English

Lower bounds on maximal determinants of binary matrices via the probabilistic method

Combinatorics 2016-10-26 v6

Abstract

Let D(n)D(n) be the maximal determinant for n×nn \times n {±1}\{\pm 1\}-matrices, and R(n)=D(n)/nn/2{\mathcal R}(n) = D(n)/n^{n/2} be the ratio of D(n)D(n) to the Hadamard upper bound. We give several new lower bounds on R(n){\mathcal R}(n) in terms of dd, where n=h+dn = h+d, hh is the order of a Hadamard matrix, and hh is maximal subject to hnh \le n. A relatively simple bound is R(n)(2πe)d/2(1d2(π2h)1/2)   for all   n1.{\mathcal R}(n) \ge \left(\frac{2}{\pi e}\right)^{d/2} \left(1 - d^2\left(\frac{\pi}{2h}\right)^{1/2}\right) \;\text{ for all }\; n \ge 1. An asymptotically sharper bound is R(n)(2πe)d/2exp(d(π2h)1/2+  O(d5/3h2/3)).{\mathcal R}(n) \ge \left(\frac{2}{\pi e}\right)^{d/2} \exp\left(d\left(\frac{\pi}{2h}\right)^{1/2} + \; O\left(\frac{d^{5/3}}{h^{2/3}}\right)\right). We also show that R(n)(2πe)d/2{\mathcal R}(n) \ge \left(\frac{2}{\pi e}\right)^{d/2} if nn0n \ge n_0 and n0n_0 is sufficiently large, the threshold n0n_0 being independent of dd, or for all n1n\ge 1 if 0d30 \le d \le 3 (which would follow from the Hadamard conjecture). The proofs depend on the probabilistic method, and generalise previous results that were restricted to the cases d=0d=0 and d=1d=1.

Keywords

Cite

@article{arxiv.1402.6817,
  title  = {Lower bounds on maximal determinants of binary matrices via the probabilistic method},
  author = {Richard P. Brent and Judy-anne H. Osborn and Warren D. Smith},
  journal= {arXiv preprint arXiv:1402.6817},
  year   = {2016}
}

Comments

37 pages, 2 tables, 59 references. Added some references in v2, fixed typos in v3 and v4, added footnote 2 on page 13 re proof of Lemma 12 in v5, revised footnote 2 in v6

R2 v1 2026-06-22T03:16:53.873Z