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)
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ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
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Question

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.

Problem:
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).
Transcribed Image Text:Problem: 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).
Representatives
3
4
5
6
7
8
Find:O(1)
9.
1
3
4
5 6
7
8 9
3
1
1
1
1
Sizes of the sets
22
Transcribed Image Text:Representatives 3 4 5 6 7 8 Find:O(1) 9. 1 3 4 5 6 7 8 9 3 1 1 1 1 Sizes of the sets 22
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