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## Stable Matching Problems with Constant Length Preference Lists

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**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.