计算多个二分图极值问题的精确解，验证了关键边界情况的上界。
Exact Zarankiewicz Values On Two Finite Frontier Slices
- 通过证书驱动的计算机辅助证明，解决多个Zarankiewicz问题
- 给出Z(12,n,3,3) = 6n（18≤n≤22）等7个精确值和1个区间
- 适合组合数学与极值图论研究者阅读
Zarankiewicz数Z(m,n,s,t)是二分图中不含Ks,t子图时的最大边数。本文通过基于证书的计算机辅助方法，给出两个有限切片及邻近前沿的精确值：Z(12,n,3,3) = 6n（18 ≤ n ≤ 22），Z(13,22,3,3) = 137，Z(13,18,3,3) = 116，Z(14,18,3,3) = 124，Z(15,18,3,3) = 132，Z(14,17,3,3) = 118，Z(15,17,3,3) = 126，且132 ≤ Z(16,17,3,3) ≤ 133。核心为12×18与13×18的轨道证书，排除所有更高边数的假设矩阵。删除引理与显式见证者解决四个邻近单元，而16×17结果因仅确认132下界与已发表133上界，故以区间形式报告。13×22证明通过约简至83个度序列，理性分离77个，其余六例经标记行同余、枚举、模Gram测试与精确Farkas证书排除。所有结论均用标准库Python与精确整数/有理数运算重演，浮点优化仅用于发现证书。
原文摘要 · Abstract (English)
The Zarankiewicz number Z(m,n,s,t) is the maximum number of edges in a bipartite graph with parts of orders m and n containing no copy of Ks,t. We give one combined, certificate-based computer-assisted proof for two finite slices and a corrected neighboring frontier: Z(12,n,3,3) = 6n (18 <= n <= 22), Z(13,22,3,3) = 137, Z(13, 18, 3, 3) = 116, Z(14, 18, 3, 3) = 124, Z(15,18,3,3) = 132, Z(14, 17, 3, 3) = 118, Z(15, 17, 3, 3) = 126, 132 <= Z(16,17,3,3) <= 133. The load-bearing new upper bounds are the exact 12 x 18 and 13 x 18 certificate packages. Their orbit certificates exclude every hypothetical matrix at the next edge count. Deletion lemmas and explicit witnesses close four neighboring cells, while the 16 x 17 entry is deliberately reported as an interval because only its 132-edge lower witness and the published 133 upper bound are certified here. Separately, the 13 x 22 proof excludes 138 ones by reducing to 83 degree profiles, rationally separating 77 of them, and eliminating the remaining six by marked-row congruences, leave enumeration, modular Gram tests, and exact Farkas certificates. All accepted claims are replayed by standard-library Python and exact integer/rational arithmetic; floating-point optimization is used only to discover certificates.
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