1. 1 Stable Marriage Problem

2. 2 Consider a society with n men (denoted by capital letters) and n women (denoted by lower case letters). A marriage M is a 1-1 correspondence between the men and women. Each person has a preference list of the members of the opposite sex organized in a decreasing order of desirability.

3. 3 A marriage is said to be unstable if there exist 2 marriage couples X-x and Y-y such that X desires y more than x and y desires X more than Y. The pair X-y is said to be “ dissatisfied. ” A marriage M is called “ stable marriage ” if there is no dissatisfied couple.

4. 4 Assume a monogamous, hetersexual society. Eg. n=4. A: abcd B: bacd C: adcb D: dcab a: ABCD b: DCBA c:ABCD d:CDAB Consider a marriage M: A-a, B-b, C-c, D-d, C-d is dissatisfied. Why?

5. 5 Is the stable marriage unique?. Eg. n=2. A: ab B: ba a: BA b: AB A-a, B-b and A-b, B-a are both stable!

6. Proposal algorithm: D. Gale and L. S. Shapley: "College Admissions and the Stability of Marriage", American Mathematical Monthly 69, 9-14, 1962.American Mathematical Monthly Assume that the men are numbered in an arbitrary order. (Yields the same result.) 1. The lowest numbered unmarried man X proposes to the most desirable woman on his list who has not already rejected him; call her x. 2 The woman x will accept the proposal if she is currently unmarried, or if her current mate Y is less desirable to her than X (Y is jilted and reverts to the unmarried state). 3. The algorithm repeats this process until every person has married. jilt: vt. ( 及物動詞 ) 拋棄 ( 情人 ) n. ( 名詞 noun) 拋棄情人者 ( 尤指女子 )

7. Proposal algorithm: D. Gale and L. S. Shapley: "College Admissions and the Stability of Marriage", American Mathematical Monthly 69, 9-14, 1962.American Mathematical Monthly David Gale (December 13, 1921 - March 7, 2008) was a distinguished American mathematician and economist. He was a Professor Emeritus at University of California, Berkeley, affiliated with departments of Mathematics, Economics, and Industrial Engineering and Operations Research. Lloyd Stowell Shapley (born June 2, 1923) is a distinguished American mathematician and economist. He is a Professor Emeritus at University of California, Los Angeles, affiliated with departments of Mathematics and Economics. Gale, David (1998) Lloyd Stowell Shapley

8. 8 This algorithm is used by hospitals in North America in the match program that assigns medical graduates to residency positions. Women are unhappy if men propose, and men are unhappy if women propose. Does it always terminate with a stable marriage?

9. 9 Observe that: An unmatched man always has at least one woman available that he can proposition. At each step the proposer will eliminate one woman on his list and the total size of the lists is n 2. Thus the algorithm uses at most n 2 proposals. I.e. it always terminates.

10. 10 Claim that the final marriage M is stable. Proof: Otherwise, let X-y be a dissatisfied pair, where in M they are paired as X-x, Y-y. Since X prefers y to x, he must have proposed to y before getting married to x.

11. 11 Since y either rejected X or accepted him only to jilt him later, her mates thereafter (including Y) must be more desirable to her than X. Therefore y must prefer Y to X,  contradicting the assumption that y is dissatisfied.

12. 12 We say a woman w is a valid partner of a man m if there is a stable marriage containing m-w. We say that w is the best valid partner of m if w is a valid partner of m and no woman whom m ranks higher than w is a valid partner of m. Denote it as best(m). Let S* = {(m, best(m)): for all man m}

13. Thm: S* = {(m, best(m)): for all man m}. Any order of execution of the proposal algorithm results in S*. Proof: By contradiction. Let S be the stable marriage and suppose some man m is matched with a woman who is not his best valid partner. Consider the first moment during the execution that m is rejected by a valid partner w, who must be best(m). Why? The rejection of m by w happened either because the current partner of w had higher rank than m, or w jilted m.

14. Thm: S* = {(m, best(m)): for all man m}. Any order of execution of the proposal algorithm results in S*. Either way, w has a partner m ’ whom she prefers to m. Since w is a valid partner of m, there is a stable marriage S ’ containing (m, w). Who is the partner of m ’ in S ’ ? Suppose it is w ’. Since the rejection of m by w was the first rejection of a man by a valid partner during the execution for S, it must be that m ’ had not been rejected by any valid partner when he was accepted by w.

15. Thm: S* = {(m, best(m)): for all man m}. Any order of execution of the proposal algorithm results in S*. Since man proposed in decreasing order of preference and w ’ is a valid partner of m ’, it must be that m ’ prefers w to w ’. I.e., m ’ : w w ’. But we know that w prefers m ’ to m, since w rejected m in favor of m ’. I.e., w : m ’ m. Since (m ’, w) is not in S ’, it follows (m ’, w) is unsatisfied in S ’. It contradicts that S ’ is stable.

16. 16 For a woman w, we say that m is a valid partner, if there is a stable marriage containing m-w. We say that m is the worst valid partner of w if m is a valid partner of w and no man whom w ranks lower than m is a valid partner of w. Theorem: In S*, each woman is matched with her worst valid partner. (HW)

17. 17 Goal: Perform an average-case analysis of this (deterministic) algorithm. For this average-case analysis, we assume that the men ’ s lists are chosen independently and uniformly; the women ’ s lists can be arbitrary but must be fixed in advance.

18. 18 T p denotes the number of proposal made during the execution of the Proposal Algorithm. The running time is proportional to T p. But it seems difficult to analyze T p.

19. 19 Principle of Deferred Decisions: The idea is to assume that the entire set of random choices is not made in advance. At each step of the process, we fix only the random choices that must be revealed to the algorithm. We use it to simplify the average-case analysis of the Proposal Algorithm.

20. 20 Suppose that men do not know their lists to start with. Each time a man has to make a proposal, he picks a random woman from the set of women not already propositioned by him, and proceeds to propose to her. The only dependency that remains is that the random choice of a woman at any step depends on the set of proposals made so far by the current proposer.

21. 21 However, we can eliminate the dependency by modifying the algorithm, i.e. a man chooses a woman uniformly at random from the set of all n women, including those to whom he has already proposed. Call this new version the Amnesiac Algorithm.

22. 22 Note that a man making a proposal to a woman who has already rejected him will be rejected again. Thus the output by the Amnesiac Algorithm is exactly the same as that of the original Proposal Algorithm.

23. 23 Let T A denote the number of proposals made by the Amnesiac Algorithm. T p >m  T A >m, i.e. T A stochastically dominates T p. Thus Pr[T p >m]  Pr[T A >m] for all m.

24. 24 It suffices to find an upper bound to analyze the distribution T A. A benefit of analyzing T A is that we need only count that total number of proposals made, since each proposal is made uniformly and independently to one of n women.

25. 25 The algorithm terminates with a stable marriage once all women have received at least one proposal each. Moreover, bounding the value of T A is a special case of the coupon collector ’ s problem.

26. 26 Thm: For any constant c  R, and m=n ln (n)+cn, The Amnesiac Algorithm terminates with a stable marriage once all women have received at least one proposal each.

27. 27 Bounding the value of T A is a special case of the coupon collector ’ s problem. Let X be a r.v. defined to be the number of trials required to collect at least one of each type of coupon.

28. 28 Define the r.v. X i, for 0  i  n-1, to be the number of trials in the i-th stage, so that Let p i denote the probability of success on any trial of the i-th stage.

29. 29 Recall that X i is geometrically distributed with p i. So Thus ln (n)+  (1) I.e. E[X]=n ln (n)+O(n)

30. 30 X i ‘ s are independent, thus   2 /6

31. 31 Let  denote the event that coupon type i is not collected in the first r trials. Thus For r =  n ln (n), e -r/n =n - ,  >1.

32. 32 Poisson Heuristic Let  denote the number of times the coupon of type i is chosen during the first r trials. is the same as the event { =0}. has the binomial distribution with parameter r and p=1/n. 

33. 33 Let  be a positive real number. Y: (non-negative integer r.v.) Y has the Poisson distribution with parameter if for any non-negative integer y,

34. 34 For proper small and as r , the Poisson distribution with = rp is a good approximation to the binomial distribution with parameter r and p. Approximate by the Poisson distribution with parameter =r/n. Thus

35. 35 Claim:, for 1  i  n, are almost independent. I.e. for any index set {j 1,..,j k } not containing i, Pf:

36. 36 Thus, Let m =  n( ln (n)+c), for any constant c. 0 e^(-e^-4)=0.981; e^(-e^-5)=0.993 ; e^(-e^-6)=0.997

37. 37 A Rigorous Analysis Lemma: Let c be a real constant, and m = n ln (n)+cn, for positive integer n. Then, for any fixed positive integer k,

38. 38 Pf: HW: Prove, for all t,n such that n  1 and |t|  n. By the above, we have

39. 39 Observe that e -km/n =n -k e -ck. Further, and

40. 40 Thm: Let the r.v. X denote the number of trials for collecting each of the n types of coupons. Then, for any constant c, and m = n ln (n)+cn, Pf: Let

41. 41 Note that the event Let denote the partial sum formed by the first k terms of this series. Principle of Inclusion-Exclusion

42. 42 We have by the Boole-Bonferroni inequality: E 1,.., E n : arbitrary events. 1. For even k: 2. For odd k:

43. 43 By symmetry, all the k-wise intersections of the events are all equally likely, i.e. More precisely, For all positive integer k, define

44. 44 Define the partial sum of P k ’ s as the first k terms of the power series expansion of. Thus I.e., for all  >0, there exists k* such that k>k*, |S k -f(c)|< .

45. 45 Since Thus for all  >0 and k, when n is sufficiently large, Thus for all  >0 and k>k*, and large enough n, we have and which implies that and

46. 46 Using the bracketing property of partial sum, we have that for any  >0 and n sufficiently large, 