An Algorithm

• He points out that by designing the algorithm, we can also answer the question "Does every set of preferences have a stable match?"
• The basic outline

  • Gale-Shapley(M, W)
    
     set everyone's status to free
     while there is a man who is free and has not proposed to all women
        select a man m who is free and has not yet proposed to all women
        let w be the highest ranked women in m's list to who he has not yet proposed
        if w is free
            add (m, w) to S
            mark m and w as engaged
        else w is currently engaged to m'
            if w prefers m to m'
               remove (m', w) from S
               mark m' as free
               add (m, w) to S
               mark m as engaged
            endif
        endif
     endwhile

• Let's trace it for
  • M: {A, B, C}
  • W: {1, 2, 3}
  • A: 1, 2, 3
  • B: 2, 1, 3
  • C 1, 2, 3
  • 1: A, B, C
  • 2: A, C, B
  • 3: A, C, B

   Start 
       A(F): 1 2 3    B(F): 2 1 3   C(F): 1 2 3
       1(F): A B C    2(F): A C B   3(F): A C B
       S:
  
  Step 1, select C  (This is a choice to exercise the algorithm.)
      The first person on C's list is 1
      1 does not have a match, (C, 1) becomes a match
  
       A(F): 1 2 3    B(F): 2 1 3   C(E): {1} 2 3
       1(E): A B C    2(F): A C B   3(F): A C B
       S: (C, 1)
  
  Step 2, Select A  (This is a choice to exercise the algorithm)
       The first person on A's list is 1
       1 Prefers A to C 
          (A, 1)  becomes a match
          Mark A as E
          Mark C as F
  
       Note (A, 1) Are effectively removed.
       A(E): {1} 2 3    B(F): 2 1 3   C(F): {1} 2 3
       1(E): A B C    2(F): A C B   3(F): A C B
       S: (A, 1)
  
  Step 3, Select B  (This is a choice to exercise the algorithm)
       B prefers 2
       2 is free so (B, 2) Becomes a match
  
       A(E): {1} 2 3    B(E): {2} 1 3   C(F): {1} 2 3
       1(E): A B C    2(E): A C B   3(F): A C B
       S: (A, 1) (B, 2)
  
  Step 4, Select C (Only choice)
       The current first person on C's list is 2
       2 is matched with B, but prefers C
       (C, 2) becomes a match.
  
       A(E): {1} 2 3    B(F): {2} 1 3   C(E): {1 2} 3
       1(E): A B C    2(E): A C B   3(F): A C B
       S: (A, 1) (C, 2)
  
  Step 5,  Select B (Only Choice)
       B selects 1, but is rejected as 1 has a better match.
  
       A(E): {1} 2 3    B(F): {2 1} 3   C(E): {1 2} 3
       1(E): A B C    2(E): A C B   3(F): A C B
       S: (A, 1) (C, 2)
  
  Step 6, Select B (Only Choice)
       B selects 3
       3 is free 
       (B, 3) Becomes a match
  
       A(E): {1} 2 3    B(E): {2 1 3}   C(E): {1 2} 3
       1(E): A B C    2(E): A C B   3(F): A C B
       S: (A, 1) (C, 2) (B, 3)
  
  There is no one left who is not engaged so the algorithm ends.