pareto optimality in house allocation problems david manlove department of computing science...
DESCRIPTION
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 projectsTRANSCRIPT
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Pareto Optimality in House Allocation
Problems
David ManloveDepartment of Computing ScienceUniversity of Glasgow
David AbrahamComputer Science DepartmentCarnegie-Mellon University
Katarína CechlárováInstitute of MathematicsPJ Safárik University in Košice
Kurt MehlhornMax-Planck-Institut fűr InformatikSaarbrűcken
Supported by Royal Society of Edinburgh/Scottish Executive Personal Research Fellowshipand Engineering and Physical Sciences Research Council grant GR/R84597/01
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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 : h3 h4 h2 a3 : h4 h3 a4 : h1 h4
Let n=r+s and let m=total length of preference lists
a1 finds h1 and h2 acceptable
a3 prefers h4 to h3
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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
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The underlying graph Weighted bipartite graph G=(V,E)
Vertex set V=AH Edge set: { ai, hj } E if and only if ai finds hj acceptable Weight of edge { ai, hj } is rank of hj in ai’s preference list
Example a1 : h2 h1
a2 : h3 h4 h2
a3 : h4 h3
a4 : h1 h4
a1
a2
a3
a4
h1
h2
h3
h4
21
1
2
3
2
112
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Definition of a matching
A matching M is a subset of A×H such that: if (ai, hj)M then ai finds hj acceptable each agent belongs to at most one pair in M each house belongs to at most one pair in M
Example a1 : h2 h1
a2 : h3 h4 h2
a3 : h4 h3
a4 : h1 h4
a1
a2
a3
a4
h1
h2
h3
h4
M={(a1, h1), (a2, h4), (a3, h3)}
M(a1)=h1
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Maximum matchings
In many applications we wish to match as many agents as possible
A maximum matching is a matching with largest size
Example
Some maximum matchings are “better” than others!
a1 : h2 h1
a2 : h3 h4 h2
a3 : h4 h3
a4 : h1 h4
a1 : h2 h1
a2 : h3 h4 h2
a3 : h4 h3
a4 : h1 h4
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Pareto optimal matchings A matching M1 is Pareto optimal if there is no
matching M2 such that:1. Some agent is better off in M2 than in M1
2. 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 : h1 h2
a3 : h3
a1 : h2 h1
a2 : h1 h2
a3 : h3
M1
M2
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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) to M(ai+1) (0ir-1)
a1 : h2 h1 a2 : h3 h4 h2 a3 : h4 h3 a4 : h1 h4
Proposition: M is Pareto optimal if and only if M is maximal, trade-in-free and coalition-free
M is not maximal due to a3 and h3
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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) to M(ai+1) (0ir-1)
a1 : h2 h1 a2 : h3 h4 h2 a3 : h4 h3 a4 : h1 h4
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 a2 and h3
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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) to M(ai) (0ir-1)
a1 : h2 h1 a2 : h3 h4 h2 a3 : h4 h3 a4 : h1 h4
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 a1, a2, a4
a1
a3
a4
h1
h3
h4
a2 h2
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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) to M(ai) (0 i r-1)
a1 : h2 h1 a2 : h3 h4 h2 a3 : h4 h3 a4 : h1 h4
Lemma: M is Pareto optimal if and only if M is maximal, trade-in-free and coalition-free
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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) to M(ai)
Example a1 : h2 h1 a2 : h3 h4 h2 a3 : h4 h3 a4 : h1 h4
M admits a coalition if and only if G(M) has a directed cycle
Theorem: we may check whether a given matching M is Pareto optimal in O(m) time
a1
a4
a2
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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 : h1 h2 h3 a2 : h1 h2 a3 : h1 h2
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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 : h1 h2 h3 a2 : h1 h2 a3 : h1 h2
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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 : h1 h2 h3 a2 : h1 h2 a3 : h1 h2
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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 : h1 h2 h3 a2 : h1 h2 a3 : h1 h2
M1={(a1,h1), (a2,h2)}
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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 : h1 h2 h3 a2 : h1 h2 a3 : h1 h2
a1 : h1 h2 h3 a2 : h1 h2 a3 : h1 h2
M1={(a1,h1), (a2,h2)}
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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 : h1 h2 h3 a2 : h1 h2 a3 : h1 h2
a1 : h1 h2 h3 a2 : h1 h2 a3 : h1 h2
M1={(a1,h1), (a2,h2)}
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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 : h1 h2 h3 a2 : h1 h2 a3 : h1 h2
a1 : h1 h2 h3 a2 : h1 h2 a3 : h1 h2
M1={(a1,h1), (a2,h2)}
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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 : h1 h2 h3 a2 : h1 h2 a3 : h1 h2
a1 : h1 h2 h3 a2 : h1 h2 a3 : h1 h2
M1={(a1,h1), (a2,h2)}
M2={(a1,h3), (a2,h2), (a3,h1)}
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Related work Rank maximal matchings
Matching M is rank maximal if, in M1. Maximum number of agents get their first-choice house;2. Subject to (1), maximum number of agents get 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
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Related work (cont) Take underlying weighted bipartite graph G
The weight of a matching M in G is the sum of the weights of the edges contained in M
where AM is the set of agents who are matched in M ranka(h) is the rank of h in a’s preference list
Find a maximum cardinality minimum weight matching M in G M is Pareto optimal For if not there exists a matching M′ such that
ranka(M′(a)) ranka(M(a)) for all aAM ranka(M′(a)) < ranka(M(a)) for some aAM
so M′ contradicts the minimum weight property of M A maximum cardinality minimum weight matching
may be found in O(nmlog n) time Gabow and Tarjan, 1989
MAa
a aMrankMwt ))(()(
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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
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Phase 1 a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Maximum matching M in G has size 8 M must be maximal No guarantee that M is trade-in-free or coalition-free
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Phase 1 a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Maximum matching M in G has size 9 M must be maximal No guarantee that M is trade-in-free or coalition-free
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Phase 1 a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
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
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Phase 2 description 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 : h4 h5 h3 h2 h1 a2 : h3 h4 h5 h9 h1 h2 a3 : h5 h4 h1 h2 h3 a4 : h3 h5 h4 a5 : h4 h3 h5 a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10 a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11 a8 : h1 h5 h4 h3 h7 h6 h8 a9 : h4 h3 h5 h9
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 , h7curra6
=10curra7
=9
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Phase 2 description
a1 : h4 h5 h3 h2 h1 a2 : h3 h4 h5 h9 h1 h2 a3 : h5 h4 h1 h2 h3 a4 : h3 h5 h4 a5 : h4 h3 h5 a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10 a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11 a8 : h1 h5 h4 h3 h7 h6 h8 a9 : h4 h3 h5 h9
while S is nonemptypop 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;
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 , h7curra6
=10curra7
=9
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Phase 2 example
a1 : h4 h5 h3 h2 h1 a2 : h3 h4 h5 h9 h1 h2 a3 : h5 h4 h1 h2 h3 a4 : h3 h5 h4 a5 : h4 h3 h5 a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10 a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11 a8 : h1 h5 h4 h3 h7 h6 h8 a9 : h4 h3 h5 h9
while S is nonemptypop 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;
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 , h7curra6
=10curra7
=9
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Phase 2 example
a1 : h4 h5 h3 h2 h1 a2 : h3 h4 h5 h9 h1 h2 a3 : h5 h4 h1 h2 h3 a4 : h3 h5 h4 a5 : h4 h3 h5 a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10 a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11 a8 : h1 h5 h4 h3 h7 h6 h8 a9 : h4 h3 h5 h9
while S is nonemptypop 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;
Lh6=(a6 , 5), (a7 , 4), (a8 , 6)
Lh7=(a7 , 5), (a8 , 5)
Lh8=(a6 , 4)
Lh10=(a7 , 7)
Lh11=(a6 , 8)
S=h6 , h10curra6
=6curra7
=9
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Phase 2 example
a1 : h4 h5 h3 h2 h1 a2 : h3 h4 h5 h9 h1 h2 a3 : h5 h4 h1 h2 h3 a4 : h3 h5 h4 a5 : h4 h3 h5 a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10 a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11 a8 : h1 h5 h4 h3 h7 h6 h8 a9 : h4 h3 h5 h9
while S is nonemptypop 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;
Lh6=(a6 , 5), (a7 , 4), (a8 , 6)
Lh7=(a7 , 5), (a8 , 5)
Lh8=(a6 , 4)
Lh10=(a7 , 7)
Lh11=(a6 , 8)
S=h6 , h10curra6
=6curra7
=9
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Phase 2 example
a1 : h4 h5 h3 h2 h1 a2 : h3 h4 h5 h9 h1 h2 a3 : h5 h4 h1 h2 h3 a4 : h3 h5 h4 a5 : h4 h3 h5 a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10 a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11 a8 : h1 h5 h4 h3 h7 h6 h8 a9 : h4 h3 h5 h9
while S is nonemptypop 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;
Lh6=(a6 , 5), (a7 , 4), (a8 , 6)
Lh7=(a7 , 5), (a8 , 5)
Lh8=(a6 , 4)
Lh10=
Lh11=(a6 , 8)
S=h6 , h11curra6
=6curra7
=7
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Phase 2 example
a1 : h4 h5 h3 h2 h1 a2 : h3 h4 h5 h9 h1 h2 a3 : h5 h4 h1 h2 h3 a4 : h3 h5 h4 a5 : h4 h3 h5 a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10 a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11 a8 : h1 h5 h4 h3 h7 h6 h8 a9 : h4 h3 h5 h9
while S is nonemptypop 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;
Lh6=(a6 , 5), (a7 , 4), (a8 , 6)
Lh7=(a7 , 5), (a8 , 5)
Lh8=(a6 , 4)
Lh10=
Lh11=(a6 , 8)
S=h6 , h11curra6
=6curra7
=7
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Phase 2 example
a1 : h4 h5 h3 h2 h1 a2 : h3 h4 h5 h9 h1 h2 a3 : h5 h4 h1 h2 h3 a4 : h3 h5 h4 a5 : h4 h3 h5 a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10 a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11 a8 : h1 h5 h4 h3 h7 h6 h8 a9 : h4 h3 h5 h9
while S is nonemptypop 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;
Lh6=(a6 , 5), (a7 , 4), (a8 , 6)
Lh7=(a7 , 5), (a8 , 5)
Lh8=(a6 , 4)
Lh10=
Lh11=
S=h6curra6
=6curra7
=7
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Phase 2 example
a1 : h4 h5 h3 h2 h1 a2 : h3 h4 h5 h9 h1 h2 a3 : h5 h4 h1 h2 h3 a4 : h3 h5 h4 a5 : h4 h3 h5 a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10 a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11 a8 : h1 h5 h4 h3 h7 h6 h8 a9 : h4 h3 h5 h9
while S is nonemptypop 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;
Lh6=(a7 , 4), (a8 , 6)
Lh7=(a7 , 5), (a8 , 5)
Lh8=(a6 , 4)
Lh10=
Lh11=
S=h7curra6
=5curra7
=7
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Phase 2 example
a1 : h4 h5 h3 h2 h1 a2 : h3 h4 h5 h9 h1 h2 a3 : h5 h4 h1 h2 h3 a4 : h3 h5 h4 a5 : h4 h3 h5 a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10 a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11 a8 : h1 h5 h4 h3 h7 h6 h8 a9 : h4 h3 h5 h9
while S is nonemptypop 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;
Lh6=(a7 , 4), (a8 , 6)
Lh7=(a7 , 5), (a8 , 5)
Lh8=(a6 , 4)
Lh10=
Lh11=
S=h7curra6
=5curra7
=7
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Phase 2 example
a1 : h4 h5 h3 h2 h1 a2 : h3 h4 h5 h9 h1 h2 a3 : h5 h4 h1 h2 h3 a4 : h3 h5 h4 a5 : h4 h3 h5 a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10 a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11 a8 : h1 h5 h4 h3 h7 h6 h8 a9 : h4 h3 h5 h9
while S is nonemptypop 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;
Lh6=(a7 , 4), (a8 , 6)
Lh7=(a8 , 5)
Lh8=(a6 , 4)
Lh10=
Lh11=
S=curra6
=5curra7
=5
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Phase 2 example
a1 : h4 h5 h3 h2 h1 a2 : h3 h4 h5 h9 h1 h2 a3 : h5 h4 h1 h2 h3 a4 : h3 h5 h4 a5 : h4 h3 h5 a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10 a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11 a8 : h1 h5 h4 h3 h7 h6 h8 a9 : h4 h3 h5 h9
while S is nonemptypop 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;
Lh6=(a7 , 4), (a8 , 6)
Lh7=(a8 , 5)
Lh8=(a6 , 4)
Lh10=
Lh11=
S=curra6
=5curra7
=5
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Once Phase 2 terminates, matching is trade-in-free
Data structures may be initialised in O(m) time Main loop takes O(m) time overall Phase 2 is O(m) Coalitions may remain…
Phase 2 termination
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Once Phase 2 terminates, matching is trade-in-free
Data structures may be initialised in O(m) time Main loop takes O(m) time overall Phase 2 is O(m) Coalitions may remain…
Phase 2 termination
Example a1 : h4 h5 h3 h2 h1 a2 : h3 h4 h5 h9 h1 h2 a3 : h5 h4 h1 h2 h3 a4 : h3 h5 h4 a5 : h4 h3 h5 a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10 a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11 a8 : h1 h5 h4 h3 h7 h6 h8 a9 : h4 h3 h5 h9
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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
Phase 3 description
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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 description
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 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
P=a1
a1
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 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
a1
P=a1, a4
EXTEND
a4
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 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
a1
EXTEND
a4
P=a1, a4 , a3 a3
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 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
a1
EXTEND
a4
P=a1, a4 , a3 , a5 a3
a5
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 1 h1
a2 0 h2
a3 1 h3
a4 2 h4
a5 1 h5
a6 0 h6
a7 0 h7
a8 0 h8
a9 0 h9
a1
EXTEND
a4
P=a1, a4 , a3 , a5 , a4 a3
a5
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 1 h1
a2 0 h2
a3 1 h3
a4 2 h4
a5 1 h5
a6 0 h6
a7 0 h7
a8 0 h8
a9 0 h9
a1
CYCLE
a4
P=a1, a4, a3 , a5 a3
a5
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 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
CYCLE
a4
a3
a5
P=a1, a4, a3 , a5
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 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
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 1 h1
a2 1 h2
a3 0 h3
a4 0 h4
a5 0 h5
a6 0 h6
a7 0 h7
a8 0 h8
a9 0 h9
a1
EXTENDP=a1 , a2
a2
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 1 h1
a2 1 h2
a3 0 h3
a4 0 h4
a5 0 h5
a6 0 h6
a7 0 h7
a8 0 h8
a9 1 h9
a1
EXTENDP=a1 , a2 , a9
a2
a9
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 1 h1
a2 1 h2
a3 0 h3
a4 0 h4
a5 0 h5
a6 0 h6
a7 0 h7
a8 0 h8
a9 1 h9
a1
BACKTRACKP=a1 , a2
a2
a9
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 1 h1
a2 1 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 , a2
a2
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 2 h1
a2 1 h2
a3 0 h3
a4 0 h4
a5 0 h5
a6 0 h6
a7 0 h7
a8 0 h8
a9 0 h9
a1
EXTENDP=a1 , a2 , a1
a2
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 2 h1
a2 1 h2
a3 0 h3
a4 0 h4
a5 0 h5
a6 0 h6
a7 0 h7
a8 0 h8
a9 0 h9
a1
CYCLEP=a1 , a2
a2
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 0 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
CYCLEP=a1 , a2
a2
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 0 h1
a2 0 h2
a3 0 h3
a4 0 h4
a5 0 h5
a6 1 h6
a7 0 h7
a8 0 h8
a9 0 h9
P=a6
a6
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 0 h1
a2 0 h2
a3 0 h3
a4 0 h4
a5 0 h5
a6 1 h6
a7 0 h7
a8 1 h8
a9 0 h9
P=a6 , a8
EXTENDa6 a8
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 0 h1
a2 0 h2
a3 0 h3
a4 0 h4
a5 0 h5
a6 1 h6
a7 1 h7
a8 1 h8
a9 0 h9
P=a6 , a8 , a7
EXTENDa6 a8
a7
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 0 h1
a2 0 h2
a3 0 h3
a4 0 h4
a5 0 h5
a6 2 h6
a7 1 h7
a8 1 h8
a9 0 h9
P=a6 , a8 , a7 , a6
EXTENDa6 a8
a7
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
a1 0 h1
a2 0 h2
a3 0 h3
a4 0 h4
a5 0 h5
a6 2 h6
a7 1 h7
a8 1 h8
a9 0 h9
P=a6 , a8 , a7
CYCLEa6 a8
a7
Agent counter House label
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 0 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
P=a6 , a8 , a7
CYCLEa6 a8
a7
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Phase 3: example a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
Agent counter House labela1 0 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
P=
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Once Phase 3 terminates, matching is coalition-free
Data structures may be initialised in O(m) time Nested loops take O(m) time overall Phase 3 is O(m)
Phase 3 termination
a1 : h4 h5 h3 h2 h1
a2 : h3 h4 h5 h9 h1 h2
a3 : h5 h4 h1 h2 h3
a4 : h3 h5 h4
a5 : h4 h3 h5
a6 : h2 h3 h5 h8 h6 h7 h1 h11 h4 h10
a7 : h1 h4 h3 h6 h7 h2 h10 h5 h11
a8 : h1 h5 h4 h3 h7 h6 h8
a9 : h4 h3 h5 h9
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Initial property rights Suppose A’A and each member of A’ owns a house
initially For each agent aA’, 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
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Minimum Pareto optimal matchings
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
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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 r1, …, rk to an augmenting path P=a1, h1, …, ak, hk where ri=rankai
(hi)
Examples: 1,3,2 1,2,2 Find a lexicographically smallest augmenting path
h1
h2
h3
h4
1
2
1
h5
a1
a2
a3
a4
23
1
12
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Summary Definition of the House Allocation problem (HA) O(m) algorithm for testing a matching for Pareto optimality O(m) algorithm for finding a Pareto optimal matching O(nm) algorithm for finding a maximum Pareto optimal
matching NP-hardness of finding a minimum Pareto optimal
matching Interpolation result
More information: D.J. Abraham, K. Cechlarova, D.F.M., K. Mehlhorn:
Pareto Optimality in House Allocation Problems, to appear atISAAC 2004: the 15th International Symposium on Algorithmsand Computation, Hong Kong, December 20-22, 2004
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Open problemsFinding 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 problemTechnical Report TR-2004-182 of the Computing Science Department of Glasgow University
Solvable in O(nm) time?