CSE - 5311 Advanced Algorithms

Instructor : Gautam Das

Submitted by Raja Rajeshwari Anugula & Srujana Tiruveedhi

Contents

RED – BLACK Trees History Properties Rotations Insertion

Hashing Union Find Algorithm

RED BLACK TREES RB trees is a Binary Tree with one extra bit of storage

per node; its color, which can be either RED or BLACK.

Each node of the tree contains fields color, key, left, right, parent.

Red-Black Trees ensure that longest path is no more than twice as long as any other path so that the tree is approximately BALANCED.

RED BLACK TREESSTRUCTURAL PROPERTIES:

Every node is colored red or black. The Root is Black. Every “leaf” (Nil) is colored black. Both children of a red node are black. Every simple path from a child of node X

to a leaf has the same number of black nodes.

RED BLACK TREESPoints to remember: This number is known as the black-height

of X(bh(X)). A RB tree with n internal nodes has a

height of almost 2log(n+1). Maximum path length is O(log n). Finding an element is real quick in RB

trees, i.e,., it takes O(log n) time. Insertion and Deletion take O(log n) time.

RED BLACK TREESROTATIONS Insertion and Deletion modify the tree, the

result may violate the properties of red black trees. To restore these properties rotations are done.

We can have either LEFT rotation or RIGHT rotation by which we must change colors of some of the nodes in the tree and also change the pointer structure.

RED BLACK TREES LEFT ROTATE

RIGHT ROTATE

a c

a b b c When we do a LEFT rotation on a node x we assume that its right child

y is not nil ,i.e x may be any node in the tree whose right child is not Nil.

It makes y the new root of the sub tree, with x as y’s left child and y’s left child as x’s right child.

x

y

y

x

RED BLACK TREES The Idea to insert is that we traverse the

tree to see where it fits , assuming that it fits at the end, and initially color the inserted node RED, then traverse up again.

Coloring rule while insertion If the father node and uncle node of the

inserted node are red then make father and uncle as BLACK and grand father as RED.

RED BLACK TREES

C

A D

B

a

b c

d e

C

DA

a

b c

d eB

After Recoloring

Case 1a: Father and Uncle are Red , problem node is right child

Recolor it to

BLACK

RED BLACK TREESCase 1b: Father and Uncle are Red , problem node is left child

C

B D

A

a b

c d e

C

B D

A

a b

cd e

After Recoloring

Recolor it to

BLACK

RED BLACK TREESCase 2a: Father red and Uncle Black, problem node is left child

C

B

A

a b

d

B

CA

d

cba

After Rotation

c

D

D

RED BLACK TREESCase 2b: Father red and Uncle Black, problem node is right child

C

A

B

a

b

d

C

A

B

dc

ba

After Rotation

c

D D

apply 2a for the above tree

RED BLACK TREES

11

2 14

1

4

7 15

5 8

11

2 14

1

4

7 15

5 8InsertNode

4

Apply

Case 1b

Example:

RED BLACK TREES

11

2 14

1

4

7 15

5 8

11

7 14

2

4

8 15

51

Apply

Case 2b

RED BLACK TREES

11

7 14

2

4

8 15

51

7

2 11

1

4 15

145 8

Apply

Case 2a

HASHING Has table is an effective data structure for

implementing dictionaries. Although searching for an element in hash table in

the worst case is Θ(n) time, under reasonable assumptions the expected time to search for an element is O(1).

With hashing this element is stored in slot h(k) i.e we use a hash function h to compute the slot

from the key k. (h maps the universe U of keys into the slots of a hash table T[0…m-1])

h :U {0,1,2…..m-1} Two keys may hash to the same slot. This is

called collision.

UNION FIND

Basics: Applications involve grouping n elements into a collection

of disjoint sets. The important operations are

MAKE-SET(x): Creates a new set UNION(x,y): Unites the dynamic sets that contain x and y into a

new set that is the union of these two sets. FIND-SET(x): Returns a pointer to the representative of the set

containing x The number of union operations is atmost n-1.

MAKE-SET OPERATION Makes a singleton set Every set should have a name which should be any

element of the set Make-Set(1) Make-Set(2) * * * * *

Make-Set(n)

UNION OPERATION

Initially each number is a set by itself. From n singleton sets gradually merge to form a set. After n-1 union operations we get a single set of n

numbers.

1 3

2

4

UNION :Merge two sets and create a new set

FIND OPERATION

Every set has a name. Thus Find(number) returns name of the

set. It can be performed any number of times

on the sets. The time taken for a find operation is O(n)

whereas for Union operation it is O(1).

LINKED LIST REPRESENTATION

Every Set is represented as linked list where the first element is the name of the set.

We have the array of elements which have pointers to the elements in the linked lists.

5 1

3 4 7

10

5 2

1 3

4 7

10

2

‘3’ is head of this set

‘5’ is head if this set

’10’ is head of this singleton set

For ‘n’ Unions and ‘m’ Finds, then time taken is n+mn.

If we have a pointer from each element in the set to the head, then the time to find operation is O(1).

NOTE: If m is large, Find : O(n+mn) Union: O(1)

LINKED LIST REPRESENTATION

‘3’ is the head

‘5’ is the head

LINKED LIST REPRESENTATION

The 2 sets are being merged by connecting 5 and 7

Each element pointing to the head i,e ‘3’ in this example

In this case the union takes O(n2+m) time.

If we assume that the smaller set is attached to the end of the larger set in Union operation, then UnionO(n) and Find O(1)

But in the AMORTIZED ANALYSIS, Average time taken for union is O(log n).

So ‘n’ Unions and ‘m’ Finds take (nlog n +m) time.

To guarantee O(log n) time for Union, instead of pointing each element to the main head, point the Heads’ of the individual sets to a main head.

If we consider path lengths to combine two trees(L1 is the path length of tree1 and L2 is the path length of tree2), then

If L1>L2 or L2>L1, path length doesn’t change i.e. It is still the longer path.

If L1=L2, then the path length is L2+1 if the head of tree1 is pointed to head of tree2 L1+1 if the head of tree2 is pointed to head of tree1