Popularity in the Capacitated House Allocation Problem David Manlove Department of Computing Science University of Glasgow Colin Sng BCTCS - 5th April 2006

Capacitated House Allocation Problem (CHA) Set of agents A = {a1, a2,…, ar} Set of houses H = {h1,h2,…,hs} Each agent ai has a strictly ranked preference list of acceptable houses; houses have no preference over agents. Each house hj has a capacity cj which indicates the maximum number of agents who may be matched to it A matching is a partial function from A to H such that Each agent ai is only matched to acceptable houses Each agent ai is matched to at most one house hj Each house hj is matched at most to cj agents Example a1: h1 h2 h1:c1=2 a2: h1 h4 h2 h2:c2=1 a3: h2 h1 h4 h3 h3:c3=2 a4: h4 h2 h3 h4:c4=1 a5: h4 h1 a6: h2 h1 h4 a7: h4 BCTCS - 5th April 2006

Popular Matchings Let M be a matching in a given instance I of CHA. Define M(ai) to be the house matched to a matched agent ai in M Define M(hj) to be the set of agents matched to a house hj in M Given two matchings M and M', agent ai prefers M to M' if either ai is matched in M but unmatched in M', or ai is matched in both M and M', but prefers M(ai) to M'(ai) We say that M is more popular than M' if there are more agents who prefer M to M' than prefer M' to M A matching M is popular if there is no other matching M' that is more popular than M. BCTCS - 5th April 2006

Popular matchings may not exist Consider the following example instance I1 of CHA ( c1=c2=c3=1) a1: h1 h2 h3 a2: h1 h2 h3 a3: h1 h2 h3 M1= {(a1,h1), (a2,h2),(a3,h3)} and M2 = {(a1,h3),(a2,h1),(a3,h2)} are two possible matchings. a1 prefers M1 to M2 (1 agent) but a2 and a3 prefers M2 to M1 (2 agents) So M2 is more popular than M1 In fact, there is no popular matching Problem: given any CHA instance, find a popular matching if one exists BCTCS - 5th April 2006

Background Known results Problem was first considered by Gardenfors (1975) in context of Stable Marriage problem (SM) D.J. Abraham, R.W. Irving, T. Kavitha, K. Melhorn (2005) considered the problem in special setting of CHA, where each house has capacity 1 (otherwise known as House Allocation problems). Falls in category of matching problems with one-sided preference lists. Examples:- Scottish Executive Teaching Induction Scheme, allocation of campus housing in American universities CHA follows as natural extension of problem setting. BCTCS - 5th April 2006

Characterisation of popular matchings in CHA (i) For each agent ai, f(ai) denotes the first-ranked house on ai’s preference list, we call f(ai) an f-house For each house hj, let f(hj) = aiA: f(ai) =hj and fj =|f(hj)| Lemma 1. Let M be any popular matching for a given CHA instance I. Then, for every f-house hj, |M(hj)∩f(hj)| = min {cj,fj} BCTCS - 5th April 2006

Execution (i) a1: h1 h2 a2: h1 h4 h2 a3: h2 h1 h4 h3 a4: h4 h2 h3 Example a1: h1 h2 a2: h1 h4 h2 a3: h2 h1 h4 h3 a4: h4 h2 h3 a5: h4 h1 a6: h2 h1 h4 a7: h4 h1:c1=2, f1 c1 h2:c2=1,f2>c2 h3:c3=2 h4:c4=1,f4>c4 Lemma 1. Let M be any popular matching for a given CHA instance I. Then, for every f-house hj, |M(hj)∩f(hj)| = min {cj,fj} a1 a2 a3 a4 a5 a6 h1 h2 h3 h4 a7 BCTCS - 5th April 2006

Execution (ii) Example a1: h1 h2 a2: h1 h4 h2 a3: h2 h1 h4 h3 h1:c1=2, f1 c1 h2:c2=1,f2>c2 h3:c3=2 h4:c4=1,f4>c4 Lemma 1. Let M be any popular matching for a given CHA instance I. Then, for every f-house hj, |M(hj)∩f(hj)| = min {cj,fj} a1 a2 a3 a4 a5 a6 h1 h2 h3 h4 a7 BCTCS - 5th April 2006

Execution (iii) Example a1: h1 h2 a2: h1 h4 h2 a3: h2 h1 h4 h3 Match a1 & a2 to h1 Lemma 1. Let M be any popular matching for a given CHA instance I. Then, for every f-house hj, |M(hj)∩f(hj)| = min {cj,fj} a1 a2 a3 a4 a5 a6 h1 h2 h3 h4 a7 BCTCS - 5th April 2006

Execution (iv) Example a1: h1 h2 a2: h1 h4 h2 a3: h2 h1 h4 h3 Lemma 1. Let M be any popular matching for a given CHA instance I. Then, for every f-house hj, |M(hj)∩f(hj)| = min {cj,fj} a1 h1 a2 h2 a3 h3 a4 a5 h4 a6 a7 BCTCS - 5th April 2006

Characterisation of popular matchings in CHA (ii) For each agent ai, s(ai) denotes the most-preferred house hj on ai’s preference list such that either (i) hj is a non f-house, or (ii) hj is an f-house such that hjf(ai) and fj < cj we call s(ai) an s-house create a last resort house l(ai) and append l(ai) to ai’s preference list BCTCS - 5th April 2006

Execution (v) Example a3: h2 h1 h4 h3 a4: h4 h2 h3 a5: h4 h1 l(a5) a6: h2 h1 h4 l(a6) a7: h4 l(a7) h1:c1=0 h2:c2=1,f2>c2 h3:c3=2 h4:c4=1,f4>c4 (f-house,s-house) a1 h1 a2 h2 a3 h3 a4 l(a5) a5 h4 a6 l(a6) a7 l(a7) BCTCS - 5th April 2006

Characterisation of popular matchings in CHA (iii) Lemma 2. Let M be any popular matching in a CHA instance. Then no agent ai can be matched in M to a house between f(ai) and s(ai) on ai’s preference list. ai: f(ai) … M(ai) …s(ai) Lemma 3. Let M be any popular matching in a CHA instance. Then no agent can be matched in M to a house worse than s(ai) on ai’s preference list. ai: f(ai) … s(ai) M(ai) BCTCS - 5th April 2006

Characterisation of popular matchings in CHA (iv) By Lemmas 2 and 3, for every agent ai, M(ai)  {f(ai), s(ai)}, so only need a subgraph G' of G to find popular matching For each agent ai in the graph G, delete all edges except those belonging to f(ai) and s(ai) Theorem 1. A matching M is popular if and only if For every f-house hj, If fjcj, then f(hj) M(hj), else If fj>cj, then |M(hj)|=cj and M(hj) f(hj) M is an agent-complete matching of the subgraph G' BCTCS - 5th April 2006

Execution (vi) Example a3: h2 h1 h4 h3 a4: h4 h2 h3 a5: h4 h1 l(a5) a6: h2 h1 h4 l(a6) a7: h4 l(a7) h1:c1=0 h2:c2=1,f2>c2 h3:c3=2 h4:c4=1,f4>c4 (f-house,s-house) a1 h1 a2 h2 a3 h3 a4 l(a5) a5 h4 a6 l(a6) a7 l(a7) BCTCS - 5th April 2006

Execution (vii) Example a3: h2 h1 h4 h3 a4: h4 h2 h3 a5: h4 h1 l(a5) a6: h2 h1 h4 l(a6) a7: h4 l(a7) h2:c2=1,f2>c2 h3:c3=2 h4:c4=1,f4>c4 (f-house,s-house) NB: Condition 1(a) of Theorem 1 already satisfied, i.e. for every f-house hj, if fjcj, then f(hj) M(hj), a1 h1 a2 h2 a3 h3 a4 l(a5) a5 h4 a6 l(a6) a7 l(a7) BCTCS - 5th April 2006

Execution (viii) Example a3: h2 h1 h4 h3 a4: h4 h2 h3 a5: h4 h1 l(a5) a6: h2 h1 h4 l(a6) a7: h4 l(a7) h2:c2=1,f2>c2 h3:c3=2 h4:c4=1,f4>c4 (f-house,s-house) Find an agent-complete matching to satisfy Condition 2 of Theorem 1 a1 h1 a2 h2 a3 h3 a4 l(a5) a5 h4 a6 l(a6) a7 l(a7) BCTCS - 5th April 2006

Execution (ix) Example a3: h2 h1 h4 h3 a4: h4 h2 h3 a5: h4 h1 l(a5) a6: h2 h1 h4 l(a6) a7: h4 l(a7) h2:c2=1,f2>c2 h3:c3=2 h4:c4=1,f4>c4 (f-house,s-house) Find an agent-complete matching to satisfy Condition 2 of Theorem 1 a1 h1 a2 h2 a3 h3 a4 l(a5) a5 h4 a6 l(a6) a7 l(a7) BCTCS - 5th April 2006

Execution (x) Example a3: h2 h1 h4 h3 a4: h4 h2 h3 a5: h4 h1 l(a5) a6: h2 h1 h4 l(a6) a7: h4 l(a7) h2:c2=1,f2>c2 h3:c3=2 h4:c4=1,f4>c4 (f-house,s-house) Ensure Condition 1(b) of Theorem 1 i.e. for every f-house hj, if fj>cj, then |M(hj)|=cj and M(hj) f(hj) a1 h1 a2 h2 a3 h3 a4 l(a5) a5 h4 a6 l(a6) a7 l(a7) BCTCS - 5th April 2006

Execution (x) Example a3: h2 h1 h4 h3 a4: h4 h2 h3 a5: h4 h1 l(a5) a6: h2 h1 h4 l(a6) a7: h4 l(a7) h2:c2=1,f2>c2 h3:c3=2 h4:c4=1,f4>c4 (f-house,s-house) Ensure Condition 1(b) of Theorem 1 i.e. for every f-house hj, if fj>cj, then |M(hj)|=cj and M(hj) f(hj) a1 h1 a2 h2 a3 h3 a4 l(a5) a5 h4 a6 l(a6) a7 l(a7) BCTCS - 5th April 2006

Execution (xi) Example a3: h2 h1 h4 h3 a4: h4 h2 h3 a5: h4 h1 l(a5) a6: h2 h1 h4 l(a6) a7: h4 l(a7) h2:c2=1,f2>c2 h3:c3=2 h4:c4=1,f4>c4 (f-house,s-house) Promote a6 to h2 Ensure Condition 1(b) of Theorem 1 i.e. for every f-house hj, if fj>cj, then |M(hj)|=cj and M(hj) f(hj) a1 h1 a2 h2 a3 h3 a4 l(a5) a5 h4 a6 l(a6) a7 l(a7) BCTCS - 5th April 2006

Execution (xii) Example a3: h2 h1 h4 h3 a4: h4 h2 h3 a5: h4 h1 l(a5) a6: h2 h1 h4 l(a6) a7: h4 l(a7) h2:c2=1,f2>c2 h3:c3=2 h4:c4=1,f4>c4 (f-house,s-house) Nothing to do for a5 & a7 Ensure Condition 1(b) of Theorem 1 i.e. for every f-house hj, if fj>cj, then |M(hj)|=cj and M(hj) f(hj) a1 h1 a2 h2 a3 h3 a4 l(a5) a5 h4 a6 l(a6) a7 l(a7) BCTCS - 5th April 2006

Execution (xiii) Example a1: h1 h2 l(a1) a2: h1 h4 h2 l(a2) a3: h2 h1 h4 h3 a4: h4 h2 h3 a5: h4 h1 l(a5) a6: h2 h1 h4 l(a6) a7: h4 l(a7) (f-house,s-house) Matching satisfies Theorem 1 Popular matching = {(a1,h1), (a2, h1),(a3,h3),(a4,h4), (a6, h2)} a1 h1 a2 h2 a3 h3 a4 a5 h4 a6 a7 BCTCS - 5th April 2006

Summary Time complexity of algorithm is O(r3/2s1/2) Algorithm can be extended to find maximum cardinality popular matching also in O(r3/2s1/2) time Have also extended findings to the case where preference list may contain ties, i.e. Capacitated House Allocation problem with ties (CHAT) O((r+s)L) algorithm for finding a maximum cardinality popular matching in an instance of CHAT, if one exists Open problem: Extend setting to include the case where houses also have preferences over agents. What is the complexity of finding a maximum cardinality popular matching if one exists? BCTCS - 5th April 2006