
AbstractA graph with n vertices that contains no triangle and no 5‐cycle and minimum degree exceeding n/4 contains an independent set with at least (3n)/7 vertices. This is best possible. The proof proceeds by producing a homomorphism to the 7‐cycle and invoking the No Homomorphism Lemma. For k ≥ 4, a graph with n vertices, odd girth 2k+1, and minimum degree exceeding n/(k+1) contains an independent set with at least kn/(2k+1) vertices; however, we suspect this is not best possible. © 1993 John Wiley & Sons, Inc.
Extremal problems in graph theory, independent set, cycle, triangle, graph homomorphisms, Paths and cycles, no homomorphism lemma, minimum degree
Extremal problems in graph theory, independent set, cycle, triangle, graph homomorphisms, Paths and cycles, no homomorphism lemma, minimum degree
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