加权有向图的邻接矩阵

Question

A) Suppose A be a adjacency matrix of weighted directed graph G with n vertex in which A[i,j] be a weight of edge i to j . if there is no such edge A[i ,i]=0 . Matrix A^K= A*A*A*...A . if we use + instead of * and use min instead of +, Slot A^k [i,j] not describe weight of path i to j with at most k edge. I want to find this problem show what things?

B) Suppose A be a adjacency matrix of weighted directed graph (wihout loop and multiple edge) G with n vertex in which A[i,j] be a weight of edge i to j . if there is no such edge A[i ,j]=infinity , and for evrey i we have A[i, i]=0 . Matrix A^K= A*A*A*...A . Slot A^k [i,j] Show what things? min weight? or...?

any idea?

Edit: i means these algorithm find which in graph? find maximum weight? min weight? find nothing?

Answer 1

Let B = A^K

B[i, j] would represent shortest path from i .. j that takes exactly k steps. How ?

Given a Matrix A what happens if we multiple A with itself ?

    // Initialize a matrix result which would be the matrix obtained by A*A
    vector< N, vector<int> (N, INF) > result;
    REP(i,0,N) REP(j,0,N) REP(k,0,N)
        result[i][j] = min(result[i][j], (A[i][k] + A[k][j]));

Initially A[i, j] had the direct cost of going to j from i . As you can see that result[i, j] is minimum of result[i, j] and A[i, k] + A[k, j] , So we conclude that result[i, j] contains path going from i .. j using one intermediate vertex k . And since we iterate this for all k's , result[i, j] contains the minimum cost path that traverses exactly two edges. We can now generalize for A^k .

What if we set A[i, i] to 0 ?

This allows self loops. In the above code when k == i ,

    result[i, j] = min(result[i,j], A[i, i] + A[i, j])
    result[i, j] = min(result[i,j], A[i,j)] // since A[i, i] = 0;

This means that now result[i, j] would denote the shortest path to reach from i .. j if we can use either one or two edges.

So Now B[i, j] would represent shortest path from i .. j that takes almost k steps.

Similarly, We can use the same concept to calculate no of paths from i .. j that take exactly k steps. What you can do is that initialize A[i, j] as 1 if we have an edge between i and j else 0 .

A^k would give you the desired result.