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Planar Graphs Without Adjacent Cycles of Length at Most Five are (2, 0, 0)-Colorable
Bulletin of the Malaysian Mathematical Sciences Society ( IF 1.2 ) Pub Date : 2020-08-28 , DOI: 10.1007/s40840-020-01004-8
Xiangwen Li , Qin Shen , Fanyu Tian

A graph G is \((d_{1},d_{2},\ldots ,d_{k})\)-colorable if the vertex set of G can be partitioned into subsets \(V_{1}, V_{2},\ldots , V_{k}\) such that the subgraph \(G[V_{i}]\) induced by \(V_{i}\) has maximum degree at most \(d_{i}\) for \(i = 1, 2, \ldots , k\). Novosibirsk’s Conjecture (Sib lektron Mat Izv 3:428–440, 2006) says that every planar graph without 3-cycles adjacent to cycles of length 3 or 5 is 3-colorable. Borodin et al. (Discrete Math 310: 167–173, 2010) asked whether every planar graph without adjacent cycles of length at most 5 is 3-colorable. Cohen-Addad et al. (J Comb Theory Ser B 122:452–456, 2017) gave a negative answer to both Novosibirsk’s conjecture and Borodin et al.’s problem. Zhang et al. (Discrete Math 339:3032–3042, 2016) asked whether every planar graph without adjacent cycles of length at most 5 is (1, 0, 0)-colorable. In this paper, we show that every planar graph without adjacent cycles of length at most five is (2, 0, 0)-colorable, which improves the result of Chen et al. (Discrete Math 339:886–905, 2016) who proved that every planar graph without cycles of length 4 and 5 is (2, 0, 0)-colorable.



中文翻译:

最多不具有五个长度相邻周期的平面图是(2,0,0)-可着色的

如果可以将G的顶点集划分为子集\(V_ {1},V_ {2,则图形G\((d_ {1},d_ {2},\ ldots,d_ {k})\)是可着色的},\ ldots,V_ {K} \) ,使得所述子图\(G [V_ {I}] \)通过诱导\(V_ {I} \)具有最大的程度至多\(D_ {I} \)\(i = 1,2,\ ldots,k \)。新西伯利亚的猜想(Sib lektron Mat Izv 3:428–440,2006年)说,每个不带3个循环且与长度3或5的循环相邻的平面图都是3色的。Borodin等。(离散数学310:167–173,2010年)询问每个没有相邻循环长度最多为5的平面图是否都是3色的。Cohen-Addad等。(J Comb Theory Ser B 122:452–456,2017)对新西伯利亚的猜想和Borodin等人的问题都给出了否定的答案。张等。(离散数学339:3032–3042,2016)询问每个不带相邻长度为5的长度不连续的平面图是否可着色(1、0、0)。在本文中,我们表明每个不带相邻的长度为5的长度周期的平面图都是(2,0,0)可着色的,这改善了Chen等人的结果。(离散数学339:886–905,

更新日期:2020-08-28
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