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Given a $0/1$ square matrix, the permanent and determinant modulo $2^k$ is in $\oplus P$ and $\oplus L$ respectively for any fixed $k$. In fact both are in $\oplus L$ (in fact in $\oplus SPACE(k^2\log n)$ by https://mbraverm.princeton.edu/files/planarCounting.pdf.

Have there been any improvements?

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  • $\begingroup$ I don't think permanent mod $2$ is complete for $\oplus P$ since it is in $P$. I was thinking $\oplus SPACE(f(k)+\log(n))$ form where $f(k)=O(k^c)$ where $c<3$ (perhaps $c=1$)? Or $\oplus SPACE(f(k)\log(n))$ form where $f(k)=O(k^c)$ where $c<2$ (perhaps $c=1$)? $\endgroup$
    – Turbo
    May 9, 2023 at 19:42
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    $\begingroup$ Note if space is not a constraint we can get the mod $2^k$ result in $n^{O(k)}$ time and I would expect at least an $\oplus SPACE(k\log n)$ algorithm to exist! $\endgroup$
    – Turbo
    May 9, 2023 at 20:57
  • $\begingroup$ @JoshuaGrochow I think not since by the result in the paper permanent mod $2^2$ is in $\oplus SPACE(\log n)$ and hence in $\oplus L$. $\endgroup$
    – Turbo
    May 9, 2023 at 22:55

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I don't think so. Theorem 5.1 of the linked paper shows that, for any fixed k, permanent mod $2^k$ is $\mathsf{\oplus L}$-complete. (I believe the same is true for det mod $2^k$.) So it won't be in a smaller class without collapsing the two classes.

Regarding parametrized complexity, Curticapean and Xia showed that permanent modulo $2^k$ is $\mathsf{\oplus W[1]}$-hard. So an algorithm of the form you mentioned in the comments, in $\mathsf{\oplus SPACE}(f(k) + \log n)$ would put it in $\mathsf{\oplus TIME}(g(k) poly(n))$.

In terms of runtime, assuming $\oplus ETH$, they also show a lower bound on the runtime of $n^{\Omega(k / \log k)}$. So one cannot do much better than what is known. (It'd be interesting if someone had closed that $\log k$ gap in the exponent, but I haven't heard of this.)

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