Stable Marriage Problem Introductory talk 　２

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Stable Marriage Problem Introductory talk 　２　. Yosuke KIKUCHI Dept Comp. and Info. Eng. Tsuyama College of Tech OKAYAMA, JAPAN. Contents. Original stable marriage problem ( conntinued ) Experimental study Scored stable marriage. Review of previous talk. Introduction of background

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### Stable Marriage Problem Introductory talk　２

Yosuke KIKUCHI

Dept Comp. and Info. Eng.

Tsuyama College of Tech

OKAYAMA, JAPAN

Contents
• Original stable marriage problem (conntinued)
• Experimental study
• Scored stable marriage
Review of previous talk
• Introduction of background
• Original stable marriage problem
• Setting
• Definition of stable
• application
• Gale-Shapley algorithm
• Features of the algorithm
• Proof of correctness of the algorithm
• Number of proposals (worst case and mean value)
Stable marriage problem
• 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 {α,β}

Man Ａ

Woman α

Preference list

Men’s

Women’s

Man Ｂ

Woman β

ＡαＢβ

Blocking pair
• Men {A,B},Women {α,β}

Man Ａ

Woman α

Preference list

Men’s

Women’s

Man Ｂ

Woman β

ＡαＢβ

It is not a blocking pair.

Blocking pair
• Men {A,B},Women {α,β}

Man Ａ

Woman α

Preference list

Men’s

Women’s

Man Ｂ

Woman β

ＡαＢβ

Example of stable and unstable

Men’s preference list

Women’s preference list

(Aα, Bβ, Cγ, Dδ)

A and βare a blocking pair.

(Aδ, Bα, Cβ, Dγ)

This is a stable 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)http://www.jrmp.jp/2008/3/26

Gale-Shapleyalgorithm
• 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

χ: 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(χ prefers X to her current partner){

engage X and χ;

X=preceding partner of χ;}

if(X != Ω) withdraw χ from X’s list;

}

k = k+1;

}

Output matching ;

Gale-Shapley algorithm

Ω α Ω β Ω γ Ω δ

D

γ

α

A

γ

B

β

C

β

δ

Men’s list

Women’s list

Gale-Shapleyalgorithm
• Features ・The algorithm always finds a stable matching.・The matching is optimal solution for men (women).・The matching is independent from the order of proposals.
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?

ΩαΩβΩγΩδ

D

γ

α

A

γ

B

β

β

C

δ

Men’s list

Women’s list

When the algorithm ends?

We address the order of proposals

Woman’s list

α

• δ, β, δ, β, γ, γ, β, δ, α, γ, δ

δ

β

δ

β

γ

γ

β

δ

When all women appear

in this sequence, we obtain

a stable matching.

coupon collector’s problem

Capsuled toy vending machine

• 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

new coupons when he

q1=1,

q2=m/n,

q3=(m/n)2,…

q1+q2+q3+...=1+(m/n)+(m/n)2+...

The mean value of boxes

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=nlnn+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=nlnn+O(n).

Man can marry with a woman who is

ranked ln n in his list on the average.

Experimental study （1/2）
• Implementing the Gale-Shapley algorithm

window.exe

Experimental study（2/2）
• Measurement of the mean of solution of the Gale-Shapley algorithm

・Pentium D:2.66GHz ・Memory:１Gbyte・OS:WindowsXP

Random number generator : Random Streamer RPG102

1000times measurements for n=5,50,75,100

• result
Structure of stable matchings

Men’s list

Women’s list

All stable matchings : (A α, B β, C γ, D δ), (A α, B β, C δ, C γ),

(A β, B α, C γ, D δ), (A β, B α, C δ, C γ), (A β, B δ, C α, D γ),

(A γ, B α, C δ, D β), (A γ, B δ, C α, D β)

Structure of stable matchings

Men’s list

Women’s list

All stable matchings : (A α, B β, C γ, D δ), (A α, B β, C δ, C γ),

(A β, B α, C γ, D δ), (A β, B α, C δ, C γ), (A β, B δ, C α, D γ),

(A γ, B α, C δ, D β), (A γ, B δ, C α, D β)

• All stable matchings : (α, β, γ, δ), (α, β, δ, γ),
• (β, α, γ, δ), (β, α, δ, γ), (β, δ, α, γ),
• (γ, α, δ, β), (γ, δ, α, β)
Structure of stable matchings

Men’s list

Women’s list

• All stable matchings : (α, β, γ, δ), (α, β, δ, γ), (β, α, γ, δ),
• (β, α, δ, γ), (β, δ, α, γ), (γ, α, δ, β), (γ, δ, α, β)

1 1 1 1

(α, β, γ, δ)

1 1 2 2

(α, β, δ, γ)

2 2 1 1

(β, α, γ, δ)

2 2 2 2

(β, α, δ, γ)

3 2 2 3

(γ, α, δ, β)

2 3 3 2

(β, δ, α, γ)

3 3 3 3

(γ, δ, α, β)

Structure of stable matchings

1 1 1 1

(α, β, γ, δ)

• Distributive lattice

1 1 2 2

(α, β, δ, γ)

2 2 1 1

(β, α, γ, δ)

2 2 2 2

(β, α, δ, γ)

2 3 3 2

(β, δ, α, γ)

3 2 2 3

(γ, α, δ, β)

3 3 3 3

(γ, δ, α, β)

Scored stable marriage problem

Men’s preference list

Woman’s preference list

How to calculate score of matching

How to calculate score

Aγ: 40　＋　5　＝　45

Aδ,Bα,Cβ,Dγ:

65 + 40 + 90 + 45 = 240

Find optimal matching

Men’s preference list

Maximum score

Assignment problemusing Hungarian method

Woman’s preference list

30 + 20

What is Hungarian method？
• This method can find n matching such that total amount is minimum in n×n matrix.

Maximum score 200-score in matrix

Finding optimal matching

The matching that be found by Hungarian method

AδBβCγDα

65 + 95 + 60 + 50 = 270

Case: score is inclined.

Men’s preference list

Woman’s preference list

More than one solution by using Hungarian method

Aα,Bδ,Cβ,Dγ : 200

Aβ,Bα,Cδ,Dγ: 200Aγ,Bδ,Cα,Dβ: 200

・

・

・

Use Gale-Shapley algorithm

Aα,Bβ,Cδ,Dγ: 200

Scored stable marriage problem

Apply Hungarian method

Apply Gale-Shapley algorithmfind stable matching

compare the score of Gale-Shapley and the core of Hungarian method.

If the score of Gale-Shapley = the core of Hungarian method, then we choose the score of Gale-Shapley

Further work
• If the preference list is incomplete and includes tie, finding the maximum stable matching is NP-hard [ Halldorsson et al, 2003].
• 1.875-approximation algorithm is developed by Iwama et al, 2007.

Can we improve the approximation ratio?

It is known that lower bound 1.105 unless P=NP .

Further work
• Stable roommates problem

This problem does not have solution

for any preference list.

This problem is a generalization of

stable marriage problem.

Further work
• Stable roommates problem
Further work
• Stable roommates problem

AB, CD

B and D are a blocking pair.

This is no solution for this list.

AC, BD

A and D are a blocking pair.

A and B are a blocking pair.

Gap between original and scored
• When the solution of Gale-Shapley algorithm and the solution of Hungarian method?

Men’s preference list

Woman’s preference list

δ

Gap between original and scored

Men’s preference list

Woman’s preference list

Stable matching is (Aδ, Bα, CβDγ) and the score is 4+30+51+40=125

The maximum score is 218 and the matching is (Aγ, Bβ , Cδ, Dα), (90+40+49+39).

But this matching is unstable, because C and βis a blocking pair.

Which is better for men ?

• Edward G. Thurber, Concerning the maximum
• number of stable matchings in the stable marriage
• problem,Discrete Mathematics 248 (2002) 195–219.
• Jan Eeckhout, On the uniqueness of stable marriage
• matchings, Economics Letters 69 (2000) 1–8

Suppose a sequence S=α1 α2 α3 ... αn

is obtained by Gale-Shapley algorithm.

The partial sequence of S,

Each alphabet αkis containd at most m

in the partial sequence A of S,

A=α1 α2 α3 ... αm(m+1)/2 where m is a

positive integer.

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

new coupons when he