Monte Carlo methods
(Redirected from Monte Carlo Simulation)
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
11l
F monte_carlo_pi(n)
V inside = 0
L 1..n
V x = random:()
V y = random:()
I x * x + y * y <= 1
inside++
R 4.0 * inside / n
print(monte_carlo_pi(1000000))- Output:
3.13775
360 Assembly
* 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- 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
Action!
INCLUDE "H6:REALMATH.ACT"
DEFINE PTR="CARD"
DEFINE REAL_SIZE="6"
BYTE ARRAY realArray(1536)
PTR FUNC RealArrayPointer(BYTE i)
PTR p
p=realArray+i*REAL_SIZE
RETURN (p)
PROC InitRealArray()
REAL r2,r255,ri,div
REAL POINTER pow
INT i
IntToReal(2,r2)
IntToReal(255,r255)
FOR i=0 TO 255
DO
IntToReal(i,ri)
RealDiv(ri,r255,div)
pow=RealArrayPointer(i)
Power(div,r2,pow)
OD
RETURN
PROC CalcPi(INT n REAL POINTER pi)
BYTE x,y
INT i,counter
REAL tmp1,tmp2,tmp3,r1,r4
REAL POINTER pow
counter=0
IntToReal(1,r1)
IntToReal(4,r4)
FOR i=1 TO n
DO
x=Rand(0)
pow=RealArrayPointer(x)
RealAssign(pow,tmp1)
y=Rand(0)
pow=RealArrayPointer(y)
RealAssign(pow,tmp2)
RealAdd(tmp1,tmp2,tmp3)
IF RealGreaterOrEqual(tmp3,r1)=0 THEN
counter==+1
FI
OD
IntToReal(counter,tmp1)
RealMult(r4,tmp1,tmp2)
IntToReal(n,tmp3)
RealDiv(tmp2,tmp3,pi)
RETURN
PROC Test(INT n)
REAL pi
PrintF("%I samples -> ",n)
CalcPi(n,pi)
PrintRE(pi)
RETURN
PROC Main()
Put(125) PutE() ;clear the screen
PrintE("Initialization of data...")
InitRealArray()
Test(10)
Test(100)
Test(1000)
Test(10000)
RETURN- Output:
Screenshot from Atari 8-bit computer
Initialization of data... 10 samples -> 3.2 100 samples -> 3.28 1000 samples -> 3.212 10000 samples -> 3.1156
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;
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
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
Arturo
Pi: function [throws][
inside: new 0.0
do.times: throws [
if 1 > hypot random 0 1.0 random 0 1.0 -> inc 'inside
]
return 4 * inside / throws
]
loop [100 1000 10000 100000 1000000] 'n ->
print [pad to :string n 8 "=>" Pi n]
- Output:
100 => 3.4
1000 => 3.112
10000 => 3.1392
100000 => 3.14368
1000000 => 3.14106
AutoHotkey
Search autohotkey.com: Carlo methods
Source: AutoHotkey forum by Laszlo
MsgBox % MontePi(10000) ; 3.154400
MsgBox % MontePi(100000) ; 3.142040
MsgBox % MontePi(1000000) ; 3.142096
MontePi(n) {
Loop %n% {
Random x, -1, 1.0
Random y, -1, 1.0
p += x*x+y*y < 1
}
Return 4*p/n
}
AWK
# --- with command line argument "throws" ---
BEGIN{ th=ARGV[1];
for(i=0; i<th; i++) cin += (rand()^2 + rand()^2) < 1
printf("Pi = %8.5f\n",4*cin/th)
}
usage: awk -f pi 2300
Pi = 3.14333
BASIC
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
BASIC256
# Monte Carlo Simulator
# Determine value of pi
# 21010513
tosses = 1000
in_c = 0
i = 0
for i = 1 to tosses
x = rand
y = rand
x2 = x * x
y2 = y * y
xy = x2 + y2
d_xy = sqr(xy)
if d_xy <= 1 then
in_c += 1
endif
next i
print float(4*in_c/tosses)
- Output:
Throws Result 1000 3.208 10000 3.142 20000 3.1388 40000 3.1452
Other solution:
print " Number of throws Ratio (Pi) Error"
for pow = 2 to 8
n = 10 ^ pow
pi_ = getPi(n)
error_ = 3.141592653589793238462643383280 - pi_
print rjust(string(int(n)), 17); " "; ljust(string(pi_), 13); " "; ljust(string(error_), 13)
next
end
function getPi(n)
incircle = 0.0
for throws = 0 to n
incircle = incircle + (rand()^2 + rand()^2 < 1)
next
return 4.0 * incircle / throws
end function
- Output:
Number of throws Ratio (Pi) Error
100 2.970297 0.17129562389
1000 3.14085914086 0.00073351273
10000 3.13208679132 0.00950586227
100000 3.14428855711 -0.00269590352
1000000 3.14041685958 0.00117579401
10000000 3.14094968591 0.000643
100000000 3.14153 0.00006264501
BBC BASIC
PRINT FNmontecarlo(1000)
PRINT FNmontecarlo(10000)
PRINT FNmontecarlo(100000)
PRINT FNmontecarlo(1000000)
PRINT FNmontecarlo(10000000)
END
DEF FNmontecarlo(t%)
LOCAL i%, n%
FOR i% = 1 TO t%
IF RND(1)^2 + RND(1)^2 < 1 n% += 1
NEXT
= 4 * n% / t%
- Output:
3.136
3.1396
3.13756
3.143624
3.1412816
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
- 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
Liberty BASIC
for pow = 2 to 6
n = 10^pow
print n, getPi(n)
next
end
function getPi(n)
incircle = 0
for throws=0 to n
scan
incircle = incircle + (rnd(1)^2+rnd(1)^2 < 1)
next
getPi = 4*incircle/throws
end function- Output:
100 2.89108911 1000 3.12887113 10000 3.13928607 100000 3.13864861 1000000 3.13945686
Locomotive Basic
10 mode 1:randomize time:defint a-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
Run BASIC
for pow = 2 to 6
n = 10 ^ pow
print n; chr$(9); getPi(n)
next
end
function getPi(n)
incircle = 0
for throws = 0 to n
incircle = incircle + (rnd(1)^2 + rnd(1)^2 < 1)
next
getPi = 4 * incircle / throws
end function
- Output:
100 3.12 1000 3.108 10000 3.1652 100000 3.14248 1000000 3.1435
True BASIC
FUNCTION getpi(throws)
LET 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
LET randx = (rnd*2)-1 !range -1 to 1
LET randy = (rnd*2)-1 !range -1 to 1
!distance from (0,0) = sqrt((x-0)^2+(y-0)^2)
LET dist = sqr(randx^2+randy^2)
IF dist < 1 then !circle with diameter of 2 has radius of 1
LET incircle = incircle+1
END IF
NEXT i
LET getpi = 4*incircle/throws
END FUNCTION
CLEAR
PRINT getpi(10000)
PRINT getpi(100000)
PRINT getpi(1000000)
PRINT getpi(10000000)
END
- Output:
3.1304 3.14324 3.141716 3.1416452
PureBasic
OpenConsole()
Procedure.d MonteCarloPi(throws.d)
inCircle.d = 0
For i = 1 To throws.d
randX.d = (Random(2147483647)/2147483647)*2-1
randY.d = (Random(2147483647)/2147483647)*2-1
dist.d = Sqr(randX.d*randX.d + randY.d*randY.d)
If dist.d < 1
inCircle = inCircle + 1
EndIf
Next i
pi.d = (4 * inCircle / throws.d)
ProcedureReturn pi.d
EndProcedure
PrintN ("'built-in' #Pi = " + StrD(#PI,20))
PrintN ("MonteCarloPi(10000) = " + StrD(MonteCarloPi(10000),20))
PrintN ("MonteCarloPi(100000) = " + StrD(MonteCarloPi(100000),20))
PrintN ("MonteCarloPi(1000000) = " + StrD(MonteCarloPi(1000000),20))
PrintN ("MonteCarloPi(10000000) = " + StrD(MonteCarloPi(10000000),20))
PrintN("Press any key"): Repeat: Until Inkey() <> ""
- Output:
'built-in' #PI = 3.14159265358979310000 MonteCarloPi(10000) = 3.17119999999999980000 MonteCarloPi(100000) = 3.14395999999999990000 MonteCarloPi(1000000) = 3.14349599999999980000 MonteCarloPi(10000000) = 3.14127720000000020000 Press any key
C
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
double pi(double tolerance)
{
double x, y, val, error;
unsigned long sampled = 0, hit = 0, i;
do {
/* don't check error every turn, make loop tight */
for (i = 1000000; i; i--, sampled++) {
x = rand() / (RAND_MAX + 1.0);
y = rand() / (RAND_MAX + 1.0);
if (x * x + y * y < 1) hit ++;
}
val = (double) hit / sampled;
error = sqrt(val * (1 - val) / sampled) * 4;
val *= 4;
/* some feedback, or user gets bored */
fprintf(stderr, "Pi = %f +/- %5.3e at %ldM samples.\r",
val, error, sampled/1000000);
} while (!hit || error > tolerance);
/* !hit is for completeness's sake; if no hit after 1M samples,
your rand() is BROKEN */
return val;
}
int main()
{
printf("Pi is %f\n", pi(3e-4)); /* set to 1e-4 for some fun */
return 0;
}
C#
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
C++
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<time.h>
using namespace std;
int main(){
int jmax=1000; // maximum value of HIT number. (Length of output file)
int imax=1000; // maximum value of random numbers for producing HITs.
double x,y; // Coordinates
int hit; // storage variable of number of HITs
srand(time(0));
for (int j=0;j<jmax;j++){
hit=0;
x=0; y=0;
for(int i=0;i<imax;i++){
x=double(rand())/double(RAND_MAX);
y=double(rand())/double(RAND_MAX);
if(y<=sqrt(1-pow(x,2))) hit+=1; } //Choosing HITs according to analytic formula of circle
cout<<""<<4*double(hit)/double(imax)<<endl; } // Print out Pi number
}
Clojure
(defn calc-pi [iterations]
(loop [x (rand) y (rand) in 0 total 1]
(if (< total iterations)
(recur (rand) (rand) (if (<= (+ (* x x) (* y y)) 1) (inc in) in) (inc total))
(double (* (/ in total) 4)))))
(doseq [x (take 5 (iterate #(* 10 %) 10))] (println (str (format "% 8d" x) ": " (calc-pi x))))
- Output:
100: 3.2
1000: 3.124
10000: 3.1376
100000: 3.14104
1000000: 3.141064
(defn experiment
[]
(if (<= (+ (Math/pow (rand) 2) (Math/pow (rand) 2)) 1) 1 0))
(defn pi-estimate
[n]
(* 4 (float (/ (reduce + (take n (repeatedly experiment))) n))))
(pi-estimate 10000)
- Output:
3.1347999572753906
Common Lisp
(defun approximate-pi (n)
(/ (loop repeat n count (<= (abs (complex (random 1.0) (random 1.0))) 1.0)) n 0.25))
(dolist (n (loop repeat 5 for n = 1000 then (* n 10) collect n))
(format t "~%~8d -> ~f" n (approximate-pi n)))
- Output:
1000 -> 3.132 10000 -> 3.1184 100000 -> 3.1352 1000000 -> 3.142072 10000000 -> 3.1420677
Crystal
def approx_pi(throws)
times_inside = throws.times.count {Math.hypot(rand, rand) <= 1.0}
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 = 3.1 10000 samples: PI = 3.1428 100000 samples: PI = 3.1454 1000000 samples: PI = 3.141012 10000000 samples: PI = 3.141148
D
import std.stdio, std.random, std.math;
double pi(in uint nthrows) /*nothrow*/ @safe /*@nogc*/ {
uint inside;
foreach (immutable i; 0 .. nthrows)
if (hypot(uniform01, uniform01) <= 1)
inside++;
return 4.0 * inside / nthrows;
}
void main() {
foreach (immutable p; 1 .. 8)
writefln("%10s: %07f", 10 ^^ p, pi(10 ^^ p));
}
- Output:
10: 3.200000
100: 3.120000
1000: 3.076000
10000: 3.140400
100000: 3.146520
1000000: 3.140192
10000000: 3.141476
- Output:
with foreach(p;1..10)
10: 3.200000
100: 3.240000
1000: 3.180000
10000: 3.150400
100000: 3.143080
1000000: 3.140996
10000000: 3.141442
100000000: 3.141439
1000000000: 3.141559
More Functional Style
void main() {
import std.stdio, std.random, std.math, std.algorithm, std.range;
immutable isIn = (int) => hypot(uniform01, uniform01) <= 1;
immutable pi = (in int n) => 4.0 * n.iota.count!isIn / n;
foreach (immutable p; 1 .. 8)
writefln("%10s: %07f", 10 ^^ p, pi(10 ^^ p));
}
- Output:
10: 3.200000
100: 3.320000
1000: 3.128000
10000: 3.140800
100000: 3.128400
1000000: 3.142836
10000000: 3.141550
Dart
From example at Dart Official Website
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;
}
- Output:
The script give in reality an output formatted in HTML
π ≅ 3.14139
Delphi
function MonteCarloPi(N: cardinal): double;
{Approximate Pi by seeing if points fall inside circle}
var I,InsideCnt: integer;
var X,Y: double;
begin
InsideCnt:=0;
for I:=1 to N do
begin
{Random X,Y = 0..1}
X:=Random;
Y:=Random;
{See if it falls in Unit Circle}
if X*X + Y*Y <= 1 then Inc(InsideCnt);
end;
{Because X and Y are squared, they only fall with 1/4 of the circle}
Result:=4 * InsideCnt / N;
end;
procedure ShowOneSimulation(Memo: TMemo; N: cardinal);
var MyPi: double;
begin
MyPi:=MonteCarloPi(N);
Memo.Lines.Add(Format('Samples: %15.0n Pi= %2.15f',[N+0.0,MyPi]));
end;
procedure ShowMonteCarloPi(Memo: TMemo);
begin
ShowOneSimulation(Memo,1000);
ShowOneSimulation(Memo,10000);
ShowOneSimulation(Memo,100000);
ShowOneSimulation(Memo,1000000);
ShowOneSimulation(Memo,10000000);
ShowOneSimulation(Memo,100000000);
end;
- Output:
Samples: 1,000 Pi= 3.156000000000000 Samples: 10,000 Pi= 3.152000000000000 Samples: 100,000 Pi= 3.142920000000000 Samples: 1,000,000 Pi= 3.140864000000000 Samples: 10,000,000 Pi= 3.141990800000000 Samples: 100,000,000 Pi= 3.141426720000000
E
This computes a single quadrant of the described square and circle; the effect should be the same since the other three are symmetric.
def pi(n) {
var inside := 0
for _ ? (entropy.nextFloat() ** 2 + entropy.nextFloat() ** 2 < 1) in 1..n {
inside += 1
}
return inside * 4 / n
}Some sample runs:
? pi(10) # value: 2.8 ? pi(10) # value: 2.0 ? pi(100) # value: 2.96 ? pi(10000) # value: 3.1216 ? pi(100000) # value: 3.13088
? pi(100000) # value: 3.13848
EasyLang
func mc n .
for i = 1 to n
x = randomf
y = randomf
if x * x + y * y < 1
hit += 1
.
.
return 4 * hit / n
.
numfmt 4 0
print mc 10000
print mc 100000
print mc 1000000
print mc 10000000
Output:
3.1292 3.1464 3.1407 3.1413
EDSAC order code
Because real numbers on EDSAC were restricted to the interval [-1,1), this solution estimates pi/10 instead of pi. With 100,000 trials the program would have taken about 3.5 hours on the original EDSAC.
[Monte Carlo solution for Rosetta Code.]
[EDSAC program, Initial Orders 2.]
[Arrange the storage]
T45K P56F [H parameter: library s/r P1 to print real number]
T46K P78F [N parameter: library s/r P7 to print integer]
T47K P210F [M parameter: main routine]
T48K P114F [& (delta) parameter: library s/r C6 (division)]
T49K P150F [L parameter: library subroutine R4 to read data]
T51K P172F [G parameter: generator for pseudo-random numbers]
[Library subroutine M3, runs at load time and is then overwritten.
Prints header; here the header sets teleprinter to figures.]
PFGKIFAFRDLFUFOFE@A6FG@E8FEZPF
*!!!!TRIALS!!EST!PI#X10@&..
PK [after header, blank tape and PK (WWG, 1951, page 91)]
[================ Main routine ====================]
E25K TM GK
[Variables]
[0] PF PF [target count: print result when count = target]
[2] PF PF [count of points]
[4] PF PF [count of hits (point inside circle)]
[6] PF PF [x-coordinate - 1/2]
[Constants]
T8#Z PF T10#Z PF [clear sandwich bits in 35-bit constants]
T8Z [resume normal loading]
[8] PD PF [35-bit constant 1]
[10] L1229F Y819F [35-bit constant 2/5 (near enough)]
[12] IF [1/2]
[13] RF [1/4]
[14] #F [figures shift]
[15] MF [dot (decimal point) in figures mode]
[16] @F [carriage return]
[17] &F [line feed]
[18] !F [space]
[Enter with acc = 0]
[19] A19@ GL [read seed for LCG into 0D]
AD T4D [pass seed to LCG in 4D]
[23] A23@ GG [initialize LCG]
T2#@ T4#@ [zero trials and hits]
[Outer loop: round target counts]
[27] TF [clear acc]
[28] A28@ GL [read next target count into 0D]
SD [acc := -target]
E85@ [exit if target = 0]
T#@ [store negated target]
[Inner loop : round points in the square]
[33] TF T4D [pass LCG range = 0 to return random real in [0,1)]
[35] A35@ G1G [call LCG, 0D := random x]
AD S12@ T6#@ [store x - 1/2 over next call]
T4D
[41] A41@ G1G [call LCG, 0D := random y]
AD S12@ TD [store y - 1/2]
H6#@ V6#@ [acc := (x - 1/2)^2]
HD VD [acc := acc := (x - 1/2)^2 + (y - 1/2)^2]
S13@ [test for point inside circle, i.e. acc < 1/4]
E56@ [skip if not]
TF A4#@ A8#@ T4#@ [inc number of hits]
[56] TF A2#@ A8#@ U2#@ [inc number of trials]
A#@ [add negated target]
G33@ [if not reached target, loop back]
A2#@ TD [pass number of trials to print s/r]
[64] A64@ GN [print number of trials]
A4#@ TD A2#@ T4D [pass hits and trials to division s/r]
[70] A70@ G& [0D := hits/trials, estimated value of pi/4]
HD V10#@ TD [times 2/5; pass estimated pi/10 to print s/r]
O18@ O18@ O8@ O15@ [print ' 0.']
[79] A79@ GH P5F [print estimated pi/10 to 5 decimals]
O16@ O17@ [print CR, LF]
E27@ [loop back for new target]
[85] O14@ [exit: print dummy character to flush printer buffer]
ZF [halt program]
[==================== Generator for pseudo-random numbers ===========]
[Linear congruential generator, same algorithm as Delphi 7 LCG.
38 locations]
E25K TG
GK G10@ G15@ T2#Z PF T2Z I514D P257F T4#Z PF T4Z PD PF T6#Z PF T6Z PF RF A6#@ S4#@ T6#@ E25F E8Z PF T8Z PF PF A3F T14@ A4D T8#@ ZF A3F T37@ H2#@ V8#@ L512F L512F L1024F A4#@ T8#@ H6#@ C8#@ T8#@ S4D G32@ TD A8#@ E35@ H4D TD V8#@ L1F TD ZF
[==================== LIBRARY SUBROUTINES ============================]
[D6: Division, accurate, fast.
36 locations, workspace 6D and 8D.
0D := 0D/4D, where 4D <> 0, -1.]
E25K T& GK
GKA3FT34@S4DE13@T4DSDTDE2@T4DADLDTDA4DLDE8@RDU4DLDA35@
T6DE25@U8DN8DA6DT6DH6DS6DN4DA4DYFG21@SDVDTDEFW1526D
[R4: Input of one signed integer at runtime.
22 storage locations; working positions 4, 5, and 6.]
E25K TL
GKA3FT21@T4DH6@E11@P5DJFT6FVDL4FA4DTDI4FA4FS5@G7@S5@G20@SDTDT6FEF
[P1: Prints non-negative fraction in 0D, without '0.']
E25K TH
GKA18@U17@S20@T5@H19@PFT5@VDUFOFFFSFL4FTDA5@A2FG6@EFU3FJFM1F
[P7, prints long strictly positive integer;
10 characters, right justified, padded left with spaces.
Even address; 35 storage locations; working position 4D.]
E25K TN
GKA3FT26@H28#@NDYFLDT4DS27@TFH8@S8@T1FV4DAFG31@SFLDUFOFFFSF
L4FT4DA1FA27@G11@XFT28#ZPFT27ZP1024FP610D@524D!FO30@SFL8FE22@
[===================================================================]
[The following, without the comments and white space, might have
been input from a separate tape.]
E25K TM GK
E19Z [define entry point]
PF [acc = 0 on entry]
[Integers supplied by user: (1) seed for LCG; (2) list of numbers of trials
for which to print result; increasing order, terminated by 0.
To be read by library subroutine R4; sign comes after value.]
987654321+100+1000+10000+100000+0+- Output:
TRIALS EST PI/10
100 0.32400
1000 0.31319
10000 0.31371
100000 0.31410
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)
- 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
Erlang
With inline test
-module(monte).
-export([main/1]).
monte(N)->
monte(N,0,0).
monte(0,InCircle,NumPoints) ->
4 * InCircle / NumPoints;
monte(N,InCircle,NumPoints)->
Xcoord = rand:uniform(),
Ycoord = rand:uniform(),
monte(N-1,
if Xcoord*Xcoord + Ycoord*Ycoord < 1 -> InCircle + 1; true -> InCircle end,
NumPoints + 1).
main(N) -> io:format("PI: ~w~n", [ monte(N) ]).
- Output:
8> [monte:main(X) || X <- [10000,100000,100000,10000000] ]. PI: 3.136 PI: 3.1464 PI: 3.1412 PI: 3.1416704 [ok,ok,ok,ok]
With test in a function
-module(monte2).
-export([main/1]).
monte(N)->
monte(N,0,0).
monte(0,InCircle,NumPoints) ->
4 * InCircle / NumPoints;
monte(N,InCircle,NumPoints)->
X = rand:uniform(),
Y = rand:uniform(),
monte(N-1, within(X,Y,InCircle), NumPoints + 1).
within(X,Y,IN)->
if X*X + Y*Y < 1 -> IN + 1;
true -> IN
end.
main(N) -> io:format("PI: ~w~n", [ monte(N) ]).
- Output:
Xcoord 6> [monte2:main(X) || X <- [10000000,1000000,100000,10000] ]. PI: 3.1424172 PI: 3.140544 PI: 3.14296 PI: 3.1252 [ok,ok,ok,ok]
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- Output:
3.136 3.1468 3.14392 3.143824 3.141514
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):
F#
There is some support and test expressions.
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
- Output:
3.164 3.122 3.1436
Factor
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
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
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
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
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
Futhark
Since Futhark is a pure language, random numbers are implemented using Sobol sequences.
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)
Go
Using standard library math/rand:
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
Using x/exp/rand:
For very careful Monte Carlo studies, you might consider the subrepository rand library. The random number generator there has some advantages such as better known statistical properties and better use of memory.
package main
import (
"fmt"
"math"
"time"
"golang.org/x/exp/rand"
)
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(uint64(time.Now().UnixNano()))
fmt.Println(getPi(10000))
fmt.Println(getPi(100000))
fmt.Println(getPi(1000000))
fmt.Println(getPi(10000000))
fmt.Println(getPi(100000000))
}
Haskell
import Control.Monad
import System.Random
getPi throws = do
results <- replicateM throws one_trial
return (4 * fromIntegral (sum 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)
- Output:
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
Or, using foldM, and dropping sqrt:
import Control.Monad (foldM, (>=>))
import System.Random (randomRIO)
import Data.Functor ((<&>))
------- APPROXIMATION TO PI BY A MONTE CARLO METHOD ------
monteCarloPi :: Int -> IO Double
monteCarloPi n =
(/ fromIntegral n) . (4 *) . fromIntegral
<$> foldM go 0 [1 .. n]
where
rnd = randomRIO (0, 1) :: IO Double
go a _ = rnd >>= ((<&>) rnd . f a)
f a x y
| 1 > x ** 2 + y ** 2 = succ a
| otherwise = a
--------------------------- TEST -------------------------
main :: IO ()
main =
mapM_
(monteCarloPi >=> print)
[1000, 10000, 100000, 1000000]
- Output:
For example:
3.244 3.1116 3.14116 3.141396
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) ! 3.1412856Icon and Unicon
- 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:
piMC=: monad define "0
4* y%~ +/ 1>: %: +/ *: <: +: (2,y) ?@$ 0
)
Tacit Solution:
piMCt=: (0.25&* %~ +/@(1 >: [: +/&.:*: _1 2 p. 0 ?@$~ 2&,))"0
Examples:
piMC 1e6
3.1426
piMC 10^i.7
4 2.8 3.24 3.168 3.1432 3.14256 3.14014
Alternative Tacit Solution:
pimct=. (4 * +/ % #)@:(1 >: |)@:(? j. ?)@:($&0)"0
(,. pimct) 10 ^ 3 + i.6
1000 3.168
10000 3.122
100000 3.13596
1e6 3.1428
1e7 3.14158
1e8 3.14154
Java
public class MC {
public static void main(String[] args) {
System.out.println(getPi(10000));
System.out.println(getPi(100000));
System.out.println(getPi(1000000));
System.out.println(getPi(10000000));
System.out.println(getPi(100000000));
}
public static double getPi(int numThrows){
int inCircle= 0;
for(int i= 0;i < numThrows;i++){
//a square with a side of length 2 centered at 0 has
//x and y range of -1 to 1
double randX= (Math.random() * 2) - 1;//range -1 to 1
double randY= (Math.random() * 2) - 1;//range -1 to 1
//distance from (0,0) = sqrt((x-0)^2+(y-0)^2)
double dist= Math.sqrt(randX * randX + randY * randY);
//^ or in Java 1.5+: double dist= Math.hypot(randX, randY);
if(dist < 1){//circle with diameter of 2 has radius of 1
inCircle++;
}
}
return 4.0 * inCircle / numThrows;
}
}
- Output:
3.1396 3.14256 3.141516 3.1418692 3.14168604
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
ES5
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
ES6
(() => {
"use strict";
// --- APPROXIMATION OF PI BY A MONTE CARLO METHOD ---
// monteCarloPi :: Int -> Float
const monteCarloPi = n =>
4 * enumFromTo(1)(n).reduce(a => {
const [x, y] = [rnd(), rnd()];
return (x ** 2) + (y ** 2) < 1 ? (
1 + a
) : a;
}, 0) / n;
// --------------------- GENERIC ---------------------
// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = m =>
n => Array.from({
length: 1 + n - m
}, (_, i) => m + i);
// rnd :: () -> Float
const rnd = Math.random;
// ---------------------- TEST -----------------------
// From 1000 samples to 10E7 samples
return enumFromTo(3)(7).forEach(x => {
const nSamples = 10 ** x;
console.log(
`${nSamples} samples: ${monteCarloPi(nSamples)}`
);
});
})();
- Output:
For example
1000 samples: 3.064 10000 samples: 3.1416 100000 samples: 3.14756 1000000 samples: 3.142536 10000000 samples: 3.142808
jq
Adapted from Wren
Works with gojq, the Go implementation of jq
jq does not have a built-in PRNG so we will use /dev/urandom as a source of entropy by invoking jq as follows:
# In case gojq is used, trim leading 0s:
function prng {
cat /dev/urandom | tr -cd '0-9' | fold -w 10 | sed 's/^0*\(.*\)*\(.\)*$/\1\2/'
}
prng | jq -nMr -f program.jq
program.jq
def lpad($len): tostring | ($len - length) as $l | (" " * $l)[:$l] + .;
def percent: "\(100000 * . | round / 1000)%";
def pi: 4* (1|atan);
def rfloat: input/1E10;
def mcPi:
. as $n
| reduce range(0; $n) as $i (0;
rfloat as $x
| rfloat as $y
| if ($x*$x + $y*$y <= 1) then . + 1 else . end)
| 4 * . / $n ;
"Iterations -> Approx Pi -> Error",
"---------- ---------- ------",
( pi as $pi
| range(1; 7)
| pow(10;.) as $p
| ($p | mcPi) as $mcpi
| ((($pi - $mcpi)|length) / $pi) as $error
| "\($p|lpad(10)) \($mcpi|lpad(10)) \($error|percent|lpad(6))" )- Output:
Iterations -> Approx Pi -> Error
---------- ---------- ------
10 2.8 10.873%
100 3.28 4.406%
1000 3.172 0.968%
10000 3.1456 0.128%
100000 3.13316 0.268%
1000000 3.139956 0.052%
Jsish
From Javascript ES5 entry, with PRNG seeded during unit testing for reproducibility.
/* Monte Carlo methods, in Jsish */
function mcpi(n) {
var x, y, m = 0;
for (var i = 0; i < n; i += 1) {
x = Math.random();
y = Math.random();
if (x * x + y * y < 1) {
m += 1;
}
}
return 4 * m / n;
}
if (Interp.conf('unitTest')) {
Math.srand(0);
; mcpi(1000);
; mcpi(10000);
; mcpi(100000);
; mcpi(1000000);
}
/*
=!EXPECTSTART!=
mcpi(1000) ==> 3.108
mcpi(10000) ==> 3.1236
mcpi(100000) ==> 3.13732
mcpi(1000000) ==> 3.142124
=!EXPECTEND!=
*/
- Output:
prompt$ jsish -u monteCarlos.jsi [PASS] monteCarlos.jsi
Julia
using Printf
function monteπ(n)
s = count(rand() ^ 2 + rand() ^ 2 < 1 for _ in 1:n)
return 4s / n
end
for n in 10 .^ (3:8)
p = monteπ(n)
println("$(lpad(n, 9)): π ≈ $(lpad(p, 10)), pct.err = ", @sprintf("%2.5f%%", 100 * abs(p - π) / π))
end
- Output:
1000: π ≈ 3.224, pct.err = 0.02623%
10000: π ≈ 3.1336, pct.err = 0.254%
100000: π ≈ 3.13468, pct.err = 0.220%
1000000: π ≈ 3.14156, pct.err = 0.001%
10000000: π ≈ 3.1412348, pct.err = 0.011%
100000000: π ≈ 3.14123216, pct.err = 0.011%
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
Kotlin
// 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
}
}
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
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 ; 3.140828LSL
To test it yourself; rez a box on the ground, and add the following as a New Script. (Be prepared to wait... LSL can be slow, but the Servers are typically running thousands of scripts in parallel so what do you expect?)
integer iMIN_SAMPLE_POWER = 0;
integer iMAX_SAMPLE_POWER = 6;
default {
state_entry() {
llOwnerSay("Estimating Pi ("+(string)PI+")");
integer iSample = 0;
for(iSample=iMIN_SAMPLE_POWER ; iSample<=iMAX_SAMPLE_POWER ; iSample++) {
integer iInCircle = 0;
integer x = 0;
integer iMaxSamples = (integer)llPow(10, iSample);
for(x=0 ; x<iMaxSamples ; x++) {
if(llSqrt(llPow(llFrand(2.0)-1.0, 2.0)+llPow(llFrand(2.0)-1.0, 2.0))<1.0) {
iInCircle++;
}
}
float fPi = ((4.0*iInCircle)/llPow(10, iSample));
float fError = llFabs(100.0*(PI-fPi)/PI);
llOwnerSay((string)iSample+": "+(string)iMaxSamples+" = "+(string)fPi+", Error = "+(string)fError+"%");
}
llOwnerSay("Done.");
}
}
- Output:
Estimating Pi (3.141593) 0: 1 = 4.000000, Error = 27.323954% 1: 10 = 4.000000, Error = 27.323954% 2: 100 = 2.880000, Error = 8.326753% 3: 1000 = 3.188000, Error = 1.477192% 4: 10000 = 3.133600, Error = 0.254414% 5: 100000 = 3.138840, Error = 0.087620% 6: 1000000 = 3.142684, Error = 0.034739% Done.
Lua
function MonteCarlo ( n_throws )
math.randomseed( os.time() )
n_inside = 0
for i = 1, n_throws do
if math.random()^2 + math.random()^2 <= 1.0 then
n_inside = n_inside + 1
end
end
return 4 * n_inside / n_throws
end
print( MonteCarlo( 10000 ) )
print( MonteCarlo( 100000 ) )
print( MonteCarlo( 1000000 ) )
print( MonteCarlo( 10000000 ) )
- Output:
3.1436 3.13636 3.14376 3.1420188
Mathematica /Wolfram Language
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 /@ (10^Range@6)
A less elegant way to solve the problem, is to imagine a (well-trained) monkey, throwing a number of darts at a dartboard.
The darts land randomly on the board, at different x and y coordinates. We want to know how many darts land inside the circle. We then guess Pi by dividing the total number of darts inside the circle by the total number of darts thrown (assuming they all hit the square board), and multiplying the whole lot by 4.
We create a function MonkeyDartsPi, which can take a variable number of throws as input:
MonkeyDartsPi[numberOfThrows_] := (
xyCoordinates = RandomReal[{0, 1}, {numberOfThrows, 2}];
InsideCircle = Length[Select[Total[xyCoordinates^2, {2}],#<=1&]] ;
4*N[InsideCircle / Length[xyCoordinates],1+Log10[numberOfThrows]])We do several runs with a larger number of throws each time, increasing by powers of 10.
Grid[Table[{n, MonkeyDartsPi[n]}, {n, 10^Range[7]} ], Alignment -> Left]We see that as the number of throws increases, we get closer to the value of Pi:
10 2.8 100 3.20 1000 3.176 10000 3.1356 100000 3.13700 1000000 3.142624 10000000 3.1416328
MATLAB
See: Monte Carlo Simulation in MATLAB for more examples
The first example given is not vectorized. MATLAB has a self-imposed memory limitation that prevents this simulation from having more than 3 decimal digits of accuracy. Because of this limitation it is best to vectorize the code as much as possible so extra memory isn't consumed by unneeded variables. Therefore, I have provided a second solution that is maximally vectorized.
Minimally Vectorized:
function piEstimate = monteCarloPi(numDarts)
%The square has a sides of length 2, which means the circle has radius
%1.
%Generate a table of random x-y value pairs in the range [0,1] sampled
%from the uniform distribution for each axis.
darts = rand(numDarts,2);
%Any darts that are in the circle will have position vector whose
%length is less than or equal to 1 squared.
dartsInside = ( sum(darts.^2,2) <= 1 );
piEstimate = 4*sum(dartsInside)/numDarts;
end
Completely Vectorized:
function piEstimate = monteCarloPi(numDarts)
piEstimate = 4*sum( sum(rand(numDarts,2).^2,2) <= 1 )/numDarts;
end
- Output:
>> monteCarloPi(7000000)
ans =
3.141512000000000
Maxima
load("distrib");
approx_pi(n):= block(
[x: random_continuous_uniform(0, 1, n),
y: random_continuous_uniform(0, 1, n),
r, cin: 0, listarith: true],
r: x^2 + y^2,
for r0 in r do if r0<1 then cin: cin + 1,
4*cin/n);
float(approx_pi(100));
MAXScript
fn monteCarlo iterations =
(
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
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 = 1000 the output is 3.152.
Nim
import math, random
randomize()
proc pi(nthrows: float): float =
var inside = 0.0
for i in 1..int64(nthrows):
if hypot(rand(1.0), rand(1.0)) < 1:
inside += 1
result = 4 * inside / nthrows
for n in [10e4, 10e6, 10e7, 10e8]:
echo pi(n)
- Output:
3.15336 3.1405116 3.14163332 3.141486144
OCaml
let get_pi throws =
let rec helper i count =
if i = throws then count
else
let rand_x = Random.float 2.0 -. 1.0
and rand_y = Random.float 2.0 -. 1.0 in
let dist = sqrt (rand_x *. rand_x +. rand_y *. rand_y) in
if dist < 1.0 then
helper (i+1) (count+1)
else
helper (i+1) count
in float (4 * helper 0 0) /. float throws
Example:
# get_pi 10000;; - : float = 3.15 # get_pi 100000;; - : float = 3.13272 # get_pi 1000000;; - : float = 3.143808 # get_pi 10000000;; - : float = 3.1421704 # get_pi 100000000;; - : float = 3.14153872
Octave
function p = montepi(samples)
in_circle = 0;
for samp = 1:samples
v = [ unifrnd(-1,1), unifrnd(-1,1) ];
if ( v*v.' <= 1.0 )
in_circle++;
endif
endfor
p = 4*in_circle/samples;
endfunction
l = 1e4;
while (l < 1e7)
disp(montepi(l));
l *= 10;
endwhile
Since it runs slow, I've stopped it at the second iteration, obtaining:
3.1560 3.1496
Much faster implementation
function result = montepi(n)
result = sum(rand(1,n).^2+rand(1,n).^2<1)/n*4;
endfunction
PARI/GP
MonteCarloPi(tests)=4.*sum(i=1,tests,norml2([random(1.),random(1.)])<1)/tests;A hundred million tests (about a minute) yielded 3.14149000, slightly more precise (and round!) than would have been expected. A million gave 3.14162000 and a thousand 3.14800000.
Pascal
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.
PascalABC.NET
##
for var i := 3 to 8 do
begin
var n := integer(10 ** i);
var count := 0;
loop n do
if random.sqr + random.sqr < 1 then count += 1;
Writeln(n:10, 4 * count / n:12:8);
end;
- Output:
1000 3.11600000
10000 3.13600000
100000 3.13792000
1000000 3.14151200
10000000 3.14198160
100000000 3.14155664
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;
- Output:
10000: 3.132000 1000000: 3.141596
Phix
with javascript_semantics integer N = 100 for i=1 to 6 do integer inside = 0 for n=1 to N do integer x = rand(N), y = rand(N) inside += (x*x+y*y<N*N) end for pp({N,4*inside/N}) N *= 10 end for
- Output:
{100,3.2}
{1000,3.116}
{10000,3.1736}
{100000,3.13996}
{1000000,3.141856}
{10000000,3.1415728}
PHP
<?
$loop = 1000000; # loop to 1,000,000
$count = 0;
for ($i=0; $i<$loop; $i++) {
$x = rand() / getrandmax();
$y = rand() / getrandmax();
if(($x*$x) + ($y*$y)<=1) $count++;
}
echo "loop=".number_format($loop).", count=".number_format($count).", pi=".($count/$loop*4);
?>
- Output:
loop=1,000,000, count=785,462, pi=3.141848
Picat
Some general Monte Carlo simulators. N is the number of runs, F is the simulation function.
Using while loop
sim1(N, F) = C =>
C = 0,
I = 0,
while (I <= N)
C := C + apply(F),
I := I + 1
end.
List comprehension
This is simpler, but slightly slower than using while loop.
sim2(N, F) = sum([apply(F) : _I in 1..N]).Recursion
sim_rec(N,F) = S =>
sim_rec(N,N,F,0,S).
sim_rec(0,_N,_F,S,S).
sim_rec(C,N,F,S0,S) :-
S1 = S0 + apply(F),
sim_rec(C-1,N,F,S1,S).Test
Of the three different MC simulators, sim_rec/2 (using recursion) is slightly faster than the other two (sim1/2 and sim2/2) which have about the same speed.
go =>
foreach(N in 0..7)
sim_pi(10**N)
end,
nl.
% The specific pi simulation
sim_pi(N) =>
Inside = sim(N,pi_f),
MyPi = 4.0*Inside/N,
Pi = math.pi,
println([n=N, myPi=MyPi, diff=Pi-MyPi]).
% The simulation function:
% returns 1 if success, 0 otherwise
pi_f() = cond(frand()**2 + frand()**2 <= 1, 1, 0).- Output:
[n = 1,myPi = 4.0,diff = -0.858407346410207] [n = 10,myPi = 3.2,diff = -0.058407346410207] [n = 100,myPi = 3.12,diff = 0.021592653589793] [n = 1000,myPi = 3.152,diff = -0.010407346410207] [n = 10000,myPi = 3.1672,diff = -0.025607346410207] [n = 100000,myPi = 3.13888,diff = 0.002712653589793] [n = 1000000,myPi = 3.14192,diff = -0.000327346410207] [n = 10000000,myPi = 3.1408988,diff = 0.000693853589793]
PicoLisp
(de carloPi (Scl)
(let (Dim (** 10 Scl) Dim2 (* Dim Dim) Pi 0)
(do (* 4 Dim)
(let (X (rand 0 Dim) Y (rand 0 Dim))
(when (>= Dim2 (+ (* X X) (* Y Y)))
(inc 'Pi) ) ) )
(format Pi Scl) ) )
(for N 6
(prinl (carloPi N)) )- Output:
3.4 3.23 3.137 3.1299 3.14360 3.140964
PowerShell
function Get-Pi ($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
Python
At the interactive prompt
Python 2.6rc2 (r26rc2:66507, Sep 18 2008, 14:27:33) [MSC v.1500 32 bit (Intel)] on win32 IDLE 2.6rc2
One use of the "sum" function is to count how many times something is true (because True = 1, False = 0):
>>> import random, math
>>> throws = 1000
>>> 4.0 * sum(math.hypot(*[random.random()*2-1
for q in [0,1]]) < 1
for p in xrange(throws)) / float(throws)
3.1520000000000001
>>> throws = 1000000
>>> 4.0 * sum(math.hypot(*[random.random()*2-1
for q in [0,1]]) < 1
for p in xrange(throws)) / float(throws)
3.1396359999999999
>>> throws = 100000000
>>> 4.0 * sum(math.hypot(*[random.random()*2-1
for q in [0,1]]) < 1
for p in xrange(throws)) / float(throws)
3.1415666400000002
As a program using a function
from random import 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
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
Quackery
[ $ "bigrat.qky" loadfile ] now!
[ [ 64 bit ] constant
dup random dup *
over random dup * +
swap dup * < ] is hit ( --> b )
[ 0 swap times
[ hit if 1+ ] ] is sims ( n --> n )
[ dup echo say " trials "
dup sims 4 *
swap 20 point$ echo$ cr ] is trials ( n --> )
' [ 10 100 1000 10000 100000 1000000 ] witheach trials- Output:
10 trials 2.8 100 trials 3.2 1000 trials 3.172 10000 trials 3.1484 100000 trials 3.1476 1000000 trials 3.142256
R
# nice but not suitable for big samples!
monteCarloPi <- function(samples) {
x <- runif(samples, -1, 1) # for big samples, you need a lot of memory!
y <- runif(samples, -1, 1)
l <- sqrt(x*x + y*y)
return(4*sum(l<=1)/samples)
}
# this second function changes the samples number to be
# multiple of group parameter (default 100).
monteCarlo2Pi <- function(samples, group=100) {
lim <- ceiling(samples/group)
olim <- lim
c <- 0
while(lim > 0) {
x <- runif(group, -1, 1)
y <- runif(group, -1, 1)
l <- sqrt(x*x + y*y)
c <- c + sum(l <= 1)
lim <- lim - 1
}
return(4*c/(olim*group))
}
print(monteCarloPi(1e4))
print(monteCarloPi(1e5))
print(monteCarlo2Pi(1e7))
Racket
#lang racket
(define (in-unit-circle? x y) (<= (sqrt (+ (sqr x) (sqr y))) 1))
;; point in ([-1,1], [-1,1])
(define (random-point-in-2x2-square) (values (* 2 (- (random) 1/2)) (* 2 (- (random) 1/2))))
;; Area of circle is (pi r^2). r is 1, area of circle is pi
;; Area of square is 2^2 = 4
;; There is a pi/4 chance of landing in circle
;; .: pi = 4*(proportion passed) = 4*(passed/samples)
(define (passed:samples->pi passed samples) (* 4 (/ passed samples)))
;; generic kind of monte-carlo simulation
(define (monte-carlo run-length report-frequency
sample-generator pass?
interpret-result)
(let inner ((samples 0) (passed 0) (cnt report-frequency))
(cond
[(= samples run-length) (interpret-result passed samples)]
[(zero? cnt) ; intermediate report
(printf "~a samples of ~a: ~a passed -> ~a~%"
samples run-length passed (interpret-result passed samples))
(inner samples passed report-frequency)]
[else
(inner (add1 samples)
(if (call-with-values sample-generator pass?)
(add1 passed) passed) (sub1 cnt))])))
;; (monte-carlo ...) gives an "exact" result... which will be a fraction.
;; to see how it looks as a decimal we can exact->inexact it
(let ((mc (monte-carlo 10000000 1000000 random-point-in-2x2-square in-unit-circle? passed:samples->pi)))
(printf "exact = ~a~%inexact = ~a~%(pi - guess) = ~a~%" mc (exact->inexact mc) (- pi mc)))
- Output:
1000000 samples of 10000000: 785763 passed -> 785763/250000 2000000 samples of 10000000: 1571487 passed -> 1571487/500000 3000000 samples of 10000000: 2356776 passed -> 98199/31250 4000000 samples of 10000000: 3141924 passed -> 785481/250000 5000000 samples of 10000000: 3927540 passed -> 196377/62500 6000000 samples of 10000000: 4713072 passed -> 98189/31250 7000000 samples of 10000000: 5498300 passed -> 54983/17500 8000000 samples of 10000000: 6283199 passed -> 6283199/2000000 9000000 samples of 10000000: 7068065 passed -> 1413613/450000 exact = 3926793/1250000 inexact = 3.1414344 (pi - guess) = 0.00015825358979304482
A little more Racket-like is the use of an iterator (in this case for/fold), which is clearer than an inner function:
#lang racket
(define (in-unit-circle? x y) (<= (sqrt (+ (sqr x) (sqr y))) 1))
;; Good idea made in another task that:
;; The proportions of hits is the same in the unit square and 1/4 of a circle.
;; point in ([0,1], [0,1])
(define (random-point-in-unit-square) (values (random) (random)))
;; generic kind of monte-carlo simulation
;; Area of circle is (pi r^2). r is 1, area of circle is pi
;; Area of square is 2^2 = 4
;; There is a pi/4 chance of landing in circle
;; .: pi = 4*(proportion passed) = 4*(passed/samples)
(define (passed:samples->pi passed samples) (* 4 (/ passed samples)))
(define (monte-carlo/2 run-length report-frequency sample-generator pass? interpret-result)
(interpret-result
(for/fold ((pass 0))
([n (in-range run-length)]
#:when (when (and (not (zero? n)) (zero? (modulo n report-frequency)))
(printf "~a samples of ~a: ~a passed -> ~a~%"
n run-length pass (interpret-result pass n)))
#:when (call-with-values sample-generator pass?))
(add1 pass))
run-length))
;; (monte-carlo ...) gives an "exact" result... which will be a fraction.
;; to see how it looks as a decimal we can exact->inexact it
(let ((mc (monte-carlo/2 10000000 1000000 random-point-in-unit-square in-unit-circle? passed:samples->pi)))
(printf "exact = ~a~%inexact = ~a~%(pi - guess) = ~a~%" mc (exact->inexact mc) (- pi mc)))
[Similar output]
Raku
(formerly Perl 6)
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];
REXX
A specific─purpose commatizer function is included to format the number of iterations.
/*REXX program computes and displays the value of pi÷4 using the Monte Carlo algorithm*/
numeric digits 20 /*use 20 decimal digits to handle args.*/
parse arg times chunk digs r? . /*does user want a specific number? */
if times=='' | times=="," then times= 5e12 /*five trillion should do it, hopefully*/
if chunk=='' | chunk=="," then chunk= 100000 /*perform Monte Carlo in 100k chunks.*/
if digs =='' | digs=="," then digs= 99 /*indicates to use length of PI - 1. */
if datatype(r?, 'W') then call random ,,r? /*Is there a random seed? Then use it.*/
/* [↓] pi meant to line─up with a SAY.*/
pi= 3.141592653589793238462643383279502884197169399375105820974944592307816406
pi= strip( left(pi, digs + length(.) ) ) /*obtain length of pi to what's wanted.*/
numeric digits length(pi) - 1 /*define decimal digits as length PI -1*/
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.*/
say /*display a blank line for separation. */
limit = 10000 - 1 /*REXX random generates only integers. */
limitSq = limit **2 /*··· so, instead of one, use limit**2.*/
accuracy= 0 /*accuracy of Monte Carlo pi (so far).*/
@reps= 'repetitions: Monte Carlo pi is' /*a handy─dandy short literal for a SAY*/
!= 0 /*!: is the accuracy of pi (so far). */
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 _<=accuracy then iterate /*Not better accuracy? Keep truckin'. */
say right(commas(reps), 20) @reps 'accurate to' _-1 "places." /*─1≡dec. point*/
accuracy= _ /*use this accuracy for next baseline. */
end /*j*/
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: procedure; arg _; do k=length(_)-3 to 1 by -3; _=insert(',',_,k); end; return _
- 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.
For more example runs using REXX, see the discussion page.
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 tOutput:
monteCarlo(1000) = 3.11600000 monteCarlo(10000) = 3.00320000 monteCarlo(100000) = 2.99536000
RPL
≪ 0
1 3 PICK START
RAND SQ RAND SQ + 1 < +
NEXT
SWAP / 4 *
≫ 'MCARL' STO
100 MCARL 1000 MCARL 10000 MCARL 100000 MCARL
- Output:
4: 3.2 3: 3.084 2: 3.1684 1: 3.14154
Ruby
def approx_pi(throws)
times_inside = throws.times.count {Math.hypot(rand, rand) <= 1.0}
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
Rust
extern crate rand;
use rand::Rng;
use std::f64::consts::PI;
// `(f32, f32)` would be faster for some RNGs (including `rand::thread_rng` on 32-bit platforms
// and `rand::weak_rng` as of rand v0.4) as `next_u64` combines two `next_u32`s if not natively
// supported by the RNG. It would less accurate however.
fn is_inside_circle((x, y): (f64, f64)) -> bool {
x * x + y * y <= 1.0
}
fn simulate<R: Rng>(rng: &mut R, samples: usize) -> f64 {
let mut count = 0;
for _ in 0..samples {
if is_inside_circle(rng.gen()) {
count += 1;
}
}
(count as f64) / (samples as f64)
}
fn main() {
let mut rng = rand::weak_rng();
println!("Real pi: {}", PI);
for samples in (3..9).map(|e| 10_usize.pow(e)) {
let estimate = 4.0 * simulate(&mut rng, samples);
let deviation = 100.0 * (1.0 - estimate / PI).abs();
println!("{:9}: {:<11} dev: {:.5}%", samples, estimate, deviation);
}
}
- Output:
Real pi: 3.141592653589793
1000: 3.212 dev: 2.24114%
10000: 3.156 dev: 0.45860%
100000: 3.14112 dev: 0.01505%
1000000: 3.14122 dev: 0.01186%
10000000: 3.1408112 dev: 0.02487%
100000000: 3.14186092 dev: 0.00854%
Scala
object MonteCarlo {
private val random = new scala.util.Random
/** Returns a random number between -1 and 1 */
def nextThrow: Double = (random.nextDouble * 2.0) - 1.0
/** Returns true if the argument point would be 'inside' the unit circle with
* center at the origin, and bounded by a square with side lengths of 2
* units. */
def insideCircle(pt: (Double, Double)): Boolean = pt match {
case (x, y) => (x * x) + (y * y) <= 1.0
}
/** Runs the simulation the specified number of times. Uses the result to
* estimate a value of pi */
def simulate(times: Int): Double = {
val inside = Iterator.tabulate (times) (_ => (nextThrow, nextThrow)) count insideCircle
inside.toDouble / times.toDouble * 4.0
}
def main(args: Array[String]): Unit = {
val sims = Seq(10000, 100000, 1000000, 10000000, 100000000)
sims.foreach { n =>
println(n+" simulations; pi estimation: "+ simulate(n))
}
}
}
- Output:
10000 simulations; pi estimation: 3.1492 100000 simulations; pi estimation: 3.1396 1000000 simulations; pi estimation: 3.14208 10000000 simulations; pi estimation: 3.1409944 100000000 simulations; pi estimation: 3.1414386
Seed7
$ include "seed7_05.s7i";
include "float.s7i";
const func float: pi (in integer: throws) is func
result
var float: pi is 0.0;
local
var integer: throw is 0;
var integer: inside is 0;
begin
for throw range 1 to throws do
if rand(0.0, 1.0) ** 2 + rand(0.0, 1.0) ** 2 <= 1.0 then
incr(inside);
end if;
end for;
pi := flt(4 * inside) / flt(throws);
end func;
const proc: main is func
begin
writeln(" 10000: " <& pi( 10000) digits 5);
writeln(" 100000: " <& pi( 100000) digits 5);
writeln(" 1000000: " <& pi( 1000000) digits 5);
writeln(" 10000000: " <& pi( 10000000) digits 5);
writeln("100000000: " <& pi(100000000) digits 5);
end func;- Output:
10000: 3.14520 100000: 3.15000 1000000: 3.14058 10000000: 3.14223 100000000: 3.14159
SequenceL
First solution is serial due to the use of random numbers. Will always give the same result for a given n and seed
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);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).
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);Sidef
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))
}
- Output:
100: 3.320000
1000: 3.120000
10000: 3.169600
100000: 3.138920
1000000: 3.142344
SparForte
As a structured script.
#!/usr/local/bin/spar
pragma annotate( summary, "monte" )
@( description, "A Monte Carlo Simulation is a way of approximating the" )
@( description, "value of a function where calculating the actual value is" )
@( description, "difficult or impossible. It uses random sampling to define" )
@( description, "constraints on the value and then makes a sort of 'best" )
@( description, "guess.'" )
@( description, "" )
@( description, "Write a function to run a simulation like this with a" )
@( description, "variable number of random points to select. Also, show the" )
@( description, "results of a few different sample sizes. For software" )
@( description, "where the number pi is not built-in, we give pi to a couple" )
@( description, "of digits: 3.141592653589793238462643383280 " )
@( see_also, "http://rosettacode.org/wiki/Monte_Carlo_methods" )
@( author, "Ken O. Burtch" );
pragma license( unrestricted );
pragma restriction( no_external_commands );
procedure monte is
function pi_estimate (throws : positive) return float is
inside : natural := 0;
begin
for throw in 1..throws loop
if numerics.random ** 2 + numerics.random ** 2 <= 1.0 then
inside := @ + 1;
end if;
end loop;
return 4.0 * float (inside) / float (throws);
end pi_estimate;
begin
? " 1_000:" & strings.image (pi_estimate ( 1_000))
@ " 10_000:" & strings.image (pi_estimate ( 10_000))
@ " 100_000:" & strings.image (pi_estimate ( 100_000))
@ " 1_000_000:" & strings.image (pi_estimate ( 1_000_000));
end monte;
Stata
program define mcdisk
clear all
quietly set obs `1'
gen x=2*runiform()
gen y=2*runiform()
quietly count if (x-1)^2+(y-1)^2<1
display 4*r(N)/_N
end
. mcdisk 10000
3.1424
. mcdisk 1000000
3.141904
. mcdisk 100000000
3.1416253
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))
- Output:
3.08 3.128 3.1548 3.149 3.142032 3.1414772 3.14166832
Tcl
proc pi {samples} {
set i 0
set inside 0
while {[incr i] <= $samples} {
if {sqrt(rand()**2 + rand()**2) <= 1.0} {
incr inside
}
}
return [expr {4.0 * $inside / $samples}]
}
puts "PI is approx [expr {atan(1)*4}]\n"
foreach runs {1e2 1e4 1e6 1e8} {
puts "$runs => [pi $runs]"
}
result
PI is approx 3.141592653589793 1e2 => 2.92 1e4 => 3.1344 1e6 => 3.141924 1e8 => 3.14167724
Ursala
#import std
#import flo
mcp "n" = times/4. div\float"n" (rep"n" (fleq/.5+ sqrt+ plus+ ~~ sqr+ minus/.5+ rand)?/~& plus/1.) 0.Here's a walk through.
mcp "n" =... defines a function namedmcpin terms of a dummy variable"n", which will be the number of iterations used in the simulationrandignores its argument and returns a uniformly distributed number between 0 and 1minus/.5is composed withrandto compute the difference between the random number and 0.5sqrsquares the difference~~says to apply the function twice and return the pair of resultspluscomposed with that adds the pair of resultssqrttakes the square root of the sumfleq/.5is floating point comparison with a fixed right side of.5, returning true if its argument is greater or equal- Everything from
fleqtorandforms 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 themcpfunction.- 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 bydiv\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>
Wren
import "random" for Random
import "./fmt" for Fmt
var rand = Random.new()
var mcPi = Fn.new { |n|
var inside = 0
for (i in 1..n) {
var x = rand.float()
var y = rand.float()
if (x*x + y*y <= 1) inside = inside + 1
}
return 4 * inside / n
}
System.print("Iterations -> Approx Pi -> Error\%")
System.print("---------- ---------- ------")
var n = 1000
while (n <= 1e8) {
var pi = mcPi.call(n)
var err = (Num.pi - pi).abs / Num.pi * 100.0
Fmt.print("$9d -> $10.8f -> $6.4f", n, pi, err)
n = n * 10
}
- Output:
Sample run:
Iterations -> Approx Pi -> Error%
---------- ---------- ------
1000 -> 3.21200000 -> 2.2411
10000 -> 3.16720000 -> 0.8151
100000 -> 3.13944000 -> 0.0685
1000000 -> 3.14048000 -> 0.0354
10000000 -> 3.14191240 -> 0.0102
100000000 -> 3.14142320 -> 0.0054
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);
]- Output:
2.92000
3.13200
3.14375
3.14192
zkl
fcn monty(n){
inCircle:=0;
do(n){
x:=(0.0).random(1); y:=(0.0).random(1);
if(x*x + y*y < 1.0) inCircle+=1;
}
4.0*inCircle/n
}Or, in a more functional style (using a reference for state info):
fcn monty(n){
4.0 * (1).pump(n,Void,fcn(r){
x:=(0.0).random(1); y:=(0.0).random(1);
if(x*x + y*y < 1.0) r.inc();
r
}.fp(Ref(0)) ).value/n;
}- Output:
T(100,1000,10000,0d100_000,0d1_000_000,0d10_000_000)
.apply2(fcn(n){"%10,d : %+f".fmt(n,monty(n)-(1.0).pi).println()})
100 : -0.061593
1,000 : +0.018407
10,000 : -0.013993
100,000 : -0.000833
1,000,000 : -0.004385
10,000,000 : +0.000619
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