A congruence obstruction to Roman's bound for Zarankiewicz numbers

Abstract

Let z(m,n;s,t) be the largest number of ones in an m x n zero-one matrix with no s x t all-ones submatrix. Roman&#39;s 1975 inequality remains the best general upper bound for s>=3, but it is not attained on a large part of the range just below the design threshold T=(t-1)C(m,s)/(s+1). The proof has two steps. First, for n=T-c with 1<=c<=sT/(s+2), Roman&#39;s bound equals the elementary counting bound (s+1)(T-c)+floor(2c/s), adding nothing beyond a budget inequality and convexity. Second, attainment forces all but at most one column to have size s+1 or s+2; each such column has a point lying in a number of s-sets divisible by d=gcd(s,C(s+1,2)), pinning the leftover coverage there to a single residue mu mod d, which a global count rules out. With r=c mod s and slack sigma(r) depending only on s, we prove z(m,T-c;s,t) <= Rom(m,T-c)-1 whenever 1<=sigma(r)<d and m*mu != s*sigma(r), or mu != 0 and m*mu > s*sigma(r). The first case is an odd-s phenomenon confined to one residue class, giving order-m^s values of n; the second needs mu != 0 but covers the whole interval once m exceeds a threshold depending only on s and mu. For s=4, t=2, m=28 it covers all 2730 values of n; when c=(s+1)/2 and an s-(m,s+1,t-1) design exists, z=(s+1)(T-c) exactly. Finally we relate the obstruction to linear programming: the relaxation over all 2^m subset variables collapses, under symmetrisation, to the counting bound, so no linear relaxation of the covering constraints alone can beat the bound of Chen, Horsley, and Mammoliti (arXiv:2310.12685, &#34;Zarankiewicz numbers near the triple system threshold&#34;). For the refined program of Davies, Gill, and Horsley, its optimum is still attained at the Roman vertex on an explicit sub-family, and we record where their program does better.