time complexity of union find

Dec. 09, 2015Wei Li

Zehao CaiIshan Sharma

Time Complexity of Union Find

1Wei/Zehao/Ishan CSCI 6212/Arora/Fall 2015

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 algorithmsupports 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)


Find(b) = c | Find(d) = f | Find(b) = fb → h → c | d → f | b → h → c → f

Quick-find & Quick-union


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


Note. In this case, rank = height.

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



Union by Rank & Path Compression

This is why we call it “union by rank” rather than “union by height”.

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


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.


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.


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


We define “bucket” here: a bucket is a set that contains vertices with particular ranks.

Proof


𝒍𝒐𝒈∗𝒏

𝑙𝑜𝑔∗𝑛 ∶= /0𝑖𝑓𝑛 ≤ 11 + 𝑙𝑜𝑔∗ 𝑙𝑜𝑔𝑛 𝑖𝑓𝑛 > 1

Definition: For all non-negative integer n, 𝑙𝑜𝑔∗𝑛is 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


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 2𝐵⁄ +𝑛 2CI9⁄ + 𝑛 2CI<⁄ +⋯ +𝑛 2<EK9 ≤ 2𝐵 −1 −𝐵 ∗ 𝑛/2𝐵⁄ ≤ n

Proof


Let F represent the list of "find" operations performed, and let

Then the total cost of m finds is T = T1 + T2 + T3

Proof


Proof


T1 = constant time cost (1) per m operations: O(m)

T2 = maximum number of different buckets: O(𝑚𝑙𝑜𝑔∗𝑛)

T3 = for all buckets ( for all notes in one bucket)

= ∑ ∑ N<O

<EK9PQC

RST∗N9

≤ 𝑙𝑜𝑔∗𝑛 2C − 1 − 𝐵 N<E

≤ 𝑙𝑜𝑔∗ 𝑛2C N<E

= 𝑛𝑙𝑜𝑔∗ 𝑛

Proof

T = T1 + T2 + T3 = O(m) + O(𝑚𝑙𝑜𝑔∗𝑛)+ O(𝑛𝑙𝑜𝑔∗𝑛)

𝑚 ≥ 𝑛 → O(𝒎𝒍𝒐𝒈∗𝒏)


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


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


https://en.wikipedia.org/wiki/Disjoint-set_data_structure

https://en.wikipedia.org/wiki/Proof_of_O(log*n)_time_complexity_of_union%E2%80%93find

http://www.ccse.kfupm.edu.sa/~wasfi/Resources/ICS353CD/Lecture17/lec17_slide01.swf

http://sarielhp.org/teach/2004/b/webpage/lec/22_uf.pdf

https://www.cs.princeton.edu/courses/archive/spring13/cos423/lectures/UnionFind.pdf

References


time complexity of union find

Software