Looking at the document Fundamentals of Computing Series, The Stable Marriage Problem.

Theorem 1.2.3 - page 12:

In a man-optimal version of stable matching, each woman has worst partner that she can have in any stable matching.


Suppose not. Let $M_0$ be the man-optimal stable matching, and suppose there is a stable matching $M’$ and a woman $w$ such that $w$ prefers $m = p_{M_0}(w)$ to $m' = p_{M'}(w)$ . But then $(m,w)$ blocks $M'$ unless $m$ prefers $p_{M'}(m)$ to $w = p_{M_0}(m)$, in contradiction of the fact that $m$ has no stable partner better than his partner in $M_0$.

I'm having trouble visualizing the definition of the problem and the proof (what is the contradiction?).

First, what is the question implying? From what I read and the fact that in most stable matching examples, all the women do not end up with the completely last person on their list ... So I'm a bit confused.

In the proof, here is what I am getting: in $M'$ we suppose $w$ prefers $m$ to $m'$. But then if there is a stable matching containing $(m,w)$ this would leave $w$ with her worst partner and that is a contradiction. Is this correct?

In addition, if $m$ did prefer $w'$ it would contradict that it is not his first pick ?

I'm new to computer science concepts so any help is appreciated.


2 Answers 2


The claim is not that every woman ends with the last man on her list. Rather, consider all stable matchings, and all partners of some woman $w$ in these stable matchings. Among them, pick the worst one (according to her view) $m$. Then in the man-optimal matching, $w$ is matched to $m$.

Now that you understand what they're trying to prove, read the proof again. (It couldn't have made sense before, because the claim as you stated it just isn't true.)

  • $\begingroup$ ok I see now, that you. In this case, is it true that with the G-S algorithm, there can only be 2 sets of stable matching (with stable matching set 1 with man-optimal and the other one coming from woman-optimal)? $\endgroup$
    – KJ.
    Jan 29, 2013 at 0:58
  • $\begingroup$ Definitely not. There could be exponentially many stable matchings. $\endgroup$ Jan 29, 2013 at 1:06
  • $\begingroup$ You can check the book by Gusfield and Irving, The Stable Marriage Problem: Structure and Algorithms, or the lecture notes by Knuth. $\endgroup$ Jan 29, 2013 at 1:12
  • $\begingroup$ does it depend on the order of men picked to propose? $\endgroup$
    – KJ.
    Jan 29, 2013 at 1:32
  • $\begingroup$ Any implementation of the Gale-Shapley algorithm produces a single stable matching. However, there are lots more stable matchings out there. $\endgroup$ Jan 29, 2013 at 2:52

There is a very nice lecture video on Youtube which describes all properties. I recommend you to watch is if you are new to the concept.


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