A graph G is (d1,d2,…,dk)\documentclass[12pt]{minimal}
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\begin{document}$$(d_{1},d_{2},\ldots ,d_{k})$$\end{document}-colorable if the vertex set of G can be partitioned into subsets V1,V2,…,Vk\documentclass[12pt]{minimal}
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\begin{document}$$V_{1}, V_{2},\ldots , V_{k}$$\end{document} such that the subgraph G[Vi]\documentclass[12pt]{minimal}
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\begin{document}$$G[V_{i}]$$\end{document} induced by Vi\documentclass[12pt]{minimal}
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\begin{document}$$V_{i}$$\end{document} has maximum degree at most di\documentclass[12pt]{minimal}
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\begin{document}$$d_{i}$$\end{document} for i=1,2,…,k\documentclass[12pt]{minimal}
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\begin{document}$$i = 1, 2, \ldots , k$$\end{document}. 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.
