# Monte Carlo methods

(Redirected from Monte Carlo Simulation)
Monte Carlo methods
You are encouraged to solve this task according to the task description, using any language you may know.

A Monte Carlo Simulation is a way of approximating the value of a function where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value and then makes a sort of "best guess."

A simple Monte Carlo Simulation can be used to calculate the value for π. If you had a circle and a square where the length of a side of the square was the same as the diameter of the circle, the ratio of the area of the circle to the area of the square would be π/4.

So, if you put this circle inside the square and select many random points inside the square, the number of points inside the circle divided by the number of points inside the square and the circle would be approximately π/4.

Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.

For software where the number π is not built-in, we give π to a couple of digits: 3.141592653589793238462643383280

## Contents

with Ada.Text_IO;                use Ada.Text_IO;with Ada.Numerics.Float_Random;  use Ada.Numerics.Float_Random; procedure Test_Monte_Carlo is   Dice : Generator;    function Pi (Throws : Positive) return Float is      Inside : Natural := 0;   begin      for Throw in 1..Throws loop         if Random (Dice) ** 2 + Random (Dice) ** 2 <= 1.0 then            Inside := Inside + 1;         end if;      end loop;      return 4.0 * Float (Inside) / Float (Throws);   end Pi;begin   Put_Line ("     10_000:" & Float'Image (Pi (     10_000)));   Put_Line ("    100_000:" & Float'Image (Pi (    100_000)));   Put_Line ("  1_000_000:" & Float'Image (Pi (  1_000_000)));   Put_Line (" 10_000_000:" & Float'Image (Pi ( 10_000_000)));   Put_Line ("100_000_000:" & Float'Image (Pi (100_000_000)));end Test_Monte_Carlo;

The implementation uses built-in uniformly distributed on [0,1] random numbers. Note that the accuracy of the result depends on the quality of the pseudo random generator: its circle length and correlation to the function being simulated.

Output:
     10_000: 3.13920E+00
100_000: 3.14684E+00
1_000_000: 3.14197E+00
10_000_000: 3.14215E+00
100_000_000: 3.14151E+00


## ALGOL 68

Works with: ALGOL 68 version Standard - no extensions to language used
Works with: ALGOL 68G version Any - tested with release mk15-0.8b.fc9.i386
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8.8d.fc9.i386
PROC pi = (INT throws)REAL:BEGIN   INT inside := 0;   TO throws DO      IF random ** 2 + random ** 2 <= 1 THEN         inside +:= 1      FI   OD;   4 * inside / throwsEND # pi #; print (("     10 000:",pi (     10 000),new line));print (("    100 000:",pi (    100 000),new line));print (("  1 000 000:",pi (  1 000 000),new line));print ((" 10 000 000:",pi ( 10 000 000),new line));print (("100 000 000:",pi (100 000 000),new line))
Output:
     10 000:+3.15480000000000e  +0
100 000:+3.12948000000000e  +0
1 000 000:+3.14169200000000e  +0
10 000 000:+3.14142040000000e  +0
100 000 000:+3.14153276000000e  +0


## AutoHotkey

Search autohotkey.com: Carlo methods
Source: AutoHotkey forum by Laszlo

 MsgBox % MontePi(10000)   ; 3.154400 MsgBox % MontePi(100000)  ; 3.142040 MsgBox % MontePi(1000000) ; 3.142096  MontePi(n) {    Loop %n% {       Random x, -1, 1.0       Random y, -1, 1.0       p += x*x+y*y < 1    }    Return 4*p/n }

## AWK

 # --- with command line argument "throws" --- BEGIN{ th=ARGV[1]; for(i=0; i<th; i++) cin += (rand()^2 + rand()^2) < 1  printf("Pi = %8.5f\n",4*cin/th)} usage: awk -f pi 2300 Pi =  3.14333

## BASIC

Works with: QuickBasic version 4.5
Translation of: Java
DECLARE FUNCTION getPi! (throws!)CLSPRINT getPi(10000)PRINT getPi(100000)PRINT getPi(1000000)PRINT getPi(10000000) FUNCTION getPi (throws)	inCircle = 0		FOR i = 1 TO throws			'a square with a side of length 2 centered at 0 has			'x and y range of -1 to 1			randX = (RND * 2) - 1'range -1 to 1			randY = (RND * 2) - 1'range -1 to 1			'distance from (0,0) = sqrt((x-0)^2+(y-0)^2)			dist = SQR(randX ^ 2 + randY ^ 2)			IF dist < 1 THEN 'circle with diameter of 2 has radius of 1				inCircle = inCircle + 1			END IF		NEXT i	getPi = 4! * inCircle / throwsEND FUNCTION
Output:
3.16
3.13648
3.142828
3.141679


## BBC BASIC

      PRINT FNmontecarlo(1000)      PRINT FNmontecarlo(10000)      PRINT FNmontecarlo(100000)      PRINT FNmontecarlo(1000000)      PRINT FNmontecarlo(10000000)      END       DEF FNmontecarlo(t%)      LOCAL i%, n%      FOR i% = 1 TO t%        IF RND(1)^2 + RND(1)^2 < 1 n% += 1      NEXT      = 4 * n% / t%
Output:
     3.136
3.1396
3.13756
3.143624
3.1412816


## C

#include <stdio.h>#include <stdlib.h>#include <math.h> double pi(double tolerance){	double x, y, val, error;	unsigned long sampled = 0, hit = 0, i; 	do {		/* don't check error every turn, make loop tight */		for (i = 1000000; i; i--, sampled++) {			x = rand() / (RAND_MAX + 1.0);			y = rand() / (RAND_MAX + 1.0);			if (x * x + y * y < 1) hit ++;		} 		val = (double) hit / sampled;		error = sqrt(val * (1 - val) / sampled) * 4;		val *= 4; 		/* some feedback, or user gets bored */		fprintf(stderr, "Pi = %f +/- %5.3e at %ldM samples.\r",			val, error, sampled/1000000);	} while (!hit || error > tolerance);              /* !hit is for completeness's sake; if no hit after 1M samples,                 your rand() is BROKEN */ 	return val;} int main(){	printf("Pi is %f\n", pi(3e-4)); /* set to 1e-4 for some fun */	return 0;}

## C++

 #include<iostream>#include<math.h>#include<stdlib.h>#include<time.h> using namespace std;int main(){    int jmax=1000; // maximum value of HIT number. (Length of output file)    int imax=1000; // maximum value of random numbers for producing HITs.    double x,y;    // Coordinates    int hit;       // storage variable of number of HITs    srand(time(0));    for (int j=0;j<jmax;j++){        hit=0;        x=0; y=0;        for(int i=0;i<imax;i++){            x=double(rand())/double(RAND_MAX);            y=double(rand())/double(RAND_MAX);        if(y<=sqrt(1-pow(x,2))) hit+=1; }          //Choosing HITs according to analytic formula of circle    cout<<""<<4*double(hit)/double(imax)<<endl; }  // Print out Pi number}

## C#

using System; class Program {    static double MonteCarloPi(int n) {        int inside = 0;        Random r = new Random();         for (int i = 0; i < n; i++) {            if (Math.Pow(r.NextDouble(), 2)+ Math.Pow(r.NextDouble(), 2) <= 1) {                inside++;            }        }         return 4.0 * inside / n;    }     static void Main(string[] args) {        int value = 1000;        for (int n = 0; n < 5; n++) {            value *= 10;            Console.WriteLine("{0}:{1}", value.ToString("#,###").PadLeft(11, ' '), MonteCarloPi(value));        }    }}
Output:
     10,000:3.1436
100,000:3.14632
1,000,000:3.139476
10,000,000:3.1424476
100,000,000:3.1413976


## Clojure

(defn calc-pi [iterations]  (loop [x (rand) y (rand) in 0 total 1]     (if (< total iterations)      (recur (rand) (rand) (if (<= (+ (* x x) (* y y)) 1) (inc in) in) (inc total))      (double (* (/ in total) 4))))) (doseq [x (take 5 (iterate #(* 10 %) 10))] (println (str (format "% 8d" x) ": " (calc-pi x))))
Output:
     100: 3.2
1000: 3.124
10000: 3.1376
100000: 3.14104
1000000: 3.141064


## Common Lisp

(defun approximate-pi (n)  (/ (loop repeat n count (<= (abs (complex (random 1.0) (random 1.0))) 1.0)) n 0.25)) (dolist (n (loop repeat 5 for n = 1000 then (* n 10) collect n))  (format t "~%~8d -> ~f" n (approximate-pi n)))
Output:
    1000 -> 3.132
10000 -> 3.1184
100000 -> 3.1352
1000000 -> 3.142072
10000000 -> 3.1420677


## D

import std.stdio, std.random, std.math; double pi(in uint nthrows) /*nothrow*/ @safe /*@nogc*/ {    uint inside;    foreach (immutable i; 0 .. nthrows)        if (hypot(uniform01, uniform01) <= 1)            inside++;    return 4.0 * inside / nthrows;} void main() {    foreach (immutable p; 1 .. 8)        writefln("%10s: %07f", 10 ^^ p, pi(10 ^^ p));}
Output:
        10: 3.200000
100: 3.120000
1000: 3.076000
10000: 3.140400
100000: 3.146520
1000000: 3.140192
10000000: 3.141476
Output:
with foreach(p;1..10):
        10: 3.200000
100: 3.240000
1000: 3.180000
10000: 3.150400
100000: 3.143080
1000000: 3.140996
10000000: 3.141442
100000000: 3.141439
1000000000: 3.141559

### More Functional Style

void main() {    import std.stdio, std.random, std.math, std.algorithm, std.range;     immutable isIn = (int) => hypot(uniform01, uniform01) <= 1;    immutable pi = (in int n)  => 4.0 * n.iota.count!isIn / n;     foreach (immutable p; 1 .. 8)        writefln("%10s: %07f", 10 ^^ p, pi(10 ^^ p));}
Output:
        10: 3.200000
100: 3.320000
1000: 3.128000
10000: 3.140800
100000: 3.128400
1000000: 3.142836
10000000: 3.141550

## E

This computes a single quadrant of the described square and circle; the effect should be the same since the other three are symmetric.

def pi(n) {    var inside := 0    for _ ? (entropy.nextFloat() ** 2 + entropy.nextFloat() ** 2 < 1) in 1..n {         inside += 1    }    return inside * 4 / n}

Some sample runs:

? pi(10)
# value: 2.8

? pi(10)
# value: 2.0

? pi(100)
# value: 2.96

? pi(10000)
# value: 3.1216

? pi(100000)
# value: 3.13088

? pi(100000)
# value: 3.13848


## Elixir

defmodule MonteCarlo do  def pi(n) do    :random.seed(:os.timestamp)    count = Enum.count(1..n, fn _ ->      x = :random.uniform      y = :random.uniform      :math.sqrt(x*x + y*y) <= 1    end)    4 * count / n  endend Enum.each([1000, 10000, 100000, 1000000, 10000000], fn n ->  :io.format "~8w samples: PI = ~f~n", [n, MonteCarlo.pi(n)]end)
Output:
    1000 samples: PI = 3.112000
10000 samples: PI = 3.127200
100000 samples: PI = 3.145440
1000000 samples: PI = 3.142904
10000000 samples: PI = 3.141124


Tacit Solution:

piMCt=: (0.25&* %~ +/@(1 >: [: +/&.:*: _1 2 p. 0 ?@$~ 2&,))"0 Examples:  piMC 1e63.1426 piMC 10^i.74 2.8 3.24 3.168 3.1432 3.14256 3.14014 ## Java public class MC { public static void main(String[] args) { System.out.println(getPi(10000)); System.out.println(getPi(100000)); System.out.println(getPi(1000000)); System.out.println(getPi(10000000)); System.out.println(getPi(100000000)); } public static double getPi(int numThrows){ int inCircle= 0; for(int i= 0;i < numThrows;i++){ //a square with a side of length 2 centered at 0 has //x and y range of -1 to 1 double randX= (Math.random() * 2) - 1;//range -1 to 1 double randY= (Math.random() * 2) - 1;//range -1 to 1 //distance from (0,0) = sqrt((x-0)^2+(y-0)^2) double dist= Math.sqrt(randX * randX + randY * randY); //^ or in Java 1.5+: double dist= Math.hypot(randX, randY); if(dist < 1){//circle with diameter of 2 has radius of 1 inCircle++; } } return 4.0 * inCircle / numThrows; }} Output: 3.1396 3.14256 3.141516 3.1418692 3.14168604  Works with: Java version 8+ package montecarlo; import java.util.stream.IntStream;import java.util.stream.DoubleStream; import static java.lang.Math.random;import static java.lang.Math.hypot;import static java.lang.System.out; public interface MonteCarlo { public static void main(String... arguments) { IntStream.of( 10000, 100000, 1000000, 10000000, 100000000 ) .mapToDouble(MonteCarlo::pi) .forEach(out::println) ; } public static double range() { //a square with a side of length 2 centered at 0 has //x and y range of -1 to 1 return (random() * 2) - 1; } public static double pi(int numThrows){ long inCircle = DoubleStream.generate( //distance from (0,0) = hypot(x, y) () -> hypot(range(), range()) ) .limit(numThrows) .unordered() .parallel() //circle with diameter of 2 has radius of 1 .filter(d -> d < 1) .count() ; return (4.0 * inCircle) / numThrows; }} Output: 3.1556 3.14416 3.14098 3.1419512 3.14160312  ## JavaScript function mcpi(n){ var x,y,m=0; for(var i = 0; i < n; i += 1) { x = Math.random(); y = Math.random(); if (x*x + y*y < 1) { m += 1; } } return 4*m/n;} console.log(mcpi(1000));console.log(mcpi(10000));console.log(mcpi(100000));console.log(mcpi(1000000));console.log(mcpi(10000000)); 3.168 3.1396 3.13692 3.140512 3.1417656  ## Julia function montepi(n) s = 0 for i = 1:n s += rand()^2 + rand()^2 < 1 end return 4*s/nend Output: julia> for n in 10.^(3:8) p = montepi(n) println("$n: π ≈ $p, |error| =$(abs(p-π))")
end
1000: π ≈ 3.188, |error| = 0.04640734641020705
10000: π ≈ 3.128, |error| = 0.013592653589793002
100000: π ≈ 3.1446, |error| = 0.003007346410206946
1000000: π ≈ 3.14242, |error| = 0.000827346410206875
10000000: π ≈ 3.1413596, |error| = 0.00023305358979319735
100000000: π ≈ 3.1417196, |error| = 0.00012694641020694064


## K

   sim:{4*(+/{~1<+/(2_draw 0)^2}'!x)%x}    sim 100003.103    sim'10^!84 2.8 3.4 3.072 3.1212 3.14104 3.14366 3.1413

## Liberty BASIC

  for pow = 2 to 6    n = 10^pow    print n, getPi(n)next end function getPi(n)    incircle = 0    for throws=0 to n        scan        incircle = incircle + (rnd(1)^2+rnd(1)^2 < 1)    next    getPi = 4*incircle/throwsend function
Output:
100           2.89108911
1000          3.12887113
10000         3.13928607
100000        3.13864861
1000000       3.13945686


## Locomotive Basic

10 mode 1:randomize time:defint a-z20 input "How many samples";n30 u=n/100+140 r=10050 for i=1 to n60 if i mod u=0 then locate 1,3:print using "##% done"; i/n*10070 x=rnd*2*r-r80 y=rnd*2*r-r90 if sqr(x*x+y*y)<r then m=m+1100 next110 pi2!=4*m/n120 locate 1,3130 print m;"points in circle"140 print "Computed value of pi:"pi2!150 print "Difference to real value of pi: ";160 print using "+#.##%"; (pi2!-pi)/pi*100

## Logo

 to square :n  output :n * :nendto trial :r  output less? sum square random :r square random :r  square :rendto sim :n :r  make "hits 0  repeat :n [if trial :r [make "hits :hits + 1]]  output 4 * :hits / :nend show sim    1000 10000  ; 3.18show sim   10000 10000  ; 3.1612show sim  100000 10000  ; 3.145show sim 1000000 10000  ; 3.140828

## LSL

To test it yourself; rez a box on the ground, and add the following as a New Script. (Be prepared to wait... LSL can be slow, but the Servers are typically running thousands of scripts in parallel so what do you expect?)

integer iMIN_SAMPLE_POWER = 0;integer iMAX_SAMPLE_POWER = 6;default {	state_entry() {		llOwnerSay("Estimating Pi ("+(string)PI+")");		integer iSample = 0;		for(iSample=iMIN_SAMPLE_POWER ; iSample<=iMAX_SAMPLE_POWER  ; iSample++) {			integer iInCircle = 0;			integer x = 0;			integer iMaxSamples = (integer)llPow(10, iSample);			for(x=0 ; x<iMaxSamples ; x++) {				if(llSqrt(llPow(llFrand(2.0)-1.0, 2.0)+llPow(llFrand(2.0)-1.0, 2.0))<1.0) {					iInCircle++;				}			}			float fPi = ((4.0*iInCircle)/llPow(10, iSample));			float fError = llFabs(100.0*(PI-fPi)/PI);			llOwnerSay((string)iSample+": "+(string)iMaxSamples+" = "+(string)fPi+", Error = "+(string)fError+"%");		}		llOwnerSay("Done.");	}}
Output:
Estimating Pi (3.141593)
0: 1 = 4.000000, Error = 27.323954%
1: 10 = 4.000000, Error = 27.323954%
2: 100 = 2.880000, Error = 8.326753%
3: 1000 = 3.188000, Error = 1.477192%
4: 10000 = 3.133600, Error = 0.254414%
5: 100000 = 3.138840, Error = 0.087620%
6: 1000000 = 3.142684, Error = 0.034739%
Done.

## Lua

function MonteCarlo ( n_throws )    math.randomseed( os.time() )     n_inside = 0    for i = 1, n_throws do    	if math.random()^2 + math.random()^2 <= 1.0 then            n_inside = n_inside + 1    	end    end         return 4 * n_inside / n_throwsend print( MonteCarlo( 10000 ) )print( MonteCarlo( 100000 ) )print( MonteCarlo( 1000000 ) )print( MonteCarlo( 10000000 ) )
Output:
3.1436
3.13636
3.14376
3.1420188

## Mathematica

We define a function with variable sample size:

  MonteCarloPi[samplesize_Integer] := N[4Mean[If[# > 1, 0, 1] & /@ Norm /@ RandomReal[1, {samplesize, 2}]]]

Example (samplesize=10,100,1000,....10000000):

  {#, MonteCarloPi[#]} & /@ (10^Range[1, 7]) // Grid

gives back:

 10		3.2100		3.161000		3.15210000		3.1228100000		3.148721000000		3.140810000000	3.14134
 monteCarloPi = 4. Mean[UnitStep[1 - Total[RandomReal[1, {2, #}]^2]]] &;monteCarloPi /@ (10^Range@6)

## MATLAB

See: Monte Carlo Simulation in MATLAB for more examples

The first example given is not vectorized. MATLAB has a self-imposed memory limitation that prevents this simulation from having more than 3 decimal digits of accuracy. Because of this limitation it is best to vectorize the code as much as possible so extra memory isn't consumed by unneeded variables. Therefore, I have provided a second solution that is maximally vectorized.

Minimally Vectorized:

function piEstimate = monteCarloPi(numDarts)     %The square has a sides of length 2, which means the circle has radius    %1.     %Generate a table of random x-y value pairs in the range [0,1] sampled    %from the uniform distribution for each axis.    darts = rand(numDarts,2);     %Any darts that are in the circle will have position vector whose    %length is less than or equal to 1 squared.    dartsInside = ( sum(darts.^2,2) <= 1 );     piEstimate = 4*sum(dartsInside)/numDarts; end

Completely Vectorized:

function piEstimate = monteCarloPi(numDarts)     piEstimate = 4*sum( sum(rand(numDarts,2).^2,2) <= 1 )/numDarts; end
Output:
>> monteCarloPi(7000000) ans =    3.141512000000000

## Maxima

load("distrib");approx_pi(n):= block(  [x: random_continuous_uniform(0, 1, n),   y: random_continuous_uniform(0, 1, n),   r, cin: 0, listarith: true],   r: x^2 + y^2,   for r0 in r do if r0<1 then cin: cin + 1,   4*cin/n); float(approx_pi(100));

## MAXScript

fn monteCarlo iterations =
(
pointsInCircle = 0
for i in 1 to iterations do
(
if length testPoint <= radius then
(
pointsInCircle += 1
)
)
4.0 * pointsInCircle / iterations
)


## МК-61/52

П0	П1	0	П4	СЧ	x^2	^	СЧ	x^2	+1	-	x<0	15	КИП4	L0	04	ИП4	4	*ИП1	/	С/П

Example: for n = 100 the output is 3.2.

## Nim

import mathrandomize() proc pi(nthrows): float =  var inside = 0  for i in 1..int64(nthrows):    if hypot(random(1.0), random(1.0)) < 1:      inc inside  return float(4 * inside) / nthrows for n in [10e4, 10e6, 10e7, 10e8]:  echo pi(n)
Output:
3.15336
3.1405116
3.14163332
3.141486144

## OCaml

let get_pi throws =  let rec helper i count =    if i = throws then count    else      let rand_x = Random.float 2.0 -. 1.0      and rand_y = Random.float 2.0 -. 1.0 in      let dist = sqrt (rand_x *. rand_x +. rand_y *. rand_y) in      if dist < 1.0 then        helper (i+1) (count+1)      else        helper (i+1) count  in float (4 * helper 0 0) /. float throws

Example:

# get_pi 10000;;
- : float = 3.15
# get_pi 100000;;
- : float = 3.13272
# get_pi 1000000;;
- : float = 3.143808
# get_pi 10000000;;
- : float = 3.1421704
# get_pi 100000000;;
- : float = 3.14153872


## Octave

function p = montepi(samples)  in_circle = 0;  for samp = 1:samples    v = [ unifrnd(-1,1), unifrnd(-1,1) ];    if ( v*v.' <= 1.0 )      in_circle++;    endif  endfor  p = 4*in_circle/samples;endfunction l = 1e4;while (l < 1e7)  disp(montepi(l));  l *= 10;endwhile

Since it runs slow, I've stopped it at the second iteration, obtaining:

 3.1560
3.1496

###  Much faster implementation

 function result = montepi(n)  result = sum(rand(1,n).^2+rand(1,n).^2<1)/n*4;endfunction

## PARI/GP

MonteCarloPi(tests)=4.*sum(i=1,tests,norml2([random(1.),random(1.)])<1)/tests;

A hundred million tests (about a minute) yielded 3.14149000, slightly more precise (and round!) than would have been expected. A million gave 3.14162000 and a thousand 3.14800000.

## Pascal

Library: Math
Program MonteCarlo(output); uses  Math; function MC_Pi(expo: integer): real;  var    x, y: real;    i, hits, samples: longint;  begin    samples := 10**expo;    hits := 0;    randomize;    for i := 1 to samples do    begin      x := random;      y := random;      if sqrt(x*x + y*y) < 1.0 then        inc(hits);    end;    MC_Pi := 4.0 * hits / samples;  end; var  i: integer;begin  for i := 4 to 8 do    writeln (10**i, ' samples give ', MC_Pi(i):7:5, ' as pi.');end.
Output:
:> ./MonteCarlo
10000 samples give 3.14480 as pi.
100000 samples give 3.14484 as pi.
1000000 samples give 3.13970 as pi.
10000000 samples give 3.14100 as pi.
100000000 samples give 3.14162 as pi.


## Perl

sub pi {  my $nthrows = shift; my$inside = 0;  foreach (1 .. $nthrows) { my$x = rand * 2 - 1,       $y = rand * 2 - 1; if (sqrt($x*$x +$y*$y) < 1) {$inside++;    }  }  return 4 * $inside /$nthrows;} printf "%9d: %07f\n", $_, pi($_) foreach 10**4, 10**6;

## Perl 6

Works with: rakudo version 2015-09-24

We'll consider the upper-right quarter of the unitary disk centered at the origin. Its area is $\pi \over 4$.

my @random_distances = ([+] rand**2 xx 2) xx *; sub approximate_pi(Int $n) { 4 * @random_distances[^$n].grep(* < 1) / $n} say "Monte-Carlo π approximation:";say "$_ iterations:  ", approximate_pi $_ for 100, 1_000, 10_000;  Output: Monte-Carlo π approximation: 100 iterations: 2.88 1000 iterations: 3.096 10000 iterations: 3.1168 We don't really need to write a function, though. A lazy list would do: my @pi = ([\+] 4 * (1 > [+] rand**2 xx 2) xx *) Z/ 1 .. *;say @pi[10, 1000, 10_000]; ## PHP <?$loop = 1000000; # loop to 1,000,000$count = 0;for ($i=0; $i<$loop; $i++) {$x = rand() / getrandmax();  $y = rand() / getrandmax(); if(($x*$x) + ($y*$y)<=1)$count++;}echo "loop=".number_format($loop).", count=".number_format($count).", pi=".($count/$loop*4);?>
Output:
loop=1,000,000, count=785,462, pi=3.141848

## PicoLisp

(de carloPi (Scl)   (let (Dim (** 10 Scl)  Dim2 (* Dim Dim)  Pi 0)      (do (* 4 Dim)         (let (X (rand 0 Dim)  Y (rand 0 Dim))            (when (>= Dim2 (+ (* X X) (* Y Y)))               (inc 'Pi) ) ) )      (format Pi Scl) ) ) (for N 6   (prinl (carloPi N)) )
Output:
3.4
3.23
3.137
3.1299
3.14360
3.140964

## PowerShell

Works with: PowerShell version 2
function Get-Pi ($Iterations = 10000) {$InCircle = 0    for ($i = 0;$i -lt $Iterations;$i++) {        $x = Get-Random 1.0$y = Get-Random 1.0        if ([Math]::Sqrt($x *$x + $y *$y) -le 1) {            $InCircle++ } }$Pi = [decimal] $InCircle /$Iterations * 4    $RealPi = [decimal] "3.141592653589793238462643383280"$Diff = [Math]::Abs(($Pi -$RealPi) / $RealPi * 100) New-Object PSObject  | Add-Member -PassThru NoteProperty Iterations$Iterations         | Add-Member -PassThru NoteProperty Pi $Pi  | Add-Member -PassThru NoteProperty "% Difference"$Diff}

This returns a custom object with appropriate properties which automatically enables a nice tabular display:

PS Home:\> 10,100,1e3,1e4,1e5,1e6 | ForEach-Object { Get-Pi $_ } Iterations Pi % Difference ---------- -- ------------ 10 3,6 14,591559026164641753596309630 100 3,40 8,225361302488828322840959090 1000 3,208 2,1138114877600474293158225800 10000 3,1444 0,0893606116311387583356211100 100000 3,14712 0,1759409006731298209938938800 1000000 3,141364 0,0072782698142600895432451100 ## PureBasic OpenConsole() Procedure.d MonteCarloPi(throws.d) inCircle.d = 0 For i = 1 To throws.d randX.d = (Random(2147483647)/2147483647)*2-1 randY.d = (Random(2147483647)/2147483647)*2-1 dist.d = Sqr(randX.d*randX.d + randY.d*randY.d) If dist.d < 1 inCircle = inCircle + 1 EndIf Next i pi.d = (4 * inCircle / throws.d) ProcedureReturn pi.d EndProcedure PrintN ("'built-in' #Pi = " + StrD(#PI,20))PrintN ("MonteCarloPi(10000) = " + StrD(MonteCarloPi(10000),20))PrintN ("MonteCarloPi(100000) = " + StrD(MonteCarloPi(100000),20))PrintN ("MonteCarloPi(1000000) = " + StrD(MonteCarloPi(1000000),20))PrintN ("MonteCarloPi(10000000) = " + StrD(MonteCarloPi(10000000),20)) PrintN("Press any key"): Repeat: Until Inkey() <> ""  Output: 'built-in' #PI = 3.14159265358979310000 MonteCarloPi(10000) = 3.17119999999999980000 MonteCarloPi(100000) = 3.14395999999999990000 MonteCarloPi(1000000) = 3.14349599999999980000 MonteCarloPi(10000000) = 3.14127720000000020000 Press any key ## Python ### At the interactive prompt Python 2.6rc2 (r26rc2:66507, Sep 18 2008, 14:27:33) [MSC v.1500 32 bit (Intel)] on win32 IDLE 2.6rc2 One use of the "sum" function is to count how many times something is true (because True = 1, False = 0): >>> import random, math>>> throws = 1000>>> 4.0 * sum(math.hypot(*[random.random()*2-1 for q in [0,1]]) < 1 for p in xrange(throws)) / float(throws)3.1520000000000001>>> throws = 1000000>>> 4.0 * sum(math.hypot(*[random.random()*2-1 for q in [0,1]]) < 1 for p in xrange(throws)) / float(throws)3.1396359999999999>>> throws = 100000000>>> 4.0 * sum(math.hypot(*[random.random()*2-1 for q in [0,1]]) < 1 for p in xrange(throws)) / float(throws)3.1415666400000002 ### As a program using a function  from random import randomfrom math import hypottry: import psyco psyco.full()except: pass def pi(nthrows): inside = 0 for i in xrange(nthrows): if hypot(random(), random()) < 1: inside += 1 return 4.0 * inside / nthrows for n in [10**4, 10**6, 10**7, 10**8]: print "%9d: %07f" % (n, pi(n))  ### Faster implementation using Numpy  import numpy as np n = input('Number of samples: ')print np.sum(np.random.rand(n)**2+np.random.rand(n)**2<1)/float(n)*4  ## R # nice but not suitable for big samples!monteCarloPi <- function(samples) { x <- runif(samples, -1, 1) # for big samples, you need a lot of memory! y <- runif(samples, -1, 1) l <- sqrt(x*x + y*y) return(4*sum(l<=1)/samples)} # this second function changes the samples number to be# multiple of group parameter (default 100).monteCarlo2Pi <- function(samples, group=100) { lim <- ceiling(samples/group) olim <- lim c <- 0 while(lim > 0) { x <- runif(group, -1, 1) y <- runif(group, -1, 1) l <- sqrt(x*x + y*y) c <- c + sum(l <= 1) lim <- lim - 1 } return(4*c/(olim*group))} print(monteCarloPi(1e4))print(monteCarloPi(1e5))print(monteCarlo2Pi(1e7)) ## Racket #lang racket (define (in-unit-circle? x y) (<= (sqrt (+ (sqr x) (sqr y))) 1));; point in ([-1,1], [-1,1])(define (random-point-in-2x2-square) (values (* 2 (- (random) 1/2)) (* 2 (- (random) 1/2)))) ;; Area of circle is (pi r^2). r is 1, area of circle is pi;; Area of square is 2^2 = 4;; There is a pi/4 chance of landing in circle;; .: pi = 4*(proportion passed) = 4*(passed/samples)(define (passed:samples->pi passed samples) (* 4 (/ passed samples))) ;; generic kind of monte-carlo simulation(define (monte-carlo run-length report-frequency sample-generator pass? interpret-result) (let inner ((samples 0) (passed 0) (cnt report-frequency)) (cond [(= samples run-length) (interpret-result passed samples)] [(zero? cnt) ; intermediate report (printf "~a samples of ~a: ~a passed -> ~a~%" samples run-length passed (interpret-result passed samples)) (inner samples passed report-frequency)] [else (inner (add1 samples) (if (call-with-values sample-generator pass?) (add1 passed) passed) (sub1 cnt))]))) ;; (monte-carlo ...) gives an "exact" result... which will be a fraction.;; to see how it looks as a decimal we can exact->inexact it(let ((mc (monte-carlo 10000000 1000000 random-point-in-2x2-square in-unit-circle? passed:samples->pi))) (printf "exact = ~a~%inexact = ~a~%(pi - guess) = ~a~%" mc (exact->inexact mc) (- pi mc))) Output: 1000000 samples of 10000000: 785763 passed -> 785763/250000 2000000 samples of 10000000: 1571487 passed -> 1571487/500000 3000000 samples of 10000000: 2356776 passed -> 98199/31250 4000000 samples of 10000000: 3141924 passed -> 785481/250000 5000000 samples of 10000000: 3927540 passed -> 196377/62500 6000000 samples of 10000000: 4713072 passed -> 98189/31250 7000000 samples of 10000000: 5498300 passed -> 54983/17500 8000000 samples of 10000000: 6283199 passed -> 6283199/2000000 9000000 samples of 10000000: 7068065 passed -> 1413613/450000 exact = 3926793/1250000 inexact = 3.1414344 (pi - guess) = 0.00015825358979304482 A little more Racket-like is the use of an iterator (in this case for/fold), which is clearer than an inner function: #lang racket(define (in-unit-circle? x y) (<= (sqrt (+ (sqr x) (sqr y))) 1));; Good idea made in another task that:;; The proportions of hits is the same in the unit square and 1/4 of a circle.;; point in ([0,1], [0,1])(define (random-point-in-unit-square) (values (random) (random)));; generic kind of monte-carlo simulation;; Area of circle is (pi r^2). r is 1, area of circle is pi;; Area of square is 2^2 = 4;; There is a pi/4 chance of landing in circle;; .: pi = 4*(proportion passed) = 4*(passed/samples)(define (passed:samples->pi passed samples) (* 4 (/ passed samples))) (define (monte-carlo/2 run-length report-frequency sample-generator pass? interpret-result) (interpret-result (for/fold ((pass 0)) ([n (in-range run-length)] #:when (when (and (not (zero? n)) (zero? (modulo n report-frequency))) (printf "~a samples of ~a: ~a passed -> ~a~%" n run-length pass (interpret-result pass n))) #:when (call-with-values sample-generator pass?)) (add1 pass)) run-length)) ;; (monte-carlo ...) gives an "exact" result... which will be a fraction.;; to see how it looks as a decimal we can exact->inexact it(let ((mc (monte-carlo/2 10000000 1000000 random-point-in-unit-square in-unit-circle? passed:samples->pi))) (printf "exact = ~a~%inexact = ~a~%(pi - guess) = ~a~%" mc (exact->inexact mc) (- pi mc))) [Similar output] ## REXX /*REXX program computes pi ÷ 4 using the Monte Carlo algorithm. */parse arg times chunks . /*does user want a specific number? */if times=='' then times=1000000000 /*one billion should do it, me thinks. */if chunks=='' then chunks=10000 /*do Monte Carlo in 10,000 chunks. */limit=10000-1 /*REXX random generates only integers. */limitSq=limit**2 /*··· so, instead of one, use limit**2.*/!=0 /*the number of "pi hits" (so far). */accur=0 /*accuracy of Monte Carlo pi (so far). */if 1=='f1'x then piChar='pi' /*if EBCDIC, then use literal. */ else piChar='e3'x /* " ASCII, " " pi glyph, */ pi=3.14159265358979323846264338327950288419716939937511 /*this, da real McCoy*/numeric digits length(pi) /*this program uses these decimal digs.*/say 'real pi='pi"+" /*we might as well brag about it. */say /*a blank line, just for the eyeballs. */ do j=1 for times%chunks do chunks /*do Monte Carlo, one chunk-at-a-time.*/ if random(0,limit)**2 + random(0,limit)**2 <=limitSq then !=!+1 end /*chunks*/ reps=chunks*j /*compute the number of repetitions. */ piX=4*!/reps /*let's see how this puppy does so far.*/ _=compare(piX,pi) /*compare apples and ··· crabapples. */ if _<=accur then iterate /*if not better accuracy, keep going. */ say right(commas(reps),20) 'repetitions: Monte Carlo' piChar, "is accurate to" _-1 'places.' /*subtract one for decimal point.*/ accur=_ /*use this accuracy for the baseline. */ end /*j*/exit /*stick a fork in it, we're all done. *//*────────────────────────────────────────────────────────────────────────────*/commas: procedure; parse arg _; n=_'.9'; #=123456789; b=verify(n,#,"M") e=verify(n,#'0',,verify(n,#"0.",'M'))-4 do j=e to b by -3; _=insert(',',_,j); end /*j*/; return _ output when using the default input: real pi=3.14159265358979323846264338327950288419716939937511+ 10,000 repetitions: Monte Carlo π is accurate to 3 places. 30,000 repetitions: Monte Carlo π is accurate to 4 places. 110,000 repetitions: Monte Carlo π is accurate to 5 places. 2,270,000 repetitions: Monte Carlo π is accurate to 6 places. 2,680,000 repetitions: Monte Carlo π is accurate to 7 places. 14,180,000 repetitions: Monte Carlo π is accurate to 9 places.  ## Ruby def approx_pi(throws) times_inside = throws.times.count {Math.hypot(rand, rand) <= 1.0} # in pre-Ruby-1.8.7: times_inside = throws.times.select {Math.hypot(rand, rand) <= 1.0}.length 4.0 * times_inside / throwsend [1000, 10_000, 100_000, 1_000_000, 10_000_000].each do |n| puts "%8d samples: PI = %s" % [n, approx_pi(n)]end Output:  1000 samples: PI = 3.2 10000 samples: PI = 3.14 100000 samples: PI = 3.13244 1000000 samples: PI = 3.145124 10000000 samples: PI = 3.1414788 ## Scala object MonteCarlo { private val random = new scala.util.Random /** Returns a random number between -1 and 1 */ def nextThrow: Double = (random.nextDouble * 2.0) - 1.0 /** Returns true if the argument point would be 'inside' the unit circle with * center at the origin, and bounded by a square with side lengths of 2 * units. */ def insideCircle(pt: (Double, Double)): Boolean = pt match { case (x, y) => (x * x) + (y * y) <= 1.0 } /** Runs the simulation the specified number of times. Uses the result to * estimate a value of pi */ def simulate(times: Int): Double = { val inside = Iterator.tabulate (times) (_ => (nextThrow, nextThrow)) count insideCircle inside.toDouble / times.toDouble * 4.0 } def main(args: Array[String]): Unit = { val sims = Seq(10000, 100000, 1000000, 10000000, 100000000) sims.foreach { n => println(n+" simulations; pi estimation: "+ simulate(n)) } }} Output: 10000 simulations; pi estimation: 3.1492 100000 simulations; pi estimation: 3.1396 1000000 simulations; pi estimation: 3.14208 10000000 simulations; pi estimation: 3.1409944 100000000 simulations; pi estimation: 3.1414386 ## Seed7 $ include "seed7_05.s7i";  include "float.s7i"; const func float: pi (in integer: throws) is func  result    var float: pi is 0.0;  local    var integer: throw is 0;    var integer: inside is 0;  begin    for throw range 1 to throws do      if rand(0.0, 1.0) ** 2 + rand(0.0, 1.0) ** 2 <= 1.0 then        incr(inside);      end if;    end for;    pi := flt(4 * inside) / flt(throws);  end func; const proc: main is func  begin    writeln("    10000: " <& pi(    10000) digits 5);    writeln("   100000: " <& pi(   100000) digits 5);    writeln("  1000000: " <& pi(  1000000) digits 5);    writeln(" 10000000: " <& pi( 10000000) digits 5);    writeln("100000000: " <& pi(100000000) digits 5);  end func;
Output:
    10000: 3.14520
100000: 3.15000
1000000: 3.14058
10000000: 3.14223
100000000: 3.14159


## Swift

Translation of: JavaScript
import Foundation func mcpi(sampleSize size:Int) -> Double {    var x = 0 as Double    var y = 0 as Double    var m = 0 as Double     for i in 0..<size {        x = Double(arc4random()) / Double(UINT32_MAX)        y = Double(arc4random()) / Double(UINT32_MAX)         if ((x * x) + (y * y) < 1) {            m += 1        }    }     return (4.0 * m) / Double(size)} println(mcpi(sampleSize: 100))println(mcpi(sampleSize: 1000))println(mcpi(sampleSize: 10000))println(mcpi(sampleSize: 100000))println(mcpi(sampleSize: 1000000))println(mcpi(sampleSize: 10000000))println(mcpi(sampleSize: 100000000))
Output:
3.08
3.128
3.1548
3.149
3.142032
3.1414772
3.14166832


## Tcl

proc pi {samples} {    set i 0    set inside 0    while {[incr i] <= $samples} { if {sqrt(rand()**2 + rand()**2) <= 1.0} { incr inside } } return [expr {4.0 *$inside / $samples}]} puts "PI is approx [expr {atan(1)*4}]\n"foreach runs {1e2 1e4 1e6 1e8} { puts "$runs => [pi \$runs]"}

result

PI is approx 3.141592653589793

1e2 => 2.92
1e4 => 3.1344
1e6 => 3.141924
1e8 => 3.14167724

## Ursala

#import std#import flo mcp "n" = times/4. div\float"n" (rep"n" (fleq/.5+ sqrt+ plus+ ~~ sqr+ minus/.5+ rand)?/~& plus/1.) 0.

Here's a walk through.

• mcp "n" = ... defines a function named mcp in terms of a dummy variable "n", which will be the number of iterations used in the simulation
• rand ignores its argument and returns a uniformly distributed number between 0 and 1
• minus/.5 is composed with rand to compute the difference between the random number and 0.5
• sqr squares the difference
• ~~ says to apply the function twice and return the pair of results
• plus composed with that adds the pair of results
• sqrt takes the square root of the sum
• fleq/.5 is floating point comparison with a fixed right side of .5, returning true if its argument is greater or equal
• Everything from fleq to rand forms the predicate for the ? conditional operator.
• If the condition is true, the identity function is applied, ~&
• If the condition is false, the plus/1. function is applied, which adds one to its argument.
• rep"n" applied to a function has the effect of composing that function with itself "n" times, with "n" in this case being the parameter to the mcp function.
• The function being repeated "n" times is applied to an argument of 0.
• A division of the result by the number "n" converted to a floating point value is performed by div\float"n".
• The result of the division is quadrupled by times/4..

test program:

#cast %eL pis = mcp* <10,100,1000,10000,100000,1000000>
Output:
<
2.800000e+00,
3.600000e+00,
3.164000e+00,
3.118800e+00,
3.144480e+00,
3.141668e+00>

## XPL0

code Ran=1, CrLf=9;code real RlOut=48; func real MontePi(N);   \Calculate pi using Monte Carlo methodint  N;                 \number of randomly selected pointsint  I, X, Y, C;def  R = 10000;         \radius of circle[C:= 0;                 \initialize count of points in circlefor I:= 0 to N-1 do        [X:= Ran(R);         Y:= Ran(R);        if X*X + Y*Y <= R*R then C:= C+1;        ];return float(C)*4.0 / float(N);   \Acir/Asqr = pi*R^2/4*R^2 = pi/4]; [RlOut(0, MontePi(        100));  CrLf(0); RlOut(0, MontePi(     10_000));  CrLf(0); RlOut(0, MontePi(  1_000_000));  CrLf(0); RlOut(0, MontePi(100_000_000));  CrLf(0);]
Output:
    2.92000
3.13200
3.14375
3.14192


## zkl

fcn monty(n){   inCircle:=0;    do(n){      x:=(0.0).random(1); y:=(0.0).random(1);      if(x*x + y*y < 1.0) inCircle+=1;   }   4.0*inCircle/n}

Or, in a more functional style (using a reference for state info):

fcn monty(n){   4.0 * (1).pump(n,Void,fcn(r){      x:=(0.0).random(1); y:=(0.0).random(1);      if(x*x + y*y < 1.0) r.inc();       r   }.fp(Ref(0)) ).value/n;}
Output:
T(100,1000,10000,0d100_000,0d1_000_000,0d10_000_000)
.apply2(fcn(n){"%10,d : %+f".fmt(n,monty(n)-(1.0).pi).println()})
100 : -0.061593
1,000 : +0.018407
10,000 : -0.013993
100,000 : -0.000833
1,000,000 : -0.004385
10,000,000 : +0.000619
`