Monte Carlo methods: Difference between revisions

====Other solution:====

<syntaxhighlight lang="freebasic">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</syntaxhighlight>

{{out}}

<pre> 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</pre>

==={{header|Run BASIC}}===

{{trans|Liberty BASIC}}

<syntaxhighlight lang="freebasic">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</syntaxhighlight>

{{out}}

<pre>100 3.12

1000 3.108

10000 3.1652

100000 3.14248

1000000 3.1435</pre>

==={{header|True BASIC}}===

{{trans|BASIC}}

<syntaxhighlight lang="qbasic">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</syntaxhighlight>

{{out}}

<pre>3.1304

3.14324

3.141716

3.1416452</pre>