In this paper, we prove that every graph contains a cycle intersecting all maximum independent sets. Using this, we further prove that every graph with stability number α is spanned by α disjoint cycles. Here, the empty set, the graph of order 1 and the path of order 2 are all considered as degenerate cycles.

We show that the total chromatic number of a simple k-chromatic graph exceeds the chromatic index by at most 18k ⅓ log ½ 3k.

A k×n array with entries from the q-letter alphabet {0, 1, …, q − 1} is said to be t-covering if each k × t submatrix has (at least one set of) qt distinct rows. We use the Lovász local lemma to obtain a general upper bound on the minimal number K = K(n, t, q) of rows for which a t-covering array exists; for t = 3 and q = 2, we are able to match the best-known such bound. Let Kλ = Kλ(n, t, q), (λ ≥ 2), denote the minimum number of rows that guarantees the existence of an array for which each set of t columns contains, amongst its rows, each of the qt possible ‘words’ of length t at least λ times. The Lovász lemma yields an upper bound on Kλ that reveals how substantially fewer rows are needed to accomplish subsequent t-coverings (beyond the first). Finally, given a random k × n array, the Stein–Chen method is employed to obtain a Poisson approximation for the number of sets of t columns that are deficient, i.e. missing at least one word.

We describe a very simple method of randomly permuting the cube {0, 1}n such that the sample space is very small, but, given any m distinct points in {0, 1}n, the images of those points under the random permutation are approximately uniformly distributed over all sequences of m distinct points.

In this paper, asymptotical estimates of the form Rn(1+o(1)) for various classes of planar valency-restricted Eulerian maps are established. It follows, in particular, that ‘almost all’ (as n → ∞) n-edged planar Eulerian maps have n/3 (1+o(1)) vertices. A brief survey of known asymptotical results (a table of values of R) for various classes of planar maps is also presented.

We show distributional results for the length of the longest matching consecutive subsequence between two independent sequences A1, A2, …, Am and B1, B2, …, Bn whose letters are taken from a finite alphabet. We assume that A1, A2, … are i.i.d. with distribution μ and B1, B2, … are i.i.d. with distribution ν. It is known that if μ and v are not too different, the Chen–Stein method for Poisson approximation can be used to establish distributional results. We extend these results beyond the region where the Chen–Stein method was previously successful. We use a combination of ‘matching by patterns’ results obtained by Arratia and Waterman [1], and the Chen–Stein method to show that the Poisson approximation can be extended. Our method explains how the matching is achieved. This provides an explanation for the formulas in Arratia and Waterman [1] and thus answers one of the questions posed in comment F19 in Aldous [2]. Furthermore, in the case where the alphabet consists of only two letters, the phase transition observed by Arratia and Waterman [1] for the strong law of large numbers extends to the distributional result. We conjecture that this phase transition on the distributional level holds for any finite alphabet.

We consider transition operators P on a countable set, which are reversible, irreducible and invariant under a group G of permutations of X with compact point stabilizers. We relate the computation of the spectral radius (norm) of P with the spectral radii of certain matrices defined over the factor set G\X. In various cases, this allows easy computation of the norm of P.

In 1964 Dirac conjectured that every graph with n vertices and at least 3n − 5 edges contains a subdivision of K5 We prove a weakened version with 7/2;n − 7 instead of 3n − 5. We prove that, for any prescribed vertex νo, the subdivision can be found such that νo is not one of the five branch vertices. This variant of Dirac's problem, which was suggested by the present author in 1973, is best possible in the sense that 7/2;n − 7 cannot be replaced by 7/2;n − 15/2;.