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--- courses:cs211:winter2012:journals:jeanpaul:chapterfour_sectionvi [2012/03/01 02:48] – mugabej
+++ courses:cs211:winter2012:journals:jeanpaul:chapterfour_sectionvi [2012/03/01 03:54] (current) – [Implementing Kruskal's Algorithm] mugabej
@@ Line 22: / Line 22: @@
     * ** Union(A, B) ** : merges sets A and B into one set
       * Goal implementation: O(log n)
+    * ** MakeUnionFind(S) **: returns the Union-Find data structure on set S where all elements are in separate sets.
+      * Example of such a set: A connected component with no edges
+      * Goal Implementation: O(n), n = |S|
   * The sets obtained will be the connected components of the graph
   * So, for a given node u, Find(u) returns the name of the component containing u
@@ Line 27: / Line 30: @@
   * If Find(u) ≠ Find(v), then Union(Find(u), Find(v)) merges the two components into one
   * The ** Union-Find ** data structure maintains components of a graph only when we add edges, not when we delete them
-  *
+  * The "name of the set" returned by Union-Find will be arbitrary
+\\
+==== A simple Data Structure  for Union-Find ====
+>>>>>>>>>>>>>>>>> Maintain an Array ** Component ** of size n that contains the name of the set currently containing each element\\
+>>>>>>>>>>>>>>>>> Let S = {1,...,n}\\
+>>>>>>>>>>>>>>>>> Component[s] = the name of set containing s\\
+>>>>>>>>>>>>>>>>> To implement ** MakeUnionFind(S) **:\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Set up an array and initialize it to Component[s] = s for all s in S\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> With this operation, Find(v) is done in O(1) time\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> But Union(A,B) can take O(n) time with this operation(-->We have to update the values of Component[s] for all elements in A and B)\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> To improve on this bound of Union(A,B):\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Maintain the list of elements in each set\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Choose the name for the union to be the name of one of the sets, so we only update Component[s] for s in the other set\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> If the other set is big, name the union to be the name of that set and update Component[s] for s in the smaller set\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Improvements:\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Maintain an Array Size of length n\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Size[A] = the size of A\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> So when a Union(A,B) operation is needed, we use the name of the larger set for the union and update Component for only the smaller set\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> However, we still have a O(n) time even after the improvements\\
+>>>>>>>>>>>>>>>>> So, overall, Find(u) takes O(1) time,MakeUnionFind(A,B) takes O(n) time and any sequence of k Union operations takes at most O(klogk) time with this simple implementation of the algorithm.\\
+\\
+==== A Better Implementation for Union-Find ====
+\\
+  * Instead of using an array, we now use a pointer-based implementation
+  * With this implementation:\\
+>>>>>>>>>>>>>>>>> We maintain a set S of size n \\
+>>>>>>>>>>>>>>>>> Unions keep the name of the largest set \\
+>>>>>>>>>>>>>>>>> A Union operation takes O(1) time \\
+>>>>>>>>>>>>>>>>> MakeUnionFind(S) takes O(n) time \\
+>>>>>>>>>>>>>>>>> Find(u) takes O(logn) time \\
+==== Implementing Kruskal's Algorithm ====
+>>>>>>>>>>>>>>>>> T = {}
+>>>>>>>>>>>>>>>>> foreach (u ∈ V) make a set containing singleton u
+>>>>>>>>>>>>>>>>> for i = 1 to m
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> (u,v) = ei
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> if (u and v are in different sets)
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> T = T ∪ {ei}
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> merge the sets containing u and v
+>>>>>>>>>>>>>>>>> return T
+( The above algorithm, is from the class discussion)
+\\
+\\
+Kruskal's algorithm implemented as above runs in O(mlogn) time.
+\\
+\\
+This section was also interesting, although it contained some long proofs. I give it an 8/10