Java中的Math.pow()背后的算法是什么

2022-09-03 12:59:31

我刚刚开始使用Java,作为第一个项目,我正在编写一个程序,用于查找给定数字的根(在本例中为立方根)。目前,我正在尝试牛顿-拉尔夫森来实现这一目标。这是代码 -

import java.util.Scanner;

import static java.lang.Math.abs;

public class newClass {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        System.out.println("Number whose cube root u wanna find:");
        Double number = input.nextDouble();
        Double epsilon = 0.0001;
        Double ans = number/2.00;
        while (abs((abs(number) - abs(Math.pow(ans,3))))>epsilon){
            System.out.println("in loop");
            ans = ans - ((Math.pow(ans,3) - number)/(3*Math.pow(ans,2)));
            System.out.println(ans);
            if ((number - ans)<=epsilon){
                System.out.println(ans);
            }
        }
        //System.out.println(Math.pow(number,1.0/3.0));


    }
}

这最多只能工作11位数字,之后它变得太大,IDE无法处理。但是,如果我只是简单地使用它,它适用于更大的数字,并立即计算出来。Math.pow(number,1.0/3.0)


那么,使用什么算法来提供即时答案?我知道我的方法依赖于猜测,我猜math.pow()实际上可能正在计算答案,但是如何计算呢?Math.pow()


答案 1

这是一个有趣的问题。如果你查看Java的Math类的源代码,你会发现它调用StrictMath.pow(double1,double2),而StrictMath的签名是public static native double pow(double a, double b);

因此,最终,这是一个真正的本机调用,可能因平台而异。但是,在某个地方有一个实现,它并不容易查看。以下是函数的描述和函数本身的代码:

注意

通过数学,试图理解它可能不可避免地导致更多的问题。但是,通过在Java数学函数源代码上搜索此Github并浏览数学摘要,您绝对可以更好地理解本机函数。快乐探索:)

方法说明

Method:  Let x =  2   * (1+f)
      1. Compute and return log2(x) in two pieces:
              log2(x) = w1 + w2,
         where w1 has 53-24 = 29 bit trailing zeros.
      2. Perform y*log2(x) = n+y' by simulating muti-precision
         arithmetic, where |y'|<=0.5.
      3. Return x**y = 2**n*exp(y'*log2)

特殊情况

      1.  (anything) ** 0  is 1
      2.  (anything) ** 1  is itself
      3.  (anything) ** NAN is NAN
      4.  NAN ** (anything except 0) is NAN
      5.  +-(|x| > 1) **  +INF is +INF
      6.  +-(|x| > 1) **  -INF is +0
      7.  +-(|x| < 1) **  +INF is +0
      8.  +-(|x| < 1) **  -INF is +INF
      9.  +-1         ** +-INF is NAN
      10. +0 ** (+anything except 0, NAN)               is +0
      11. -0 ** (+anything except 0, NAN, odd integer)  is +0
      12. +0 ** (-anything except 0, NAN)               is +INF
      13. -0 ** (-anything except 0, NAN, odd integer)  is +INF
      14. -0 ** (odd integer) = -( +0 ** (odd integer) )
      15. +INF ** (+anything except 0,NAN) is +INF
      16. +INF ** (-anything except 0,NAN) is +0
      17. -INF ** (anything)  = -0 ** (-anything)
      18. (-anything) ** (integer) is (-1)**(integer)*(+anything**integer)
      19. (-anything except 0 and inf) ** (non-integer) is NAN

准确性

       pow(x,y) returns x**y nearly rounded. In particular
                      pow(integer,integer)
       always returns the correct integer provided it is
       representable.

源码

#ifdef __STDC__
        double __ieee754_pow(double x, double y)
#else
        double __ieee754_pow(x,y)
        double x, y;
#endif
{
        double z,ax,z_h,z_l,p_h,p_l;
        double y1,t1,t2,r,s,t,u,v,w;
        int i0,i1,i,j,k,yisint,n;
        int hx,hy,ix,iy;
        unsigned lx,ly;

        i0 = ((*(int*)&one)>>29)^1; i1=1-i0;
        hx = __HI(x); lx = __LO(x);
        hy = __HI(y); ly = __LO(y);
        ix = hx&0x7fffffff;  iy = hy&0x7fffffff;

    /* y==zero: x**0 = 1 */
        if((iy|ly)==0) return one;

    /* +-NaN return x+y */
        if(ix > 0x7ff00000 || ((ix==0x7ff00000)&&(lx!=0)) ||
           iy > 0x7ff00000 || ((iy==0x7ff00000)&&(ly!=0)))
                return x+y;

    /* determine if y is an odd int when x < 0
     * yisint = 0       ... y is not an integer
     * yisint = 1       ... y is an odd int
     * yisint = 2       ... y is an even int
     */
        yisint  = 0;
        if(hx<0) {
            if(iy>=0x43400000) yisint = 2; /* even integer y */
            else if(iy>=0x3ff00000) {
                k = (iy>>20)-0x3ff;        /* exponent */
                if(k>20) {
                    j = ly>>(52-k);
                    if((j<<(52-k))==ly) yisint = 2-(j&1);
                } else if(ly==0) {
                    j = iy>>(20-k);
                    if((j<<(20-k))==iy) yisint = 2-(j&1);
                }
            }
        }

    /* special value of y */
        if(ly==0) {
            if (iy==0x7ff00000) {       /* y is +-inf */
                if(((ix-0x3ff00000)|lx)==0)
                    return  y - y;      /* inf**+-1 is NaN */
                else if (ix >= 0x3ff00000)/* (|x|>1)**+-inf = inf,0 */
                    return (hy>=0)? y: zero;
                else                    /* (|x|<1)**-,+inf = inf,0 */
                    return (hy<0)?-y: zero;
            }
            if(iy==0x3ff00000) {        /* y is  +-1 */
                if(hy<0) return one/x; else return x;
            }
            if(hy==0x40000000) return x*x; /* y is  2 */
            if(hy==0x3fe00000) {        /* y is  0.5 */
                if(hx>=0)       /* x >= +0 */
                return sqrt(x);
            }
        }

        ax   = fabs(x);
    /* special value of x */
        if(lx==0) {
            if(ix==0x7ff00000||ix==0||ix==0x3ff00000){
                z = ax;                 /*x is +-0,+-inf,+-1*/
                if(hy<0) z = one/z;     /* z = (1/|x|) */
                if(hx<0) {
                    if(((ix-0x3ff00000)|yisint)==0) {
                        z = (z-z)/(z-z); /* (-1)**non-int is NaN */
                    } else if(yisint==1)
                        z = -1.0*z;             /* (x<0)**odd = -(|x|**odd) */
                }
                return z;
            }
        }

        n = (hx>>31)+1;

    /* (x<0)**(non-int) is NaN */
        if((n|yisint)==0) return (x-x)/(x-x);

        s = one; /* s (sign of result -ve**odd) = -1 else = 1 */
        if((n|(yisint-1))==0) s = -one;/* (-ve)**(odd int) */

    /* |y| is huge */
        if(iy>0x41e00000) { /* if |y| > 2**31 */
            if(iy>0x43f00000){  /* if |y| > 2**64, must o/uflow */
                if(ix<=0x3fefffff) return (hy<0)? huge*huge:tiny*tiny;
                if(ix>=0x3ff00000) return (hy>0)? huge*huge:tiny*tiny;
            }
        /* over/underflow if x is not close to one */
            if(ix<0x3fefffff) return (hy<0)? s*huge*huge:s*tiny*tiny;
            if(ix>0x3ff00000) return (hy>0)? s*huge*huge:s*tiny*tiny;
        /* now |1-x| is tiny <= 2**-20, suffice to compute
           log(x) by x-x^2/2+x^3/3-x^4/4 */
            t = ax-one;         /* t has 20 trailing zeros */
            w = (t*t)*(0.5-t*(0.3333333333333333333333-t*0.25));
            u = ivln2_h*t;      /* ivln2_h has 21 sig. bits */
            v = t*ivln2_l-w*ivln2;
            t1 = u+v;
            __LO(t1) = 0;
            t2 = v-(t1-u);
        } else {
            double ss,s2,s_h,s_l,t_h,t_l;
            n = 0;
        /* take care subnormal number */
            if(ix<0x00100000)
                {ax *= two53; n -= 53; ix = __HI(ax); }
            n  += ((ix)>>20)-0x3ff;
            j  = ix&0x000fffff;
        /* determine interval */
            ix = j|0x3ff00000;          /* normalize ix */
            if(j<=0x3988E) k=0;         /* |x|<sqrt(3/2) */
            else if(j<0xBB67A) k=1;     /* |x|<sqrt(3)   */
            else {k=0;n+=1;ix -= 0x00100000;}
            __HI(ax) = ix;

        /* compute ss = s_h+s_l = (x-1)/(x+1) or (x-1.5)/(x+1.5) */
            u = ax-bp[k];               /* bp[0]=1.0, bp[1]=1.5 */
            v = one/(ax+bp[k]);
            ss = u*v;
            s_h = ss;
            __LO(s_h) = 0;
        /* t_h=ax+bp[k] High */
            t_h = zero;
            __HI(t_h)=((ix>>1)|0x20000000)+0x00080000+(k<<18);
            t_l = ax - (t_h-bp[k]);
            s_l = v*((u-s_h*t_h)-s_h*t_l);
        /* compute log(ax) */
            s2 = ss*ss;
            r = s2*s2*(L1+s2*(L2+s2*(L3+s2*(L4+s2*(L5+s2*L6)))));
            r += s_l*(s_h+ss);
            s2  = s_h*s_h;
            t_h = 3.0+s2+r;
            __LO(t_h) = 0;
            t_l = r-((t_h-3.0)-s2);
        /* u+v = ss*(1+...) */
            u = s_h*t_h;
            v = s_l*t_h+t_l*ss;
        /* 2/(3log2)*(ss+...) */
            p_h = u+v;
            __LO(p_h) = 0;
            p_l = v-(p_h-u);
            z_h = cp_h*p_h;             /* cp_h+cp_l = 2/(3*log2) */
            z_l = cp_l*p_h+p_l*cp+dp_l[k];
        /* log2(ax) = (ss+..)*2/(3*log2) = n + dp_h + z_h + z_l */
            t = (double)n;
            t1 = (((z_h+z_l)+dp_h[k])+t);
            __LO(t1) = 0;
            t2 = z_l-(((t1-t)-dp_h[k])-z_h);
        }

    /* split up y into y1+y2 and compute (y1+y2)*(t1+t2) */
        y1  = y;
        __LO(y1) = 0;
        p_l = (y-y1)*t1+y*t2;
        p_h = y1*t1;
        z = p_l+p_h;
        j = __HI(z);
        i = __LO(z);
        if (j>=0x40900000) {                            /* z >= 1024 */
            if(((j-0x40900000)|i)!=0)                   /* if z > 1024 */
                return s*huge*huge;                     /* overflow */
            else {
                if(p_l+ovt>z-p_h) return s*huge*huge;   /* overflow */
            }
        } else if((j&0x7fffffff)>=0x4090cc00 ) {        /* z <= -1075 */
            if(((j-0xc090cc00)|i)!=0)           /* z < -1075 */
                return s*tiny*tiny;             /* underflow */
            else {
                if(p_l<=z-p_h) return s*tiny*tiny;      /* underflow */
            }
        }
    /*
     * compute 2**(p_h+p_l)
     */
        i = j&0x7fffffff;
        k = (i>>20)-0x3ff;
        n = 0;
        if(i>0x3fe00000) {              /* if |z| > 0.5, set n = [z+0.5] */
            n = j+(0x00100000>>(k+1));
            k = ((n&0x7fffffff)>>20)-0x3ff;     /* new k for n */
            t = zero;
            __HI(t) = (n&~(0x000fffff>>k));
            n = ((n&0x000fffff)|0x00100000)>>(20-k);
            if(j<0) n = -n;
            p_h -= t;
        }
        t = p_l+p_h;
        __LO(t) = 0;
        u = t*lg2_h;
        v = (p_l-(t-p_h))*lg2+t*lg2_l;
        z = u+v;
        w = v-(z-u);
        t  = z*z;
        t1  = z - t*(P1+t*(P2+t*(P3+t*(P4+t*P5))));
        r  = (z*t1)/(t1-two)-(w+z*w);
        z  = one-(r-z);
        j  = __HI(z);
        j += (n<<20);
        if((j>>20)<=0) z = scalbn(z,n); /* subnormal output */
        else __HI(z) += (n<<20);
        return s*z;
}

答案 2