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Jul 11, 2023 •228 likes •1.45k views

Summer School on Matching Problems, Markets and Mechanisms David Manlove Outline 1. The Hospitals / Residents problem and its variants 2. The House Allocation problem 3. Kidney exchange 1 Tutorial 2 The House Allocation Problem with

  1. A faster algorithm l Three-phase algorithm with O( √ nm ) overall complexity • [Abraham, Cechlárová, M and Mehlhorn, 2004] l Phase 1 – O( √ nm ) time – Find a maximum matching in G – Classical O( √ nm ) augmenting path algorithm • [Hopcroft and Karp, 1973] l Phase 2 – O( m ) time – Enforce trade-in-free property l Phase 3 – O( m ) time – Enforce coalition-free property – Extension of Gale’s Top-Trading Cycles (TTC) algorithm • [Shapley and Scarf, 1974] 24

  2. Phase 1 a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 25

  3. Phase 1 a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 l Maximum matching M in G has size 9 l M must be maximal l No guarantee that M is trade-in-free or coalition-free 26

  4. Phase 1 M not coalition-free a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 M not trade-in-free a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 l Maximum matching M in G has size 9 l M must be maximal l No guarantee that M is trade-in-free or coalition-free 27

  5. Phase 2 description l For each house h , maintain a list L h , initially containing those pairs ( a , r ) such that: – a is a matched applicant who prefers h to M ( a ) – r is the rank of h in a ’ s list l Maintain a stack S of unmatched houses h whose list L h is nonempty l Each matched applicant a maintains a pointer curr a to the rank of M ( a ) l Example a 1 : h 4 h 5 h 3 h 2 h 1 L h 6 = 〈 ( a 6 , 5), ( a 7 , 4), ( a 8 , 6) 〉 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 L h 7 = 〈 ( a 6 , 6), ( a 7 , 5), ( a 8 , 5) 〉 a 3 : h 5 h 4 h 1 h 2 h 3 L h 8 = 〈 ( a 6 , 4) 〉 a 4 : h 3 h 5 h 4 L h 10 = 〈 ( a 7 , 7) 〉 a 5 : h 4 h 3 h 5 L h 11 = 〈 ( a 6 , 8) 〉 S = 〈 h 6 , h 7 〉 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 curr a 6 =10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 curr a 7 =9 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 28

  6. Phase 2 description while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h ’ = M(a); remove (a,h ’ ) from M and add (a,h) to M; curr a := r; h := h ’ ; push h onto the stack if L h is nonempty; a 1 : h 4 h 5 h 3 h 2 h 1 L h6 = 〈 ( a 6 , 5), ( a 7 , 4), ( a 8 , 6) 〉 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 L h7 = 〈 ( a 6 , 6), ( a 7 , 5), ( a 8 , 5) 〉 a 3 : h 5 h 4 h 1 h 2 h 3 L h8 = 〈 ( a 6 , 4) 〉 a 4 : h 3 h 5 h 4 L h10 = 〈 ( a 7 , 7) 〉 a 5 : h 4 h 3 h 5 L h11 = 〈 ( a 6 , 8) 〉 S = 〈 h 6 , h 7 〉 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 curr a6 =10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 curr a7 =9 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 29

  7. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h ’ = M(a); remove (a,h ’ ) from M and add (a,h) to M; curr a := h; h := h ’ ; push h onto the stack if L h is nonempty; a 1 : h 4 h 5 h 3 h 2 h 1 L h6 = 〈 ( a 6 , 5), ( a 7 , 4), ( a 8 , 6) 〉 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 L h7 = 〈 ( a 6 , 6), ( a 7 , 5), ( a 8 , 5) 〉 a 3 : h 5 h 4 h 1 h 2 h 3 L h8 = 〈 ( a 6 , 4) 〉 a 4 : h 3 h 5 h 4 L h10 = 〈 ( a 7 , 7) 〉 a 5 : h 4 h 3 h 5 L h11 = 〈 ( a 6 , 8) 〉 S = 〈 h 6 , h 7 〉 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 curr a6 =10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 curr a7 =9 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 30

  8. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h ’ = M(a); remove (a,h ’ ) from M and add (a,h) to M; curr a := h; h := h ’ ; push h onto the stack if L h is nonempty; a 1 : h 4 h 5 h 3 h 2 h 1 L h6 = 〈 ( a 6 , 5), ( a 7 , 4), ( a 8 , 6) 〉 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 L h7 = 〈 ( a 7 , 5), ( a 8 , 5) 〉 a 3 : h 5 h 4 h 1 h 2 h 3 L h8 = 〈 ( a 6 , 4) 〉 a 4 : h 3 h 5 h 4 L h10 = 〈 ( a 7 , 7) 〉 a 5 : h 4 h 3 h 5 L h11 = 〈 ( a 6 , 8) 〉 S = 〈 h 6 , h 10 〉 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 curr a6 =6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 curr a7 =9 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 31

  9. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h ’ = M(a); remove (a,h ’ ) from M and add (a,h) to M; curr a := h; h := h ’ ; push h onto the stack if L h is nonempty; a 1 : h 4 h 5 h 3 h 2 h 1 L h6 = 〈 ( a 6 , 5), ( a 7 , 4), ( a 8 , 6) 〉 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 L h7 = 〈 ( a 7 , 5), ( a 8 , 5) 〉 a 3 : h 5 h 4 h 1 h 2 h 3 L h8 = 〈 ( a 6 , 4) 〉 a 4 : h 3 h 5 h 4 L h10 = 〈 ( a 7 , 7) 〉 a 5 : h 4 h 3 h 5 L h11 = 〈 ( a 6 , 8) 〉 S = 〈 h 6 , h 10 〉 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 curr a6 =6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 curr a7 =9 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 32

  10. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h ’ = M(a); remove (a,h ’ ) from M and add (a,h) to M; curr a := h; h := h ’ ; push h onto the stack if L h is nonempty; a 1 : h 4 h 5 h 3 h 2 h 1 L h6 = 〈 ( a 6 , 5), ( a 7 , 4), ( a 8 , 6) 〉 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 L h7 = 〈 ( a 7 , 5), ( a 8 , 5) 〉 a 3 : h 5 h 4 h 1 h 2 h 3 L h8 = 〈 ( a 6 , 4) 〉 a 4 : h 3 h 5 h 4 L h10 = 〈 〉 a 5 : h 4 h 3 h 5 L h11 = 〈 ( a 6 , 8) 〉 S = 〈 h 6 , h 11 〉 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 curr a6 =6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 curr a7 =7 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 33

  11. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h ’ = M(a); remove (a,h ’ ) from M and add (a,h) to M; curr a := h; h := h ’ ; push h onto the stack if L h is nonempty; a 1 : h 4 h 5 h 3 h 2 h 1 L h6 = 〈 ( a 6 , 5), ( a 7 , 4), ( a 8 , 6) 〉 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 L h7 = 〈 ( a 7 , 5), ( a 8 , 5) 〉 a 3 : h 5 h 4 h 1 h 2 h 3 L h8 = 〈 ( a 6 , 4) 〉 a 4 : h 3 h 5 h 4 L h10 = 〈 〉 a 5 : h 4 h 3 h 5 L h11 = 〈 ( a 6 , 8) 〉 S = 〈 h 6 , h 11 〉 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 curr a6 =6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 curr a7 =7 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 34

  12. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h ’ = M(a); remove (a,h ’ ) from M and add (a,h) to M; curr a := h; h := h ’ ; push h onto the stack if L h is nonempty; a 1 : h 4 h 5 h 3 h 2 h 1 L h6 = 〈 ( a 6 , 5), ( a 7 , 4), ( a 8 , 6) 〉 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 L h7 = 〈 ( a 7 , 5), ( a 8 , 5) 〉 a 3 : h 5 h 4 h 1 h 2 h 3 L h8 = 〈 ( a 6 , 4) 〉 a 4 : h 3 h 5 h 4 L h10 = 〈 〉 a 5 : h 4 h 3 h 5 L h11 = 〈 〉 S = 〈 h 6 〉 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 curr a6 =6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 curr a7 =7 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 35

  13. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h ’ = M(a); remove (a,h ’ ) from M and add (a,h) to M; curr a := h; h := h ’ ; push h onto the stack if L h is nonempty; a 1 : h 4 h 5 h 3 h 2 h 1 L h6 = 〈 ( a 7 , 4), ( a 8 , 6) 〉 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 L h7 = 〈 ( a 7 , 5), ( a 8 , 5) 〉 a 3 : h 5 h 4 h 1 h 2 h 3 L h8 = 〈 ( a 6 , 4) 〉 a 4 : h 3 h 5 h 4 L h10 = 〈 〉 a 5 : h 4 h 3 h 5 L h11 = 〈 〉 S = 〈 h 7 〉 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 curr a6 =5 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 curr a7 =7 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 36

  14. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h ’ = M(a); remove (a,h ’ ) from M and add (a,h) to M; curr a := h; h := h ’ ; push h onto the stack if L h is nonempty; a 1 : h 4 h 5 h 3 h 2 h 1 L h6 = 〈 ( a 7 , 4), ( a 8 , 6) 〉 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 L h7 = 〈 ( a 7 , 5), ( a 8 , 5) 〉 a 3 : h 5 h 4 h 1 h 2 h 3 L h8 = 〈 ( a 6 , 4) 〉 a 4 : h 3 h 5 h 4 L h10 = 〈 〉 a 5 : h 4 h 3 h 5 L h11 = 〈 〉 S = 〈 h 7 〉 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 curr a6 =5 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 curr a7 =7 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 37

  15. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h ’ = M(a); remove (a,h ’ ) from M and add (a,h) to M; curr a := h; h := h ’ ; push h onto the stack if L h is nonempty; a 1 : h 4 h 5 h 3 h 2 h 1 L h6 = 〈 ( a 7 , 4), ( a 8 , 6) 〉 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 L h7 = 〈 ( a 8 , 5) 〉 a 3 : h 5 h 4 h 1 h 2 h 3 L h8 = 〈 ( a 6 , 4) 〉 a 4 : h 3 h 5 h 4 L h10 = 〈 〉 a 5 : h 4 h 3 h 5 L h11 = 〈 〉 S = 〈 〉 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 curr a6 =5 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 curr a7 =5 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 38

  16. Phase 2 example while S is nonempty pop a house h from S; remove the first pair (a,r) from the head of L h ; if r < curr a // a prefers h to M(a) let h ’ = M(a); remove (a,h ’ ) from M and add (a,h) to M; curr a := h; h := h ’ ; push h onto the stack if L h is nonempty; a 1 : h 4 h 5 h 3 h 2 h 1 L h6 = 〈 ( a 7 , 4), ( a 8 , 6) 〉 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 L h7 = 〈 ( a 8 , 5) 〉 a 3 : h 5 h 4 h 1 h 2 h 3 L h8 = 〈 ( a 6 , 4) 〉 a 4 : h 3 h 5 h 4 L h10 = 〈 〉 a 5 : h 4 h 3 h 5 L h11 = 〈 〉 S = 〈 〉 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 curr a6 =5 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 curr a7 =5 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 39

  17. Phase 2 termination l Once Phase 2 terminates, matching is trade-in-free l Data structures may be initialised in O( m ) time l Main loop takes O( m ) time overall l ∴ Phase 2 is O( m ) l Coalitions may remain … 40

  18. Phase 2 termination l Once Phase 2 terminates, matching is trade-in-free l Data structures may be initialised in O( m ) time l Main loop takes O( m ) time overall l ∴ Phase 2 is O( m ) l Coalitions may remain … l Example a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 41

  19. Phase 3 description l Once Phase 2 terminates, matching is trade-in-free l Still need to eliminate potential coalitions l Repeatedly finding and eliminating coalitions takes O( m 2 ) time – Cycle detection in G ( M ) takes O( m ) time – O( m ) coalitions in the worst case l Faster method: extension of TTC algorithm – An applicant matched to his/her first-choice house cannot be in a coalition • Such an applicant can be removed from consideration – Houses matched to such applicants are no longer exchangeable • Such a house can be removed from consideration – This rule can be recursively applied until either • No applicant remains (matching is coalition-free) • A coalition exists, which can be found and removed 42

  20. Phase 3 description l Build a path P of applicants (represented by a stack) l Each house is initially unlabelled l Each applicant a has a pointer p ( a ) pointing to M ( a ) or the first unlabelled house on a ’s preference list (whichever comes first) l Keep a counter c ( a ) for each applicant a (initially c ( a )=0 ) – This represents the number of times a appears on the stack l Outer loop iterates over each matched applicant a such that p ( a ) ≠ M ( a ) l Initialise P to contain applicant a l Inner loop iterates while P is nonempty – Pop an applicant 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 ’’ ) 43

  21. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 1 h 1 a 2 0 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 0 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 0 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 1 P = 〈 a 1 〉 44

  22. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 1 h 1 a 2 0 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 1 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 0 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 4 a 1 EXTEND P = 〈 a 1 , a 4 〉 45

  23. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 1 h 1 a 2 0 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 1 h 3 a 4 : h 3 h 5 h 4 a 4 1 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 0 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 4 a 1 EXTEND a 3 P = 〈 a 1 , a 4 , a 3 〉 46

  24. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 1 h 1 a 2 0 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 1 h 3 a 4 : h 3 h 5 h 4 a 4 1 h 4 a 5 : h 4 h 3 h 5 a 5 1 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 0 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 4 a 1 EXTEND a 3 P = 〈 a 1 , a 4 , a 3 , a 5 〉 a 5 47

  25. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 1 h 1 a 2 0 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 1 h 3 a 4 : h 3 h 5 h 4 a 4 2 h 4 a 5 : h 4 h 3 h 5 a 5 1 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 0 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 4 a 1 EXTEND a 3 P = 〈 a 1 , a 4 , a 3 , a 5 , a 4 〉 a 5 48

  26. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 1 h 1 a 2 0 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 1 h 3 a 4 : h 3 h 5 h 4 a 4 2 h 4 a 5 : h 4 h 3 h 5 a 5 1 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 0 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 4 a 1 CYCLE a 3 P = 〈 a 1 , a 4 , a 3 , a 5 〉 a 5 49

  27. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 1 h 1 a 2 0 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 0 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 0 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 4 a 1 CYCLE a 3 P = 〈 a 1 , a 4 , a 3 , a 5 〉 a 5 50

  28. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 1 h 1 a 2 0 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 0 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 0 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 1 P = 〈 a 1 〉 51

  29. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 1 h 1 a 2 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 0 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 0 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 1 a 2 EXTEND P = 〈 a 1 , a 2 〉 52

  30. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 1 h 1 a 2 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 0 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 0 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 1 h 9 a 9 : h 4 h 3 h 5 h 9 a 1 a 2 EXTEND P = 〈 a 1 , a 2 , a 9 〉 a 9 53

  31. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 1 h 1 a 2 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 0 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 0 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 1 h 9 a 9 : h 4 h 3 h 5 h 9 a 1 a 2 BACKTRACK P = 〈 a 1 , a 2 〉 a 9 54

  32. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 1 h 1 a 2 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 0 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 0 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 1 a 2 P = 〈 a 1 , a 2 〉 55

  33. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 2 h 1 a 2 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 0 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 0 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 1 a 2 EXTEND P = 〈 a 1 , a 2 , a 1 〉 56

  34. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 2 h 1 a 2 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 0 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 0 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 1 a 2 CYCLE P = 〈 a 1 , a 2 〉 57

  35. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 0 h 1 a 2 0 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 0 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 0 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 1 a 2 CYCLE P = 〈 a 1 , a 2 〉 58

  36. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 0 h 1 a 2 0 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 0 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 1 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 6 P = 〈 a 6 〉 59

  37. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 0 h 1 a 2 0 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 0 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 1 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 1 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 6 a 8 EXTEND P = 〈 a 6 , a 8 〉 60

  38. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 0 h 1 a 2 0 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 0 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 1 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 1 h 7 a 8 1 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 6 a 8 EXTEND P = 〈 a 6 , a 8 , a 7 〉 a 7 61

  39. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 0 h 1 a 2 0 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 0 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 2 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 1 h 7 a 8 1 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 6 a 8 EXTEND P = 〈 a 6 , a 8 , a 7 , a 6 〉 a 7 62

  40. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 0 h 1 a 2 0 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 0 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 2 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 1 h 7 a 8 1 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 6 a 8 CYCLE P = 〈 a 6 , a 8 , a 7 〉 a 7 63

  41. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 0 h 1 a 2 0 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 0 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 0 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 a 6 a 8 CYCLE P = 〈 a 6 , a 8 , a 7 〉 64

  42. Phase 3: example applicant counter house label a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 1 0 h 1 a 2 0 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 3 0 h 3 a 4 : h 3 h 5 h 4 a 4 0 h 4 a 5 : h 4 h 3 h 5 a 5 0 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 6 0 h 6 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 7 0 h 7 a 8 0 h 8 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 0 h 9 a 9 : h 4 h 3 h 5 h 9 P = 〈〉 〈〉 65

  43. Phase 3 termination l Once Phase 3 terminates, matching is coalition-free l Data structures may be initialised in O( m ) time l Nested loops take O( m ) time overall l ∴ Phase 3 is O( m ) a 1 : h 4 h 5 h 3 h 2 h 1 a 2 : h 3 h 4 h 5 h 9 h 1 h 2 a 3 : h 5 h 4 h 1 h 2 h 3 a 4 : h 3 h 5 h 4 a 5 : h 4 h 3 h 5 a 6 : h 2 h 3 h 5 h 8 h 6 h 7 h 1 h 11 h 4 h 10 a 7 : h 1 h 4 h 3 h 6 h 7 h 2 h 10 h 5 h 11 a 8 : h 1 h 5 h 4 h 3 h 7 h 6 h 8 a 9 : h 4 h 3 h 5 h 9 l Theorem: given an instance of HA, a maximum Pareto optimal matching can be found in O( √ nm ) time. 66

  44. Housing markets l Each applicant owns a house initially – their initial endowment M 0 l Any matching M must be individually rational – for each applicant a , either M ( a ) =M 0 ( a ) or a prefers M to M 0 l An individually rational matching M is in the strict core if there is no other matching M' that weakly blocks with respect to the coalition S – the members of S can only improve by exchanging their own resources – some member of S prefers M' to M – no member of S prefers M to M' l Every Housing Market admits a unique strict core matching (using the TTC algorithm) – [Roth and Postlewaite, 1977] – [Shapley and Scarf, 1974] l The TTC algorithm is strategy-proof – [Roth, 1982a] 67

  45. Tutorial Outline 2.1: Pareto optimal matchings 2.2: Popular matchings 2.3: Profile-based optimal matchings 68

  46. Popular matchings l A matching M is popular if there is no matching M' such that more applicants prefer M' to M than prefer M to M' l Define the relation ← on matchings by M ← M' if more applicants prefer M to M' than prefer M' to M l A popular matching is a minimal element in the relation ← l A popular matching need not exist, e.g., a 1 : h 1 h 2 h 3 a 2 : h 1 h 2 h 3 a 3 : h 1 h 2 h 3 69

  47. Popular matchings may not exist l A matching M is popular if there is no matching M' such that more applicants prefer M' to M than prefer M to M' l Define the relation ← on matchings by M ← M' if more applicants prefer M to M' than prefer M' to M l A popular matching is a minimal element in the relation ← l A popular matching need not exist, e.g., a 1 : h 1 h 2 h 3 a 2 : h 1 h 2 h 3 a 3 : h 1 h 2 h 3 l The blue matching shown is unique up to symmetry 70

  48. Popular matchings may not exist l A matching M is popular if there is no matching M' such that more applicants prefer M' to M than prefer M to M' l Define the relation ← on matchings by M ← M' if more applicants prefer M to M' than prefer M' to M (i.e., M' is more popular than M ) l A popular matching is a minimal element in the relation ← l A popular matching need not exist, e.g., a 1 : h 1 h 2 h 3 a 2 : h 1 h 2 h 3 a 3 : h 1 h 2 h 3 l The blue matching shown is unique up to symmetry l The green matching M' satisfies M' ← M 71

  49. Popular matchings can have different sizes a 1 : h 1 h 4 a 2 : h 2 h 5 a 3 : h 3 h 4 h 6 a 4 : h 1 a 5 : h 2 a 6 : h 3 72

  50. Popular matchings can have different sizes a 1 : h 1 h 4 a 2 : h 2 h 5 a 3 : h 3 h 4 h 6 a 4 : h 1 a 5 : h 2 a 6 : h 3 l The black matching is the unique maximum matching 73

  51. Popular matchings can have different sizes a 1 : h 1 h 4 a 2 : h 2 h 5 a 3 : h 3 h 4 h 6 a 4 : h 1 a 5 : h 2 a 6 : h 3 l The black matching is the unique maximum matching l It isn ’ t popular: the green matching M' satisfies M' ← M 74

  52. Popular matchings can have different sizes a 1 : h 1 h 4 a 2 : h 2 h 5 a 3 : h 3 h 4 h 6 a 4 : h 1 a 5 : h 2 a 6 : h 3 l In fact the green matching is itself popular (size 5 ) 75

  53. Popular matchings can have different sizes a 1 : h 1 h 4 a 2 : h 2 h 5 a 3 : h 3 h 4 h 6 a 4 : h 1 a 5 : h 2 a 6 : h 3 l In fact the green matching is itself popular (size 5 ) l So is the blue matching (size 3 ) 76

  54. Popular matchings can have different sizes a 1 : h 1 h 4 a 2 : h 2 h 5 a 3 : h 3 h 4 h 6 a 4 : h 1 a 5 : h 2 a 6 : h 3 l There is no popular matching of size 6 l So the green matching is a maximum popular matching 77

  55. Summary of results l Is there a polynomial-time algorithm to determine whether a popular matching exists, and if so to find one? l Problem introduced by [Gärdenfors, 1975] in the context of two-sided preferences l Neat (and surprising) characterisation of popular matchings l Case of strict preferences: – O( n + m ) algorithm to determine whether a popular matching exists, and if so to find a largest one l Ties in the preference lists: – O( √ nm ) algorithm for the same problem l [Abraham, Irving, Kavitha, Mehlhorn, 2005] 78

  56. Graph representation l Bipartite graph G , applicant vertices A = { a 1 , . . ., a r } , house vertices H = { h 1 , . . ., h s } l Edge { a i , h j } of weight k if h j is the k th choice of a i l E k is the set of all edges of weight k l Example: a 1 h 1 a 1 : h 1 h 3 h 4 E 1 in red h 2 a 2 a 2 : h 1 h 4 h 2 E 2 in blue a 3 h 3 a 3 : h 2 h 1 h 3 E 3 in green h 4 a 4 : h 2 h 3 h 5 a 4 a 5 : h 3 h 5 h 1 h 5 a 5 79

  57. Last resorts l For each applicant a append a unique house l ( a ) to the end of his list – his last resort – This ensures every maximal matching is applicant-complete – The size of a matching is the number of applicants not matched to their last resort Example: l ( a 1 ) h 1 E 1 in red a 1 l (a 1 ) a 1 : h 1 h 3 h 4 l ( a 2 ) h 2 E 2 in blue a 2 l (a 2 ) a 2 : h 1 h 4 h 2 l ( a 3 ) h 3 a 3 l (a 3 ) a 3 : h 2 h 1 h 3 E 3 in green h 4 l ( a 4 ) l (a 4 ) a 4 : h 2 h 3 h 5 a 4 E 4 in black a 5 : h 3 h 5 h 1 l (a 5 ) h 5 l ( a 5 ) a 5 80

  58. f -houses and s -houses l For applicant a , f ( a ) denotes his first choice house l House h is an f -house if h = f ( a ) for some a l For applicant a , s ( a ) denotes the first non f -house on his list – s ( a ) is bound to exist (because of l ( a ) ) 81

  59. f -houses and s -houses l For applicant a , f ( a ) denotes his first choice house l House h is an f -house if h = f ( a ) for some a l For applicant a , s ( a ) denotes the first non f -house on his list – s ( a ) is bound to exist (because of l ( a ) ) Example: l (a 1 ) a 1 : h 1 h 3 h 4 a 2 : h 1 h 4 h 2 l (a 2 ) a 3 : h 2 h 1 h 3 l ( a 3 ) l ( a 4 ) a 4 : h 2 h 3 h 5 l ( a 5 ) a 5 : h 3 h 5 h 1 – f -houses in green – f -houses are h 1 , h 2 and h 3 82

  60. f -houses and s -houses l For applicant a , f ( a ) denotes his first choice house l House h is an f -house if h = f ( a ) for some a l For applicant a , s ( a ) denotes the first non f -house on his list – s ( a ) is bound to exist (because of l ( a ) ) Example: l (a 1 ) a 1 : h 1 h 3 h 4 a 2 : h 1 h 4 h 2 l (a 2 ) a 3 : h 2 h 1 h 3 l ( a 3 ) l ( a 4 ) a 4 : h 2 h 3 h 5 l ( a 5 ) a 5 : h 3 h 5 h 1 – f -houses in green – f -houses are h 1 , h 2 and h 3 – s -houses in blue – s -houses are l ( a 3 ) , h 4 and h 5 83

  61. Lemmas and Theorem l Lemma 1 : If M is a popular matching, and h is an f -house then M ( h ) = a for some a such that h = f ( a ) . l Lemma 2 : If M is a popular matching, then M ( a ) cannot lie between f ( a ) and s ( a ) on a ’ s preference list. l Lemma 3 : If M is a popular matching, then M ( a ) is never below s ( a ) on a ’ s preference list. l Theorem 1 : M is a popular matching if and only if (i) every f -house is matched in M , and (ii) for each applicant a , M ( a ) = f ( a ) or M ( a ) = s ( a ) . 84

  62. The reduced graph G' l For each applicant vertex a i , delete all incident edges except those to f ( a i ) and s ( a i ) Example l ( a 1 ) h 1 a 1 l (a 1 ) a 1 : h 1 h 3 h 4 l ( a 2 ) h 2 a 2 l (a 2 ) a 2 : h 1 h 4 h 2 l ( a 3 ) h 3 a 3 a 3 : h 2 h 1 h 3 l (a 3 ) h 4 l ( a 4 ) a 4 : h 2 h 3 h 5 l (a 4 ) a 4 a 5 : h 3 h 5 h 1 l (a 5 ) l ( a 5 ) h 5 a 5 f ( a i ) in green ( a i , f ( a i )) in green ( a i , s ( a i )) in blue s ( a i ) in blue 85

  63. The reduced graph G' l Theorem 2 : Matching M is popular if and only if (i) every f -house is matched in M ; and (ii) M is an applicant-complete matching in the reduced graph G' Example l ( a 1 ) h 1 a 1 l (a 1 ) a 1 : h 1 h 3 h 4 l ( a 2 ) h 2 a 2 l (a 2 ) a 2 : h 1 h 4 h 2 l ( a 3 ) h 3 a 3 a 3 : h 2 h 1 h 3 l (a 3 ) h 4 l ( a 4 ) a 4 : h 2 h 3 h 5 l (a 4 ) a 4 l ( a 5 ) l (a 5 ) h 5 a 5 : h 3 h 5 h 1 a 5 86

  64. The reduced graph G' l Theorem 2 : Matching M is popular if and only if (i) every f -house is matched in M ; and (ii) M is an applicant-complete matching in the reduced graph G' Example l ( a 1 ) h 1 a 1 l (a 1 ) a 1 : h 1 h 3 h 4 l ( a 2 ) h 2 a 2 l (a 2 ) a 2 : h 1 h 4 h 2 l ( a 3 ) h 3 a 3 a 3 : h 2 h 1 h 3 l (a 3 ) h 4 l ( a 4 ) a 4 : h 2 h 3 h 5 l (a 4 ) a 4 l ( a 5 ) l (a 5 ) h 5 a 5 : h 3 h 5 h 1 a 5 87

  65. Popular matching algorithm form G = (V = A ∪ H, E), the graph of the instance; form G', the reduced graph of G; if G' admits an applicant-complete matching M (*) for each f-house h unmatched in M a = any applicant in f(h); promote a to h in M; else no popular matching exists; l Complexity: – all steps except (*) are clearly O( m + n ) • O( m ) if n = O( m ) – standard matching algorithm for (*) no better than O( n 3/2 ) 88

  66. Finding an applicant-complete matching in G ' M = ∅ ; while some house h has degree 1 a = unique applicant adjacent to h; M = M ∪ {(a,h)}; G' = G' – {a, h}; // remove a and h from G' while some house h has degree 0 G' = G' – {h}; // every surviving house has degree at least 2 // every surviving applicant also has degree 2 if |surviving houses| < |surviving applicants| no applicant-complete matching exists; else // G' is a family of disjoint cycles return M ∪ any maximum matching of G'; l all steps are O( n + m ) 89

  67. Example execution l (a 1 ) l ( a 1 ) a 1 : h 1 h 3 h 4 h 1 a 1 l (a 2 ) a 2 : h 1 h 4 h 2 l ( a 2 ) h 2 a 2 l (a 3 ) a 3 : h 2 h 1 h 3 l ( a 3 ) h 3 a 3 l (a 4 ) a 4 : h 2 h 3 h 5 h 4 l ( a 4 ) l (a 5 ) a 4 a 5 : h 3 h 5 h 1 l ( a 5 ) h 5 a 5 l M = ∅ – house l ( a 3 ) has degree 1 ; add ( a 3 , l ( a 3 )) to M – remove a 3 and l ( a 3 ) from G' 90

  68. Example execution l (a 1 ) l ( a 1 ) a 1 : h 1 h 3 h 4 h 1 a 1 l (a 2 ) a 2 : h 1 h 4 h 2 l ( a 2 ) h 2 a 2 l (a 3 ) a 3 : h 2 h 1 h 3 h 3 l (a 4 ) a 4 : h 2 h 3 h 5 h 4 l ( a 4 ) l (a 5 ) a 4 a 5 : h 3 h 5 h 1 l ( a 5 ) h 5 a 5 l M = {( a 3 , l ( a 3 ))} – house h 2 has degree 1 ; add ( a 4 , h 2 ) to M – remove a 4 and h 2 from G' 91

  69. Example execution l (a 1 ) l ( a 1 ) a 1 : h 1 h 3 h 4 h 1 a 1 l (a 2 ) a 2 : h 1 h 4 h 2 l ( a 2 ) a 2 l (a 3 ) a 3 : h 2 h 1 h 3 h 3 l (a 4 ) a 4 : h 2 h 3 h 5 h 4 l ( a 4 ) l (a 5 ) a 5 : h 3 h 5 h 1 l ( a 5 ) h 5 a 5 l M = {( a 3 , l ( a 3 )), ( a 4 , h 2 )} – house h 5 has degree 1 ; add ( a 5 , h 5 ) to M – remove a 5 and h 5 from G' 92

  70. Example execution l (a 1 ) l ( a 1 ) a 1 : h 1 h 3 h 4 h 1 a 1 l (a 2 ) a 2 : h 1 h 4 h 2 l ( a 2 ) a 2 l (a 3 ) a 3 : h 2 h 1 h 3 h 3 l (a 4 ) a 4 : h 2 h 3 h 5 h 4 l ( a 4 ) l (a 5 ) a 5 : h 3 h 5 h 1 l ( a 5 ) l M = {( a 3 , l ( a 3 )), ( a 4 , h 2 ), ( a 5 , h 5 )} – remove degree zero houses h 3 , l ( a 1 ) , l ( a 2 ) , l ( a 4 ) , l ( a 5 ) 93

  71. Example execution l (a 1 ) a 1 : h 1 h 3 h 4 h 1 a 1 l (a 2 ) a 2 : h 1 h 4 h 2 a 2 l (a 3 ) a 3 : h 2 h 1 h 3 l (a 4 ) a 4 : h 2 h 3 h 5 h 4 l (a 5 ) a 5 : h 3 h 5 h 1 l M = {( a 3 , l ( a 3 )), ( a 4 , h 2 ), ( a 5 , h 5 )} – no. of surviving houses = no. of surviving applicants – G' consists of just one cycle; – traverse the cycle adding alternate edges to M ; 94

  72. Example execution l (a 1 ) a 1 : h 1 h 3 h 4 h 1 a 1 l (a 2 ) a 2 : h 1 h 4 h 2 a 2 l (a 3 ) a 3 : h 2 h 1 h 3 l (a 4 ) a 4 : h 2 h 3 h 5 h 4 l (a 5 ) a 5 : h 3 h 5 h 1 l M = {( a 1 , h 1 ), ( a 2 , h 4 ), ( a 3 , l ( a 3 )), ( a 4 , h 2 ), ( a 5 , h 5 )} 95

  73. Example execution l (a 1 ) l ( a 1 ) a 1 : h 1 h 3 h 4 h 1 a 1 l (a 2 ) a 2 : h 1 h 4 h 2 l ( a 2 ) h 2 a 2 l (a 3 ) a 3 : h 2 h 1 h 3 l ( a 3 ) h 3 a 3 l (a 4 ) a 4 : h 2 h 3 h 5 h 4 l ( a 4 ) l (a 5 ) a 4 a 5 : h 3 h 5 h 1 l ( a 5 ) h 5 a 5 l M = {( a 1 , h 1 ), ( a 2 , h 4 ), ( a 3 , l ( a 3 )), ( a 4 , h 2 ), ( a 5 , h 5 )} – h 3 is an unmatched f -house in M ; – promote a 5 to h 3 96

  74. Example execution l (a 1 ) l ( a 1 ) a 1 : h 1 h 3 h 4 h 1 a 1 l (a 2 ) a 2 : h 1 h 4 h 2 l ( a 2 ) h 2 a 2 l (a 3 ) a 3 : h 2 h 1 h 3 l ( a 3 ) h 3 a 3 l (a 4 ) a 4 : h 2 h 3 h 5 h 4 l ( a 4 ) l (a 5 ) a 4 a 5 : h 3 h 5 h 1 l ( a 5 ) h 5 a 5 l M = {( a 1 , h 1 ), ( a 2 , h 4 ), ( a 3 , l ( a 3 )), ( a 4 , h 2 ), ( a 5 , h 3 )} l Matching M with last resort edges removed is now popular 97

  75. Example execution l (a 1 ) l ( a 1 ) a 1 : h 1 h 3 h 4 h 1 a 1 l (a 2 ) a 2 : h 1 h 4 h 2 l ( a 2 ) h 2 a 2 l (a 3 ) a 3 : h 2 h 1 h 3 l ( a 3 ) h 3 a 3 l (a 4 ) a 4 : h 2 h 3 h 5 h 4 l ( a 4 ) l (a 5 ) a 4 a 5 : h 3 h 5 h 1 l ( a 5 ) h 5 a 5 l M = {( a 1 , h 1 ), ( a 2 , h 4 ), ( a 4 , h 2 ), ( a 5 , h 3 )} l Matching M is now popular l Theorem: There is an O( n+m ) algorithm that finds a popular matching or reports that none exists, given an instance of HA. l Straightforward extension to find a maximum popular matching l Non-trivial extension to the case of ties in the preference lists ( O( √ nm ) time) – [Abraham, Irving, Kavitha and Mehlhorn, 2005] 98

  76. Tutorial Outline 2.1: Pareto optimal matchings 2.2: Popular matchings 2.3: Profile-based optimal matchings 99

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