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Estimating Pi by Monte Carlo using mXparser

Using built-in random variable [Uni] from uniform continuous distribution over [0; 1] interval we sample two numbers (x,y) then check if point is inside the circle. Operation is repeated n-times, then proportion falling into circle is being measured. This is only 1/4 of the circle (positive x and positive y) with radius 1, so 4 times proportion should give estimation of pi.

Argument n = new Argument("n = 100000");
Expression e = new Expression("4 * sum(i, 1, n, if( [Uni]^2 + [Uni]^2 <= 1; 1; 0) ) / n", n);
mXparser.consolePrintln("Res. : " + e.getExpressionString() + " = " + e.calculate());
[mXparser-v.3.0.0] Res. : 4 * sum(i, 1, n, if( [Uni]^2 + [Uni]^2 <= 1; 1; 0) ) / n = 3.14748

Best regards

mXparser - 1000 downloads!

mXparser – 1000 downloads !

mXparser 1000 downloads
I am happy to announce that first milestone was achieved – as of today mXparser has 1000 unique downloads, and the number is growing! This is very exciting, especially considering questions you asked and feedback I gained.  mXparser seems fits your needs providing the same API for JAVA, Android and .NET (including MONO). Single framework / technology days have  gone away, and are not  coming back, meaning cross-platform apps  are with us for good.
MathParser.org launch
mXparser downloads were significantly boosted after launching http://MathParser.org site, amazing! Thank you Google!
mXparser / MathParser.org users location
 mXparser / MathParser.org users location
Best regards,
Mariusz Gromada