## Monte-Carlo Method to Approximate PI

Here are a couple simple examples illustrating how to use Monte-Carlo Method to calculate an approximation of PI. Straight from Wikipedia the algorithm works per the following:

- Draw a square on the ground, then inscribe a circle within it.
- Uniformly scatter some objects of uniform size (grains of rice or sand) over the square.
- Count the number of objects inside the circle and the total number of objects.
- The ratio of the two counts is an estimate of the ratio of the two areas, which is π/4. Multiply the result by 4 to estimate π.

As to why the ratio is π/4:

```
A_circle = πr^2
A_square = (2r)^2
circle transcribed in a square, r_c = r_s = r
A_square / A_circle (r = 1) = ratio of areas between square and circle
= 4 / π
```

GitHub Links: Haskell Code Java Code

**Haskell Example**

```
import System.Random
import Data.List
inCircle :: (Double,Double) -> Bool
inCircle p = x * x + y * y < 1.0
where x = fst p
y = snd p
monteCarloPi :: [(Double, Double)] -> Double
monteCarloPi ps = ratio * 4.0
where ratio = (fromIntegral count) / (fromIntegral samples)
count = length $ filter inCircle ps
samples = length ps
randList :: Int -> StdGen -> [Double]
randList n rng = scale $ take n (randoms rng :: [Double])
where scale l = map (\x -> 2 * x - 1) l -- scales doubles in range of 0,1 to random -1,1
main = do
rng <- newStdGen
rng2 <- newStdGen
let xs = randList 1000000 rng
let ys = randList 1000000 rng2
let ps = zip xs ys
print $ monteCarloPi ps
```

**Java Example**

```
import java.util.Random;
/**
* Created by kenny on 5/29/14.
*/
public class EvaulatePie {
private static final Random RANDOM = new Random();
/*
A_circle = πr^2
A_square = (2r)^2
circle transcribed in a square, r_c = r_s = r
A_square / A_circle (r = 1) = ratio of areas between square and circle
= 4 / π
*/
public static void main(String[] args) {
monteCarlo(10);
monteCarlo(100);
monteCarlo(1000);
monteCarlo(10000);
monteCarlo(100000);
monteCarlo(1000000);
monteCarlo(10000000);
monteCarlo(100000000);
}
/*
make n random guesses, take the ratio of guesses in the circle,
and divide that by the total guesses (all of which will be in the square)
this will give you the A_square / A_circle ratio, 4/π
*/
public static void monteCarlo(final int randomSamples) {
int inCircle = 0;
for(int i = 0; i < randomSamples; i++) {
// generate random x/y variables between -1, 1 (which are guaranteed to be within the square
final double x = 2 * RANDOM.nextDouble() - 1;
final double y = 2 * RANDOM.nextDouble() - 1;
if(inCircle(x, y)) {
inCircle++;
}
}
final double ratio = (double) inCircle / randomSamples;
final double piApprox = ratio * 4; // ratio is 4/π, so multiply by 4 to get π approximation
System.out.println("Random Samples: " + randomSamples);
System.out.println("In circle: " + inCircle + ", In square: " + randomSamples);
System.out.println("PI Approximation: " + piApprox + ", Error: " + Math.abs(Math.PI - piApprox));
}
/*
eq of circle centered at origin: x^2 + y^2 = r^2
if (x)^2 + (y)^2 < r^2 then is within circle where r = 1
*/
private static boolean inCircle(final double x, final double y) {
// System.out.println(x + ", " + y);
return x * x + y * y < 1.0;
}
}
```

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