Pareto Optimality in House Allocation Problems David Manlove Department of Computing Science University of Glasgow David Abraham Computer Science Department.

Pareto Optimality in House Allocation Problems David Manlove Department of Computing Science University of Glasgow David Abraham Computer Science Department Carnegie-Mellon University Katarína Cechlárová Institute of Mathematics PJ Safárik University in Košice Kurt Mehlhorn Max-Planck-Institut fűr Informatik Saarbrűcken Supported by Royal Society of Edinburgh/Scottish Executive Personal Research Fellowship and Engineering and Physical Sciences Research Council grant GR/R84597/01

2 House Allocation problem (HA) Set of agents A={a 1, a 2, …, a r } Set of houses H={h 1, h 2, …, h s } Each agent a i has an acceptable set of houses A i  H a i ranks A i in strict order of preference Example: a 1 : h 2 h 1 a 2 : h 3 h 4 h 2 a 3 : h 4 h 3 a 4 : h 1 h 4 Let n = r + s and let m =total length of preference lists  a 1 finds h 1 and h 2 acceptable a 3 prefers h 4 to h 3

3 Applications House allocation context: Large-scale residence exchange in Chinese housing markets Yuan, 1996 Allocation of campus housing in American universities, such as Carnegie-Mellon, Rochester and Stanford Abdulkadiroğlu and Sönmez, 1998 Other matching problems: US Naval Academy: students to naval officer positions Roth and Sotomayor, 1990 Scottish Executive Teacher Induction Scheme Assigning students to projects

4 The underlying graph Weighted bipartite graph G=(V,E) Vertex set V=A  H Edge set: { a i, h j }  E if and only if a i finds h j acceptable Weight of edge { a i, h j } is rank of h j in a i ’s preference list Example a 1 : h 2 h 1 a 2 : h 3 h 4 h 2 a 3 : h 4 h 3 a 4 : h 1 h 4 a1a1 a2a2 a3a3 a4a4 h1h1 h2h2 h3h3 h4h4     2 1 1 2 3 2 1 1 2

5 Matchings in the underlying graph A matching M in G is a subset of E such that each vertex of G is incident to at most one edge of M Each agent is assigned to at most one house Each house is assigned at most one agent An agent is only ever assigned to an acceptable house Example a 1 : h 2 h 1 a 2 : h 3 h 4 h 2 a 3 : h 4 h 3 a 4 : h 1 h 4 a1a1 a2a2 a3a3 a4a4 h1h1 h2h2 h3h3 h4h4     2 1 1 2 3 2 1 1 2 M={(a 1, h 1 ), (a 2, h 4 ), (a 3, h 3 )} M(a 1 )=h 1

6 A matching M in G is a subset of E such that each vertex of G is incident to at most one edge of M Each agent is assigned to at most one house Each house is assigned at most one agent An agent is only ever assigned to an acceptable house Example a 1 : h 2 h 1 a 2 : h 3 h 4 h 2 a 3 : h 4 h 3 a 4 : h 1 h 4 a1a1 a2a2 a3a3 a4a4 h1h1 h2h2 h3h3 h4h4     2 1 1 2 3 2 1 1 2 M={(a 1, h 2 ), (a 2, h 3 ), (a 3, h 4 ), (a 4, h 1 )} Matchings in the underlying graph

7 Pareto optimal matchings A matching M 1 is Pareto optimal if there is no matching M 2 such that: 1. Some agent is better off in M 2 than in M 1 2. No agent is worse off in M 2 than in M 1 Example M 1 is not Pareto optimal since a 1 and a 2 could swap houses – each would be better off M 2 is Pareto optimal a 1 : h 2 h 1 a 2 : h 1 h 2 a 3 : h 3 a 1 : h 2 h 1 a 2 : h 1 h 2 a 3 : h 3 M1M1 M2M2

8 Testing for Pareto optimality A matching M is maximal if there is no agent a and house h, each unmatched in M, such that a finds h acceptable A matching M is trade-in-free if there is no matched agent a and unmatched house h such that a prefers h to M(a) A matching M is coalition-free if there is no coalition, i.e. a sequence of matched agents  a 0,a 1,…,a r-1  such that a i prefers M(a i ) to M(a i+1 ) (0  i  r-1) a 1 : h 2 h 1 a 2 : h 3 h 4 h 2 a 3 : h 4 h 3 a 4 : h 1 h 4 Proposition: M is Pareto optimal if and only if M is maximal, trade-in-free and coalition-free M is not maximal due to a 3 and h 3

9 Testing for Pareto optimality A matching M is maximal if there is no agent a and house h, each unmatched in M, such that a finds h acceptable A matching M is trade-in-free if there is no matched agent a and unmatched house h such that a prefers h to M(a) A matching M is coalition-free if there is no coalition, i.e. a sequence of matched agents  a 0,a 1,…,a r-1  such that a i prefers M(a i ) to M(a i+1 ) (0  i  r-1) a 1 : h 2 h 1 a 2 : h 3 h 4 h 2 a 3 : h 4 h 3 a 4 : h 1 h 4 Proposition: M is Pareto optimal if and only if M is maximal, trade-in-free and coalition-free M is not trade-in-free due to a 2 and h 3

10 Testing for Pareto optimality A matching M is maximal if there is no agent a and house h, each unmatched in M, such that a finds h acceptable A matching M is trade-in-free if there is no matched agent a and unmatched house h such that a prefers h to M(a) A matching M is coalition-free if there is no coalition, i.e. a sequence of matched agents  a 0,a 1,…,a r-1  such that a i prefers M(a i+1 ) to M(a i ) (0  i  r-1) a 1 : h 2 h 1 a 2 : h 3 h 4 h 2 a 3 : h 4 h 3 a 4 : h 1 h 4 Proposition: M is Pareto optimal if and only if M is maximal, trade-in-free and coalition-free M is not coalition-free due to  a 1, a 2, a 4  a1a1 a3a3 a4a4 h1h1 h3h3 h4h4 a2a2 h2h2

11 Testing for Pareto optimality A matching M is maximal if there is no agent a and house h, each unmatched in M, such that a finds h acceptable A matching M is trade-in-free if there is no matched agent a and unmatched house h such that a prefers h to M(a) A matching M is coalition-free if there is no coalition, i.e. a sequence of matched agents  a 0,a 1,…,a r-1  such that a i prefers M(a i+1 ) to M(a i ) (0  i  r-1) Lemma: M is Pareto optimal if and only if M is maximal, trade-in-free and coalition-free Theorem: we may check whether a given matching M is Pareto optimal in O(m ) time

12 Finding a Pareto optimal matching Simple greedy algorithm, referred to as the serial dictatorship mechanism by economists for each agent a in turn if a has an unmatched house on his list match a to the most-preferred such house; else report a as unmatched; Theorem: The serial dictatorship mechanism constructs a Pareto optimal matching in O(m ) time Abdulkadiroğlu and Sönmez, 1998 Example a 1 : h 1 h 2 h 3 a 2 : h 1 h 2 a 3 : h 1 h 2 a 1 : h 1 h 2 h 3 a 2 : h 1 h 2 a 3 : h 1 h 2 M 1 ={(a 1,h 1 ), (a 2,h 2 )} M 2 ={(a 1,h 3 ), (a 2,h 2 ), (a 3,h 1 )}

13 Related work Rank maximal matchings Matching M is rank maximal if, in M 1. Maximum number of agents obtain their first-choice house; 2. Subject to (1), maximum number of agents obtain their second-choice house; etc. Irving, Kavitha, Mehlhorn, Michail, Paluch, SODA 04 A rank maximal matching is Pareto optimal, but need not be of maximum size Popular matchings Matching M is popular if there is no other matching M ’ such that: more agents prefer M ’ to M than prefer M to M ’ Abraham, Irving, Kavitha, Mehlhorn, SODA 05 A popular matching is Pareto optimal, but need not exist Maximum cardinality minimum weight matchings Such a matching M may be found in G in O(  nmlog n) time Gabow and Tarjan, 1989 M is a maximum Pareto optimal matching

14 Faster algorithm for finding a maximum Pareto optimal matching Three-phase algorithm with O(  nm) overall complexity Phase 1 – O(  nm) time Find a maximum matching in G Classical O(  nm) augmenting path algorithm Hopcroft and Karp, 1973 Phase 2 – O(m) time Enforce trade-in-free property Phase 3 – O(m) time Enforce coalition-free property Extension of Gale’s Top-Trading Cycles (TTC) algorithm Shapley and Scarf, 1974

15 Phase 1 a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Maximum matching M in G has size 8 M must be maximal No guarantee that M is trade-in-free or coalition-free

16 Phase 1 a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Maximum matching M in G has size 9 M must be maximal No guarantee that M is trade-in-free or coalition-free

17 Phase 1 a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Maximum matching M in G has size 9 M must be maximal No guarantee that M is trade-in-free or coalition-free M not coalition-free M not trade-in-free

18 Phase 2 outline Repeatedly search for a matched agent a and an unmatched house h such that a prefers h to h ’= M(a) Promote a to h h ’ is now unmatched Example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9

22 Phase 2 termination Once Phase 2 terminates, matching is trade-in-free With suitable data structures, Phase 2 is O(m) Coalitions may remain… a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9

23 Build a path P of agents (represented by a stack) Each house is initially unlabelled Each agent a has a pointer p(a) pointing to M(a) or the first unlabelled house on a ’s preference list (whichever comes first) Keep a counter c(a) for each agent a (initially c(a)=0 ) This represents the number of times a appears on the stack Outer loop iterates over each matched agent a such that p(a)  M(a) Initialise P to contain agent a Inner loop iterates while P is nonempty Pop an agent a ’ from P If c(a ’ )=2 we have a coalition (CYCLE) Remove by popping the stack and label the houses involved Else if p(a ’ )=M(a ’ ) we reach a dead end (BACKTRACK) Label M(a ’ ) Else add a ’’ where p(a ’ )=M(a ’’ ) to the path (EXTEND) Push a ’ and a ’’ onto the stack Increment c(a ’’ ) Phase 3 outline

24 Once Phase 3 terminates, matching is coalition-free Phase 3 termination a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Phase 3 is O(m) Theorem: A maximum Pareto optimal matching can be found in O(  nm) time

25 Minimum Pareto optimal matchings Theorem: Problem of finding a minimum Pareto optimal matching is NP-hard Result holds even if all preference lists have length 3 Reduction from Minimum Maximal Matching Problem is approximable within a factor of 2 Follows since any Pareto optimal matching is a maximal matching in the underlying graph G Any two maximal matchings differ in size by at most a factor of 2 Korte and Hausmann, 1978

26 Conclusions Open problems - finding a maximum Pareto optimal matching Ties in the preference lists Solvable in O(  nmlog n) time Solvable in O(  nm) time? One-many case (houses may have capacity >1 ) Non-bipartite case Solvable in O(  (n  (m, n))mlog 3/2 n) time D.J. Abraham, D.F. Manlove Pareto optimality in the Roommates problem Technical Report TR-2004-182 of the Computing Science Department of Glasgow University Solvable in O(  nm) time? Further details D.J. Abraham, K. Cechlárová, D.F. Manlove and K. Mehlhorn, Pareto Optimality in House Allocation Problems, In Proceedings of ISAAC 2004, vol 3341 of Lecture Notes in Computer Science

28 The envy graph Straightforward to check a given matching M for the maximality and trade-in-free properties in O(m) time To check for the existence of a coalition: Form the envy graph of M, denoted by G(M) Vertex for each matched agent Edge from a i to a j if and only if a i prefers M(a j ) to M(a i ) Example a 1 : h 2 h 1 a 2 : h 3 h 4 h 2 a 3 : h 4 h 3 a 4 : h 1 h 4 M admits a coalition if and only if G(M) has a directed cycle Proposition: we may check whether a given matching M is Pareto optimal in O(m ) time a1a1 a4a4 a2a2

29 Phase 2 outline For each house h, maintain a list L h, initially containing those pairs ( a, r ) such that: a is a matched agent who prefers h to M(a) r is the rank of h in a ’s list Maintain a stack S of unmatched houses h whose list L h is nonempty Each matched agent a maintains a pointer curr a to the rank of M(a) Example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 L h 6 =  (a 6, 5), (a 7, 4), (a 8, 6)  L h 7 =  (a 6, 6), (a 7, 5), (a 8, 5)  L h 8 =  (a 6, 4)  L h 10 =  (a 7, 7)  L h 11 =  (a 6, 8)  S =  h 6, h 7  curr a 6 =10 curr a 7 =9

30 Phase 2 outline a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curr a := h; h := h’; push h onto the stack if L h is nonempty; L h 6 =  (a 6, 5), (a 7, 4), (a 8, 6)  L h 7 =  (a 6, 6), (a 7, 5), (a 8, 5)  L h 8 =  (a 6, 4)  L h 10 =  (a 7, 7)  L h 11 =  (a 6, 8)  S =  h 6, h 7  curr a 6 =10 curr a 7 =9

31 Phase 2 example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curr a := h; h := h’; push h onto the stack if L h is nonempty; L h 6 =  (a 6, 5), (a 7, 4), (a 8, 6)  L h 7 =  (a 6, 6), (a 7, 5), (a 8, 5)  L h 8 =  (a 6, 4)  L h 10 =  (a 7, 7)  L h 11 =  (a 6, 8)  S =  h 6, h 7  curr a 6 =10 curr a 7 =9

32 Phase 2 example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curr a := h; h := h’; push h onto the stack if L h is nonempty; L h 6 =  (a 6, 5), (a 7, 4), (a 8, 6)  L h 7 =  (a 7, 5), (a 8, 5)  L h 8 =  (a 6, 4)  L h 10 =  (a 7, 7)  L h 11 =  (a 6, 8)  S =  h 6, h 10  curr a 6 =6 curr a 7 =9

33 Phase 2 example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curr a := h; h := h’; push h onto the stack if L h is nonempty; L h 6 =  (a 6, 5), (a 7, 4), (a 8, 6)  L h 7 =  (a 7, 5), (a 8, 5)  L h 8 =  (a 6, 4)  L h 10 =  (a 7, 7)  L h 11 =  (a 6, 8)  S =  h 6, h 10  curr a 6 =6 curr a 7 =9

34 Phase 2 example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curr a := h; h := h’; push h onto the stack if L h is nonempty; L h 6 =  (a 6, 5), (a 7, 4), (a 8, 6)  L h 7 =  (a 7, 5), (a 8, 5)  L h 8 =  (a 6, 4)  L h 10 =  L h 11 =  (a 6, 8)  S =  h 6, h 11  curr a 6 =6 curr a 7 =7

35 Phase 2 example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curr a := h; h := h’; push h onto the stack if L h is nonempty; L h 6 =  (a 6, 5), (a 7, 4), (a 8, 6)  L h 7 =  (a 7, 5), (a 8, 5)  L h 8 =  (a 6, 4)  L h 10 =  L h 11 =  (a 6, 8)  S =  h 6, h 11  curr a 6 =6 curr a 7 =7

36 Phase 2 example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curr a := h; h := h’; push h onto the stack if L h is nonempty; L h 6 =  (a 6, 5), (a 7, 4), (a 8, 6)  L h 7 =  (a 7, 5), (a 8, 5)  L h 8 =  (a 6, 4)  L h 10 =  L h 11 =  S =  h 6  curr a 6 =6 curr a 7 =7

37 Phase 2 example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curr a := h; h := h’; push h onto the stack if L h is nonempty; L h 6 =  (a 7, 4), (a 8, 6)  L h 7 =  (a 7, 5), (a 8, 5)  L h 8 =  (a 6, 4)  L h 10 =  L h 11 =  S =  h 7  curr a 6 =5 curr a 7 =7

38 Phase 2 example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curr a := h; h := h’; push h onto the stack if L h is nonempty; L h 6 =  (a 7, 4), (a 8, 6)  L h 7 =  (a 7, 5), (a 8, 5)  L h 8 =  (a 6, 4)  L h 10 =  L h 11 =  S =  h 7  curr a 6 =5 curr a 7 =7

39 Phase 2 example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curr a := h; h := h’; push h onto the stack if L h is nonempty; L h 6 =  (a 7, 4), (a 8, 6)  L h 7 =  (a 8, 5)  L h 8 =  (a 6, 4)  L h 10 =  L h 11 =  S =  curr a 6 =5 curr a 7 =5

40 Phase 2 example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curr a := h; h := h’; push h onto the stack if L h is nonempty; L h 6 =  (a 7, 4), (a 8, 6)  L h 7 =  (a 8, 5)  L h 8 =  (a 6, 4)  L h 10 =  L h 11 =  S =  curr a 6 =5 curr a 7 =5

41 Phase 2 termination a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9

42 Once Phase 2 terminates, matching is trade-in-free Phase 2 is O(m) Coalitions may remain… Phase 2 termination a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9

43 Elimination of coalitions Repeatedly finding and eliminating coalitions takes O(m 2 ) time Cycle detection in G(M) takes O(m) time O(m) coalitions in the worst case Faster method: extension of TTC algorithm An agent matched to his/her first-choice house cannot be in a coalition Such an agent can be removed from consideration Houses matched to such agents are no longer exchangeable Such a house can be removed from consideration This rule can be recursively applied until either No agent remains (matching is coalition-free) A coalition exists, which can be found and removed Phase 3 description

44 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 1h 1 a 2 0h 2 a 3 0h 3 a 4 0h 4 a 5 0h 5 a 6 0h 6 a 7 0h 7 a 8 0h 8 a 9 0h 9 P=a1P=a1 a1a1

45 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 1h 1 a 2 0h 2 a 3 0h 3 a 4 1h 4 a 5 0h 5 a 6 0h 6 a 7 0h 7 a 8 0h 8 a 9 0h 9 a1a1 P=a1, a4P=a1, a4 EXTEND a4a4

46 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 1h 1 a 2 0h 2 a 3 1h 3 a 4 1h 4 a 5 0h 5 a 6 0h 6 a 7 0h 7 a 8 0h 8 a 9 0h 9 a1a1 EXTEND a4a4 P=  a 1, a 4, a 3  a3a3

47 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 1h 1 a 2 0h 2 a 3 1h 3 a 4 1h 4 a 5 1h 5 a 6 0h 6 a 7 0h 7 a 8 0h 8 a 9 0h 9 a1a1 EXTEND a4a4 P=  a 1, a 4, a 3, a 5  a3a3 a5a5

48 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 1h 1 a 2 0h 2 a 3 1h 3 a 4 2h 4 a 5 1h 5 a 6 0h 6 a 7 0h 7 a 8 0h 8 a 9 0h 9 a1a1 EXTEND a4a4 P=  a 1, a 4, a 3, a 5, a 4  a3a3 a5a5

49 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 1h 1 a 2 0h 2 a 3 1h 3 a 4 2h 4 a 5 1h 5 a 6 0h 6 a 7 0h 7 a 8 0h 8 a 9 0h 9 a1a1 CYCLE a4a4 P=  a 1, a 4, a 3, a 5  a3a3 a5a5

50 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 1h 1 a 2 0h 2 a 3 0h 3 a 4 0h 4 a 5 0h 5 a 6 0h 6 a 7 0h 7 a 8 0h 8 a 9 0h 9 a1a1 CYCLE a4a4 a3a3 a5a5 P=  a 1, a 4, a 3, a 5 

51 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 1h 1 a 2 0h 2 a 3 0h 3 a 4 0h 4 a 5 0h 5 a 6 0h 6 a 7 0h 7 a 8 0h 8 a 9 0h 9 a1a1 P=a1P=a1

52 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 1h 1 a 2 1h 2 a 3 0h 3 a 4 0h 4 a 5 0h 5 a 6 0h 6 a 7 0h 7 a 8 0h 8 a 9 0h 9 a1a1 EXTEND P=  a 1, a 2  a2a2

53 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 1h 1 a 2 1h 2 a 3 0h 3 a 4 0h 4 a 5 0h 5 a 6 0h 6 a 7 0h 7 a 8 0h 8 a 9 1h 9 a1a1 EXTEND P=  a 1, a 2, a 9  a2a2 a9a9

54 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 1h 1 a 2 1h 2 a 3 0h 3 a 4 0h 4 a 5 0h 5 a 6 0h 6 a 7 0h 7 a 8 0h 8 a 9 1h 9 a1a1 BACKTRACK P=  a 1, a 2  a2a2 a9a9

55 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 1h 1 a 2 1h 2 a 3 0h 3 a 4 0h 4 a 5 0h 5 a 6 0h 6 a 7 0h 7 a 8 0h 8 a 9 0h 9 a1a1 P=  a 1, a 2  a2a2

56 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 2h 1 a 2 1h 2 a 3 0h 3 a 4 0h 4 a 5 0h 5 a 6 0h 6 a 7 0h 7 a 8 0h 8 a 9 0h 9 a1a1 EXTEND P=  a 1, a 2, a 1  a2a2

57 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 2h 1 a 2 1h 2 a 3 0h 3 a 4 0h 4 a 5 0h 5 a 6 0h 6 a 7 0h 7 a 8 0h 8 a 9 0h 9 a1a1 CYCLE P=  a 1, a 2  a2a2

58 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 0h 1 a 2 0h 2 a 3 0h 3 a 4 0h 4 a 5 0h 5 a 6 0h 6 a 7 0h 7 a 8 0h 8 a 9 0h 9 a1a1 CYCLE P=  a 1, a 2  a2a2

59 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 0h 1 a 2 0h 2 a 3 0h 3 a 4 0h 4 a 5 0h 5 a 6 1h 6 a 7 0h 7 a 8 0h 8 a 9 0h 9 P=a6P=a6 a6a6

60 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 0h 1 a 2 0h 2 a 3 0h 3 a 4 0h 4 a 5 0h 5 a 6 1h 6 a 7 0h 7 a 8 1h 8 a 9 0h 9 P=  a 6, a 8  EXTEND a6a6 a8a8

61 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 0h 1 a 2 0h 2 a 3 0h 3 a 4 0h 4 a 5 0h 5 a 6 1h 6 a 7 1h 7 a 8 1h 8 a 9 0h 9 P=  a 6, a 8, a 7  EXTEND a6a6 a8a8 a7a7

62 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 0h 1 a 2 0h 2 a 3 0h 3 a 4 0h 4 a 5 0h 5 a 6 2h 6 a 7 1h 7 a 8 1h 8 a 9 0h 9 P=  a 6, a 8, a 7, a 6  EXTEND a6a6 a8a8 a7a7

63 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 a 1 0h 1 a 2 0h 2 a 3 0h 3 a 4 0h 4 a 5 0h 5 a 6 2h 6 a 7 1h 7 a 8 1h 8 a 9 0h 9 P=  a 6, a 8, a 7  CYCLE a6a6 a8a8 a7a7 Agent counter House label

64 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 0h 1 a 2 0h 2 a 3 0h 3 a 4 0h 4 a 5 0h 5 a 6 0h 6 a 7 0h 7 a 8 0h 8 a 9 0h 9 P=  a 6, a 8, a 7  CYCLE a6a6 a8a8 a7a7

65 Phase 3: example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 Agent counter House label a 1 0h 1 a 2 0h 2 a 3 0h 3 a 4 0h 4 a 5 0h 5 a 6 0h 6 a 7 0h 7 a 8 0h 8 a 9 0h 9 P= 

66 Initial property rights Suppose A ’  A and each member of A ’ owns a house initially For each agent a  A ’, denote this house by h(a) Truncate a ’s list at h(a) Form matching M by pre-assigning a to h(a) Use Hopcroft-Karp algorithm to augment M to a maximum cardinality matching M ’ in restricted HA instance Then proceed with Phases 2 and 3 as before Constructed matching M ’ is individually rational If A ’= A then we have a housing market TTC algorithm finds the unique matching that belongs to the core Shapley and Scarf, 1974 Roth and Postlewaite, 1977

67 Interpolation of Pareto optimal matchings Given an HA instance I, p - (I) and p + (I) denote the sizes of a minimum and maximum Pareto optimal matching Theorem: I admits a Pareto optimal matching of size k, for each k such that p - (I)  k  p + (I) Given a Pareto optimal matching of size k, O ( m ) algorithm constructs a Pareto optimal matching of size k+1 or reports that k = p + (I) Based on assigning a vector  r 1, …, r k  to an augmenting path P =  a 1, h 1, …, a k, h k  where r i =rank a i (h i ) Examples:  1,3,1   1,2,2  Find a lexicographically smallest augmenting path h1h1 h2h2 h3h3 h4h4 1 2 1 h5h5 a1a1 a2a2 a3a3 a4a4 2 3 1 2 1

Pareto Optimality in House Allocation Problems David Manlove Department of Computing Science University of Glasgow David Abraham Computer Science Department.

Similar presentations

Presentation on theme: "Pareto Optimality in House Allocation Problems David Manlove Department of Computing Science University of Glasgow David Abraham Computer Science Department."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Pareto Optimality in House Allocation Problems David Manlove Department of Computing Science University of Glasgow David Abraham Computer Science Department.

Similar presentations

Presentation on theme: "Pareto Optimality in House Allocation Problems David Manlove Department of Computing Science University of Glasgow David Abraham Computer Science Department."— Presentation transcript:

Similar presentations

About project

Feedback