In the stable marriage problem, is it possible to find an instance with $2^{n -1}$ stable matchings when $n$ is a power of 2 (or just even)? If yes, how? I know how to build an instance in which $2^{n/2}$ stable matchings can be obtained, but was wondering if the aforementioned number of stable matchings ($2^{n -1}$) can be obtained too.


1 Answer 1


Yes. Thurber showed [1,Theorem 5] that for all $n\geq 1$, the maximum number of stable matchings is at least $\frac{(2.28)^n}{(1+\sqrt{3})^{1+\log_2 n}}$.

If I'm not mistaken this is strictly greater than $2^n$ for all $n\geq 52$ (and of course asymptotically it's an exponential factor more).

[1]Thurber, Edward G., Concerning the maximum number of stable matchings in the stable marriage problem, Discrete Math. 248, No. 1-3, 195-219 (2002). ZBL0997.05002.

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    $\begingroup$ He also mentions that examples meeting the specific question of the OP (instances with $2^{n-1}$ stable matchings when $n$ is a power of $2$) were previously constructed in P. W. Irving, R. Leather, The complexity of counting stable marriages, SIAM J. Comput. 15 (1986), 655--667. $\endgroup$ Commented Sep 16, 2022 at 8:16
  • $\begingroup$ I've heard of this gem of a paper (soon to be updated) giving a $3.55^n$ upper bound: arxiv.org/abs/2011.00915 $\endgroup$
    – domotorp
    Commented Sep 18, 2022 at 18:25
  • $\begingroup$ @domotorp Nice! I had read the proof of the previous exponential upper bound (with the huge base) but was not aware of this improvement. I'll give this one a read as well, curious to see how you achieved that. $\endgroup$
    – Tassle
    Commented Sep 18, 2022 at 19:28
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    $\begingroup$ You can also watch it: youtube.com/… $\endgroup$
    – domotorp
    Commented Sep 18, 2022 at 19:56

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