我正在寻找固定点 16.16 数字的最佳平方根逆算法。下面的代码是我到目前为止所拥有的（但基本上它取平方根并除以原始数字，我想得到没有除法的平方根倒数）。如果它改变了任何东西，代码将为 armv5te 编译。
uint32_t INVSQRT(uint32_t n)
{
uint64_t op, res, one;
op = ((uint64_t)n<<16);
res = 0;
one = (uint64_t)1 << 46;
while (one > op) one >>= 2;
while (one != 0)
{
if (op >= res + one)
{
op -= (res + one);
res += (one<<1);
}
res >>= 1;
one >>= 2;
}
res<<=16;
res /= n;
return(res);
}
诀窍是将牛顿法应用于问题 x - 1/y^2 = 0。因此，给定 x，使用迭代方案求解 y。
Y_(n+1) = y_n * (3 - x*y_n^2)/2
除以 2 只是移位，或者最坏的情况是乘以 0.5。该方案完全按照要求收敛到 y=1/sqrt(x)，并且根本没有任何真正的除法。
唯一的问题是您需要一个不错的 y 起始值。我记得迭代收敛的估计 y 是有限制的。
ARMv5TE 处理器提供快速整数乘法器和“计数前导零”指令。它们通常还带有中等大小的缓存。基于此，最适合高性能实现的方法似乎是查找表以获取初始近似值，然后进行两次 Newton-Raphson 迭代以获得完全准确的结果。我们可以通过将额外的预计算合并到表中来进一步加快这些迭代中的第一次，这是 40 年前 Cray 计算机使用的一种技术。
下面的函数fxrsqrt()实现了这种方法。r它从参数的倒数平方根的8 位近似值开始a，但不是存储r，而是每个表元素存储 3r（在 32 位条目的低 10 位中）和 r 3（在高 22 位中）的 32 位条目）。这允许将第一次迭代快速计算为 r 1 = 0.5 * (3 * r - a * r 3 )。然后以常规方式计算第二次迭代，即 r 2 = 0.5 * r 1 * (3 - r 1 * (r 1 * a))。
为了能够准确地执行这些计算，无论输入的大小如何，参数a都会在计算开始时进行归一化，本质上将其表示为2.32定点数乘以比例因子 2 scal。在计算结束时，根据公式 1/sqrt(2 2n ) = 2 -n对结果进行非规范化。通过舍入最高有效丢弃位为 1 的结果，提高了准确性，导致几乎所有结果都被正确舍入。详尽的测试报告：results too low: 639 too high: 1454 not correctly rounded: 2093
该代码使用两个辅助函数：__clz()确定非零 32 位参数中前导零位的数量。__umulhi()计算两个无符号 32 位整数的完整 64 位乘积的 32 个最高有效位。这两个函数都应该通过编译器内部函数或使用一些内联汇编来实现。在下面的代码中，我展示了非常适合 ARM CPU 的可移植实现以及适用于 x86 平台的内联汇编版本。在 ARMv5TE 平台上__clz()应该映射到CLZ指令，并且__umulhi()应该映射到UMULL.
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <math.h>
#define USE_OWN_INTRINSICS 1
#if USE_OWN_INTRINSICS
__forceinline int __clz (uint32_t a)
{
int r;
__asm__ ("bsrl %1,%0\n\t" : "=r"(r): "r"(a));
return 31 - r;
}
uint32_t __umulhi (uint32_t a, uint32_t b)
{
uint32_t r;
__asm__ ("movl %1,%%eax\n\tmull %2\n\tmovl %%edx,%0\n\t"
: "=r"(r) : "r"(a), "r"(b) : "eax", "edx");
return r;
}
#else // USE_OWN_INTRINSICS
int __clz (uint32_t a)
{
uint32_t r = 32;
if (a >= 0x00010000) { a >>= 16; r -= 16; }
if (a >= 0x00000100) { a >>= 8; r -= 8; }
if (a >= 0x00000010) { a >>= 4; r -= 4; }
if (a >= 0x00000004) { a >>= 2; r -= 2; }
r -= a - (a & (a >> 1));
return r;
}
uint32_t __umulhi (uint32_t a, uint32_t b)
{
return (uint32_t)(((uint64_t)a * b) >> 32);
}
#endif // USE_OWN_INTRINSICS
/*
* For each sub-interval in [1, 4), use an 8-bit approximation r to reciprocal
* square root. To speed up subsequent Newton-Raphson iterations, each entry in
* the table combines two pieces of information: The least-significant 10 bits
* store 3*r, the most-significant 22 bits store r**3, rounded from 24 down to
* 22 bits such that accuracy is optimized.
*/
uint32_t rsqrt_tab [96] =
{
0xfa0bdefa, 0xee6af6ee, 0xe5effae5, 0xdaf27ad9,
0xd2eff6d0, 0xc890aec4, 0xc10366bb, 0xb9a71ab2,
0xb4da2eac, 0xadce7ea3, 0xa6f2b29a, 0xa279a694,
0x9beb568b, 0x97a5c685, 0x9163027c, 0x8d4fd276,
0x89501e70, 0x8563da6a, 0x818ac664, 0x7dc4fe5e,
0x7a122258, 0x7671be52, 0x72e44a4c, 0x6f68fa46,
0x6db22a43, 0x6a52623d, 0x67041a37, 0x65639634,
0x622ffe2e, 0x609cba2b, 0x5d837e25, 0x5bfcfe22,
0x58fd461c, 0x57838619, 0x560e1216, 0x53300a10,
0x51c72e0d, 0x50621a0a, 0x4da48204, 0x4c4c2e01,
0x4af789fe, 0x49a689fb, 0x485a11f8, 0x4710f9f5,
0x45cc2df2, 0x448b4def, 0x421505e9, 0x40df5de6,
0x3fadc5e3, 0x3e7fe1e0, 0x3d55c9dd, 0x3d55d9dd,
0x3c2f41da, 0x39edd9d4, 0x39edc1d4, 0x38d281d1,
0x37bae1ce, 0x36a6c1cb, 0x3595d5c8, 0x3488f1c5,
0x3488fdc5, 0x337fbdc2, 0x3279ddbf, 0x317749bc,
0x307831b9, 0x307879b9, 0x2f7d01b6, 0x2e84ddb3,
0x2d9005b0, 0x2d9015b0, 0x2c9ec1ad, 0x2bb0a1aa,
0x2bb0f5aa, 0x2ac615a7, 0x29ded1a4, 0x29dec9a4,
0x28fabda1, 0x2819e99e, 0x2819ed9e, 0x273c3d9b,
0x273c359b, 0x2661dd98, 0x258ad195, 0x258af195,
0x24b71192, 0x24b6b192, 0x23e6058f, 0x2318118c,
0x2318718c, 0x224da189, 0x224dd989, 0x21860d86,
0x21862586, 0x20c19183, 0x20c1b183, 0x20001580
};
/* This function computes the reciprocal square root of its 16.16 fixed-point
* argument. After normalization of the argument if uses the most significant
* bits of the argument for a table lookup to obtain an initial approximation
* accurate to 8 bits. This is followed by two Newton-Raphson iterations with
* quadratic convergence. Finally, the result is denormalized and some simple
* rounding is applied to maximize accuracy.
*
* To speed up the first NR iteration, for the initial 8-bit approximation r0
* the lookup table supplies 3*r0 along with r0**3. A first iteration computes
* a refined estimate r1 = 1.5 * r0 - x * r0**3. The second iteration computes
* the final result as r2 = 0.5 * r1 * (3 - r1 * (r1 * x)).
*
* The accuracy for all arguments in [0x00000001, 0xffffffff] is as follows:
* 639 results are too small by one ulp, 1454 results are too big by one ulp.
* A total of 2093 results deviate from the correctly rounded result.
*/
uint32_t fxrsqrt (uint32_t a)
{
uint32_t s, r, t, scal;
/* handle special case of zero input */
if (a == 0) return ~a;
/* normalize argument */
scal = __clz (a) & 0xfffffffe;
a = a << scal;
/* initial approximation */
t = rsqrt_tab [(a >> 25) - 32];
/* first NR iteration */
r = (t << 22) - __umulhi (t, a);
/* second NR iteration */
s = __umulhi (r, a);
s = 0x30000000 - __umulhi (r, s);
r = __umulhi (r, s);
/* denormalize and round result */
r = ((r >> (18 - (scal >> 1))) + 1) >> 1;
return r;
}
/* reference implementation, 16.16 reciprocal square root of non-zero argment */
uint32_t ref_fxrsqrt (uint32_t a)
{
double arg = a / 65536.0;
double rsq = sqrt (1.0 / arg);
uint32_t r = (uint32_t)(rsq * 65536.0 + 0.5);
return r;
}
int main (void)
{
uint32_t arg = 0x00000001;
uint32_t res, ref;
uint32_t err, lo = 0, hi = 0;
do {
res = fxrsqrt (arg);
ref = ref_fxrsqrt (arg);
err = 0;
if (res < ref) {
err = ref - res;
lo++;
}
if (res > ref) {
err = res - ref;
hi++;
}
if (err > 1) {
printf ("!!!! arg=%08x res=%08x ref=%08x\n", arg, res, ref);
return EXIT_FAILURE;
}
arg++;
} while (arg);
printf ("results too low: %u too high: %u not correctly rounded: %u\n",
lo, hi, lo + hi);
return EXIT_SUCCESS;
}
我有一个解决方案，我将其描述为“快速逆 sqrt，但适用于 32 位定点”。没有表格，没有参考，只是直截了当的猜测。
如果需要，请跳转到下面的源代码，但要注意一些事情。
(x * y)>>16可以替换为您想要的任何定点乘法方案。int fxisqrt(int input){
if(input <= 65536){
return 1;
}
long xSR = input>>1;
long pushRight = input;
long msb = 0;
long shoffset = 0;
long yIsqr = 0;
long ysqr = 0;
long fctrl = 0;
long subthreehalf = 0;
while(pushRight >= 65536){
pushRight >>=1;
msb++;
}
shoffset = (16 - ((msb)>>1));
yIsqr = 1<<shoffset;
//y = (y * (98304 - ( ( (x>>1) * ((y * y)>>16 ) )>>16 ) ) )>>16; x2
//Incantation 1
ysqr = (yIsqr * yIsqr)>>16;
fctrl = (xSR * ysqr)>>16;
subthreehalf = 98304 - fctrl;
yIsqr = (yIsqr * subthreehalf)>>16;
//Incantation 2 - Increases precision greatly, but may not be neccessary
ysqr = (yIsqr * yIsqr)>>16;
fctrl = (xSR * ysqr)>>16;
subthreehalf = 98304 - fctrl;
yIsqr = (yIsqr * subthreehalf)>>16;
return yIsqr;
}