Monte Carlo methods

From Rosetta Code
(Redirected from Monte Carlo Simulation)
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 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

Ada[edit]

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[edit]

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 / 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))
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[edit]

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[edit]

 
# --- 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[edit]

Works with: QuickBasic version 4.5
Translation of: Java
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
Output:
3.16
3.13648
3.142828
3.141679

BBC BASIC[edit]

      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[edit]

#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++[edit]

 
#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#[edit]

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[edit]

(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[edit]

(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[edit]

import std.stdio, std.random, std.math;
 
double pi(in uint nthrows) /*nothrow*/ @safe [email protected]*/ {
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[edit]

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[edit]

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[edit]

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
end
end
 
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

ERRE[edit]

 
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
Output:
 3.136
 3.1468
 3.14392
 3.143824
 3.141514

Euler Math Toolbox[edit]

 
>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):
 

Test.png

Factor[edit]

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

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 ;

Example use:

10000 pi .
3.1412

Fantom[edit]

 
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)}")
}
}
}
 
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[edit]

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[edit]

Works with: Fortran version 90 and later
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
Output:
        10000     3.12120
       100000     3.13772
      1000000     3.13934
     10000000     3.14114
    100000000     3.14147

Go[edit]

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))
}
Output:
3.1164
3.1462
3.142892
3.1419692
3.14149596

Haskell[edit]

 
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)
 

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[edit]

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

Icon and Unicon[edit]

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

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[edit]

Explicit Solution:

piMC=: monad define "0
4* y%~ +/ 1>: %: +/ *: <: +: (2,y) ?@$ 0
)

Tacit Solution:

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

Examples:

   piMC 1e6
3.1426
piMC 10^i.7
4 2.8 3.24 3.168 3.1432 3.14256 3.14014

Java[edit]

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[edit]

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[edit]

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

   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

Liberty BASIC[edit]

 
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
 
 
Output:
100           2.89108911
1000          3.12887113
10000         3.13928607
100000        3.13864861
1000000       3.13945686

Locomotive Basic[edit]

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

Monte Carlo, 200 points, Locomotive BASIC.png Monte Carlo, 5000 points, Locomotive BASIC.png

[edit]

 
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
 

LSL[edit]

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[edit]

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 ) )
Output:
3.1436
3.13636
3.14376
3.1420188

Mathematica[edit]

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 3.2
100 3.16
1000 3.152
10000 3.1228
100000 3.14872
1000000 3.1408
10000000 3.14134
 
 
monteCarloPi = 4. Mean[UnitStep[1 - Total[RandomReal[1, {2, #}]^2]]] &;
monteCarloPi /@ ([email protected])
 

MATLAB[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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)
Output:
3.15336
3.1405116
3.14163332
3.141486144

OCaml[edit]

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[edit]

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[edit]

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

PARI/GP[edit]

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[edit]

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[edit]

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[edit]

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 .

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[edit]

<?
$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[edit]

(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[edit]

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[edit]

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[edit]

At the interactive prompt[edit]

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[edit]

 
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))
 

Faster implementation using Numpy[edit]

 
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[edit]

# 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[edit]

#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[edit]

/*REXX program computes and displays the value of  pi÷4  using the Monte Carlo algorithm*/
pi=3.141592653589793238462643383279502884197169399375105820974944592307816406 /*true pi.*/
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=1000000000 /*one billion should do it, hopefully. */
if chunk=='' | chunk=="." then chunk= 10000 /*perform Monte Carlo in 10k chunks.*/
limit=10000-1 /*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(0,limit)**2 + random(0,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 /*if not better accuracy, keep trukin'.*/
say right(commas(reps),20) @reps 'accurate to' _-1 "places." /*-1 for dec. pt.*/
accur=_ /*use this accuracy for next 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 inputs:

                  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.
              20,000 repetitions:  Monte Carlo  pi  is accurate to 4 places.
             630,000 repetitions:  Monte Carlo  pi  is accurate to 5 places.
             700,000 repetitions:  Monte Carlo  pi  is accurate to 7 places.
          20,950,000 repetitions:  Monte Carlo  pi  is accurate to 8 places.
          26,130,000 repetitions:  Monte Carlo  pi  is accurate to 10 places.
 Error 4 running "c:\montecpi.rex", line 18: Program interrupted

Ring[edit]

 
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
 

Output:

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

Ruby[edit]

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
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[edit]

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[edit]

$ 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[edit]

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[edit]

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[edit]

#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[edit]

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);
]
Output:
    2.92000
    3.13200
    3.14375
    3.14192

zkl[edit]

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