The Gale-Shapley Algorithm

Objectives

1. Introduce and understand the Gale-Shapley Algorithm

Notes

• Published in 1962, but had been discovered/solved earlier.
• Big idea:
  • Boys try to get their best match, but might have to propose multiple times.
  • Girls take what is offered, but might get to "upgrade"
  • Couples are "engaged" until the end of the algorithm.

• STABLE-MATCH(M, W)
  
   Mark all m ∈ M and w ∈ W as free
   S ← {}
   While there is a m ∈ M who is free and has not proposed to every woman
       Chose a free m ∈ M
       w ← the highest ranking w on m's list that m has not proposed to
       if w is free
          S ← S  ∪ (m,w) 
          mark m and w as engaged
       // w is engaged to m' and (m',w) is in the set S.
       else if w prefers m to m' 
          S ← S - {(m',w)}
          mark m' as free
          S ← S ∪ (m,w)
          mark m as engaged
    return S
  

• What does this algorithm do
  • For women, it represents a sequence of increasingly better matches.
  • For men it represents a sequence of matches that are less desirable.
  • But changing the loop in line 3 to women would change the situation.
• Let's try this with the data from before

•  
  
       
  
            
  1
  C
  B
  E
  A
  D
  free
            
  2
  A
  B
  E
  C
  D
  free
            
  3
  D
  C
  B
  A
  E
  free
            
  4
  A
  C
  D
  B
  E
  free
            
  5
  A
  B
  D
  E
  C
  free
            
  
            
  A
  3
  5
  2
  1
  4
  free
            
  B
  5
  2
  1
  4
  3
  free
            
  C
  4
  3
  5
  1
  2
  free
            
  D
  1
  2
  3
  4
  5
  free
            
  E
  2
  3
  4
  1
  5
  free
        
  S = {}
  1 is free so 
  m = 1
     Note, the choice of 1 is arbitrary, I could choose any man to start with.
  
  C is 1's first choice and  she is free so
     S = {(1,C)}
     We will mark both as engaged and remove C from 1's list
   
  
       
  
            
  1
   
  B
  E
  A
  D
  engaged
            
  2
  A
  B
  E
  C
  D
  free
            
  3
  D
  C
  B
  A
  E
  free
            
  4
  A
  C
  D
  B
  E
  free
            
  5
  A
  B
  D
  E
  C
  free
            
  
            
  A
  3
  5
  2
  1
  4
  free
            
  B
  5
  2
  1
  4
  3
  free
            
  C
  4
  3
  5
  1
  2
  engaged
            
  D
  1
  2
  3
  4
  5
  free
            
  E
  2
  3
  4
  1
  5
  free
        
  
  
  m = 2
      Note: I could chose any free man at this point. 
  A is 2's first choice and she is free so 
     S = {(1,C), (2,A)}
     We will mark both as engaged and remove A from 2's list
   
  
       
  
            
  1
   
  B
  E
  A
  D
  engaged
            
  2
   
  B
  E
  C
  D
  engaged
            
  3
  D
  C
  B
  A
  E
  free
            
  4
  A
  C
  D
  B
  E
  free
            
  5
  A
  B
  D
  E
  C
  free
            
  
            
  A
  3
  5
  2
  1
  4
  engaged
            
  B
  5
  2
  1
  4
  3
  free
            
  C
  4
  3
  5
  1
  2
  engaged
            
  D
  1
  2
  3
  4
  5
  free
            
  E
  2
  3
  4
  1
  5
  free
        
  
  m = 3
  D is 3's first choice and she is free so 
     S = {(1,C), (2,A), (3,D)}
     We will mark both as engaged and remove D from 3's list
   
  
       
  
            
  1
   
  B
  E
  A
  D
  engaged
            
  2
   
  B
  E
  C
  D
  engaged
            
  3
   
  C
  B
  A
  E
  engaged
            
  4
  A
  C
  D
  B
  E
  free
            
  5
  A
  B
  D
  E
  C
  free
            
  
            
  A
  3
  5
  2
  1
  4
  engaged
            
  B
  5
  2
  1
  4
  3
  free
            
  C
  4
  3
  5
  1
  2
  engaged
            
  D
  1
  2
  3
  4
  5
  engaged
            
  E
  2
  3
  4
  1
  5
  free
        
  
  m = 4
  A is 4's first choice, but she is engaged to 2,
     A prefers 2 to 4 so 4 is rejected.
  C is 4's second choice, but she is engaged to 1
     C prefers 4 to 1
     C breaks the engagement with 1, who becomes free
     C becomes engaged to 4.
     S = {(4,C), (2,A), (3,D)}
     Mark 4 as engaged
     Remove A and C from 4's list
     Mark 1 as free
   
  
       
  
            
  1
   
  B
  E
  A
  D
  free
            
  2
   
  B
  E
  C
  D
  engaged
            
  3
   
  C
  B
  A
  E
  engaged
            
  4
   
   
  D
  B
  E
  engaged
            
  5
  A
  B
  D
  E
  C
  free
            
  
            
  A
  3
  5
  2
  1
  4
  engaged
            
  B
  5
  2
  1
  4
  3
  free
            
  C
  4
  3
  5
  1
  2
  engaged
            
  D
  1
  2
  3
  4
  5
  engaged
            
  E
  2
  3
  4
  1
  5
  free
        
    
  m = 1
     Again, this could be 5.
  B is 1's  current first choice and she is free so 
     S = {(4,C), (2,A), (3,D), (1,B)}
     We will mark both as engaged and remove B from 1's list
   
  
       
  
            
  1
   
   
  E
  A
  D
  engaged
            
  2
   
  B
  E
  C
  D
  engaged
            
  3
   
  C
  B
  A
  E
  engaged
            
  4
   
   
  D
  B
  E
  engaged
            
  5
  A
  B
  D
  E
  C
  free
            
  
            
  A
  3
  5
  2
  1
  4
  engaged
            
  B
  5
  2
  1
  4
  3
  engaged
            
  C
  4
  3
  5
  1
  2
  engaged
            
  D
  1
  2
  3
  4
  5
  engaged
            
  E
  2
  3
  4
  1
  5
  free
        
  
  m = 5
  A is 5's first choice, but she is engaged to 2,
     A prefers 5 to 2
     C breaks the engagement with 2, who becomes free
     C becomes engaged to 5.
     S = {(4,C), (5,A), (3,D), (1,B)}
     Mark 5 as engaged
     Remove A from 5's list
     Mark 2 as free
  
   
  
       
  
            
  1
   
   
  E
  A
  D
  engaged
            
  2
   
  B
  E
  C
  D
  free
            
  3
   
  C
  B
  A
  E
  engaged
            
  4
   
   
  D
  B
  E
  engaged
            
  5
   
  B
  D
  E
  C
  engaged
            
  
            
  A
  3
  5
  2
  1
  4
  engaged
            
  B
  5
  2
  1
  4
  3
  engaged
            
  C
  4
  3
  5
  1
  2
  engaged
            
  D
  1
  2
  3
  4
  5
  engaged
            
  E
  2
  3
  4
  1
  5
  free
        
  
  m =  2
  B is 2's current first choice, but she is engaged to 1,
     B prefers 2 to 1
     B breaks the engagement with 1, who becomes free
     B becomes engaged to 2.
     S = {(4,C), (5,A), (3,D), (2,B)}
     Mark 2 as engaged
     Remove B from 2's list
     Mark 1 as free
   
  
       
  
            
  1
   
   
  E
  A
  D
  free
            
  2
   
   
  E
  C
  D
  engaged
            
  3
   
  C
  B
  A
  E
  engaged
            
  4
   
   
  D
  B
  E
  engaged
            
  5
   
  B
  D
  E
  C
  engaged
            
  
            
  A
  3
  5
  2
  1
  4
  engaged
            
  B
  5
  2
  1
  4
  3
  engaged
            
  C
  4
  3
  5
  1
  2
  engaged
            
  D
  1
  2
  3
  4
  5
  engaged
            
  E
  2
  3
  4
  1
  5
  free
        
  
  m = 1
  E is 1's  current first choice and she is free so 
     S = {(4,C), (5,A), (3,D), (2,B), (1,E)}
     We will mark both as engaged and remove E from 1's list
   
  
       
  
            
  1
   
   
   
  A
  D
  engaged
            
  2
   
   
  E
  C
  D
  engaged
            
  3
   
  C
  B
  A
  E
  engaged
            
  4
   
   
  D
  B
  E
  engaged
            
  5
   
  B
  D
  E
  C
  engaged
            
  
            
  A
  3
  5
  2
  1
  4
  engaged
            
  B
  5
  2
  1
  4
  3
  engaged
            
  C
  4
  3
  5
  1
  2
  engaged
            
  D
  1
  2
  3
  4
  5
  engaged
            
  E
  2
  3
  4
  1
  5
  engaged
        
  
  All men are engaged, so the algorithm is done.

• Is this stable?

  • 1 C B E A D free 2 A B E C D free 3 D C B A E free 4 A C D B E free 5 A B D E C free

    A 3 5 2 1 4 free B 5 2 1 4 3 free C 4 3 5 1 2 free D 1 2 3 4 5 free E 2 3 4 1 5 free

  • S = {(5,A), (2,B), (4,C), (3,D), (1,E)}
    • Unknown {A, B, C, D, E, 1, 2, 3, 4, 5}
    • Might Swap {}
    • (5,A)
      • 5 preferred A so he will not swap ever
      • A preferred 3 to 5, so there is a chance here.
    • Unknown {B, C, D, E, 1, 2, 3, 4}
    • Might Swap {A:3}
    • (3,D)
      • 3 preferred D so he will not swap ever.
        • So this rules out a swap for A.
      • D preferred 1 and 2 to 3 so there is a chance here.
    • Unknown {B, C, E, 1, 2, 4}
    • Might Swap {D:1,2}
    • (4,C)
      • 4 preferred A, but a is not willing to swap for anyone but 3.
      • C is 4's second choice and since he can not get A, he will not swap for anyone else.
      • C preferred 4 so she will not swap.
    • Unknown {B, E, 1, 2}
    • Might Swap {D:1,2}
    • (2,B)
      • 2 preferred A to B, but she will not swap for him so he will not swap for anyone else.
      • B preferred 5 to 2, but he will not swap so she will not swap for anyone else.
    • Unknown {E, 1}
    • Might Swap {D:1}
    • (1,E)
      • 1 preferred C and B to E, but they will not swap.
      • Also D is the only possible swap left for 1, and they were last on 1's list.
  • So the solution is correct.