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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

Stable Marriage Problem Introductory talk 2 

Yosuke KIKUCHI

Dept Comp. and Info. Eng.

Tsuyama College of Tech

OKAYAMA, JAPAN

contents
Contents
  • Original stable marriage problem (conntinued)
  • Experimental study
  • Scored stable marriage
review of previous talk
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
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 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
Blocking pair
  • Men {A,B},Women {α,β}

Man A

Woman α

Preference list

Men’s

Women’s

Man B

Woman β

AαBβ

blocking pair1
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 pair2
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
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
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 shapley algorithm
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.
slide13

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 algorithm1
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 shapley algorithm2
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
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
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
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
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
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 value1
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 value2
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 st ud y 1 2
Experimental study (1/2)
  • Implementing the Gale-Shapley algorithm

window.exe

experimental study 2 2
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
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 matchings1
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 matchings2
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 matchings3
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
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
Find optimal matching

Men’s preference list

Maximum score

Assignment problemusing Hungarian method

Woman’s preference list

30 + 20

what is hungarian method
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
Finding optimal matching

The matching that be found by Hungarian method

AδBβCγDα

65 + 95 + 60 + 50 = 270

case score is inclined
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 problem1
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
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 work1
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 work2
Further work
  • Stable roommates problem
further work3
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.

AD, BC

A and B are a blocking pair.

gap between original and scored
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 scored1
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 ?

further reading
Further reading

Please consult Google.

  • 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
about the women sequence
About the women sequence

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 value3
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+・・・+qm+1=1+(m/n)+(m/n)2+ ・・・+(m/n)m

Can we obtain more strict upper

bound of the number of proposals?