# Count all possible walks from a source to a destination with exactly k edges

Given a directed graph and two vertices ‘u’ and ‘v’ in it, count all possible walks from ‘u’ to ‘v’ with exactly k edges on the walk.

The graph is given as adjacency matrix representation where value of graph[i][j] as 1 indicates that there is an edge from vertex i to vertex j and a value 0 indicates no edge from i to j.

For example consider the following graph. Let source ‘u’ be vertex 0, destination ‘v’ be 3 and k be 2. The output should be 2 as there are two walk from 0 to 3 with exactly 2 edges. The walks are {0, 2, 3} and {0, 1, 3}

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

A simple solution is to start from u, go to all adjacent vertices and recur for adjacent vertices with k as k-1, source as adjacent vertex and destination as v. Following is C++ implementation of this simple solution.

## C++

```// C++ program to count walks from u to v with exactly k edges
#include <iostream>
using namespace std;

// Number of vertices in the graph
#define V 4

// A naive recursive function to count walks from u to v with k edges
int countwalks(int graph[][V], int u, int v, int k)
{
// Base cases
if (k == 0 && u == v)      return 1;
if (k == 1 && graph[u][v]) return 1;
if (k <= 0)                return 0;

// Initialize result
int count = 0;

// Go to all adjacents of u and recur
for (int i = 0; i < V; i++)
if (graph[u][i] == 1)  // Check if is adjacent of u
count += countwalks(graph, i, v, k-1);

return count;
}

// driver program to test above function
int main()
{
/* Let us create the graph shown in above diagram*/
int graph[V][V] = { {0, 1, 1, 1},
{0, 0, 0, 1},
{0, 0, 0, 1},
{0, 0, 0, 0}
};
int u = 0, v = 3, k = 2;
cout << countwalks(graph, u, v, k);
return 0;
}
```

## Java

```// Java program to count walks from u to v with exactly k edges
import java.util.*;
import java.lang.*;
import java.io.*;

class KPaths
{
static final int V = 4; //Number of vertices

// A naive recursive function to count walks from u
// to v with k edges
int countwalks(int graph[][], int u, int v, int k)
{
// Base cases
if (k == 0 && u == v)           return 1;
if (k == 1 && graph[u][v] == 1) return 1;
if (k <= 0)                     return 0;

// Initialize result
int count = 0;

// Go to all adjacents of u and recur
for (int i = 0; i < V; i++)
if (graph[u][i] == 1)  // Check if is adjacent of u
count += countwalks(graph, i, v, k-1);

return count;
}

// Driver method
public static void main (String[] args) throws java.lang.Exception
{
/* Let us create the graph shown in above diagram*/
int graph[][] =new int[][] { {0, 1, 1, 1},
{0, 0, 0, 1},
{0, 0, 0, 1},
{0, 0, 0, 0}
};
int u = 0, v = 3, k = 2;
KPaths p = new KPaths();
System.out.println(p.countwalks(graph, u, v, k));
}
}//Contributed by Aakash Hasija
```

Output:
`2`

The worst case time complexity of the above function is O(Vk) where V is the number of vertices in the given graph. We can simply analyze the time complexity by drawing recursion tree. The worst occurs for a complete graph. In worst case, every internal node of recursion tree would have exactly n children.
We can optimize the above solution using Dynamic Programming. The idea is to build a 3D table where first dimension is source, second dimension is destination, third dimension is number of edges from source to destination, and the value is count of walks. Like other Dynamic Programming problems, we fill the 3D table in bottom up manner.

## C++

```// C++ program to count walks from u to v with exactly k edges
#include <iostream>
using namespace std;

// Number of vertices in the graph
#define V 4

// A Dynamic programming based function to count walks from u
// to v with k edges
int countwalks(int graph[][V], int u, int v, int k)
{
// Table to be filled up using DP. The value count[i][j][e] will
// store count of possible walks from i to j with exactly k edges
int count[V][V][k+1];

// Loop for number of edges from 0 to k
for (int e = 0; e <= k; e++)
{
for (int i = 0; i < V; i++)  // for source
{
for (int j = 0; j < V; j++) // for destination
{
// initialize value
count[i][j][e] = 0;

// from base cases
if (e == 0 && i == j)
count[i][j][e] = 1;
if (e == 1 && graph[i][j])
count[i][j][e] = 1;

// go to adjacent only when number of edges is more than 1
if (e > 1)
{
for (int a = 0; a < V; a++) // adjacent of source i
if (graph[i][a])
count[i][j][e] += count[a][j][e-1];
}
}
}
}
return count[u][v][k];
}

// driver program to test above function
int main()
{
/* Let us create the graph shown in above diagram*/
int graph[V][V] = { {0, 1, 1, 1},
{0, 0, 0, 1},
{0, 0, 0, 1},
{0, 0, 0, 0}
};
int u = 0, v = 3, k = 2;
cout << countwalks(graph, u, v, k);
return 0;
}```

## Java

```// Java program to count walks from u to v with exactly k edges
import java.util.*;
import java.lang.*;
import java.io.*;

class KPaths
{
static final int V = 4; //Number of vertices

// A Dynamic programming based function to count walks from u
// to v with k edges
int countwalks(int graph[][], int u, int v, int k)
{
// Table to be filled up using DP. The value count[i][j][e]
// will/ store count of possible walks from i to j with
// exactly k edges
int count[][][] = new int[V][V][k+1];

// Loop for number of edges from 0 to k
for (int e = 0; e <= k; e++)
{
for (int i = 0; i < V; i++)  // for source
{
for (int j = 0; j < V; j++) // for destination
{
// initialize value
count[i][j][e] = 0;

// from base cases
if (e == 0 && i == j)
count[i][j][e] = 1;
if (e == 1 && graph[i][j]!=0)
count[i][j][e] = 1;

// go to adjacent only when number of edges
// is more than 1
if (e > 1)
{
for (int a = 0; a < V; a++) // adjacent of i
if (graph[i][a]!=0)
count[i][j][e] += count[a][j][e-1];
}
}
}
}
return count[u][v][k];
}

// Driver method
public static void main (String[] args) throws java.lang.Exception
{
/* Let us create the graph shown in above diagram*/
int graph[][] =new int[][] { {0, 1, 1, 1},
{0, 0, 0, 1},
{0, 0, 0, 1},
{0, 0, 0, 0}
};
int u = 0, v = 3, k = 2;
KPaths p = new KPaths();
System.out.println(p.countwalks(graph, u, v, k));
}
}//Contributed by Aakash Hasija
```

Output:
`2`

Time complexity of the above DP based solution is O(V3K) which is much better than the naive solution.

We can also use Divide and Conquer to solve the above problem in O(V3Logk) time. The count of walks of length k from u to v is the [u][v]’th entry in (graph[V][V])k. We can calculate power of by doing O(Logk) multiplication by using the divide and conquer technique to calculate power. A multiplication between two matrices of size V x V takes O(V3) time. Therefore overall time complexity of this method is O(V3Logk).

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