如何实现平方根功能?
答案 0 :(得分:32)
来源here。
问题陈述:给定x> 0,找到y使得y ^ 2 = x => y = x / y(这是关键步骤)。
double test(double x, double g) {
if closeEnough(x/g, g)
return g;
else
return test(x, betterGuess(x, g));
}
boolean closeEnough(double a, double b) {
return (Math.abs(a - b) < .001);
}
double betterGuess(double x, double g) {
return ((g + x/g) / 2);
}
sqrt(2) | Guess g x / g | New guess, (g + x / g) / 2
----------------|------------------------------|-------------------------------
test(2, 1) | 1 2 / 1 = 2 | (2 + 1) / 2 = 1.5
test(2, 1.5) | 1.5 2 / 1.5 = 1.3333 | (1.3333 + 1.5) / 2 = 1.4167
test(2, 1.4167) | 1.4167 2 / 1.4167 = 1.4118 | (1.4167 + 1.4118) / 2 = 1.4142
test(2, 1.4142) | 1.4142 ... | ...
答案 1 :(得分:10)
使用Binary Search和C ++进行简单实现
double root(double n){
double lo = 0, hi = n, mid;
for(int i = 0 ; i < 1000 ; i++){
mid = (lo+hi)/2;
if(mid*mid == n) return mid;
if(mid*mid > n) hi = mid;
else lo = mid;
}
return mid;
}
请注意,while
循环在二进制搜索中最常见,但我个人更喜欢在处理十进制数时使用for
,它可以节省一些特殊情况处理并从小循环中获得非常准确的结果1000
甚至500
(两者都会为几乎所有数字提供相同的结果,但只是为了安全)。
修改:查看此Wikipedia article了解各种专门用于计算平方根的方法。
答案 2 :(得分:6)
在Intel硬件上,它通常在硬件SQRT指令之上实现。有些图书馆只是直接使用结果,有些图书馆可能会通过几轮牛顿优化来使其在角落情况下更准确。
答案 3 :(得分:4)
FDLIBM(可自由分发的LIBM)有一个非常好的文档版本的sqrt。 e_sqrt.c
有一个版本使用整数运算和一次修改一位的递推公式。
另一种方法使用牛顿方法。它从一些黑魔法和一个查找表开始,得到前8位,然后应用递推公式
y_{i+1} = 1/2 * ( y_i + x / y_i)
其中x是我们开始的数字。这是Heron方法的Babylonian method。它可以追溯到公元一世纪的亚历山德拉英雄。
还有另一种称为Fast inverse square root或reciproot的方法。它使用一些“邪恶的浮点位黑客”来找到1 / sqrt(x)的值。 i = 0x5f3759df - ( i >> 1 );
它使用mantisse和exponent来利用float的二进制表示。如果我们的数字x是(1 + m)* 2 ^ e,其中m是尾数,e是指数,结果y = 1 / sqrt(x)=(1 + n)* 2 ^ f。记录日志
lg(y) = - 1/2 lg(x)
f + lg(1+n) = -1/2 e - 1/2 lg(1+m)
因此我们看到结果的指数部分是-1/2数字的指数。黑魔法基本上对指数进行逐位移位,并在尾数上使用线性近似。
一旦你有了一个好的第一次近似,你可以使用牛顿的方法来获得更好的结果,最后用一些比特级别来修复最后一个数字。
答案 4 :(得分:2)
这是Newton算法的实现,请参阅https://tour.golang.org/flowcontrol/8。
int: s ==> Size
int: w ==> Width
bool: c ==> Crop
hex: c ==> BorderColor
bool: d ==> Download
int: h ==> Height
bool: s ==> Stretch
bool: h ==> Html
bool: p ==> SmartCrop
bool: pa ==> PreserveAspectRatio
bool: pd ==> Pad
bool: pp ==> SmartCropNoClip
bool: pf ==> SmartCropUseFace
int: p ==> FocalPlane
bool: n ==> CenterCrop
int: r ==> Rotate
bool: r ==> SkipRefererCheck
bool: fh ==> HorizontalFlip
bool: fv ==> VerticalFlip
bool: cc ==> CircleCrop
bool: ci ==> ImageCrop
bool: o ==> Overlay
str: o ==> EncodedObjectId
str: j ==> EncodedFrameId
int: x ==> TileX
int: y ==> TileY
int: z ==> TileZoom
bool: g ==> TileGeneration
bool: fg ==> ForceTileGeneration
bool: ft ==> ForceTransformation
int: e ==> ExpirationTime
str: f ==> ImageFilter
bool: k ==> KillAnimation
int: k ==> FocusBlur
bool: u ==> Unfiltered
bool: ut ==> UnfilteredWithTransforms
bool: i ==> IncludeMetadata
bool: ip ==> IncludePublicMetadata
bool: a ==> EsPortraitApprovedOnly
int: a ==> SelectFrameint
int: m ==> VideoFormat
int: vb ==> VideoBegin
int: vl ==> VideoLength
bool: lf ==> LooseFaceCrop
bool: mv ==> MatchVersion
bool: id ==> ImageDigest
int: ic ==> InternalClient
bool: b ==> BypassTakedown
int: b ==> BorderSize
str: t ==> Token
str: nt0 ==> VersionedToken
bool: rw ==> RequestWebp
bool: rwu ==> RequestWebpUnlessMaybeTransparent
bool: rwa ==> RequestAnimatedWebp
bool: nw ==> NoWebp
bool: rh ==> RequestH264
bool: nc ==> NoCorrectExifOrientation
bool: nd ==> NoDefaultImage
bool: no ==> NoOverlay
str: q ==> QueryString
bool: ns ==> NoSilhouette
int: l ==> QualityLevel
int: v ==> QualityBucket
bool: nu ==> NoUpscale
bool: rj ==> RequestJpeg
bool: rp ==> RequestPng
bool: rg ==> RequestGif
bool: pg ==> TilePyramidAsProto
bool: mo ==> Monogram
bool: al ==> Autoloop
int: iv ==> ImageVersion
int: pi ==> PitchDegrees
int: ya ==> YawDegrees
int: ro ==> RollDegrees
int: fo ==> FovDegrees
bool: df ==> DetectFaces
str: mm ==> VideoMultiFormat
bool: sg ==> StripGoogleData
bool: gd ==> PreserveGoogleData
bool: fm ==> ForceMonogram
int: ba ==> Badge
int: br ==> BorderRadius
hex: bc ==> BackgroundColor
hex: pc ==> PadColor
hex: sc ==> SubstitutionColor
bool: dv ==> DownloadVideo
bool: md ==> MonogramDogfood
int: cp ==> ColorProfile
bool: sm ==> StripMetadata
int: cv ==> FaceCropVersion
以下是魔术线的数学解释。假设您要查找多项式的根$ f(x)= x ^ 2 - a $。通过Newton的方法,您可以从初始猜测$ x_0 = 1 $开始。下一个猜测是$ x_1 = x_0 - f(x_0)/ f'(x_0)$,其中$ f'(x)= 2x $。因此,您的新猜测是
$ x_1 = x_0 - (x_0 ^ 2 - a)/ 2x_0 $
答案 5 :(得分:1)
SQRT();功能在幕后。
它始终检查图表中的中点。示例:sqrt(16)= 4; SQRT(4)= 2;
现在,如果您在16或4之内提供任何输入,例如sqrt(10)==?
它找到2和4的中点,即= x,然后再次找到x和4的中点(它排除了此输入中的下限)。它一次又一次地重复这个步骤,直到它得到完美答案,即sqrt(10)== 3.16227766017。它位于b / w 2和4.所有这个内置函数都是使用微积分,微分和积分创建的。
答案 6 :(得分:1)
Python实现: 根值的下限是此函数的输出。 示例:8的平方根是2.82842 ...,此函数将给出输出'2'
def mySqrt(x):
# return int(math.sqrt(x))
if x==0 or x==1:
return x
else:
start = 0
end = x
while (start <= end):
mid = int((start + end) / 2)
if (mid*mid == x):
return mid
elif (mid*mid < x):
start = mid + 1
ans = mid
else:
end = mid - 1
return ans
答案 7 :(得分:1)
Formula: root(number, <root>, <depth>) == number^(root^(-depth))
Usage: root(number,<root>,<depth>)
Example: root(16,2) == sqrt(16) == 4
Example: root(16,2,2) == sqrt(sqrt(16)) == 2
Example: root(64,3) == 4
Implementation in C#:
static double root(double number, double root = 2f, double depth = 1f)
{
return Math.Pow(number, Math.Pow(root, -depth));
}
答案 8 :(得分:1)
到目前为止,解决方案主要是浮点运算……并且还假定除法指令可用且快速。
这是一个简单的简单例程,不使用FP或除法。除了第一个if语句可在输入较小时加快例程运行速度之外,每一行都会在结果中另外计算一位。
constexpr unsigned int root(unsigned int x) {
unsigned int i = 0;
if (x >= 65536) {
if ((i + 32768) * (i + 32768) <= x) i += 32768;
if ((i + 16384) * (i + 16384) <= x) i += 16384;
if ((i + 8192) * (i + 8192) <= x) i += 8192;
if ((i + 4096) * (i + 4096) <= x) i += 4096;
if ((i + 2048) * (i + 2048) <= x) i += 2048;
if ((i + 1024) * (i + 1024) <= x) i += 1024;
if ((i + 512) * (i + 512) <= x) i += 512;
if ((i + 256) * (i + 256) <= x) i += 256;
}
if ((i + 128) * (i + 128) <= x) i += 128;
if ((i + 64) * (i + 64) <= x) i += 64;
if ((i + 32) * (i + 32) <= x) i += 32;
if ((i + 16) * (i + 16) <= x) i += 16;
if ((i + 8) * (i + 8) <= x) i += 8;
if ((i + 4) * (i + 4) <= x) i += 4;
if ((i + 2) * (i + 2) <= x) i += 2;
if ((i + 1) * (i + 1) <= x) i += 1;
return i;
}
答案 9 :(得分:0)
计算平方根(不使用内置的math.sqrt函数):
<强> SquareRootFunction.java 强>
public class SquareRootFunction {
public double squareRoot(double value,int decimalPoints)
{
int firstPart=0;
/*calculating the integer part*/
while(square(firstPart)<value)
{
firstPart++;
}
if(square(firstPart)==value)
return firstPart;
firstPart--;
/*calculating the decimal values*/
double precisionVal=0.1;
double[] decimalValues=new double[decimalPoints];
double secondPart=0;
for(int i=0;i<decimalPoints;i++)
{
while(square(firstPart+secondPart+decimalValues[i])<value)
{
decimalValues[i]+=precisionVal;
}
if(square(firstPart+secondPart+decimalValues[i])==value)
{
return (firstPart+secondPart+decimalValues[i]);
}
decimalValues[i]-=precisionVal;
secondPart+=decimalValues[i];
precisionVal*=0.1;
}
return(firstPart+secondPart);
}
public double square(double val)
{
return val*val;
}
}
<强> MainApp.java 强>
import java.util.Scanner;
public class MainApp {
public static void main(String[] args) {
double number;
double result;
int decimalPoints;
Scanner in = new Scanner(System.in);
SquareRootFunction sqrt=new SquareRootFunction();
System.out.println("Enter the number\n");
number=in.nextFloat();
System.out.println("Enter the decimal points\n");
decimalPoints=in.nextInt();
result=sqrt.squareRoot(number,decimalPoints);
System.out.println("The square root value is "+ result);
in.close();
}
}
答案 10 :(得分:0)
long long int floorSqrt(long long int x)
{
long long r = 0;
while((long)(1<<r)*(long)(1<<r) <= x){
r++;
}
r--;
long long b = r -1;
long long ans = 1 << r;
while(b >= 0){
if(((long)(ans|1<<b)*(long)(ans|1<<b))<=x){
ans |= (1<<b);
}
b--;
}
return ans;
}
答案 11 :(得分:0)
有些东西叫做巴比伦方法。
static float squareRoot(float n)
{
/*We are using n itself as
initial approximation This
can definitely be improved */
float x = n;
float y = 1;
// e decides the accuracy level
double e = 0.000001;
while(x - y > e)
{
x = (x + y)/2;
y = n/x;
}
return x;
}
有关更多信息的链接:https://www.geeksforgeeks.org/square-root-of-a-perfect-square/
答案 12 :(得分:0)
因此,如果没有关于是否使用内置ceil或round函数的规范,这是Java中的一种递归方法,它使用Newton-Raphson方法来查找无符号数的平方根。 / p>
public class FindSquareRoot {
private static double newtonRaphson(double N, double X, double oldX) {
if(N <= 0) return 0;
if (Math.round(X) == Math.ceil(oldX))
return X;
return newtonRaphson(N, X - ((X * X) - N)/(2 * X), X);
}
//Driver method
public static void main (String[] args) {
System.out.println("Square root of 48.8: " + newtonRaphson(48.8, 10, 0));
}
}
答案 13 :(得分:0)
我也在做一个sqrt函数,100000000次迭代需要14秒,与sqrt的1秒相比还是没什么
double mysqrt(double n)
{
double x = n;
int it = 4;
if (n >= 90)
{
it = 6;
}
if (n >= 5000)
{
it = 8;
}
if (n >= 20000)
{
it = 10;
}
if (n >= 90000)
{
it = 11;
}
if (n >= 200000)
{
it = 12;
}
if (n >= 900000)
{
it = 13;
}
if (n >= 3000000)
{
it = 14;
}
if (n >= 10000000)
{
it = 15;
}
if (n >= 30000000)
{
it = 16;
}
if (n >= 100000000)
{
it = 17;
}
if (n >= 300000000)
{
it = 18;
}
if (n >= 1000000000)
{
it = 19;
}
for (int i = 0; i < it; i++)
{
x = 0.5*(x+n/x);
}
return x;
}
但是最快的实现是:
float Q_rsqrt( float number )
{
long i;
float x2, y;
const float threehalfs = 1.5F;
x2 = number * 0.5F;
y = number;
i = * ( long * ) &y; // evil floating point bit level hacking
i = 0x5f3759df - ( i >> 1 ); // what the fuck?
y = * ( float * ) &i;
y = y * ( threehalfs - ( x2 * y * y ) ); // 1st iteration
// y = y * ( threehalfs - ( x2 * y * y ) ); // 2nd iteration, this can be removed
return y;
}
float mysqrt(float n) {return 1/Q_rsqrt(n);}
答案 14 :(得分:0)
在Golang中解决我的问题。
package main
import (
"fmt"
)
func Sqrt(x float64) float64 {
z := 1.0 // initial guess to be 1
i := 0
for int(z*z) != int(x) { // until find the first approximation
// Newton root algorithm
z -= (z*z - x) / (2 * z)
i++
}
return z
}
func main() {
fmt.Println(Sqrt(8900009870))
}
遵循经典/通用解决方案。
package main
import (
"fmt"
"math"
)
func Sqrt(num float64) float64 {
const DIFF = 0.0001 // To fix the precision
z := 1.0
for {
z1 := z - (((z * z) - num) / (2 * z))
// Return a result when the diff between the last execution
// and the current one is lass than the precision constant
if (math.Abs(z1 - z) < DIFF) {
break
}
z = z1
}
return z
}
func main() {
fmt.Println(Sqrt(94339))
}
有关更多信息,请检查here
答案 15 :(得分:-1)
用法:root(数字,root,深度)
示例:root(16,2)== sqrt(16)== 4
示例:root(16,2,2)== sqrt(sqrt(16))== 2
示例:root(64,3)== 4
在C#中的实现:
double root(double number, double root, double depth = 1f)
{
return number ^ (root ^ (-depth));
}
用法:Sqrt(数字,深度)
示例:Sqrt(16)== 4
示例:Sqrt(8,2)== sqrt(sqrt(8))
double Sqrt(double number, double depth = 1) return root(number,2,depth);
通过: Imk0tter