Karatsuba Algorithm for fast Multiplication of Large Decimal Numbers represented as Strings - GeeksforGeeks

15 Oct, 2021
Given two numeric strings A and B, the task is to find the product of the two numeric strings efficiently.
Example:
Input: A = 5678, B = 1234Output: 7006652
Input: A = 74638463789, B = 35284567382Output: 2633585904851937530398
Approach: The given problem can be solved using Karastuba’s Algorithm for Fast Multiplication, the idea is to append zeroes in front of the integers such that both the integers have an equal and even number of digits n. Thereafter, divide the numbers in the following way:
A =  Al * 10n/2 + Ar    [Al and Ar contain leftmost and rightmost n/2 digits of A] B =  Bl * 10n/2 + Br    [Bl and Br contain leftmost and rightmost n/2 digits of B]
Therefore, the product A * B can also be represented as follows:
A * B = (Al * 10n/2 + Ar) * (Bl * 10n/2 + Br)=> A * B = 10n*Al*Bl + 10n/2*(Al*Br + Bl*Ar) + Ar*Br=> A * B = 10n*Al*Bl + 10n/2*((Al + Ar)*(Bl + Br) – Al*Bl – Ar*Br) + Ar*Br  [since Al*Br + Bl*Ar = (Al + Ar)*(Bl + Br) – Al*Bl – Ar*Br]
Notice that the above expression only requires three multiplications Al*Bl, Ar*Br, and (Al + Ar)*(Bl + Br), instead of the standard four. Hence, the recurrence becomes T(n) = 3T(n/2) + O(n) and solution of this recurrence is O(n1.59). This idea has been discussed more thoroughly in this article. Therefore the above problem can be solved using the steps below:
Create a function findSum(), which finds the sum of two large numbers represented as strings. Similarly, create a function findDiff(), which finds the difference of two large numbers represented as strings.
In the recursive function multiply(A, B), which multiplies the numbers using Karatsuba’s Algorithm, firstly append zeroes in front of A and B to make their digit count equal and even.
Then calculate the values of Al, Ar, Bl, and Br as defined above and recursively find the values of multiply(Al, Bl), multiply(Ar, Br), and multiply((Al + Ar), (Bl + Br)) and plug their values in the above-derived equation.
Print the answer to the equation after removing the leading zeroes from it.
Below is the implementation of the above approach:
C++
// C++ progrram for the above approach  #include <bits/stdc++.h>using namespace std;  // Function to find the sum of larger// numbers represented as a stringstring findSum(string str1, string str2){    // Before proceeding further, make    // sure length of str2 is larger    if (str1.length() > str2.length())        swap(str1, str2);      // Stores the result    string str = "";      // Calculate length of both string    int n1 = str1.length();    int n2 = str2.length();      // Reverse both of strings    reverse(str1.begin(), str1.end());    reverse(str2.begin(), str2.end());      int carry = 0;    for (int i = 0; i < n1; i++) {          // Find the sum of the current        // digits and carry        int sum            = ((str1[i] - '0')               + (str2[i] - '0')               + carry);        str.push_back(sum % 10 + '0');          // Calculate carry for next step        carry = sum / 10;    }      // Add remaining digits of larger number    for (int i = n1; i < n2; i++) {        int sum = ((str2[i] - '0') + carry);        str.push_back(sum % 10 + '0');        carry = sum / 10;    }      // Add remaining carry    if (carry)        str.push_back(carry + '0');      // Reverse resultant string    reverse(str.begin(), str.end());      return str;}  // Function to find difference of larger// numbers represented as stringsstring findDiff(string str1, string str2){    // Stores the result of difference    string str = "";      // Calculate length of both string    int n1 = str1.length(), n2 = str2.length();      // Reverse both of strings    reverse(str1.begin(), str1.end());    reverse(str2.begin(), str2.end());      int carry = 0;      // Run loop till small string length    // and subtract digit of str1 to str2    for (int i = 0; i < n2; i++) {          // Compute difference of the        // current digits        int sub            = ((str1[i] - '0')               - (str2[i] - '0')               - carry);          // If subtraction < 0 then add 10        // into sub and take carry as 1        if (sub < 0) {            sub = sub + 10;            carry = 1;        }        else            carry = 0;          str.push_back(sub + '0');    }      // Subtract the remaining digits of    // larger number    for (int i = n2; i < n1; i++) {        int sub = ((str1[i] - '0') - carry);          // If the sub value is -ve,        // then make it positive        if (sub < 0) {            sub = sub + 10;            carry = 1;        }        else            carry = 0;          str.push_back(sub + '0');    }      // Reverse resultant string    reverse(str.begin(), str.end());      // Return answer    return str;}  // Function to remove all leading 0s// from a given stringstring removeLeadingZeros(string str){    // Regex to remove leading 0s    // from a string    const regex pattern("^0+(?!$)");      // Replaces the matched value    // with given string    str = regex_replace(str, pattern, "");    return str;}  // Function to multiply two numbers// using Karatsuba algorithmstring multiply(string A, string B){    if (A.length() > B.length())        swap(A, B);      // Make both numbers to have    // same digits    int n1 = A.length(), n2 = B.length();    while (n2 > n1) {        A = "0" + A;        n1++;    }      // Base case    if (n1 == 1) {          // If the length of strings is 1,        // then return their product        int ans = stoi(A) * stoi(B);        return to_string(ans);    }      // Add zeros in the beginning of    // the strings when length is odd    if (n1 % 2 == 1) {        n1++;        A = "0" + A;        B = "0" + B;    }      string Al, Ar, Bl, Br;      // Find the values of Al, Ar,    // Bl, and Br.    for (int i = 0; i < n1 / 2; ++i) {        Al += A[i];        Bl += B[i];        Ar += A[n1 / 2 + i];        Br += B[n1 / 2 + i];    }      // Recursively call the function    // to compute smaller product      // Stores the value of Al * Bl    string p = multiply(Al, Bl);      // Stores the value of Ar * Br    string q = multiply(Ar, Br);      // Stores value of ((Al + Ar)*(Bl + Br)    // - Al*Bl - Ar*Br)    string r = findDiff(        multiply(findSum(Al, Ar),                 findSum(Bl, Br)),        findSum(p, q));      // Multiply p by 10^n    for (int i = 0; i < n1; ++i)        p = p + "0";      // Multiply s by 10^(n/2)    for (int i = 0; i < n1 / 2; ++i)        r = r + "0";      // Calculate final answer p + r + s    string ans = findSum(p, findSum(q, r));      // Remove leading zeroes from ans    ans = removeLeadingZeros(ans);      // Return Answer    return ans;}  // Driver Codeint main(){    string A = "74638463789";    string B = "35284567382";      cout << multiply(A, B);      return 0;}
2633585904851937530398

Time Complexity: O(Nlog 3) or O(N1.59), where N is the maximum among the lengths given strings A and B.Auxiliary Space: O(N2)
large-numbers
Algorithms
Divide and Conquer
Mathematical
Strings
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Divide and Conquer
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