$$(1,0,0)$$ ( 1, 0, 0 ) -Colorability of planar graphs without prescribed short cycles

Yuehua Bu, Jinghan Xu and Yingqian Wang ()
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Yuehua Bu: Zhejiang Normal University
Jinghan Xu: Zhejiang Normal University
Yingqian Wang: Zhejiang Normal University

Journal of Combinatorial Optimization, 2015, vol. 30, issue 3, No 14, 627-646

Abstract: Abstract Let $$d_1, d_2,\ldots ,d_k$$ d 1 , d 2 , … , d k be $$k$$ k non-negative integers. A graph $$G$$ G is $$(d_1,d_2,\ldots ,d_k)$$ ( d 1 , d 2 , … , d k ) -colorable, if the vertex set of $$G$$ G can be partitioned into subsets $$V_1,V_2,\ldots ,V_k$$ V 1 , V 2 , … , V k such that the subgraph $$G[V_i]$$ G [ V i ] induced by $$V_i$$ V i has maximum degree at most $$d_i$$ d i for $$i=1,2,\ldots ,k$$ i = 1 , 2 , … , k . Let $$\digamma $$ ϝ be the family of planar graphs with cycles of length neither 4 nor 8. In this paper, we prove that a planar graph in $$\digamma $$ ϝ is $$(1,0,0)$$ ( 1 , 0 , 0 ) -colorable if it has no cycle of length $$k$$ k for some $$k\in \{7,9\}$$ k ∈ { 7 , 9 } . Together with other known related results, this completes a neat conclusion on the $$(1,0,0)$$ ( 1 , 0 , 0 ) -colorability of planar graphs without prescribed short cycles, more precisely, for every triple $$(4,i,j)$$ ( 4 , i , j ) , planar graphs without cycles of length 4, $$i$$ i or $$j$$ j are $$(1,0,0)$$ ( 1 , 0 , 0 ) -colorable whenever $$4

Keywords: Planar graph; Steinberg conjecture; Improper coloring; Cycle (search for similar items in EconPapers)
Date: 2015
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DOI: 10.1007/s10878-013-9653-5

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