Complete Implementation of Kruskal's

I have finished the implementation completly and tested it on
the example graph given in wikipedia, with 10 vertex's and 21 edges.
The image of the graph and MST is also being added to the Repo, along
with the text file of the I/O during testing.

Signed-off-by: Aditya Prasad <[email protected]>

Author: anantham

Kruskal's.py

@@ -6,6 +6,9 @@
 # entering too many nodes and edges.The adjacency matrix is a good implementation
 # for a graph when the number of edges is large.So i wont be using that here
 
+# for sorting purpose
+import operator
+
 # Vertex, which will represent each vertex in the graph.Each Vertex uses a dictionary
 # to keep track of the vertices to which it is connected, and the weight of each edge. 
 class Vertex:
@@ -130,6 +133,8 @@ def __iter__(self):
 
 # step 1 : Take all edges and sort them
 
+'''
+not required as of now
 #  Time Complexity of Solution:
 #  Best Case O(n+k); Average Case O(n+k); Worst Case O(n+k),
 #  where n is the size of the input array and k means the
@@ -151,6 +156,7 @@ def counting_sort(weights,max_weight):
             ndx += 1
             # reset the counter back to the set of zero's
             counter[i] -= 1
+'''
 
 # now we have a optimal sorting function in hand, lets sort the list of edges.
 # a dictionary with weights of an edge and the vertexes involved in that edge.
@@ -159,18 +165,17 @@ def counting_sort(weights,max_weight):
 for ver1 in the_graph:
     # take every vertex ver1 is connected to = ver2
     for ver2 in ver1.getConnections():
-        # make the dictionary with the weights and the 2 vertex's involved with the edge (thier key)
-        vrwght[ver1.connectedTo[ver2]]=[ver1.getId(),ver2.getId()]
+        # make the dictionary with the weights and the 2 vertex's involved with the
+        # edge (thier key) use the pair of vertex's id as the key to avoid uniqueness
+        # problems in the dictionary, mutliple edges might have the SAME weight 
+        vrwght[ver1.getId(),ver2.getId()]=[ver1.connectedTo[ver2]]
 
 print "\nThe edges with thier unsorted weights are"
 print vrwght
 
-temp_weights=vrwght.keys()
-counting_sort(temp_weights,max_weight)
 
-sorted_weights={}
-for weight in temp_weights:
-    sorted_weights[weight]=vrwght[weight]
+
+sorted_weights=sorted(vrwght.items(), key=operator.itemgetter(1))
 
 print "\nAfter sorting"
 print sorted_weights
@@ -265,28 +270,44 @@ def union(self, *objects):
     # execute FIND for all the vertex's in the_graph
     X[the_graph.getVertex(vertex_key)]
 
+
+
+
+for i in range(len(sorted_weights)):
+    if(X[the_graph.getVertex(sorted_weights[i][0][0])]==X[the_graph.getVertex(sorted_weights[i][0][1])]):
+        pass
+    else:
+        MST[sorted_weights[i][0]]=sorted_weights[i][1]
+        X.union(the_graph.getVertex(sorted_weights[i][0][0]),the_graph.getVertex(sorted_weights[i][0][1]))
+        
+'''
 # now the UNION.
-for weight in sorted_weights:
+for vertex_pair in sorted_weights:
+    print vertex_pair
     # here sorted_weights[weight] gives the set of 2 vertex's involved in the that edge
-    if(X[the_graph.getVertex(sorted_weights[weight][0])]==X[the_graph.getVertex(sorted_weights[weight][1])]):
+    if(X[the_graph.getVertex(vertex_pair[0][0])]==X[the_graph.getVertex(vertex_pair[0][1])]):
         # if both vertices have the same parent (name) then they are in the same set, so ignore this edge
         pass
     else:
         # else as they belong to different sets we can ADD this edge to the MST (MST will be a subset of sorted_weights)
-        MST[weight]=sorted_weights[weight]
+        MST[vertex_pair[0]]=sorted_weights[vertex_pair[0]]
         # and merge the sets these two vertices belong to thus we call union on them.
-        X.union(the_graph.getVertex(sorted_weights[weight][0]),the_graph.getVertex(sorted_weights[weight][1]))
+        X.union(the_graph.getVertex(vertex_pair[0]),the_graph.getVertex(vertex_pair[1]))
+'''
 
 # thus we have the MST done
 
 print " \n\nIn the graph with these vertex's"
 print the_graph.getVertices()
 
-print "\n With these edges between the vertexes given above, we obtain a Minimal Spanning Tree\n"
+print "\n With these "+str(len(MST))+" edges between the vertexes given above, we obtain a Minimal Spanning Tree\n"
 print MST
 
 print "\n Please note this is a dictionary with  key as the weight of the edge and value as the key's of the two vertex's involved in this edge"
-        
+
+# I HAVE TESTED THIS IMPLEMENTATION WITH THE SAMPLE PROBLEM GIVEN IN WIKIPEDIA
+# THE IMAGE OF THE GRAPH AND THE ONLY MST IS INCLUDED IN THE REPO, ALONG WITH THE
+# COMMANDLINE I/O OF TESTING, BOTH ARE SAVED AS Kruskal_test (.jpg and .txt respectively)