Every Planar Graph With Neither Adjacent 3-cycles Nor 5-cycles Is (2,0,0)-colorable

Let G=(V(G), E(G)). A k-coloring of G is a mapping φ:V(G)â†'{1,2,..., k) satisfies that for every i,1≤i≤k, G[Vi] is edgeless, where G[Vi] denotes the subgraph induced by the vertices colored i. We say that a graph G is k-colorable if there exists a k-coloring of G; A (c1, C2,..., ck)-coloring of G is a mapping φ:V(G) â†'{1,2,..., k} satisfies that for every i,1< i< k, the maximum degree of G[Vi]is at most ci. A graph G is (ci, C2,..., Ck)-colorable if there exists a (c1, C2,..., ck)-coloring of G.In recent years, the classic four-color problem has been researched by many scholars. In 1977, Appel and Haken proved it through accumulation with the help of computer. However, the proof of this problem purely by elegant math-analysis has been still unfounded. In order to find a proof purely by elegant math-analysis to solve this problem, many experts and scholars start to do some profound research on the three-color problem of planar graph. In 2003, Borodin and Raspaud conjectured that every planar graph with neither 5-cycles nor adjacent 3-cycles is 3-colorable. We proved in this paper that every planar graph with neither adjacent 3-cycles nor 5-cycles is (2,0,0)-colorable.In chapter 1, we introduce the background and the significance of the research, and also the problem we have solved in this paper systematically.In chapter 2, we introduce some basic concepts and symbols in detail.In chapter 3, we find some reducible configurations of G and give the detailed proof of them.In chapter 4, we make the discharging rules, and prove that the final charge of vertices and faces are positive in G.In chapter 5, we give a summary of the paper and make a prospect of the research in 3-coloring in the future.