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Union by Rank

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структуры данных

Калифорнийский университет в Сан-Диего

4.7 (оценок: 2,096) | Зарегистрировано учащихся: 91K

Курс 1 из 6 — Структуры и алгоритмы данных Специализация

A good algorithm usually comes together with a set of good data structures that allow the algorithm to manipulate the data efficiently. In this course, we consider the common data structures that are used in various computational problems. You will learn how these data structures are implemented in different programming languages and will practice implementing them in our programming assignments. This will help you to understand what is going on inside a particular built-in implementation of a data structure and what to expect from it. You will also learn typical use cases for these data structures. A few examples of questions that we are going to cover in this class are the following: 1. What is a good strategy of resizing a dynamic array? 2. How priority queues are implemented in C++, Java, and Python? 3. How to implement a hash table so that the amortized running time of all operations is O(1) on average? 4. What are good strategies to keep a binary tree balanced? You will also learn how services like Dropbox manage to upload some large files instantly and to save a lot of storage space!

Просмотреть программу курса

Получаемые навыки

Binary Search Tree, Priority Queue, Hash Table, Stack (Abstract Data Type), List

Рецензии

4.7 (оценок: 2,096)

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VK

Jul 16, 2018

Amazing course.\n\nInteresting exercises help with learning and understanding of internal mechanisms basic data structures, how some of them work, the pros and cons of each.\n\nThanks!

SC

Feb 27, 2019

Perfect course for learning more about fundamental data structures, except for presentations on few difficult topics like splay trees, where explanations can be made more elaborate!

Из урока

Priority Queues and Disjoint Sets

We start this module by considering priority queues which are used to efficiently schedule jobs, either in the context of a computer operating system or in real life, to sort huge files, which is the most important building block for any Big Data processing algorithm, and to efficiently compute shortest paths in graphs, which is a topic we will cover in our next course. For this reason, priority queues have built-in implementations in many programming languages, including C++, Java, and Python. We will see that these implementations are based on a beautiful idea of storing a complete binary tree in an array that allows to implement all priority queue methods in just few lines of code. We will then switch to disjoint sets data structure that is used, for example, in dynamic graph connectivity and image processing. We will see again how simple and natural ideas lead to an implementation that is both easy to code and very efficient. By completing this module, you will be able to implement both these data structures efficiently from scratch.

Trees for Disjoint Sets7:54

Union by Rank9:41

Path Compression6:07

Analysis (Optional)18:17

Преподаватели

Alexander S. Kulikov
Visiting Professor
Michael Levin
Lecturer
Daniel M Kane
Assistant Professor
Neil Rhodes
Adjunct Faculty

Текст видео

Okay, when merging two trees we're going to hang the shorter one under the root

of a taller one.

This means that when merging two trees we need

a way to quickly find the height of both trees.

Instead of just computing them we're going to keep the height of each

possible subtree in our forest in a separate array called rank.

Name the rank of i is equal to the height of the subtree rooted at i.

The reason we call it rank will become clear a little bit later.

Let me also mention that this way of merging two trees,

based on the height is called the union by rank heuristic.

To keep the rank, we need a small addition to our MakeSet implementation,

namely when creating a single set, we also set rank of i to be equal to zero.

This reflects the fact that it is currently just a root containing one node,

that just a tree containing one node, that is a tree of height zero.

We do not need to change Find.

So the Find operation doesn't need to change rank, and

it also doesn't use rank in any way.

To merge two trees containing the given two elements i and j, we do the following.

We first find the roots of the point in two trees by calling the Find

operation two times.

We store this root in variables i_id and j_id.

We then check whether i_id is equal to j_id.

If they are equal, this means that elements i and

j already lie in the same set.

So we just return in this case.

So this is done in the following if loop.

We then check whether the height of the tree containing element i

is larger than the height of the tree containing element j.

If it is larger, then we hang the tree with the root j_id,

and this root of element i_id.

This is done as follows.

Parent of j_id is set to i_id.

Otherwise, we do the opposite thing.

We just assigned parent of i_id to be equal to j_id.

So the last thing is that we need to check whether

the height of the corresponding two threes are just equal.

Let me illustrate this again with a small example.

Assume that we are merging the following two trees.

In this case the height of these two elements and

this element is zero and this height of this element is 1.

So in this case roots are equal.

The ranks of the corresponding roots are equal.

To merge these two trees we do the following.

We just hang the left tree under the root of the right tree.

If you can see, in this case,

the height of the resulting tree actually increases and this is the only case

when the union operation increases the height of this tree.

So in this case, initially, the longest path contained just one edge.

In this case we go the path that can contain two edges.

So we need to update this rank and this is done in the last check.

So if initially the ranks of our two trees that are going to be merged

are equal we hang one of them under the root of the other one and

increase the rank of the resulting tree by one.

Let's consider a small example, in this case we have six elements.

Let's call MakeSet for each of these elements.

These fields have a data structure as follows.

So currently, each element is its own parent, right?

So its current set is just a single one set.

Also, the height of each sub-tree in our data structure is currently equal to 0.

Now let's call Union(2,4).

In this case, the rank of the subtree rooted at 2 is equal to 0.

The height of the subtree rooted at 4 is equal to 0.

So it doesn't mean which one to hang under the root of the other one, so

let's hang 2 under 4.

This changes the data structure as follows.

Now it's a parent of 2 is 4 and

the rank of the subtree rooted at 4 is equal to 1.

Okay, now let's call Union(5,2).

In this case the height of

the tree that contains the element number 2 is equal to 1, right?

While the height of the tree that contains element number 5 is equal to 0.

So, in this case we're going to hang 5 under 4.

We do this as follows.

So, this change the data structure, only this changes only this cell.

So now 4 is the parent of 5, and

it doesn't change any rank in our sub tree, in our forest.

Okay, now lets call Union(3,1).

This is done as follows, now 1 is rank 1, and

now the parent of 3 is equal to 1, okay?

Now, let's call Union(2,3), and again, in this case, 2

lies in a set in the tree whose root is 4.

And currently, the rank of 4 is equal to 1.

Also, 3 lies in a set whose root is 1.

And currently, rank of 1 is equal to 1.

Which means that after merging these two trees we will get a tree of height 2.

So we do this as follows, now 1 is the root of the resulting tree and

its rank is equal to 2.

Finally we call Union(2,6) and this will just attach 6 to 1, as follows.

In our current implementation, we maintain the following important invariant.

At any point of time and for any node i, rank of i is equal

to the height of the subtree rooted at this node, i, right?

We will use this invariant to prove the following lemma.

The height of any tree in our forest is at most binary logarithm of n.

This will immediately imply that the running time of all operations

with our data structure is at most logarithmic, right?

To prove this lemma we will prove another lemma shown here on this slide.

We're going to prove that if we have a tree in our forest whose height

is k then this tree contains at least two to the k nodes.

This will imply the first lemma as follows.

I assume that some tree has height with more,

strictly greater than binany logarithm of n.

Using the second lemma it will be possible to show then

that this tree contains more than n nodes, right?

Which would lead to a contradiction with the fact that we

only have n objects in our data structure.

Here we are going to prove the second lemma by induction on k.

Recall that we proved that any tree of height k

in our forest contains at least 2 to the k nodes.

We're going to prove this by induction on k.

When k is equal to zero this means that we have a tree just of height 0,

which means that it contains just one node.

So, in this case, the statement clearly holds.

Now, to prove the induction step, let's recall that the only way to get a height,

to get a tree of height k, is to merge two trees, whose height is equal to k- 1.

I mean to merge both trees such that the height of the first tree is equal

to k- 1 and the height of the second tree is equal to k-1.

By the induction hypothesis the both of these two trees

contain at least 2 to the k-1 node.

Which means that our resulting tree contains

at least 2 to the k- 1 + 2 to the k- 1 nodes,

which is exactly equal to 2 to the k, right?

Which means that the lemma is proved.

To conclude, the running time of both Union and

Find operations in our current implementation is at most logarithmic.

Why is so?

Well, just because we keep our trees shallow, so

that the height of any tree in our forest is at most logarithmic.

This immediately implies the time of any

Find operation is also big O of logarithm of n.

Recall also that the Union operation consists of two calls to the Find

operation and also a few constant time operations,

which means also that the running time of Union is also big O of log n.

In the next video, we will see another beautiful heuristic which will just

decrease the running time of both these operations to nearly a constant.

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