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May 10, 2023 at 4:06 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Apr 10, 2023 at 3:53 answer added Alejandro Quinche timeline score: -1
Mar 25, 2015 at 13:01 review Close votes
Mar 25, 2015 at 14:56
Mar 25, 2015 at 12:53 comment added Emil Jeřábek Btw, your previous question is here: mathoverflow.net/q/172909 . (The question is not the same, but the underlying confusion is.)
Mar 25, 2015 at 12:32 comment added joro @EmilJeřábek you might be right, thanks :) btw, for Q=2^k, you can compute the permanent modulo Q in O(n^4k), but this is another story.
Mar 25, 2015 at 11:55 comment added Emil Jeřábek You cannot test whether perm(B) is nonzero modulo Q. You can only test whether it is nonzero, and that does not tell you anything about A.
Mar 25, 2015 at 11:49 comment added joro @EmilJeřábek $perm(B) \not \equiv 0 \pmod{Q}$ implies $perm(A) \ne 0$.
Mar 25, 2015 at 11:42 comment added joro @EmilJeřábek I don't think this is deja vu :). $A$ is the main matrix and $perm(A)=X$ and $B$ is (0,1) with $perm(B)=0$. We have $0 \equiv X \pmod{Q}$. So $X=Q N$. Q is greater than the upper bound, $Q > X$.
Mar 25, 2015 at 11:00 comment added Emil Jeřábek I have a deja vu. Didn't you ask this already some time ago? The answer is still the same: $\mathrm{perm}(A)=0$ doesn't imply $\mathrm{perm}(B)=0$. In fact, the way the reduction works, the permanent of $B$ is always positive (and fairly large).
Mar 25, 2015 at 10:04 comment added joro @DimaPasechnik I am using digraph algorithms for ZP01 and have seen the reduction to (0,1) in other papers.
Mar 25, 2015 at 10:03 comment added Dima Pasechnik my understanding is that there is no good formula for permanent...
Mar 25, 2015 at 9:59 history asked joro CC BY-SA 3.0