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The permanent of an $n$-by- $n$ matrix $A=\left(a_{i j}\right)$ is defined as $$ \operatorname{perm}(A)=\sum_{\sigma \in S_{n}} \prod_{i=1}^{n} a_{i, \sigma(i)} $$ The sum here extends over all elements $\sigma$ of the symmetric group $S_{n}$ i.e. over all permutations of the numbers $1,2, \ldots, n$. $$ \operatorname{perm}\left(\begin{array}{ll} a & b \\ c & d \end{array}\right)=a d+b c $$ The best known general exact algorithm of computing the permanent is due to H.J. Ryser. Ryser's formula can be evaluated using $\mathrm{O}\left(2^{n-1} n^{2}\right)$ arithmetic operations $$ \operatorname{perm}(A)=(-1)^{n} \sum_{S \subseteq\{1, \ldots, n\}}(-1)^{|S|} \prod_{i=1}^{n} \sum_{j \in S} a_{i j} $$

Is there any method or algorithm to detect a permanent of a square matrix $A(1,-1,0)$ with large even order(1000-1600) zero or nonzero by (super) computer? ( for instance $\left.A_{1000 \times 1000}, A_{1600 \times 1600} \cdot\right)$. It would be better if it could be calculated out by (super) computer. Thanks in advance.

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    $\begingroup$ Isn't this #P-complete? $\endgroup$ Commented Oct 4, 2021 at 2:34
  • $\begingroup$ Computing the permanent of a (0,1)-matrix is #P-complete. If given a matrix A(1,-1,0), can we detect it zero or nonzero without computing it out by some methods? $\endgroup$ Commented Oct 4, 2021 at 2:42
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    $\begingroup$ We can sort out some cases by noticing that pemanent and determinant are congruent modulo 2. So, if determinant is odd, then permanent must be nonzero. $\endgroup$ Commented Oct 4, 2021 at 4:17
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    $\begingroup$ Here's a relevant answer from the cstheory stack exchange. cstheory.stackexchange.com/questions/32885/… Of course, this doesn't answer your question. $\endgroup$ Commented Oct 4, 2021 at 4:50
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    $\begingroup$ Also, if permanent of the 0-1 matrix with elements of $A$ taken by absolute value is zero (which is easy to test), then permanent of $A$ is zero as well. $\endgroup$ Commented Oct 4, 2021 at 12:46

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