Recommandé
Contenu connexe
Tendances
Tendances (20)
En vedette
En vedette (20)
Similaire à Time complexity of union find
Similaire à Time complexity of union find (20)
Dernier
Dernier (20)
Time complexity of union find
- 1. Dec. 09, 2015 Wei Li Zehao Cai Ishan Sharma Time Complexity of Union Find 1Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015
- 2. Algorithm Definition Disjoint-set data structure is a data structure that keeps track of a set of elements partitioned into a number of disjoint (non-overlapping) subsets. Union find algorithm supports three operations on a set of elements: • MAKE-SET(x). Create a new set containing only element x. • FIND(x). Return a canonical element in the set containing x. • UNION(x, y). Merge the sets containing x and y. Implementation: Linked-list, Tree(Often) 2Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015
- 3. Find(b) = c | Find(d) = f | Find(b) = f b → h → c | d → f | b → h → c → f Quick-find & Quick-union 3Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015
- 4. Definition: The rank of a node x is similar to the height of x. When performing the operation Union(x, y), we compare rank(x) and rank(y): • If rank(x) < rank(y), make y the parent of x. • If rank(x) > rank(y), make x the parent of y. • If rank(x) = rank(y), make y the parent of x and increase the rank of y by one. First Optimization: Union By Rank Heuristic 4Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015 Note. In this case, rank = height.
- 5. During the execution of Find(e), e and all intermediate vertices on the path from e to the root are made children of the root x. Second Optimization: Path Compression 5Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015
- 6. 6Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015 Union by Rank & Path Compression This is why we call it “union by rank” rather than “union by height”.
- 7. Algorithms Worst-case time Quick-find 𝑚𝑛 Quick-union 𝑚𝑛 QU + Union by Rank 𝑛 + 𝑚𝑙𝑜𝑔𝑛 QU + Path compression 𝑛 + 𝑚𝑙𝑜𝑔𝑛 QU + Union by rank + Path compression 𝒏 + 𝒎𝒍𝒐𝒈∗ 𝒏 m union-find operations on a set of n objects. Time Complexity 7Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015
- 8. Lemma 1: as the find function follows the path along to the root, the rank of node it encounters is increasing. Union: a tree with smaller rank will be attached to a tree with greater rank, rather than vice versa. Find: all nodes visited along the path will be attached to the root, which has larger rank than its children. 8Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015
- 9. Lemma 2: A node u which is root of a sub-tree with rank r has at least 2r nodes. Proof: Initially when each node is the root of its own tree, it's trivially true. Assume that a node u with rank r has at least 2r nodes. Then when two tree with rank r Unions by Rank and form a tree with rank r + 1, the new node has at least 2r + 2r = 2r + 1 nodes. 9Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015
- 10. Lemma 3: The maximum number of nodes of rank r is at most n/2r. Proof: From lemma 2, we know that a node u which is root of a sub-tree with rank r has at least 2r nodes. We will get the maximum number of nodes of rank r when each node with rank r is the root of a tree that has exactly 2r nodes. In this case, the number of nodes of rank r is n / 2r 10Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015
- 11. We define “bucket” here: a bucket is a set that contains vertices with particular ranks. Proof 11Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015
- 12. 𝒍𝒐𝒈∗ 𝒏 𝑙𝑜𝑔∗ 𝑛 ∶= / 0 𝑖𝑓 𝑛 ≤ 1 1 + 𝑙𝑜𝑔∗ 𝑙𝑜𝑔𝑛 𝑖𝑓 𝑛 > 1 Definition: For all non-negative integer n, 𝑙𝑜𝑔∗ 𝑛 is defined as We have 𝑙𝑜𝑔∗ 𝑛 ≤ 5 unless n exceeds the atoms in the universe. 𝑙𝑜𝑔∗ 29 = 1 + 𝑙𝑜𝑔∗ 2: = 1 𝑙𝑜𝑔∗ 16 = 𝑙𝑜𝑔∗ 2<= = 1 + 𝑙𝑜𝑔∗ 2< = 3 𝑙𝑜𝑔∗ 65536 = 𝑙𝑜𝑔∗ 2<== = 1 + 𝑙𝑜𝑔∗ 2<= = 4 𝑙𝑜𝑔∗2@AAB@ = 𝑙𝑜𝑔∗2<=== = 1 + 𝑙𝑜𝑔∗2<== = 5 𝑙𝑜𝑔∗ 4 = 𝑙𝑜𝑔∗ 2< = 1 + 𝑙𝑜𝑔∗ 29 = 2 12Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015
- 13. We can make two observations about the buckets. The total number of buckets is at most 𝒍𝒐𝒈∗ 𝒏. Proof: When we go from one bucket to the next, we add one more two to the power, that is, the next bucket to [B, 2B − 1] will be [2C ,2<E − 1 ] The maximum number of elements in bucket [B, 2B – 1] is at most 𝒏. Proof: The maximum number of elements in bucket [B, 2B – 1] is at most 𝑛 2 𝐵⁄ + 𝑛 2CI9⁄ + 𝑛 2CI<⁄ + ⋯ + 𝑛 2<EK9 ≤ 2 𝐵 − 1 − 𝐵 ∗ 𝑛/2 𝐵⁄ ≤ n Proof 13Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015
- 14. Let F represent the list of "find" operations performed, and let Then the total cost of m finds is T = T1 + T2 + T3 Proof 14Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015
- 16. T1 = constant time cost (1) per m operations: O(m) T2 = maximum number of different buckets: O(𝑚 𝑙𝑜𝑔∗ 𝑛) T3 = for all buckets ( for all notes in one bucket) = ∑ ∑ N <O <E K9 PQC RST∗ N 9 ≤ 𝑙𝑜𝑔∗ 𝑛 2C − 1 − 𝐵 N <E ≤ 𝑙𝑜𝑔∗ 𝑛 2C N <E = 𝑛 𝑙𝑜𝑔∗ 𝑛 Proof T = T1 + T2 + T3 = O(m) + O(𝑚𝑙𝑜𝑔∗ 𝑛) + O(𝑛𝑙𝑜𝑔∗ 𝑛) 𝑚 ≥ 𝑛 → O(𝒎𝒍𝒐𝒈∗ 𝒏) 16Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015
- 17. Algorithms Worst-case time Quick-find 𝑚𝑛 Quick-union 𝑚𝑛 QU + Union by Rank 𝑛 + 𝑚𝑙𝑜𝑔𝑛 QU + Path compression 𝑛 + 𝑚𝑙𝑜𝑔𝑛 QU + Union by rank + Path compression 𝒏 + 𝒎𝒍𝒐𝒈∗ 𝒏 m union-find operations on a set of n objects. Time Complexity 17Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015
- 18. Algorithm & Time Complexity • Simple data structure, algorithm easy to implement. • Complex to prove time complexity. (Proved in 1975, Tarjan, Robert Endre ) • Time complexity is near linear. Applications • Keep track of the connected components of an undirected graph; • Find minimum spanning tree of a graph. Conclusions 18Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015