Stable Marriage Problem Introductory talk Yosuke KIKUCHI Dept Comp. and Info. Eng. Tsuyama College of Tech OKAYAMA, JAPAN

Contents Original stable marriage problem Experimental study Scored stable marriage

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Stable marriage problem Gale and Shapley studied the problem in And they proposed efficient algorithm to solve it. Gale There are Ｎ -women and N-men. Each parson has a preference list of opposite sex. Then decide N-matching between women and men.

A matching is stable A matching is unstable if a man A and a woman a, not married to each other, but they mutually prefer each other to their partners. These pair A and a is called blocking pair. A matching is stable if it does not contain a blocking pair.

Blocking pair Men {A,B}, Women {a,b} Preference list Men’sWomen’s Ａ a Ｂ b Man Ａ Man Ｂ Woman a Woman b 12 Aba B ba 12 aBA bAB

Setting of stable marriage problem Case Ｎ＝４ (Men {A,B,C,D}, Women {a,b,c,d}) 1234 Acbda Bbacd Cbdac Dcadb 1234 aABDC bCADB cCBDA dBACD Men’s preference listWomen’s preference list

Simple approach 1234 Acbda Bbacd Cbdac Dcadb 1234 aABDC bCADB cCBDA dBACD Men’s preference listWomen’s preference list (Aa, Bb, Cc, Dd) A and b are a blocking pair. (Ab, Ba, Cc, Dd) b and C are a blocking pair. (Ac, Ba, Cb, Dd) c and D are a blocking pair. (Ad, Ba, Cb, Dc) This is a stable matching. Is this approach valid for any preference list ?

Simple approach Case Ｎ＝ 3(Men {A,B,C}, Women {a,b,c}) 123 Abac Bcab Cabc 123 aACB bCAB cCBA Men’s preference listWomen’s preference list (Aa, Bb, Cc) (Ab, Ba, Cc) A and b are a blocking pair. b and C are a blocking pair.A and a are a blocking pair. a and C are a blocking pair. (Ac, Ba, Cb) (Ac, Bb, Ca) This approach does not end. But (Ab, Bc, Ca) and (Aa, Bc, Cb) are stable matcihg. Thus this approach can not obtain stable matching for any lists. This is initial matching!

Applications of stable matching Distribution of students among laboratory Distribution interns into hospitals [1] － reference － [1] 医師臨床研修マッチング協議会 (JAPAN INTERNS MATCHING CONFERENCE(?)) URL) /3/26http://

Distribution n interns into m hospitals n k : capacity of k-th hospital n 1 +n 2 + ・・・ +n m =n The number of interns > the capacities of hospitals ⇒ fictitious hospital (the worst choice for inters) (the capacity of fictitious hospital = The number of interns - the capacities of hospitals) The number of interns < the capacities of hospitals ⇒ fictitious interns (the worst choice for hospitals) (the number of fictitious interns = The capacities of hospitals - the number of interns )

Distribution n interns into m hospitals Interns I 1 I 2. I n Hospitals H 1 (capacity n 1 ) H 2 (capacity n 2 ). H m (capacity n m

Distribution n interns into m hospitals Interns I 1 I 2. I n Hospitals H 11 (capacity 1) H 12 (capacity 1) H 1n 1 (capacity 1) H 21 (capacity 1) H 22 (capacity 1) H 2n 2 (capacity 1) … … … Same list Same list Then we modify this model to original stable marriage model.

Incomplete lists 123 Aa Bcab Cca 123 aCA bBAC cBC Men’s preference listWomen’s preference list In these reference lists, only (Aa, Bb, Cc) is matching. But this matching is not stable matching. It is known that there exists a stable matching for complete lists.

Incomplete lists 123 Aaωω Bcab Ccaω 123 aCAΩ bBAC cBCΩ Men’s preference listWomen’s preference list We add new man Ω and new woman ω. Ω is the worst choice for women. ω is the worst choice for men.

Gale-Shapley algorithm This algorithm can find a stable matching. The matching is optimal for men(women). The algorithm contains following four steps. – 1. man proposes to his desirable woman one by one. – 2. woman decides whether accept or reject. – 3. If a man is refused, then he remove her name from his preference list. – 4. Repeat the steps above, until every man is accepted by a woman.

Pseudo code of Gale-Shapley algorithm Variables and constants n: number of women = number of men k: number of couples X: suitor x: woman toward whom the suitor makes advances Ω: (undesirable) imaginary man K=0; all the women are engaged to Ω; while(k < n){ X=(k+1)-st man; while(X != Ω){ x=the best choice remaining on X’s list; if(x prefers X to her current partner){ engage X and x; X=preceding partner of x;} if(X != Ω) withdraw x from X’s list; } k = k+1; } Output matching ;

cb Gale-Shapley algorithm Men’s listWomen’s list AB C b a D c d 1234 Acbda Bbacd Cbdac Dcadb 1234 aABDC bCADB cCBDA dBACD Ω a Ω b Ω c Ω d

Gale-Shapley algorithm Features ・ The algorithm can find a stable matching. ・ The matching is optimal solution for men (women). ・ The matching is independent from the order of proposals.

The algorithm can find a stable matching Man Ａ Man Woman Woman a Suppose A is not married to a and if A prefers a to his partner in the matching obtained. Then a has rejected A’s proposal and is married to someone she prefers to A. Then A and a is not blocking pair. Thus the matching is stable.

The matching is optimal solution for men (women) A man can not marry with a woman who ranks higher in his list than his partner. If the women make the advances, then the matching obtained is optimal for women.

Man Ａ Woman a Man Ａ Man B Woman a Woman b If one stable matching contains Aa, and another contains Ab and Ba, then either A prefers b to a and a prefers A to B or A prefers a to b and a prefers B to A.

Man Ａ Woman a Man Ａ Man B Woman a Woman b If one stable matching contains Aa, and another contains Ab and Ba, then either A prefers b to a and a prefers A to B or A prefers a to b and a prefers B to A. Every other stable matching is better for one of the spouses and worse for the other.

There are two stbale matching. One is the (Aa, …), and another is the (Ab, Ba, …). Then it is hold that either A prefers b to a and a prefers A to B, or A prefers a to b and a prefers B to A.

Proof. We prove the situations of A and a can not both worsen in the second matching. We prove that they can not improve for the two at the same time. Notation: bAa ⇒ A prefers b to a. AaB ⇒ a prefers A to B.

A=X 0, a=x 0, b=x 1, and suppose bAa then x 1 X 0 x 0. The matching (Aa, …) is a worse choice for X 0, x 1 marries with X 1 and X 1 x 1 X 0. The matching is better choice for x 1. The matching (Ab, Ba, …) is a worse choice for x 1. X 1 marries with x 2 and x 2 X 1 x 1.

We obtain the sequence X 0 x 0 X 1 x 1 X 2 x 2 … in the matching (Aa, …), X 0 x 1 X 1 x 2 X 2 x 3 … in the matching (Ab, Ba, …) where x k+1 X k x k and X k+1 x k+1 X k for all k ≧ 0. Since the number of person is finite, there exist integers j and k, j <k, such that X j = X k. Let j be the smallest integer having this property and for this j, let k be the smallest integer such that X j = X k and k>j. Then x j = x k.

If j=0 since otherwise X k-1 x k =X k-1 x j would appear in the matching (Ab, Ba, …) as well as X j-1 x j, from which X j-1 =X k-1, contradicting the fact that j is the smallest integer with X j =X k. Thus X k-1 x 0 appears in the matching (Ab, Ba, …). But x 0 =a. Thus X k-1 =B. Given that X k x k X k-1, we have proved AaB.

Number of proposals Worst case O(n 2 ) times (The maximum number of rejection is n(n-1)/2. ) The complexity of the algorithm is O(n 2 ). Best case n times (each man is not rejected. Thus each man’s partner is the really best partner for him. ) Then we estimate the mean number of proposals. We will find an upper bound for the mean number of proposals.

cb When the algorithm is end? Men’s listWomen’s list AB C b a D c d 1234 Acbda Bbacd Cbdac Dcadb 1234 aABDC bCADB cCBDA dBACD Ω a Ω b Ω c Ω d

We address the order of proposals When the algorithm ends? Woman’s list Ａ 1234 aABDC bCADB cCBDA d Ｃ A Ｂ D When all women appear in this sequence, we obtain a stable matching. d,b,d,b,c,c,b,d,a,c,dda b d bccbd 1234 Adb Bbc Cd Dcbda

coupon collector’s problem There are ｎ distinct coupons and that each time one buys a box of detergent one obtains a coupon at random. How many boxes must be bought, on average, to obtain all n coupons? Ｃ Ｂ Ｃ Ａ Ｃ Ｂ Ｄ Ｄ Case n=4 ｛ A,B,C,D ｝ →we stop to buy boxes, when we obtain 4 kinds of coupon.

mean value q k : the probability that at least k boxes are necessary m: number of coupons that we have. n : number of kinds of coupons Mean number of boxes one must buy to obtain new coupons when he already have m coupons. The mean value of boxes one must buy to obtain all n coupons. q 1 =1, q 2 =m/n, q 3 =(m/n) 2,… q 1 +q 2 +q =1+(m/n)+(m/n)

mean value where H n is the sum of the first n terms of the harmonic series 1+1/2+1/3+ ・・・. H n =ln n+γ+1/(2n)-1/(12n 2 )+ε where 0 < ε < 1/(120n 4 ), γ is Euler’s constant. An upper bound for the mean number E(N) of proposals to obtain a stable matching by Gale-Shapley algorithm is nH n =nln n+O(n).

mean value An upper bound for the mean number E(N) of proposals to obtain a stable matching by Gale-Shapley algorithm is nH n =nln n+O(n). Man can marry with a woman who is ranked ln n in his list on the average.