Floyd-Warshall Algorithm: Difference between revisions

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Line 18: Line 18:
    
    
                             │ 0 if i == j
                             │ 0 if i == j
  for all i,j ∈ V, A[i,j,0] = │ c<sub>ij</sub> if (i, j)∈E
  for all i,j ∈ V, A[i,j,0] = │ c<sub>ij</sub> if (i, j) ∈ E
                             │ +∞ if i≠j and (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])
 
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Revision as of 23:29, 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 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.

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])