Coloring Graphs with Few Odd Cycles: 11086

Abstract

Solution by Richard Stong, Rice University, Houston, TX. We prove by induction on the number of vertices that if no vertex of G lies on (?J) odd cycles, then G has a proper /c-coloring (that is, x (G) < k). The claim is trivial with at most k vertices. For a vertex v of G the induction hypothesis yields a proper /c-coloring of G ? v. Since there are (??) pairs of colors and at most (*) ? 1 odd cycles through v, there is a pair of colors, say red and blue, such that no odd cycle through v joins v to both a red neighbor and a blue neighbor. Let G' be the subgraph of G ? v induced by the red and blue vertices. If any component of G' contains both a red neighbor and a blue neighbor of v, then adding v to the path connecting them in G' forms an odd cycle in G joining v to red and blue neighbors, contradicting our choice of colors. Therefore, switching red and blue in each component of G' containing a red neighbor of v removes red from the neighborhood of v. Now coloring v red extends the proper ^-coloring to G.