[Received 4 July 1983âRevised 21 May 1984]. ABSTRACT. It is a well-known conjecture that if a regular graph G of order 2n has degree d(G) satisfying.
REGULAR GRAPHS OF HIGH DEGREE ARE 1-FACTORIZABLE A. G. CHETWYND and A. J. W. HILTON [Received 4 July 1983—Revised 21 May 1984]
ABSTRACT
It is a well-known conjecture that if a regular graph G of order 2n has degree d(G) satisfying d(G) ^ n, then G is the union of edge-disjoint 1-factors. It is well known that this conjecture is true for d(G) equal to 2n —1 or 2n — 2. We show here that it is true for d(G) equal to2n — 3, In — 4, or2n — 5. We also show that it is true for $|K(G)|.
1. Introduction The graphs we shall consider will be simple, that is, they will have no multiple edges or loops. An edge-colouring of a graph G is a map ^ , where ^ is a set of colours and E(G) is the set of edges of G, such that no two incident edges receive the same colour. The chromatic index x'(G) of G is the least value of | # | for which an edgecolouring of G exists. A well-known theorem of Vizing [13] states that
where A(G) is the maximum degree of G. Graphs for which A(G) = #'(G) are said to be Class 1, and otherwise they are Class 2. A regular Class 1 graph is often called 1-factorizable, as it is the union of edge-disjoint 1-factors. For further information on edge-colourings of graphs see the book [9] by Fiorini and Wilson. A well-known but, up to recently, apparently intractable conjecture, is: CONJECTURE 1. A regular graph of order 2n and degree d(G) satisfying
is Class 1. (Note that 2^n+ 1)J - 1 ^ n.) We do not know to whom this conjecture should be attributed, but it is certainly a very obvious conjecture and has occurred independently to many people. If it could be proved, it would have many applications outside the immediate area of graph theory. The lower bound 2[J(n+l)J — 1 is the best possible. A connected regular graph of order 2n and degree 2[^(«+ 1)J — 2 which is of Class 2 can be formed for n = 2m-f 1, with m ^ 2 , from two copies of K2m + 1 by removing one edge (say albl and a2b2) from each, and replacing them by edges axa2 and b1b2. The Petersen graph is an example of a connected regular Class 2 graph of order 2n and degree 2[i{«+ 1)J — 3. It is well known that K2n is Class 1, and a trivial consequence is that a regular graph of order 2n and degree 2n — 2 is Class 2 (as any such graph can be formed by removing A.M.S. (1980) subject classification: primary 05C70. Proc. London Math. Soc. (3), 50 (1985), 193-206. 5388.3.50
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A. G. CHETWYND AND A. J. W. HILTON
a 1-factor from K2n). It is probably known that the conjecture is true if d(G) = 2 n - 3 , but the authors do not have a reference for this, and provide a short proof. Rosa and Wallis [12] recently proved the case when d{G) = 2n — 4 under the special circumstance that G (the complement of G) is Class 1 also. Haggkvist kindly showed us recently a sketch of a proof of the conjecture in the case when d{G) ^ jf| | V{G)\, and it seems likely that his method could be worked up to give a somewhat better result. He has even more recently announced that he can prove that, given e > 0, there exists N such that if | V{G)| ^ N and | V{G)| is even, and G is regular with d{G) > (J + e)| V{G)\, then G is 1-factorizable. Our method and his bear no resemblance to each other. We prove here three main theorems. The first was conjectured by Chetwynd and Hilton in [3]. THEOREM 1. Let G be a connected graph with three vertices of maximum degree. Then G is Class 2 if and only if G has three vertices of degree \ V(G) \ — 1 and the remainder have degree | K(G)|-2 {this implies that | V{G)\ is odd).
Further results concerning the edge-chromatic class of graphs with a given number of vertices of maximum degree will appear in a forthcoming paper by Chetwynd and Hilton [4]. The proof of Theorem 1 uses in its most simple form the essential and novel idea which enabled us to prove Theorems 2 and 3. Theorems 2 and 3 are both special cases of the conjecture. THEOREM 2. Let G be a regular graph of order In and degree d{G) equal to 2 n - 3 , 2n-4, or 2 n - 5 . Let d{G) ^ 2\%n+ 1)J — 1. Then G is Class 1. As the proof in the case where d{G) = 2n — 5 is rather long, we give only a sketch of it. The next case after this would be when d(G) = 2n — 6. We feel we ought to point out that this case appears at the moment to be too difficult to cope with without some further idea. The reason is that in the case where d(G) = 2n — 5, one can deduce from Lemma 6 that A(G) ^ \ \ V{G) \, and the number \ \ V(G) \ marries (more or less) with the applications we have to make of Dirac's theorem (Lemma 9) in which \ \ V(G) | also features. However, in the case where d(G) = In — 6, one can only deduce from Lemma 6 that A(G) ^ 11 V(G)\, which does not marry with Dirac's theorem. THEOREM
3. Let G be a regular graph of order 2n and degree 2n — k, where ^Sf(k-l){2n-k)\t9k 2n— 1 J \
2
Then G is Class 1. A slightly simpler, slightly weaker form of Theorem 3 is given in the following corollary. COROLLARY
1. Let G be a regular graph of order 2n whose degree satisfies
Then G is Class 1.
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Proof. lfd(G) ^ f | K(G) | then 2n-k>\2n/l inequality of Theorem 3 is satisfied.
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and so 2n 2* 7fc = f/c+ffc. Clearly the
The inequality d(G) ^ f | V(G)| is approximately d(G) ^ 0-8571 V(G)\. Theorem 3 is actually a bit sharper than this, and yields d{G) ^ 0-8491 V(G)\. Corollary 1 has an application on the subject of 'Intricacy' about which an interesting paper has recently been written by Opencomb [11]. Briefly, suppose that we have a set of edge-disjoint 1-factors of Kln. It may well be that this set of 1-factors cannot be completed to give a 1-factorization of K2n. In that case, for some integer j = j(n), it is certainly possible to partition the given set of edge-disjoint 1-factors into j parts in such a way that the set of 1-factors in each part of the partition can be extended to a 1-factorization of K2n. The intricacy of this problem is the least j for which there always exists a partition into j parts, each of which can be extended to a 1-factorization of K2n. The conjecture would imply that the intricacy of this problem was 2. Corollary 1 implies that it is no more than 7. In the notation of [11], we have: COROLLARY 2. For n ^ 3,
2 < /c(Pack2 x'(G\e) (or i'(G) > x'{G\v), respectively). The graph G is critical if G is Class 2 and each edge of G is critical. For v e V(G), let d*(v) be the number of vertices of maximum degree to which v is adjacent (v itself is not included). An accessible proof of the first lemma, Vizing's adjacency lemma [13], may be found in [9]. LEMMA
1. Let G be a critical graph. Let u,w e V(G) and let u be adjacent to w. Then
An immediate corollary of Lemma 1 is: LEMMA 2. Let G be a critical graph. Then each vertex is adjacent to at least two vertices of maximum degree (that is, d*(v) ^ 2 (for all v ^ V(G))). LEMMA
3. / / G has at most two vertices of maximum degree, then G is Class 1.
Proof. Let u (and possibly w) be a vertex (vertices) of maximum degree. Apply Vizing's argument to G\w, and then to G, taking u as the pivot vertex in G\w, and then w as the pivot vertex in G, and colouring the edge uv (if uv is an edge) last. LEMMA 4. / / G is a Class 2 graph, then G contains a critical subgraph of the same maximum degree.
Proof. From G remove edges el,e2,-.- according to the following rule: if G\{el,...,ei\ is Class 2 and there exists an edge ceG\{e 1 ) ... ) g j } such that is also Class 2, let ei+l be such an edge. If there is no such edge, then G\{e1,...,ehe] G\{ei,...,ei} is critical. When this process stops, the critical graph obtained has the same maximum degree, for, by Lemma 4, the removal of the edge ei + l cannot lower the maximum degree. For our purposes, the following lemma will be very useful.
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LEMMA 5. For a graph G,let e e E(G) and w e V(G), and let e and w be incident. Let d*(w) ^ 1. Then A(G\e) = A(G) => X'(G\e) = X'(G), and A(G\w) = A(G) => X'(G\w) = X'(G). Proof. If G is Class 1, then we have A(G) = X'(G) > x'(G\e) > A(G\e) = A(G), and so x'(G) = x'(G\e). Similarly, x'(G\w) = x'(G). If G is Class 2, let G* be a critical subgraph of G with A(G) = A(G*). Then, in view of Lemma 2, e $ E(G*), because d£.(w) ^ d%(w) ^ 1. Similarly, w £ V(G*). There is an alternative proof of Lemma 5 which does not depend on Vizing's adjacency lemma, nor on the notion of critical graphs. It does, however, depend on knowledge of the original proof of Vizing's theorem that x\G) ^ A(G)+ 1 (this proof is the one presented in [9]). Alternative proof of Lemma 5. If d*(w) ^ 1 and A(G\w) = A(G), then Vizing's argument may be applied to extend the edge-colouring of G\w without increasing the chromatic index, provided that w is the pivot vertex, and that the edge (if there is one) joining w to a vertex of degree A(G) is coloured last. If d*(w) ^ 1, A(G\e) = A(G), and A(G\w) = A(G), then X'(G\e) = x\G), since x'(G) > X'(G\e) > X'(G\w) = X'(G); if, however, d*(w) ^ 1 and A(G\e) = A(G) but A(G\w) ^ A(G), then G and G\e have just one (the same) vertex of maximum degree other than, possibly, w. It is easy to show by Vizing's argument that both are Class 1, provided an edge on this vertex of maximum degree is coloured last (if d(w) = A, colour the edge joining the two vertices of maximum degree last.) LEMMA
6. Let G be a critical graph. If G has r vertices of degree A(G), then
Proof. Let u be a vertex with d(u) =
