Stable Matching Problems with Constant Length Preference Lists

Stable Matching Problems with Constant Length Preference Lists Rob Irving, David Manlove, Gregg O’Malley University Of Glasgow Department of Computing Science

SMTI Formalisation • Setofn1 men SM = {m1 , m2 , …., mn1} • Setofn2 women SW = {w1 , w2 , …., wn2} • Each man mi ranks a subset of SW in preference order, and mi’s list may contain ties. • Each woman wj ranks a subset of SM in preference order, and wj’s list may contain ties. • A matching M is a set of (man , woman) pairs (m,w)such that each of m and w appear in at most one pair, and m and w are on each other’s list. • We say a (man, woman) pair (m , w)blocksM if: • Either m is unmatchedormstrictly prefers w to his partner in M,and • Either w is unmatchedorw strictly prefers m to her partner in M. • A matching that admits no blocking pair is said to be stable • Can’t improve by making an arrangement outside the matching.

Properties • When no ties are allowed in a participant’s list: • A stable matching for an instance of SMI can always be found using a slightly modified version of an algorithm known as the Gale/Shapley algorithm (1962). • Gale and Sotomayor also proved in 1985 that for an instance of SMI all stable matchings have the same size. • When ties are allowed in a participant’s list: • Again we can always find a stable matching for an instance of SMTI by breaking the ties arbitrarily and running the Gale/Shapley algorithm. • However stable matchings may have different sizes in this case.

SMTI Example • Two possible stable matchings are: M = {(m1 , w1)} M’ = {( m1 , w2) , (m2 , w1)} m1: w1 w2 w1: (m1 m2) m2: w1 w2: m1 Men’s preferences Women’s preferences

The History • A natural problem to consider is finding a stable matching that matches the largest number of men and women. We denote this problem by MAX-SMTI. • MAX-SMTI was shown to be NP-hard by Iwamaet al. in 1999. • A further natural restriction of MAX-SMTI is finding a maximum stable matching when the preference lists are of a constant length. • This has applications for the matching of graduating medical students to hospitals posts in many countries – as typically student’s lists are small and of fixed length. • The above problem is the one-to-many generalisation of SMTI called the Hospitals/Residents problem with Ties (HRT).

The History cont.. • The following table shows a list of the known results for the case of constant length preference lists. The numbers indicate the upper bounds on the length of the preference lists.

Our Contribution. • We show MAX-SMTI is polynomial-time solvable where men’s lists are of size 2 and contain no ties, and the women’s lists are of unbounded length and may contain ties.

(2,n)-MAX-SMTI • The algorithm is presented in 3 phases. • Phase 1 : adapted Gale/Shapley algorithm. • Phase 2 : network flow stage. • Phase 3 : continuation of phase 1. • An allocation (similar to matching only women can be multiply assigned) is produced by phase 1. • Phase 2 attempts to move men from multiply assigned women to unassigned women. • Phase 3 may or may not be need, it reallocates men still assigned to multiply assigned women.

Phase 1 • men “propose” to the women; • women “hold” proposals; • if some woman wj receives a proposal from man mi, then she deletes all strict successors of mi from her list; • Terminates with an allocation A1

Phase 1 Example m1: w1 w2 w1: (m1 m2 m3) m5 m2: w1 w4 w2: (m1 m4) (m3 m5) m3: w1 w2 w3: m4 m4: w2 w3 w4: m2 m5: w1 w2 Men’s preferences Women’s preferences

Phase 1 Example m1: w1 w2 w1: ( m1 m2 m3 ) m5 m2: w1 w4 w2: ( m1 m4 ) ( m3 m5 ) m3: w1 w2 w3: m4 m4: w2 w3 w4: m2 m5: w1 w2 Men’s preferences Women’s preferences Allocation A1 output by phase 1: A1 = { ( m1 , w1 ) , ( m2 , w1 ) , ( m3 , w1 ) , ( m4 , w2 ) } • We note here that w1 is multiply assigned.

Phase 2 : Network Construction • Add source node s and sink node t. • For each multiply assigned woman wj, $ edge (s , wj) with capacity 1 less than the number of assignees to wj. • For each unassigned woman wj, $ edge (wj , t) of capacity 1. • Let mi be a man with 2 women left on his list. Let wj be mi’s first-choice and wk be mi’s second-choice. Add the edges (wj , mi) and (mi , wk) with capacity 1. • Women with only 1 partner may be represented by a vertex as a result of this step.

Phase 2 : Network w2 m1 m4 w3 1 1 1 1 1 w1 2 s t 1 1 w4 m2 1

Phase 2 : Lists m1: w1 w2 w1: ( m1 m2 m3 ) m5 m2: w1 w4 w2: ( m1 m4 ) ( m3 m5 ) m3: w1 w2 w3: m4 m4: w2 w3 w4: m2 m5: w1 w2 Men’s preferences Women’s preferences

Phase 2 : Max Flow w2 m1 m4 w3 1 1 1 1 1 w1 2 s t 1 1 w4 m2 1 • The maximum (saturating) flow gives rise to the following: • m1 being moved from w1 to w2; • m4 being moved from w2 to w3; • m2 being moved from w1 to w4.

Phase 2 : Allocation m1: w1 w2 w1: ( m1 m2 m3 ) m5 m2: w1 w4 w2: ( m1 m4 ) ( m3 m5 ) m3: w1 w2 w3: m3 m4: w2 w3 w4: m4 m5: w1 w2 Men’s preferences Women’s preferences • In this case we have found the maximum stable matching, namely: M = {( m1 , w2 ) , ( m2 ,w4 ) , ( m3 ,w1 ) , ( m4 ,w3 )}

Phase 3 • In general there may still be women who are multiply assigned after phase 2. • It can proven, however, that if the remaining ties are broken arbitrarily and we continue with phase 1, a stable matching of maximum size is obtained.

Open Problems • Is (2,n)-MAX-HRT polynomial-time solvable? • Is the generalisation of (2,n)-MAX-SMTI and (2,n)-MAX-HRT in which both sides preference lists contain ties polynomial-time solvable? • Finding the exact boundary between P and NP-hard cases, i.e. when both men and women have preference lists of size at most 3 and their lists contain ties.

Stable Matching Problems with Constant Length Preference Lists