ldots,{c}_{j}$ so that uv _i has color c _i , $1 \leq i \leq j,$ and leave uv _{j + 1} uncolored (Fig. 7.9). Let $S = (H\, \cup \, (u{v}_{1})) - u{v}_{j+1}.$ Then S and H have the same number of edges.

Fig. 7.9

Another graph for proof of Theorem 7.6.5

Now consider ${S}_{c{c}_{j}},$ the subgraph of S defined by the edges of S with colors c and c _j . Clearly, each component of ${S}_{c{c}_{j}}$ is either an even cycle or a path in which the adjacent edges alternate with colors c and c _j .

Now, c is represented at each of the vertices ${v}_{1},{v}_{2},\ldots,{v}_{k},$ and in particular at v _{j + 1} and v _k . But c _j is not represented at v _{j + 1} and v _k , since we have just moved c _j to uv _j , and ${c}_{j} = {c}_{k}$ is not represented at v _k . Hence, in ${S}_{c{c}_{j}},$ the degrees of v _{j + 1} and v _k are both equal to 1. Moreover, c _j is represented at u, but c is not. Therefore, u also has degree 1 in ${S}_{c{c}_{j}}.$ As each component of ${S}_{c{c}_{j}}$ is either a path or an even cycle, not all of u, ${v}_{j+1},$ and v _k can be in the same component of S _{cc j} (since a nontrivial path has only two vertices of degree 1).

If u and v _{j + 1} are in different components of ${S}_{c{c}_{j}},$ interchange the colors c and c _{j + 1} in the component containing v _{j + 1}. Then c is not represented at both u and v _{j + 1}, and so we can color the edge uv _{j + 1} with c. This gives a $(1 + \Delta )$-edge coloring to the graph $S\, \cup \, (u{v}_{j+1}).$

Suppose then u and v _{j + 1} are in the same components of ${S}_{c{c}_{j}}.$ Then, necessarily, v _k is not in this component. Interchange c and c _j in the component containing v _k . In this case, further cascade the colors so that uv _i has color c _i , $1 \leq i \leq k - 1.$ Now color uv _k with color c.

Thus, we have extended our edge coloring of S with $1 + \Delta$ colors to one more edge of G. This contradiction proves that H = G, and thus ${\chi }^{{\prime }}(G) \leq 1 + \Delta.$

Actually, Vizing proved a more general result than the one given above. Let G be any loopless graph and let μ denote the maximum number of edges joining two vertices in G. Then the generalized Vizing’s theorem states that $\Delta \leq {\chi }^{{\prime }}\leq \Delta + \mu.$ This theorem is the best possible in that there are graphs with ${\chi }^{{\prime }} = \Delta + \mu.$ For example, let G be the graph of Fig. 7.10. Since any two edges of G are adjacent, ${\chi }^{{\prime }} = m(G) = 3\mu = \Delta + \mu.$ For a proof of the generalized Vizing’s theorem, see Yap [194].

Fig. 7.10

Graph illustrating the generalized Vizing’s theorem

Definition 7.6.6.

Graphs for which ${\chi }^{{\prime }} = \Delta$ are called Class 1 graphs and those for which ${\chi }^{{\prime }} = 1 + \Delta$ are called Class 2 graphs.

Example 7.6.7.

Bipartite graphs are of class 1 (see Theorem 7.6.3), whereas the Petersen graph (see Exercise 6.3) and any simple cubic graph with a cut edge (see Exercise 6.6) are of class 2.

For details relating to graphs of class 1 and class 2, see [62, 194].

Exercise 6.9.

Let G be a simple Δ-edge-chromatic critical graph [i.e., G is of class 1 and for every edge e of G, ${\chi }^{{\prime }}(G - e) < {\chi }^{{\prime }}(G)$]. Prove that if uv ∈ E(G), then $d(u) + d(v) \geq \Delta + 2.$

We now return to the timetable problem. Following are some examples of such a problem.

Problem 1.

In a social health checkup scheme, specialist physicians are to visit various health centers. Given the places each physician has to visit and also the time interval of his or her visit, how can we fit in an itinerary? The assumption is that each health center can accommodate only one doctor at a time.

Problem 2.

Mobile laboratories are to visit various schools in a city. Given the places each lab has to visit and also the time interval (period) of visits in a day, how can we fit in a timetable for the laboratories?

Problem 3.

In an educational institution, as is well known, teachers have to instruct various classes. Given the various classes each teacher has to instruct in a day, how can we fit in a timetable? It is presumed that a teacher can teach only one class at a time and that each class could be taught by only one teacher at a time!

We shall now discuss Problem 3. Let ${x}_{1},{x}_{2},\ldots,{x}_{n}$ denote the teachers and ${y}_{1},{y}_{2},\ldots,{y}_{m}$ the classes. Let t _ij denote the number of periods for which teacher x _i has to meet class y _j . How can we draw up a timetable? If there are constraints on the availability of classrooms, what is the minimum number of periods required to implement a timetable? If the number of periods in a day is specified, what is the minimum number of rooms required to implement the timetable? All these problems could be analyzed by using a suitable graph.

Let G(T, C) be a bipartite graph formed with $T =\{ {x}_{1},{x}_{2},\ldots,{x}_{p}\}$ and $C =\{ {y}_{1},{y}_{2},\ldots,{y}_{q}\}$ as the bipartition and in which there are t _ij parallel edges with x _i and y _j as their common ends. If T denotes the set of teachers and C the set of classrooms, a teaching assignment for a period determines a matching in the bipartite graph G. Conversely, any matching in G corresponds to a teaching assignment for one period. The edges of G could be partitioned into $\Delta$ edge-disjoint matchings (see Theorem 7.6.3). Corresponding to the Δ matchings, a Δ-period timetable can be drawn up.

Let N be the total number of periods to be taught by all teachers put together. Then, on average, N ∕ Δ classes are to be taught per period. Hence, at least $\left \lceil N/\Delta \right \rceil$ rooms are necessary to implement a Δ-period timetable. We present below a method for drawing up such a timetable. For this, we need Lemma 7.6.8.

Lemma 7.6.8.

Let M and N be disjoint matchings of a graph G with |M| > |N|. Then there are disjoint matchings M′ and N′ of G with |M′| = |M|− 1 and |N′| = |N| + 1 and with $M\prime\cup N\prime= M \cup N.$

Proof.

Consider the subgraph $H = G[M \cup N].$ Each component of H is either an even cycle or a path with edges alternating between M and N. Since $\vert M\vert \, >\, \vert N\vert,$ some path component P of H must have its initial and terminal edges in M. Let $P = {v}_{0}{e}_{1}{v}_{1}{e}_{2}{v}_{2}\ldots {e}_{2r+1}{v}_{2r+1}.$

Now set

$$M\prime= (M\setminus \{{e}_{1},{e}_{3},\ldots,{e}_{2r+1}\})\, \cup \,\{ {e}_{2},{e}_{4},\ldots,{e}_{2r}\}$$

and

$$N\prime= (N\setminus \{{e}_{2},{e}_{4},\ldots,{e}_{2r}\})\, \cup \,\{ {e}_{1},{e}_{3},\ldots,{e}_{2r+1}\}.$$

Then M′ and N′ are disjoint matchings of G satisfying the conditions of the lemma.

Theorem 7.6.9.

If G is a bipartite graph (with m edges), and if $m\, \geq \, t\, \geq \, \Delta,$ then there exist t disjoint matchings ${M}_{1},{M}_{2},\ldots,{M}_{t}$ of G such that

$$E = {M}_{1} \cup {M}_{2} \cup \ldots \cup {M}_{t}$$

and, for $1 \leq i\, \leq \, t,$

$$\left \lfloor m/t\right \rfloor \,\leq \,\vert {M}_{i}\vert \leq \left \lceil m/t\right \rceil.$$

(In other words, any connected bipartite graph G is equitably t-edge-colorable, where $m\, \geq \, t\, \geq \, \Delta.$)

Proof.

By Theorem 7.6.3, $\chi \prime= \Delta.$ Hence, E(G) can be partitioned into Δ matchings ${M\prime }_{1},{M\prime }_{2},\ldots,{M\prime }_{\Delta }.$ So for $t\, \geq \, \Delta,$ there exist disjoint matchings ${M\prime }_{1},{M\prime }_{2},\ldots,{M\prime }_{t},$ where M′ _i = ∅ for $\Delta + 1 \leq i\, \leq \, t,$ and

$$E = {M\prime }_{1} \cup {M\prime }_{2} \cup \ldots {M\prime }_{t}.$$

Now repeatedly apply Lemma 7.6.8 to pairs of matchings that differ by more than one in size. This would eventually result in matchings ${M}_{1},{M}_{2},\ldots,{M}_{t}$ of G satisfying the condition stated in the theorem.

Coming back to our timetable problem, if the number of rooms available, say r, is less than $N/\Delta$ (so that $N/r > \Delta ),$ then the number of periods is to be correspondingly increased. Hence, starting with an edge partition of E(G) into matchings ${M\prime }_{1},{M\prime }_{2},\ldots,{M\prime }_{\Delta },$ we apply Lemma 7.6.8 repeatedly to get an edge partition of E(G) into disjoint matchings ${M}_{1},{M}_{2},\ldots,{M}_{\left \lceil N/r\right \rceil }.$ This partition gives a $\left \lceil N/r\right \rceil$-period timetable that uses r rooms.

Illustration.

The teaching assignments of five professors, x ₁, x ₂, x ₃, x ₄, x ₅, in the mathematics department of a particular university are given by the following array:

The bipartite graph G corresponding to the above problem is shown in Fig. 7.11. Each of the sets of edges drawn by the ordinary lines, dashed lines, and thick lines gives a matching in G. The three matchings cover the edges of G. Hence, they can be the basis of a three-period timetable. The corresponding timetable is given in Table 7.1.

Fig. 7.11

Bipartite graph corresponding to Problem 1

Table 2

Table 7.1 Timetable

In each period, four classes are to be met. Hence, at least four rooms are needed to implement this timetable. Here Δ = 3 and N = 12. Consequently, G could be covered by three matchings each containing $\left \lfloor 12/3\right \rceil$ or $\left \lceil 12/3\right \rceil$ edges, that is, exactly four edges. This gives the edge partition

$$M\prime=\{ {M\prime }_{1},{M\prime }_{2},{M\prime }_{3}\},$$

where

$$\begin{array}{rlrlrl} {M\prime }_{1} & =\{ {x}_{1}{y}_{1},{x}_{2}{y}_{2},{x}_{3}{y}_{4},{x}_{5}{y}_{3}\}, & & \\ {M\prime }_{2} & =\{ {x}_{1}{y}_{2},{x}_{2}{y}_{3},{x}_{3}{y}_{1},{x}_{5}{y}_{4}\}, & & \end{array}$$

and

$${M\prime }_{3} =\{ {x}_{1}{y}_{2},{x}_{2}{y}_{1},{x}_{3}{y}_{4}.{x}_{4}{y}_{3}\}.$$

Now, take $M\prime \prime=\{ {M\prime }_{1},{M\prime }_{2},{M\prime }_{3},{M\prime }_{4} = {\o} \},$ and apply Lemma 7.6.8. This gives an edge partition $M =\{ {M}_{1},{M}_{2},{M}_{3},{M}_{4}\},$ where ${M}_{1} =\{ {x}_{1}{y}_{1},{x}_{2}{y}_{2},{x}_{3}{y}_{4}\},$ ${M}_{2} =\{ {x}_{1}{y}_{2},{x}_{2}{y}_{3},{x}_{5}{y}_{4}\},$ ${M}_{3} =\{ {x}_{2}{y}_{1},{x}_{3}{y}_{4},{x}_{4}{y}_{3}\},$ and ${M}_{4} =\{ {x}_{5}{y}_{3},{x}_{3}{y}_{1},{x}_{1}{y}_{2}\}.$ The above partition yields a four-period timetable using three rooms.

7.7 Snarks

A consequence of the Vizing–Gupta theorem is that if G is a simple cubic graph ${\chi }^{{\prime }}(G) = 3$ or 4. By Exercise 6.6, if G is a simple cubic graph with a cut edge ${\chi }^{{\prime }}(G) = 4.$ So the natural question is: Are there 2-edge-connected, simple cubic graphs that are 4-edge-chromatic? Such graphs are important in their own right, since their existence is related to the four-color problem (see Chap. 8). The search for such graphs has led to the study of snarks.

Definition 7.7.1.

A snark is a cyclically 4-edge-connected cubic graph of girth at least 5 that has chromatic index 4.

Exercise 7.1.

Prove that no snark can be Hamiltonian.

Clearly, the Petersen graph is a snark. In fact, Theorem 7.7.2 is an interesting result.

Theorem 7.7.2.

The Petersen graph P is the smallest snark and it is the unique snark on 10 vertices.

Proof.

Let G be a snark and A a cyclical edge cut of G. Then G − A has two components, each having a cycle of length at least 5 (since G is of girth at least 5). Hence, $\left \vert V (G)\right \vert \geq 10.$ Thus, P is a smallest snark since $\left \vert V (P)\right \vert = 10.$

We now show that any snark G on 10 vertices must be isomorphic to P. Let A be a cyclical edge cut of G. If $\left \vert A\right \vert = 4,$ then each component of G − A is a 5-cycle. But this will not account for all the 15 edges of G. If $\left \vert A\right \vert > 5,$ then $\left \vert E(G)\right \vert > 5 + 5 + 5 = 15,$ a contradiction. Hence, $\left \vert A\right \vert = 5,$ and let $A =\{ {u}_{i}{v}_{i}\, :\, 1 \leq i \leq 5\}.$ Then G − A consists of two 5-cycles. Let one of these cycles be $\{{u}_{1},{u}_{2},{u}_{3},{u}_{4},{u}_{5}\}.$ Let v _i be the third neighbor of u _i not belonging to the set $\{{u}_{1},{u}_{2},{u}_{3},{u}_{4},{u}_{5}\}$ for each i. If v ₁ v ₂ or v ₁ v ₅ is an edge of G, then G contains a 4-cycle (see Fig. 7.12).

Fig. 7.12

Graph for proof of Theorem 7.7.2

Since G is cubic, ${v}_{1}{v}_{3} \in E(G)$ and ${v}_{1}{v}_{4} \in E(G).$ Similarly, v ₂ v ₄, v ₂ v ₅, and v ₃ v ₅ are edges of G and hence $G\, \simeq \, P.$

The construction of snarks is not easy. In 1975, Isaacs constructed two infinite classes of snarks. Prior to that, only four kinds of snarks were known: (1) the Petersen graph on 10 vertices, (2) Blanus$\check{\mbox{ a}}$’s graphs on 18 vertices, (3) Szekeres’ graph on 50 vertices, and (4) Blanche Descartes’ graph on 210 vertices.

7.8 Kirkman’s Schoolgirl Problem

Kirkman’s schoolgirl problem was introduced in 1850 by Reverend Thomas J. Kirkman as “query 6” in page 48 of the Ladies and Gentlemen’s Diary. The problem is the following: A teacher would like to take 15 schoolgirls out for a walk, the girls being arranged in five rows of three. The teacher would like to ensure equal chances of friendship between any two girls. Hence, it is desirable to find different row arrangements for the seven days of the week such that any pair of girls walk in the same row on exactly one day of the week.

Kirkman’s 15-schoolgirl problem has a solution. In fact, one of the possible schedules is given in Fig. 7.13.

Fig. 7.13

Kirkman’s schoolgirl problem

In the general case, one wants to arrange 6n + 3 girls in 2n + 1 rows of three. The problem is to find different row arrangements for 3n + 1 different days in such a way that any pair of girls walks in the same row on exactly one day out of the 3n + 1 days. The existence of such an arrangement was proved by Ray-Chaudhuri and R. M. Wilson [164]. In graph-theoretic terminology, Kirkman’s schoolgirl problem corresponds to an edge coloring $\mathcal{C}\, :\, E({K}_{6n+3}) \rightarrow \{ {c}_{1},\ldots,{c}_{3n+1}\}$ of the complete graph G = K _{6n + 3} with 3n + 1 colors such that if E _i denotes the set of all edges receiving the color c _i and ${G}_{i} = G[{E}_{i}],$ then G _i has 2n + 1 components, each component being a triangle.

The general problem can be tackled as follows: Consider the triangle graphT of ${K}_{6n+3}$ defined as follows: The vertex set of T is the set of all triads of $V ({K}_{6n+3}),$ and two distinct vertices of T are joined by an edge in T if and only if the corresponding triads have two elements in common. Let S be any independent set of T. Each vertex of S gives rise to three pairs of vertices of K _{6n + 3}, and each such pair belongs to at most one vertex of S. Hence, we have $3\,\left \vert S\right \vert \leq \begin{pmatrix}{6n + 3}\\ {2}\end{pmatrix},$ that is, $\left \vert S\right \vert \leq (2n + 1)(3n + 1).$ We must then find an independent set ${S}^{{\prime }}$ of cardinality $\left \vert {S}^{{\prime }}\right \vert = (2n + 1)(3n + 1).$ Such a set exists since every solution of the Kirkman’s schoolgirl problem yields an independent set of T with (2n + 1)(3n + 1) vertices. We observe that ${S}^{{\prime }}$ covers each pair of V (K _{6n + 3}) exactly once. Having found a maximum independent set S ^′ in T, we form a new graph T ^′ as follows: We take S ^′ as its vertex set and join two vertices of T ^′ by an edge if and only if the corresponding triads have exactly one vertex in common. We note that each independent set of T ^′ is a partition of a subset of V (K _{6n + 3}) into subsets of cardinality 3, and hence each independent set of T ^′ has at most (2n + 1) vertices. If the chromatic number of T ^′ is 3n + 1, then there is a partition $({V }_{1},{V }_{2},\ldots,{V }_{3n+1})$ of $V ({T}^{{\prime }})$ into parts of size 2n + 1 each. This partition is a solution to the Kirkman’s schoolgirl problem, and conversely, each solution to the Kirkman’s schoolgirl problem yields such a partition.

Exercise 8.1.

Let $m\, \geq \, n + 2$ and let there exist edge partitions $\mathcal{F}$ and $\mathcal{G}$ of K _n and K _m , respectively, into triangles with $\mathcal{F}\subset \mathcal{G}.$ Prove that $m \geq 2n + 1.$

7.9 Chromatic Polynomials

In 1946, Birkhoff and Lewis [23] introduced the chromatic polynomial of a graph in their attempt to tackle the four-color problem (see Chap. 8) through algebraic techniques.

For a graph G and a given set of $\lambda$ colors, the function f(G; λ) is defined to be the number of ways of (vertex) coloring G properly using the λ colors. Hence, f(G; λ) = 0 when G has no proper λ-coloring. Clearly, the minimum λ for which f(G; λ) > 0 is the chromatic number χ(G) of G.

It is easy to see that $f({K}_{n};\lambda ) = \lambda (\lambda - 1)\ldots (\lambda - n + 1)$ for $\lambda \geq n.$ This is because any vertex of K _n can be colored by any one of the given λ colors. After coloring a vertex of K _n , a second vertex of K _n can be colored by any one of the remaining (λ − 1) colors, and so on. In particular, $f({K}_{3};\lambda ) = \lambda (\lambda - 1)(\lambda - 2).$ Also, $f({K}_{n}^{c};\lambda ) = {\lambda }^{n}.$

Let e = uv be any edge of G. Recall (see Sect. 4.3, Chap. 4) that the graph $G \circ e$ is obtained from G by contracting the edge e. Theorem 7.9.1 presents a simple reduction formula to compute $f(G;\lambda ).$

Theorem 7.9.1.

Let G be any graph. Then $f(G;\lambda ) = f(G - e;\lambda ) - f(G \circ e;\lambda )$ for any edge e of G.

Proof.

$f(G - e;\lambda )$ denotes the number of proper colorings of G − e using $\lambda$ colors. Hence, it is the sum of the number of proper colorings of G − e in which u and v receive the same color and the number of proper colorings of G − e in which u and v receive distinct colors. The former number is $f(G \circ e;\lambda ),$ and the latter number is $f(G;\lambda ).$

Exercise 9.1.

If G and H are disjoint graphs, show that

$$f(G\, \cup \, H;\lambda ) = f(G;\lambda )f(H;\lambda ).$$

Theorem 7.9.1 could be used recursively to determine $f(G;\lambda )$ for graphs of small size by taking the given graph on n vertices as G and successively deleting edges until we end up with the totally disconnected graph ${K}_{n}^{c}.$ It can also be determined by taking the given graph as G − e and recursively adding a new edge e until we end up with the complete graph K _n . For a fixed n, when m(G), the number of edges of G is small, the first method is preferable, and when it is large, the second method is preferable. These two methods are illustrated for the graph C ₄. [Here the function $f(G;\lambda )$ is represented by the graph itself.]

Method 1

Method 2

The function $f({C}_{4};\lambda )$ computed above is a monic polynomial with integer coefficients of degree n = 4 in which the coefficient of ${\lambda }^{3} = -4 = -m,$ the constant term is zero, and the coefficients alternate in sign. That this is the case with all such functions $f(G;\lambda )$ is the content of Theorem 7.9.2. For this reason, the function $f(G;\lambda )$ is called the chromatic polynomial of the graph G.

Theorem 7.9.2.

For a simple graph G of order n and size m, f(G;λ) is a monic polynomial of degree n in λ with integer coefficients and constant term zero. In addition, its coefficients alternate in sign and the coefficient of ${\lambda }^{n-1}$ is − m.

Proof.

The proof is by induction on m. If m = 0, G is ${K}_{n}^{c}$ and $f({K}_{n}^{c};\lambda ) = {\lambda }^{n},$ and if m = 1, G is K ₂ and $f({K}_{2};\lambda ) = {\lambda }^{2} - \lambda,$ and the statement of the theorem is trivially true in these cases. Suppose now that the theorem holds for all graphs with fewer than m edges, where m ≥ 2. Let G be any simple graph of order n and size m, and let e be any edge of G. Both G − e and $G \circ e$ (after removal of multiple edges, if necessary) are simple graphs with at most m − 1 edges, and hence, by the induction hypothesis,

$$f(G - e;\lambda ) = {\lambda }^{n} - {a}_{ 0}{\lambda }^{n-1} + {a}_{ 1}{\lambda }^{n-2} -\ldots + {(-1)}^{n-1}{a}_{ n-2}\lambda,$$

and

$$f(G \circ e;\lambda ) = {\lambda }^{n-1} - {b}_{ 1}{\lambda }^{n-2} + \ldots + {(-1)}^{n-2}{b}_{ n-2}\lambda,$$

where ${a}_{0},\ldots,{a}_{n-2};$ ${b}_{1},\ldots,{b}_{n-2}$ are nonnegative integers (so that the coefficients alternate in sign), and a ₀ is the number of edges in G − e, which is m − 1. By Theorem 7.9.2, $f(G;\lambda ) = f(G - e;\lambda ) - f(G \circ e;\lambda ),$ and hence

$$f(G;\lambda ) = {\lambda }^{n} - ({a}_{ 0} + 1){\lambda }^{n-1} + ({a}_{ 1} + {b}_{1}){\lambda }^{n-2} -\ldots + {(-1)}^{n-1}({a}_{ n-2} + {b}_{n-2})\lambda.$$

Since a ₀ + 1 = m, $f(G;\lambda )$ has all the stated properties.

Theorem 7.9.3.

A simple graph G on n vertices is a tree if and only if $f(G;\lambda ) = \lambda {(\lambda - 1)}^{n-1}.$

Proof.

Let G be a tree. We prove that $f(G;\lambda ) = \lambda {(\lambda - 1)}^{n-1}$ by induction on n. If n = 1, the result is trivial. So assume the result for trees with at most n − 1 vertices, n ≥ 2. Let G be a tree with n vertices, and e be a pendent edge of G. By Theorem 7.9.1, $f(G;\lambda ) = f(G - e;\lambda ) - f(G \circ e;\lambda ).$ Now, G − e is a forest with two component trees of orders n − 1 and 1, and hence $f(G - e;\lambda ) = (\lambda {(\lambda - 1)}^{n-2})\lambda$ (see Exercise 9.1). Since $G \circ e$ is a tree with n − 1 vertices, $f(G \circ e;\lambda ) = \lambda {(\lambda - 1)}^{n-2}.$ Thus, $f(G;\lambda ) = (\lambda {(\lambda - 1)}^{n-2})\lambda - \lambda {(\lambda - 1)}^{n-2} = \lambda {(\lambda - 1)}^{n-1}.$

Conversely, assume that G is a simple graph with $f(G;\lambda ) = \lambda {(\lambda - 1)}^{n-1} = {\lambda }^{n} - (n - 1){\lambda }^{n-1} + \ldots + {(-1)}^{n-1}\lambda.$ Hence, by Theorem 7.9.2, G has n vertices and n − 1 edges. Further, the last term, ${(-1)}^{n-1}\lambda,$ ensures that Gn is connected (see Exercise 9.2). Hence, G is a tree (see Theorem 4.2.4).n

Remark 7.9.4.

Theorem 7.9.3 shows that the chromatic polynomial of a graph G does not fix the graph uniquely up to isomorphism. For example, even though the graphs K _{1, 3} and P ₄ are not isomorphic, they have the same chromatic polynomial, namely, $\lambda {(\lambda - 1)}^{3}.$

Exercise 9.2.

If G has ω components, show that ${\lambda }^{\omega }$ is a factor of $f(G;\lambda ).$

Exercise 9.3.

Show that there exists no graph with the following polynomials as chromatic polynomial (i) ${\lambda }^{5} - 4{\lambda }^{4} + 8{\lambda }^{3} - 4{\lambda }^{2} + \lambda ;$ (ii) ${\lambda }^{4} - 3{\lambda }^{3} + {\lambda }^{2};$ (iii) ${\lambda }^{7} - {\lambda }^{6} + 1.$

Exercise 9.4.

Find a graph G whose chromatic polynomial is ${\lambda }^{5} - 6{\lambda }^{4} + 11{\lambda }^{3} - 6{\lambda }^{2}.$

Exercise 9.5.

Show that for the cycle C _n of length n, $f({C}_{n};\lambda ) = {(\lambda - 1)}^{n} + {(-1)}^{n}(\lambda - 1),$ n ≥ 3.

Exercise 9.6.

Show that for any graph G, $f(G \vee {K}_{1};\lambda ) = \lambda f(G;\lambda - 1),$ and hence prove that $f({W}_{n};\lambda ) = \lambda {(\lambda - 2)}^{n} + {(-1)}^{n}\lambda (\lambda - 2).$

7.10 Notes

A good reference for graph colorings is the book by Jensen and Toft [116]. The book by Fiorini and Wilson [62] concentrates on edge colorings. Theorem 7.5.7 (Mycielski’s theorem) has also been proved independently by Blanche Descartes [50] as well as by Zykov [195]. For a complete description of graph homomorphisms, see [105].

The proof of Brooks’ theorem given in this chapter is based on the proof given by Fournier [67] (see also references [27] and [106]).

The term “snark” was given to the snark graph by Martin Gardner after the unusual creature that is described in Lewis Carroll’s poem, The Hunting of the Snark. A detailed account of the snarks, including their constructions, can be found in the interesting book by Holton and Sheehan [106].