Fastest Way to Multiply N Continuous Numbers in Python

Fastest Way to multiply two Numbers

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Two numbers are given as a binary string, our task is to find the result of multiplication for those numbers in a faster and efficient way.

Using the Divide and Conquer strategy, we can solve the problem, in a very efficient manner. We will split the numbers into two halves.

let Xleft and Xright are two parts of first number X, and Yleft, Yright are two parts of second number Y. So the product;

To make it simple, we can perform this operation

Input and Output

Input: Two binary numbers: 1101 and 0111 Output: The result is: 91

Algorithm

addBitString(num1, num2)

Input:Two numbers to add.

Output: The result after addition.

Begin    adjust num1 and num2 lengths    length := length of num1    carry := 0     for i := length -1 down to 0, do       num1Bit := num1[i]       num2Bit := num2[i]       sum := num1Bit XOR num2Bit XOR carry       finalSum := sum + finalSum       carry := (num1Bit AND num2Bit) OR (num2Bit AND carry) OR (num1Bit AND carry)    done     if carry ≠ 0, then       finalSum := 1 + finalSum    return finalSum End

multiply(num1, num2)

Input: Two numbers to multiply.

Output: The result after multiplication.

Begin    adjust num1 and num2 lengths    length := length of num1    if n = 0, then       return 0    if n = 1, then       return (num1[0] * num2[0])    firstHalf := n/2    secondHalf := (n - firstHalf)     n1Left := substring of (0 to firstHalf) from num1    n1Right := substring of (firstHalf to secondHalf) from num1    n2Left := substring of (0 to firstHalf) from num2    n2Right := substring of (firstHalf to secondHalf) from num2     p1 := multiply(n1Left, n2Left)    p2 := multiply(n1Right, n2Right)     add1 := addBitString(n1Left, n1Right)    add2 := addBitString(n2Left, n2Right)    p3 := multiply(add1, add2)     mask1 := shift 1 to left for 2*secondHalf bits    mask2 := shift 1 to left for secondHalf bits    return P1*mask2 + (p3 – p1 – p2)*mask2 + p2 End

Example

#include<iostream> using namespace std;  int lengthAdjust(string &num1, string &num2) {    //adjust length of binary string and send length of string    int len1 = num1.size();    int len2 = num2.size();     if (len1 < len2) {       for (int i = 0 ; i < len2 - len1 ; i++)          num1 = '0' + num1; //add 0 before the first string    } else if (len1 > len2) {       for (int i = 0 ; i < len1 - len2 ; i++)          num2 = '0' + num2; //add 0 before the second string    }    return num1.size(); }  string addBitStrings(string num1, string num2) {    string finalSum;     int length = lengthAdjust(num1, num2);    //adjust and update number lengths and store length    int carry = 0;     // Initialize carry     for (int i = length-1 ; i >= 0 ; i--) {       int num1Bit = num1[i] - '0';       int num2Bit = num2[i] - '0';        int sum = (num1Bit ^ num2Bit ^ carry)+'0';    //we know sum = A XOR B XOR C        finalSum = (char)sum + finalSum;       //the carry = (A AND B) OR (B AND C) OR (C AND A)       carry = (num1Bit&num2Bit) | (num2Bit&carry) | (num1Bit&carry);    }     if (carry)   //when carry is present       finalSum = '1' + finalSum; //add carry as MSb    return finalSum; }  long int multiply(string num1, string num2) {    int n = lengthAdjust(num1, num2);    //find length after adjusting them    if (n == 0)    //when there is 0 length string, return 0       return 0;    if (n == 1)       return (num1[0] - '0')*(num2[0] - '0');    //perform single bit muliplication     int firstHalf = n/2;   // First half range    int secondHalf = (n-firstHalf);    // Second half range     string num1Left = num1.substr(0, firstHalf);    //first half of number 1    string num1Right = num1.substr(firstHalf, secondHalf);    //second half of number 1    string num2Left = num2.substr(0, firstHalf);    string num2Right = num2.substr(firstHalf, secondHalf);     // find left right multiplication, and multiply after adding left and right part    long int P1 = multiply(num1Left, num2Left);    long int P2 = multiply(num1Right, num2Right);    long int P3 = multiply(addBitStrings(num1Left, num1Right), addBitStrings(num2Left, num2Right));     return P1*(1<<(2*secondHalf)) + (P3 - P1 - P2)*(1<<secondHalf) + P2; }  int main() {    string num1, num2;    cout << "Enter First number in Binary: "; cin >>num1;    cout << "Enter Second number in Binary: "; cin >>num2;    cout << "The result is: " << multiply(num1, num2); }

Output

Enter First number in Binary: 1101 Enter Second number in Binary: 0111 The result is: 91

Updated on 17-Jun-2020 09:35:14

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