advanced data structures and algorithms · advanced data structures and algorithms department of...

Advanced Data Structures and

Algorithms

Department of Computer Science _ UHD 1

Computer Science Department

College of Science and Technology

University of Human Development,

2015 – 2016

Associate Professor Dr. Raed Ibraheem Hamed

Randomized Data Structures Department of Computer Science _ UHD 2

I flipped a coin 30 times:

HTHTHHHTTTHHHTHTTTTTTHHHHHHTTH

One may almost suggest such a sequence is not random;

what is the probability of getting 6 tails and then 6 heads?

– Unfortunately, this is exactly what we got with the first

attempt to generate such a sequence, and is as random a

sequence as we can generate.

Random Numbers

Randomized Data Structures Department of Computer Science _ UHD

Randomized Data Structures

3

One application of randomized algorithms in the area

of data structures, specifically, Treaps and lists.


Randomized Data Structures

Skip listsTreaps


First introduced in 1989 by Aragon and Seidel

Randomized Binary Search Tree

A combination of a binary search tree and a heap (a “tree -

heap”)

Each node contains

◦ Data / Key (comparable)

◦ Priority (random integer number)

◦ Left, right child reference

Nodes are in BST order by data/key and in heap order by priority

random integer number.

Every subtree of a treap is a treap

Treaps


A treap is a binary search tree in which every node

has both a search key and a priority, where the

inorder sequence of search keys is sorted and each

node’s priority is smaller than the priorities of its

children.

Treaps Definition


The search tree has the structure that would result if elements

were inserted in the order of their priorities.

Tree + Heap = Treaps

Definition: A treap is a binary tree. Each node contains one element x with

key(x) ∈ U and prio(x) ∈ R. The following properties hold. �

Search tree property For each element x:

• elements y in the left subtree of x satisfy: key(y) < key(x)

• elements y in the right subtree of x satisfy : key(y) > key(x)

Heap property For all elements x, y:

• If y is a child of x, then prio(y) > prio(x).

• All priorities are pairwise distinct.


Treap : Dictionary Data Structure

Treap is a binary search tree

binary tree property

search tree property

Treap is also a heap

heap-order property

randomly assigned priorities

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Heap in Yellow; Search Tree in Blue

Legend

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A B C D E F G H I J K L M N O P Q R S T U V W X Y Z


There is only one possible treap for a given set of Keys and

priorities Proof:

• by heap property: the key k, with the highest priority must

be the root of the treap

• by BST property: all keys < k must be in the left subtree of the

root and all keys > k must be in the right subtree

• Inductively, the subtrees of the root must be constructed in the

same manner

i.e. by heap property: the root of the LST, k1, must have the

highest priority of any key in the LS and by BST: all keys in LST

of k1 are < k1 and all keys in RST of k1 are > k1

Treaps


Treap Insert

Choose a random priority

Insert as in normal BST

Rotate up until heap order is restored

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P Q

P

Let P be Q's left child. Set P to be the new root." Basically that's

the description of the rotation to the right or clockwise:

Binary Tree Rotation (Left or Right)


The right rotation operation as shown in the image to the

left is performed with Q as the root and hence is a right

rotation on, or rooted at, Q. This operation results in a

rotation of the tree in the clockwise direction. The inverse

operation is the left rotation, which results in a movement

in a counter-clockwise direction (the left rotation shown

above is rooted at P). The key to understanding how a

rotation functions is to understand its constraints.



T1, T2 and T3 are subtrees of the tree rooted with y (on left side)

or x (on right side)

y x

/ \ Right Rotation / \

x T3 – – – – – – – > T1 y

/ \ < - - - - - - - / \

T1 T2 Left Rotation T2 T3

Keys in both of the above trees follow the following order

keys(T1) < key(x) < keys(T2) < key(y) < keys(T3)

So BST property is not violated anywhere.



Tree + Heap… Why Bother?

Insert data in sorted order into a treap …

What shape tree comes out?

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The shape of the tree is

fully specified by what the

keys are and what their

random priorities are!


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Insert J with priority 2

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z


Find the key

Increase its value to

When X is found, rotate with child that has the smaller

priority

If X is a leaf, just delete it

If X is not a leaf, recursively remove X from its new

subtree

Snip it off

Treap Delete Strategy


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Remove K

Step 1 - Rotate K with

Right Child ( S )


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Right Child ( N )




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After Rotating K with N


Right Child ( L )


After Rotating K with L


Step 4 - K is a leaf

delete K

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Delete (9)



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Thank you

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