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Floyd's Algorithm For The All-Pairs Shortest-Path Problem
Floyd's Algorithm for the All-Pairs Shortest-Path Problem:
Â
           The Floyd's algorithm is invented by R. Floyd. This algorithm is applied to both directed and undirected weighted graphs provided that they do not contain a cycle of a negative length.
The Floyd's algorithm is also used to find the shortest path in order to find the distances.
Â
Algorithm Floyd (W[1..n,1...n])
//The input given is a weighted matrix W of a graph with no negative length cycle
//The outputs will a distance matrix of the shortest paths lengths
for k ß1 to n do
           for i ß 1 to n do
                       for j ß 1 to n do
                                   D[i, j] ß min { D[i, j], D[i, k]+D[k, j]}
return D
Â
Consider the following matrix.
Â
           a         b         c         d
a         0         âˆÂ        3         âˆ
b         2         0         âˆÂ        âˆ
c         âˆÂ        7         0         1
d         6         âˆÂ        âˆÂ        0
Â
Steps to solve the Floyds algorithm.
Note: The iteration should start with ‘j' for example k=1;i=1;j=1; and continue till k=1;i=4;j=4 and the iteration should end when all the three variables reaches iteration as 4, That is k=4;i=4;j=4.
We need to take only the minimum value as per the algorithm. We need to find D(1) matrix till D(4) matrix.
Â
Step 1:
                       D[i, j]  D[i, k]+D[k, j]     min { D[i, j], D[i, k]+D[k, j]}  Result
K=1; i=1; j=1Â Â 0Â Â Â Â Â Â Â Â Â 0+0 = 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0
K=1; i=1; j=2  âˆÂ        0+∠= âˆÂ                     âˆÂ                                            âˆ
K=1; i=1; j=3Â Â 3Â Â Â Â Â Â Â Â Â 0+3 = 3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 3
K=1; i=1;j=4   âˆÂ        0+∠= âˆÂ                     âˆÂ                                            âˆ
K=1; i=2; j=1  2         2+∠= ∠                    2                                             2
K=1; i=2;j=2   0         2+∠= âˆÂ                     0                                             0
K=1; i=2; j=3  âˆÂ        2+3 =  5                     5                                             5
K=1; i=2; j=4  âˆÂ        2+∠= âˆÂ                     âˆÂ                                            âˆ
K=1; i=3; j=1  âˆÂ        âˆ+0 = âˆÂ                     âˆÂ                                            âˆ
K=1; i=3; j=2  7         âˆ+∠= âˆÂ                    7                                             7
K=1; i=3; j=3  0         âˆ+3 = âˆÂ                     0                                             0
K=1; i=3; j=4  1         âˆ+∠= âˆÂ                    1                                             1
K=1; i=4; j=1Â Â 6Â Â Â Â Â Â Â Â Â 6+0 = 6Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 6Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 6
K=1; i=4; j=2  âˆÂ        6+∠= âˆÂ                     âˆÂ                                            âˆ
K=1; i=4; j=3  âˆÂ        6+3 = 9                       9                                             9
K=1; i=4; j=4  0         6+∠= âˆÂ                     0                                             0
Â
So, the matrix will be
          Â
D(1) =        a         b         c         d
a         0         âˆÂ        3         âˆ
b         2         0         5         âˆ
c         âˆÂ        7         0         1
d         6         âˆÂ        9         0
Now the position of (2, 3) is changed from ∠to 5 and (4, 3) is changed from ∠to 9.
Â
Step 2:
                       D[i, j]  D[i, k]+D[k, j]     min { D[i, j], D[i, k]+D[k, j]}  Result
K=2; i=1; j=1  0         âˆ+2 = âˆÂ                     0                                             0
K=2; i=1; j=2  âˆÂ        âˆ+0 = âˆÂ                     âˆÂ                                            âˆ
K=2; i=1; j=3  3         âˆ+5 = âˆÂ                     3                                             3
K=2; i=1;j=4   âˆÂ        âˆ+∠= âˆÂ                    âˆÂ                                            âˆ
K=2; i=2; j=1Â Â 2Â Â Â Â Â Â Â Â Â 0+2 = 2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2
K=2; i=2;j=2Â Â Â 0Â Â Â Â Â Â Â Â Â 0+0 = 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0
K=2; i=2; j=3 Â 5Â Â Â Â Â Â Â Â Â 0+5 =Â Â 5Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 5Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 5
K=2; i=2; j=4  âˆÂ        0+∠= âˆÂ                     âˆÂ                                            âˆ
K=2; i=3; j=1  âˆÂ        7+2 = 9                       9                                             9
K=2; i=3; j=2Â Â 7Â Â Â Â Â Â Â Â Â 7+0 = 7Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 7Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 7
K=2; i=3; j=3 Â 0Â Â Â Â Â Â Â Â Â 7+5 = 12Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0
K=2; i=3; j=4  âˆÂ        7+∠= âˆÂ                     1                                             âˆ
K=2; i=4; j=1  6         âˆ+∠= âˆÂ                    6                                             6
K=2; i=4; j=2  âˆÂ        âˆ+0 = âˆÂ                     âˆÂ                                            âˆ
K=2; i=4; j=3  9         âˆ+5= âˆÂ                      9                                             9
K=2; i=4; j=4  0         âˆ+∠= âˆÂ                    0                                             0
Â
          Â
Now,
D(2) =                           a         b         c         d
a         0         âˆÂ        3         âˆ
b         2         0         5         âˆ
c         9         7         0         1
d         6         âˆÂ        9         0
In this matrix the position of (3,1) is changed from ∠to 9.
Â
Like wise if we perform the iteration in third step.
In third step the iteration of K is set to 3 and performs the previous steps.
K=3;i=4;j=4
Â
The resultant matrix
Â
D(3) =                           a         b         c         d
a         0         10      3         4
b         2         0         5         6
c         9         7         0         1
d         6         16      9         0
The resultant matrix shows that the positions were infinity (âˆ) is changed to the minimum cost.
The final matrix will be, when you perform K=4; i=4; and j=4.
Â
D(4) =                           a         b         c         d
a         0         10      3         4
b         2         0         5         6
c         7         7         0         1
d         6         16      9         0
This is the steps for solving the Floyds algorithm
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About the Author
Karthick.R, Lecturer
Karpagam Institute of Technology,
Coimbatore-21
ph:9944641416











































