{"paper":{"title":"A new approach to the results of K\\\"ovari, S\\'os, and Tur\\'an concerning rectangle-free subsets of the grid","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"math.CO","authors_text":"Jacob Manske, Jeremy F. Alm","submitted_at":"2012-06-06T02:17:23Z","abstract_excerpt":"For positive integers $m$ and $n$, define $f(m,n)$ to be the smallest integer such that any subset $A$ of the $m \\times n$ integer grid with $|A| \\geq f(m,n)$ contains a rectangle; that is, there are $x\\in [m]$ and $y \\in [n]$ and $d_{1},d_{2} \\in \\mathbb{Z}^{+}$ such that all four points $(x,y)$, $(x+d_{1},y)$, $(x,y+d_{2})$, and $(x+d_{1},y+d_{2})$ are contained in $A$. In \\cite{kovarisosturan}, K\\\"ovari, S\\'os, and Tur\\'an showed that $\\dlim_{k \\to \\infty}\\dfrac{f(k,k)}{k^{3/2}} = 1$. They also showed that whenever $p$ is a prime number, $f(p^{2},p^{2}+p) = p^{2}(p+1)+1$. We recover their a"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1206.1107","kind":"arxiv","version":2},"verdict":{"id":null,"model_set":{},"created_at":null,"strongest_claim":"","one_line_summary":"","pipeline_version":null,"weakest_assumption":"","pith_extraction_headline":""},"references":{"count":0,"sample":[],"resolved_work":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57","internal_anchors":0},"formal_canon":{"evidence_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"author_claims":{"count":0,"strong_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"builder_version":"pith-number-builder-2026-05-17-v1"}