On a set of subsets, the three operations amount to: - Create a set of n disjoint subsets with every node in its own subset - Test whether A and B are in the same subset - If A and B are in the same subset, then do nothing, else unify the subsets to which A and B belong The most efficient way to implement the operation is: - For every A it is possible to ask for the index of the subset to which A belongs - This will be denoted as find(A). And we have A~B == find(A) == find(B). The operation of adding a relation between A and B will be denoted by union(A, B) (even though nothing needs to happen). Thus, union(A, B) one performs Solution: - Keep track of the size of each equivalence class. - Let every element point to the representative of its set. On “union", change the name of the smaller equivalence class to the larger. - Thus "union(A,B)" takes O(min(|A], |B|) time. The total time spent for k "union" operations is O(k logk) C O(k logN). (Why ?) - In other words, the Amortized time complexity of "union" is O(log N).
On a set of subsets, the three operations amount to: - Create a set of n disjoint subsets with every node in its own subset - Test whether A and B are in the same subset - If A and B are in the same subset, then do nothing, else unify the subsets to which A and B belong The most efficient way to implement the operation is: - For every A it is possible to ask for the index of the subset to which A belongs - This will be denoted as find(A). And we have A~B == find(A) == find(B). The operation of adding a relation between A and B will be denoted by union(A, B) (even though nothing needs to happen). Thus, union(A, B) one performs Solution: - Keep track of the size of each equivalence class. - Let every element point to the representative of its set. On “union", change the name of the smaller equivalence class to the larger. - Thus "union(A,B)" takes O(min(|A], |B|) time. The total time spent for k "union" operations is O(k logk) C O(k logN). (Why ?) - In other words, the Amortized time complexity of "union" is O(log N).
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Prove that the amortized complexity of the union operation in the solution is O(log n). Consider the number of times a given element may change its set.
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