Monte Carlo methods: Difference between revisions

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if times=='' | times=="," then times=5000000000 /*five billion should do it, hopefully.*/
if times=='' | times=="," then times=5000000000 /*five billion should do it, hopefully.*/
if chunk=='' | chunk=="." then chunk= 10000 /*perform Monte Carlo in 10k chunks.*/
if chunk=='' | chunk=="." then chunk= 10000 /*perform Monte Carlo in 10k chunks.*/
limit=10000-1 /*REXX random generates only integers. */
limit=10000 /*REXX random generates only integers. */
limitSq=limit**2 /*··· so, instead of one, use limit**2.*/
limitSq=limit**2 /*··· so, instead of one, use limit**2.*/
accur=0 /*accuracy of Monte Carlo pi (so far). */
accur=0 /*accuracy of Monte Carlo pi (so far).*/
!=0; @reps= 'repetitions: Monte Carlo pi is' /*pi decimal digit accuracy (so far).*/
!=0; @reps= 'repetitions: Monte Carlo pi is' /*pi decimal digit accuracy (so far).*/
say /*a blank line, just for the eyeballs.*/
say /*a blank line, just for the eyeballs.*/
do j=1 for times % chunk
do j=1 for times % chunk
do chunk /*do Monte Carlo, one chunk at-a-time.*/
do chunk /*do Monte Carlo, one chunk at-a-time.*/
if random(0, limit)**2 + random(0, limit)**2 <= limitSq then !=!+1
if random(, limit)**2 + random(, limit)**2 <= limitSq then !=! + 1
end /*chunk*/
end /*chunk*/
reps=chunk * j /*calculate the number of repetitions. */
reps=chunk * j /*calculate the number of repetitions. */
_=compare(4*! / reps, pi) /*compare apples and ··· crabapples. */
_=compare(4*! / reps, pi) /*compare apples and ··· crabapples. */
if _<=accur then iterate /*if not better accuracy, keep trukin'.*/
if _<=accur then iterate /*Not better accuracy? Keep truckin'. */
say right(comma(reps), 20) @reps 'accurate to' _-1 "places." /*-1 for dec. pt.*/
say right(comma(reps), 20) @reps 'accurate to' _-1 "places." /*─1 ≡ dec. point*/
accur=_ /*use this accuracy for next baseline. */
accur=_ /*use this accuracy for next baseline. */
end /*j*/
end /*j*/

Revision as of 22:55, 20 June 2017

Task
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 .

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 .

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 .


Task

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 is not built-in, we give as a number of digits: 

            3.141592653589793238462643383280



360 Assembly

<lang 360asm>* Monte Carlo methods 08/03/2017 MONTECAR CSECT 

        USING  MONTECAR,R13       base register
        B      72(R15)            skip savearea
        DC     17F'0'             savearea
        STM    R14,R12,12(R13)    save previous context
        ST     R13,4(R15)         link backward
        ST     R15,8(R13)         link forward
        LR     R13,R15            set addressability
        LA     R8,1000            isamples=1000
        LA     R6,4               i=4
      DO WHILE=(C,R6,LE,=F'7')    do i=4 to 7
        MH     R8,=H'10'            isamples=isamples*10
        ZAP    HITS,=P'0'           hits=0
        LA     R7,1                 j=1
      DO WHILE=(CR,R7,LE,R8)        do j=1 to isamples
        BAL    R14,RNDPK              call random
        ZAP    X,RND                  x=rnd
        BAL    R14,RNDPK              call random
        ZAP    Y,RND                  y=rnd
        ZAP    WP,X                   x
        MP     WP,X                   x**2
        DP     WP,ONE                 ~
        ZAP    XX,WP(8)               x**2   normalized
        ZAP    WP,Y                   y
        MP     WP,Y                   y**2
        DP     WP,ONE                 ~
        ZAP    YY,WP(8)               y**2   normalized
        AP     XX,YY                  xx=x**2+y**2
      IF CP,XX,LT,ONE THEN            if x**2+y**2<1 then
        AP     HITS,=P'1'               hits=hits+1
      ENDIF    ,                      endif
        LA     R7,1(R7)               j++
      ENDDO    ,                    enddo j
        CVD    R8,PSAMPLES          psamples=isamples
        ZAP    WP,=P'4'             4
        MP     WP,ONE               ~
        MP     WP,HITS              *hits
        DP     WP,PSAMPLES          /psamples
        ZAP    MCPI,WP(8)           mcpi=4*hits/psamples
        XDECO  R6,WC                edit i
        MVC    PG+4(1),WC+11        output i
        MVC    WC,MASK              load mask
        ED     WC,PSAMPLES          edit psamples
        MVC    PG+6(8),WC+8         output psamples
        UNPK   WC,MCPI              unpack mcpi
        OI     WC+15,X'F0'          zap sign
        MVC    PG+31(1),WC+6        output mcpi
        MVC    PG+33(6),WC+7        output mcpi decimals
        XPRNT  PG,L'PG              print buffer
        LA     R6,1(R6)             i++
      ENDDO    ,                  enddo i
        L      R13,4(0,R13)       restore previous savearea pointer
        LM     R14,R12,12(R13)    restore previous context
        XR     R15,R15            rc=0
        BR     R14                exit

RNDPK EQU * ---- random number generator 

        ZAP    WP,RNDSEED         w=seed    
        MP     WP,RNDCNSTA        w*=cnsta
        AP     WP,RNDCNSTB        w+=cnstb
        MVC    RNDSEED,WP+8       seed=w mod 10**15
        MVC    RND,=PL8'0'        0<=rnd<1
        MVC    RND+3(5),RNDSEED+3 return rnd
        BR     R14           ---- return

PSAMPLES DS 0D,PL8 F(15,0) RNDSEED DC PL8'613058151221121' linear congruential constant RNDCNSTA DC PL8'944021285986747' " RNDCNSTB DC PL8'852529586767995' " RND DS PL8 fixed(15,9) ONE DC PL8'1.000000000' 1 fixed(15,9) HITS DS PL8 fixed(15,0) X DS PL8 fixed(15,9) Y DS PL8 fixed(15,9) MCPI DS PL8 fixed(15,9) XX DS PL8 fixed(15,9) YY DS PL8 fixed(15,9) PG DC CL80'10**x xxxxxxxx samples give Pi=x.xxxxxx' buffer MASK DC X'40202020202020202020202020202120' mask CL16 15num WC DS PL16 character 16 WP DS PL16 packed decimal 16 

        YREGS
        END    MONTECAR</lang>
Output:
10**4    10000 samples give Pi=3.129200
10**5   100000 samples give Pi=3.145000
10**6  1000000 samples give Pi=3.141180
10**7 10000000 samples give Pi=3.141677


Ada

<lang ada>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;</lang> 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

<lang algol68>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 / throws

END # 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))</lang>

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 <lang autohotkey> 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

} </lang>

AWK

<lang AWK>

  1. --- 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

</lang>

BASIC

Works with: QuickBasic version 4.5
Translation of: Java

<lang qbasic>DECLARE FUNCTION getPi! (throws!) CLS PRINT 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 / throws END FUNCTION</lang>

Output:
3.16
3.13648
3.142828
3.141679

BBC BASIC

<lang bbcbasic> 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%</lang>
Output:
     3.136
    3.1396
   3.13756
  3.143624
 3.1412816

C

<lang C>#include <stdio.h>

  1. include <stdlib.h>
  2. 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; }</lang>

C++

<lang cpp>

  1. include<iostream>
  2. include<math.h>
  3. include<stdlib.h>
  4. 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

} </lang>

C#

<lang csharp>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));
       }
   }

}</lang>

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

<lang lisp>(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))))</lang>

Output:
     100: 3.2
    1000: 3.124
   10000: 3.1376
  100000: 3.14104
 1000000: 3.141064

Common Lisp

<lang 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)))</lang>
Output:
    1000 -> 3.132
   10000 -> 3.1184
  100000 -> 3.1352
 1000000 -> 3.142072
10000000 -> 3.1420677

D

<lang 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));

}</lang>

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

<lang d>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));

}</lang>

Output:
        10: 3.200000
       100: 3.320000
      1000: 3.128000
     10000: 3.140800
    100000: 3.128400
   1000000: 3.142836
  10000000: 3.141550

Dart

From example at Dart Official Website

<lang dart> import 'dart:async'; import 'dart:html'; import 'dart:math' show Random;

// We changed 5 lines of code to make this sample nicer on // the web (so that the execution waits for animation frame, // the number gets updated in the DOM, and the program ends // after 500 iterations).

main() async { 

 print('Compute π using the Monte Carlo method.');
 var output = querySelector("#output");
 await for (var estimate in computePi().take(500)) {
   print('π ≅ $estimate');
   output.text = estimate.toStringAsFixed(5);
   await window.animationFrame;
 }

}

/// Generates a stream of increasingly accurate estimates of π. Stream<double> computePi({int batch: 100000}) async* { 

 var total = 0;
 var count = 0;
 while (true) {
   var points = generateRandom().take(batch);
   var inside = points.where((p) => p.isInsideUnitCircle);
   total += batch;
   count += inside.length;
   var ratio = count / total;
   // Area of a circle is A = π⋅r², therefore π = A/r².
   // So, when given random points with x ∈ <0,1>,
   // y ∈ <0,1>, the ratio of those inside a unit circle
   // should approach π / 4. Therefore, the value of π
   // should be:
   yield ratio * 4;
 }

}

Iterable<Point> generateRandom([int seed]) sync* { 

 final random = new Random(seed);
 while (true) {
   yield new Point(random.nextDouble(), random.nextDouble());
 }

}

class Point { 

 final double x, y;
 const Point(this.x, this.y);
 bool get isInsideUnitCircle => x * x + y * y <= 1;

} </lang>

Output:

The script give in reality an output formatted in HTML 

π ≅ 3.14139

E

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

<lang e>def pi(n) { 

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

}</lang>

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

<lang elixir>defmodule MonteCarlo do 

 def pi(n) do
   count = Enum.count(1..n, fn _ ->
     x = :rand.uniform
     y = :rand.uniform
     :math.sqrt(x*x + y*y) <= 1
   end)
   4 * count / n
 end

end

Enum.each([1000, 10000, 100000, 1000000, 10000000], fn n -> 

 :io.format "~8w samples: PI = ~f~n", [n, MonteCarlo.pi(n)]

end)</lang>

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

ERRE

<lang ERRE> PROGRAM RANDOM_PI

! ! for rosettacode.org !

!$DOUBLE

PROCEDURE MONTECARLO(T->RES) 

     LOCAL I,N
     FOR I=1 TO T DO
       IF RND(1)^2+RND(1)^2<1 THEN N+=1 END IF
     END FOR
     RES=4*N/T

END PROCEDURE

BEGIN 

     RANDOMIZE(TIMER) ! init rnd number generator
     MONTECARLO(1000->RES)     PRINT(RES)
     MONTECARLO(10000->RES)    PRINT(RES)
     MONTECARLO(100000->RES)   PRINT(RES)
     MONTECARLO(1000000->RES)  PRINT(RES)
     MONTECARLO(10000000->RES) PRINT(RES)

END PROGRAM</lang>

Output:
 3.136
 3.1468
 3.14392
 3.143824
 3.141514

Euler Math Toolbox

<lang Euler Math Toolbox> >function map MonteCarloPI (n,plot=false) ... $ X:=random(1,n); $ Y:=random(1,n); $ if plot then $ plot2d(X,Y,>points,style="."); $ plot2d("sqrt(1-x^2)",color=2,>add); $ endif $ return sum(X^2+Y^2<1)/n*4; $endfunction >MonteCarloPI(10^(1:7)) 

[ 3.6  2.96  3.224  3.1404  3.1398  3.141548  3.1421492 ]

>pi 

3.14159265359

>MonteCarloPI(10000,true): </lang>

F#

There is some support and test expressions.

<lang fsharp> let print x = printfn "%A" x

let MonteCarloPiGreco niter = 

   let eng = System.Random()
   let action () =
       let x: float = eng.NextDouble()
       let y: float = eng.NextDouble()
       let res: float = System.Math.Sqrt(x**2.0 + y**2.0)
       if res < 1.0 then
           1
       else
           0
   let res = [ for x in 1..niter do yield action() ]
   let tmp: float = float(List.reduce (+) res) / float(res.Length)
   4.0*tmp

MonteCarloPiGreco 1000 |> print MonteCarloPiGreco 10000 |> print MonteCarloPiGreco 100000 |> print </lang>

Output:
3.164
3.122
3.1436

Factor

Since Factor lets the user choose the range of the random generator, we use 2^32.

<lang factor>USING: kernel math math.functions random sequences ;

limit ( -- n ) 2 32 ^ ; inline
in-circle ( x y -- ? ) limit [ sq ] tri@ [ + ] [ <= ] bi* ;
rand ( -- r ) limit random ;
pi ( n -- pi ) [ [ drop rand rand in-circle ] count ] keep / 4 * >float ;</lang>

Example use:

<lang factor>10000 pi . 3.1412</lang>

Fantom

<lang fantom> class MontyCarlo { 

 // assume square/circle of width 1 unit
 static Float findPi (Int samples)
 {
   Int insideCircle := 0
   samples.times 
   {
     x := Float.random
     y := Float.random
     if ((x*x + y*y).sqrt <= 1.0f) insideCircle += 1
   }
   return insideCircle * 4.0f / samples
 }

 public static Void main () 
 {
   [100, 1000, 10000, 1000000, 10000000].each |sample|
   {
     echo ("Sample size $sample gives PI as ${findPi(sample)}")
   }
 }

} </lang>

Output:
Sample size 100 gives PI as 3.2
Sample size 1000 gives PI as 3.132
Sample size 10000 gives PI as 3.1612
Sample size 1000000 gives PI as 3.139316
Sample size 10000000 gives PI as 3.1409272

Forth

Works with: GNU Forth
include random.fs

10000 value r

: hit? ( -- ? )
  r random dup *
  r random dup * +
  r dup * < ;

: sims ( n -- hits )
  0 swap 0 do hit? if 1+ then loop ;

1000 sims 4 * . 3232  ok
10000 sims 4 * . 31448  ok
100000 sims 4 * . 313704  ok
1000000 sims 4 * . 3141224  ok
10000000 sims 4 * . 31409400  ok

Fortran

Works with: Fortran version 90 and later

<lang fortran>MODULE Simulation 

  IMPLICIT NONE

  CONTAINS

  FUNCTION Pi(samples)
    REAL :: Pi
    REAL :: coords(2), length
    INTEGER :: i, in_circle, samples
 
    in_circle = 0
    DO i=1, samples
      CALL RANDOM_NUMBER(coords)
      coords = coords * 2 - 1
      length = SQRT(coords(1)*coords(1) + coords(2)*coords(2))
      IF (length <= 1) in_circle = in_circle + 1
    END DO
    Pi = 4.0 * REAL(in_circle) / REAL(samples)
  END FUNCTION Pi

END MODULE Simulation
 
PROGRAM MONTE_CARLO

  USE Simulation 
  
  INTEGER :: n = 10000

  DO WHILE (n <= 100000000)
    WRITE (*,*) n, Pi(n)
    n = n * 10
  END DO
    
END PROGRAM MONTE_CARLO</lang>
Output:
        10000     3.12120
       100000     3.13772
      1000000     3.13934
     10000000     3.14114
    100000000     3.14147
Works with: Fortran version 2008 and later

<lang fortran> 

       program mc
       integer :: n,i
       real(8) :: pi
       n=10000
       do i=1,5
         print*,n,pi(n)
         n = n * 10
       end do
       end program

       function  pi(n)
       integer :: n
       real(8) :: x(2,n),pi
       call random_number(x)
       pi = 4.d0 * dble( count( hypot(x(1,:),x(2,:)) <= 1.d0 ) ) / n
       end function

</lang>

FreeBASIC

<lang freebasic>' version 23-10-2016 ' compile with: fbc -s console

Randomize Timer 'seed the random function

Dim As Double x, y, pi, error_ Dim As UInteger m = 10, n, n_start, n_stop = m, p

Print Print " Mumber of throws Ratio (Pi) Error" Print

Do 

   For n = n_start To n_stop -1
       x = Rnd
       y = Rnd
       If (x * x + y * y) <= 1 Then p = p +1
   Next
   Print Using "    ############,  "; m ;
   pi = p * 4 / m
   error_ = 3.141592653589793238462643383280 - pi
   Print RTrim(Str(pi),"0");Tab(35); Using "##.#############"; error_
   m = m * 10
   n_start = n_stop
   n_stop = m

Loop Until m > 1000000000 ' 1,000,000,000


' empty keyboard buffer While Inkey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>

Output:
 Mumber of throws  Ratio (Pi)     Error

               10  3.2            -0.0584073464102
              100  3.16           -0.0184073464102
            1,000  3.048           0.0935926535898
           10,000  3.1272          0.0143926535898
          100,000  3.13672         0.0048726535898
        1,000,000  3.14148         0.0001126535898
       10,000,000  3.1417668      -0.0001741464102
      100,000,000  3.14141         0.0001826535898
    1,000,000,000  3.14169192     -0.0000992664102

Futhark

Since Futhark is a pure language, random numbers are implemented using Sobol sequences.

<lang Futhark> import "futlib/math"

default(f32)

fun dirvcts(): [2][30]i32 = 

   [
           [
               536870912, 268435456, 134217728, 67108864, 33554432, 16777216, 8388608, 4194304, 2097152, 1048576, 524288, 262144, 131072, 65536, 32768, 16384, 8192, 4096, 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1
           ],
           [
               536870912, 805306368, 671088640, 1006632960, 570425344, 855638016, 713031680, 1069547520, 538968064, 808452096, 673710080, 1010565120, 572653568, 858980352, 715816960, 1073725440, 536879104, 805318656, 671098880, 1006648320, 570434048, 855651072, 713042560, 1069563840, 538976288, 808464432, 673720360, 1010580540, 572662306, 858993459
           ]
   ]


fun grayCode(x: i32): i32 = (x >> 1) ^ x

--- Sobol Generator

fun testBit(n: i32, ind: i32): bool = 

   let t = (1 << ind) in (n & t) == t

fun xorInds(n: i32) (dir_vs: [num_bits]i32): i32 = 

   let reldv_vals = zipWith (\ dv i  ->
                               if testBit(grayCode n,i)
                               then dv else 0)
                            dir_vs (iota num_bits)
   in reduce (^) 0 reldv_vals

fun sobolIndI (dir_vs: [m][num_bits]i32, n: i32): [m]i32 = 

   map (xorInds n) dir_vs

fun sobolIndR(dir_vs: [m][num_bits]i32) (n: i32 ): [m]f32 = 

   let divisor = 2.0 ** f32(num_bits)
   let arri    = sobolIndI( dir_vs, n )
   in map (\ (x: i32): f32  -> f32(x) / divisor) arri

fun main(n: i32): f32 = 

   let rand_nums = map (sobolIndR (dirvcts())) (iota n)
   let dists     = map (\xy ->
                          let (x,y) = (xy[0],xy[1]) in f32.sqrt(x*x + y*y))
                       rand_nums

   let bs        = map (\d -> if d <= 1.0f32 then 1 else 0) dists

   let inside    = reduce (+) 0 bs
   in 4.0f32*f32(inside)/f32(n)

</lang>

Go

<lang go>package main

import ( 

   "fmt"
   "math"
   "math/rand"
   "time"

)

func getPi(numThrows int) float64 { 

   inCircle := 0
   for 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
       randX := rand.Float64()*2 - 1 //range -1 to 1
       randY := rand.Float64()*2 - 1 //range -1 to 1
       //distance from (0,0) = sqrt((x-0)^2+(y-0)^2)
       dist := math.Hypot(randX, randY)
       if dist < 1 { //circle with diameter of 2 has radius of 1
           inCircle++
       }
   }
   return 4 * float64(inCircle) / float64(numThrows)

}

func main() { 

   rand.Seed(time.Now().UnixNano())
   fmt.Println(getPi(10000))
   fmt.Println(getPi(100000))
   fmt.Println(getPi(1000000))
   fmt.Println(getPi(10000000))
   fmt.Println(getPi(100000000))

}</lang>

Output:
3.1164
3.1462
3.142892
3.1419692
3.14149596

Haskell

<lang haskell> import System.Random import Control.Monad

get_pi throws = do results <- replicateM throws one_trial 

                  return (4 * fromIntegral (foldl (+) 0 results) / fromIntegral throws)
 where
   one_trial = do rand_x <- randomRIO (-1, 1)
                  rand_y <- randomRIO (-1, 1)
                  let dist :: Double
                      dist = sqrt (rand_x*rand_x + rand_y*rand_y)
                  return (if dist < 1 then 1 else 0)

</lang> Example: 

Prelude System.Random Control.Monad> get_pi 10000
3.1352
Prelude System.Random Control.Monad> get_pi 100000
3.15184
Prelude System.Random Control.Monad> get_pi 1000000
3.145024

HicEst

<lang HicEst>FUNCTION Pi(samples) 

  inside = 0
  DO i = 1, samples
     inside = inside + ( (RAN(1)^2 + RAN(1)^2)^0.5 <= 1)
  ENDDO
  Pi = 4 * inside / samples

END 

  WRITE(ClipBoard) Pi(1E4) ! 3.1504
  WRITE(ClipBoard) Pi(1E5) ! 3.14204
  WRITE(ClipBoard) Pi(1E6) ! 3.141672
  WRITE(ClipBoard) Pi(1E7) ! 3.1412856</lang>

Icon and Unicon

<lang Icon>procedure main() 

 every t := 10 ^ ( 5 to 9 ) do
    printf("Rounds=%d Pi ~ %r\n",t,getPi(t))

end

link printf

procedure getPi(rounds) 

  incircle := 0. 
  every 1 to rounds do 
     if 1 > sqrt((?0 * 2 - 1) ^ 2 + (?0 * 2 - 1) ^ 2) then 
        incircle +:= 1
  return 4 * incircle / rounds

end</lang>

Library: Icon Programming Library

printf.icn provides printf

Output:
Rounds=100000 Pi ~ 3.143400
Rounds=1000000 Pi ~ 3.141656
Rounds=10000000 Pi ~ 3.140437
Rounds=100000000 Pi ~ 3.141375
Rounds=1000000000 Pi ~ 3.141604

J

Explicit Solution: <lang j>piMC=: monad define "0 

 4* y%~ +/ 1>: %: +/ *: <: +: (2,y) ?@$ 0

)</lang>

Tacit Solution: <lang j>piMCt=: (0.25&* %~ +/@(1 >: [: +/&.:*: _1 2 p. 0 ?@$~ 2&,))"0</lang>

Examples: <lang j> piMC 1e6 3.1426 

  piMC 10^i.7

4 2.8 3.24 3.168 3.1432 3.14256 3.14014</lang>

Java

<lang 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; } }</lang>

Output:
3.1396
3.14256
3.141516
3.1418692
3.14168604
Works with: Java version 8+

<lang java>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;
 }

}</lang>

Output:
3.1556
3.14416
3.14098
3.1419512
3.14160312

JavaScript

ES5

<lang 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));</lang> 

3.168
3.1396
3.13692
3.140512
3.1417656

ES6

<lang JavaScript>(() => { 

   'use strict';

   // monteCarloPi :: Int -> Float
   const monteCarloPi = n =>
       4 * range(1, n)
       .reduce(a => {
           const [x, y] = [rnd(), rnd()];
           return x * x + y * y < 1 ? a + 1 : a;
       }, 0) / n;



   // GENERIC FUNCTIONS

   // range :: Int -> Int -> [Int]
   const range = (m, n) =>
       Array.from({
           length: Math.floor(n - m) + 1
       }, (_, i) => m + i);

   // rnd :: () -> Float
   const rnd = Math.random;



   // TEST with from 1000 samples to 10E8 samples
   return range(3, 8)
       .map(x => monteCarloPi(Math.pow(10, x)));

   // e.g. -> [3.14, 3.1404, 3.13304, 3.142408, 3.1420304, 3.14156788]

})(); </lang>

Output:

(5 sample runs with increasing sample sizes)

<lang JavaScript>[3.14, 3.1404, 3.13304, 3.142408, 3.1420304, 3.14156788]</lang>

Julia

<lang julia>function montepi(n) 

   s = 0
   for i = 1:n
       s += rand()^2 + rand()^2 < 1
   end
   return 4*s/n

end</lang>

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

<lang K> sim:{4*(+/{~1<+/(2_draw 0)^2}'!x)%x} 

  sim 10000

3.103 

  sim'10^!8

4 2.8 3.4 3.072 3.1212 3.14104 3.14366 3.1413</lang>

Kotlin

<lang scala>// version 1.1.0

fun mcPi(n: Int): Double { 

   var inside = 0
   (1..n).forEach {
       val x = Math.random()
       val y = Math.random()
       if (x * x + y * y <= 1.0) inside++
   }
   return 4.0 * inside / n

}

fun main(args: Array<String>) { 

   println("Iterations -> Approx Pi  -> Error%")
   println("----------    ----------    ------")
   var n = 1_000
   while (n <= 100_000_000) {
       val pi = mcPi(n)
       val err = Math.abs(Math.PI - pi) / Math.PI * 100.0
       println(String.format("%9d  -> %10.8f -> %6.4f", n, pi, err))
       n *= 10
   }

}</lang> Sample output:

Output:
Iterations -> Approx Pi  -> Error%
----------    ----------    ------
     1000  -> 3.12800000 -> 0.4327
    10000  -> 3.15040000 -> 0.2803
   100000  -> 3.14468000 -> 0.0983
  1000000  -> 3.13982000 -> 0.0564
 10000000  -> 3.14182040 -> 0.0072
100000000  -> 3.14160244 -> 0.0003

Liberty BASIC

<lang lb> 

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/throws

end function 

</lang>
Output:
100           2.89108911
1000          3.12887113
10000         3.13928607
100000        3.13864861
1000000       3.13945686

Locomotive Basic

<lang locobasic>10 mode 1:randomize time:defint a-z 20 input "How many samples";n 30 u=n/100+1 40 r=100 50 for i=1 to n 60 if i mod u=0 then locate 1,3:print using "##% done"; i/n*100 70 x=rnd*2*r-r 80 y=rnd*2*r-r 90 if sqr(x*x+y*y)<r then m=m+1 100 next 110 pi2!=4*m/n 120 locate 1,3 130 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</lang>

<lang logo> to square :n 

 output :n * :n

end to trial :r 

 output less? sum square random :r square random :r  square :r

end to sim :n :r 

 make "hits 0
 repeat :n [if trial :r [make "hits :hits + 1]]
 output 4 * :hits / :n

end

show sim 1000 10000  ; 3.18 show sim 10000 10000  ; 3.1612 show sim 100000 10000  ; 3.145 show sim 1000000 10000  ; 3.140828 </lang>

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?) <lang LSL>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."); } }</lang>

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

<lang 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_throws

end

print( MonteCarlo( 10000 ) ) print( MonteCarlo( 100000 ) ) print( MonteCarlo( 1000000 ) ) print( MonteCarlo( 10000000 ) )</lang>

Output:
3.1436
3.13636
3.14376
3.1420188

Mathematica

We define a function with variable sample size: <lang Mathematica> 

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

</lang> Example (samplesize=10,100,1000,....10000000): <lang Mathematica> 

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

</lang> gives back: <lang Mathematica> 10 3.2 100 3.16 1000 3.152 10000 3.1228 100000 3.14872 1000000 3.1408 10000000 3.14134 </lang>

<lang Mathematica> monteCarloPi = 4. Mean[UnitStep[1 - Total[RandomReal[1, {2, #}]^2]]] &; monteCarloPi /@ (10^Range@6) </lang>

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: <lang MATLAB>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 </lang>

Completely Vectorized: <lang MATLAB>function piEstimate = monteCarloPi(numDarts) 

   piEstimate = 4*sum( sum(rand(numDarts,2).^2,2) <= 1 )/numDarts;

end</lang>

Output:

<lang MATLAB>>> monteCarloPi(7000000)

ans = 

  3.141512000000000</lang>

Maxima

<lang 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));</lang>

MAXScript

fn monteCarlo iterations =
(
    radius = 1.0
    pointsInCircle = 0
    for i in 1 to iterations do
    (
        testPoint = [(random -radius radius), (random -radius radius)]
        if length testPoint <= radius then
        (
            pointsInCircle += 1
        )
    )
    4.0 * pointsInCircle / iterations
)

МК-61/52

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

Example: for n = 100 the output is 3.2.

Nim

<lang nim>import math randomize()

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)</lang>
Output:
3.15336
3.1405116
3.14163332
3.141486144

OCaml

<lang 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</lang>

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

<lang 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</lang>

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

 3.1560
 3.1496

Much faster implementation

<lang octave> function result = montepi(n) 

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

endfunction </lang>

PARI/GP

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

<lang pascal>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. </lang>

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

<lang perl>sub pi { 

 my $nthrows = shift;
 my $inside = 0;
 foreach (1 .. $nthrows) {
   my $x = rand() * 2 - 1;
   my $y = rand() * 2 - 1;
   if (sqrt($x*$x + $y*$y) < 1) {
     $inside++;
   }
 }
 return 4 * $inside / $nthrows;

}

printf "%9d: %07f\n", $_, pi($_) for 10**4, 10**6;</lang>

Output:
    10000: 3.132000
  1000000: 3.141596

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 . <lang Perl 6>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;

</lang>

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:

<lang perl6>my @pi = ([\+] 4 * (1 > [+] rand**2 xx 2) xx *) Z/ 1 .. *; say @pi[10, 1000, 10_000];</lang>

Phix

<lang Phix>integer N = 100 for i=1 to 6 do 

   integer inside = 0
   for i=1 to N do
       integer x = rand(N),
               y = rand(N)
       inside += (x*x+y*y<N*N)
   end for
   ?{N,4*inside/N}
   N *= 10

end for</lang>

Output:
{100,3.2}
{1000,3.116}
{10000,3.1736}
{100000,3.13996}
{1000000,3.141856}
{10000000,3.1415728}

PHP

<lang 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); ?></lang>

Output:
loop=1,000,000, count=785,462, pi=3.141848

PicoLisp

<lang 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)) )</lang>
Output:
3.4
3.23
3.137
3.1299
3.14360
3.140964

PowerShell

Works with: PowerShell version 2

<lang powershell>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

}</lang> 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

<lang 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() <> "" </lang>

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): <lang python>>>> 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</lang>

As a program using a function

<lang python> from random import random from math import hypot try: 

   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))

</lang>

Faster implementation using Numpy

<lang python> 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 </lang>

R

<lang 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)

}

  1. this second function changes the samples number to be
  2. 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))</lang>

Racket

<lang 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)))</lang>
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>#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)))</lang>

[Similar output]

REXX

A specific-purpose commatizer function is included to format the number of iterations. <lang rexx>/*REXX program computes and displays the value of pi÷4 using the Monte Carlo algorithm*/ /*true pi*/ pi=3.141592653589793238462643383279502884197169399375105820974944592307816406 say ' 1 2 3 4 5 6 7 ' say 'scale: 1·234567890123456789012345678901234567890123456789012345678901234567890123' say /* [↑] a two-line scale for showing pi*/ say 'true pi= ' pi"+" /*we might as well brag about true pi.*/ numeric digits length(pi) - 1 /*this program uses these decimal digs.*/ parse arg times chunk . /*does user want a specific number? */ if times== | times=="," then times=5000000000 /*five billion should do it, hopefully.*/ if chunk== | chunk=="." then chunk= 10000 /*perform Monte Carlo in 10k chunks.*/ limit=10000 /*REXX random generates only integers. */ limitSq=limit**2 /*··· so, instead of one, use limit**2.*/ accur=0 /*accuracy of Monte Carlo pi (so far).*/ !=0; @reps= 'repetitions: Monte Carlo pi is' /*pi decimal digit accuracy (so far).*/ say /*a blank line, just for the eyeballs.*/ 

     do j=1  for times % chunk
                      do chunk                  /*do Monte Carlo,  one chunk at-a-time.*/
                      if random(, limit)**2 + random(, limit)**2 <= limitSq  then !=! + 1
                      end   /*chunk*/
     reps=chunk * j                             /*calculate the number of repetitions. */
     _=compare(4*! / reps, pi)                  /*compare apples and  ···  crabapples. */
     if _<=accur  then iterate                  /*Not better accuracy?  Keep truckin'. */
     say right(comma(reps), 20) @reps  'accurate to'  _-1  "places."  /*─1 ≡ dec. point*/
     accur=_                                    /*use this accuracy for next baseline. */
     end   /*j*/

exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ comma: procedure; arg _; do k=length(_) to 1 by -3; _=insert(',',_,k); end; return _</lang>

output   when using the default input:
                    1         2         3         4         5         6         7
scale:    1·234567890123456789012345678901234567890123456789012345678901234567890123

true pi=  3.141592653589793238462643383279502884197169399375105820974944592307816406+

             10,000, repetitions:  Monte Carlo  pi  is accurate to 3 places.
             50,000, repetitions:  Monte Carlo  pi  is accurate to 4 places.
            850,000, repetitions:  Monte Carlo  pi  is accurate to 5 places.
            890,000, repetitions:  Monte Carlo  pi  is accurate to 6 places.
          5,130,000, repetitions:  Monte Carlo  pi  is accurate to 7 places.
          8,620,000, repetitions:  Monte Carlo  pi  is accurate to 8 places.
         10,390,000, repetitions:  Monte Carlo  pi  is accurate to 9 places.

Ring

<lang ring> decimals(8) see "monteCarlo(1000) = " + monteCarlo(1000) + nl see "monteCarlo(10000) = " + monteCarlo(10000) + nl see "monteCarlo(100000) = " + monteCarlo(100000) + nl

func monteCarlo t 

    n=0
    for i = 1 to t
        if sqrt(pow(random(1),2) + pow(random(1),2)) <= 1 n += 1 ok
    next
    t = (4 * n) / t
    return t

</lang> Output: 

monteCarlo(1000) = 3.11600000
monteCarlo(10000) = 3.00320000
monteCarlo(100000) = 2.99536000

Ruby

<lang 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 / throws

end

[1000, 10_000, 100_000, 1_000_000, 10_000_000].each do |n| 

     puts "%8d samples: PI = %s" % [n, approx_pi(n)]

end</lang>

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

<lang 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))
   }
 }

}</lang>

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

<lang 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;</lang>
Output:
    10000: 3.14520
   100000: 3.15000
  1000000: 3.14058
 10000000: 3.14223
100000000: 3.14159

SequenceL

First solution is serial due to the use of random numbers. Will always give the same result for a given n and seed <lang sequenceL> import <Utilities/Random.sl>; import <Utilities/Conversion.sl>;

main(args(2)) := monteCarlo(stringToInt(args[1]), stringToInt(args[2]));

monteCarlo(n, seed) := let totalHits := monteCarloHelper(n, seedRandom(seed), 0); in (totalHits / intToFloat(n))*4.0;

monteCarloHelper(n, generator, result) := let xRand := getRandom(generator); x := xRand.Value/(generator.RandomMax + 1.0); yRand := getRandom(xRand.Generator); y := yRand.Value/(generator.RandomMax + 1.0);

newResult := result + 1 when x^2 + y^2 < 1.0 else result; in result when n < 0 else monteCarloHelper(n - 1, yRand.Generator, newResult); </lang>

The second solution will run in parallel. It will also always give the same result for a given n and seed. (Note, the function monteCarloHelper is the same in both versions).

<lang sequenceL> import <Utilities/Random.sl>; import <Utilities/Conversion.sl>;

main(args(2)) := monteCarlo(stringToInt(args[1]), stringToInt(args[2]));

chunks := 100; monteCarlo3(n, seed) := let newSeeds := getRandomSequence(seedRandom(seed), chunks).Value; totalHits := monteCarloHelper(n / chunks, seedRandom(newSeeds), 0); in (sum(totalHits) / intToFloat((n / chunks)*chunks))*4.0;

monteCarloHelper(n, generator, result) := let xRand := getRandom(generator); x := xRand.Value/(generator.RandomMax + 1.0); yRand := getRandom(xRand.Generator); y := yRand.Value/(generator.RandomMax + 1.0);

newResult := result + 1 when x^2 + y^2 < 1.0 else result; in result when n < 0 else monteCarloHelper(n - 1, yRand.Generator, newResult); </lang>

Sidef

<lang ruby>func monteCarloPi(nthrows) { 

   4 * (^nthrows -> count_by {
       hypot(1.rand(2) - 1, 1.rand(2) - 1) < 1
   }) / nthrows

}

for n in [1e2, 1e3, 1e4, 1e5, 1e6] { 

   printf("%9d: %07f\n", n, monteCarloPi(n))

}</lang>

Output:
      100: 3.320000
     1000: 3.120000
    10000: 3.169600
   100000: 3.138920
  1000000: 3.142344

Swift

Translation of: JavaScript

<lang Swift>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))</lang>

Output:
3.08
3.128
3.1548
3.149
3.142032
3.1414772
3.14166832

Tcl

<lang 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]"

}</lang> result 

PI is approx 3.141592653589793

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

Ursala

<lang Ursala>#import std

  1. import flo

mcp "n" = times/4. div\float"n" (rep"n" (fleq/.5+ sqrt+ plus+ ~~ sqr+ minus/.5+ rand)?/~& plus/1.) 0.</lang> 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: <lang Ursala>#cast %eL

pis = mcp* <10,100,1000,10000,100000,1000000></lang>

Output:
<
   2.800000e+00,
   3.600000e+00,
   3.164000e+00,
   3.118800e+00,
   3.144480e+00,
   3.141668e+00>

XPL0

<lang XPL0>code Ran=1, CrLf=9; code real RlOut=48;

func real MontePi(N); \Calculate pi using Monte Carlo method int N; \number of randomly selected points int I, X, Y, C; def R = 10000; \radius of circle [C:= 0; \initialize count of points in circle for 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);

]</lang>

Output:
    2.92000
    3.13200
    3.14375
    3.14192

zkl

<lang 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

}</lang> Or, in a more functional style (using a reference for state info): <lang zkl>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;

}</lang>

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
Retrieved from "https://rosettacode.org/wiki/Monte_Carlo_methods?oldid=82626"