Notes On The Stable Matching Problem

• 0. The main loop of the Gale-Shapley algorithm is configured to continue while
  
  "some man is unmatched and hasn't proposed to every woman."
  
  We can write this a little more formally as
  
  "∃ man m | m is unmatched and m has not proposed to every woman."
  
  The logical negative of that statement is the condition that makes the loop stop, and it can be expressed as
  
  "∀ man m | m is engaged or m has proposed to every woman."
  
  
• 1. In other words, the stopping condition of the main loop is this:
  
  Every man m is either engaged, or m has proposed to every woman (or both).
  
  
• 2. In each iteration of the loop, there is a proposal of a man to a woman. Because each man goes down his preference list, each proposal is new, meaning that the same man will never again propose to the same woman. If there are N men and N women, there cannot be more than N² of these new proposals during the execution of the algorithm, simply because N² is how many proposals there will be if there are no repeats and every man proposes to every woman.
  
  
• 3. According to (1) the loop stops if every man has proposed to every woman. According to (2), after N² iterations of the loop, every conceivable proposal of a man to a woman would have happened, so every man would have proposed to every woman. Therefore, no matter what the input is, the main loop of the algorithm stops after no more than N² iterations. (In actual fact, the loop always stops after fewer than N² iterations.)
  
  (This proves that the loop never becomes an infinite loop, and moreover it tells us something about how efficient the algorithm may be. ( O(N²) loop iterations )
  
  
• 4. "Once a woman has become engaged, she remains engaged." What does this mean? After becoming engaged for the first time, a woman may separate from her current fiancé, but she will immediately become engaged to someone else. After becoming engaged for the first time, thereafter she will be engaged at the end of each iteration of the main loop. She will be engaged to someone, though she may change from one fiancé to another by "trading up."
  
  
• 5. It is a logical consequence of (4) that once all women have received a proposal, all N women are engaged.
  
  
• 6. Claim: If all N women are engaged, then all N men are engaged.
  [Proof: The way the algorithm works, a man can only be engaged to one woman at a time. So when the N women are all engaged, each is engaged to a different man. Because the total number of men is N, the N different men to whom the women are engaged are all the men.]
  
  
• 7. Claim: If all N men are engaged, then all N women are engaged. [Proof: The way the algorithm works, a woman can only be engaged to one man at a time, so the proof of this claim is just like the previous proof of (6).]
  
  
• 8. Claim: When the Gale-Shapley algorithm stops, all the men and women are engaged, proving that the Gale-Shapley algorithm always creates a perfect matching of the men to the women.
  
  [Proof: First of all, we can tell from the way that the algorithm works that it always creates a match (a one-to-one and onto mapping) between a set of some of the men, and a set of some of the women. Therefore, to prove our claim above, it will be enough to prove that all of the men and all of the women are engaged when the algorithm stops.
  
  In (1) we showed that the following is true when the loop stops: "Every man is either engaged, or has proposed to every woman (or both)."
  
  Because it turns out to make the proof a little easier, let's begin by asking this question: What if one of the men has proposed to every woman when the loop stops? If so, then by (5), all the women are engaged, and so by (6) all the men are engaged. So, in that case the claim we wish to prove is true.
  
  What if none of the men has proposed to all of the women when the loop stops? The stopping condition says that every man is either engaged or has proposed to every woman. So, when the loop stops, if none of the men has proposed to every women, then all of the men must be engaged. According to (7), then all the women are engaged too.
  
  We've shown that all the men and all the women are engaged when the loop stops, so that completes the proof.]
  
  The alert reader may have noticed that in proving the claim above, we've also demonstrated that the loop condition we were given is "over constrained." It should be clear now that the condition for continuing in the loop could be simply that there is a free man. We now know that if a man is free, he has not proposed to every woman.
  
  The next thing we want to find out is whether this perfect matching that we get from the Gale-Shapely algorithm is guaranteed to be stable. Here is the idea: Suppose that
  
  
• 9. Alice is matched to Bill, and
  
  
• 10. Charlotte is matched to David.
  
  Is it possible that (David, Alice) is an unstable pair? In other words, is it possible that both the following things are true?
  
  
• 11. Does David like Alice better than his match, Charlotte?
  
  
• 12. Does Alice like David better than her match, Bill?
  
  If so then Alice and David both have an incentive to 'run off' with each other. This would be an instability in the matching.
  
  
• 13. Well, suppose that 9, 10, and 11 are true. Consider the following logical consequences of that.
  
  
  • 14. Because 11 is true (David likes Alice better than Charlotte), Alice is higher on David's list than Charlotte, and so David must have proposed to Alice before he proposed to Charlotte.
    
    
  • 15. By 9, Alice is not with David now, and because of the way the algorithm works, she must have rejected David. Either she turned David down when he proposed to her (because she was already engaged to someone higher on her list), or she was once engaged to David, and then dumped him to "trade up."
    
    
  • 16. Either way, after Alice rejected David, she was engaged to someone, mr. X, whom she liked more than David. After that, she never "traded down." The only possibility is that she did some more "trading up."
    
    
  • 17. Because of the reasoning in step 16, above, we can conclude that Bill, the guy with whom Alice ended up, is someone she likes more than David.
    
    
  • 18. To sum up the logic, if 9, 10, and 11 are true, then "Alice does not like David more than Bill." In other words, if 9, 10, and 11 are true, then 12 is false.
  
  
• By showing that "if 9, 10, and 11 are true then 12 is false," we have shown that 9, 10, 1l, and 12 can't all be true. This proves that the algorithm creates a matching with no instabilities.
  
  
• So, we've reached an important milestone! We have now proved that the Gale-Shapley algorithm really works. No matter what the inputs are, the algorithm always outputs a stable matching, which, by definition is a perfect matching with no unstable pairs.
  
  

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• Our text goes on to define what it means for a man and woman (m, w) to be valid partners. It means there exists a stable matching in which m and w are one of the matches.
  
  It is important to understand that there may be stable matchings that cannot be created by the Gale-Shapley algorithm. Keep that in mind when you think about valid partners. A man and woman (m,w) are valid partners if there is any stable matching that matches those two with each other.
  
  
• After defining valid partners, the text proves that
  
  19. The Gale-Shapley algorithm always matches each man with his "best valid partner" - the valid partner who is highest on his preference list.
  
  
• Make sure you understand the idea of a "best valid partner." It definitely does not mean "the first choice." Given a set of n men and n women, and all their preference lists, there are many different ways to make a perfect matching. We know that at least one of those prefect matchings is stable (the output of the Gale-Shapley algorithm). There could be more than one stable matching for those men and women and their preferences. How many stable matchings there are depends on exactly what the preferences are. It's possible that a given pair (m,w) are not matched in any of the stable matchings that may exist for the 2n people and their preferences. In that case, we say that m and w are not valid partners. For every man m, there is at least one valid partner (the one he gets from the Gale-Shapley algorithm). Maybe there are more valid partners of m that he gets in other stable matchings. The best valid partner of m is whichever of his valid partners who is highest on his list. That best valid partner could be the actual first woman on m's list, but she could be a woman further down the list, because the first woman on the list may not be a valid partner for m. (Maybe there's no stable matching in which m is matched up with the first woman on his list.) It's even possible that m's best valid partner is the last woman on his list! Depending on what exactly the preferences are, it could work out that way!
  
  
• Anyway, it seems pretty amazing that the Gale-Shapley algorithm simultaneously gives all the men the "best" match they could hope to receive in any stable matching. Also this fact shows that the algorithm always gives the same matching as its output, no matter what particular men are chosen during the execution to make the next proposal! No matter what, the algorithm will match each guy to his one and only best valid partner!
  
  The central idea of the proof is to demonstrate that when the algorithm runs no man is rejected by a valid partner.
  
  
• The text also proves that
  
  20. The G-S algorithm pairs each woman with her worst valid partner.
  
  
• I made proofs of 19 and 20 that are somewhat different from the proofs in our text. Those proofs are included in one of the other documents in my set of notes for this chapter. At the time I'm writing this, the title of that document is "BVP_pf.html."