1992
DOI: 10.1016/0012-365x(92)90152-6
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Contributions to the theory of graphic sequences

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Cited by 39 publications
(38 citation statements)
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“…(4) gives in k m , is attained at k m = 1 2 (a + b + 1 ± 1) and is equal to k(n − 1) + 1 4 ((a + b + 1) 2 − 1) − bn. Thus the inequality (a + b + 1) 2 ≤ 4bn + 1 implies r k ≤ k(n − 1) and so the sequence d is graphic by [8,Theorem 3]. In case (II), the (unique) maximal value of the right-hand side of Eq.…”
Section: Lemma 6 For a Strong Index Kmentioning
confidence: 94%
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“…(4) gives in k m , is attained at k m = 1 2 (a + b + 1 ± 1) and is equal to k(n − 1) + 1 4 ((a + b + 1) 2 − 1) − bn. Thus the inequality (a + b + 1) 2 ≤ 4bn + 1 implies r k ≤ k(n − 1) and so the sequence d is graphic by [8,Theorem 3]. In case (II), the (unique) maximal value of the right-hand side of Eq.…”
Section: Lemma 6 For a Strong Index Kmentioning
confidence: 94%
“…Sufficiency in case (I) follows immediately from Theorem 1. For the other cases, we employ similar ideas to those of [8].…”
Section: Proof Of Sufficiencymentioning
confidence: 96%
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“…Proof. The first term in this bound comes by substituting α 1 + 1 for the maximum degree into the graphic bound given by Zverovich and Zverovich (Theorem 6, [10]); it follows that if the above bound holds then ⊕ 1,2 α is graphic. The second term uses a degree sequence packing result.…”
Section: Forced and Forbidden Edgesmentioning
confidence: 99%
“…In Section 3, we give sufficient conditions based on the maximum degree in E 12 for a solution to exist for the case of non oriented complete bipartite graphs and complete graphs (see [18] for additional results on the sequences of degrees of a graph). This exhibits a solvable case of the basic image reconstruction problem with k = 3 colors.…”
Section: Introductionmentioning
confidence: 99%