# 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 adjacency matrix representation where the 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 walks from 0 to 3 with exactly 2 edges. The walks are {0, 2, 3} and {0, 1, 3}

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Simple Approach: Create a recursive function that takes the current vertex, destination vertex, and the count of the vertex. Call the recursive function with all adjacent vertices of a current vertex with the value of k as k-1. When the value of k is 0, then check whether the current vertex is the destination or not. If destination, then the output answer is 1.

The following is the implementation of this simple solution.

## C++

 // C++ program to count walks from u to// v with exactly k edges#include 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 edgesint 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 functionint 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 edgesimport 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

## Python3

 # Python3 program to count walks from# u to v with exactly k edges # Number of vertices in the graphV = 4 # A naive recursive function to count# walks from u to v with k edgesdef countwalks(graph, u, v, k):     # Base cases    if (k == 0 and u == v):        return 1    if (k == 1 and graph[u][v]):        return 1    if (k <= 0):        return 0         # Initialize result    count = 0         # Go to all adjacents of u and recur    for i in range(0, V):                 # Check if is adjacent of u        if (graph[u][i] == 1):             count += countwalks(graph, i, v, k-1)         return count # Driver Code # Let us create the graph shown in above diagramgraph = [[0, 1, 1, 1, ],         [0, 0, 0, 1, ],         [0, 0, 0, 1, ],         [0, 0, 0, 0] ] u = 0; v = 3; k = 2print(countwalks(graph, u, v, k)) # This code is contributed by Smitha Dinesh Semwal.

## C#

 // C# program to count walks from u to// v with exactly k edgesusing System; class GFG {     // Number of vertices    static int V = 4;     // A naive recursive function to    // count walks from u to v with    // k edges    static 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++)             // Check if is adjacent of u            if (graph[u, i] == 1)                count += countwalks(graph, i, v, k - 1);         return count;    }     // Driver method    public static void Main()    {         /* 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;         Console.Write(            countwalks(graph, u, v, k));    }} // This code is contributed by nitin mittal.

## PHP



## Javascript



Output
2

Complexity Analysis:

• Time Complexity: O(Vk).
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 a recursion tree. The worst occurs for a complete graph. In the worst case, every internal node of the recursion tree would have exactly n children.
• Auxiliary Space: O(V).
To store the stack space and the visited array O(V) space is needed.

Efficient Approach: The solution can be optimized using Dynamic Programming. The idea is to build a 3D table where the first dimension is the source, the second dimension is the destination, the third dimension is the number of edges from source to destination, and the value is the count of walks. Like others, Dynamic Programming problems, fill the 3D table in a bottom-up manner.

## C++

 // C++ program to count walks from// u to v with exactly k edges#include 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 edgesint 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 the                // 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 functionint 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 edgesimport 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

## Python3

 # Python3 program to count walks from# u to v with exactly k edges # Number of verticesV = 4 # A Dynamic programming based function# to count walks from u to v with k edges  def countwalks(graph, u, v, 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    count = [[[0 for k in range(k + 1)]              for i in range(V)]             for j in range(V)]     # Loop for number of edges from 0 to k    for e in range(0, k + 1):        # For Source        for i in range(V):            # For Destination            for j in range(V):                # Initialize value                # count[i][j][e] = 0                 # From base cases                if (e == 0 and i == j):                    count[i][j][e] = 1                if (e == 1 and 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 a in range(V):                         # Adjacent of i                        if (graph[i][a] != 0):                            count[i][j][e] += count[a][j][e - 1]     return count[u][v][k]  # Driver codeif __name__ == '__main__':     # Let us create the graph shown    # in above diagram    graph = [[0, 1, 1, 1],             [0, 0, 0, 1],             [0, 0, 0, 1],             [0, 0, 0, 0]]     u = 0    v = 3    k = 2     print(countwalks(graph, u, v, k)) # This code is contributed by Rajput-Ji

## C#

 // C# program to count walks from u to v// with exactly k edgesusing System; class GFG {    static int V = 4; // Number of vertices     // A Dynamic programming based function    // to count walks from u to v with k edges    static 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 source            for (int i = 0; i < V; i++) {                 // for destination                for (int j = 0; j < V; j++) {                    // 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) {                        // adjacent of i                        for (int a = 0; a < V; a++)                            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()    {        /* Let us create the graph shown in         above diagram*/        int[, ] graph = { { 0, 1, 1, 1 },                          { 0, 0, 0, 1 },                          { 0, 0, 0, 1 },                          { 0, 0, 0, 0 } };        int u = 0, v = 3, k = 2;         Console.WriteLine(            countwalks(graph, u, v, k));    }} // This is Code Contributed by anuj_67.

## Javascript



## PHP

 1) {                    for (\$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/* Let us create the graph shown in above diagram*/\$graph = array(array(0, 1, 1, 1),                array(0, 0, 0, 1),                array(0, 0, 0, 1),                array(0, 0, 0, 0)); \$u = 0;\$v = 3;\$k = 2;echo countwalks(\$graph, \$u, \$v, \$k); // This code is contributed by rajsanghavi9.?>

Output
2

Complexity Analysis:

• Time Complexity: O(V3).
Three nested loops are needed to fill the DP table, so the time complexity is O(V3).
• Auxiliary Space: O(V2K).
To store the DP table O(V2K) space is needed.

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 the power 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.

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