Transitive closure of a graph using Floyd Warshall Algorithm

Last Updated : 23 Jul, 2025

Given a directed graph, determine if a vertex j is reachable from another vertex i for all vertex pairs (i, j) in the given graph. Here reachable means that there is a path from vertex i to j. The reach-ability matrix is called the transitive closure of a graph.

Example:

Input: Graph = [[0, 1, 1, 0], [0, 0, 1, 0], [1, 0, 0, 1], [0, 0, 0, 0]]
Output:
1 1 1 1
1 1 1 1
1 1 1 1
0 0 0 1

Approach:

The idea is to use the Floyd-Warshall algorithm to find the transitive closure of a directed graph. We start by initializing a result matrix with the original adjacency matrix and set all diagonal elements to 1 (since every vertex can reach itself). Then for each intermediate vertex k, we check if vertex i can reach vertex j either directly or through vertex k. If a path exists from i to k and from k to j, then we mark that i can reach j in our result matrix.

Step by step approach:

Initialize result matrix with input graph and set all diagonal elements to 1.
For each intermediate vertex, check all possible vertex pairs.
If vertex i can reach intermediate vertex and intermediate vertex can reach vertex j, mark that i can reach j. This iteratively builds paths of increasing length through intermediate vertices.
Return the final matrix where a value of 1 indicates vertex i can reach vertex j.

C++

// C++ program to find Transitive closure of
// a graph using Floyd Warshall Algorithm
#include <bits/stdc++.h>
using namespace std;

vector<vector<int>> transitiveClosure(vector<vector<int>> graph) {
    int n = graph.size();
    
    vector<vector<int>> ans(n, vector<int>(n, 0));
    
    // Copy the graph into resultant matrix
    for (int i=0; i<n; i++) {
      for (int j=0; j<n; j++) {
        ans[i][j] = graph[i][j];
      }
    }
    
    // Transtive closure of (i, i) will always be 1 
    for (int i=0; i<n; i++) ans[i][i] = 1;
    
    // Apply floyd Warshall Algorithm
    // For each intermediate node k
    for (int k=0; k<n; k++) {
        for (int i=0; i<n; i++) {
            for (int j=0; j<n; j++) {
              
                // Check if a path exists between i to k and 
                // between k to j.
                if (ans[i][k]==1 && ans[k][j]==1) {
                    ans[i][j] = 1;
                }
            }
        }
    }
    
    return ans;
}

int main() {
    int n = 4;
    vector<vector<int>> graph = 
    {{0, 1, 1, 0}, {0, 0, 1, 0}, {1, 0, 0, 1}, {0, 0, 0, 0}};
    vector<vector<int>> ans = transitiveClosure(graph);
    
    for (int i=0; i<n; i++) {
      for (int j=0; j<n; j++) {
        cout << ans[i][j] << " ";
      }
      cout << endl;
    }

    return 0;
}

Java

// Java program to find Transitive closure of
// a graph using Floyd Warshall Algorithm
import java.util.*;

class GfG {
    
    static ArrayList<ArrayList<Integer>> transitiveClosure(int[][] graph) {
        int n = graph.length;
        ArrayList<ArrayList<Integer>> ans = new ArrayList<>();
        
        // Copy the graph into resultant matrix
        for (int i = 0; i < n; i++) {
            ArrayList<Integer> row = new ArrayList<>();
            for (int j = 0; j < n; j++) {
                row.add(graph[i][j]);
            }
            ans.add(row);
        }

        // Transtive closure of (i, i) will always be 1 
        for (int i = 0; i < n; i++) ans.get(i).set(i, 1);

        // Apply floyd Warshall Algorithm
        // For each intermediate node k
        for (int k = 0; k < n; k++) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {

                    // Check if a path exists between i to k and 
                    // between k to j.
                    if (ans.get(i).get(k) == 1 && ans.get(k).get(j) == 1) {
                        ans.get(i).set(j, 1);
                    }
                }
            }
        }

        return ans;
    }

    public static void main(String[] args) {
        int n = 4;
        int[][] graph = {{0, 1, 1, 0}, {0, 0, 1, 0}, {1, 0, 0, 1}, {0, 0, 0, 0}};
        ArrayList<ArrayList<Integer>> ans = transitiveClosure(graph);

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                System.out.print(ans.get(i).get(j) + " ");
            }
            System.out.println();
        }
    }
}

Python

# Python program to find Transitive closure of
# a graph using Floyd Warshall Algorithm

# Copy the graph into resultant matrix
def transitiveClosure(graph):
    n = len(graph)
    ans = [[graph[i][j] for j in range(n)] for i in range(n)]

    # Transtive closure of (i, i) will always be 1 
    for i in range(n):
        ans[i][i] = 1

    # Apply floyd Warshall Algorithm
    # For each intermediate node k
    for k in range(n):
        for i in range(n):
            for j in range(n):

                # Check if a path exists between i to k and 
                # between k to j.
                if ans[i][k] == 1 and ans[k][j] == 1:
                    ans[i][j] = 1

    return ans

if __name__ == "__main__":
    n = 4
    graph = [[0, 1, 1, 0], [0, 0, 1, 0], [1, 0, 0, 1], [0, 0, 0, 0]]
    ans = transitiveClosure(graph)

    for i in range(n):
        for j in range(n):
            print(ans[i][j], end=" ")
        print()

// C# program to find Transitive closure of
// a graph using Floyd Warshall Algorithm
using System;
using System.Collections.Generic;

class GfG {

    static List<List<int>> transitiveClosure(int[,] graph) {
        int n = graph.GetLength(0);
        List<List<int>> ans = new List<List<int>>();
        
        // Copy the graph into resultant matrix
        for (int i = 0; i < n; i++) {
            List<int> row = new List<int>();
            for (int j = 0; j < n; j++) {
                row.Add(graph[i, j]);
            }
            ans.Add(row);
        }

        // Transtive closure of (i, i) will always be 1 
        for (int i = 0; i < n; i++) ans[i][i] = 1;

        // Apply floyd Warshall Algorithm
        // For each intermediate node k
        for (int k = 0; k < n; k++) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {

                    // Check if a path exists between i to k and 
                    // between k to j.
                    if (ans[i][k] == 1 && ans[k][j] == 1) {
                        ans[i][j] = 1;
                    }
                }
            }
        }

        return ans;
    }

    static void Main() {
        int n = 4;
        int[,] graph = new int[,] {
            {0, 1, 1, 0},
            {0, 0, 1, 0},
            {1, 0, 0, 1},
            {0, 0, 0, 0}
        };
        List<List<int>> ans = transitiveClosure(graph);

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                Console.Write(ans[i][j] + " ");
            }
            Console.WriteLine();
        }
    }
}

JavaScript

// JavaScript program to find Transitive closure of
// a graph using Floyd Warshall Algorithm

function transitiveClosure(graph) {
    let n = graph.length;
    let ans = new Array(n).fill(0).map(() => new Array(n).fill(0));
    
    // Copy the graph into resultant matrix
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++) {
            ans[i][j] = graph[i][j];
        }
    }

    // Transtive closure of (i, i) will always be 1 
    for (let i = 0; i < n; i++) ans[i][i] = 1;

    // Apply floyd Warshall Algorithm
    // For each intermediate node k
    for (let k = 0; k < n; k++) {
        for (let i = 0; i < n; i++) {
            for (let j = 0; j < n; j++) {

                // Check if a path exists between i to k and 
                // between k to j.
                if (ans[i][k] === 1 && ans[k][j] === 1) {
                    ans[i][j] = 1;
                }
            }
        }
    }

    return ans;
}

let n = 4;
let graph = [[0, 1, 1, 0], [0, 0, 1, 0], [1, 0, 0, 1], [0, 0, 0, 0]];
let ans = transitiveClosure(graph);

for (let i = 0; i < n; i++) {
    let row = "";
    for (let j = 0; j < n; j++) {
        row += ans[i][j] + " ";
    }
    console.log(row);
}