AI 中文总结
本文研究增强Zarankiewicz问题，证明从最大C4-自由矩阵开始可能降低最优值，给出z_A与z_L的差距下界及z_A的精确二阶渐近，并确定密度阈值c≥1/12时有限密度趋于1/3。
AI 中文摘要
我们研究了增强Zarankiewicz问题，在该问题中，将不相交的单元格对添加到没有全1的$2\times2$子矩阵的二元矩阵中。这些对必须满足相容性条件，目标是对每个原始占用单元格和每个添加的对各计数一次。我们表明，从最大$C_4$-自由矩阵开始可能降低最终最优值，回答了Qi、Cui和Xu的一个问题。设${z_A}(m,n)$为所有$C_4$-自由初始矩阵上的最优值，${z_L}(m,n)$为初始矩阵必须具有最大占用单元格数时的最优值。当$n\to\infty$且$n\le m=o(n^2)$时，我们证明\\[ {z_A}(m,n)-{z_L}(m,n)\ge\left(\frac1{30}-o(1)\right)mn \\]并确定尖锐的二阶项：\\[ {z_A}(m,n)=\frac{mn}{3}+\left(\frac1{\sqrt6}+o(1)\right)n\sqrt m. \\]一个显式构造在$m=n=1893$处给出了分离。我们还找到了一个尖锐的密度阈值：当$n\to\infty$且$m/n^2\to c>0$时，有限密度${z_L}(m,n)/(mn)$趋于$1/3$当且仅当$c\ge1/12$。证明结合了密度和稳定性估计、组合构造以及一个精确的多项式证书。
英文摘要
We study the augmented Zarankiewicz problem, in which disjoint pairs of cells are added to a binary matrix with no all-one $2\times2$ submatrix. The pairs must satisfy compatibility conditions, and the objective counts each original occupied cell and each added pair once. We show that starting with a maximum $C_4$-free matrix can lower the final optimum, answering a question of Qi, Cui, and Xu. Let ${z_A}(m,n)$ be the optimum over all $C_4$-free initial matrices, and ${z_L}(m,n)$ the optimum when the initial matrix must have the maximum number of occupied cells. As $n\to\infty$ with $n\le m=o(n^2)$, we prove \[ {z_A}(m,n)-{z_L}(m,n)\ge\left(\frac1{30}-o(1)\right)mn \] and determine the sharp second-order term: \[ {z_A}(m,n)=\frac{mn}{3}+\left(\frac1{\sqrt6}+o(1)\right)n\sqrt m. \] An explicit construction gives a separation at $m=n=1893$. We also find a sharp density threshold: when $n\to\infty$ and $m/n^2\to c>0$, the limited density ${z_L}(m,n)/(mn)$ tends to $1/3$ if and only if $c\ge1/12$. The proofs combine density and stability estimates, combinatorial constructions, and an exact polynomial certificate.
Comments41 pages, 1 figure; exact computational certificates and Lean 4 companion files included
