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  …

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 men Given preferences of n women Find 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  Amy: Harry>Greg>Ian  Bertha: Greg>Harry>Ian  Clare: Greg>Harry>Ian Men do ok, women less well

Stable marriage Another 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 Women do ok, men less well

Gale Shapley algorithm 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 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

Gale Shapley algorithm 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 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 Greg: Amy>Bertha>Clare Harry: Bertha>Amy>Clare Ian: Amy>Bertha>Clare Amy: Harry>Greg>Ian Bertha: Greg>Harry>Ian Clare: Greg>Harry>Ian

Gale Shapley algorithm  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!

Gale Shapley algorithm 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!

Gale Shapley algorithm 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!

Gale Shapley algorithm Each of n men can make at most (n-1) proposals Hence GS runs in O(n 2 ) time There may be more than one stable marriage GS finds man optimal solution There is no stable matching in which any man does better GS finds woman pessimal solution In all stable marriages, every woman does at least as well or better

Gale Shapley algorithm 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!

Gale Shapley algorithm 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!

Gale Shapley algorithm 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 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 Greg: Amy>Bertha>Clare Harry: Bertha>Amy>Clare Ian: Amy>Bertha>Clare Amy: Harry>Greg>Ian Bertha: Greg>Harry>Ian Clare: Greg>Harry>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 Greg: Amy>Bertha>Clare Harry: Bertha>Amy>Clare Ian: Amy>Bertha>Clare Amy: Harry>Greg>Ian Bertha: Greg>Harry>Ian Clare: Greg>Harry>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 more Existence  Strongly stable marriage may not exist O(n 4 ) 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: Harry>Greg>Ian  Bertha: Greg>Harry>Ian  Clare: Greg>Harry>Ian  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)

Impossibility of strategy proofness  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

Impossibility of strategy proofness  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(n 2 )  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, …