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Let $A$ be an $m$ by $n$ $(0,1)$-matrix. For $1\leq i \leq m$ and $1\leq j \leq n$, let $f(A,i,j)$ be the number of entries in $A$ not in row $i$, not in column $j$, and not equal to $a_{ij}$.

I would like a proof or counterexample to the following conjecture:

If $A$ is not all 1's or all 0's, then there exist $i$ and $j$ such that $f(A,i,j)\geq \frac{(m-1)(n-1)-1}{2}$.

Example 1: For $A=\begin{bmatrix} 1 & 0 & 1 & 0\\0 & 1 & 0 & 1 \\1 & 0 & 1 & 0\\0 & 1 & 0 & 1 \\\end{bmatrix}$, we have $f(A,1,1)=4\geq\frac{3\cdot3-1}{2}$.

Example 2: For $A=\begin{bmatrix} 0 & 1 & 0 & 0 & 1\\1 & 0 & 0 & 1 & 0 \\1 & 0 & 1 & 0 & 1\\0 & 0 & 0 & 0 & 1\\\end{bmatrix}$, we have $f(A,1,2)=6\geq\frac{3\cdot 4-1}{2}$.

Jeremy K
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  • It suffices to show that the average $(\sum_{i,j}f(A,i,j))/(mn)$ satisfies the inequality because then the maximum will, too. Also notice that each (0,1) pair from different rows and columns contribute twice to the sum. – RobPratt Aug 09 '19 at 05:36
  • @RobPratt unfortunately this stronger statement doesn't seem to be true (assuming I did my coding correctly). – user113102 Aug 09 '19 at 15:01
  • Yes, I just found the same. This average can be calculated by solving a minimum cut problem on a graph with one node per matrix entry $(i,j)$ and an edge between two entries that do not share a row or column. For $m \le n \le 10$, the average is $2(m-1)(n-1)/(mn)$. – RobPratt Aug 09 '19 at 16:01
  • @RobPratt Looking at the average does seem like the right approach. Then perhaps induction on $m$ and $n$ may do the trick. I'll post a solution if I get it to work. Thanks! – Jeremy K Aug 09 '19 at 18:06
  • I (annoyingly) can't even do this for the 2x$n$ case. Induction seems kinda tricky. Admittedly I'm trying some naive things. – user113102 Aug 09 '19 at 19:21
  • Any $A$ with only one 1 or only one 0 has an average $f(A,i,j)$ value of $2(m-1)(n-1)/(mn)$, which is not always larger than the desired lower bound. – RobPratt Aug 09 '19 at 19:38

1 Answers1

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For $m \le n \le 10$, the following stronger conjecture is true: $$\min_A \max_{i,j} f(A,i,j) = \left\lceil \frac{(m-1)(n-1)-1}{2} \right\rceil$$

You can solve this problem via mixed integer linear programming by minimizing $z$ subject to \begin{align} 1 \le \sum_{i,j} A_{i,j} &\le mn-1 \\ \sum_{k \not= i} \sum_{\ell \not=j} A_{k,\ell} - z &\le (m-1)(n-1)A_{i,j} \\ \sum_{k \not= i} \sum_{\ell \not=j} (1-A_{k,\ell}) - z &\le (m-1)(n-1)(1-A_{i,j}) \\ A_{i,j} &\in\{0,1\} \end{align} The first constraint avoids all 0 or all 1. The second constraint enforces $z \ge f(A,i,j)$ if $A_{i,j}=0$ and is otherwise redundant. The third constraint enforces $z \ge f(A,i,j)$ if $A_{i,j}=1$ and is otherwise redundant.

RobPratt
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