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All vertex pairs connected with exactly k edges in a graph

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  • Difficulty Level : Medium
  • Last Updated : 16 Jun, 2022

Given a directed graph represented as an adjacency matrix and an integer ‘k’, the task is to find all the vertex pairs that are connected with exactly ‘k’ edges. 
Also, find the number of ways in which the two vertices can be linked in exactly k edges.

Examples :  

Input : k = 3 and graph :
0 1 0 0 0 
0 0 1 0 0 
0 0 0 1 1 
1 0 0 0 0 
0 0 1 0 0 
Output :
1 -> 4 in 1 way(s)
1 -> 5 in 1 way(s)
2 -> 1 in 1 way(s)
2 -> 3 in 1 way(s)
3 -> 2 in 1 way(s)
3 -> 4 in 1 way(s)
3 -> 5 in 1 way(s)
4 -> 3 in 1 way(s)
5 -> 1 in 1 way(s)
5 -> 3 in 1 way(s)

Input : k = 2 and graph :
0 0 0 
1 0 1 
0 1 0 
Output :
2 -> 2 in 1 way(s)
3 -> 1 in 1 way(s)
3 -> 3 in 1 way(s)

Approach :  

  • We will multiply the adjacency matrix with itself ‘k’ number of times.
  • In the resultant matrix, res[i][j] will be the number of ways in which vertex ‘j’ can be reached from vertex ‘i’ covering exactly ‘k’ edges.

Below is the implementation of the above approach :  

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to multiply two square matrices
vector<vector<int>> multiplyMatrices(
  vector<vector<int>> arr1,
  vector<vector<int>> arr2)
{
  int order = arr1.size();
  vector<vector<int>> ans(order, vector<int>(order));
   
  for(int i = 0; i < order; i++)
  {
    for(int j = 0; j < order; j++)
    {
      for(int k = 0; k < order; k++)
      {
        ans[i][j] += arr1[i][k] * arr2[k][j];
      }
    }
  }
  return ans;
}
 
// Function to find all the pairs that
// can be connected with exactly 'k' edges
void solve(vector<vector<int>> arr, int k)
{
  vector<vector<int>> res(
    arr.size(), vector<int>(arr[0].size()));
   
  // Copying arr to res,
  // which is the result for k=1
  for(int i = 0; i < res.size(); i++)
    for(int j = 0; j < res.size(); j++)
      res[i][j] = arr[i][j];
 
  // Multiplying arr with itself
  // the required number of times
  for(int i = 2; i <= k; i++)
    res = multiplyMatrices(res, arr);
 
  for(int i = 0; i < res.size(); i++)
    for(int j = 0; j < res.size(); j++)
       
      // If there is a path between 'i'
      // and 'j' in exactly 'k' edges
      if (res[i][j] > 0)
      {
        cout << i << " -> " << j
             << " in " << res[i][j]
             << " way(s)" << endl;
      }
}
 
// Driver code
int main(int argc, char const *argv[])
{
  vector<vector<int>> arr(5, vector<int>(5));
  arr[0][1] = 1;
  arr[1][2] = 1;
  arr[2][3] = 1;
  arr[2][4] = 1;
  arr[3][0] = 1;
  arr[4][2] = 1;
  int k = 3;
   
  solve(arr, k);
}
 
// This code is contributed by sanjeev2552

Java




// Java implementation of the approach
public class KPaths {
 
    // Function to multiply two square matrices
    static int[][] multiplyMatrices(int[][] arr1, int[][] arr2)
    {
        int order = arr1.length;
        int[][] ans = new int[order][order];
        for (int i = 0; i < order; i++) {
            for (int j = 0; j < order; j++) {
                for (int k = 0; k < order; k++) {
                    ans[i][j] += arr1[i][k] * arr2[k][j];
                }
            }
        }
        return ans;
    }
 
    // Function to find all the pairs that
    // can be connected with exactly 'k' edges
    static void solve(int[][] arr, int k)
    {
        int[][] res = new int[arr.length][arr[0].length];
 
        // copying arr to res,
        // which is the result for k=1
        for (int i = 0; i < res.length; i++)
            for (int j = 0; j < res.length; j++)
                res[i][j] = arr[i][j];
 
        // multiplying arr with itself
        // the required number of times
        for (int i = 2; i <= k; i++)
            res = multiplyMatrices(res, arr);
 
        for (int i = 0; i < res.length; i++)
            for (int j = 0; j < res.length; j++)
 
                // if there is a path between 'i'
                // and 'j' in exactly 'k' edges
                if (res[i][j] > 0)
                    System.out.println(i + " -> " + j + " in " + res[i][j] + " way(s)");
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int[][] arr = new int[5][5];
        arr[0][1] = 1;
        arr[1][2] = 1;
        arr[2][3] = 1;
        arr[2][4] = 1;
        arr[3][0] = 1;
        arr[4][2] = 1;
        int k = 3;
        solve(arr, k);
    }
}

Python3




# Python3 implementation of the approach
 
# Function to multiply two square matrices
def multiplyMatrices(arr1, arr2):
 
    order = len(arr1)
    ans = [[0 for i in range(order)] for j in range(order)]
 
    for i in range(order):
        for j in range(order):
            for k in range(order):
                ans[i][j] += arr1[i][k] * arr2[k][j]
 
    return ans
 
 
# Function to find all the pairs that
# can be connected with exactly 'k' edges
def solve(arr, k):
    res = [[0 for i in range(len(arr))] for j in range(len(arr))]
 
    # Copying arr to res,
    # which is the result for k=1
    for i in range(len(arr)):
        for j in range(len(arr)):
            res[i][j] = arr[i][j]
 
    # Multiplying arr with itself
    # the required number of times
    for i in range(2, k+1):
        res = multiplyMatrices(res, arr)
 
    for i in range(len(arr)):
        for j in range(len(arr)):
 
            # If there is a path between 'i'
            # and 'j' in exactly 'k' edges
            if (res[i][j] > 0):
                print(i, "->", j, "in", res[i][j], "way(s)")
 
 
# Driver Code
if __name__ == "__main__":
 
    arr = [1, 2, 3, 4]
    arr = [[0 for i in range(5)] for j in range(5)]
    arr[0][1] = 1
    arr[1][2] = 1
    arr[2][3] = 1
    arr[2][4] = 1
    arr[3][0] = 1
    arr[4][2] = 1
    k = 3
 
    solve(arr, k)
     
# This code is contributed by kirtishsurangalikar

C#




// C# implementation of the approach
using System;
 
class KPaths
{
 
// Function to multiply two square matrices
static int[,] multiplyMatrices(int[,] arr1,
                               int[,] arr2)
{
    int order = arr1.GetLength(0);
    int[,] ans = new int[order, order];
    for (int i = 0; i < order; i++)
    {
        for (int j = 0; j < order; j++)
        {
            for (int k = 0; k < order; k++)
            {
                ans[i, j] += arr1[i, k] *
                             arr2[k, j];
            }
        }
    }
    return ans;
}
 
// Function to find all the pairs that
// can be connected with exactly 'k' edges
static void solve(int[,] arr, int k)
{
    int[,] res = new int[arr.GetLength(0),
                         arr.GetLength(1)];
 
    // copying arr to res,
    // which is the result for k = 1
    for (int i = 0; i < res.GetLength(0); i++)
        for (int j = 0; j < res.GetLength(1); j++)
            res[i, j] = arr[i, j];
 
    // multiplying arr with itself
    // the required number of times
    for (int i = 2; i <= k; i++)
        res = multiplyMatrices(res, arr);
 
    for (int i = 0; i < res.GetLength(0); i++)
        for (int j = 0; j < res.GetLength(1); j++)
 
            // if there is a path between 'i'
            // and 'j' in exactly 'k' edges
            if (res[i,j] > 0)
                Console.WriteLine(i + " -> " + j + " in " +
                                    res[i, j] + " way(s)");
}
 
// Driver code
public static void Main(String[] args)
{
    int[,] arr = new int[5, 5];
    arr[0, 1] = 1;
    arr[1, 2] = 1;
    arr[2, 3] = 1;
    arr[2, 4] = 1;
    arr[3, 0] = 1;
    arr[4, 2] = 1;
    int k = 3;
    solve(arr, k);
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// javascript implementation of the approach
 
// Function to multiply two square matrices
function multiplyMatrices(arr1, arr2)
{
    var order = arr1.length;
    var ans = Array(order).fill(0).map(x => Array(order).fill(0));
    for (var i = 0; i < order; i++) {
        for (var j = 0; j < order; j++) {
            for (var k = 0; k < order; k++) {
                ans[i][j] += arr1[i][k] * arr2[k][j];
            }
        }
    }
    return ans;
}
 
// Function to find all the pairs that
// can be connected with exactly 'k' edges
function solve(arr , k)
{
    var res = Array(arr.length).fill(0).map(x => Array(arr[0].length).fill(0));
 
    // copying arr to res,
    // which is the result for k=1
    for (var i = 0; i < res.length; i++)
        for (var j = 0; j < res.length; j++)
            res[i][j] = arr[i][j];
 
    // multiplying arr with itself
    // the required number of times
    for (var i = 2; i <= k; i++)
        res = multiplyMatrices(res, arr);
 
    for (var i = 0; i < res.length; i++)
        for (var j = 0; j < res.length; j++)
 
            // if there is a path between 'i'
            // and 'j' in exactly 'k' edges
            if (res[i][j] > 0)
                document.write(i + " -> " + j + " in " + res[i][j] + " way(s)<br>");
}
 
// Driver code
var arr = Array(5).fill(0).map(x => Array(5).fill(0));
arr[0][1] = 1;
arr[1][2] = 1;
arr[2][3] = 1;
arr[2][4] = 1;
arr[3][0] = 1;
arr[4][2] = 1;
var k = 3;
solve(arr, k);
 
// This code is contributed by shikhasingrajput
</script>

Output

0 -> 3 in 1 way(s)
0 -> 4 in 1 way(s)
1 -> 0 in 1 way(s)
1 -> 2 in 1 way(s)
2 -> 1 in 1 way(s)
2 -> 3 in 1 way(s)
2 -> 4 in 1 way(s)
3 -> 2 in 1 way(s)
4 -> 0 in 1 way(s)
4 -> 2 in 1 way(s)

The time complexity of the above code can be reduced for large values of k by using matrix exponentiation. The complexity can be changed from O(n^3 * k) to O(n^3 * log k) 

C++




#include <iostream>
#include <vector>
using namespace std;
 
// Function to multiply two square matrices
static vector<vector<int> >
multiplyMatrices(vector<vector<int> > &arr1,
                 vector<vector<int> > &arr2)
{
    int order = arr1.size();
    vector<vector<int> > ans(order, vector<int>(order));
 
    for (int i = 0; i < order; i++) {
        for (int j = 0; j < order; j++) {
            for (int k = 0; k < order; k++) {
                ans[i][j] += arr1[i][k] * arr2[k][j];
            }
        }
    }
    return ans;
}
 
vector<vector<int> > identity(int n)
{
    vector<vector<int> > r(n, vector<int>(n));
    for (int i = 0; i < n; i++)
        r[i][i] = 1;
 
    return r;
}
vector<vector<int> > power(vector<vector<int> >& x, int y,
                           int n)
{
    vector<vector<int> > res = identity(n);
    while (y > 0) {
 
        if ((y & 1) == 1)
            res = multiplyMatrices(res, x);
 
        // y must be even now
        // y = y / 2
        y = y >> 1;
        x = multiplyMatrices(x, x);
    }
    return res;
}
 
// Function to find all the pairs that
// can be connected with exactly 'k' edges
void solve(vector<vector<int> > &arr, int k)
{
    vector<vector<int> > res(arr.size(),
                             vector<int>(arr[0].size()));
    res = power(arr, k, arr[0].size());
    for (int i = 0; i < res.size(); i++) {
        for (int j = 0; j < res.size(); j++) {
            // if there is a path between 'i' and 'j' in
            // exactly 'k' edges
            if (res[i][j] > 0) {
                cout << i << " -> " << j << " in "
                     << res[i][j] << " way(s)" << endl;
            }
        }
    }
}
 
int main()
{
    vector<vector<int> > arr(5, vector<int>(5));
    arr[0][1] = 1;
    arr[1][2] = 1;
    arr[2][3] = 1;
    arr[2][4] = 1;
    arr[3][0] = 1;
    arr[4][2] = 1;
    int k = 3;
 
    solve(arr, k);
    return 0;
}
 
// This code is contributed by Tapesh (tapeshdua420)

Java




class KPaths {
 
    // Function to multiply two square matrices
    static int[][] multiplyMatrices(int[][] arr1, int[][] arr2)
    {
        int order = arr1.length;
        int[][] ans = new int[order][order];
        for (int i = 0; i < order; i++) {
            for (int j = 0; j < order; j++) {
                for (int k = 0; k < order; k++) {
                    ans[i][j] += arr1[i][k] * arr2[k][j];
                }
            }
        }
        return ans;
    }
 
    // Function to find all the pairs that
    // can be connected with exactly 'k' edges
    static void solve(int[][] arr, int k)
    {
        int[][] res = new int[arr.length][arr[0].length];
 
        res = power(arr, k, arr[0].length);
        for (int i = 0; i < res.length; i++)
            for (int j = 0; j < res.length; j++)
 
                // if there is a path between 'i'
                // and 'j' in exactly 'k' edges
                if (res[i][j] > 0)
                    System.out.println(i + " -> " + j + " in " + res[i][j] + " way(s)");
    }
 
    static int[][] power(int x[][], int y, int n)
    {
        // MATRIX EXPONENTIATION
        // Initialize result
        int res[][] = identity(n);
 
        while (y > 0) {
 
            if ((y & 1) == 1)
                res = multiplyMatrices(res, x);
 
            // y must be even now
            // y = y / 2
            y = y >> 1;
            x = multiplyMatrices(x, x);
        }
        return res;
    }
    static int[][] identity(int n)
    {
        // returns identity matrix of order n
        int r[][] = new int[n][n];
        for (int i = 0; i < n; i++)
            r[i][i] = 1;
 
        return r;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int[][] arr = new int[5][5];
        arr[0][1] = 1;
        arr[1][2] = 1;
        arr[2][3] = 1;
        arr[2][4] = 1;
        arr[3][0] = 1;
        arr[4][2] = 1;
        int k = 3;
        solve(arr, k);
    }
}

Python3




# Function to multiply two square matrices
def multiplyMatrices(arr1, arr2):
    order = len(arr1)
    ans = [[0 for i in range(order)] for j in range(order)]
    for i in range(order):
        for j in range(order):
            for k in range(order):
                ans[i][j] += arr1[i][k] * arr2[k][j]
    return ans
 
 
def identity(n):
    r = [[0 for i in range(n)] for j in range(n)]
    for i in range(n):
        r[i][i] = 1
    return r
 
 
def power(x, y, n):
    res = identity(n)
    while y > 0:
        if ((y & 1) == 1):
            res = multiplyMatrices(res, x)
             
        # y must be even now
        y = y >> 1
        x = multiplyMatrices(x, x)
    return res
 
# Function to find all the pairs that
# can be connected with exactly 'k' edges
 
 
def solve(arr, k):
    res = [[0 for i in range(len(arr))] for j in range(len(arr))]
    res = power(arr, k, len(arr[0]))
    for i in range(len(res)):
        for j in range(len(res)):
            # if there is a path between 'i' and 'j' in
            # exactly 'k' edges
            if res[i][j] > 0:
                print("{} -> {} in {} way(s)".format(i, j, res[i][j]))
 
 
if __name__ == "__main__":
    arr = [[0 for i in range(5)] for j in range(5)]
    arr[0][1] = 1
    arr[1][2] = 1
    arr[2][3] = 1
    arr[2][4] = 1
    arr[3][0] = 1
    arr[4][2] = 1
    k = 3
 
    solve(arr, k)
     
# This code is contributed by Tapesh (tapeshdua420)

C#




// C# implementation of the above approach:
using System;
 
class KPaths
{
 
    // Function to multiply two square matrices
    static int[,] multiplyMatrices(int[,] arr1,
                                   int[,] arr2)
    {
        int order = arr1.GetLength(0);
        int[,] ans = new int[order,order];
        for (int i = 0; i < order; i++)
        {
            for (int j = 0; j < order; j++)
            {
                for (int k = 0; k < order; k++)
                {
                    ans[i, j] += arr1[i, k] *
                                 arr2[k, j];
                }
            }
        }
        return ans;
    }
 
    // Function to find all the pairs that
    // can be connected with exactly 'k' edges
    static void solve(int[,] arr, int k)
    {
        int[,] res = new int[arr.GetLength(0),
                             arr.GetLength(1)];
 
        res = power(arr, k, arr.GetLength(0));
        for (int i = 0; i < res.GetLength(0); i++)
            for (int j = 0; j < res.GetLength(1); j++)
 
                // if there is a path between 'i'
                // and 'j' in exactly 'k' edges
                if (res[i, j] > 0)
                    Console.WriteLine(i + " -> " + j + 
                      " in " + res[i, j] + " way(s)");
    }
 
    static int[,] power(int [,]x, int y, int n)
    {
         
        // MATRIX EXPONENTIATION
        // Initialize result
        int [,]res = identity(n);
 
        while (y > 0)
        {
 
            if ((y & 1) == 1)
                res = multiplyMatrices(res, x);
 
            // y must be even now
            // y = y / 2
            y = y >> 1;
            x = multiplyMatrices(x, x);
        }
        return res;
    }
     
    static int[,] identity(int n)
    {
        // returns identity matrix of order n
        int [,]r = new int[n, n];
        for (int i = 0; i < n; i++)
            r[i, i] = 1;
 
        return r;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int[,] arr = new int[5, 5];
        arr[0, 1] = 1;
        arr[1, 2] = 1;
        arr[2, 3] = 1;
        arr[2, 4] = 1;
        arr[3, 0] = 1;
        arr[4, 2] = 1;
        int k = 3;
        solve(arr, k);
    }
}
 
// This code is contributed by PrinciRaj1992

Output

0 -> 3 in 1 way(s)
0 -> 4 in 1 way(s)
1 -> 0 in 1 way(s)
1 -> 2 in 1 way(s)
2 -> 1 in 1 way(s)
2 -> 3 in 1 way(s)
2 -> 4 in 1 way(s)
3 -> 2 in 1 way(s)
4 -> 0 in 1 way(s)
4 -> 2 in 1 way(s)

Complexity Analysis: 

Time Complexity: O(n3 * logk)

Space Complexity: O(n2)


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