挑战

您必须以最短的长度计算pi。欢迎加入任何语言，您可以使用任何公式来计算pi。它必须能够计算pi到至少5个小数位。最短，将以字符为单位。比赛持续48小时。开始。

^{注意：这个类似的问题指出，必须使用4 *（1 – 1/3 + 1/5 – 1/7 +…）级数来计算PI。这个问题并没有这个限制，而事实上很多答案在这里（包括最有可能胜出）将是无效的，其他的问题。因此，这不是重复的。}

Python3、7

在交互式外壳中运行

355/113

输出：3.1415929203539825，精确到小数点后6位

最后，我有一个击败APL的解决方案！

哦，如果您想知道，该比率称为密率（字面上的“精确比率”），由中国数学家祖崇智（公元429-500）提出。可以在此处找到相关的维基百科文章。祖还给出了22/7的比率作为“粗略比率”，他是第一个提出3.1415926 <= pi <= 3.1415927的数学家。

PHP — 132 127 125 124字节

基本的蒙特卡洛模拟。每进行10M次迭代，它将显示当前状态：

for($i=1,$j=$k=0;;$i++){$x=mt_rand(0,1e7)/1e7;$y=mt_rand(0,1e7)/1e7;$j+=$x*$x+$y*$y<=1;$k++;if(!($i%1e7))echo 4*$j/$k."\n";}

感谢cloudfeet和zamnuts的建议！

样本输出：

$ php pi.php
3.1410564
3.1414008
3.1413388
3.1412641
3.14132568
3.1413496666667
3.1414522857143
3.1414817
3.1415271111111
3.14155092
...
3.1415901754386
3.1415890482759
3.1415925423731

J 6

{:*._1

说明：*.给出复数的长度和角度。-1的角度是pi。{:占据列表的末尾[长度，角度]

莱比尼兹（Leibniz）系列仅适用于21个字节的缓慢收敛的系列装饰论者：

      +/(4*_1&^%>:@+:)i.1e6
 3.14159

Perl，42个字节

map{$a+=(-1)**$_/(2*$_+1)}0..9x6;print$a*4

它使用Leibniz公式计算π ：

999999用作最大的n，以获取五个十进制数字的精度。

结果： 3.14159165358977

Piet，许多密码

不是我的答案，但这是我所见过的最佳解决方案：

我的理解是，它会将圆上的像素相加，然后除以半径，然后再除一次。那是：

A = πr²  # solve for π
π = A/r²
π = (A/r)/r

在我看来，更好的方法是程序以任意大小生成此图像，然后通过Piet解释器运行它。

资料来源：http : //www.dangermouse.net/esoteric/piet/samples.html

从技术上讲，我正在计算，9

0+3.14159

从技术上讲，我仍然会计算，10

PI-acos(1)

我计算很困难，8

acos(-1)

我意外的是PI，12岁

"3.14"+"159"

从技术上讲，这个答案很臭。

杀伤人员地雷-6

2ׯ1○1

输出3.141592654。它计算的反正弦是1的两倍。

13个字符的解决方案是：

--/4÷1-2×⍳1e6

这3.141591654为我输出，符合要求的精度。
它使用简单的+ 4/1 - 4/3 + 4/5 - 4/7 ...序列进行计算。

J-5个字节

|^._1

这意味着|log(-1)|。

Google计算器，48岁

stick of butter*(26557.4489*10^-9)/millimeters^3

取一小块黄油，进行高级计算，然后从中制成pi。我想，既然其他所有人都在做简单的数学答案，那么我会添加一个更独特的答案。

例

八度，31

quad(inline("sqrt(4-x^2)"),0,2)

通过数值积分计算半径为2的四分之一圆的面积。

octave:1> quad(inline("sqrt(4-x^2)"),0,2)
ans =     3.14159265358979

Mathematica 6

180N@° 
-->3.14159

Python, 88

解决方案：

l=q=d=0;t,s,n,r=3.,3,1,24
while s!=l:l,n,q,d,r=s,n+q,q+8,d+r,r+32;t=(t*n)/d;s+=t
print s

Python shell中的示例输出：

>>> print s
3.14159265359

设法避免任何进口。可以轻松地交换以使用任意精度的Decimal库；只需替换3.为Decimal('3')，然后在前后设置精度，然后一元加结果即可转换精度。

与这里的许多答案不同，实际上是计算 π而不是依赖内置常量或数学伪造函数math.acos(-1)，例如math.radians(180)，。

x86 assembly language (5 characters)

fldpi

Whether this loads a constant from ROM or actually calculates the answer depends on the processor though (but on at least some, it actually does a calculation, not just loading the number from ROM). To put things in perspective, it's listed as taking 40 clock cycles on a 387, which is rather more than seems to make sense if it were just loading the value from ROM.

If you really want to ensure a calculation you could do something like:

fld1
fld1
fpatan
fimul f

f dd 4

[for 27 characters]

bc -l, 37 bytes

for(p=n=2;n<7^7;n+=2)p*=n*n/(n*n-1);p

I don't see any other answers using the Wallis product, so since its named after my namesake (my History of Mathematics lecturer got a big kick out of that), I couldn't resist.

Turns out its a fairly nice algorithm from the golfing perspective, but its rate of convergence is abysmal - approaching 1 million iterations just to get 5 decimal places:

$ time bc -l<<<'for(p=n=2;n<7^7;n+=2)p*=n*n/(n*n-1);p'
3.14159074622629555058

real    0m3.145s
user    0m1.548s
sys 0m0.000s
$ 

bc -l, 15 bytes

Alternatively, we can use Newton-Raphson to solve sin(x)=0, with a starting approximation of 3. Because this converges in so few iterations, we simply hard-code 2 iterations, which gives 10 decimal places:

x=3+s(3);x+s(x)

The iterative formula according to Newton-Raphson is:

x[n+1] = x[n] - ( sin(x[n]) / sin'(x[n]) )

sin' === cos and cos(pi) === -1, so we simply approximate the cos term to get:

x[n+1] = x[n] + sin(x[n])

Output:

$ bc -l<<<'x=3+s(3);x+s(x)'
3.14159265357219555873
$ 

python - 47 45

pi is actually being calculated without trig functions or constants.

a=4
for i in range(9**6):a-=(-1)**i*4/(2*i+3)

result:

>>> a
3.1415907719167966

C, 99

Directly computes area / r^2 of a circle.

double p(n,x,y,r){r=10000;for(n=x=0;x<r;++x)for(y=1;y<r;++y)n+=x*x+y*y<=r*r;return(double)n*4/r/r;}

This function will calculate pi by counting the number of pixels in a circle of radius r then dividing by r*r (actually it just calculates one quadrant). With r as 10000, it is accurate to 5 decimal places (3.1415904800). The parameters to the function are ignored, I just declared them there to save space.

Javascript, 43 36

x=0;for(i=1;i<1e6;i++){x+=1/i/i};Math.sqrt(6*x)

x becomes zeta(2)=pi^2/6 so sqrt(6*x)=pi. (47 characters)

After using the distributive property and deleting the curly brackets from the for loop you get:

x=0;for(i=1;i<1e6;i++)x+=6/i/i;Math.sqrt(x)

(43 characters)

It returns:

3.14159169865946

Edit:

I found an even shorter way using the Wallis product:

x=i=2;for(;i<1e6;i+=2)x*=i*i/(i*i-1)

(36 characters)

It returns:

3.141591082792245

Python, Riemann zeta (58 41 char)

(6*sum(n**-2for n in range(1,9**9)))**0.5

Or spare two characters, but use scipy

import scipy.special as s
(6*s.zeta(2,1))**0.5

Edit: Saved 16 (!) characters thanks to amcgregor

Javascript: 99 characters

Using the formula given by Simon Plouffe in 1996, this works with 6 digits of precision after the decimal point:

function f(k){return k<2?1:f(k-1)*k}for(y=-3,n=1;n<91;n++)y+=n*(2<<(n-1))*f(n)*f(n)/f(2*n);alert(y)

This longer variant (130 characters) has a better precision, 15 digits after the decimal point:

function e(x){return x<1?1:2*e(x-1)}function f(k){return k<2?1:f(k-1)*k}for(y=-3,n=1;n<91;n++)y+=n*e(n)*f(n)*f(n)/f(2*n);alert(y)

I made this based in my two answers to this question.

Ruby, 54 50 49

p (0..9**6).map{|e|(-1.0)**e/(2*e+1)*4}.reduce :+

Online Version for testing.

Another version without creating an array (50 chars):

x=0;(0..9**6).each{|e|x+=(-1.0)**e/(2*e+1)*4}; p x

Online Version for testing.

TI CAS, 35

lim(x*(1/(tan((180-360/x)/2))),x,∞)

Perl - 35 bytes

$\=$\/(2*$_-1)*$_+2for-46..-1;print

Produces full floating point precision. A derivation of the formula used can be seen elsewhere.

Sample usage:

$ perl pi.pl
3.14159265358979

Arbitrary Precision Version

use bignum a,99;$\=$\/(2*$_-1)*$_+2for-329..-1;print

Extend as needed. The length of the iteration (e.g. -329..-1) should be adjusted to be approximately log₂(10) ≈ 3.322 times the number of digits.

3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211707

Or, using bigint instead:

use bigint;$\=$\/(2*$_-1)*$_+2e99for-329..-1;print

This runs noticably faster, but doesn't include a decimal point.

3141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067

C# 192

class P{static void Main(){var s=(new System.Net.WebClient()).DownloadString("http://www.ctan.org/pkg/tex");System.Console.WriteLine(s.Substring(s.IndexOf("Ver­sion")+21).Split(' ')[0]);}}

Outputs:

3.14159265

No math involved. Just looks up the current version of TeX and does some primitive parsing of the resulting html. Eventually it will become π according to Wikipedia.

Python 3 Monte Carlo (103 char)

from random import random as r
sum(1 for x,y in ((r(),r()) for i in range(2**99)) if x**2+y**2<1)/2**97

Game Maker Language, 34

Assumes all uninitialized variables as 0. This is default in some versions of Game Maker.

for(i=1;i<1e8;i++)x+=6/i/i;sqrt(x)

Result:

3.14159169865946

Java - 83 55

Shorter version thanks to Navin.

class P{static{System.out.print(Math.toRadians(180));}}

Old version:

class P{public static void main(String[]a){System.out.print(Math.toRadians(180));}}

R: 33 characters

sqrt(8*sum(1/seq(1,1000001,2)^2))
[1] 3.141592

Hopefully this follows the rules.

Ruby, 82

q=1.0
i=0
(0.0..72).step(8){|k|i+=1/q*(4/(k+1)-2/(k+4)-1/(k+5)-1/(k+6))
q*=16}
p i

Uses some formula I don't really understand and just copied down. :P

Output: 3.1415926535897913

Ruby, 12

p 1.570796*2

I am technically "calculating" pi an approximation of pi.

JavaScript - 19 bytes

Math.pow(29809,1/9)

Calculates the 9th root of 29809.

3.1415914903890925