Lecture 10

Disjoint Set ADT

Preliminary Definitions

A set is a collection of objects.

Set A is a subset of set B if all elements of A are in B.

Subsets are sets

Union of two sets A and B is a set C which consists of all elements in A and B

Two sets are mutually disjoint if they do not have a common element.

A partition of a set is a collection of subsets such that

Union of all these subsets is the set itself

Any two subsets are mutually disjoint

S = {1,2,3,4}, A = {1,2}, B = {3,4}, C = {2,3,4}, D = {4}

Is A, B a partition of S?

Is A, C partition of S?

Yes

Is A, D partition of S?

No

No

Union and Find Operations

Operations on partitions.

Union

Need to form union of two different sets of a partition

Find

Need to find out which set an element belongs to

Every set in the partition has a number.

The numbers can be anything as long as different sets have distinct numbers.

Find(a) returns the number of the set containing a.

Can two different sets contain the same element?

No, the sets in a partition are disjoint

Disjoint Set Data Structure

The set is represented by the root of the tree.

The number assigned to a set is the number of the root element.

Every element has a number.

Elements of a set are stored in a tree (not necessarily binary)

B = {3, 4}

3

4

B is assigned number 3

Are the numbers distinct for different sets?

No two sets have the same root as they are disjoint, thus they have distinct numbers

Find(a) returns the number of the root node of the tree containing a.

Union operation makes one tree sub-tree of another

Root of one tree becomes child of the root of another.

B = {3, 4}

3

4

Find(4) returns? 3

Find(3) returns? 3

B = {3, 4}

3

4

A = {1,2}

1

2

Want to do A union B

We have:

1

23

4

Tree Representation

Will use an array based representation, array S

Let the elements be 1,2,….N

S[j] contains the number for the parent of j

S[j] = 0 if j is the root.

Initially all trees are singletons

Trees build up with unions.

Note that we don’t use any pointers here.

B = {3, 4}

3

4

A = {1,2}

1

2

Want to do A union B

We have:

1

23

4

0 1 0 3

S 0 1 1 3

S

Pseudo Code for Find

Find(a) {

If S[a] = 0, return a;

else Find(S[a]);

return;

}

Complexity? O(N)

Pseudo-Code for Union

Union(root 1, root 2)

{

S[root2] = root1;

}

Complexity? O(1)

More Efficient UnionWill improve the worst case find complexity to log N

When we do a union operation make the smaller tree a subtree of the bigger one

Thus the root of the smaller subtree becomes a child of the root of the bigger one.

A = {1,2,3} B = {4}

Root of B is root of A after union, B is subtree of A

Alternatively, union operation can be done by height as well

Tree of lesser height is made subtree of the other.

We consider only size here.

Array storage changes somewhat.

If j is a root, S[j] = - size of tree rooted at j

If j is not a root, S[j] = parent of j

Why is S[j] not equal to the size of tree j, if j is a root?

Initially, what is the content of array S?

All elements are -1

Size of tree j is an integer, if S[j]=size of tree j and j is root, then it would look like root of j is another element, thus j is not root

Pseudo-Code for Union

Union(root 1, root 2)

{

If S[root2] < S[root1], S[root1] = root2;

else S[root2]=root1;

}

Complexity? O(1)

Pseudo Code for Find

Find(a) {

If S[a] < 0, return a;

else Find(S[a]);

return;

}

Complexity Analysis for Find Operation

If the depth (distance from root) of a node A increases, then the earlier tree consisting the node A becomes a subtree of another.

Since only a smaller tree becomes a subtree of another, total size of the combined tree must be at least twice the previous one consisting A.

Each time depth of a node increases, the size of the tree increases by at least a factor of 2.

At first every node has depth 0

Next time depth is 1, tree size is at least 2,

depth is 2, tree size is at least 4…

depth is k, tree size is at least 2k

We know that 2k <= N

Thus k <= log N

Complexity of Find operation is O(log N)

Complexity of any M operations is O(MlogN)

Depth of any tree is at most log N

Path Compression

Makes all operations almost linear in the worst case.

Whenever you do Find(j) make S[k]=Find(j) for all elements on the path of j to the root, except the root.

All nodes on the path of j now point to the root directly

1

23

4

Do Find(5)

Find(5) encounters 5, 4 and 3 before reaching root 1

After Find(5):

5

1

2 3 4

5

Later Find operations will have lower costs as their depths have been reduced.

Any Find operation reduces the cost of future ones.

Pseudo Code for New Find

Find(a) {

If S[a] < 0, return a;

else S[a]=Find(S[a]);

return;

}

Complexity Analysis

Any M operations take O(Mlog*N) if M is (N)

log*N is the number of times we take loglog….logN so as to get a number less than or equal to 1 (log base 2, even otherwise asymptotic order remains the same).

log*N grows very slowly with N and is less than 4 or 5 for all practical values of N,

log*232 is less than 5

Thus the worst case complexity is linear for all practical purposes.

Reading Assignment

Chapter 8, till section 8.61. (i.e. section 8.6.1 onwards can be omitted).