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Possible walks from a source to a destination with exactly k edges
A directed graph is given. Another two vertices u and v are also given, u is the starting vertex, and v is the ending vertex. Our task is to find a number of walks from vertex u to v with exactly k edges. The value of k is also provided in the algorithm.
By using dynamic programming, we need to create a 3D table, Where the row will point the values of u, columns will point the values v and depth will be used to track the number of edges from start to end.
Input and Output
Input: The adjacency matrix of the graph: The destination vertex is 3. K = 2 0 1 1 1 0 0 0 1 0 0 0 1 0 0 0 0 Output: There are 2 Possible Walks, from 0 to 3 with 2 edges.
Algorithm
numberOdWalks(u, v, k)
Input: The start vertex u, the end vertex v, the number of edges k.
Output: Number of possible walks with k edges.
Begin define 3D array count of order (n x n x k+1) //n is number of vertices for edge in range 0 to k, do for i in range 0 to n-1, do for j in range 0 to n-1, do count[i, j, edge] := 0 if edge = 0 and i = j, then count[i, j, edge] := 1 if edge = 1 and (i, j) is connected, then count[i, j, edge] := 1 if edge > 1, then for a in range 0 to n, and adjacent with i do count[i, j, edge] := count[i, j, edge] + count[a, j, edge - 1] done done done done return count[u, v, k] End
Example
#include <iostream>
#define NODE 7
using namespace std;
int graph[NODE][NODE] = {
{0, 1, 1, 1},
{0, 0, 0, 1},
{0, 0, 0, 1},
{0, 0, 0, 0}
};
int numberOfWalks(int u, int v, int k) {
int count[NODE][NODE][k+1];
for (int edge = 0; edge <= k; edge++) { //for k edges (0..k)
for (int i = 0; i < NODE; i++) {
for (int j = 0; j < NODE; j++) {
count[i][j][edge] = 0; //initially all values are 0
if (edge == 0 && i == j) //when e is 0 and ith and jth node are same, only one path
count[i][j][edge] = 1;
if (edge == 1 && graph[i][j]) //when e is 0 and direct path from (i to j), one path
count[i][j][edge] = 1;
if (edge >1) { //for more than one edges
for (int a = 0; a < NODE; a++) // adjacent of source i
if (graph[i][a])
count[i][j][edge] += count[a][j][edge-1];
}
}
}
}
return count[u][v][k];
}
int main() {
int u = 0, v = 3, k = 2;
cout << "There are "<< numberOfWalks(u, v, k)<<" Possible Walks, from ";
cout <<u<<" to "<<v<<" with "<<k<<" edges.";
}Output
There are 2 Possible Walks, from 0 to 3 with 2 edges.
Updated on: 17-Jun-2020
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