Research

# Stable Marriage Problem

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).
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 the implementation of the above algorithm.

```if __name__ == '__main__':
# Python3 program for stable marriage problem

# Number of Men or Women
# N = 4
N = 3

# This function returns true if woman 'w' prefers man 'm1' over man 'm'
def wPrefersM1OverM(preferences, 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 preferences[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 preferences[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(preferences):

# Stores partner of women.
# This is our output array that stores pairing 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 not mFree[m]:
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 = preferences[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 not wPrefersM1OverM(preferences, w, m, m1):
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", "\t\tMan")
for i in range(N):
print(" ", i + N, "\t\t", wPartner[i])

# Driver Code
# preferences = [[3,4,5],
#                [3,5,4],
#                [5,4,3],
#                [2,1,0],
#                [2,0,1],
#                [0,1,2]]

preferences = [[5,4,3],
[5,3,4],
[3,4,5],
[0,1,2],
[0,2,1],
[2,1,0]]

stableMarriage(preferences)
```

Output:

```Woman   Man
4      2
5      1
6      3
7      0
```