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Asked: 2026-06-04T23:42:55+00:00

I have a connected directed weighted graph. The edge weights represent probabilities of moving

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I have a connected directed weighted graph. The edge weights represent probabilities of moving between vertices; weights for all edges emanating from a vertex sum up to one. The graph contains two sinks: A and B. For each vertex in the graph, I want to know the probability that a walk originating there will reach A and the same for B. What kind of problem is this? How do I solve it?

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  1. Editorial Team

    This problem is of the algebra kind. For a path starting at a vertex, the probability of reaching A is the average of probabilities of reaching A from each of its neighbouring vertices, weighted by the edge weights. Let’s put this into more concrete terms.

    Let P be the adjacency matrix for the graph. That is, Pi,j is the probability of moving from vertex i to vertex j. Set PA,A = 1. If we take a vector of probabilities assigned to each vertex and multiply it by P, then the resulting vector contains a weighted average of each vertex’s neighbours. What we are looking for is a vector v, such that P v = v and vA = 1.

    This vector v is the eigenvector of P corresponding to the eigenvalue of 1. Does P always have such an eigenvalue? Fortunately, the Perron-Frobenius theorem tells us that it does, and that this is the largest eigenvalue of P. The solution is then to form the adjacency matrix P and find the eigenvector corresponding to its largest eigenvalue.

    There is also an approximate solution. If we take a vector x of vertex probabilities, with xA = 1, and the other elements set to 0, then Pk x will converge to v as k goes to infinity. Pk might be easier to compute for small values of k than the eigenvector.

    Example

    Let’s look at the following simple graph:

    diagram

    If we order the vertices alphabetically, then the matrix P corresponding to the graph is:

    enter image description here

    This matrix has an eigenvalue equal to 1, and the corresponding eigenvector is: [1 0 70/79 49/79]. That is, the exact probability of reaching A from C is 70/79, and from D it is 49/79. If you work out the answer for B, it comes out to 9/79 and 30/79, which is exactly what we expect.

    The value of P16 [1 0 0 0] is approximately [1 0 0.886 0.62] and is correct to 6 decimal places.

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