Abstract

For a graph G , denote by f k ( G ) the smallest number of vertices that must be deleted from G so that the remaining induced subgraph has its maximum degree shared by at least k vertices. It is not difficult to prove that there are graphs for which already f 2 ( G ) ≥ n ( 1 − o ( 1 ) ) , where n is the number of vertices of G . It is conjectured that f k ( G ) = Θ ( n ) for every fixed k . We prove this for k = 2 , 3 . While the proof for the case k = 2 is easy, already the proof for the case k = 3 is considerably more difficult. The case k = 4 remains open. A related parameter, s k ( G ) , denotes the maximum integer m so that there are k vertex-disjoint subgraphs of G , each with m vertices, and with the same maximum degree. We prove that for every fixed k , s k ( G ) ≥ n / k − o ( n ) . The proof relies on probabilistic arguments.

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