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The Karatsuba algorithm allows multiplication of large numbers using three multiplications of smaller numbers having half the digits, plus some additions and shifts. It uses divide and conquer by dividing the numbers into halves. The product is written as the sum of multiple terms involving the halves multiplied and added in different ways. A tricky expression combines two middle terms into one multiplication, reducing the recurrence relation from four to three subproblems, resulting in a more efficient O(n1.59) time complexity compared to O(n2) for standard multiplication.

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65 views10 pages

Daa 5 PDF

The Karatsuba algorithm allows multiplication of large numbers using three multiplications of smaller numbers having half the digits, plus some additions and shifts. It uses divide and conquer by dividing the numbers into halves. The product is written as the sum of multiple terms involving the halves multiplied and added in different ways. A tricky expression combines two middle terms into one multiplication, reducing the recurrence relation from four to three subproblems, resulting in a more efficient O(n1.59) time complexity compared to O(n2) for standard multiplication.

Uploaded by

Kk
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Karatsuba Algorithm

•The basic step of Karatsuba's algorithm


is a formula that allows one to compute
the product of two large
numbers x and y using
•three multiplications of smaller
numbers, each with about half as many
digits as x or y,
•plus some additions and digit shifts.
Karatsuba Algorithm
Karatsuba Algorithm
• Using Divide and Conquer, we can multiply
two integers in less time complexity.
• We divide the given numbers in two
halves.
• X = Xl * 2 ^ n/2 + Xr
[Xl and Xr contain leftmost and rightmost
n/2 bits of X]
• Y = Yl * 2 ^ n/2 + Yr
[Yl and Yr contain leftmost and rightmost
n/2 bits of Y]
Karatsuba Algorithm
• If x = 1234
• y = 4321
• how do we represent them using 2 digit
numbers?
Karatsuba Algorithm
• The product XY can be written as following.
XY = (Xl* 2^n/2 + Xr)(Yl*2^n/2 + Yr)
= 2^n XlYl + 2^n/2 (XlYr + XrYl) + XrYr
• There are four multiplications of size n/2,
• so we basically divided the problem of size n
into four sub-problems of size n/2.
• But that doesn’t help because solution of
recurrence
• T(n) = 4T(n/2) + O(n) is O(n^2)
Karatsuba Algorithm
• The tricky part of this algorithm is to
change the middle two terms to some other
form so that only one extra multiplication
would be sufficient.
• The following is tricky expression for
middle two terms.
• XlYr + XrYl = (Xl + Xr)(Yl + Yr) - XlYl- XrYr
• So the final value of XY becomes
• XY = 2^n XlYl + 2^n/2 * [(Xl + Xr)(Yl + Yr) -
XlYl - XrYr] + XrYr
Karatsuba Algorithm
• With above trick, the recurrence
becomes
• T(n) = 3T(n/2) + O(n)
• and solution of this recurrence is
O(n ^ log23) = O(n ^1.59).
Karatsuba Algorithm
Karatsuba Algorithm
Karatsuba Algorithm: Time
complexity

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