Explain the Karatsuba-Ofman algorithm to multiply 2 n-bit integers. Derive a recurrence relation for its complexity and solve this recurrence relation.
Using Divide and Conquer, we can multiply two integers in less time complexity. We divide the given numbers in two halves. Let the given numbers be X and Y.
For simplicity let us assume that n is even
X = Xl*2n/2 + Xr [Xl and Xr contain leftmost and rightmost n/2 bits of X] Y = Yl*2n/2 + Yr [Yl and Yr contain leftmost and rightmost n/2 bits of Y]
The product XY can be written as following.
XY = (Xl*2n/2 + Xr)(Yl*2n/2 + Yr) = 2n XlYl + 2n/2(XlYr + XrYl) + XrYr
If we take a look at the above formula, 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). 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 = 2n XlYl + 2n/2 * [(Xl + Xr)(Yl + Yr) - XlYl - XrYr] + XrYr
With above trick, the recurrence becomes T(n) = 3T(n/2) + O(n) and solution of this recurrence is O(n1.59).
Explain the Karatsuba-Ofman algorithm to multiply 2 n-bit integers. Derive a recurrence relation for its complexity and solve this recurrence relation.
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Algorithm Question: Problem 3. Solve the recurrence relation T(n) = 2T(n/2) + lg n, T(1) 0.
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Consider the following recursive algorithm for computing the sum of the first n cubes: S(n) = 13 + 23 + … + n3. (a) Set up a recurrence relation for the number of multiplications made by this algorithm. (b) Provide an initial condition for the recurrence relation you develop at the question (a). (c) Solve the recurrence relation of the question (a) and present the time complexity as described at the question number 1. Algorithm S n) Input: A positive...