How Can I Efficiently Square Large Integers in C Using Integer Arithmetic?-C++-php.cn

How Can I Efficiently Square Large Integers in C Using Integer Arithmetic?

Mary-Kate Olsen

Release： 2024-12-23 07:28:54

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How Can I Efficiently Square Large Integers in C Using Integer Arithmetic?

Fast bignum square computation

Problem:

How do I compute y = x^2 as fast as possible without precision loss using C and integer arithmetics (32bit with Carry)?

Solution:

The problem can be solved using Karatsuba multiplication, which has a complexity of O(N^(log2(3))), where N is the number of digits.

Implementation:

Here is an implementation of Karatsuba multiplication in C :

void karatsuba(int *a, int *b, int n, int *c) {
  if (n <= 1) {
    c[0] = a[0] * b[0];
    return;
  }
  int half = n / 2;
  int *a0 = new int[half];
  int *a1 = new int[half];
  int *b0 = new int[half];
  int *b1 = new int[half];
  for (int i = 0; i < half; i++) {
    a0[i] = a[i];
    a1[i] = a[i + half];
    b0[i] = b[i];
    b1[i] = b[i + half];
  }
  int *c0 = new int[half];
  int *c1 = new int[half];
  int *c2 = new int[n];
  karatsuba(a0, b0, half, c0);
  karatsuba(a1, b1, half, c1);
  for (int i = 0; i < n; i++)
    c2[i] = 0;
  for (int i = 0; i < half; i++)
    for (int j = 0; j < half; j++)
      c2[i + j] += a0[i] * b1[j];
  for (int i = 0; i < half; i++)
    for (int j = 0; j < half; j++)
      c2[i + j + half] += a1[i] * b0[j];
  for (int i = 0; i < n; i++)
    c[i] = c0[i] + c1[i] + c2[i];
  delete[] a0;
  delete[] a1;
  delete[] b0;
  delete[] b1;
  delete[] c0;
  delete[] c1;
  delete[] c2;
}

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This implementation has a complexity of O(N^(log2(3))), which is significantly faster than the naive O(N^2) algorithm.

Conclusion:

Using Karatsuba multiplication, it is possible to compute y = x^2 much faster than using the naive O(N^2) algorithm.

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