3.Disjoint Sets

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Disjoint Sets

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Equivalence Relation

An equivalence relation is a relation operator that

observes three properties:

Reflexive: (a R a), for all a

Symmetric: (a R b) if and only if (b R a)

Transitive: (a R b) and (b R c) implies (a R c)

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Example

The relation


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Dynamic Equivalence Problem

The equivalence class of an element a S is the

subset of S that contains all the elements related

to a.

The equivalence classes form a partition of S (Si

Sj =); i.e., the sets are disjoint.

Find (X) returns the name of the set (equivalence

class) containing a given element X.

Union (X, Y) returns the new root4


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Basic Data Structure

Tree data structure is used to implement the set.

The name of a set is given by the node at the

root.

The array implementation of tree represents that

P[i] is the parent of ith element.

Ifi is the root P[i]=0.

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Operations on Disjoint Sets

Union (Add operation (e.g., add relation a~b))

Check if a and b are already related: if they are in

the same equivalence class.

If not, merge the two equivalence classes

containing a and b into a new equivalence class.

Find

Return the name (pointer or index of representative)

of the set containing a given element

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AnithaPadmanabanDynamic Equivalence Problem

Eight elements, initially in different sets.

1 2 3 4 5 6 7 8

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After union(5,6)

1 2 3 4 5

6

7 8

8


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After union(7,8)

1 2 3 4 5

6

7

8

9


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After union(5,7)

1 2 3 4 5

6 7

8

10


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After union(3,4)

1 2 3

4

5

6 7

8

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After union(4,5)

1 2 3

4 5

6 7

8

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Smart Union Algorithms

The previous union algorithm arbitrarily makes

the second tree a subtree of the first.

Heuristics to reduce the depth of the tree.

Union operations is performed by three ways:

Arbitrary Union

Union by Size

Union by Height

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Arbitrary Union

This can be performed by making the second tree

a subtree of the first

1 2 3 4 5 6 7

Arbitrary Union14


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Union (2,3) and Union (6,7)

1 2 3 4 5 6 7

1 2

3

4 5 6

7

0 0 2 0 0 0 6

1 2 3 4 5 6 7

Arbitrary Union15


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1 2

3

4 5

6

7

0 0 2 0 0 2 6

1 2 3 4 5 6 7

Arbitrary Union16


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Union by Size

This can be performed by making the smaller tree

a subtree of the larger

1 2 3 4 5 6 7

Union by Size17


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Union (3,4) and Union(5,6)

1 2 3 4 5 6 7

1 2 3

4

5

6

7

Union by Size18


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Union (3,5)

1 2 3

4

5

6

7

1 2 3

4

5

6

7

Union by Size19


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Union (2,3)

Here first tree is smaller, therefore it becomessubtree of the second tree

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2

3

4

5

6

7

Union by Size20


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Implementation of Union by Size

Initially, the size of the tree is -1 for all tree

Parent Array

Eight elements, initially in different sets.

1 2 3 4 5 6 7 8

-1 -1 -1 -1 -1 -1 -1 -1

1 2 3 4 5 6 7 8

Union by Size21


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After union(5,6)

1 2 3 4 5

6

7 8

-1 -1 -1 -1 -1 5 -1 -1

1 2 3 4 5 6 7 8

Union by Size22


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After union(7,8)

1 2 3 4 5

6

7

8

-1 -1 -1 -1 -1 5 -1 7

1 2 3 4 5 6 7 8

Union by Size23


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After union(5,7)

1 2 3 4 5

6 7

8

-1 -1 -1 -1 -1 5 5 7

1 2 3 4 5 6 7 8

Union by Size24


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After union(3,4)

1 2 3

4

5

6 7

8

-1 -1 -1 3 -1 5 5 7

1 2 3 4 5 6 7 8

Union by Size25


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After union(4,5)

1 2 3

4 5

6 7

8

-1 -1 -1 3 3 5 5 7

1 2 3 4 5 6 7 8

Union by Size26


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Find(8) 1 2 3

4 5

6 7

8

-1 -1 -1 3 3 5 5 7

1 2 3 4 5 6 7 8

Union by Size27

E l


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AnithaPadmanabanExample

Union (3,4)

P[3]=P[3]+P[4]

= -1 + -1

= -2 (Parent)

P[4]= 3 (Parent)

-1 -1 -1 -1 -1 -1 -1 -1

1 2 3 4 5 6 7 8

-1 -1 -2 3 -1 -1 -1 -1

1 2 3 4 5 6 7 8

Union by Size28

E l


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Union (5,6)

P[5]=P[5]+P[6]

= -1 + -1

= -2 (Parent)

P[6]= 5 (Parent)

-1 -1 -2 3 -2 5 -1 -1

1 2 3 4 5 6 7 8

-1 -1 -2 3 -1 -1 -1 -1

1 2 3 4 5 6 7 8

Union by Size29

E l


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Union(3,5)

P[3]=P[3]+P[5]

= -2 + -2

= -4 (Parent)

P[5]= 3 (Parent)

-1 -1 -4 3 3 5 -1 -1

1 2 3 4 5 6 7 8

-1 -1 -2 3 -2 5 -1 -1

1 2 3 4 5 6 7 8

Union by Size30

P d BExample


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Union(2,3)

P[3]=P[2]+P[3]

= -1 + -4

= -5 (Parent)

P[2]= 3 (Parent)

-1 3 -5 3 3 5 -1 -1

1 2 3 4 5 6 7 8

-1 -1 -4 3 3 5 -1 -1

1 2 3 4 5 6 7 8

Union by Size31

P d B


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Advantages Union by Size

Easy to implement

Requires no extra spaces

Fast execution

Sequence of M operations requires O(M)average

time

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P d B


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Union by Height

Instead of size, here the height of the tree is

considered

This can be implemented by making the shallowtree a subtree of the deeper tree

Initially the array is set to 0

0 0 0 0 0 0 0 0

1 2 3 4 5 6 7 8

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Example

Union(3,4)

1 2 3 4 5 6 7

1 2 3

4

5 6 7

0 0 -1 3 0 0 0 01 2 3 4 5 6 7 8

If (S[R1] = = S[R2]) then

S[R] = S[R1]-1

S[3] = S[4] then

S[3] = 0 1

S[3] = 134

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Here,


S[R] = S[R1]-1

ie., S[5] = = S[6]

S[5] = S[5] 1

= 0 1

= 1

0 0 -1 3 -1 5 0 0

1 2 3 4 5 6 7 8

0 0 -1 3 0 0 0 0

1 2 3 4 5 6 7 8

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Here,


S[R] = S[R1]-1

ie., S[3] = = S[5]

S[3] = S[3] 1

= 1 1

= 2

0 0 -1 3 -1 5 0 0

1 2 3 4 5 6 7 8

0 0 -2 3 -1 5 0 0

1 2 3 4 5 6 7 8

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Here,

S[R2] < S[R1]

i.e, S[3] < S[2]

therefore, S[2] = 3

0 0 -2 3 -1 5 0 0

1 2 3 4 5 6 7 8

0 3 -2 3 -1 5 0 0

1 2 3 4 5 6 7 8

If S[R1] is equal to S[R2] thenboth S[R1] and S[R2] is updated

else

S[R1] is alone updated

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Path Compression

Path compression is performed during a Find

operation and is independent of the strategy used

to perform unions

Modify Find (X) such that every node on the path

from X to the root has its parent changed to the

root

Effectively reduce the length of the path

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3.Disjoint Sets