浮点平方根算法 [英] Floating point square root algorithms

问题描述

我想知道如果任何人都可以向我提供的浮点平方根算法的一些例子，可以利用硬件除法。

I am wondering if anybody can provide me with some examples of floating point square root algorithms which can utilize a hardware divider.

多余的细节： 我有一个浮点单元我开发了具有硬件浮点IEEE-754 32位乘法器，加法器，和分频器。我已经实现用牛顿迭代法只用乘法和加法/减法平方根，但现在我想，如果我有一个硬件除法提供给我来比较平方根的吞吐量。

Extra details: I have a floating point unit I am developing which has a hardware floating-point IEEE-754 32-bit multiplier, adder, and divider. I already implemented square root using the Newton-Raphson method using only multiplication and addition/subtraction, but now I want to compare the throughput of square root if I have a hardware divider available to me.

1特定输入是难以计算准确是0x7F7FFFFF的平方根（3.4028234663852886E38）。

1 particular input that is difficult to compute accurately is the square root of 0x7F7FFFFF (3.4028234663852886E38).

推荐答案

解决方案通过@tmyklebu提供肯定会出现，以满足您的要求。

Solution provide by @tmyklebu certainly appears to meet your requirements.

r = input value
s(0) = initial estimate of sqrt(r).  Example: r with its exponent halved.
s(n) = sqrt(r)

S＆其中; - （S + R / S）/ 2

s <- (s + r/s)/2

据二次收敛，执行请求的鸿沟。 N = 3或4应该这样做了32位浮点。

It has quadratic convergence, performs the requested divide. N = 3 or 4 should do it for 32 bit float.

[每OP要求编辑]

// Initial estimate
static double S0(double R) {
  double OneOverRoot2 = 0.70710678118654752440084436210485;
  double Root2        = 1.4142135623730950488016887242097;
  int Expo;
  // Break R into mantissa and exponent parts.
  double Mantissa = frexp(R, &Expo);
  int j;
  printf("S0 %le %d %le\n", Mantissa, Expo, frexp(sqrt(R), &j));
  // If exponent is odd ...
  if (Expo & 1) {
    // Pretend the mantissa [0.5 ... 1.0) is multiplied by 2 as Expo is odd, 
    //   so it now has the value [1.0 ... 2.0)
    // Estimate the sqrt(mantissa) as [1.0 ... sqrt(2))
    // IOW: linearly map (0.5 ... 1.0) to (1.0 ... sqrt(2))
    Mantissa = (Root2 - 1.0)/(1.0 - 0.5)*(Mantissa - 0.5) + 1.0;
  }
  else {
    // The mantissa is in range [0.5 ... 1.0)
    // Estimate the sqrt(mantissa) as [1/sqrt(2) ... 1.0)
    // IOW: linearly map (0.5 ... 1.0) to (1/sqrt(2) ... 1.0)
    Mantissa = (1.0 - OneOverRoot2)/(1.0 - 0.5)*(Mantissa - 0.5) + OneOverRoot2;
  }
  // Form initial estimate by using the above mantissa estimate and exponent/2
  return ldexp(Mantissa, Expo/2);
}

// S = (S + R/S)/2 method
double Sqrt(double R) {
  double S = S0(R);
  int i = 5;  // May be reduced to 3 or 4 for double and 2 for float
  do {
    printf("S  %u %le %le\n", 5-i, S, (S-sqrt(R))/sqrt(R));
    S = (S + R/S)/2;
  } while (--i);
  return S;
}

void STest(double x) {
  printf("T  %le %le %le\n", x, Sqrt(x), sqrt(x));
}

int main(void) {
  STest(612000000000.0);
  return 0;
}

---

经过3次迭代的双重收敛。

---

Converges after 3 iterations for double.

S0 5.566108e-01 40 7.460635e-01
秒0 7.762279e + 05 -7.767318e-03
第1条7.823281e + 05 3.040175e-05
S 2 7.823043e + 05 4.621193e-10
第3 7.823043e + 05 0.000000e + 00
S 4 7.823043e + 05 0.000000e + 00
牛逼6.120000e + 11 7.823043e + 05 7.823043e + 05

S0 5.566108e-01 40 7.460635e-01
S 0 7.762279e+05 -7.767318e-03
S 1 7.823281e+05 3.040175e-05
S 2 7.823043e+05 4.621193e-10
S 3 7.823043e+05 0.000000e+00
S 4 7.823043e+05 0.000000e+00
T 6.120000e+11 7.823043e+05 7.823043e+05

这篇关于浮点平方根算法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！