arXiv subjects

Ankan Sadhu

Publications and source records attributed to Ankan Sadhu.

3 recordsLinked to original sources

arXiv2026

Albertson's Conjecture Holds for r at Most 26

Albertson conjectured that every graph with chromatic number r has crossing number at least cr(K_r). The conjecture was verified for r <= 12 by Albertson, Cranston and Fox, for r <= 16 by Bar'at and T'oth, for r <= 18 by Ackerman, and recently for r <= 24 by Cranston, who reduced the remaining cases r in {25, 26} to three orders. We settle those three orders, so that Albertson's Conjecture holds for all r <= 26. Only published results are used, and an appendix reproves the range 19 <= r <= 24 so that the case r <= 26 does not rest on unpublished work. We also show that if chi(G) = 27 and cr(G) < cr(K_27), then G has a 27-critical subgraph of order 53 or 54 whose complement is connected.

math.CO↗

arXiv2026

A congruence obstruction to Roman's bound for Zarankiewicz numbers

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'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'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) 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, "Zarankiewicz numbers near the triple system threshold"). 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.

math.CO↗

arXiv2026

An improved upper bound for the Zarankiewicz number z(43;2) <= 294 via structural rigidity, including hexary packing number pa(5;6) = 31

The Zarankiewicz number z(n;2) is the largest number of ones in an n x n zero-one matrix with no all-one 2x2 submatrix. At n=43, the Kovari-Sos-Turan bound, sharpened by Reiman, gives z(43;2) <= 301 with no rounding slack; equality would force a projective plane of order six, which Tarry ruled out in 1900. That fact alone says nothing about how far below 301 the truth lies. We prove z(43;2) <= 294. The argument is local and geometric. Writing D = 301 - E for the deficiency of a configuration with E ones, a crossing count paired with its dual forces every line to at most eight points and every point to at most eight lines whenever D <= 6. A deficiency count then produces a clean point of degree seven, collinear with all others, unless the configuration has both a line of eight points and a point of degree eight. The seven lines through a clean point partition the remaining 42 points into groups that every other line meets at most once, so the number of (line, missed group) incidences equals D. At D=6 this forces group sizes 6,6,6,6,6,6,6 or 7,6,6,6,6,6,5; the second forces a TD(4,6), and the first yields 32 blocks of a partial TD(5,6), whose refutation needs an unknown packing number. We determine it. pa(5;6) and pa(6;6) are listed as open in the Handbook of Combinatorial Designs at 30 <= N <= 34; a class holds at most six words, so at most one full class gives at most 6 + 5*5 = 31, and what was missing was refuting the two-full-class case, done here by exhaustive search. Hence pa(5;6) = pa(6;6) = 31. With no clean point, the configuration is rigid enough to complete to a projective plane of order six. With a known 290-one configuration, this gives 290 <= z(43;2) <= 294. We also derive z(43;2) <= 299 via Totten's classification, and rule out configurations with 290+ ones from deleting rows/columns of PG(2,7).

math.CO↗