2000
DOI: 10.1002/(sici)1097-0118(200004)33:4<237::aid-jgt4>3.3.co;2-1
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Decomposing graphs with girth at least five under degree constraints

Abstract: We prove that the vertex set of a simple graph with minimum degree at least s + t − 1 and girth at least 5 can be decomposed into two parts, which induce subgraphs with minimum degree at least s and t, respectively, where s, t are positive integers ≥ 2.In [2], M. Stiebitz showed that the vertex set of a graph with minimum degree at least s + t + 1 can be decomposed into two parts, which induce subgraphs with minimum degree at least s and t, respectively. In [1], A. Kaneko showed that this result holds for tria… Show more

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Cited by 11 publications
(16 citation statements)
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“…That this holds was conjectured in and proved by Kaneko . Divan proved that if s,t2, then δ(G)s+t2 for every graph GG¯δfalse(s,tfalse) containing no cycles of order 3 or 4.…”
Section: Introductionmentioning
confidence: 86%
See 1 more Smart Citation
“…That this holds was conjectured in and proved by Kaneko . Divan proved that if s,t2, then δ(G)s+t2 for every graph GG¯δfalse(s,tfalse) containing no cycles of order 3 or 4.…”
Section: Introductionmentioning
confidence: 86%
“…That this holds was conjectured in [20] and proved by Kaneko [9]. Divan [4] proved that if s, t ≥ 2, then δ(G) ≤ s + t − 2 for every graph G ∈ G δ (s, t ) containing no cycles of order 3 or 4. Let ρ be a monotone increasing graph parameter.…”
Section: Introductionmentioning
confidence: 88%
“…In [6], Stiebitz showed that every graph with minimum degree at least s + t + 1 admits an (s, t)-bipartition. Kaneko [4] and Diwan [2] strengthened this result, proving that it suffices to assume that the minimum degree is at least s + t or s + t − 1 (s, t ≥ 2) if G contains no cycles shorter than four or five, respectively.…”
Section: Introductionmentioning
confidence: 89%
“…that G admits a(2,2)-bipartition S , T such that | |S | − |T | | ≤ (n − 1)/2 − (n − 1)/2 + 5. If there exists x ∈ V(G) such that δ(G − x) ≥ 4, then δ(G ) ≥ 4 and G contains no K 3,r , as G ⊆ G, where G = G − x.…”
mentioning
confidence: 99%
“…Kaneko proved that any triangle‐free graph with minimum degree at least s+t can already force a partition (A,B) as above. The minimum degree condition was further sharpen by Diwan , when cycles of length four are also forbidden. To be precise, Diwan proved that, assuming s,t2, any graph of girth at least five with minimum degree at least s+t1 has a partition (A,B) such that the subgraphs induced on A and B have minimum degree at least s and t , respectively.…”
Section: Introductionmentioning
confidence: 99%