Abstract

Extending a theorem of Alon, we prove a conjecture of Katchalski that every graph of order n and minimal degree at least n/k > 1 contains a cycle of length at least n/(k - 1). The result is best possible for all values of n and k (2 ≤ k < n).

Introduction

A well-known result of Erdös and Gallai [4] states that, for n ≥ k ≥ 3, a graph of order n and size greater than ½(k - 1)(n - 1) has circumference at least k, that is, it contains a cycle of length at least k. According to Dirac's [3] classical theorem, every graph of order n ≥ 3 and minimal degree at least ½n is Hamiltonian. What can one say about the circumference of a graph of order n and minimal degree at least d ≥ 2? Recently Alon [1] came close to giving a complete answer to this question when he proved that for 2 ≤ k < n every graph of order n and minimal degree at least n/k has circumference at least [n/(k - 1)]. Our aim here is to improve on this slightly, namely to show that the assertion holds without the integer sign, as conjectured by Katchalski. Although this seems to need a surprising amount of work, we feel it is worth it since the new result is best possible for all values of n and k (2 ≤ k < n) and implies a complete answer to the question above concerning the minimal circumference of a graph of order n and minimal degree d ≥ 2.