an extension of vizing's theorem

An Extension of Vizing'sTheorem

K. H. ChewSCHOOL OF MATHEMATICS

UNIVERSITY OF NEW SOUTH WALESSYDNEY, 2052, AUSTRALIA

ABSTRACT

Let G = (V (G), E(G)) be a simple graph of maximum degree ∆ ≤ D such that thegraph induced by vertices of degree D is either a null graph or is empty. We give anupper bound on the number of colours needed to colour a subset S of V (G) ∪ E(G)such that no adjacent or incident elements of S receive the same colour. In particular,if S = E(G), we have the chromatic index χ′(G) ≤ D whereas if S = V (G) ∪E(G) andD ≥ |V (G)|

k − k + 1 for some positive integer k, we have the total chromatic numberχT (G) ≤ D + k. c© 1997 John Wiley & Sons, Inc.

1. INTRODUCTION

The graphs we consider are finite and simple. We denote the maximum degree of a graph G by∆(G) (or simply ∆) and denote the degree of a vertex x in G by dG(x) (or simply d(x)). Thenotation used in this paper is standard (see [2]).

A colouring of a subset S of V (G)∪E(G) is a function π: S → C where C is a set of colourssuch that no adjacent or incident elements of S receive the same colour. So throughout thispaper, all colourings are proper. A colouring of V (G), E(G) and V (G)∪E(G) is called a vertexcolouring, edge colouring and total colouring respectively. The minimum number of colours forwhich there exists a vertex, edge and total colouring of G is called the chromatic number χ(G),chromatic index χ′(G) and total chromatic number χT (G), respectively. It is well-known thatχ(G) ≤ ∆(G) + 1 and χ′(G) ≤ ∆(G) + 1. The result that χ′(G) ≤ ∆(G) + 1 is the famoustheorem of Vizing which we shall show is a special case of results obtained in this paper.

Behzad [1] and Vizing [15] conjectured that for any graph G,

χT (G) ≤ ∆(G) + 2.

Journal of Graph Theory Vol. 24, No. 4, 347 355 (1997)c© 1997 John Wiley & Sons, Inc. CCC 0364-9024/97/040347-09

348 JOURNAL OF GRAPH THEORY

This conjecture is known as the total colouring conjecture, which has been shown to be true forvarious classes of graphs (see [3, 9] for surveys). There are three good upper bounds for thetotal chromatic number. Two of them are due to Hind [10, 11]. The first [10] is that χT (G) ≤χ′(G) + 2d√χ(G)e and the second [11] is that if ∆ ≥ |V (G)|

k for some positive integer k, thenχT (G) ≤ ∆ + 2k+ 1. A third upper bound due independently to McDiarmid and Reed [13] andChetwynd and Haggkvist [4] is that if |V (G)| < k! then χT (G) ≤ χ′(G) + k. We shall showthat if ∆(G) is relatively large compared to |V (G)|, then a good upper bound can be obtained forχT (G), which is better than any of the above upper bounds.

2. PRELIMINARY RESULTS

We give here further definitions and notation and some preliminary results.In a given colouring π of a subset S of E(G) ∪ V (G), a set of elements of S consisting of all

those edges and vertices assigned the same colour is called a colour class. A colour class withelements assigned colour α is called the α-colour class. The following theorem, which we shalluse, says that a vertex colouring of a graph G can be chosen so that any two colour classes differin size by at most one.

Theorem A (Hajnal and Szemeredi [8]). Let G be a graph of order n. Then G has a vertexcolouring φ that uses ∆ + 1 colours, with each colour class having either d n

∆+1e or b n∆+1c

elements.

In a colouring π of S, the set of colours that appear on vertex x and on edges incident withvertex x will be denoted by Cπ(x), that is,

Cπ(x) = {π(x) ∪ π(xv)|xv ∈ E(G)}.

Let C − Cπ(x) = Cπ(x). If colour α ∈ Cπ(x), we say α is missing from x.Our proofs here depend on a coloured subgraph which is either a path or a cycle. To refer to

this coloured subgraph, we use the term chain for a path or cycle. More precisely, an x-y chainof G is a finite, alternating sequence

x = x0, e1, x1, e2, . . . , xn−1, en, xn = y

of vertices and edges beginning with vertex x and ending with vertex y such that ei = xi−1xi for1 ≤ i ≤ n and the vertices are distinct except possibly for x and y. The x-y chain is a cycle if xand y coincide, otherwise it is a path.

An x-y chain with edges e1, e2, . . . , en satisfying

π(ei) =

{α if i is odd,β otherwise,

is denoted by Hπ(α, β, x, y). Colours α and β in such a chain must necessarily be distinct in acolouring. A single vertex x forms the trivial x-x path or Hπ(α, β, x, x). Note that if vertex xhas no incident edge which is coloured α even though x may have an incident edge coloured β,Hπ(α, β, x, y) is the single vertex x. The vertices in Hπ(α, β, x, y) cannot be coloured with α orβ except possibly for vertices x and y. If both vertices x and y are not coloured with α or β, thenthe chain Hπ(α, β, x, y) is said to be changeable as we can interchange the colours α and β of theedges of Hπ(α, β, x, y) without producing an improper colouring. A chain Hπ(α, β, x, y) is said

AN EXTENSION OF VIZING'S THEOREM 349

to be maximal if there does not exist a chain Hπ(α, β, x, v) which has more vertices (or edges)than Hπ(α, β, x, y). A maximal chain Hπ(α, β, x, y) will be denoted by Hπ(α, β, x). Notethat if a maximal chain Hπ(α, β, x) has at least one edge and is not a cycle, then it must end at avertex with exactly one incident edge coloured with α or β.

Let π be a colouring of a subset S of V (G)∪E(G−xy1) where xy1 is an edge of G. A fan Fat x is a sequence of distinct edges xy1, xy2, . . . , xyn such that for each i ∈ {1, . . . , n− 1}, theedge xyi+1 is assigned with a colour βi missing from yi. We write F = (xy1, . . . , xyn). Vertexx is said to be the pivot vertex and y1 is the initial vertex. The endvertices different from x of alledges which are in at least one fan at x are called x-vertices. So yi, where i ≥ 1, is an x-vertex.Note that in a fan F = (xy1, . . . , xyn), if there is no colour missing from yn, then clearly wecannot extend the fan F . However, if there is a colour β missing from yn, then either β 6∈ Cπ(x)or β ∈ Cπ(x), in which case, β = π(x) or β = π(xy′) for some x-vertex y′ and furthermore ify′ 6∈ {y2, . . . , yn−1}, we can extend F .

Two chains H1 with vertices x1, . . . , xn (possibly x1 = xn) and H2 with vertices y1, . . . , ym(possibly y1 = ym) are said to be equivalent, denoted by H1 ≡ H2, if n = m and either (i)xi = yi or (ii) xi = yn+1−i for each i = 1, . . . , n.

The following result shows that under certain conditions, we can modify a colouring of asubset S of V (G) ∪ E(G − e), where e is an edge of G, to a colouring of S ∪ {e} withoutusing additional colours and without altering the colours of vertices. This is done by changingthe colours of edges and the uncoloured edge in such a way that a colour is missing from each ofthe two endvertices of the uncoloured edge, thereby allowing us to colour that uncoloured edge,as in the proof of Vizing's theorem given in [6]. The proof involves an interchange of colours ofedges in a maximal chain with edges coloured alternatively with α and β. Interchange of coloursof vertices is complicated because a component of the two coloured subgraph induced by verticescoloured with α or β need not be a path or cycle. So throughout this paper, in particular in Lemma1 and Theorems 1 and 2, the colour of each vertex remains unchanged.

Lemma 1. Let G be a graph with an edge xy1 6∈ S. Suppose for some subset S of V (G) ∪E(G− xy1), there is a colouring π: S → C. Let F = (xy1, xy2, . . . , xyt) be a fan such that

(1) α ∈ Cπ(yt);or

(2) α ∈ Cπ(v) for each x-vertex v, either β = π(xyt) or β ∈ Cπ(yt), and Hπ(α, β, yt) ischangeable and does not contain vertex x and furthermore if β = π(xyt), Hπ(α, β, yt)does not contain yt−1.

Then for somep ∈ {1, 2, . . . , t}, there is a colouringπ′: S∪{xy1}−{xyp} → C withα ∈ Cπ′(yp)and where the colour of each vertex remains unchanged. If in addition, α ∈ Cπ(x), π can beextended to a colouring π′′: S ∪ {xy1} → C with no change in the colour of each vertex.

Proof. Note that conditions (1) and (2) cannot hold at the same time. To prove that colouringπ′ exists, we first suppose (1) holds or (2) holds with β = π(xyt). We uncolour each edge xyjfor 2 ≤ j ≤ t and assign the colour of edge xyj+1 to each edge xyj for 1 ≤ j ≤ t − 1, and forbrevity, we shall say that we recolour the fan F . This leaves edge xyt uncoloured and producesa new colouring θ of S ∪ {xy1} − {xyt} from π. Note that if t = 1, then θ = π. Since onlycolours of edges incident to x have been changed, θ(e) = π(e) for all edges e incident to yt excepte = xyt. Thus if (1) holds, then α ∈ Cθ(yt), and θ is the desired colouring π′ with p = t.

We now suppose (2) holds with β = π(xyt). Observe that since α ∈ Cπ(v) for each x-vertexv and β ∈ Cπ(yt−1), α and β must be distinct colours. (Note if β ∈ Cπ(yt), then also α 6= β.)


From (2), Hπ(α, β, yt) does not contain x and yt−1, which implies that Hθ(α, β, yt) does notcontain x and yt−1. This implies that under θ, we can interchange the colours α and β of theedges of Hθ(α, β, yt) to obtain a colouring π′ of S ∪{xy1}− {xyt} without altering the coloursof incident edges to x which are coloured α or β under θ (and so Cπ(x) = Cθ(x) = Cπ′(x)) andα ∈ Cπ′(yt). This is the desired colouring π′ with p = t.

We next suppose (2) holds with β ∈ Cπ(yt). Then we can choose yp, 1 ≤ p ≤ t, such thatβ ∈ Cπ(yp) and Hπ(α, β, yp) is changeable and does not contain x and Hπ(α, β, yp) does notend at yi for each i < p. If Hπ(α, β, yp) does end at yj for some j < p, then Hπ(α, β, yp) andHπ(α, β, yj) are equivalent chains, then we can replace p with j. From the original colouringπ, we interchange the colours α and β of the edges of Hπ(α, β, yp) to obtain a colouring θ of Swithout altering the colours of incident edges to x. It is possible that Hπ(α, β, yp) contains yjfor some j < p without ending at yj . However, we still have Cπ(yj) = Cθ(yj). This means thatthe original fan F ′ = (xy1, . . . , xyp) under π, remains a fan under θ and so we can recolour F ′

from θ to obtain π′ and leave xyp uncoloured and α 6∈ Cπ′(yp). This proves that π′ exists in (1)and (2).

Finally, if α ∈ Cπ(x) (= Cπ′(x)), then we can extend π′ to a colouring π′′: S ∪ {xy1} → Cby colouring the edge xyp with α. This completes the proof of the lemma.

The vertex yt, in condition (2) of Lemma 1, is said to be α-removable with respect to π orsimply α-removable if it is clear from the context what the colouring is. We shall use Lemma 1extensively in the proofs of our main results.

3. THE MAIN RESULTS

We now come to the main results.

Theorem 1. Let G be a graph with an edge xy such that the following hold:

(1) For some subset S of V (G) ∪ E(G− xy), there is a colouring π: S → C with |C| = λ.

(2) Each colour class of π has at most k vertices, where k is some fixed non-negative integer≤ λ− ∆(G) and if k ≥ 2, π has an α-colour class that has no vertices.

(3) Each vertex adjacent to x has degree ≤ λ− 1 − k in G.

Then π can be extended to a colouring π′: S ∪ {xy} → C with no change in the colour of eachvertex.

Proof. We shall proceed in two main steps. We apply Step 1 if there is a colouring θ: S ∪{xy} − {uv} → C with α ∈ Cθ(u) where if k ≥ 2, there is no vertex coloured with α and ifk ∈ {0, 1}, α is taken to be any colour missing from u. We apply Step 2 if under π: S → C, α ∈Cπ(x) where k ≥ 2. We shall show that Step 2 leads back to Step 1.

Step 1. Suppose that by suitably changing the colours of the edges of G, there is an edge uv ofG such that there is a colouring θ: S ∪ {xy} − {uv} → C with α ∈ Cθ(u). Let S′ = S ∪ {xy}−{uv}. If k ∈ {0, 1}, we take u = x and v = y and since λ ≥ ∆ + k, we may take α tobe any colour missing from u and so θ = π. Our aim in Step 1 is to extend θ to a colouringπ′: S′ ∪ {uv} → C where S′ ∪ {uv} = S ∪ {xy}.

Let the pivot vertex and initial vertex be u and v respectively. If there is a u-vertex v′ withα ∈ Cθ(v

′), we can apply Lemma 1 to G with uncoloured edge uv, changing only the coloursof edges under θ, to obtain a colouring of π′: S′ ∪ {uv} → C. Thus we assume α ∈ Cθ(v

′) foreach u-vertex v′.


LetE(θ, u) be the set of colours of all edges different from uv which are in at least one fan at u,that isE(θ, u) = {θ(uv′)|v′ is a u vertex under θ} and suppose |E(θ, u)| = q. Since |E(θ, u)| =q, there are altogether q + 1 u-vertices since we also include vertex v. Let B(θ, u) = ∪v′Cθ(v

′)where v′ is any u-vertex. Clearly E(θ, u) ⊆ B(θ, u). If there is a fan F = (uv1, . . . , uvn)with v = v1 and µ missing from both vn and u, we recolour F from θ and leave edge uvnuncoloured, then colour uvn with µ, thereby completing the proof of Step 1. Thus we nowsuppose if µ ∈ Cθ(vn) (⊆B(θ, u)), then µ = θ(u) or µ = θ(uv′) for some u-vertex v′. Thus wehave B(θ, u) ⊆ E(θ, u) ∪ {θ(u)}. It follows that |B(θ, u)| ≤ q + 1 and |B(θ, u)| = q if θ(u)does not exist.

For each µ ∈ B(θ, u), let S(θ, µ) be the number of u-vertices which have colour µ missingunder θ. Note that edge uv is uncoloured and so for each u-vertex v′, we have

|Cθ(v′)| ≤

{d(v′) if v′ = v or θ(v′) does not exist,d(v′) + 1 if v′ 6= v and θ(v′) exists.

It follows that there are at least λ− d(v) colours missing from vertex v and at least λ− |Cθ(v′)|

colours missing from each remaining u-vertex v′ and so we have∑µ

S(θ, µ) ≥ λ− d(v) +∑v′ 6=v

λ− |Cθ(v′)|. (∗)

If k ≥ 2, then since d(v′) ≤ ∆ and λ ≥ ∆ + k, (∗) reduces to∑µ

S(θ, µ) ≥ k + q(k − 1).

If k ∈ {0, 1}, then u = x and v = y and d(v′) ≤ λ − 1 − k for each u-vertex v′. If k = 1,|Cθ(v

′)| ≤ d(v′) + 1 and so λ − |Cθ(v′)| ≥ k ≥ 1, and if k = 0, |Cθ(v

′)| ≤ d(v′) and soλ− |Cθ(v

′)| ≥ 1 and thus (∗) reduces to∑µ

S(θ, µ) ≥ 1 + k + q.

Since |B(θ, u)| ≤ q + 1 and |B(θ, u)| = q if k = 0, by the pigeonhole principle, there exists acolour β ∈ B(θ, u) such that

S(θ, β) ≥{

2 if k ∈ {0, 1},k if k ≥ 2.

(If k = 0, then actually we need not require that d(y) ≤ λ− 1 to conclude that there is a colourβ ∈ B(θ, u) with S(θ, β) ≥ 2. This is because there are at least λ−∆ + 1(≥1) colours missingfrom y.)

Since there is no change in colour of vertices, each colour class of θ has at most k verticesand if k ≥ 2, the α-colour class contains no vertices. It follows that there are at most max{2, k}pairwise vertex-disjoint, maximal chains Hθ(α, β, v

′) where each chain is not changeable. Wewill need this fact in the discussion that immediately follows.

We have β ∈ B(θ, u) ⊆ E(θ, u) ∪ {θ(u)} and so if β 6= θ(u), then there is an edge uw of Gcoloured with β. Let S(θ, β) = s. We claim that there is an α-removable vertex and we dividethe proof into three cases.

Case 1. Suppose β = θ(u). We claim that there is a u-vertex vt with β 6∈ Cθ(vt), and suchthat Hθ(α, β, vt) is changeable. Suppose not. Then there are s u-vertices zi with β 6∈ Cθ(zi)


for 1 ≤ i ≤ s and each chain Hθ(α, β, zi) is not changeable. Each chain Hθ(α, β, zi) doesnot contain u as α 6∈ Cθ(u). If Hθ(α, β, zi) and Hθ(α, β, zj) have some vertices in commonwhere i 6= j, then Hθ(α, β, zi), being a maximal chain, must end at zj , and so Hθ(α, β, zi) ischangeable, a contradiction. Let zs+1 = u. Then there are altogether s + 1 pairwise vertex-disjoint chains Hθ(α, β, zi) (1 ≤ i ≤ s + 1), where each chain is not changeable. Sinces + 1 ≥ max{2, k} + 1, this is a contradiction. This proves that there is a u-vertex vt withβ 6∈ Cθ(vt), and such that Hθ(α, β, vt) is changeable. It follows that Hθ(α, β, vt) does notcontain u and so vt is α-removable.

Case 2. Suppose β 6= θ(u) and Hθ(α, β, w) ends at a vertex coloured with β. Note thatthe edge uw exists. Using the same reasoning in Case 1, and since s ≥ max{2, k}, there is au-vertex vt with β 6∈ Cθ(vt), and such that Hθ(α, β, vt) is changeable. Then Hθ(α, β, vt) doesnot contain u and so vt is α-removable.

Case 3. Suppose β 6= θ(u) and Hθ(α, β, w) does not end at a vertex coloured with β. Sinces ≥ 2, there are at least two u-vertices which have colour β missing. This means that we canchoose au-vertex vt withβ 6∈ Cθ(vt), and such thatHθ(α, β, vt) does not containu. Furthermore,if Hθ(α, β, w) begins or ends at a vertex coloured with α, and so k must necessarily be 1, vtis chosen so that Hθ(α, β, vt) is changeable and so vt is α-removable. Thus we assume thatHθ(α, β, w) does not begin or end at a vertex coloured with α. It follows that θ(w) 6∈ {α, β}and that Hθ(α, β, w) is changeable. Since the initial edge of Hθ(α, β, w) must be coloured withα and α 6∈ Cθ(u), Hθ(α, β, w) does not contain u.

Subcase 3a. Suppose w = vt+1 with β 6∈ Cθ(vt). If Hθ(α, β, w) contains vt, then sinceβ 6∈ Cθ(vt), Hθ(α, β, w) must end at vt. It follows that

Hθ(α, β, vt, w) ≡ Hθ(α, β, w).

Note that Hθ(α, β, vt) contains the edge uw, which contradicts the fact that Hθ(α, β, vt) doesnot contain u. Thus Hθ(α, β, w) does not contain vt, an so w = vt+1 is α-removable.

Subcase 3b. Suppose w = vr for some r ∈ {2, . . . , t− 1}. So far, we know that Hθ(α, β, w)is changeable and does not contain u and so if Hθ(α, β, w) does not contain vr−1 as well, thenw = vr is α-removable. So we assume that Hθ(α, β, w) contains vr−1 and as in Subcase 3a,Hθ(α, β, vr−1, w) and Hθ(α, β, w) are equivalent. We now uncolour each uvj for r ≤ j ≤ t,assign colour θ(uvj+1) to each edge uvj for r ≤ j ≤ t− 1 and assign colour β to edge uvt. Thisproduces a colouring θ′ of S′ from θ with

Hθ′(α, β, vr−1, w) ≡ Hθ(α, β, vr−1, w)

which is changeable. Note that β 6∈ Cθ′(w) and so Hθ′(α, β, vr−1, w) is the maximal chainHθ′(α, β, vr−1) which does not contain u. Therefore vr−1 is α-removable, in this case, withrespect to θ′.

Therefore, we have shown that, in all cases, there is an α-removable vertex. We now applyLemma 1 to G with uncoloured edge uv and with α missing from u, to obtain a colouringπ′: S′ ∪ {uv} → C.

In particular, we have proved the theorem for k ∈ {0, 1}. We will now show that for k ≥ 2,Step 2 leads us back to Step 1, thereby completing the proof of the theorem.

Step 2. Given that there is a colouring π of S with the edge xy 6∈ S, suppose there is an edgexz of G coloured with α. If such an edge xz does not exist, then we proceed by Step 1 withθ = π. Note that there is no vertex coloured with α as k ≥ 2. Let the pivot vertex be x and theinitial vertex be y = y1. We replace u with x and θ with π accordingly in all the terms definedin Step 1.


If there is an x-vertex y′ with α 6∈ Cπ(y′), we can apply Lemma 1 to G with xy 6∈ S, to obtaina colouring π′ of S ∪ {xy} − {xy′′} with α 6∈ Cπ′(y′′) for some x-vertex y′′. (Actually, fromthe proof of Lemma 1, we can take y′′ to be y′.) We can then proceed by Step 1 with y′′ as thepivot vertex and x as the initial vertex. Thus we assume α ∈ Cπ(y′) for all x-vertices y′. As inStep 1, we can also suppose that

E(π, x) ⊆ B(π, x) ⊆ E(π, x) ∪ {π(x)}.We are given that each vertex adjacent to x has degree ≤ λ− 1− k in G. This will always be

true if λ > ∆ + k. Then under π, there are at least 1 + k colours missing from vertex y and atleast k colours missing from the each of the remaining x-vertices. Thus

∑µ

S(π, µ) ≥ 1 + k + qk,

where |E(π, x)| = q. Since |B(π, x)| ≤ q + 1, by the pigeonhole principle, there exists a colourβ ∈ B(π, x) such that S(π, β) ≥ k + 1. Since each colour class contains at most k vertices,there are at most k pairwise vertex-disjoint, maximal chains Hπ(α, β, y′) where each chain is notchangeable. This fact is implicitly used in the discussion that immediately follows.

We have β ∈ B(π, x) ⊆ E(π, x) ∪ {π(x)} and so if β 6= π(x), then there is an edge xw ofG coloured with β. We claim that there is an α-removable vertex. Let S(π, β) = s. There ares x-vertices zi with β 6∈ Cπ(zi) for 1 ≤ i ≤ s and with the possible exception of say i = i0,Hπ(α, β, zi) does not contain x (and therefore also w and z). We label the vertices zi so thati0 = s.

If Hπ(α, β, z) ends at a vertex coloured with β, where π(xz) = α, then as in Step 1, sinces − 1 ≥ k, there is an x-vertex yt ∈ {z1, . . . , zs−1} such that Hπ(α, β, yt) is changeable,and so yt is α-removable. Next assume Hπ(α, β, z) does not end at a vertex coloured withβ. Then Hπ(α, β, w) also does not end at a vertex coloured with β and so Hπ(α, β, w) ischangeable. If Hπ(α, β, w) is a cycle, then Hπ(α, β, zs) also does not contain x and so we canchoose yt ∈ {z1, . . . , zs} such that Hπ(α, β, yt) is changeable, in which case, yt is α-removable.Therefore, we assume that Hπ(α, β, w) is not a cycle which implies that Hπ(α, β, w) does notcontain x. The rest of the proof is similar to Subcase 3a and Subcase 3b of Step 1 by replacingvt with a vertex yt ∈ {z1, . . . , zs−1}.

Therefore, we have shown that in all cases, there is an α-removable vertex y′. We now applyLemma 1 to G with uncoloured edge xy, to obtain a colouring π′: S ∪ {xy} − {xy′′} → C forsome x-vertex y′′ with α ∈ Cπ′(y′′). We then proceed by Step 1 with y′′ as the pivot vertex andx as the initial vertex. This completes the proof of the theorem.

Let GD be the subgraph of G induced by vertices of degree D in G.

Theorem 2. Let G be a graph of maximum degree ∆(G) ≤ D such that GD is either a nullgraph or is empty. Let φ be a colouring of a subset V of vertices of G that uses at most D + kcolours, k ≥ 0, with each colour class of φ having at most k vertices, and if k ≥ 2, there is acolour class that has no vertices. Then φ can be extended to a colouring π of V ∪E(G) with nochange in the colour of each vertex and without using any additional colours.

Proof. We first note that D ≥ 2, otherwise D = ∆(G) = 1 and so GD is non-empty andis not a null graph, a contradiction. Let E = {uv|uv ∈ E(G), dG(u) = ∆(G)}. To extend φ tothe required colouring π, we shall use Theorem 1 to first colour the edges of E(G)−E, and thenthe edges of E. Following this order of colouring edges, we assume that for some subgraph H ofG with vertex set V (H) = V (G) and edge xy ∈ E(H), φ has been extended to a colouring φ′


of V ∪E(H − xy) with no change in the colour of each vertex and without using any additionalcolours. Each colour class of φ′ has at most k vertices, and if k ≥ 2, φ′ has a colour class thathas no vertices.

Let dG(y) ≤ dG(x). To apply Theorem 1 on graph H , we now just need to show that eachvertex u adjacent to x in H has degree dH(u) ≤ D − 1. This is clearly true if D ≥ ∆(G) + 1and so we assume now D = ∆(G). If dG(x) = ∆(G), then since G∆ is a null graph, eachvertex u adjacent to x has degree dH(u) ≤ ∆(G) − 1. If dG(x) ≤ ∆(G) − 1, then we havedG(y) ≤ ∆(G)−1 and xy ∈ E(G)−E. All edges (including xy) ofE(G)−E must be colouredbefore edges of E, so H −xy does not contain any edge of E and thus ∆(H −xy) ≤ ∆(G)− 1.It follows that each vertex u adjacent to x in H has degree dH(u) ≤ D − 1. By Theorem 1 withD = λ− k, φ′ can be extended to a colouring φ′′ of V ∪ E(H) with no change in the colour ofeach vertex and without using any additional colours. By colouring the edges one by one in thisway, we can obtain the desired colouring π.

Setting k = 0 in Theorem 2 gives us the following corollary.

Corollary 3. Let G be a graph of maximum degree ∆ ≤ D such that GD is either a null graphor is empty. Then χ′(G) ≤ D.

With D = ∆ + 1, we clearly have Vizing's theorem [14] (clearly our proof here essentiallyreduces to Vizing's proof) and with D = ∆, we have a special case of Fournier's theorem [7]which says that if G∆ is a forest, then χ′(G) ≤ ∆. Some extension of Fournier's theorem can befound in Hoffman and Rodger [12].

We now present a significant improvement of Hind's second upper bound [11], mentioned inthe introduction. The author was not aware of Hind's paper [11] when this paper was first written.

Theorem 4. Let G be a graph of maximum degree ∆ ≤ D such that GD is either a null graphor is empty and D ≥ |V (G)|

k − k + 1 for some positive integer k. Then χT (G) ≤ D + k.

Proof. Let |V (G)| = n. We first show that G has a vertex colouring φ that uses D + k − 1colours, with each colour class of φ having at most k vertices. If D + k ≥ n+ 1, then we assignour colour to one vertex, thereby obtaining the desired vertex colouring φ.

Assume D+k ≤ n. Since n ≥ D+k ≥ nk +1, we must have k ≥ 2. Let p = D+k−∆−2

which satisfies 0 ≤ p ≤ n − ∆ − 1. Then there are p distinct vertices v1, v2, . . . , vp such thateach vi is non-adjacent to a single vertex v of degree ∆. Add an edge joining v to each vi for1 ≤ i ≤ p, to form a graph G∗ which has maximum degree D+k− 2(≥∆). By Theorem A, G∗

has a vertex colouring φ that uses D + k − 1 colours, with each colour class of φ having eitherd nD+k−1e or b k

D+k−1c vertices. Since D+k−1 ≥ nk , we have d n

D+k−1e ≤ k. Any two verticeslike in G must also be adjacent in G∗ and hence will be assigned different colours under φ. Itfollows that φ is also a vertex colouring of G and with each colour class of φ having at most kvertices.

Note that there is still one colour available for colouring only the edges. Thus by Theorem 2,χT (G) ≤ D + k.

We now present an important corollary of Theorem 4.

Corollary 5. Let G be a graph of maximum degree ∆ such that G∆ is a null graph and ∆ ≥|V (G)|

2 − 1. Then χT (G) ≤ ∆ + 2.


REMARKS AND ACKNOWLEDGMENTS

Using similar argument, the results in this paper have been extended to multigraphs in the au-thor's doctoral thesis. The author would like to thank the referees for helpful comments and hissupervisors for helpful discussion

References

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Received August 22, 1995

an extension of vizing's theorem

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