Defective 2-colorings Of Planar Graphs

Firstly,every planar graph is simple graph,for any planar graph G,the set of vertex denoted by V(G),the set of edge denoted by E(G).A k-coloring of a graph G is a mapping ?:E(G)?{1,2,...,k},such that ?(?)?(?)for every pair of vertex.The vertex-chromatical number of a graph G is the least integer k such that G has a k-vertex-coloring,denoted by ?(G).In this paper,we prove this result:for any planar graph G,without the distance of k-cycles are at least 3(k=3,4)and 6-cycles is(3,6)-colorable.To prove this theorem,All graphs mentioned in this paper are finite simple planar graphs.Let G be a counterexample such that|V(G)|+|E(G)|is minimized.If we can prove that the smallest counterexample does not exist,then our theorem can prove it.Prove that the theorem is mainly divided into two parts:The.first part is to determine the structure of graph G by proving that the structure of graph G does not exist.The second part define an initial weight function ?(?)= 2d(?)-6 for each??V(G),and ?(f)=d(f)-6 for each f? F(G).From the Euler's formula,we obtain (?).We will design several discharging rules and reassign weight without change in the sum of weight.Then we arrive at a new weight func-tion ?*(x)and show that for each (?).This leads to the following contradiction:(?)The contradiction shows the nonexistence of G,which completes the proof of Theo-rem.