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Pareto Optimality in House Allocation Problems

David Abraham Computer Science Department Carnegie-Mellon University. Pareto Optimality in House Allocation Problems. Katar í na Cechl á rov á Institute of Mathematics PJ Saf á rik University in Ko š ice. David Manlove Department of Computing Science University of Glasgow. Kurt Mehlhorn

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Pareto Optimality in House Allocation Problems

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  1. David Abraham Computer Science Department Carnegie-Mellon University Pareto Optimality in House Allocation Problems Katarína Cechlárová Institute of Mathematics PJ Safárik University in Košice David Manlove Department of Computing Science University of Glasgow 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={a1, a2, …, ar} • Set of houses H={h1, h2, …, hs} • Each agent ai has an acceptable set of houses Ai H • ai ranks Ai in strict order of preference • Example: • a1 : h2 h1 • a2 :h3h4 h2 • a3 : h4h3 • a4 : h1h4 • Let n=r+s and let m=total length of preference lists a1 finds h1 and h2 acceptable a3 prefers h4 to h3

  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:{ai,hj}E if and only if ai findshjacceptable • Weight of edge {ai,hj} is rank of hjinai’spreference list • Example • a1 : h2 h1 • a2 :h3h4 h2 • a3 : h4h3 • a4 : h1h4 a1  h1 2 1 a2 h2  3 1 a3 2 h3  2 1 1 a4 h4  2

  5. Matchings in the underlying graph • A matchingM in Gis 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 • a1 : h2 h1 • a2 :h3h4h2 • a3 : h4h3 • a4 : h1h4 a1  h1 2 M(a1)=h1 1 a2 h2  3 1 a3 2 h3  2 1 1 M={(a1, h1), (a2, h4), (a3, h3)} a4 h4  2

  6. Matchings in the underlying graph • A matchingM in Gis 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 • a1 : h2h1 • a2 : h3h4 h2 • a3 : h4h3 • a4 : h1h4 a1  h1 2 1 a2 h2  3 1 a3 2 h3  2 1 1 M={(a1, h2), (a2, h3), (a3, h4), (a4, h1)} a4 h4  2

  7. Pareto optimal matchings • A matchingM1 is Pareto optimal if there is no matching M2 such that: • Some agent is better off in M2 than in M1 • No agent is worse off in M2 than in M1 • Example • M1 is not Pareto optimal since a1 and a2 could swap houses – each would be better off • M2 is Pareto optimal • a1 : h2 h1 • a2 :h1h2 • a3 : h3 • a1 : h2h1 • a2 : h1h2 • a3 : h3 M1 M2

  8. Testing for Pareto optimality • A matchingM 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 a0 ,a1 ,…,ar-1 such that ai prefers M(ai) toM(ai+1) (0ir-1) • a1 : h2 h1 • a2 :h3h4 h2 • a3 : h4h3 • a4 : h1h4 • Proposition: M is Pareto optimal if and only if M is maximal, trade-in-free and coalition-free Mis not maximal due to a3 and h3

  9. Testing for Pareto optimality • A matchingM 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 a0 ,a1 ,…,ar-1 such that ai prefers M(ai) toM(ai+1) (0ir-1) • a1 : h2 h1 • a2 :h3h4 h2 • a3 : h4h3 • a4 : h1h4 • Proposition: M is Pareto optimal if and only if M is maximal, trade-in-free and coalition-free Mis not trade-in-free due to a2 and h3

  10. Testing for Pareto optimality • A matchingM 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 a0 ,a1 ,…,ar-1 such that ai prefers M(ai+1) toM(ai) (0ir-1) • a1 : h2 h1 • a2 :h3h4 h2 • a3 : h4h3 • a4 : h1h4 • Proposition: M is Pareto optimal if and only if M is maximal, trade-in-free and coalition-free a1 h1 Mis not coalition-free due to a1, a2, a4 a2 h2 a3 h3 a4 h4

  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 a0 ,a1 ,…,ar-1 such that ai prefers M(ai+1) toM(ai) (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 • a1 : h1h2h3 • a2 : h1h2 • a3 : h1h2 M1={(a1,h1), (a2,h2)} • a1 : h1 h2h3 • a2 : h1h2 • a3 : h1h2 M2={(a1,h3), (a2,h2), (a3,h1)}

  13. Related work • Rank maximal matchings • Matching M is rank maximal if, in M • Maximum number of agents obtain their first-choice house; • 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 matchingis 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 • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 • 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 • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 • 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 Mnot coalition-free • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 • Maximum matching M in G has size 9 • M must be maximal • No guarantee that M is trade-in-free or coalition-free Mnot 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 • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9

  19. 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 • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9

  20. 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 • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9

  21. 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 • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9

  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… • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9

  23. Phase 3 outline • 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’’ wherep(a’)=M(a’’)to the path (EXTEND) • Push a’ and a’’ onto the stack • Increment c(a’’)

  24. Phase 3 termination • Once Phase 3 terminates, matching is coalition-free • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5: h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 • 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))mlog3/2 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

  27. 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 ai to aj if and only if ai prefers M(aj) toM(ai) • Example • a1 : h2 h1 • a2 :h3h4 h2 • a3 : h4h3 • a4 : h1h4 • 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 a2 a1 a4

  28. Phase 2 outline • For each house h, maintain a list Lh , 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 Lh is nonempty • Each matched agent a maintains a pointer curra to the rank of M(a) • Example • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 Lh6=(a6 , 5), (a7 , 4), (a8 , 6) Lh7=(a6 , 6), (a7 , 5), (a8 , 5) Lh8=(a6 , 4) Lh10=(a7 , 7) Lh11=(a6 , 8) S=h6 , h7 curra6=10 curra7=9

  29. Phase 2 outline while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of Lh; if r < curra// a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curra := h; h := h’; push h onto the stack if Lh is nonempty; • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 Lh6=(a6 , 5), (a7 , 4), (a8 , 6) Lh7=(a6 , 6), (a7 , 5), (a8 , 5) Lh8=(a6 , 4) Lh10=(a7 , 7) Lh11=(a6 , 8) S=h6 , h7 curra6=10 curra7=9

  30. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of Lh; if r < curra// a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curra := h; h := h’; push h onto the stack if Lh is nonempty; • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 Lh6=(a6 , 5), (a7 , 4), (a8 , 6) Lh7=(a6 , 6), (a7 , 5), (a8 , 5) Lh8=(a6 , 4) Lh10=(a7 , 7) Lh11=(a6 , 8) S=h6 , h7 curra6=10 curra7=9

  31. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of Lh; if r < curra// a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curra := h; h := h’; push h onto the stack if Lh is nonempty; • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 Lh6=(a6 , 5), (a7 , 4), (a8 , 6) Lh7=(a7 , 5), (a8 , 5) Lh8=(a6 , 4) Lh10=(a7 , 7) Lh11=(a6 , 8) S=h6 , h10 curra6=6 curra7=9

  32. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of Lh; if r < curra// a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curra := h; h := h’; push h onto the stack if Lh is nonempty; • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 Lh6=(a6 , 5), (a7 , 4), (a8 , 6) Lh7=(a7 , 5), (a8 , 5) Lh8=(a6 , 4) Lh10=(a7 , 7) Lh11=(a6 , 8) S=h6 , h10 curra6=6 curra7=9

  33. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of Lh; if r < curra// a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curra := h; h := h’; push h onto the stack if Lh is nonempty; • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 Lh6=(a6 , 5), (a7 , 4), (a8 , 6) Lh7=(a7 , 5), (a8 , 5) Lh8=(a6 , 4) Lh10= Lh11=(a6 , 8) S=h6 , h11 curra6=6 curra7=7

  34. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of Lh; if r < curra// a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curra := h; h := h’; push h onto the stack if Lh is nonempty; • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 Lh6=(a6 , 5), (a7 , 4), (a8 , 6) Lh7=(a7 , 5), (a8 , 5) Lh8=(a6 , 4) Lh10= Lh11=(a6 , 8) S=h6 , h11 curra6=6 curra7=7

  35. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of Lh; if r < curra// a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curra := h; h := h’; push h onto the stack if Lh is nonempty; • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 Lh6=(a6 , 5), (a7 , 4), (a8 , 6) Lh7=(a7 , 5), (a8 , 5) Lh8=(a6 , 4) Lh10= Lh11= S=h6 curra6=6 curra7=7

  36. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of Lh; if r < curra// a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curra := h; h := h’; push h onto the stack if Lh is nonempty; • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 Lh6=(a7 , 4), (a8 , 6) Lh7=(a7 , 5), (a8 , 5) Lh8=(a6 , 4) Lh10= Lh11= S=h7 curra6=5 curra7=7

  37. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of Lh; if r < curra// a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curra := h; h := h’; push h onto the stack if Lh is nonempty; • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 Lh6=(a7 , 4), (a8 , 6) Lh7=(a7 , 5), (a8 , 5) Lh8=(a6 , 4) Lh10= Lh11= S=h7 curra6=5 curra7=7

  38. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of Lh; if r < curra// a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curra := h; h := h’; push h onto the stack if Lh is nonempty; • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 Lh6=(a7 , 4), (a8 , 6) Lh7=(a8 , 5) Lh8=(a6 , 4) Lh10= Lh11= S= curra6=5 curra7=5

  39. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of Lh; if r < curra// a prefers h to M(a) let h’= M(a); remove (a,h’) from M and add (a,h) to M; curra := h; h := h’; push h onto the stack if Lh is nonempty; • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 Lh6=(a7 , 4), (a8 , 6) Lh7=(a8 , 5) Lh8=(a6 , 4) Lh10= Lh11= S= curra6=5 curra7=5

  40. Phase 2 termination • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9

  41. Phase 2 termination • Once Phase 2 terminates, matching is trade-in-free • Phase 2 is O(m) • Coalitions may remain… • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9

  42. Phase 3 description • Elimination of coalitions • Repeatedly finding and eliminating coalitions takes O(m2) 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

  43. Phase 3: example Agent counter House label • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 a1 1 h1 a2 0 h2 a3 0 h3 a4 0 h4 a5 0 h5 a6 0 h6 a7 0 h7 a8 0 h8 a9 0 h9 a1 P=a1

  44. Phase 3: example Agent counter House label • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 a1 1 h1 a2 0 h2 a3 0 h3 a4 1 h4 a5 0 h5 a6 0 h6 a7 0 h7 a8 0 h8 a9 0 h9 a4 a1 EXTEND P=a1,a4

  45. Phase 3: example Agent counter House label • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 a1 1 h1 a2 0 h2 a3 1 h3 a4 1 h4 a5 0 h5 a6 0 h6 a7 0 h7 a8 0 h8 a9 0 h9 a4 a1 EXTEND a3 P=a1,a4 , a3

  46. Phase 3: example Agent counter House label • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 a1 1 h1 a2 0 h2 a3 1 h3 a4 1 h4 a5 1 h5 a6 0 h6 a7 0 h7 a8 0 h8 a9 0 h9 a4 a1 EXTEND a3 P=a1,a4 , a3 , a5  a5

  47. Phase 3: example Agent counter House label • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 a1 1 h1 a2 0 h2 a3 1 h3 a4 2h4 a5 1 h5 a6 0 h6 a7 0 h7 a8 0 h8 a9 0 h9 a4 a1 EXTEND a3 P=a1,a4 , a3 , a5 ,a4 a5

  48. Phase 3: example Agent counter House label • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5 : h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 a1 1 h1 a2 0 h2 a3 1 h3 a42 h4 a5 1 h5 a6 0 h6 a7 0 h7 a8 0 h8 a9 0 h9 a4 a1 CYCLE a3 P=a1,a4, a3 , a5 a5

  49. Phase 3: example Agent counter House label • a1 : h4h5h3h2h1 • a2 : h3h4h5h9h1h2 • a3 : h5h4h1h2h3 • a4 : h3h5h4 • a5: h4h3h5 • a6 : h2h3h5h8h6h7h1h11h4h10 • a7 : h1h4h3h6h7h2h10h5h11 • a8 : h1h5h4h3h7h6h8 • a9 : h4 h3h5h9 a1 1 h1 a2 0 h2 a30h3 a40 h4 a50h5 a6 0 h6 a7 0 h7 a8 0 h8 a9 0 h9 a4 a1 CYCLE a3 P=a1,a4, a3 , a5 a5

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