牛顿·拉普逊和塞——谁能给我解释一下这三条线吗-NewtonRaphsonwithSSE2-cansomeoneexplainmethese3lines

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Given the Newton iteration , it should be quite straight forward to see this in the source code.

考虑到牛顿迭代，在源代码中看到这一点应该是非常直接的。

 __m128 nr   = _mm_rsqrt_ps( x );                  // The initial approximation y_0
 __m128 muls = _mm_mul_ps( _mm_mul_ps( x, nr ), nr ); // muls = x*nr*nr == x(y_n)^2
 result = _mm_mul_ps(
               _mm_sub_ps( three, muls )    // this is 3.0 - mul;
   /*multiplied by */ __mm_mul_ps(half,nr)  // y_0 / 2 or y_0 * 0.5
 );

And to be precise, this algorithm is for the inverse square root.

准确地说，这个算法是求平方根的倒数。

Note that this still doesn't give fully a fully accurate result. rsqrtps with a NR iteration gives almost 23 bits of accuracy, vs. sqrtps's 24 bits with correct rounding for the last bit.

注意，这仍然不能给出完全准确的结果。具有NR迭代的rsqrtps提供了近23位的精度，而sqrtps的24位具有最后一位的正确四舍五入。

The limited accuracy is an issue if you want to truncate the result to integer. (int)4.99999 is 4. Also, watch out for the x == 0.0 case if using sqrt(x) ~= x * sqrt(x), because 0 * +Inf = NaN.

如果要将结果截断为整数，那么精度有限是一个问题。(int)4.99999是4。另外，如果使用sqrt(x) ~= x *√(x)，要注意x = 0.0，因为0 * +Inf = NaN。

3  

To compute the inverse square root of a, Newton's method is applied to the equation 0=f(x)=a-x^(-2) with derivative f'(x)=2*x^(-3) and thus the iteration step

计算逆平方根,牛顿法应用于方程0 = f(x)= ax)^(2)与导数f(x)= x ^ 2 *(3),因此迭代步骤

N(x) = x - f(x)/f'(x) = x - (a*x^3-x)/2 
     = x/2 * (3 - a*x^2)

This division-free method has -- in contrast to the globally converging Heron's method -- a limited region of convergence, so you need an already good approximation of the inverse square root to get a better approximation.

这种无分割的方法——与全局收敛的Heron方法相反——有一个有限的收敛区域，所以你需要一个很好的逆平方根的近似来得到更好的近似。