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Floyd's Algorithm For The All-Pairs Shortest-Path Problem

Floyd's Algorithm for the All-Pairs Shortest-Path Problem:

 

            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.

 

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

 

Consider the following matrix.

 

            a          b          c          d

a          0          âˆÂ Â Â Â Â Â Â Â  3          âˆ

b          2          0          âˆÂ Â Â Â Â Â Â Â  âˆ

c          âˆÂ Â Â Â Â Â Â Â  7          0          1

d          6          âˆÂ Â Â Â Â Â Â Â  âˆÂ Â Â Â Â Â Â Â  0

 

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.

 

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=1   0          0+0 = 0                        0                                              0

K=1; i=1; j=2   âˆÂ Â Â Â Â Â Â Â  0+∠= âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  âˆÂ 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  âˆ

K=1; i=1; j=3   3          0+3 = 3                        3                                              3

K=1; i=1;j=4    âˆÂ Â Â Â Â Â Â Â  0+∠= âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  âˆÂ 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  âˆ

K=1; i=2; j=1   2          2+∠= ∠                     2                                              2

K=1; i=2;j=2    0          2+∠= âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  0                                              0

K=1; i=2; j=3   âˆÂ Â Â Â Â Â Â Â  2+3 =   5                      5                                              5

K=1; i=2; j=4   âˆÂ Â Â Â Â Â Â Â  2+∠= âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  âˆÂ 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  âˆ

K=1; i=3; j=1   âˆÂ Â Â Â Â Â Â Â  âˆ+0 = âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  âˆÂ 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  âˆ

K=1; i=3; j=2   7          âˆ+∠= âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  7                                              7

K=1; i=3; j=3   0          âˆ+3 = âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  0                                              0

K=1; i=3; j=4   1          âˆ+∠= âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  1                                              1

K=1; i=4; j=1   6          6+0 = 6                        6                                              6

K=1; i=4; j=2   âˆÂ Â Â Â Â Â Â Â  6+∠= âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  âˆÂ 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  âˆ

K=1; i=4; j=3   âˆÂ Â Â Â Â Â Â Â  6+3 = 9                        9                                              9

K=1; i=4; j=4   0          6+∠= âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  0                                              0

 

So, the matrix will be

           

D(1) =        a          b          c          d

a          0          âˆÂ Â Â Â Â Â Â Â  3          âˆ

b          2          0          5          âˆ

c          âˆÂ Â Â Â Â Â Â Â  7          0          1

d          6          âˆÂ Â Â Â Â Â Â Â  9          0

Now the position of (2, 3) is changed from ∠to 5 and (4, 3) is changed from ∠to 9.

 

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 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  0                                              0

K=2; i=1; j=2   âˆÂ Â Â Â Â Â Â Â  âˆ+0 = âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  âˆÂ 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  âˆ

K=2; i=1; j=3   3          âˆ+5 = âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  3                                              3

K=2; i=1;j=4    âˆÂ Â Â Â Â Â Â Â  âˆ+∠= âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  âˆÂ 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  âˆ

K=2; i=2; j=1   2          0+2 = 2                        2                                              2

K=2; i=2;j=2    0          0+0 = 0                        0                                              0

K=2; i=2; j=3   5          0+5 =   5                      5                                              5

K=2; i=2; j=4   âˆÂ Â Â Â Â Â Â Â  0+∠= âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  âˆÂ 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  âˆ

K=2; i=3; j=1   âˆÂ Â Â Â Â Â Â Â  7+2 = 9                        9                                              9

K=2; i=3; j=2   7          7+0 = 7                        7                                              7

K=2; i=3; j=3   0          7+5 = 12                      0                                              0

K=2; i=3; j=4   âˆÂ Â Â Â Â Â Â Â  7+∠= âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  1                                              âˆ

K=2; i=4; j=1   6          âˆ+∠= âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  6                                              6

K=2; i=4; j=2   âˆÂ Â Â Â Â Â Â Â  âˆ+0 = âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  âˆÂ 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  âˆ

K=2; i=4; j=3   9          âˆ+5= âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  9                                              9

K=2; i=4; j=4   0          âˆ+∠= âˆÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  0                                              0

 

           

Now,

D(2) =                              a          b          c          d

a          0          âˆÂ Â Â Â Â Â Â Â  3          âˆ

b          2          0          5          âˆ

c          9          7          0          1

d          6          âˆÂ Â Â Â Â Â Â Â  9          0

In this matrix the position of (3,1) is changed from ∠to 9.

 

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

 

The resultant matrix

 

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.

 

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

 

 

 

 

 

About the Author

Karthick.R, Lecturer

Karpagam Institute of Technology,

Coimbatore-21

ph:9944641416

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