Combined Linear Congruential Generator

In computer science, Union Find is an algorithm for doing certain operations on sets. This page is about proof of O(log^*n) amortized time of Union Find

Statement: If m operations, either Union or Find, are applied to n elements, the total run time is O(m log^*n), where log^* is the iterated logarithm.

Proof

<increasing rank lemma>...</increasing rank lemma>Lemma 1: As the find function follows the path along to the root, the rank of node it encounters is strictly increasing

Proof: claim that as Find and Union operations are applied to the data set, this fact remains true over time. Initially when each node is the root of its own tree, it's trivially true. The only case when the rank of a node might be changed is when the Union by Rank operation is applied. In this case, a tree with smaller rank will be attached to a tree with greater rank, rather than vice versa. And during the find operation, all nodes visited along the path will be attached to the root, which has larger rank than its children, so this operation won't change this fact either.

<min subtree size lemma>...</min subtree size lemma>Lemma 2: A node u which is root of a subtree with rank r has at least 2^r 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 2^r 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 2^r + 2^r = 2^{r + 1} nodes.

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Lemma 3: The maximum number of nodes of rank r is at most n/2^r.

Proof: From lemma 2, we know that a node u which is root of a subtree with rank r has at least 2^r 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 2^r nodes. In this case, the number of nodes of rank r is n/2^r

For convenience, we define "bucket" here: a bucket is a set that contains vertices with particular ranks.

We create some buckets and put vertices into buckets according to their ranks. That is, vertices with rank 0 goes into the zeroth bucket, vertices with rank 1 goes into the first bucket, vertices with rank 2 and 3 goes into the second bucket, and vertices with ranks B to 2^B − 1 go into the Bth bucket.

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We can make two observations about the buckets.

1. <max buckets>...</max 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, 2^B − 1] will be [2^B, 2^2B − 1]

2. <max bucket size>...</max bucket size>The maximum number of elements in bucket [B, 2^B – 1] is at most 2n/2^B

   Proof: The maximum number of elements in bucket [B, 2^B – 1] is at most n/2^B + n/2^B+1 + n/2^B+2 + … + n/2^{2^B – 1} ≤ 2n/2^B

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

$T_{1}=\sum _{F}{\text{(link to the root)}}$

$T_{2}=\sum _{F}{\text{(number of links traversed where the buckets are different)}}$

$T_{3}=\sum _{F}{\text{(number of links traversed where the buckets are the same).}}$

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

Since each find operation makes exactly one traversal that leads to a root, we have T₁ = O(m).

Also, from the bound above on the number of buckets, we have T₂ = O(mlog^*n).

For T₃, suppose we are traversing from u to v, where u and v have rank in the bucket [B, 2^B − 1]. From lemma 1, we know that the number of times we traversed a link (u,v) where u and v were in the same bucket is at most 2^B − 1 − B, which is at most 2^B.

Therefore, $T_{3}=\sum _{[B,2^{B}-1]}\sum _{u}2^{B}$

From Observations 1 and 2, we can conclude that $T_{3}\leq \sum _{B}2^{B}{\frac {2n}{2^{B}}}\leq 2n\log ^{*}n.$

Therefore, T = T₁ + T₂ + T₃ = O(m log^*n).