Upper bounds on the permanent of dense 0-1 matrix - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T00:41:18Z http://mathoverflow.net/feeds/question/64229 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64229/upper-bounds-on-the-permanent-of-dense-0-1-matrix Upper bounds on the permanent of dense 0-1 matrix Maciej S. 2011-05-07T19:27:25Z 2011-05-08T08:34:21Z <p>I am looking for some result useful in deriving following (conjecture?):</p> <p>Let $A$ be an $n\times n$ matrix with $0-1$ entries. Suppose, that exactly $k\leqslant n$ entries are equal to $0$. Then $\mathrm{Per} A \leqslant n!\left(1-\frac{k}{2n}\right)$</p> <p>Unfortunately, trying to derive this inequality from other estimates fails in general. For example, the inequality given in <a href="http://146.163.152.131/research/papers/sdrbounds.pdf" rel="nofollow">"A note on some upper bounds for permanents of (0,1)-matrices" </a> failed with $k=4$. </p> http://mathoverflow.net/questions/64229/upper-bounds-on-the-permanent-of-dense-0-1-matrix/64238#64238 Answer by Richard Stanley for Upper bounds on the permanent of dense 0-1 matrix Richard Stanley 2011-05-07T22:15:02Z 2011-05-08T01:15:20Z <p>In fact, the maximum is achieved when no two of the 0's are in the same row or column, giving the maximum value $$S=\sum_{i=0}^k (-1)^i {k\choose i}(n-i)!.$$ <em>Proof.</em> Suppose that there are two 0's in the same row, say the first. Then some row, say the second, has all 1's. Expand the permanent by the first two rows. One of the $2\times 2$ submatrices is $\pmatrix{0&amp;0\cr1&amp;1\cr}$. Changing this to $\pmatrix{1&amp;0\cr0&amp;1\cr}$ does not decrease any of the $2\times 2$ submatrices from the first two rows, so the permanent of the entire matrix does not decrease. We can iterate this procedure (possibly using columns instead of rows) until no two 0's are in the same row and column, so the proof follows.</p> <p>We now need to show that $S$ is at most $n!\left( 1 -\frac{k}{2n}\right)$. This is trivial for $k=1$, so assume $k>1$. By the Bonferroni inequalities (http://en.wikipedia.org/wiki/Boole's_inequality), $S$ is bounded by the first three terms, i.e., $$S\leq n!-k(n-1)!+{k\choose 2}(n-2)! = n!\left(1-\frac kn+\frac{{k\choose 2}}{n(n-1)}\right).$$ Since $k\leq n$ we have $$\frac{{k\choose 2}}{n(n-1)}\leq \frac{k}{2n},$$ and the result follows.</p> http://mathoverflow.net/questions/64229/upper-bounds-on-the-permanent-of-dense-0-1-matrix/64267#64267 Answer by Maciej S. for Upper bounds on the permanent of dense 0-1 matrix Maciej S. 2011-05-08T08:25:49Z 2011-05-08T08:34:21Z <p>Thanks for help. By the way, I have checked, that solutions of more general extremal problems, using exactly the same ideas (changing a row or column that contains more than one zero) one can found in the paper <a href="http://www.sciencedirect.com/science/article/B6WHS-4D7CYGG-DM/2/d6f318c578b8d1102b7ff9862ffaa44b" rel="nofollow">Maximum permanents of matrices of zeros and ones</a></p>