Abstract
For a graph G, denote by fk (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 f2 (G) ≥ sqrt(n) (1 - o (1)), where n is the number of vertices of G. It is conjectured that fk (G) = Θ (sqrt(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, sk (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, sk (G) ≥ n / k - o (n). The proof relies on probabilistic arguments.
Original language | English |
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Pages (from-to) | 742-747 |
Number of pages | 6 |
Journal | Discrete Mathematics |
Volume | 310 |
Issue number | 4 |
DOIs | |
State | Published - 28 Feb 2010 |
Keywords
- Induced subgraph
- Maximum degree
ASJC Scopus subject areas
- Theoretical Computer Science
- Discrete Mathematics and Combinatorics