Moscow 5

Assignment 1

This is a Pi calculator that calculates pi up to n digits using the arctan formula pi=16arctan(1/5)-4arctan(1/239) as the main formula along with another variation of the formula based on the same method.

We will be optimizing the following functions:

The arctan method:

void arctan(int multiplier, int denom, int sign)
{
      INDEXER x;
      LONG remain, temp, divisor, denom2;
      SHORT NotZero = 1;
      INDEXER adv;

      for (x = 0; x < size; x++)
            powers[x] = 0;

      divisor = 1;
      denom2 = (LONG)denom;denom2 *= denom2;
      adv = 0;

      remain = (LONG)multiplier * denom;
      while (NotZero)
      {
            for (x = adv; x < size; x++)
            {
                  temp = (LONG)powers[x] + remain;
                  powers[x] = (SHORT)(temp / denom2);
                  remain = (temp - (denom2 * (LONG)powers[x])) * BASE;
            }

            remain = 0;
            for (x = adv; x < size; x++)
            {
                  temp = (LONG)powers[x] + remain;
                  term[x] = (SHORT)(temp / divisor);
                  remain = (temp - (divisor * (LONG)term[x])) * BASE;
            }
            remain = 0;

            if (sign > 0)
            {
                  LONG carry, sum;

                  carry = 0;
                  for (x = size - 1; x >=0; x--)
                  {
                        sum = (LONG)pi[x] + (LONG)term[x] + carry;
                        carry = 0;
                        if (sum >= BASE)
                        {
                              carry = 1;
                              sum -= BASE;
                        }
                        pi[x] = (SHORT)sum;
                  }
            }
            else
            {
                  LONG borrow, sum;

                  borrow = 0;
                  for (x = size - 1; x >= 0; x--)
                  {
                        sum = (LONG)pi[x] - (LONG)term[x] - borrow;
                        borrow = 0;
                        if (sum < 0)
                        {
                              borrow = 1;
                              sum += BASE;
                        }
                        pi[x] = (SHORT)sum;
                  }
            }

            sign = -sign;
            divisor += 2;
            NotZero = 0;
            for (x = adv; x < size; x++)
            {
                  if (powers[x])
                  {
                        NotZero = 1;
                        break;
                  }
            }

            if (NotZero)
            {
                  while (powers[adv] == 0)
                        adv++;
            }
            /* We can skip ones that are already 0 */
      }
}