Stable Marriage Problem Introductory talk

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

My lab. member

Stable marriage problem • Gale and Shapley studied the problem in 1962. And they proposed efficient algorithm to solve it. • 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} Man Ａ Woman a Preference list Men’s Women’s Man Ｂ Woman b ＡaＢb

Setting of stable marriage problem • Case Ｎ＝４(Men {A,B,C,D}, Women {a,b,c,d}) Men’s preference list Women’s preference list

Simple approach Men’s preference list Women’s preference list (Aa, Bb, Cc, Dd) (Ac, Ba, Cb, Dd) A and b are a blocking pair. c and D are a blocking pair. (Ad, Ba, Cb, Dc) This is a stable matching. (Ab, Ba, Cc, Dd) b and C are a blocking pair. Is this approach valid for any preference list ?

Simple approach • Case Ｎ＝3(Men {A,B,C}, Women {a,b,c}) Men’s preference list Women’s preference list 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. a and C are a blocking pair. A and b are a blocking pair. (Ac, Ba, Cb) (Aa, Bb, Cc) This is initial matching! b and C are a blocking pair. A and a are a blocking pair. (Ac, Bb, Ca) (Ab, Ba, Cc)

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)http://www.jrmp.jp/2008/3/26

Distribution n interns into m hospitals nk: capacity of k-th hospital n1+n2+・・・+nm=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 I1 I2 . . . In Hospitals H1 (capacity n1) H2 (capacity n2) . . . Hm (capacity nm

Distribution n interns into m hospitals Interns I1 I2 . . . In Hospitals H11 (capacity 1) H12 (capacity 1) H1n1 (capacity 1) H21 (capacity 1) H22 (capacity 1) H2n2 (capacity 1) Same list … Same list … … Then we modify this model to original stable marriage model.

Incomplete lists Men’s preference list Women’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 Men’s preference list Women’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-Shapleyalgorithm • 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 ;

Gale-Shapley algorithm Ω a Ω b Ω c Ω d D c a A c d B b b C Men’s list Women’s list

Gale-Shapleyalgorithm • 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 Woman 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. Man Woman a

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 Woman a Man B 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 Woman a Man B 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=X0, a=x0, b=x1 , and suppose bAa then x1X0x0. The matching (Aa, …) is a worse choice for X0, x1 marries with X1 and X1x1X0 . The matching is better choice for x1. The matching (Ab, Ba, …) is a worse choice for x1. X1 marries with x2 and x2X1x1.

We obtain the sequence X0x0 X1x1 X2x2 … in the matching (Aa, …), X0x1 X1x2 X2x3 … in the matching (Ab, Ba, …) where xk+1Xkxk andXk+1xk+1Xk for all k ≧0. Since the number of person is finite, there exist integers j and k, j <k, such that Xj= Xk. Let j be the smallest integer having this property and for this j, let k be the smallest integer such that Xj= Xk and k>j. Then xj= xk .

If j=0 since otherwise Xk-1xk=Xk-1xj would appear in the matching (Ab, Ba, …) as well as Xj-1xj, from which Xj-1=Xk-1 , contradicting the fact that j is the smallest integer with Xj=Xk . Thus Xk-1x0 appears in the matching (Ab, Ba, …). But x0=a. Thus Xk-1=B. Given that XkxkXk-1, we have proved AaB.

Number of proposals Worst case O(n2) times (The maximum number of rejection is n(n-1)/2. ) The complexity of the algorithm is O(n2). 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.

When the algorithm is end? Ω a Ω b Ω c Ω d D c a A c d B b b C Men’s list Women’s list

When the algorithm ends? We address the order of proposals Woman’s list b d d,b,d,b,c,c,b,d,a,c,d d b c c b d a When all women appear in this sequence, we obtain a stable matching. Ａ

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? Case n=4 ｛A,B,C,D｝Ｃ　Ｂ　Ｃ　Ａ　Ｃ　Ｂ　Ｄ　Ｄ →we stop to buy boxes, when we obtain 4 kinds of coupon.

mean value qk : 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. q1=1, q2=m/n, q3=(m/n)2,… q1+q2+q3+...=1+(m/n)+(m/n)2+... The mean value of boxes one must buy to obtain all n coupons.

mean value where Hn is the sum of the first n terms of the harmonic series 1+1/2+1/3+・・・. Hn=ln n+γ+1/(2n)-1/(12n2)+ε where 0 < ε < 1/(120n4), 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 nHn=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 nHn=nln n+O(n). Man can marry with a woman who is ranked ln n in his list on the average.

Stable Marriage Problem Introductory talk