By Ronald Larsen

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Then the graph G - w has only two major vertices and thus by VAL, G - w is of class I. 1, G is 1-factorizable. Suppose deg G = 2n - 4. implies that n > 4. valency A(G) > 4. 1, G is 1-factorizable. 8 1. Let G be the graph obtained from two copies of K2m+1, where m > 2, by deleting one edge (say albl and a2b2) from each, and joining them by the two edges ala2 and blb2. Prove that G is of class 2 (Chetwynd and Hilton [85]). 2 Suppose G is a regular graph of order 2n and deg G = 2n - 5 > 2 [n21] - 1.

We shall call this construction the 10-construction. 6 (HJ-construction) Let G and H be two A-critical graphs and K be a graph obtained from G and H by identifying u e V(G) and v e V(H) such that dG(u) + dH(v) < A + 2, removing edges uz e C and vz' e H and joining the vertices z and z'. 30 Then K is also A-critical. It is clear that K is connected and A(K) = A. Proof. We first prove that K is of class 2. a A-colouring of K and let nr(zz') = 1. Suppose otherwise. Let v be Then there is an edge e of G incident with u such that a(e) = 1 otherwise X'(G) = A.

1 Suppose G is a regular graph of order 2n and G ยข K2n. Then G is of class 1 if and only if for any w e V(G), G - w is of class 1. 52 Proof. Necessity. Since G is regular and G # R2n, A(C - w) = A(G) for Hence A(G - w) < x'(G - w) < XI(G) = A(G) = A(G - w), any w c V(G). from which it follows that X'(G - w) = A(G - w) and G - w is of class 1. Let A = A(G - w) and let it be a A-colouring of G - w. Sufficiency. If there is a colour, colour i say, which is absent at more than one vertex in N(w), then there is a colour, colour j say, which is present at all the vertices in N(w) and thus colour j is present at every vertex in G - w.