Say we are given two polynomials of degree n - 1 as their arrays of co-efficient

ID: 3693626 • Letter: S

Question

Say we are given two polynomials of degree n - 1 as their arrays of co-efficients, a_0..a_n-1 and b_0,.. b_n-1, so that a(x) = sigma^i=n-1_i=0 a_i x^i and b(x) = sigma^i=n-1_i=0 b_i x^i and we want to compute the description of a(x)b(x), their product. For all parts, you are not required to prove correctness of your algorithms, but are required to give a time analysis in O form. Give the straight-forward, iterative algorithm for this problem, proof. Give a simple divide-and-conquer algorithm for this problem. Use the Karatsuba method to give an improved divide-and-conquer algorithm for this problem..

Explanation / Answer

The product XY can be written as following.

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 for 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.

So the final value of XY becomes

With above trick, the recurrence becomes T(n) = 3T(n/2) + O(n) and solution of this recurrence is O(n1.59).

What if the lengths of input strings are different and are not even? To handle the different length case, we append 0’s in the beginning. To handle odd length, we put floor(n/2) bits in left half and ceil(n/2) bits in right half. So the expression for XY changes to following.

The above algorithm is called Karatsuba algorithm and it can be used for any base.

Following is C++ implementation of above algorithm.

// C++ implementation of Karatsuba algorithm for bit string multiplication.

#include<iostream>

#include<stdio.h>

using namespace std;

// FOLLOWING TWO FUNCTIONS ARE COPIED FROM http://goo.gl/q0OhZ

// Helper method: given two unequal sized bit strings, converts them to

// same length by adding leading 0s in the smaller string. Returns the

// the new length

int makeEqualLength(string &str1, string &str2)

{

int len1 = str1.size();

int len2 = str2.size();

if (len1 < len2)

{

for (int i = 0 ; i < len2 - len1 ; i++)

str1 = '0' + str1;

return len2;

}

else if (len1 > len2)

{

for (int i = 0 ; i < len1 - len2 ; i++)

str2 = '0' + str2;

}

return len1; // If len1 >= len2

}

// The main function that adds two bit sequences and returns the addition

string addBitStrings( string first, string second )

{

string result; // To store the sum bits

// make the lengths same before adding

int length = makeEqualLength(first, second);

int carry = 0; // Initialize carry

// Add all bits one by one

for (int i = length-1 ; i >= 0 ; i--)

{

int firstBit = first.at(i) - '0';

int secondBit = second.at(i) - '0';

// boolean expression for sum of 3 bits

int sum = (firstBit ^ secondBit ^ carry)+'0';

result = (char)sum + result;

// boolean expression for 3-bit addition

carry = (firstBit&secondBit) | (secondBit&carry) | (firstBit&carry);

}

// if overflow, then add a leading 1

if (carry) result = '1' + result;

return result;

}

// A utility function to multiply single bits of strings a and b

int multiplyiSingleBit(string a, string b)

{ return (a[0] - '0')*(b[0] - '0'); }

// The main function that multiplies two bit strings X and Y and returns

// result as long integer

long int multiply(string X, string Y)

{

// Find the maximum of lengths of x and Y and make length

// of smaller string same as that of larger string

int n = makeEqualLength(X, Y);

// Base cases

if (n == 0) return 0;

if (n == 1) return multiplyiSingleBit(X, Y);

int fh = n/2; // First half of string, floor(n/2)

int sh = (n-fh); // Second half of string, ceil(n/2)

// Find the first half and second half of first string.

// Refer http://goo.gl/lLmgn for substr method

string Xl = X.substr(0, fh);

string Xr = X.substr(fh, sh);

// Find the first half and second half of second string

string Yl = Y.substr(0, fh);

string Yr = Y.substr(fh, sh);

// Recursively calculate the three products of inputs of size n/2

long int P1 = multiply(Xl, Yl);

long int P2 = multiply(Xr, Yr);

long int P3 = multiply(addBitStrings(Xl, Xr), addBitStrings(Yl, Yr));

// Combine the three products to get the final result.

return P1*(1<<(2*sh)) + (P3 - P1 - P2)*(1<<sh) + P2;

}

// Driver program to test aboev functions

int main()

{

printf ("%ld ", multiply("1100", "1010"));

printf ("%ld ", multiply("110", "1010"));

printf ("%ld ", multiply("11", "1010"));

printf ("%ld ", multiply("1", "1010"));

printf ("%ld ", multiply("0", "1010"));

printf ("%ld ", multiply("111", "111"));

printf ("%ld ", multiply("11", "11"));

}