A matching in a Bipartite Graph is a set of the edges chosen in such a way that no two edges share an endpoint. A maximum matching is a matching of maximum size (maximum number of edges). In a maximum matching, if any edge is added to it, it is no longer a matching. There can be more than one maximum matchings for a given Bipartite Graph.

**Why do we care?**

There are many real world problems that can be formed as Bipartite Matching. For example, consider the following problem:

*There are M job applicants and N jobs. Each applicant has a subset of jobs that he/she is interested in. Each job opening can only accept one applicant and a job applicant can be appointed for only one job. Find an assignment of jobs to applicants in such that as many applicants as possible get jobs.*

We strongly recommend to read the following post first.

Ford-Fulkerson Algorithm for Maximum Flow Problem

**Maximum Bipartite Matching and Max Flow Problem**

**M**aximum **B**ipartite **M**atching (**MBP**) problem can be solved by converting it into a flow network (See this video to know how did we arrive this conclusion). Following are the steps.

**1) Build a Flow Network**

There must be a source and sink in a flow network. So we add a source and add edges from source to all applicants. Similarly, add edges from all jobs to sink. The capacity of every edge is marked as 1 unit.

2) Find the maximum flow.

We use Ford-Fulkerson algorithm to find the maximum flow in the flow network built in step 1. The maximum flow is actually the MBP we are looking for.

**How to implement the above approach?**

Let us first define input and output forms. Input is in the form of Edmonds matrix which is a 2D array ‘bpGraph[M][N]’ with M rows (for M job applicants) and N columns (for N jobs). The value bpGraph[i][j] is 1 if i’th applicant is interested in j’th job, otherwise 0.

Output is number maximum number of people that can get jobs.

A simple way to implement this is to create a matrix that represents adjacency matrix representation of a directed graph with M+N+2 vertices. Call the fordFulkerson() for the matrix. This implementation requires O((M+N)*(M+N)) extra space.

Extra space can be be reduced and code can be simplified using the fact that the graph is bipartite and capacity of every edge is either 0 or 1. The idea is to use DFS traversal to find a job for an applicant (similar to augmenting path in Ford-Fulkerson). We call bpm() for every applicant, bpm() is the DFS based function that tries all possibilities to assign a job to the applicant.

In bpm(), we one by one try all jobs that an applicant ‘u’ is interested in until we find a job, or all jobs are tried without luck. For every job we try, we do following.

If a job is not assigned to anybody, we simply assign it to the applicant and return true. If a job is assigned to somebody else say x, then we recursively check whether x can be assigned some other job. To make sure that x doesn’t get the same job again, we mark the job ‘v’ as seen before we make recursive call for x. If x can get other job, we change the applicant for job ‘v’ and return true. We use an array maxR[0..N-1] that stores the applicants assigned to different jobs.

If bmp() returns true, then it means that there is an augmenting path in flow network and 1 unit of flow is added to the result in maxBPM().

## C++

// A C++ program to find maximal // Bipartite matching. #include <iostream> #include <string.h> using namespace std; // M is number of applicants // and N is number of jobs #define M 6 #define N 6 // A DFS based recursive function // that returns true if a matching // for vertex u is possible bool bpm(bool bpGraph[M][N], int u, bool seen[], int matchR[]) { // Try every job one by one for (int v = 0; v < N; v++) { // If applicant u is interested in // job v and v is not visited if (bpGraph[u][v] && !seen[v]) { // Mark v as visited seen[v] = true; // If job 'v' is not assigned to an // applicant OR previously assigned // applicant for job v (which is matchR[v]) // has an alternate job available. // Since v is marked as visited in // the above line, matchR[v] in the following // recursive call will not get job 'v' again if (matchR[v] < 0 || bpm(bpGraph, matchR[v], seen, matchR)) { matchR[v] = u; return true; } } } return false; } // Returns maximum number // of matching from M to N int maxBPM(bool bpGraph[M][N]) { // An array to keep track of the // applicants assigned to jobs. // The value of matchR[i] is the // applicant number assigned to job i, // the value -1 indicates nobody is // assigned. int matchR[N]; // Initially all jobs are available memset(matchR, -1, sizeof(matchR)); // Count of jobs assigned to applicants int result = 0; for (int u = 0; u < M; u++) { // Mark all jobs as not seen // for next applicant. bool seen[N]; memset(seen, 0, sizeof(seen)); // Find if the applicant 'u' can get a job if (bpm(bpGraph, u, seen, matchR)) result++; } return result; } // Driver Code int main() { // Let us create a bpGraph // shown in the above example bool bpGraph[M][N] = {{0, 1, 1, 0, 0, 0}, {1, 0, 0, 1, 0, 0}, {0, 0, 1, 0, 0, 0}, {0, 0, 1, 1, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1}}; cout << "Maximum number of applicants that can get job is " << maxBPM(bpGraph); return 0; }

## Java

// A Java program to find maximal // Bipartite matching. import java.util.*; import java.lang.*; import java.io.*; class GFG { // M is number of applicants // and N is number of jobs static final int M = 6; static final int N = 6; // A DFS based recursive function that // returns true if a matching for // vertex u is possible boolean bpm(boolean bpGraph[][], int u, boolean seen[], int matchR[]) { // Try every job one by one for (int v = 0; v < N; v++) { // If applicant u is interested // in job v and v is not visited if (bpGraph[u][v] && !seen[v]) { // Mark v as visited seen[v] = true; // If job 'v' is not assigned to // an applicant OR previously // assigned applicant for job v (which // is matchR[v]) has an alternate job available. // Since v is marked as visited in the // above line, matchR[v] in the following // recursive call will not get job 'v' again if (matchR[v] < 0 || bpm(bpGraph, matchR[v], seen, matchR)) { matchR[v] = u; return true; } } } return false; } // Returns maximum number // of matching from M to N int maxBPM(boolean bpGraph[][]) { // An array to keep track of the // applicants assigned to jobs. // The value of matchR[i] is the // applicant number assigned to job i, // the value -1 indicates nobody is assigned. int matchR[] = new int[N]; // Initially all jobs are available for(int i = 0; i < N; ++i) matchR[i] = -1; // Count of jobs assigned to applicants int result = 0; for (int u = 0; u < M; u++) { // Mark all jobs as not seen // for next applicant. boolean seen[] =new boolean[N] ; for(int i = 0; i < N; ++i) seen[i] = false; // Find if the applicant 'u' can get a job if (bpm(bpGraph, u, seen, matchR)) result++; } return result; } // Driver Code public static void main (String[] args) throws java.lang.Exception { // Let us create a bpGraph shown // in the above example boolean bpGraph[][] = new boolean[][]{ {false, true, true, false, false, false}, {true, false, false, true, false, false}, {false, false, true, false, false, false}, {false, false, true, true, false, false}, {false, false, false, false, false, false}, {false, false, false, false, false, true}}; GFG m = new GFG(); System.out.println( "Maximum number of applicants that can"+ " get job is "+m.maxBPM(bpGraph)); } }

## Python

# Python program to find # maximal Bipartite matching. class GFG: def __init__(self,graph): # residual graph self.graph = graph self.ppl = len(graph) self.jobs = len(graph[0]) # A DFS based recursive function # that returns true if a matching # for vertex u is possible def bpm(self, u, matchR, seen): # Try every job one by one for v in range(self.jobs): # If applicant u is interested # in job v and v is not seen if self.graph[u][v] and seen[v] == False: # Mark v as visited seen[v] = True '''If job 'v' is not assigned to an applicant OR previously assigned applicant for job v (which is matchR[v]) has an alternate job available. Since v is marked as visited in the above line, matchR[v] in the following recursive call will not get job 'v' again''' if matchR[v] == -1 or self.bpm(matchR[v], matchR, seen): matchR[v] = u return True return False # Returns maximum number of matching def maxBPM(self): '''An array to keep track of the applicants assigned to jobs. The value of matchR[i] is the applicant number assigned to job i, the value -1 indicates nobody is assigned.''' matchR = [-1] * self.jobs # Count of jobs assigned to applicants result = 0 for i in range(self.ppl): # Mark all jobs as not seen for next applicant. seen = [False] * self.jobs # Find if the applicant 'u' can get a job if self.bpm(i, matchR, seen): result += 1 return result bpGraph =[[0, 1, 1, 0, 0, 0], [1, 0, 0, 1, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 1, 1, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1]] g = GFG(bpGraph) print ("Maximum number of applicants that can get job is %d " % g.maxBPM()) # This code is contributed by Neelam Yadav

## C#

// A C# program to find maximal // Bipartite matching. using System; class GFG { // M is number of applicants // and N is number of jobs static int M = 6; static int N = 6; // A DFS based recursive function // that returns true if a matching // for vertex u is possible bool bpm(bool [,]bpGraph, int u, bool []seen, int []matchR) { // Try every job one by one for (int v = 0; v < N; v++) { // If applicant u is interested // in job v and v is not visited if (bpGraph[u, v] && !seen[v]) { // Mark v as visited seen[v] = true; // If job 'v' is not assigned to // an applicant OR previously assigned // applicant for job v (which is matchR[v]) // has an alternate job available. // Since v is marked as visited in the above // line, matchR[v] in the following recursive // call will not get job 'v' again if (matchR[v] < 0 || bpm(bpGraph, matchR[v], seen, matchR)) { matchR[v] = u; return true; } } } return false; } // Returns maximum number of // matching from M to N int maxBPM(bool [,]bpGraph) { // An array to keep track of the // applicants assigned to jobs. // The value of matchR[i] is the // applicant number assigned to job i, // the value -1 indicates nobody is assigned. int []matchR = new int[N]; // Initially all jobs are available for(int i = 0; i < N; ++i) matchR[i] = -1; // Count of jobs assigned to applicants int result = 0; for (int u = 0; u < M; u++) { // Mark all jobs as not // seen for next applicant. bool []seen = new bool[N] ; for(int i = 0; i < N; ++i) seen[i] = false; // Find if the applicant // 'u' can get a job if (bpm(bpGraph, u, seen, matchR)) result++; } return result; } // Driver Code public static void Main () { // Let us create a bpGraph shown // in the above example bool [,]bpGraph = new bool[,] {{false, true, true, false, false, false}, {true, false, false, true, false, false}, {false, false, true, false, false, false}, {false, false, true, true, false, false}, {false, false, false, false, false, false}, {false, false, false, false, false, true}}; GFG m = new GFG(); Console.Write( "Maximum number of applicants that can"+ " get job is "+m.maxBPM(bpGraph)); } } //This code is contributed by nitin mittal.

**Output :**

Maximum number of applicants that can get job is 5

You may like to see below also:

Hopcroft–Karp Algorithm for Maximum Matching | Set 1 (Introduction)

Hopcroft–Karp Algorithm for Maximum Matching | Set 2 (Implementation)

**References:**

http://www.cs.cornell.edu/~wdtseng/icpc/notes/graph_part5.pdf

http://www.youtube.com/watch?v=NlQqmEXuiC8

http://en.wikipedia.org/wiki/Maximum_matching

http://www.stanford.edu/class/cs97si/08-network-flow-problems.pdf

http://www.cs.princeton.edu/courses/archive/spring13/cos423/lectures/07NetworkFlowII-2×2.pdf

http://www.ise.ncsu.edu/fangroup/or766.dir/or766_ch7.pdf

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