Test the pseudo random generator

[Ellen Zhao]

It's dangerous to blindingly trust your system's pseudo random generator as far as encryption, simulation or statistics are concerned. This test is based on a theorem given by Ernesto Cesaro ( proof can be found in[KUNTH: the art of computer programming, 1998]).

<font size="2"> /*
 * Created on 12-Sep-2003
 *
 * To change the template for this generated file go to
 * Window - Preferences - Java - Code Generation - Code and Comments
 */
import java.util.Random;
/**
 * This class tests the reliability of the random generator of a certain
 *  computer system or vm. If the result is very close to the 
 * value of Math.PI, congratulations. If not, please be careful
 * when you use your random generator to do some serious task.
 * 
 * @author Ellen Zhao
 * @version 1.0
 */
public class RandomGeneratorTest {

// Counter of pairs of mutual prim positive integers
private static int primCounter = 0;
// Caculates the largest common divisor of two positive random integers.
private static int EuclidLCD(int int1, int int2) {
int d = int1 % int2;
if (d != 0) {
return EuclidLCD(int2, d);
}
return int2;
}
// Does a test according to Ernesto Cesaro Theorem
private static double ErnestoCesaroTest(int primCounter) {
// Here is the 6 is a constant and 100000 is a variable.
// 100000 is the value of the max i in the for loop in 
// the main method. If the upper boundary of i is 1000000
// then please modify the 100000 to 1000000
int temp = 6 * 100000 / primCounter;
return Math.sqrt(temp);
}
public static void main(String[] args) {
Random random = new Random();
for (int i = 1; i <= 100000; i++) {
// Generates two different positive integers
int int1 = Math.abs(random.nextInt());
int int2 = Math.abs(random.nextInt());
// Calls the Euclid largest common divisor method
if (int1 >= int2) {
if (EuclidLCD(int1, int2) == 1) {
primCounter++;
}
} else {
if (EuclidLCD(int2, int1) == 1) {
primCounter++;
}
}
}

// Prints the result of ErnestoCesaroTest
System.out.println(ErnestoCesaroTest(primCounter));
}
}
 </font>

I was testing it on Java SDK 1.4.2, always got an output of 3.0. But I caculated using the primCounter which given by this program with my pocket Caculator, the value is rather close to 3.1416....... Why didn't the program above print a longer double, but always simply 3.0? Thank you very much in advance.
Regards,
Ellen
[ September 12, 2003: Message edited by: Ellen Zhao ]

[Ernest Friedman-Hill]

I think because you're truncating the value. In this expression:
int temp = 6 * 100000 / primCounter;
by using "int" for temp, you're effectively limiting the possible results to square roots of integers. Pi^2 is 9.7, rather far from an integer; it's getting truncated to 9, and so the printed result is 3.
To fix, use this:
double temp = 6.0 * 100000 / primCounter;
(note the 6 being changed to 6.0, to make the numerator of this fraction be a double.) You should then get values which are, indeed, close to Pi!