Pi Calculation Code Golf [closed]

Question

The Challenge

You must calculate pi in the shortest length you can. Any language is welcome to join and you can use any formula to calculate pi. It must be able to calculate pi to at least 5 decimal places. Shortest, would be measured in characters. Competition lasts for 48 hours. Begin.

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^{Note: This similar question states that PI must be calculated using the series 4 * (1 – 1/3 + 1/5 – 1/7 + …). This question does not have this restriction, and in fact a lot of answers here (including the most likely to win) would be invalid in that other question. So, this is not a duplicate.}

Answer 1

Python3, 7

Runs in the interactive shell

355/113

Output: 3.1415929203539825, correct to 6 decimal places

And finally I have a solution that beats APL!

Oh, and in case you are wondering, this ratio is called the 密率 (literally "precise ratio"), and is proposed by the Chinese mathematician Zu Chongzhi (429-500 AD). A related wikipedia article can be found here. Zu also gave the ratio 22/7 as the "rough ratio", and he is known to be the first mathematician to propose that 3.1415926 <= pi <=3.1415927

Answer 2

PHP — 132 127 125 124 bytes

Basic Monte-Carlo simulation. Every 10M iterations, it prints the current status:

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";}

Thanks to cloudfeet and zamnuts for suggestions!

Sample output:

$ 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

Answer 3

J 6

{:*._1

Explanation: *. gives length and angle of a complex number. The angle of -1 is pi. {: takes the tail of the list [length, angle]

Just for the slowly-converging-series-fettishists, for 21 bytes, a Leibniz series:

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

Answer 4

Perl, 42 bytes

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

It's calculates π using the Leibniz formula:

999999 is used as largest n to get the precision of five decimal digits.

Result: 3.14159165358977

Answer 5

TECHNICALLY I'M CALCULATING, 9

0+3.14159

TECHNICALLY I'M STILL CALCULATING, 10

PI-acos(1)

I'M CALCULATING SO HARD, 8

acos(-1)

I ACCIDENTALLY PI, 12

"3.14"+"159"

And technically, this answer stinks.

Answer 6

Piet, many codels

Not my answer, but this is the best solution I've seen to this problem:

My understanding is that it adds up the pixels in the circle and divides by the radius, and then once again. That is:

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

A better approach in my mind is a program that generates this image at an arbitrary size and then runs it through a Piet interpreter.

Source: http://www.dangermouse.net/esoteric/piet/samples.html

Answer 7

APL - 6

2ׯ1○1

Outputs 3.141592654. It computes twice the arcsine of 1.

A 13-char solution would be:

--/4÷1-2×⍳1e6

This outputs 3.141591654 for me, which fits the requested precision.
It uses the simple + 4/1 - 4/3 + 4/5 - 4/7 ... series to calculate though.

Answer 8

J - 5 bytes

|^._1

This means |log(-1)|.

Answer 9

Octave, 31

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

Calculates the area of one quarter of a circle with radius 2, through numerical integration.

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

Answer 10

Mathematica 6

180N@° 
-->3.14159

Answer 11

Python, 88

Solution :

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

Sample output in Python shell :

>>> print s
3.14159265359

Manages to avoid any imports. Can easily be swapped to use the arbitrary precision Decimal library; just replace 3. with Decimal('3'), set the precision before and after, then unary plus the result to convert precision.

And unlike a whole lot of the answers here, actually computes π instead of relying on built-in constants or math fakery, i.e. math.acos(-1), math.radians(180), etc.

Answer 12

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]

Answer 13

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
$ 

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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
$ 

Answer 14

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

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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

Answer 15

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

Answer 16

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.

Answer 17

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

Answer 18

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

Answer 19

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.

Answer 20

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.

Answer 21

TI CAS, 35

lim(x*(1/(tan((180-360/x)/2))),x,∞)

Answer 22

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.

Answer 23

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

Answer 24

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

Answer 25

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));}}

Answer 26

R: 33 characters

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

Hopefully this follows the rules.

Answer 27

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

Answer 28

Ruby, 12

p 1.570796*2

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

Answer 29

JavaScript - 19 bytes

Math.pow(29809,1/9)

Calculates the 9th root of 29809.

3.1415914903890925

Answer 30

R

A few years back I was taking a math course and the instructor asked the class how we might compute pi from scratch. A guy at the back of the class suggested drawing a circle of diameter 1 and then laying a piece of string around it.

I couldn't figure out how to do that in R. I decided the second most primitive approach would be to approximate the circle with a regular polygons. A 4096-gon gets us 5 digits. The polygons are approximated with a simple binary search using only the midpoint and distance formulae (i.e. no trigonometric functions are used).

a <- c(0,1); b <- c(0,0); c <-c(1,0)
eps <- 0.000000001
mid  <- function(a,b) { c(mean(c(a[1],b[1])), mean(c(a[2],b[2])))}
dist <- function(a,b) { sqrt((a[1]-b[1])^2 + (a[2]-b[2])^2)}
for (i in 1:10)
{
    ab1 <- mid(a,b)
    ab2 <- mid(b, ab1)
    bc1 <- mid(b,c)
    bc2 <- mid(b, bc1)
    repeat
    {
        newab <- mid(ab1,ab2)
        newbc <- mid(bc1, bc2)
        corner <- dist(newab,newbc)
        side   <- dist(a,b) - 2*dist(newab,b)
        dif <- side - corner
        if (abs(dif) < eps)
        {
            a <- c(0,newab[2]+dist(newab,newbc))
            b <- newab
            c <- newbc
            break
        }
        if (dif > 0){ab2 <- newab;bc2 <- newbc}
        if (dif < 0){ab1 <- newab;bc1 <- newbc}
    }
}
print((2^(i+2))*dist(newab,newbc))