The Stable Marriage Problem states that 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. Time Complexity of Gale-Shapley Algorithm is O(n2).
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 the implementation of the above algorithm.
C++
// 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 engagement m1 for (int i = 0; i < N; i++) { // If m1 comes before m in list of w, then w prefers her // current engagement, don't do anything if (prefer[w][i] == m1) return true; // If m comes 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 numbered // as N to 2N-1.void stableMarriage(int prefer[2*N][N]){ // Stores partner of women. This is our output array that // stores passing 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 functionsint 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;} |
Java
// Java program for stable marriage problem import java.util.*;class GFG{ // Number of Men or Women static int N = 4; // This function returns true if woman // 'w' prefers man 'm1' over man 'm' static boolean wPrefersM1OverM(int prefer[][], int w, int m, int m1) { // Check if w prefers m over // her current engagement m1 for (int i = 0; i < N; i++) { // If m1 comes before m in list of w, // then w prefers her current engagement, // don't do anything if (prefer[w][i] == m1) return true; // If m comes before m1 in w's list, // then free her current engagement // and engage her with m if (prefer[w][i] == m) return false; } return false;} // Prints stable matching for N boys and// N girls. Boys are numbered as 0 to // N-1. Girls are numbered as N to 2N-1. static void stableMarriage(int prefer[][]) { // Stores partner of women. This is our // output array that stores passing 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[] = new int[N]; // An array to store availability of men. // If mFree[i] is false, then man 'i' is // free, otherwise engaged. boolean mFree[] = new boolean[N]; // Initialize all men and women as free Arrays.fill(wPartner, -1); 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 System.out.println("Woman Man"); for (int i = 0; i < N; i++) { System.out.print(" "); System.out.println(i + N + " " + wPartner[i]);}} // Driver Codepublic static void main(String[] args) { int prefer[][] = new int[][]{{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); }} // This code is contributed by Prerna Saini |
Python3
# Python3 program for stable marriage problem# Number of Men or WomenN = 4# This function returns true if # woman 'w' prefers man 'm1' over man 'm'def wPrefersM1OverM(prefer, w, m, m1): # Check if w prefers m over her # current engagement m1 for i in range(N): # If m1 comes before m in list of w, # then w prefers her current engagement, # don't do anything if (prefer[w][i] == m1): return True # If m comes 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 numbered as N to 2N-1.def stableMarriage(prefer): # Stores partner of women. This is our output # array that stores passing 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 wPartner = [-1 for i in range(N)] # An array to store availability of men. # If mFree[i] is false, then man 'i' is free, # otherwise engaged. mFree = [False for i in range(N)] freeCount = N # While there are free men while (freeCount > 0): # Pick the first free man (we could pick any) m = 0 while (m < N): if (mFree[m] == False): break m += 1 # One by one go to all women according to # m's preferences. Here m is the picked free man i = 0 while i < N and mFree[m] == False: 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 -= 1 else: # If w is not free # Find current engagement of w 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 i += 1 # End of Else # End of the for loop that goes # to all women in m's list # End of main while loop # Print solution print("Woman ", " Man") for i in range(N): print(i + N, "\t", wPartner[i])# Driver Codeprefer = [[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)# This code is contributed by Mohit Kumar |
C#
// C# program for stable marriage problem using System;using System.Collections.Generic;class GFG { // Number of Men or Women static int N = 4; // This function returns true if woman // 'w' prefers man 'm1' over man 'm' static bool wPrefersM1OverM(int [,]prefer, int w, int m, int m1) { // Check if w prefers m over // her current engagement m1 for (int i = 0; i < N; i++) { // If m1 comes before m in list of w, // then w prefers her current engagement, // don't do anything if (prefer[w, i] == m1) return true; // If m comes before m1 in w's list, // then free her current engagement // and engage her with m if (prefer[w, i] == m) return false; } return false; } // Prints stable matching for N boys and // N girls. Boys are numbered as 0 to // N-1. Girls are numbered as N to 2N-1. static void stableMarriage(int [,]prefer) { // Stores partner of women. This is our // output array that stores passing 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 = new int[N]; // An array to store availability of men. // If mFree[i] is false, then man 'i' is // free, otherwise engaged. bool []mFree = new bool[N]; // Initialize all men and women as free for (int i = 0; i < N; i++) wPartner[i] = -1; 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 Console.WriteLine("Woman Man"); for (int i = 0; i < N; i++) { Console.Write(" "); Console.WriteLine(i + N + " " + wPartner[i]); } } // Driver Code public static void Main(String[] args) { int [,]prefer = new int[,]{{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); } } // This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript program for stable marriage problem// Number of Men or WomenN = 4;// This function returns true if woman 'w' prefers man 'm1' over man 'm'function wPrefersM1OverM( prefer, w, m, m1){ // Check if w prefers m over her current engagement m1 for (var i = 0; i < N; i++) { // If m1 comes before m in list of w, then w prefers her // current engagement, don't do anything if (prefer[w][i] == m1) return true; // If m comes 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 numbered as N to 2N-1.function stableMarriage( prefer){ // Stores partner of women. This is our output array that // stores passing 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 var wPartner = new Array(N); // An array to store availability of men. If mFree[i] is // false, then man 'i' is free, otherwise engaged. mFree = new Array(N); // Initialize all men and women as free wPartner.fill(-1); mFree.fill(false); var freeCount = N; // While there are free men while (freeCount > 0) { // Pick the first free man (we could pick any) var 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 (var i = 0; i < N && mFree[m] == false; i++) { var 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 var 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 document.write("Woman Man" +"<br>"); for (var i = 0; i < N; i++) document.write(" " + (i+N) + " " + wPartner[i] +"<br>");}var prefer = [ [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);// This code is contributed by SoumikMondal</script> |
Output
Woman Man 4 2 5 1 6 3 7 0
Time Complexity : O(N2) , where n is number of men/women.
Auxiliary Space : O(N2)
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