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How to implement Floyd-Warshall algorithm in C by Greedy Method

S Sesha Savanth

Floyd-Warshall

The Floyd-Warshall algorithm is used to find all pairs to the shortest path. This algorithm is used to find the shortest path between every pair of vertices in a given edge graph.

Let G = (V,E) be a directed graph with n vertices. Let cost be a cost adjacency matrix for G such that cost(i,i) = 0, 1<=i<=n.

Cost(i,j) = length or cost of edge (i,j), if(i,j) ∈ E(G) and cost(i,j)= ∞ if (i,j) ∉ E(G)

All-pairs shortest path problems is used to determine a matrix A such that A(i,j) is the length of a shortest path from i to j.

If k is the highest intermediate vertex between i to j path, then the i to k path will be the shortest path in graph G going through no vertex with an index greater than k-1. Similarly, the k to j path is the shortest path in graph G going through no vertex with index greater than k-1.

First, we need to find highest intermediate vertex k. Then, we need to find the two shortest paths from i to k and k to j.

Then, we will use AkA^{k}(i,j) to represent the length of the shortest path from i to j going through no vertex with an index of greater than k. This will give us:

A0A^{0}(i,j)=cost(i,j)

If it goes through the highest intermediate vertex k, then:

AkA^{k}(i,j) = Ak1A^{k-1}(i,k)+Ak1A^{k-1}(k,j)

If it does not go through the highest intermediate vertex k, then the highest intermediate vertex is:

AkA^{k}(i,j) = Ak1A^{k-1}(i,j)

To get a recurrence for AkA^{k}(i,j), we need to combine:

AkA^{k}(i,j) =min{ Ak1A^{k-1}(i,j), Ak1A^{k-1}(i,k)+Ak1A^{k-1}(k,j)}, where k>=1

Example:

Note: For a selected intermediate vertex, the path that belongs to that vertex remains the same.

A01234008521302203410
This is for A0 Matrix

By taking the above matrix we can get the A1A^{1} matrix:

A1A^{1}(2,3) = min{A0A^{0}(2,3),A0A^{0}(2,1)+A0A^{0}(1,3)}

  • A1A^{1}(2,3) = min{2,8+∞} = 2

A1A^{1}(2,4) = min{A0A^{0}(2,4),A0A^{0}(2,1)+A0A^{0}(1,4)}

  • A1A^{1}(2,4) = min{∞,8+7} = 15

A1A^{1}(3,2) = min{A0A^{0}(3,2),A0A^{0}(3,1)+A0A^{0}(1,2)}

  • A1A^{1}(3,2) = min{∞,5+3} = 8

A1A^{1}(3,4) = min{A0A^{0}(4,3),A0A^{0}(3,1)+A0A^{0}(1,4)}

  • A1A^{1}(3,4) = min{1,5+7} = 1

A1A^{1}(4,2) = min{A0A^{0}(4,2),A0A^{0}(4,1)+A0A^{0}(1,2)}

  • A1A^{1}(4,2) = min{∞,2+3} = 2

A1A^{1}(4,3) = min{A0A^{0}(4,3),A0A^{0}(4,1)+A0A^{0}(1,3)}

  • A1A^{1}(4,3) = min{∞,2+∞} = 2
A112341085223085320441510
This is for A1 Matrix

Similarly

A21234108522308535207441510
This is for A2 Matrix
A3123410752230853520744310.
This is for A3 Matrix
A4123410532230653520744310
This is for A4 Matrix

Algorithm

for(k=0;k<n;k++)
{
  for(i=0;i<n;i++)
  {
    for(j=0;j<n;j++)
    {
      if(a[i][j]>a[i][k]+a[k][j])
      {
        a[i][j] = a[i][k]+a[k[j];
      }
    }
  }
}
#include<stdio.h>
void floyd(int a[4][4], int n)
{
	for(int k=0;k<n;k++)
	{
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<n;j++)
			{
				if(a[i][j]>a[i][k]+a[k][j])
				{
					a[i][j]=a[i][k]+a[k][j];
				}
		    }
	    }
	}
	printf("All Pairs Shortest Path is :\n");
		for(int i=0;i<n;i++)
	    {
	    	for(int j=0;j<n;j++)
	    	{
	    		printf("%d ",a[i][j]);
			}
	    	printf("\n");
		}
}
int main()
{
	int cost[4][4] = {{0, 3, 999, 4}, {8, 0, 2, 999}, {5, 999, 0, 1}, {2, 999, 999, 0}}; 
	int n = 4;

	floyd(cost,n);
}

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