为Goldschmidt部门挑选良好的初步估计

Question

我正在使用Goldschmidt部门在Q22.10中计算固定点倒数,用于我在ARM上的软件光栅化器.

这是通过将分子设置为1来完成的,即分子成为第一次迭代的标量.说实话,我在这里盲目地遵循维基百科算法.文章说如果分母在半开放范围内缩放(0.5,1.0),那么一个好的初步估计可以仅基于分母:让F为估计的标量,D为分母,则F = 2 - D.

但是当这样做时,我会失去很多精确度.如果我想找到512.00002f的倒数.为了缩小数字,我在分数部分失去了10位精度,它被移出.所以,我的问题是:

• 有没有办法选择一个不需要标准化的更好估计？为什么？为什么不？为什么这是或不可能的数学证明将是伟大的.
• 此外,是否可以预先计算第一个估计值,以便系列收敛得更快？现在,它平均在第4次迭代后收敛.在ARM上,这是大约50个周期的最坏情况,并且没有考虑到clz/bsr的仿真,也没有考虑内存查找.如果可能的话,我想知道这样做是否会增加错误,以及增加错误.

这是我的测试用例.注:该软件实现clz上线13从我的岗位在这里.如果需要,可以用内在替换它.clz应返回前导零的数量,并返回值为0的32.

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

const unsigned int BASE = 22ULL;

static unsigned int divfp(unsigned int val, int* iter)
{
  /* Numerator, 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 numerator 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;
}

Answer 1

我无法忍受花一个小时解决你的问题......

该算法在Jean-Michel Muller的"Arithmetique des ordinateurs"第5.5.2节(法语)中描述.它实际上是Newton迭代的一个特例,以1为起点.本书给出了算法计算N/D的简单公式,其中D归一化范围[1/2,1 [:

e = 1 - D
Q = N
repeat K times:
  Q = Q * (1+e)
  e = e*e

每次迭代时正确位的数量加倍.在32位的情况下,4次迭代就足够了.您也可以迭代直到e变得太小而无法修改Q.

使用归一化是因为它提供了结果中的最大有效位数.当输入处于已知范围内时,计算错误和迭代次数也更容易.

一旦您的输入值被标准化,您就不需要使用BASE的值,直到您有反向.您只需要在范围0x80000000到0xFFFFFFFF范围内归一化的32位数X,并计算Y = 2 ^ 64/X(Y最多为2 ^ 33)的近似值.

可以为您的Q22.10表示实现此简化算法,如下所示:

// Fixed point inversion
// EB Apr 2010

#include <math.h>
#include <stdio.h>

// Number X is represented by integer I: X = I/2^BASE.
// We have (32-BASE) bits in integral part, and BASE bits in fractional part
#define BASE 22
typedef unsigned int uint32;
typedef unsigned long long int uint64;

// Convert FP to/from double (debug)
double toDouble(uint32 fp) { return fp/(double)(1<<BASE); }
uint32 toFP(double x) { return (int)floor(0.5+x*(1<<BASE)); }

// Return inverse of FP
uint32 inverse(uint32 fp)
{
  if (fp == 0) return (uint32)-1; // invalid

  // Shift FP to have the most significant bit set
  int shl = 0; // normalization shift
  uint32 nfp = fp; // normalized FP
  while ( (nfp & 0x80000000) == 0 ) { nfp <<= 1; shl++; } // use "clz" instead

  uint64 q = 0x100000000ULL; // 2^32
  uint64 e = 0x100000000ULL - (uint64)nfp; // 2^32-NFP
  int i;
  for (i=0;i<4;i++) // iterate
    {
      // Both multiplications are actually
      // 32x32 bits truncated to the 32 high bits
      q += (q*e)>>(uint64)32;
      e = (e*e)>>(uint64)32;
      printf("Q=0x%llx E=0x%llx\n",q,e);
    }
  // Here, (Q/2^32) is the inverse of (NFP/2^32).
  // We have 2^31<=NFP<2^32 and 2^32<Q<=2^33
  return (uint32)(q>>(64-2*BASE-shl));
}

int main()
{
  double x = 1.234567;
  uint32 xx = toFP(x);
  uint32 yy = inverse(xx);
  double y = toDouble(yy);

  printf("X=%f Y=%f X*Y=%f\n",x,y,x*y);
  printf("XX=0x%08x YY=0x%08x XX*YY=0x%016llx\n",xx,yy,(uint64)xx*(uint64)yy);
}

如代码中所述,乘法不是完整的32x32-> 64位.E将变得越来越小,最初适合32位.Q将始终为34位.我们只采用高32位的产品.

推导64-2*BASE-shl是留给读者的练习:-).如果它变为0或负数,则结果不可表示(输入值太小).

编辑.作为我的评论的后续内容,这是第二个版本,在Q上隐含第32位.E和Q现在都存储在32位:

uint32 inverse2(uint32 fp)
{
  if (fp == 0) return (uint32)-1; // invalid

  // Shift FP to have the most significant bit set
  int shl = 0; // normalization shift for FP
  uint32 nfp = fp; // normalized FP
  while ( (nfp & 0x80000000) == 0 ) { nfp <<= 1; shl++; } // use "clz" instead
  int shr = 64-2*BASE-shl; // normalization shift for Q
  if (shr <= 0) return (uint32)-1; // overflow

  uint64 e = 1 + (0xFFFFFFFF ^ nfp); // 2^32-NFP, max value is 2^31
  uint64 q = e; // 2^32 implicit bit, and implicit first iteration
  int i;
  for (i=0;i<3;i++) // iterate
    {
      e = (e*e)>>(uint64)32;
      q += e + ((q*e)>>(uint64)32);
    }
  return (uint32)(q>>shr) + (1<<(32-shr)); // insert implicit bit
}