反向sqrt为固定点

时间:2011-06-08 23:23:43

标签: math fixed-point inverse sqrt

我正在寻找固定点16.16数的最佳逆平方根算法。下面的代码是我到目前为止(但基本上它取平方根并除以原始数字,我想得到平方根没有除法)。如果它改变了什么,代码将被编译为armv5te。

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);
}

3 个答案:

答案 0 :(得分:3)

诀窍是将牛顿方法应用于问题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只是稍微偏移,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),并且没有任何真正的除法。

唯一的问题是你需要一个合适的y起始值。我记得迭代收敛的估计y有限制。

答案 1 :(得分:3)

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初始近似的表查找,然后是两次Newton-Raphson迭代以获得完全准确的结果。我们可以通过附加到表中的额外预计算来进一步加速这些迭代中的第一次,这是四十年前Cray计算机使用的一种技术。

下面的函数fxrsqrt()实现了这种方法。它以参数r的倒数平方根的8位近似值a开始,但不是存储r,而是每个表元素存储3r(在低十位中) 32位条目)和r 3 (在32位条目的高22位)。这允许快速计算第一次迭代 r 1 = 1.5 * r - a * r 3 。然后以常规方式计算第二次迭代,因为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在计算开始时被标准化,实质上将其表示为2.32定点数字乘以比例因子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。通过舍入最重要的丢弃位为1的结果,提高了准确性,导致几乎所有结果都被正确舍入。

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在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的便携式实现以及x86平台的内联汇编版本。在ARMv5TE平台上,__clz()应映射到CLZ指令,__umulhi()应映射到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, 0x9163067c, 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, 1457 results are too big by one ulp.
 * A total of 2096 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;
}

答案 2 :(得分:1)

我有一个解决方案,我将其特征描述为“快速逆sqrt,但适用于32位固定点”。没有桌子,没有参考,只是直截了当。

如果需要,请跳到下面的源代码,但要注意一些事项。

  1. (x * y)>>16可以替换为所需的任何定点乘法方案。
  2. 这不需要64位[长字],我只是为了便于演示而使用它。长字用于防止乘法溢出。定点数学库将具有定点乘法功能,可以更好地处理此问题。
  3. 最初的猜测非常好,因此您在第一个咒语中可获得相对准确的结果。
  4. 该代码比演示所需要的更为冗长。
  5. 不能使用小于65536(<1)且大于32767 << 16的值。
  6. 如果您的硬件具有除法功能,这通常不会比使用平方根表和除法更快。如果不是这样,则可以避免分裂。
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;
}