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Floyd Warshall Algorithm

Any shortest path from i to j either never touches vertex k, or it does — and if it does, it is exactly the shortest i-to-k path followed by the shortest k-to-j path, both already known once k has had its turn. So looping k from 0 to n-1 and, for every pair, checking whether routing through k beats the current best, builds the true all-pairs answer in place — no extra storage, no per-source restart. The badge on each vertex below is the current shortest distance in the row being narrated; the highlighted arrows show a two-hop improvement actually just found. Animated on: 4 vertices, directed weighted edges 0->1(5), 0->3(10), 1->2(3), 2->3(1) — find the shortest distance between every pair of vertices..

All-pairs shortest paths: give every vertex a turn as the waypoint

time O(V^3)space O(V^2)step 1 / 11

Row 0 initialized directly from its edges

line 7

Before any vertex is allowed as a waypoint, dist[i][j] is just the direct edge weight, 0 on the diagonal, or infinity where no edge exists. Row 0 starts at [0, 5, inf, 10] — only the two edges leaving vertex 0 are known yet.

Pseudocode
1FUNCTION floydWarshall(n, edges):
2 FOR i FROM 0 TO n - 1
3 FOR j FROM 0 TO n - 1
4 dist[i][j] <- INFINITY
5 dist[i][i] <- 0
6 FOR EACH (a, b, w) IN edges
7 dist[a][b] <- w
8 FOR k FROM 0 TO n - 1
9 FOR i FROM 0 TO n - 1
10 FOR j FROM 0 TO n - 1
11 IF dist[i][k] + dist[k][j] < dist[i][j]
12 dist[i][j] <- dist[i][k] + dist[k][j]
13 RETURN dist

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