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Math.log背后的算法 - Java

javaalgorithmmathlogarithm
5

最近我在编写一些程序时使用了Math.log()，现在我想知道这个方法是如何工作的。计算机是如何计算对数的呢？

谢谢解答。

可能是重复的问题，参考如何手动计算对数。 - Alejandro
可能是标准数学函数算法在哪里找？的重复问题。 - phuclv
3个回答
6

openjdk中Math.log的定义如下：

public static double log(double a) {
    return StrictMath.log(a); // default impl. delegates to StrictMath
}

这促使我查看StrictMath的源代码，其中log被声明为：

public static native double log(double a);

本地关键字 表示JNI。换句话说，它会到libc源代码中查找我们需要的内容。由于我使用NetBSD，下面是它对log的定义：

/* __ieee754_log(x)
 * Return the logrithm of x
 *
 * Method :
 *   1. Argument Reduction: find k and f such that
 *          x = 2^k * (1+f),
 *     where  sqrt(2)/2 < 1+f < sqrt(2) .
 *
 *   2. Approximation of log(1+f).
 *  Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s)
 *       = 2s + 2/3 s**3 + 2/5 s**5 + .....,
 *           = 2s + s*R
 *      We use a special Reme algorithm on [0,0.1716] to generate
 *  a polynomial of degree 14 to approximate R The maximum error
 *  of this polynomial approximation is bounded by 2**-58.45. In
 *  other words,
 *              2      4      6      8      10      12      14
 *      R(z) ~ Lg1*s +Lg2*s +Lg3*s +Lg4*s +Lg5*s  +Lg6*s  +Lg7*s
 *      (the values of Lg1 to Lg7 are listed in the program)
 *  and
 *      |      2          14          |     -58.45
 *      | Lg1*s +...+Lg7*s    -  R(z) | <= 2
 *      |                             |
 *  Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2.
 *  In order to guarantee error in log below 1ulp, we compute log
 *  by
 *      log(1+f) = f - s*(f - R)    (if f is not too large)
 *      log(1+f) = f - (hfsq - s*(hfsq+R)). (better accuracy)
 *
 *  3. Finally,  log(x) = k*ln2 + log(1+f).
 *              = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo)))
 *     Here ln2 is split into two floating point number:
 *          ln2_hi + ln2_lo,
 *     where n*ln2_hi is always exact for |n| < 2000.
 *
 * Special cases:
 *  log(x) is NaN with signal if x < 0 (including -INF) ;
 *  log(+INF) is +INF; log(0) is -INF with signal;
 *  log(NaN) is that NaN with no signal.
 *
 * Accuracy:
 *  according to an error analysis, the error is always less than
 *  1 ulp (unit in the last place).
 *
 * Constants:
 * The hexadecimal values are the intended ones for the following
 * constants. The decimal values may be used, provided that the
 * compiler will convert from decimal to binary accurately enough
 * to produce the hexadecimal values shown.
 */

#include "math.h"
#include "math_private.h"

static const double
ln2_hi  =  6.93147180369123816490e-01,  /* 3fe62e42 fee00000 */
ln2_lo  =  1.90821492927058770002e-10,  /* 3dea39ef 35793c76 */
two54   =  1.80143985094819840000e+16,  /* 43500000 00000000 */
Lg1 = 6.666666666666735130e-01,  /* 3FE55555 55555593 */
Lg2 = 3.999999999940941908e-01,  /* 3FD99999 9997FA04 */
Lg3 = 2.857142874366239149e-01,  /* 3FD24924 94229359 */
Lg4 = 2.222219843214978396e-01,  /* 3FCC71C5 1D8E78AF */
Lg5 = 1.818357216161805012e-01,  /* 3FC74664 96CB03DE */
Lg6 = 1.531383769920937332e-01,  /* 3FC39A09 D078C69F */
Lg7 = 1.479819860511658591e-01;  /* 3FC2F112 DF3E5244 */

static const double zero   =  0.0;

double
__ieee754_log(double x)
{
    double hfsq,f,s,z,R,w,t1,t2,dk;
    int32_t k,hx,i,j;
    u_int32_t lx;

    EXTRACT_WORDS(hx,lx,x);

    k=0;
    if (hx < 0x00100000) {          /* x < 2**-1022  */
        if (((hx&0x7fffffff)|lx)==0)
        return -two54/zero;     /* log(+-0)=-inf */
        if (hx<0) return (x-x)/zero;    /* log(-#) = NaN */
        k -= 54; x *= two54; /* subnormal number, scale up x */
        GET_HIGH_WORD(hx,x);
    }
    if (hx >= 0x7ff00000) return x+x;
    k += (hx>>20)-1023;
    hx &= 0x000fffff;
    i = (hx+0x95f64)&0x100000;
    SET_HIGH_WORD(x,hx|(i^0x3ff00000)); /* normalize x or x/2 */
    k += (i>>20);
    f = x-1.0;
    if((0x000fffff&(2+hx))<3) { /* |f| < 2**-20 */
        if(f==zero) { if(k==0) return zero;  else {dk=(double)k;
                   return dk*ln2_hi+dk*ln2_lo;}
        }
        R = f*f*(0.5-0.33333333333333333*f);
        if(k==0) return f-R; else {dk=(double)k;
                 return dk*ln2_hi-((R-dk*ln2_lo)-f);}
    }
    s = f/(2.0+f);
    dk = (double)k;
    z = s*s;
    i = hx-0x6147a;
    w = z*z;
    j = 0x6b851-hx;
    t1= w*(Lg2+w*(Lg4+w*Lg6));
    t2= z*(Lg1+w*(Lg3+w*(Lg5+w*Lg7)));
    i |= j;
    R = t2+t1;
    if(i>0) {
        hfsq=0.5*f*f;
        if(k==0) return f-(hfsq-s*(hfsq+R)); else
             return dk*ln2_hi-((hfsq-(s*(hfsq+R)+dk*ln2_lo))-f);
    } else {
        if(k==0) return f-s*(f-R); else
             return dk*ln2_hi-((s*(f-R)-dk*ln2_lo)-f);
    }
}

如果您需要更多帮助，请留下评论，我会进一步解释。
这些 Lg1...Lg7 常量是什么？它们肯定不是 log(1)...log(7)。 - cat
计算对数的Reme算法连续调用的结果 - hd1
3
当你有疑问时，可以查看源代码（使用Java 8进行开发）。
public static double log(double a) {
    return StrictMath.log(a); // default impl. delegates to StrictMath
}

你看到它调用了一个静态方法StrictMath.java。
static native double log(double a);

什么是native关键字，你可以在这里阅读这里。基本上，它表示该方法是用除了Java之外的另一种语言实现的。我在GitHub上找到了包含数学函数的文件夹jdk/src/share/native/java/lang/fdlibm/src/。算法是在e_log.c中实现的。里面发生了什么样的魔法，我不知道。
所以这一切都取决于本地引用C#代码。感谢您的帮助。 - tzuxi
1
对数函数（以及许多其他函数，例如sin(x)）可以用级数来近似计算： https://en.wikipedia.org/wiki/Natural_logarithm#Series 如果您需要足够好的近似（但不是完全精确），则可以在级数的前几次迭代后停止。这就是早期游戏/ demo 场编程中接近三角函数的方法。 查找表也是一种好方法，关键在于找到适当的<强>分布查找值（在线性情况下可能不理想）。
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