Floyd Warshall Algorithm Given a graph in adjacency matrix from. Find shortest distances between every pair of vertices in a given edge weighted directed Graph. If there is no path from a source to a destination, print -1 at that place. Input Format: First line - n, the number of nodes Then there are n lines each containing n integers, weight from source(i.e. row number) to destination(i.e. column number). Output Format: Print shortest distances between every pair of vertices. If there is no path from a source to a destination, print -1 at that place. Sample Input: 4 0 5 INF 10 INF 0 3 INF INF INF 0 1 INF INF INF 0 Sample Output: 0 5 8 9 INF 0 3 4 INF INF 0 1 INF INF INF 0 give a pyhon code

Here is a Python code for the Floyd Warshall Algorithm:

# Number of vertices in the graph
V = 4

# Define infinity as the large enough value. This value will be
# used for vertices not connected to each other
INF = 99999

# Solves all pair shortest path via Floyd Warshall Algorithm
def floydWarshall(graph):

    dist = list(map(lambda i : list(map(lambda j : j , i)) , graph) )

    for k in range(V):

        # pick all vertices as source one by one
        for i in range(V):

            # Pick all vertices as destination for the
            # above picked source
            for j in range(V):

                # If vertex k is on the shortest path from
                # i to j, then update the value of dist[i][j]
                dist[i][j] = min(dist[i][j] , dist[i][k]+ dist[k][j])

    printSolution(dist)

# A utility function to print the solution
def printSolution(dist):
    print("Following matrix shows the shortest distances between every pair of vertices")
    for i in range(V):
        for j in range(V):
            if(dist[i][j] == INF):
                print("%7s" %("INF")),
            else:
                print("%7d\t" %(dist[i][j])),
            if j == V-1:
                print("")

# Driver program to test the above program
# Let us create the following weighted graph
"""
            10
       (0)------->(3)
        |         /|\
      5 |          |
        |          | 1
       \|/         |
       (1)------->(2)
            3
"""
graph = [[0,5,INF,10],
         [INF,0,3,INF],
         [INF,INF,0,1],
         [INF,INF,INF,0]
        ]
# Print the solution
floydWarshall(graph)

This code first initializes the solution matrix same as the input graph matrix. Then it updates the solution matrix by considering all vertices as an intermediate vertex. The idea is to one by one pick all vertices and update all shortest paths which include the picked vertex as an intermediate vertex in the shortest path. When we pick vertex number k as an intermediate vertex, we already have considered vertices {0, 1, 2, .. k-1} as intermediate vertices. For every pair (i, j) of the source and destination vertices respectively, there are two possible cases:

1. k is not an intermediate vertex in shortest path from i to j. We keep the value of dist[i][j] as it is.
2. k is an intermediate vertex in shortest path from i to j. We update the value of dist[i][j] to dist[i][k] + dist[k][j] if dist[i][j] > dist[i][k] + dist[k][j].