Stable Marriage Problem

Given N men and N women, where each person has ranked all members of the opposite sex in order of preference, marry the men and women together such that there are no two people of opposite sex who would both rather have each other than their current partners. If there are no such people, all the marriages are “stable” (Source Wiki).

Consider the following example.

Let there be two men m1 and m2 and two women w1 and w2.
Let m1‘s list of preferences be {w1, w2}
Let m2‘s list of preferences be {w1, w2}
Let w1‘s list of preferences be {m1, m2}
Let w2‘s list of preferences be {m1, m2}

The matching { {m1, w2}, {w1, m2} } is not stable because m1 and w1 would prefer each other over their assigned partners. The matching {m1, w1} and {m2, w2} is stable because there are no two people of opposite sex that would prefer each other over their assigned partners.

It is always possible to form stable marriages from lists of preferences (See references for proof). Following is Gale–Shapley algorithm to find a stable matching:
The idea is to iterate through all free men while there is any free man available. Every free man goes to all women in his preference list according to the order. For every woman he goes to, he checks if the woman is free, if yes, they both become engaged. If the woman is not free, then the woman chooses either says no to him or dumps her current engagement according to her preference list. So an engagement done once can be broken if a woman gets better option.
Following is complete algorithm from Wiki

Initialize all men and women to free
while there exist a free man m who still has a woman w to propose to 
{
    w = m's highest ranked such woman to whom he has not yet proposed
    if w is free
       (m, w) become engaged
    else some pair (m', w) already exists
       if w prefers m to m'
          (m, w) become engaged
           m' becomes free
       else
          (m', w) remain engaged    
}

Input & Output: Input is a 2D matrix of size (2*N)*N where N is number of women or men. Rows from 0 to N-1 represent preference lists of men and rows from N to 2*N – 1 represent preference lists of women. So men are numbered from 0 to N-1 and women are numbered from N to 2*N – 1. The output is list of married pairs.

Following is C++ implementation of the above algorithm.

// C++ program for stable marriage problem #include <iostream> #include <string.h> #include <stdio.h> using namespace std;   // Number of Men or Women #define N  4   // This function returns true if woman 'w' prefers man 'm1' over man 'm' bool wPrefersM1OverM(int prefer[2*N][N], int w, int m, int m1) {     // Check if w prefers m over her current engagment m1     for (int i = 0; i < N; i++)     {         // If m1 comes before m in lisr of w, then w prefers her         // cirrent engagement, don't do anything         if (prefer[w][i] == m1)             return true;           // If m cmes before m1 in w's list, then free her current         // engagement and engage her with m         if (prefer[w][i] == m)            return false;     } }   // Prints stable matching for N boys and N girls. Boys are numbered as 0 to // N-1. Girls are numbereed as N to 2N-1. void stableMarriage(int prefer[2*N][N]) {     // Stores partner of women. This is our output array that     // stores paing information.  The value of wPartner[i]     // indicates the partner assigned to woman N+i.  Note that     // the woman numbers between N and 2*N-1. The value -1     // indicates that (N+i)'th woman is free     int wPartner[N];       // An array to store availability of men.  If mFree[i] is     // false, then man 'i' is free, otherwise engaged.     bool mFree[N];       // Initialize all men and women as free     memset(wPartner, -1, sizeof(wPartner));     memset(mFree, false, sizeof(mFree));     int freeCount = N;       // While there are free men     while (freeCount > 0)     {         // Pick the first free man (we could pick any)         int m;         for (m = 0; m < N; m++)             if (mFree[m] == false)                 break;           // One by one go to all women according to m's preferences.         // Here m is the picked free man         for (int i = 0; i < N && mFree[m] == false; i++)         {             int w = prefer[m][i];               // The woman of preference is free, w and m become             // partners (Note that the partnership maybe changed             // later). So we can say they are engaged not married             if (wPartner[w-N] == -1)             {                 wPartner[w-N] = m;                 mFree[m] = true;                 freeCount--;             }               else  // If w is not free             {                 // Find current engagement of w                 int m1 = wPartner[w-N];                   // If w prefers m over her current engagement m1,                 // then break the engagement between w and m1 and                 // engage m with w.                 if (wPrefersM1OverM(prefer, w, m, m1) == false)                 {                     wPartner[w-N] = m;                     mFree[m] = true;                     mFree[m1] = false;                 }             } // End of Else         } // End of the for loop that goes to all women in m's list     } // End of main while loop         // Print the solution     cout << "Woman   Man" << endl;     for (int i = 0; i < N; i++)        cout << " " << i+N << "\t" << wPartner[i] << endl; }   // Driver program to test above functions int main() {     int prefer[2*N][N] = { {7, 5, 6, 4},         {5, 4, 6, 7},         {4, 5, 6, 7},         {4, 5, 6, 7},         {0, 1, 2, 3},         {0, 1, 2, 3},         {0, 1, 2, 3},         {0, 1, 2, 3},     };     stableMarriage(prefer);       return 0; }

Output:

Woman   Man
 4      2
 5      1
 6      3
 7      0