# C# Monte Carlo RNG Example

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```using System;

using CenterSpace.NMath.Core;

namespace CenterSpace.NMath.Core.Examples.CSharp
{
/// <summary>
/// A .NET example in C# showing how to calculate an approximation for Pi using a Monte
/// Carlo method and the uniform random number generator class RandGenUniform.
///
/// Imagine a 2 x 2 square with corners at (1,1), (1,-1), (-1,-1) and (-1,1)
/// and a unit circle, centered on the origin, inscribed within it. Generate
/// random points inside the square and let M be the number of points
/// that fall within the unit circle and N be the total number of points
/// generated. As the number N gets large, the quantity M/N should approximate the
/// ratio of the area of the circle to the square, which is pi/4. Hence, we can
/// use the ratio 4*M/N to approximate Pi.
/// </summary>
class MonteCarloRNGExample
{

static void Main( string[] args )
{
// Construct a random number generator that generates random deviates
// distributed uniformly over the interval [-1,1]
var rng = new RandGenUniform( -1.0, 1.0 );

// We'll approximate pi to within 5 digits.
double tolerance = 1e-5;

double piApproximation = 0;
int total = 0;
int numInCircle = 0;
double x, y; // Coordinates of the random point.

// Generate random points until our approximation within
// the desired tolerance.
while ( Math.Abs( Math.PI - piApproximation ) > tolerance )
{
x = rng.Next();
y = rng.Next();
if ( x * x + y * y <= 1.0 ) // Is the point in the circle?
{
++numInCircle;
}
++total;
piApproximation = 4.0 * ( (double) numInCircle / (double) total );
}

Console.WriteLine();

Console.WriteLine( "Pi calculated to within {0} digits with {1} random points.",
-Math.Log10( tolerance ), total );
Console.WriteLine( "Approximated Pi = {0}", piApproximation );

Console.WriteLine();
Console.WriteLine( "Press Enter Key" );