matching problems

Matching problems

Toby Walsh

NICTA and UNSW

Motivation

Agents may express preferences for issues other than a collective decision Preferences for a spouse Preferences for a room-mate Preferences for a work assignment …

All examples of matching problems Husbands with Wives Students with Rooms Doctors with Hospitals …

Stable marriage

What do I know? Ask me again after

June 21st

Mathematical abstraction Idealized model All men can totally

order all women …

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Stable marriage

Given preferences of n men Greg: Amy>Bertha>Clare Harry: Bertha>Amy>Clare Ian: Amy>Bertha>Clare

Given preferences of n women Amy: Harry>Greg>Ian Bertha: Greg>Harry>Ian Clare: Greg>Harry>Ian

Stable marriage

Given preferences of n men Greg: Amy>Bertha>Clare Harry: Bertha>Amy>Clare Ian: Amy>Bertha>Clare

Given preferences of n women Amy: Harry>Greg>Ian Bertha: Greg>Harry>Ian Clare: Greg>Harry>Ian

Find a stable marriage

Stable marriage

Given preferences of n menGiven preferences of n womenFind a stable marriage

Assignment of men to women (or equivalently of women to men)

Idealization: everyone marries at the same time No pair (man,woman) not married to each other

would prefer to run off together Idealization: assumes no barrier to divorce!

Stable marriage

Unstable solution Greg: Amy>Bertha>Clare Harry: Bertha>Amy>Clare Ian: Amy>Bertha>Clare

Amy: Harry>Greg>Ian Bertha: Greg>Harry>Ian Clare: Greg>Harry>Ian

Bertha & Greg would prefer to elope

Stable marriage

One solution Greg: Amy>Bertha>Clare Harry: Bertha>Amy>Clare Ian: Amy>Bertha>Clare


Men do ok, women less well

Stable marriage

Another solution Greg: Amy>Bertha>Clare Harry: Bertha>Amy>Clare Ian: Amy>Bertha>Clare


Women do ok, men less well

Gale Shapley algorithm

Initialize every person to be freeWhile exists a free man

Find best woman he hasn’t proposed to yet If this woman is free, declare them engaged

Else if this woman prefers this proposal to her current fiance then declare them engaged (and “free” her current fiance)

Else this woman prefers her current fiance and she rejects the proposial


Initialize every person to be free While exists a free man

Find best woman he hasn’t proposed to yet

If this woman is free, declare them engaged



Greg: Amy>Bertha>Clare

Harry: Bertha>Amy>Clare

Ian: Amy>Bertha>Clare

Amy: Harry>Greg>Ian

Bertha: Greg>Harry>Ian

Clare: Greg>Harry>Ian


Terminates with everyone matched Suppose some man is unmatched at the end Then some woman is also unmatched But once a woman is matched, she only “trades” up Hence this woman was never proposed to

• But if a man is unmatched, he has proposed to and been rejected by every woman

This is a contradiction as he has never proposed to the unmatched woman!


Terminates with perfect matching Suppose there is an unstable pair in the final

matching Case 1. This man never proposed to this

woman As men propose to women in preference order, man

must prefer his current fiance Hence current pairing is stable!


Terminates with perfect matching Suppose there is an unstable pair in the final

matching Case 1. This man never proposed to this

woman Case 2. This man had proposed to this woman

But the woman rejected him (immediately or later) However, women only ever trade up Hence the woman prefers her current partner So the current pairing is stable!


Each of n men can make at most (n-1) proposalsHence GS runs in O(n2) time

There may be more than one stable marriageGS finds man optimal solutionThere is no stable matching in which any man does betterGS finds woman pessimal solutionIn all stable marriages, every woman does at least as well or better


GS finds male optimal solution Suppose some man is engaged to someone who is not

the best possible woman Then they have proposed and been rejected by this woman Consider first such man A, who is rejected by X in favour

ultimately of marrying B• There exists (some other) stable marriage with A married to X

and B to Y By assumption, B has not yet been rejected by his best

possible woman Hence B must prefer X at least as much as his best possible

woman So (A,X) (B,Y) is not a stable marriage as B and X would

prefer to elope!


GS finds woman pessimal solution Suppose some womman is engaged to someone who is

not the worst possible man Let (A,X) be married but A is not worst possible man for X There exists a stable marriage with (B,X) (A,Y) and B

worse than A for X By male optimality, A prefers X to Y Then (A,Y) is unstable!


Initialize every person to be free While exists a free man

Find best woman he hasn’t proposed to yet

If this woman is free, declare them engaged






Amy: Harry>Greg>Ian



Gale Shapley algorithm woman optimal

Initialize every person to be free While exists a free woman

Find best man she hasn’t proposed to yet

If this man is free, declare them engaged

Else if this man prefers this proposal to his current fiance then declare them engaged (and “free” his current fiance)

Else this man prefers his current fiance and he rejects the proposial




Amy: Harry>Greg>Ian



Extensions: ties

Cannot always make up our minds Angelina or Jennifer? Either would be equally

good! Stability

(weak) no couple strictly prefers each other

(strong) no couple such that one strictly prefers the other, and the other likes them as much or more

Extensions: ties

Stability (weak) no couple strictly prefers each other (strong) no couple such that one strictly prefers the

other, and the other likes them as much or moreExistence

Strongly stable marriage may not exist O(n4) algorithm for deciding existence

Weakly stable marriage always exists Just break ties aribtrarily Run GS, resulting marriage is weakly stable!

Extensions: incomplete preferences

There are some people we may be unwilling to marry I’d prefer to remain single

than marry Margaret (m,w) unstable iff

m and w do not find each other unacceptable

m is unmatched or prefers w to current fiance

w is unmatched or prefers w to current fiance

Extensions: incomplete prefs

GS algorithm Extends easily

Men and woman partition into two sets Those who have partners in all stable marriages Those who do not have partners in any stable

marriage

Extensions: ties & incomplete prefs

Weakly stable marriages may be different sizes Unlike with just ties where they are all complete

Finding weakly stable marriage of max. cardinality is NP-hard Even if only women declare ties

Extensions: unequal numbers

For instance, more men than woman See China!

Matching unstable if pair (m,w) m and w do not find each other unacceptable m is unmatched or prefers w to current fiance w is unmatched or prefers w to current fiance

Extensions: unequal numbers

GS algorithm Extends easily

If |men|>|women| then all woman are married in a stable solution Men partition into two sets

Those who have partners in all stable marriages Those who do not have partners in any stable

marriage

Strategy proofness

GS is strategy proof for men Assuming male optimal algorithm No man can do better than the male optimal

solution

However, women can profit from lying Assuming male optimal algorithm is run And they know complete preference lists

Strategy proofness

Greg: Amy>Bertha>Clare Harry: Bertha>Amy>Clare Ian: Amy>Bertha>Clare


Amy lies

Amy: Harry>Ian>Greg Bertha: Greg>Harry>Ian Clare: Greg>Harry>Ian

Impossibility of strategy proofness

[Roth 82] No matching procedure for which stating the truth

is a dominant strategy for all agents when preference lists can be incomplete

Consider Greg: Amy>Bertha Amy: Harry>Greg Harry: Bertha>Amy Bertha: Greg>Harry

Two stable marriages: (Greg,Amy)(Harry,Bertha) or (Greg,Bertha)(Harry,Amy)


Consider Greg: Amy>Bertha Amy: Harry>Greg Harry: Bertha>Amy Bertha: Greg>Harry

Two stable marriages: (Greg,Amy)(Harry,Bertha) or (Greg,Bertha)(Harry,Amy)

Suppose we get male optimal solution (Greg,Amy)(Harry,Bertha) If Amy lies and says Harry is only acceptable partner Then we must get (Harry,Amy)(Greg,Bertha) as this is the

only stable marriage Other cases can be manipulated in a similar way


Strategy proofness is hard to achieve [Roth and Sotomayor 90] With any matching procedure, if

preference lists are strict, and there is more than one stable marriage, then at least one agent can profitably lie assuming the other agents tell the truth

But one side can have no incentive to lie [Dubins and Freedman 81] With a male-proposing

matching algorithm, it is a weakly-dominant strategy for the men to tell the truth

Weakly-dominant=???

Some lessons learnt?

Historically men have in fact proposed to woman Men: propose early and

often Men: don’t lie Women: ask out the

guys (Bad news) Women:

lying and turning down proposals can be to your advantage!

Hospital residents problem

Matching of residents to hospitals Hospitals express preferences over resident Hospitals declare how many residents they take Residents express preferences over hospitals

Matching (h,r) unstable iff They are acceptable to each other r is unmatched or r prefers h to current hospital h is not full or h prefers r to one its current residents

Stable roommate

2n agents Each ranks every other agent Pair up agents according to preferences

No stable matching may exist Adam: Bob>Chris>Derek Bob: Chris>Adam>Derek Chris: Adam>Bob>Derek Derek: Adam>Bob>Chris

Conclusions

Preferences turn up in matching problems Stable marriage Roommate Hospital-residents problem

We may wish to represent Ties Incompatability (aka “incomplete preference lists) ..

Complexity depends on this Stable marriage on total orders is O(n2) Stable marriage with ties and incomplete preference lists

is NP-hard

Conclusions

Many different formalisms for representing preferences CP nets, soft constraints, utilities, …

Many different dimensions to analyse these formalisms along Expressiveness, succinctness, …

Many interesting computational problems Computing optimal, ordering outcomes, manipulating

result, deciding when to terminate preference elicitation, …

matching problems

Documents

unmatched woman

free manfind best woman

current fianceelse

timeno pair man

womanas men

unstable pair

perfect matchingsuppose

preference order