From 15409f222e2312be7b2fdfca0329a50896cb3bce Mon Sep 17 00:00:00 2001
From: Francesco Franco <130352141+Francesco601@users.noreply.github.com>
Date: Tue, 30 May 2023 14:25:25 +0200
Subject: [PATCH] add Floyd -Warshall algorithm to PR

---
 floyd_warshall.py | 101 ++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 101 insertions(+)
 create mode 100644 floyd_warshall.py

diff --git a/floyd_warshall.py b/floyd_warshall.py
new file mode 100644
index 00000000..8f516cae
--- /dev/null
+++ b/floyd_warshall.py
@@ -0,0 +1,101 @@
+# Python implementation  of Floyd Warshall Algorithm for solving "all pairs"
+# shortest path problems in a graph. The algorithm finds the shortest path
+# in a weighted digraph with positive or negative edge weights (but no negative
+# cycles). Although it does not return details of the paths themselves, it
+# is possible to reconstruct the paths with simple modifications of to the algorithm.
+# Versions of the program can also be used for finding  the "transitive closure" of
+# a relation R, or widest paths between all pairs of vertices in a weighted graph.
+# Floyd-Warshall is an example of dynamic programming. The time complexity is
+# O(V^3) even though there may be O(V^2) edges in the graph. 
+
+# Number of vertices in the graph
+V = 4
+
+# Define infinity to be a large
+# enough integer value. This value will be
+# used for vertices not connected to each other
+INF = 99999
+
+# Solves all pairs 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 = list(map(lambda i: list(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("The 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"), end=" ")
+			else:
+				print("%5d\t" % (dist[i][j]), end=' ')
+			if j == V-1:
+				print()
+
+
+# Driver's code
+if __name__ == "__main__":
+	"""
+				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]
+			]
+	# Function call
+	floydWarshall(graph)
+