Floyd-Warshall Algorithm: Difference between revisions

From NovaOrdis Knowledge Base
Jump to navigation Jump to search
Line 10: Line 10:
* [[Johnson's Algorithm]]
* [[Johnson's Algorithm]]
=Overview=
=Overview=
The Floyd-Warshall algorithm computes all-pairs shortest path in a directed graph with arbitrary edge length, include negative lengths. The only case that does not make sense is whether the graph has negative edge cycles. In this case there is no shortest path, because the cycle can be followed an infinite number of times, leading to negative infinity length paths. If this is the case, the Floyd-Warshall will report the situation and stop.
The Floyd-Warshall algorithm computes all-pairs shortest path in a directed graph with arbitrary edge length, include negative lengths. The only case that does not make sense is whether the graph has negative cost cycles. In this case there is no shortest path, because the cycle can be followed an infinite number of times, leading to negative infinity length paths. If this is the case, the Floyd-Warshall will report the situation and stop.


The running time is O(n<sup>3</sup>).
The running time is O(n<sup>3</sup>).

Revision as of 23:30, 24 November 2021

External

Internal

Overview

The Floyd-Warshall algorithm computes all-pairs shortest path in a directed graph with arbitrary edge length, include negative lengths. The only case that does not make sense is whether the graph has negative cost cycles. In this case there is no shortest path, because the cycle can be followed an infinite number of times, leading to negative infinity length paths. If this is the case, the Floyd-Warshall will report the situation and stop.

The running time is O(n3).

Algorithm

We assume that the vertices are labeled from 1 to n.

Let A = 3D array, indexed by i, j, k
# Initialize the tridimensional matrix with the trivial cases
 
                            │ 0 if i == j
for all i,j ∈ V, A[i,j,0] = │ cij if (i, j) ∈ E
                            │ +∞ if i≠j and (i,j) ∉ E

for k = 1 to n:
  for i = 1 to n:
    for j = 1 to n:
      A[i,j,k] = max(A[i,j,k-1], A[i,k,k-1] + A[k,j,k-1])
 

If the graph has a negative cost cycle.