S. and }. .. For-every-j-?-{s, applied to the s-set S \ {x s } ? {x j } yields This means that the subgraphs H 1 , x m })] are disjoint, H 1 consisting of the union of s ? 1 disjoint subgraphs K (1,(n?t)/s) Notice that |N G c (x)?N G c (x ? )| ? 1 for every two distinct x, x ? ? {x s , . . . , x m }. Otherwise the set S ? = (S \ {x s?1 , x s }) ? {x, x ? } satisfies (s ? 2)(n ? t)/s + 2((n ? t)/s + 1) ? 2 = n ? t, contradicting again Remark 3.1. Observe also that every cycle of H 2 , if any, has length at least 2(s + 1), because if C is such a cycle of length 2l with 2 ? l ? s, then the s-subset S ? of X consisting of l vertices of C and any s ? l vertices in V (H 1 ) ? X ? {x 1 , . . . , x s?1 } satisfies |N G c (S ? )| = (s ? l)(n ? t)/s + l(n ? t)/s = n ? t, against Remark 3.1. As a consequence, H 2 is either acyclic, or each of its cycles has length at least 2(s + 1) This finishes the proof, ) = ? such that Z(m, n; s, t) is guaranteed to be nonempty (see our comment after Lemma 3.3 and the hypotheses of the theorem)

T. ?. We-have-r-t-y-+-r-t-z-+-r-y-z-?-n, ?. , and |. |. , If we denote by r T Y , r T Z and r Y Z the number of edges of the subgraphs G c Let us denote by Y ? and Z ? , respectively, the subsets of Y and Z whose vertices have degree t + n in G. It is clear that, Proof: Assume that e(G) ? n 2 + 2nt ?

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