How to resolve the algorithm Random numbers step by step in the BASIC programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Random numbers step by step in the BASIC programming language
Table of Contents
Problem Statement
Generate a collection filled with 1000 normally distributed random (or pseudo-random) numbers with a mean of 1.0 and a standard deviation of 0.5 Many libraries only generate uniformly distributed random numbers. If so, you may use one of these algorithms.
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Step by Step solution about How to resolve the algorithm Random numbers step by step in the BASIC programming language
Source code in the basic programming language
# Generates normally distributed random numbers with mean 0 and standard deviation 1
function randomNormal()
return cos(2.0 * pi * rand) * sqr(-2.0 * log(rand))
end function
dim r(1000)
sum = 0.0
# Generate 1000 normally distributed random numbers
# with mean 1 and standard deviation 0.5
# and calculate their sum
for i = 0 to 999
r[i] = 1.0 + randomNormal() / 2.0
sum += r[i]
next i
mean = sum / 1000.0
sum = 0.0
# Now calculate their standard deviation
for i = 0 to 999
sum += (r[i] - mean) ^ 2.0
next i
sd = sqr(sum/1000.0)
print "Mean is "; mean
print "Standard Deviation is "; sd
end
DIM array(999)
FOR number% = 0 TO 999
array(number%) = 1.0 + 0.5 * SQR(-2*LN(RND(1))) * COS(2*PI*RND(1))
NEXT
mean = SUM(array()) / (DIM(array(),1) + 1)
array() -= mean
stdev = MOD(array()) / SQR(DIM(array(),1) + 1)
PRINT "Mean = " ; mean
PRINT "Standard deviation = " ; stdev
10 ' Random numbers
20 randomize timer
30 dim r(999)
40 sum = 0
50 for i = 0 to 999
60 r(i) = 1+randomnormal()/2
70 sum = sum+r(i)
80 next
90 mean = sum/1000
100 sum = 0
110 for i = 0 to 999
120 sum = sum+(r(i)-mean)^2
130 next
140 sd = sqr(sum/1000)
150 print "Mean is ";mean
160 print "Standard Deviation is ";sd
170 print
180 end
500 sub randomnormal()
510 randomnormal = cos(2*pi*rnd(1))*sqr(-2*log(rnd(1)))
520 end sub
10 DIM AR(999): DIM DE(999)
20 FOR N = 0 TO 999
30 AR(N)= 0 + SQR(-1.3*LOG(RND(1))) * COS(1.2*PI*RND(1))
40 NEXT N
50 :
60 REM SUM
70 LET SU = 0
80 FOR N = 0 TO 999
90 LET SU = SU + AR(N)
100 NEXT N
110 :
120 REM MEAN
130 LET ME= 0
140 LET ME = SU/1000
150 :
160 REM DEVIATION
170 FOR N = 0 TO 999
180 T = AR(N)-ME: REM SUBTRACT MEAN FROM NUMBER
190 T = T * T: REM SQUARE THE RESULT
200 DE(N) = T : REM STORE IN DEVIATION ARRAY
210 NEXT N
220 LET DS=0: REM SUM OF DEVIATION ARRAY
230 FOR N = 0 TO 999
240 LET DS = DS + DE(N)
250 NEXT N
260 LET DM=0: REM MEAN OF DEVIATION ARRAY
270 LET DM = DS / 1000
280 LET DE = 0:
290 LET DE = SQR(DM)
300 :
310 PRINT "MEAN = "ME
320 PRINT "STANDARD DEVIATION ="DE
330 END
' FB 1.05.0 Win64
Const pi As Double = 3.141592653589793
Randomize
' Generates normally distributed random numbers with mean 0 and standard deviation 1
Function randomNormal() As Double
Return Cos(2.0 * pi * Rnd) * Sqr(-2.0 * Log(Rnd))
End Function
Dim r(0 To 999) As Double
Dim sum As Double = 0.0
' Generate 1000 normally distributed random numbers
' with mean 1 and standard deviation 0.5
' and calculate their sum
For i As Integer = 0 To 999
r(i) = 1.0 + randomNormal/2.0
sum += r(i)
Next
Dim mean As Double = sum / 1000.0
Dim sd As Double
sum = 0.0
' Now calculate their standard deviation
For i As Integer = 0 To 999
sum += (r(i) - mean) ^ 2.0
Next
sd = Sqr(sum/1000.0)
Print "Mean is "; mean
Print "Standard Deviation is"; sd
Print
Print "Press any key to quit"
Sleep
window 1
local fn RandomZeroToOne as double
double result
cln result = (double)( (rand() % 100000 ) * 0.00001 );
end fn = result
local fn RandomGaussian as double
double r = fn RandomZeroToOne
end fn = 1 + .5 * ( sqr( -2 * log(r) ) * cos( 2 * pi * r ) )
long i
double mean, std, a(1000)
for i = 1 to 1000
a(i) = fn RandomGaussian
mean += a(i)
next
mean = mean / 1000
for i = 1 to 1000
std += ( a(i) - mean )^2
next
std = std / 1000
print " Average: "; mean
print "Standard Deviation: "; std
HandleEvents
dim a(1000)
mean =1
sd =0.5
for i = 1 to 1000 ' throw 1000 normal variates
a( i) =mean +sd *( sqr( -2 * log( rnd( 0))) * cos( 2 * pi * rnd( 0)))
next i
10 REM Random numbers
20 LET P = 4*ATN(1)
30 RANDOMIZE
40 DEF FNN = COS(2*P*RND)*SQR(-2*LOG(RND))
50 DIM R(999)
60 LET S = 0
70 FOR I = 0 TO 999
80 LET R(I) = 1+FNN/2
90 LET S = S+R(I)
100 NEXT I
110 LET M = S/1000
120 LET S = 0
130 FOR I = 0 TO 999
140 LET S = S+(R(I)-M)^2
150 NEXT I
160 LET D = SQR(S/1000)
170 PRINT "Mean is "; M
180 PRINT "Standard Deviation is"; D
190 PRINT
200 END
Procedure.f RandomNormal()
; This procedure can return any real number.
Protected.f x1, x2
; random numbers from the open interval ]0, 1[
x1 = (Random(999998)+1) / 1000000 ; must be > 0 because of Log(x1)
x2 = (Random(999998)+1) / 1000000
ProcedureReturn Sqr(-2*Log(x1)) * Cos(2*#PI*x2)
EndProcedure
Define i, n=1000
Dim a.q(n-1)
For i = 0 To n-1
a(i) = 1 + 0.5 * RandomNormal()
Next
dim a(1000)
pi = 22/7
for i = 1 to 1000
a( i) = 1 + .5 * (sqr(-2 * log(rnd(0))) * cos(2 * pi * rnd(0)))
next i
10 RANDOMIZE 0 : REM seeds random number generator based on uptime
20 DIM a(1000)
30 CLS
40 FOR i = 1 TO 1000
50 LET a(i) = 1 + SQR(-2 * LN(RND)) * COS(2 * PI * RND)
60 NEXT i
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