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I need to demonstrate that a graph that doesn't have odd disjunctive circuits is a five color graph. This is indeed for a homework. I need some suggestions on how to approach this problem. Any help is welcomed.

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I assume by "doesn't have odd disjunctive circuits" you that the graph "does not contain two vertex-disjoint odd circuits"? –  Chris Godsil Jan 22 '13 at 22:23
    
and "is a five colour graph" means its vertices can be coloured with five colours in such a way that adjacent vertices have different colours? –  Chris Eagle Jan 22 '13 at 22:26
    
@ChrisEagle Yes, that's what it means –  Bogdan Jan 22 '13 at 22:29
    
@ChrisGodsil Yes, it doesn't contain two vertex-disjoint odd circuits –  Bogdan Jan 22 '13 at 22:29
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1 Answer

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If there is an odd cycle, select a shortest one and remove all its vertices. The remaining graph is bipartite and can be coloured with $A$ and $B$. Put the removed cycle back in and colour it with $C$, $D$ and $E$. Why can there be no colour conflict?

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Is it because the shortest odd cycle is made out of 3 vertices and that cycle can be colored with 3 colors different from the 2 colored remaining bipartite graph? –  Bogdan Jan 22 '13 at 23:01
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