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Data Structures and Algorithms

(CS210/ESO207/ESO211)

Lecture 13 Ram model of computationfurther refinement

Algorithm paradigms

Algorithm paradigm of Divide and Conquer

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Word RAM model of computation

Further refinements

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word RAM : a model of computation

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RAM

Processor

Data

Program

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Execution of a instruction(fetching the operands, arithmetic/logical operation, storing the result back into RAM)

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RAM

Processor

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word RAM model of computation:

Characteristics

Word is the basic storageunit of RAM. Word is a collection of few bytes.

Each input item (number, name) is stored in binary format.

RAM can be viewed as a huge array of words. Any arbitrary location ofRAM can be accessedin the same time irrespectiveof the location.

Data as well as Program reside fully in RAM.

Each arithmetic or logical operation (+,-,*,/,or, xor,) involving a constant

number of words takes a constant number of steps by the CPU.

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Each arithmetic or logical operation (+,-,*,/,or, xor,) involving O( log n) bitstake a constant number of steps by the CPU, where nis the number of bits of

input instance.

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Justification for the extension

Question: How many bits are needed to access an input item from RAM ?

Answer:At least log n.(kbits can be used to create at most different addresses)

Current-state-of-the-art computers:

RAM of size 4GB

Hence 32 bits to address any item in RAM.

Support for 64-bit arithmeticAbility to perform arithmetic/logical operations on any two 64-bit numbers.

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Algorithm Paradigms

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Algorithm Paradigm

Motivation:

Many problems whose algorithms are based on a common approach.

A need of a systematic study of the characteristics of such widely used

approaches.

Algorithm Paradigms:

Divide and Conquer

Greedy Strategy

Dynamic Programming Local Search

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Divide and Conquer paradigm for

Algorithm Design

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Example 1

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Sorting

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A Homework given in 5thLecture

Merging two sorted arrays:

Given two sorted arrays A andBstoring nelements each, Design an O(n) timealgorithm to output a sorted array Ccontaining all elements of A andB.

Example: If A={1,5,17,19} B={4,7,9,13}, then output is

C={1,4,5,7,9,13,17,19}.

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BA

Mergingtwo sorted arrays A and B

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1 5 139717 19 4

C 5 7 9 131 4

17 19

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Pesudo-code for Merging two sorted arrays

Merge(A[0..n-1],B[0..m-1])

{

i0; j0; k0;

While(i

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Divide and Conquer based sorting algorithm

MSort(A,i,j)

{ If(i

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Analysis of Merge Sort

Time complexity:

If n= 1,

T(n) = cfor some constant c

If n> 1,T(n)= cn+ 2T(n/2)

= cn+ cn+ T(n/)

= cn+ cn+ cn+ T(n/)

= cn+ . (Log n terms) .+ cn

= O(nlog n)

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Analysis of Merge Sort

Correctness:

Question: What is to be proved ?

Answer: MSort(A,i,j) sorts the subarray A[i..j]

Question: How to prove ?

Answer:

By induction on the length (j-i+1) of the subarray.

Use correctness of the algorithm Merge.

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Example 2

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Faster algorithm for multiplying two

integers

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Addition is fasterthan multiplication

Given:any two -bit numbers Xand Y

Question: how many bit-operationsare required to compute X+Y?

Answer: O()

Question: how many bit-operationsare required to compute X*?

Answer: O() [left shift the number Xby places, (do it carefully)]

Question: how many bit-operationsare required to compute X*Y?Answer: O(2) (why ??)

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Can we compute

X*Y faster ??

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PursuingDivide and Conquer approach

Question: how to express X*Y in terms of multiplication/addition of{A,B,C,D} ?

Hint:First Express X and Y in terms of {A,B,C,D}.X = A* /+ B and Y = C* /+ D

Hence

X*Y =(A*C)* ?? + (A*D + B*C)* ?? + B*D

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X

Y

MSB LSB

n-1 n-2......n/2.............................2 1 0

A B

C D

/

4 multiplications

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PursuingDivide and Conquer approach

X*Y =(A*C)* ?? + (A*D + B*C)* ?? + B*D

LetT(n) : time complexity of multiplying X and Y using the above equation.

T(n) = cn+ 4T(n/2) for some constant c

= cn+ 2cn+ T(n/)

= cn+ 2cn+ cn +T(n/)

=cn+ 2cn+ cn + 8cn + +

= cn+ 2cn+ cn + 8cn + +

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X

Y

MSB LSB

n-1 n-2......n/2.............................2 1 0

A B

C D

/

O() time algo

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PursuingDivide and Conquer approach

X*Y =(A*C)* ?? + (A*D + B*C)* ?? + B*D

Observation:A*D+ B*C = (A+B)*(C+D)A*CB*D

Question:How many multiplications do we need nowto compute X*Y ?Answer: 3multiplications :

A*C

B*D

(A+B)*(C+D) . 22

X

Y

MSB LSB

n-1 n-2......n/2.............................2 1 0

A B

C D

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PursuingDivide and Conquer approach

X*Y =(A*C)* ?? + ((A+B)*(C+D) - A*CB*D)* ?? + B*D

LetT(n) : time complexity of the new algo for multiplying two n-bit numbers

T(n) = cn+ 3T(n/2) for some constant c= cn+ 3/2cn+ T(n/)

= cn+ 3/2cn+ (9 )cn + +

= O() = O(.)

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X

Y

MSB LSB

n-1 n-2......n/2.............................2 1 0

A B

C D

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Conclusion

Theorem:There is a divide and conquer based algorithm formultiplying any two n-bit numbers inO(.) time(bit operations).

Note:The fastest algorithm for this problem runs in a little bit more than O(log )

time. One such algorithm was designed in 2008by Dey, Kurur, Saha, and

Saptharishi (CSE, IIT Kanpur).

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