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Let $f(G)$ give the number of perfect matchings of a graph $G$.

Consider set $\mathcal N_{2n}=\{0,1,2,\dots,n!-1,n!\}$.

Consider collection of all $2n$ vertex balanced bipartite graph to be $\mathcal G_{2n}$.

For every $m\in\mathcal N_{2n}$ is there a $G\in \mathcal G_{2n}$ such that $f(G)=m$?

Tony Huynh
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Turbo
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  • @how may numbers we miss? he gave a bound. – Turbo May 01 '16 at 00:42
  • @Yemon Choi Well, the question of how many numbers are missed was only edited in after I answered the initial question. Thanks for trying to get my answer accepted. For me, it's fine that the question was refined instead of a new question being asked. – Tony Huynh May 17 '16 at 22:47

2 Answers2

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No. For all $n \geq 3$, there is no $G \in \mathcal{G}_{2n}$ with $f(G)=n!-1$. To see this, note that $f(K_{n,n})=n!$ and $f(K_{n,n}-e)=n!-(n-1)!$, where $K_{n,n}-e$ is $K_{n,n}$ minus an edge. Thus, there are no graphs $G$ in $\mathcal{G}_{2n}$ with $n!-(n-1)!+1, n!-(n-1)!+2 \dots$, or $n!-1$ perfect matchings

Tony Huynh
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  • Interesting how many integers do we miss in $\mathcal N_{2n}$? – Turbo Apr 20 '16 at 08:42
  • $n=3$ $n!-(n-1)!+2=3!-2!+2=6$ for which we have an example. – Turbo Apr 20 '16 at 08:48
  • @Turbo The sequence runs from $n!-(n-1)!+1$ to $n!-1$. For $n=3$, we have $n!-(n-1)!+2>n!-1$, so the sequence stops before reaching $n!-(n-1)!+2$ and contains exactly one term (which happens to be $5$). Note that we miss around $(n-1)!$ values of $f(G)$ just from this observation. – Tony Huynh Apr 20 '16 at 09:35
  • @TonyHuynh http://mathoverflow.net/questions/239152/missing-count-in-number-of-perfect-matchings. –  May 18 '16 at 10:52
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Consider $A$ the order $n$ adjacency matrix which has 1 in the $i,j$th entry if there is an edge in the bipartite graph $G$ from top vertex $i$ to bottom vertex $j$, 0 if not. Then the permanent of $A$ counts the number of perfect matchings in $G$. One now sees that any such bipartite graph missing $k\lt n$ edges is a multiple of $(n-k)!$, which has some significance when $k$ is small. In particular when $k=2$, the number of matchings is a multiple of $(n-2)!$. So the feasible values near $n!$ are indeed few in number.

Elsewhere (Number of unique determinants for an NxN (0,1)-matrix) I talk about the determinant spectrum problem for order $n$ $0-1$ matrices; a similar problem for permanents can be and probably has been asked, for which I would like to see references.

Gerhard "Ask Me About Multilinear Functions" Paseman, 2016.05.18.

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Gerhard Paseman
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