Always the same random numbers in HashSet

Question

I have the following class:

public class MyClass
{
    private Random rand;
    private HashSet<Pair<Integer, Integer>> set;

    public MyClass()
    {
       rand = new Random(Double.doubleToLongBits(Math.random()));
       set = new HashSet<Pair<Integer, Integer>>();
    }


    public void doSomething(int len)
    {   
        set.clear();

        for (int i = 0; i < 1000; i++)
        {
            int index = rand.nextInt(len - 1) + 1;

            int min = 1 - index;
            int max = len - index - 1;
            int j = rand.nextInt(max - min + 1) + min;

            if (j != 0)
            {   
               set.add(new Pair<Integer, Integer>(index, j));
            }
        }
    }
}

Pair is a custom class where I can store two integers. The problem is that every time I call doSomething() the HashSet contains always the same values.

How is it possible? How can I fix this problem?

EDIT:

this is my Pair: https://stackoverflow.com/a/677248/1395740

What are the the `equals` and `hashCode` methods on your `Pair` class? — Boris the Spider, Dec 09 '13 at 15:32
@Gladiator I use only one instance of MyClass, so the old one... — Nick, Dec 09 '13 at 15:33
What int value are you passing as argument to the doSomething() call? — NickDK, Dec 09 '13 at 15:37
Can you show us the code of Pair (if is not just 2 values + getters/setters)? — Arnaud Denoyelle, Dec 09 '13 at 15:39
@NickDK Always the same, it represents the length of an array — Nick, Dec 09 '13 at 15:39
@Nick I tried your code with a dummy implementation for Pair and it works : my pairs are different. Are you sure that you don't have a problem in the code you use to check the values? — Arnaud Denoyelle, Dec 09 '13 at 15:42
you could try to do `rand = new Random(i);` inside your for loop and the 2 calls could just be `my_random_value = rand.nextInt();` — Leggy7, Dec 09 '13 at 16:08

score 0 · Answer 1 · answered Dec 09 '13 at 15:33

0

From the documentation:

An instance of this class is used to generate a stream of pseudorandom numbers. The class uses a 48-bit seed, which is modified using a linear congruential formula. (See Donald Knuth, The Art of Computer Programming, Volume 3, Section 3.2.1.)

If two instances of Random are created with the same seed, and the same sequence of method calls is made for each, they will generate and return identical sequences of numbers.

answered Dec 09 '13 at 15:33

Joachim Rohde

5,915
2
29
46

That's what I thought about but the seed is itself random (`new Random(Double.doubleToLongBits(Math.random()))`). I looked at the implementation of Random and `new Random()` uses `System.nanotime()`. – Arnaud Denoyelle Dec 09 '13 at 15:34

score 0 · Answer 2 · answered Dec 09 '13 at 15:35

0

Math.random returns a double and Random receives a long as its seed. So probably you are getting 0 as the seed everytime. Try to change to System.getNanoTime() just for a test and see what happens.

answered Dec 09 '13 at 15:35

prmottajr

1,816
1
13
22

2

That is not what the OP is doing, the OP uses `Double.doubleToLongBits`. – Boris the Spider Dec 09 '13 at 15:36

score 0 · Answer 3 · answered Dec 09 '13 at 15:58

Regarding to your code, when run it with 5 as len parameter it's result this: [(4, -2), (4, -1), (4, -3), (2, -1), (3, -1), (3, -2), (1, 1), (2, 1), (1, 2), (3, 1), (2, 2), (1, 3)] [(4, -2), (3, -1), (4, -1), (4, -3), (3, -2), (2, -1), (1, 1), (2, 1), (1, 2), (3, 1), (2, 2), (1, 3)]

if you trace your code, it's not same pair, every time doSomething method generate new random numbers, but in loop , all numbers that lead j !=0 will ignored, so you will have that pairs,

your problem in this code

    int index = rand.nextInt(len - 1) + 1;
    int min = 1 - index;
    int max = len - index - 1;
    int j = rand.nextInt(max - min + 1) + min;

    if (j != 0)
    {   
       set.add(new Pair<Integer, Integer>(index, j));
    }

i mean this code leads to got same number regardless what random number be.

this code will generate all probabilities of two integers (pair) where it's less than len (method parameter), so every time you will have same pair with deference order , solution will be regarding to your business , — Gladiator, Dec 11 '13 at 11:51

Always the same random numbers in HashSet

3 Answers3