Approximating Maximum Independent Sets by Excluding Subgraphs 1

Limitation results for odd cyclesOur next goal is to show that our cycle-based algorithm is close to optimal for graphs withindependence ratio near 1 2 .We need the following structural result on graphs without short odd cycles.Theorem 6 For every positive integer k, if (H) 1 + 1length 2k 0 1 or less, then i(H) k2k+1 .4k+2, and H contains no odd cycles ofProof. By induction on the number of vertices in H. If there is a vertex v of degree 0 or 1, thenremove v and its neighbor from the graph. By induction, the remaining graph has independencekratio at least . But adding v to the largest independent set of the remaining graph shows2k+1kthat H itself has independence ratio greater than . 2k+1Thus, assume that every vertex has degree 2 or more. Suppose there is a cycle passingthrough only vertices of degree exactly 2. Since H has no odd cycles of length 2k 0 1 or less,kthe independence ratio of this cycle is at least . Then we could apply induction to the2k+1remainder of the graph and be nished.Thus, assume there are no such cycles. Let H 0 be the subgraph induced by the vertices ofdegree exactly 2. The subgraph H 0 must be the disjoint union of paths, so i(H 0 ) 1. Since2(H) 1 + 1 , the subgraph H 0 contains at least a fraction 1 0 1 of all the vertices.4k+2 2k+1Therefore i(H) is at least 1(1 0 1 ) = k , which completes the proof.2 2k+1 2k+1Finally, we can prove the following limitation result.Corollary 3 For every positive integer k, if i(H)