1. Specialised Binary Constraint for the Stable Marriage Problem with Ties and incomplete preference lists By Chris Unsworth and Patrick Prosser

2. Contents The Problem Representing Ties Specialised Binary Constraint Computational Comparison Conclusion Questions

3. The Stable Marriage Problem Men Women Bob Ian Jon : Ian Jon Bob : Jon Ian Bob : Bob Jon Ian : Sue Jan Liz : Liz Jan Sue : Jan Sue Liz Jan Liz Sue We have n menand n women Each man ranks the n women And each woman ranks the n men Objective : To find a matching of men to women Such that the matching is Stable

4. The Stable Marriage Problem Men Women Bob Ian Jon : Ian Jon Bob : Jon Ian Bob : Bob Jon Ian : Sue Jan Liz : Liz Jan Sue : Jan Sue Liz Jan Liz Sue A Matching But not a stable one  Bob and Sue would rather be matched to each other than to there assigned partners In this matching Bob and Sue are a Blocking pair A matching is only stable iff it contains no Blocking pairs

5. Ties and incomplete Preference lists Men Women Alf Bob Tom Ian Jim : Tom Alf Bob Ian : Ian (Alf Bob Jim) : (Alf Ian) Tom Bob : Tom (Jim Ian Bob) Alf : Ian Jim (Tom Bob) : Zoe (Ann Liz) Joe : Liz Jes (Ann Zoe) : (Ann Jes Liz Zoe) : Ann Jes Liz Zoe Joe : Joe Zoe Jes Ann Joe Liz Zoe Jes Ties indicate indifference (shown in brackets) Incomplete list indicate some potential partners are unacceptable

6. Representing Ties Each entry represented as a Triple Absolute potion in the list (ties broken arbitrarily) Absolute potion of the first person in the tie Absolute potion of the last person in the tie Zoe : Tom (Jim Ian Bob) Alf {1,1,1} {2,2,4} {2,3,4} {2,4,4} {5,5,5}

7. Problem properties Different levels of stability Here we use weak stability (details in the paper) All instances have contain a weakly stable matching can be found in polynomial time Different size matching To find the largest is NP-Hard

8. Previous Constraint Encoding 2n Integer variables (z 1..z n, z n+1..z 2n ) Initial domains of 1.. L (L = length of preference list) Domain values represent preferences i.e. Z i = 3 : person i matched to 3 rd choice O(n 2 ) constraints one for each z i,z j pair where 1 ≤ i ≤ n < j ≤ 2n Explicit list of allowed tuples Each O(n 2 ) in size Propagates in O(n 4 ) time Takes O(n 4 ) space

9. New constraint Encoding Same 2n variables O(n 2 ) specialised constraints one for each z i,z j pair where 1 ≤ i ≤ n < j ≤ 2n Two methods init() Called when initialised (details in paper) remVal(z,a) Called when a is removed from z (details in paper) Propagates in O(n 3 ) time Takes O(n 2 ) space

10. Conclusion & Future work New specialised binary constraint for SMTI Significant performance increase over previous constraint solutions n-ary constraint Empirical comparison

11. Questions? Thank you