java se8 rand gauss secure?

Question

Hi everybody! Its been a long time since i have been here, i would like u to kno that i app reciate u guys, (joke).

i was wondering if the java se8 random gaussian was cryptographically secure, and why or why not, as opposed to fortuna, or cha, or dh,

import java.util.*;

public class Jrandgauss{Jrandgauss(){}

long t1=0;long t2=0;int r=0;double r1=0;double r2=0;int bits=0;int seed=0;
Random rnd = new Random();

public void Random(long seed){rnd.setSeed(seed);}

public void setSeed(long seed){t1=seed^0x5DEECE66DL;}
//& ((1L << 48) - 1);}

protected long next(long bits){t2=seed*0x5DEECE66DL+0xBL;return t2;}
//& ((1L << 48) - 1);}

//return((int)(seed >>> (48 - bits)));

public double nextDouble() {rnd.setSeed(seed);
return (((long)next(26) << 27) + next(27))
/ (double)(1L << 53);}

private double nextNextGauss;
private boolean haveNextNextGauss = false;

public double nextGauss() {if (haveNextNextGauss) {
haveNextNextGauss= false;
return nextNextGauss;} else {double v1, v2, s; do {
v1 = 2 * nextDouble() - 1; // between -1.0 and 1.0
v2 = 2 * nextDouble() - 1; // between -1.0 and 1.0
s = v1 * v1 + v2 * v2;} while (s >= 1 || s == 0);
double mult = StrictMath.sqrt(-2 * StrictMath.log(s)/s);
nextNextGauss= v2 * mult;haveNextNextGauss= true;
return v1 * mult;}}

//::

//mult=((-2)*((ln(int^((maxlong/1M)-1))/(int^((maxlong/1M)-1))^(1/2);

public static void main(String[] args) {}
//for (s in range(r)){r2=(mult-(mult-s));
//return r2;
}//Cleaned version.

Answer 1

i was also wondering how you would get an original int back from any of these.