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6.7.2 Floyds Algorithm for the APSP Problem
In finding all-pairs shortest path, Dijkstras algorithm may be used with every pair of sources and destinations. However, it is not the best solution there is for the APSP
problem. A more elegant approach is to use the algorithm created by Floyd.
The algorithm makes use of cost adjacency matrix representation of a graph. It has a dimension of n x n for n vertices. In this algorithm, let COST be the given adjacency
matrix. Let A be the matrix to contain the cost of the shortest path, initially equal to COST. Another n x n matrix, PATH, is needed to contain the immediate vertices along
the shortest path:
PATH i,j = 0 initially, indicates that the shortest path between i and j is the edge
i,j if it exists = k
if, including k in the path from i to j at the k
th
iteration, yields a shorter path
The algorithm is as follows: 1. Initialize A to be equal to COST:
Ai, j = COSTi, j, 1 ≤
i, j ≤
n 2. If the cost of passing through the intermediate vertex k from vertex i to vertex j will
cost less than direct access from i to j, replace Ai,j with this cost and update PATHi,j as well:
For k = 1, 2, 3, ..., n a Ai, j = minimum [ A
k-1
i, j, A
k-1
i, k + A
k-1
k, j ] , 1 ≤
i, j ≤
n b If Ai, j == A
k-1
i, k + A
k-1
k, j set PATHi, j = k For example, solve the APSP problem on the following graph using Floyds algorithm:
Figure 1.101 A Weighted Directed Graph for APSP Example Figure 1.102
Since no self-referencing is allowed, there is no need to compute for Aj, j, for 1 ≤
j ≤
n.
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Also, for the k
th
iteration, there will be no changes for the k
th
row and column in A and PATH since it will add only 0 to the current value. For example, if k = 2:
A2, 1 = minimum A2, 1, A2,2 + A2,1 since A2, 2 = 0, there will always be no change to k
th
row and column. The following shows the execution of Floyds algorithm:
1 2
3 4
1 2
3 4
1
2 ∞
12
1 2
8 ∞
7
2 3
5 10
7
3 4
∞ ∞
1
4
A PATH
For the first iteration k=1: A2, 3 = minimum A2, 3 , A2, 1 + A1, 3 = minimum∞, ∞ = ∞
A2, 4 = minimum A2, 4 , A2, 1 + A1, 4 = minimum12, 20 = 12 A3, 2 = minimum A3, 2 , A3, 1 + A1, 2 = minimum10, 7 = 7
A3, 4 = minimum A3, 4 , A3, 1 + A1, 4 = minimum7, 17 = 7 A4, 2 = minimum A4, 2 , A4, 1 + A1, 2 = minimum∞, ∞ = ∞
A4, 3 = minimum A4, 3 , A4, 1 + A1, 3 = minimum1, ∞ = 1
1 2
3 4
1 2
3 4
1
2 ∞
12
1
k = 1
2
8 ∞
7
2 3
5 7
7
3
1
4
∞ ∞
1
4
A PATH
For k=2: A1, 3 = minimum A1, 3 , A1, 2 + A2, 3 = minimum∞, ∞ = ∞
A1, 4 = minimum A1, 4 , A1, 2 + A2, 4 = minimum12, 9 = 9 A3, 1 = minimum A3, 1 , A3, 2 + A2, 1 = minimum5, 15 = 5
A3, 4 = minimum A3, 4 , A3, 2 + A2, 4 = minimum7, 12 = 7 A4, 1 = minimum A4, 1 , A4, 2 + A2, 1 = minimum∞, ∞ = ∞
A4, 3 = minimum A4, 3 , A4, 2 + A2, 3 = minimum1, ∞ = ∞
1 2
3 4
1 2
3 4
1
2 ∞
9
1
2
k = 2
2
8 ∞
7
2 3
5 7
7
3
1
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1 2
3 4
1 2
3 4
4
∞ ∞
1
4
A PATH
For k=3: A1, 2 = minimum A1, 2 , A1, 3 + A3, 2 = minimum2, ∞ = ∞
A1, 4 = minimum A1, 4 , A1, 3 + A3, 4 = minimum9, ∞ = ∞ A2, 1 = minimum A2, 1 , A2, 3 + A3, 1 = minimum8, ∞ = ∞
A2, 4 = minimum A2, 4 , A2, 3 + A3, 4 = minimum7, ∞ = ∞ A4, 1 = minimum A4, 1 , A4, 3 + A3, 1 = minimum∞, 6 = 6
A4, 2 = minimum A4, 2 , A4, 3 + A3, 2 = minimum∞, 8 = 8
1 2
3 4
1 2
3 4
1
2 ∞
9
1
2 k = 3
2
8 ∞
7
2 3
5 7
7
3
1
4
6 8
1
4
3 3
A PATH
For k=3: A1, 2 = minimum A1, 2 , A1, 4 + A4, 2 = minimum2, 17 = 2
A1, 3 = minimum A1, 3 , A1, 4 + A4, 3 = minimum∞, 10 = 10 A2, 1 = minimum A2, 1 , A2, 4 + A4, 1 = minimum8, 13 = 8
A2, 3 = minimum A2, 3 , A2, 4 + A4, 3 = minimum∞, 8 = 8 A3, 1 = minimum A3, 1 , A3, 4 + A4, 1 = minimum5, 13 = 5
A3, 2 = minimum A3, 2 , A3, 4 + A4, 2 = minimum7, 15 = 7
1 2
3 4
1 2
3 4
1
2 10
9
1
4 2
k = 4
2
8 8
7
2
4
3
5 7
7
3
1
4
6 8
1
4
3 3
A PATH
After the n
th
iteration, A contains the minimum cost while PATH contains the path of the
minimum cost. To illustrate how to use the resulting matrices, let us find the shortest path from vertex 1 to vertex 4:
A1, 4 = 9 PATH1, 4 = 2 -- Since not 0, we have to get PATH2, 4:
PATH2, 4 = 0
Therefore, the shortest path from vertex 1 to vertex 4 is 1 -- 2 -- 4 with cost 9. Even if there is a direct edge from 1 to 4 with cost 12, the algorithm returned another path.
This example shows that it is not always the direct connection that is returned in getting
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the shortest path in a weighted, directed graph.
6.8 Summary
Kategori