Picking good first estimates for Goldschmidt division

Posted by Mads Elvheim on Stack Overflow See other posts from Stack Overflow or by Mads Elvheim
Published on 2010-04-18T08:01:13Z Indexed on 2010/04/18 8:03 UTC
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I'm calculating fixedpoint reciprocals in Q22.10 with Goldschmidt division for use in my software rasterizer on ARM.

This is done by just setting the nominator to 1, i.e the nominator becomes the scalar on the first iteration. To be honest, I'm kind of following the wikipedia algorithm blindly here. The article says that if the denominator is scaled in the half-open range (0.5, 1.0], a good first estimate can be based on the denominator alone: Let F be the estimated scalar and D be the denominator, then F = 2 - D.

But when doing this, I lose a lot of precision. Say if I want to find the reciprocal of 512.00002f. In order to scale the number down, I lose 10 bits of precision in the fraction part, which is shifted out. So, my questions are:

  • Is there a way to pick a better estimate which does not require normalization?
  • Also, is it possible to pre-calculate the first estimates so the series converges faster? Right now, it converges after the 4th iteration on average. On ARM this is about ~50 cycles worst case, and that's not taking emulation of clz/bsr into account, nor memory lookups.

Here is my testcase. Note: The software implementation of clz on line 13 is from my post here. You can replace it with an intrinsic if you want.

#include <stdio.h>
#include <stdint.h>

const unsigned int BASE = 22ULL;

static unsigned int divfp(unsigned int val, int* iter)
{
  /* Nominator, denominator, estimate scalar and previous denominator */
  unsigned long long N,D,F, DPREV;
  int bitpos;

  *iter = 1;
  D = val;
  /* Get the shift amount + is right-shift, - is left-shift. */
  bitpos = 31 - clz(val) - BASE;
  /* Normalize into the half-range (0.5, 1.0] */
  if(0 < bitpos)
    D >>= bitpos;
  else
    D <<= (-bitpos);

  /* (FNi / FDi) == (FN(i+1) / FD(i+1)) */
  /* F = 2 - D */
  F = (2ULL<<BASE) - D;
  /* N = F for the first iteration, because the nominator is simply 1.
     So don't waste a 64-bit UMULL on a multiply with 1 */
  N = F;
  D = ((unsigned long long)D*F)>>BASE;

  while(1){
    DPREV = D;
    F = (2<<(BASE)) - D;
    D = ((unsigned long long)D*F)>>BASE;
    /* Bail when we get the same value for two denominators in a row.
      This means that the error is too small to make any further progress. */
    if(D == DPREV)
      break;
    N = ((unsigned long long)N*F)>>BASE;
    *iter = *iter + 1;
  }
  if(0 < bitpos)
    N >>= bitpos;
  else
    N <<= (-bitpos);
  return N;
}


int main(int argc, char* argv[])
{
  double fv, fa;
  int iter;
  unsigned int D, result;

  sscanf(argv[1], "%lf", &fv);

  D = fv*(double)(1<<BASE);
  result = divfp(D, &iter); 

  fa = (double)result / (double)(1UL << BASE);
  printf("Value: %8.8lf 1/value: %8.8lf FP value: 0x%.8X\n", fv, fa, result);
  printf("iteration: %d\n",iter);

  return 0;
}

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