Optimize Floyd-Warshall for a symmetric adjacency matrix

Question

Is there an optimization that reduces the constant Floyd-Warshall runtime if you have a symmetric adjacency matrix guaranteed?

Answer 1

After some thought, I came up with:

for (int k = 0; k < N; ++k)
    for (int i = 0; i < N; ++i)
        for (int j = 0; j <= i; ++j)
            dist[j][i] = dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);

Now, of course, we need to show it correctly and faster.

Correctness is more difficult to prove, since it is based on the Floyd-Warsall proof, which is nontrivial. Here's a pretty good proof: Floyd-Warsall proof

symmetric. -, , 2 . , , .

dist[i][j][k] i j, {0, ..., k} i j.

dist[i][j][k-1] i j. , .

, :

dist[i][j][k] = min(dist[i][j][k-1], dist[i][k][k-1] + dist[k][j][k-1])

dist[i][k][k] ( dist[k][i][k]):

dist[i][k][k] = min(dist[i][k][k-1], dist[i][k][k-1] + dist[k][k][k-1])

, dist[k][k][k-1] ( ), , dist[i][k][k] = dist[i][k][k-1]. dist[k][k][k-1] = 0, , min().

, dist[i][k][k] = dist[i][k][k-1] dist[i][j][k] , dist[i][k] dist[k][j] k . dist[i][j][k-1] dist[i][j][k], dist[i][j] dist[i][j][k-1] , dist[i][j][k]. i j k, .

, .

, dist[i][j] = dist[j][i] .

, dist[a][b] = dist[b][a] a b.

dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])
           = min(dist[j][i], dist[k][i] + dist[j][k])
           = min(dist[j][i], dist[j][k] + dist[k][i])
           = dist[j][i]

, dist[a][b] = dist[b][a]. dist[i][j] = dist[j][i]

, .

. , , . , , , min() - , .

- , , , .

EDIT:

, :

for (int k = 0; k < N; ++k) {
    for (int i = 0; i < k; ++i)
        for (int j = 0; j <= i; ++j)
            dist[i][j] = min(dist[i][j], dist[i][k] + dist[j][k]);
    for (int i = k; i < N; ++i) {
        for (int j = 0; j < k; ++j)
            dist[i][j] = min(dist[i][j], dist[k][i] + dist[j][k]);
        for (int j = k; j <= i; ++j)
            dist[i][j] = min(dist[i][j], dist[k][i] + dist[k][j]);
    }
}

, - , , , .

.

Answer 2

( ) ( ), edgeCost , . , .

( d (i, j) = d (j, i)), , , cache, .