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floyd_warshall.py
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# Python Program for 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[][] will be the output matrix that will finally
have the shortest distances between every pair of vertices """
""" initializing the solution matrix same as input graph matrix
OR we can say that the initial values of shortest distances
are based on shortest paths considering no
intermediate vertices """
dist = map(lambda i : map(lambda j : j , i) , graph)
""" Add all vertices one by one to the set of intermediate
vertices.
---> Before start of an iteration, we have shortest distances
between all pairs of vertices such that the shortest
distances consider only the vertices in the set
{0, 1, 2, .. k-1} as intermediate vertices.
----> After the end of a iteration, vertex no. k is
added to the set of intermediate vertices and the
set becomes {0, 1, 2, .. k}
"""
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);