SMTI Formalisation SMI Formulation Stable Matching Problems alongside Constant Length Preference Lists

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SMTI Formalisation SMI Formulation Stable Matching Problems alongside Constant Length Preference Lists

Atlanta Christian College, US has reference to this Academic Journal, Stable Matching Problems alongside Constant Length Preference Lists Rob Irving, David Manlove, Gregg O?Malley University Of Glasgow Department of Computing Science SMI Formulation Set n1 men SM = {m1 , m2 , ?., mn1} Set n2 women SW = {w1 , w2 , ?., wn2} Each man ranks a subset of SW in strict order of preference. Each woman ranks a subset of SM in strict order of preference. A matching M is a bijection between the men in addition to women. We say a (man, woman) pair (m,w) blocks M if: m prefers w so that his partner in M, in addition to w prefers m so that her partner in M. A matching that admits no blocking pair is said so that be stable Can?t improve by making an arrangement outside the matching. We can always find a stable matching in consideration of an instance of SMI by an adaptation of an algorithm known as the Gale/Shapley algorithm (1962). Gale in addition to Sotomayor also proved in 1985 that in consideration of an instance of SMI all stable matchings have the same size. SMTI Formalisation Set of n1 men SM = {m1 , m2 , ?., mn1} Set of n2 women SW = {w1 , w2 , ?., wn2} Each man mi ranks a subset of SW in preference order, in addition to mi?s list may contain ties. Each woman wj ranks a subset of SM in preference order, in addition to wj?s list may contain ties. A matching M is a set of (man , woman) pairs (m,w) such that each of m in addition to w appear in at most one pair, in addition to m in addition to w are on each other?s list. We say a (man, woman) pair (m , w) blocks M if: Either m is unmatched or m strictly prefers w so that his partner in M, in addition to Either w is unmatched or w strictly prefers m so that her partner in M. A matching that admits no blocking pair is said so that be stable Can?t improve by making an arrangement outside the matching.

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Properties When no ties are allowed in a participant?s list: A stable matching in consideration of an instance of SMI can always be found using a slightly modified version of an algorithm known as the Gale/Shapley algorithm (1962). Gale in addition to Sotomayor also proved in 1985 that in consideration of 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 in consideration of an instance of SMTI by breaking the ties arbitrarily in addition to 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 so that consider is finding a stable matching that matches the largest number of men in addition to women. We denote this problem by MAX-SMTI. MAX-SMTI was shown so that be NP-hard by Iwama et 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 in consideration of the matching of graduating medical students so that hospitals posts in many countries ? as typically student?s lists are small in addition to of fixed length. The above problem is the one-to-many generalisation of SMTI called the Hospitals/Residents problem alongside Ties (HRT).

The History cont. The following table shows a list of the known results in consideration of 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 in addition to contain no ties, in addition to the women?s lists are of unbounded length in addition to 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 so that matching only women can be multiply assigned) is produced by phase 1. Phase 2 attempts so that move men from multiply assigned women so that unassigned women. Phase 3 may or may not be need, it reallocates men still assigned so that multiply assigned women.

ReVirt: Enabling Intrusion Analysis Through Virtual-Machine Logging in addition to Replay Outline Attacks Current Systems UMLinux Trusted Computing Base (TCB) ReVert: Details Deterministic / Non-Deterministic Events Cooperative Logging Analysis of Attacks Evaluation Virtualization Overhead Correctness Logging in addition to Replay Overhead Conclusion Discussion

Phase 1 men ?propose? so that 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 alongside 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 Allocation A1 output by phase 1: A1 = { ( m1 , w1 ) , ( m2 , w1 ) , ( m3 , w1 ) , ( m4 , w2 ) } 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 We note here that w1 is multiply assigned.

Phase 2 : Network Construction Add source node s in addition to sink node t. For each multiply assigned woman wj, $ edge (s , wj) alongside capacity 1 less than the number of assignees so that wj. For each unassigned woman wj, $ edge (wj , t) of capacity 1. Let mi be a man alongside 2 women left on his list. Let wj be mi?s first-choice in addition to wk be mi?s second-choice. Add the edges (wj , mi) in addition to (mi , wk) alongside capacity 1. Women alongside only 1 partner may be represented by a vertex as a result of this step. Phase 2 : Network s t w1 w2 w3 w4 m2 m1 m4 2 1 1 1 1 1 1 1 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 s t w1 w2 w3 w4 m2 m1 m4 2 1 1 1 1 1 1 1 1 The maximum (saturating) flow gives rise so that the following: m1 being moved from w1 so that w2; m4 being moved from w2 so that w3; m2 being moved from w1 so that 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 in addition to we continue alongside 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 in addition to (2,n)-MAX-HRT in which both sides preference lists contain ties polynomial-time solvable? Finding the exact boundary between P in addition to NP-hard cases, i.e. when both men in addition to women have preference lists of size at most 3 in addition to their lists contain ties.

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