Question

有没有任何已知的公式在java中更快地平方根，我使用eclipse这是我的普通代码。我找到了一种不准确的循环方式，我将它用于elo计算器。

 Elo = Elo - (Math.sqrt(pointslot)/(elop-eloo)*2)

Answer 1

这是Java Math.sqrt()的来源。优化。看起来很简单。

static const double one = 1.0, tiny=1.0e-300;

double z;
int sign = (int) 0x80000000; 
unsigned r, t1, s1, ix1, q1;
int ix0, s0, q, m, t, i;

ix0 = __HI(x); /* high word of x */
ix1 = __LO(x); /* low word of x */

/* take care of Inf and NaN */
if ((ix0 & 0x7ff00000) == 0x7ff00000) {            
    return x*x+x; /* sqrt(NaN) = NaN, 
                     sqrt(+inf) = +inf,
                     sqrt(-inf) = sNaN */
} 

/* take care of zero */
if (ix0 <= 0) {
    if (((ix0&(~sign)) | ix1) == 0) {
        return x; /* sqrt(+-0) = +-0 */
    } else if (ix0 < 0) {
        return (x-x) / (x-x); /* sqrt(-ve) = sNaN */
    }
}

/* normalize x */
m = (ix0 >> 20);
if (m == 0) { /* subnormal x */
    while (ix0==0) {
        m -= 21;
        ix0 |= (ix1 >> 11); ix1 <<= 21;
    }
    for (i=0; (ix0&0x00100000)==0; i++) {
        ix0 <<= 1;
    }
    m -= i-1;
    ix0 |= (ix1 >> (32-i));
    ix1 <<= i;
}

m -= 1023; /* unbias exponent */
ix0 = (ix0&0x000fffff)|0x00100000;
if (m&1) { /* odd m, double x to make it even */
    ix0 += ix0 + ((ix1&sign) >> 31);
    ix1 += ix1;
}

m >>= 1; /* m = [m/2] */

/* generate sqrt(x) bit by bit */
ix0 += ix0 + ((ix1 & sign)>>31);
ix1 += ix1;
q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
r = 0x00200000; /* r = moving bit from right to left */

while (r != 0) {
    t = s0 + r; 
    if (t <= ix0) { 
        s0 = t+r; 
        ix0 -= t; 
        q += r; 
    } 
    ix0 += ix0 + ((ix1&sign)>>31);
    ix1 += ix1;
    r>>=1;
}

r = sign;
while (r != 0) {
    t1 = s1+r; 
    t = s0;
    if ((t<ix0) || ((t == ix0) && (t1 <= ix1))) { 
        s1 = t1+r;
        if (((t1&sign) == sign) && (s1 & sign) == 0) {
            s0 += 1;
        }
        ix0 -= t;
        if (ix1 < t1) {
            ix0 -= 1;
        }
        ix1 -= t1;
        q1  += r;
    }
    ix0 += ix0 + ((ix1&sign) >> 31);
    ix1 += ix1;
    r >>= 1;
}

/* use floating add to find out rounding direction */
if((ix0 | ix1) != 0) {
    z = one - tiny; /* trigger inexact flag */
    if (z >= one) {
        z = one+tiny;
        if (q1 == (unsigned) 0xffffffff) { 
            q1=0; 
            q += 1;
        }
    } else if (z > one) {
        if (q1 == (unsigned) 0xfffffffe) {
            q+=1;
        }
        q1+=2; 
    } else
        q1 += (q1&1);
    }
}

ix0 = (q>>1) + 0x3fe00000;
ix1 =  q 1>> 1;
if ((q&1) == 1) ix1 |= sign;
ix0 += (m <<20);
__HI(z) = ix0;
__LO(z) = ix1;
return z;

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