Stabiele huwelijken

Stabiele huwelijken

The Stable Marriage Problem:More Marital Happiness Than

Reality TV

Dave Toth

Reality TV

I had to do some digging to get this, so please correct me if I’m wrong...

• Who Wants to Marry a Millionaire– lasted a few weeks

• Joe Millionaire– lasted even shorter time

Reality TV

I had to do some digging to get this, so please correct me if I’m wrong...

• Who Wants to Marry a Millionaire– lasted a few weeks

• Joe Millionaire– lasted even shorter time

• Married by America– I give it 2 days– didn’t get passed the altar

• Mr. Personality ???

Disclaimer• The method to solve the Stable Marriage Problem is inherently sexist and

not politically correct!

• The Stable Marriage Problem assumes people are heterosexual. The Stable Roommates Problem…

• To be consistent with terminology in the field and to simplify explanations, we always assume men propose to women.

• Results can be reversed if the women propose to men.

Definition of the Stable Marriage Problem

• Given:

– n men and n women

– preference list for each person (no ties)• Definitions:

– marriage: a complete 1-to-1 matching

– blocking pair: a man and woman in different couples in a marriage who prefer each other to their assigned partner

– stable marriage: a marriage with no blocking pairs

– rank: the position in a person’s list of his partner– happiness: increases as rank decreases

How Many Marriages Exist?

• for n = 1

• for n = 2

• for n = 3

• for n = x

Is There A Doctor In The House?

• Many-to-1 vs. 1-to-1

• Original context was college admissions

• Problems with residents/hospitals:

- Offers 2 years in advance

- Acceptance window of 12 hours

• In use from 1952 by National Resident Matching Program (NRMP), early participation 95%, later 85%

• Now if you don’t like the results, just sue!

From SIAM News April 2003

• “Are Medical Students Meeting Their (Best Possible) Match?”

• 3 people have filed a law suit against the NRMP and the hospitals sponsoring it.

Matchmaker, Matchmaker, Make Me A Match!

Men (1-4) Women (A-D) 1: B, D, A, C A: 2, 1, 4, 32: C, A, D, B B: 4, 3, 1, 23: B, C, A, D C: 1, 4, 3, 24: D, A, C, B D: 2, 1, 4, 3

Given this instance, what’s a marriage?Is it stable?How would you find a stable one?

The Gale Shapley Algorithm• parallel

1. all people begin unengaged

2. while there are unengaged men, each proposes until a woman accepts (men →)

3. unengaged women accept 1st proposal they get

4. if an engaged woman receives a proposal she likes better, she breaks old engagement and accepts new proposal; dumped man begins proposing where he left off (women ←)

Example

Men (1-4) Women (A-D)

1: B, D, A, C A: 2, 1, 4, 3

2: C, A, D, B B: 4, 3, 1, 2

3: B, C, A, D C: 1, 4, 3, 2

4: D, A, C, B D: 2, 1, 4, 3

Example


1: B, D, A, C A: 2, 1, 4, 3

2: C, A, D, B B: 4, 3, 1, 2

3: B, C, A, D C: 1, 4, 3, 2

4: D, A, C, B D: 2, 1, 4, 3

1 proposes to B, she accepts: (1, B)

Example


1: B, D, A, C A: 2, 1, 4, 3

2: C, A, D, B B: 4, 3, 1, 2

3: B, C, A, D C: 1, 4, 3, 2

4: D, A, C, B D: 2, 1, 4, 3


2 proposes to C, she accepts: (1, B) (2, C)

Example


1: B, D, A, C A: 2, 1, 4, 3

2: C, A, D, B B: 4, 3, 1, 2

3: B, C, A, D C: 1, 4, 3, 2

4: D, A, C, B D: 2, 1, 4, 3



3 proposes to B, she accepts & dumps 1: (2, C) (3, B)

Example


1: B, D, A, C A: 2, 1, 4, 3

2: C, A, D, B B: 4, 3, 1, 2

3: B, C, A, D C: 1, 4, 3, 2

4: D, A, C, B D: 2, 1, 4, 3




1 proposes to D, she accepts: (1, D) (2, C) (3, B)

Example


1: B, D, A, C A: 2, 1, 4, 3

2: C, A, D, B B: 4, 3, 1, 2

3: B, C, A, D C: 1, 4, 3, 2

4: D, A, C, B D: 2, 1, 4, 3





4 proposes to D, she rejects: (1, D) (2, C) (3, B)

Example


1: B, D, A, C A: 2, 1, 4, 3

2: C, A, D, B B: 4, 3, 1, 2

3: B, C, A, D C: 1, 4, 3, 2

4: D, A, C, B D: 2, 1, 4, 3





4 proposes to D, she rejects: (1, D) (2, C) (3, B)

4 proposes to A, she accepts: (1, D) (2, C) (3, B) (4, A)

It Takes Two to Tango

• Result of GS is always a stable marriage– it requires two people of opposite sex in

different couples to break up a marriage– if a man wants to leave for some woman, then

he already proposed to her and she rejected him, so she won’t leave her husband for him

Opmerkingen

In de loop van het algoritme zullen sommige paren scheiden en onmiddellijk hertrouwen met andere partners•de vrouw van het scheidende koppel zal zich hierdoor verbeteren•en de man is er steeds slechter aan toe

Bij het beëindigen van het algoritme:Een vrouw zal nooit haar huidige partner verlaten voor een man X die haar eerder heeft gevraagdEen man X wil enkel scheiden voor een “betere vrouw” Y, maar elke betere vrouw Y heeft hij in de loop van het algoritme reeds aangesproken

en hij maakt dus geen enkele kans meer bij zo een vrouw Y

Besluit:geen enkele man kan een ander huwelijk kapot makenen om een huwelijk kapot te maken moet je met 2 zijner kunnen dus geen huwelijken kapot gaan

The Party’s Over - Features of the Gale Shapley Algorithm

• Gale Shapley (GS) always ends– n men, each make at most n proposals for a

worst case total of n2

• When men propose, we obtain M0

• When women propose, we obtain Mz

3 Wild Claims about M0

1. Each man has the best partner he can have in any stable marriage.

2. Each woman has the worst partner she can have in any stable marriage.

3. GS always produces same stable marriage - order of proposals is irrelevant.

Illustration of Irrelevance of Order

Men Women

1: A B C A: 1 2 3

2: A B C B: 1 2 3

3: A B C C: 1 2 3


Men Women

1: A B C A: 1 2 3

2: A B C B: 1 2 3

3: A B C C: 1 2 3

1 proposes to A, she accepts: (1, A)


Men Women

1: A B C A: 1 2 3

2: A B C B: 1 2 3

3: A B C C: 1 2 3


2 proposes to A, she rejects him: (1, A)


Men Women

1: A B C A: 1 2 3

2: A B C B: 1 2 3

3: A B C C: 1 2 3



2 proposes to B, she accepts: (1, A), (2, B)


Men Women

1: A B C A: 1 2 3

2: A B C B: 1 2 3

3: A B C C: 1 2 3




3 proposes to A, she rejects him: (1, A), (2, B)


Men Women

1: A B C A: 1 2 3

2: A B C B: 1 2 3

3: A B C C: 1 2 3





3 proposes to B, she rejects him: (1, A), (2, B)


Men Women

1: A B C A: 1 2 3

2: A B C B: 1 2 3

3: A B C C: 1 2 3





3 proposes to B, she rejects him: (1, A), (2, B)

3 proposes to C, she accepts: (1, A), (2, B), (3, C)


Men Women

1: A B C A: 1 2 3

2: A B C B: 1 2 3

3: A B C C: 1 2 3


Men Women

1: A B C A: 1 2 3

2: A B C B: 1 2 3

3: A B C C: 1 2 3



Men Women

1: A B C A: 1 2 3

2: A B C B: 1 2 3

3: A B C C: 1 2 3


2 proposes to A, she accepts and dumps 3: (2, A)


Men Women

1: A B C A: 1 2 3

2: A B C B: 1 2 3

3: A B C C: 1 2 3





Men Women

1: A B C A: 1 2 3

2: A B C B: 1 2 3

3: A B C C: 1 2 3




1 proposes to A, she accepts and dumps 2: (1, A), (3, B)


Men Women

1: A B C A: 1 2 3

2: A B C B: 1 2 3

3: A B C C: 1 2 3





2 proposes to B, she accepts and dumps 3: (1, A), (2, B)


Men Women

1: A B C A: 1 2 3

2: A B C B: 1 2 3

3: A B C C: 1 2 3





2 proposes to B, she accepts and dumps 3: (1, A), (2, B)

3 proposes to C, she accepts: (1, A), (2, B), (3, C)

Proof of claims 1 and 3 (by contradiction) Claim 1: Each man has the best partner he can have in any stable marriage.

Claim 3: GS always produces same stable marriage – order of proposals is irrelevant.

• Arbitrary execution E of GS produces the stable marriage M with couple (mx, wa).

• A different arbitrary execution E’ produces the stable marriage M2 with couple (mx, wb).

• Without loss of generality, we say mx likes wb better than wa, so in E, mx proposed to wb and she rejected him in favor of some other man. wb’s final partner in M is some man my, thus M contains (my, wb)

• wb has preference list ...my...mx...

M: (mx, wa), (my, wb) M2: (mx, wb)

Proof of claims 1 and 3 (by contradiction) Claim 1: Each man has the best partner he can have in any stable marriage.

Claim 3: GS always produces same stable marriage - order of proposals is irrelevant.

• Arbitrary execution E of GS produces the stable marriage M with couple (mx, wa).

• A different arbitrary execution E’ produces the stable marriage M2 with couple (mx, wb).

• Without loss of generality, we say mx likes wb better than wa, so in E, mx proposed to wb and she rejected him in favor of some other man. wb’s final partner in M is some man my, thus M contains (my, wb)

• wb has preference list ...my...mx...

M: (mx, wa), (my, wb) M2: (mx, wb), (my, q)

• because of how GS works, my has no better partner x than wb or he’d be with x in M, so in M2, my has a partner he likes less than wb

• In M2, both my and wb want to dump their partners for each other, so M2 is not a stable marriage

• Since no other stable marriage is produced by GS, order irrelevant □

Proof of 2: Each woman has worst partner she can have in any stable

marriage. (by contradiction)

M0 M2

(m, w) (m, w’)

(m’, w)

• Assume w prefers m in M0 to m’ in M2

• If m prefers w to w’, then M2 not stable because (m, w) is a blocking pair

• If m does not prefer w to w’, then m has a better partner in matching other than M0, which contradicts claim 1. □

Life’s Not Fair• Clearly neither M0 or Mz are fair to both

sexes (except when they are the same), so we want some better stable marriage.

• In the worst case, the number of stable marriages grows exponentially with n, so exhaustive search is not practical for large n (proved by Knuth).

• The sex equal solution is NP complete.

• Egalitarian (optimal) solution which can be found in O(n3) time produces the maximum total happiness.

Other Directions• Couples• Incomplete Preference Lists• Lying (related to economics & game theory)

– Not possible for a set of men to lie in such a way that they all get a better partner than they would in M0

– If all men lie by saying any woman worse than their partner in M0 is unacceptable, running GS with women proposing produces M0

• Ties in the Preference Lists• Weighted Variations

New Research: The Good of the Many vs. the Good of the One

• How much happiness is gained when we allow 1 blocking pair?

• How much happiness is gained when we allow k blocking pairs?

The Infinite Case

• Research being done at WPI

• Turns out that in the infinite case, not everybody gets married!

• Come to Discrete Math Day, May 3 at WPI to learn more

http://users.wpi.edu/~bdonovan/DMD/

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