Pareto Optimality in House Allocation Problems

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

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 : 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

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=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

2

1

1

2

3

2

11

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 a1 : h2 h1

a2 : h3 h4 h2

a3 : h4 h3

a4 : h1 h4

a1

a2

a3

a4

h1

h2

h3

h4

2

1

1

2

3

2

11

2M={(a1, h1), (a2, h4), (a3, h3)}

M(a1)=h1

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 a1 : h2 h1

a2 : h3 h4 h2

a3 : h4 h3

a4 : h1 h4

a1

a2

a3

a4

h1

h2

h3

h4

2

1

1

2

3

2

11

2M={(a1, h2), (a2, h3), (a3, h4), (a4, h1)}

Matchings in the underlying graph

7

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

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 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

9




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


M is not trade-in-free due to a2 and h3

10




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


M is not coalition-free due to a1, a2, a4

a1

a3

a4

h1

h3

h4

a2 h2

11




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)

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 : 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)}

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

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

16

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


17

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


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 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

19

Phase 2 outline




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

20

Phase 2 outline




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

21

Phase 2 outline




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

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 : 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

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

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

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 problemTechnical 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 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

Proposition: we may check whether a given matching M is Pareto optimal in O(m) time

a1

a4

a2

29

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)


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

=10

curra7=9

30

Phase 2 outline

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

=10

curra7=9

31

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




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

=10

curra7=9

32

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




Lh6=(a6 , 5), (a7 , 4), (a8 , 6)

Lh7=(a7 , 5), (a8 , 5)

Lh8=(a6 , 4)

Lh10=(a7 , 7)

Lh11=(a6 , 8)

S=h6 , h10curra6

=6

curra7=9

33

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




Lh6=(a6 , 5), (a7 , 4), (a8 , 6)

Lh7=(a7 , 5), (a8 , 5)

Lh8=(a6 , 4)

Lh10=(a7 , 7)

Lh11=(a6 , 8)

S=h6 , h10curra6

=6

curra7=9

34

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




Lh6=(a6 , 5), (a7 , 4), (a8 , 6)

Lh7=(a7 , 5), (a8 , 5)

Lh8=(a6 , 4)

Lh10=

Lh11=(a6 , 8)

S=h6 , h11curra6

=6

curra7=7

35

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




Lh6=(a6 , 5), (a7 , 4), (a8 , 6)

Lh7=(a7 , 5), (a8 , 5)

Lh8=(a6 , 4)

Lh10=

Lh11=(a6 , 8)

S=h6 , h11curra6

=6

curra7=7

36

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




Lh6=(a6 , 5), (a7 , 4), (a8 , 6)

Lh7=(a7 , 5), (a8 , 5)

Lh8=(a6 , 4)

Lh10=

Lh11=

S=h6curra6

=6

curra7=7

37

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




Lh6=(a7 , 4), (a8 , 6)

Lh7=(a7 , 5), (a8 , 5)

Lh8=(a6 , 4)

Lh10=

Lh11=

S=h7curra6

=5

curra7=7

38

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




Lh6=(a7 , 4), (a8 , 6)

Lh7=(a7 , 5), (a8 , 5)

Lh8=(a6 , 4)

Lh10=

Lh11=

S=h7curra6

=5

curra7=7

39

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




Lh6=(a7 , 4), (a8 , 6)

Lh7=(a8 , 5)

Lh8=(a6 , 4)

Lh10=

Lh11=

S=curra6

=5

curra7=5

40

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




Lh6=(a7 , 4), (a8 , 6)

Lh7=(a8 , 5)

Lh8=(a6 , 4)

Lh10=

Lh11=

S=curra6

=5

curra7=5

41

Phase 2 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

42

Once Phase 2 terminates, matching is trade-in-free Phase 2 is O(m) Coalitions may remain…

Phase 2 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

43

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

44

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

45


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


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

46


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


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

47


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


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

48


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


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

49


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


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

50


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


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

51


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


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

52


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


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

EXTEND

P=a1 , a2

a2

53


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


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

EXTEND

P=a1 , a2 , a9

a2

a9

54


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


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

BACKTRACK

P=a1 , a2

a2

a9

55


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


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

56


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


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

EXTEND

P=a1 , a2 , a1

a2

57


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


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

CYCLE

P=a1 , a2

a2

58


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


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

P=a1 , a2

a2

59


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


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

60


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


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

61


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


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

62


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


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

EXTEND

a6 a8

a7

63


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

CYCLE

a6 a8

a7

Agent counter House label

64


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


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

CYCLE

a6 a8

a7

65


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


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=

66

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

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 r1, …, rk to an augmenting path P=a1, h1, …, ak, hk where ri=rankai

(hi)

Examples: 1,3,1 1,2,2

Find a lexicographically smallest augmenting path

h1

h2

h3

h4

1

2

1

h5

a1

a2

a3

a4

231

21

Pareto Optimality in House Allocation Problems

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