How to resolve the algorithm Monte Carlo methods step by step in the BASIC programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Monte Carlo methods step by step in the BASIC programming language
Table of Contents
Problem Statement
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
{\displaystyle \pi }
. 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
{\displaystyle \pi /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
{\displaystyle \pi /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
{\displaystyle \pi }
is not built-in, we give
π
{\displaystyle \pi }
as a number of digits:
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Monte Carlo methods step by step in the BASIC programming language
Source code in the basic programming language
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
# 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)
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
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%
' 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
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
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
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
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
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() <> ""
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