Contributed by sinan
on Aug 09, 2000 at 00:33 UTC
Q&A
> math
Description: I need to generate random numbers; but they have to follow
the standard distribution. The usual random number generation
routines do not do this.
In other words, if I have a range of 1 to 100, I want the
numbers around 50 to occur frequently, and the numbers
about 1 or about 100 to occur with a very low frequency.
With the usual random number generation routines, all
numbers have the same frequency.
There must be a simple algorithm that create standard
distribution numbers using the usual generator. (i.e.
An algorithm that converts such numbers)
Does anybody know any function that does this? Can anybody
write a simple algorithm for this purpose?
Please do not tell me any nonstandard modules because
I have trouble installing them.
TIA, Sinan Answer: How do I get random numbers that follow standard distribution? contributed by tilly If you don't need a perfect normal distribution then you can get a very good approximation by choosing several random numbers and adding them together. You can then multiply then add by
appropriate constants to get the desired average
and variance.
For most purposes, adding a dozen is good enough.
If you want better than that, try installing Math::Random.
DISCLAIMER: I am one dissertation short of a PhD in math. (It is handwritten, needed an income so I never cleaned it up and typed it.) Therefore for all intents and purposes, I am a mathematician. :)  Answer: How do I get random numbers that follow standard distribution? contributed by lhoward How about something like this:
sub f{
my $t=0;
$t+=rand() for(1..5);
return int($t*20+1);
}
You can vary the number of iterations (1..5) and
scaling factor (20) based on the exact
distribution curve you want.
There are several other ways to approach this
problem. If efficiency were of paramount importance,
I would try using statistical method to derive
a function that would aproximate the curve of
how in frequency you would map numbers from 0 to 0.9999 (the range of perl's rand function)
into the range you want (1 to 100).  Answer: How do I get random numbers that follow standard distribution? contributed by grackle The question posted asks for a nonmodule solution,
and an excellent one was provided by lhoward and
tilly. Here is
a solution using a CPAN module.
I looked for Math::Random on CPAN and couldn't
find it. There are modules for random
numbers, such as Crypt::TrulyRandom, but I believe
they will only give you better uniformly
distributed random variables (correct me if I'm
wrong). For a normal distribution, you
can use Math::CDF
(Cumulative Distribution
Functions). The relevant function
is Math::CDF::qnorm(), which will give you
what you want
when fed a random number x, 0<x<1, with a uniform
distribution (i.e., what you get from rand()).
(It supposedly returns a value for the inputs x=1
and x=0, but I don't know what those values could
be, you'll have to try it).
So the function you want is:
$myrand = qnorm(rand());
(or replace rand() with your function of choice).
One caveat: I don't know how the Math::CDF
module calculates these values. If it provides
more precision
than you need, it could be a big waste
of processing time. In that case
lhoward and tilly's solution, summing several
random variables, works fine. It would be nice to
have bounds for the errors, but I can't help you
there. Calculate, experiment, or find a reference....
Perhaps this would be a good place to post the
results.
 Answer: How do I get random numbers that follow standard distribution? contributed by Anonymous Monk sub norm{
my $v1,$v2,$r;
if( defined $v2 ){
$v1 = $v2;
undef $v2;
}else{
do{
$v1=rand(2)1;
$v2=rand(2)1;
$r = $v1*$v1+$v2*$v2;
}while( $r >= 1  $r == 0 );
$r = sqrt(2*(log $r)/$r);
$v1 *= $r;
$v2 *= $r;
}
return $v1;
}
 Answer: How do I get random numbers that follow standard distribution? contributed by Roy Johnson The Perl Cookbook provides this code:
sub gaussian_rand {
my ($u1, $u2); # uniformly distributed random numbers
my $w; # variance, then a weight
my ($g1, $g2); # gaussiandistributed numbers
do {
$u1 = 2 * rand()  1;
$u2 = 2 * rand()  1;
$w = $u1*$u1 + $u2*$u2;
} while ( $w >= 1 );
$w = sqrt( (2 * log($w)) / $w );
$g2 = $u1 * $w;
$g1 = $u2 * $w;
# return both if wanted, else just one
return wantarray ? ($g1, $g2) : $g1;
}
and notes:
The gaussian_rand function [generates] two numbers with a mean of 0 and a standard deviation of 1 (i.e., a Gaussian distribution). To generate numbers with a different mean and standard deviation, multiply the output of gaussian_rand by the new standard deviation, and then add the new mean
 Answer: How do I get random numbers that follow standard distribution? contributed by jlistf i think you're going to have to use statistics. i don't believe Perl has anything builtin to do it, though there might be a module on CPAN. Perl's rand($n) will give you numbers (more or less) uniformly distributed between 0 and 1. its up to you to convert this to a normalized distribution.  Answer: How do I get random numbers that follow standard distribution? contributed by sinan I finally did something that solves the problem partially.
The two following functions do that. The first function,
rands will return a "standard" randomized integer.
It has two input values: The first input
variables denote the range of the output. (i.e.
0 to input) The second one will denote the
precision: Higher it is, closer the result will
be to the normal distribution. 5 is more
than enough for many purposes.
The second function (random) is a
little bit more tricky.
The first input is the range, just like in the first
function. The second input is a string which
can either be "low" or "high". The third input
is a limit. If the second input is "low", then
the third input denotes the lowest possible
number. If it is high, then it denotes the
highest possible number. All the results that are
outside either the lowest or the highest frequency
will be ignored and "rerolled".
Here is the snippet:
sub rands{
my $t=0;
$t = 0;
$t+= ( rand()* ($_[0]/$_[1]) ) for(1..$_[1]);
return int($t+1);
}
sub random{
my ($no,$type,$limit) = @_;
$t = rands($no,5);
if($type eq "low"){
while ($t<$limit){
$t = rands($no,5)
}
}
elsif($type eq "high"){
while ($t>$limit){
$t = rands($no,5)
}
}
}
I was not able to "skew" my results, so I used
this _harsh_ method of cutting off instead.
One final note: You can set the second input
variable of rands to a low number (2,3) to get a
more homogenous distribution, i.e. a distribution
with "high variance".
Sinan  Answer: How do I get random numbers that follow standard distribution? contributed by chas By "standard distribution", I think you mean "standard Normal distribution". One way to get a
random sample from such a distribution is as follows: Get a random sample x_1,...x_n from a uniform distribution on the unit interval. (this is easily
done using the usual random number generators you refer to.) Then for each x_i, compute
z_i satisfying x_i=(1/sqrt(2 pi))\int_{\infty}^z_i e^{t^2/2}dt. (I.e. x_i=F(z_i) where F is the distribution function for a N(0,1).) Then z_1,...z_n will be a random sample from a N(0,1). You have to write a bit of
code to solve for z_i, of course. I've done similar things in c, but not perl so I can't display perl code to achieve this.
The previous suggestion to add the results of several uniformly distributed random numbers is
interesting; one needs independence to make use
of the Central Limit Theorem, so I'm not sure exactly how to set that up, but it should be possible.
The method I described first is pretty standard.
chas 
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