Random numbers: Difference between revisions

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list[i] = 1.0 + 0.5 * rng.nextGaussian()
list[i] = 1.0 + 0.5 * rng.nextGaussian()
}
}

==[[Perl]]==
[[Category:Perl]]
map {.5 + rand} 1..1000


==[[Python]]==
==[[Python]]==

Revision as of 01:53, 8 June 2007

Task
Random numbers
You are encouraged to solve this task according to the task description, using any language you may know.

The goal of this task is to generate a 1000-element array (vector, list, whatever it's called in your language) filled with normally distributed random numbers with a mean of 1.0 and a standard deviation of 0.5

C++

#include <cstdlib>   // for rand
#include <cmath>     // for atan, sqrt, log, cos
#include <algorithm> // for generate_n

double const pi = 4*std::atan(1.0);

// simple functor for normal distribution
class normal_distribution
{
public:
  normal_distribution(double m, double s): mu(m), sigma(s) {}
  double operator() // returns a single normally distributed number
  {
    double r1 = (std::rand() + 1.0)/(RAND_MAX + 1.0); // gives equal distribution in (0, 1]
    double r2 = (std::rand() + 1.0)/(RAND_MAX + 1.0);
    return mu + sigma * std::sqrt(-2*std::log(r1))*std::cos(2*pi*r2);
  }
private:
  double mu, sigma;
};

int main()
{
  double array[1000];
  std::generate_n(array, 1000, normal_distribution(1.0, 0.5));
}

IDL

result = 1.0 + 0.5*randomn(seed,1000)

Java

double[] list = new double[1000];
Random rng = new Random();
for(int i = 0;i<list.length;i++) {
  list[i] = 1.0 + 0.5 * rng.nextGaussian()
}

Perl

map {.5 + rand} 1..1000

Python

Interpreter: Python 2.5

import random
randList = [random.gauss(1, .5) for i in range(1000)]

Tcl

proc nrand {} {return [expr sqrt(-2*log(rand()))*cos(4*acos(0)*rand())]}
for {set i 0} {$i < 1000} {incr i} {lappend result [expr 1+.5*nrand()]}