Finding the number of paths of given length in a undirected unweighted graph

Question

\'Length\' of a path is the number of edges in the path.

Given a source and a destination vertex, I want to find the number of paths form the s

Answer 1

So, here's a nifty graph theory trick that I remember for this one.

Make an adjacency matrix A. where A[i][j] is 1 if there is an edge between i and j, and 0 otherwise.

Then, the number of paths of length k between i and j is just the [i][j] entry of A^k.

So, to solve the problem, build A and construct A^k using matrix multiplication (the usual trick for doing exponentiation applies here). Then just look up the necessary entry.

EDIT: Well, you need to do the modular arithmetic inside the matrix multiplication to avoid overflow issues, but that's a much smaller detail.

Answer 2

Actually the [i][j] entry of A^k shows the total different "walk", not "path", in each simple graph. We can easily prove it by "mathematical induction". However, the major question is to find total different "path" in a given graph. We there are a quite bit of different algorithm to solve, but the upper band is as follow:

(n-2)(n-3)...(n-k) which "k" is the given parameter stating length of path.

Answer 3

Let me add some more content to above answers (as this is the extended problem I faced). The extended problem is

Find the number of paths of length k in a given undirected tree.

The solution is simple for the given adjacency matrix A of the graph G find out A^k-1 and A^k and then count number of the 1s in the elements above the diagonal (or below).

Let me also add the python code.

import numpy as np

def count_paths(v, n, a):
    # v: number of vertices, n: expected path length
    paths = 0    
    b = np.array(a, copy=True)

    for i in range(n-2):
        b = np.dot(b, a)

    c = np.dot(b, a)
    x = c - b

    for i in range(v):
        for j in range(i+1, v):
            if x[i][j] == 1:
                paths = paths + 1

    return paths

print count_paths(5, 2, np.array([
                np.array([0, 1, 0, 0, 0]),
                np.array([1, 0, 1, 0, 1]),
                np.array([0, 1, 0, 1, 0]),
                np.array([0, 0, 1, 0, 0]),
                np.array([0, 1, 0, 0, 0])
            ])