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CS301 – Data Structures
Lecture No. 35
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Data Structures
Lecture No. 35
Reading Material
Data Structures and Algorithm Analysis in C++
Chapter. 8
8.2, 8.3
Summary
•
Dynamic Equivalence Problem
•
Example 1
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Parent Array
o
Initialization
o
Find (i)
o
Union (i, j)
•
Example 2
o
Initialization
o
union operation
o
find Operation
•
Running Time Analysis
Before talking about the data structure implementation in detail, we will recap the
things discussed in the previous lecture. The concepts that came under discussion in
the last lecture included the implementation of the disjoint set. It was observed that in
case of data elements, we assign a unique number to each element to make sets from
these numbers. Moreover, techniques of merger and searching in these sets by using
the operations union and find were, talked about. It was seen that if we send a set item
or a number to
find
method, it tells us the set of which this item is a member. It will
return the name of the set. This name of set may also be a number. We talked about
the
union
method in which members of two sets join together to form a new set. The
union
method returns the name or number of this new set.
Now we will discuss the data structure used to implement the disjoint sets, besides
ascertaining whether this data structure implements the find and union operations
efficiently.
Dynamic Equivalence Problem
We are using sets to store the elements. For this purpose, it is essential to remember
which element belongs to which set. We know that the elements in a set are unique
and a tree is used to represent a set. Each element in a tree has the same root, so the
root can be used to name the set. The item (number) in the root will be unique as the
set has unique values of items. We use this root as the name of set for our
convenience. Otherwise, we can use any name of our choice. So the element in the
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View Full DocumentCS301 – Data Structures
Lecture No. 35
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root (the number) is used as the name of the set. The
find
operation will return this
name. Due to the presence of many sets, there will be a collection of trees. In this
collection, each tree will be representing one set. If there are ten elements, we will
have ten sets initially. Thus there will be ten trees at the beginning. In general, we
have N elements and initially there will be N trees. Each tree will have one element.
Thus there will be N trees of one node. Now here comes a definition for this
collection of trees, which states that a collection of trees is called a
forest
.
The trees used for the sets are not necessarily binary trees. That means it is not
necessary that each node should have a maximum of two children nodes. Here a node
may have more than two children nodes.
To execute the union operation in two sets, we merge the two trees of these sets in
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 Spring '10
 Dr.Naveed Malik
 Data Structures

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