Java Random giving negative numbers

Question

I'm having trouble with Javas Random class, if i do this:

Random rng = new Random(seed) // seed == 29 in this example

String ss = "";
        for(int i = 0; i < 10; i++)
        {
            int s = rng.nextInt();
            ss += Integer.toString(s);
            ss +="\n";
        }

This is what i get back:

-1169335537
-2076183625
1478047223
1914482305
722089687
2094672350
-1234724057
-1614953544
-321574001
1000360613

From what I have read this should only be returning positive numbers for a start?

This may be a bit far fetched but it couldnt have anything to do with running a 64 bit machine on Windows 7 64 bit?

Any help at all would be awesome need to get this finished for an assignment hand in today!

only believe what you read in the javadocs. And (of course) read the javadocs. — Stephen C
– Stephen C, Commented Apr 29, 2011 at 4:07
Note Math.abs wont work one time in 2<sup>32</sup>. (And good luck testing that. Hint: Don't use a static mutable object.) — Tom Hawtin - tackline
– Tom Hawtin - tackline, Commented Apr 29, 2011 at 9:53

Ted Hopp · Accepted Answer · 2017-10-27 19:24:03Z

85

From the Java docs for nextInt():

All 2³² possible int values are produced with (approximately) equal probability.

One approach is to use the following transform:

s =  rng.nextInt() & Integer.MAX_VALUE; // zero out the sign bit

The reason something like this is needed (as opposed to using absolute value or negation) is that Integer.MIN_VALUE is too large in absolute value to be turned into a positive integer. That is, due to overflow, Math.abs(Integer.MIN_VALUE) == Integer.MIN_VALUE and Integer.MIN_VALUE == -Integer.MIN_VALUE. The above transformation preserves the approximately uniform distribution property: if you wrote a generate-and-test loop that just threw away Integer.MIN_VALUE and returned the absolute value of everything else, then the positive integers would be twice as likely as zero. By mapping Integer.MIN_VALUE to zero, that brings the probability of zero into line with the positive integers.

Here is another approach, which may actually be a tiny bit faster (although I haven't benchmarked it):

int s = rng.next(Integer.SIZE - 1); // Integer.SIZE == 32

This will generate an integer with 31 random low-order bits (and 0 as the 32^nd bit, guaranteeing a non-negative value). However (as pointed out in the comment by jjb), since next(int) is a protected method of Random, you'll have to subclass Random to expose the method (or to provide a suitable proxy for the method):

public class MyRandom extends Random {
    public MyRandom() {}
    public MyRandom(int seed) { super(seed); }

    public int nextNonNegative() {
        return next(Integer.SIZE - 1);
    }
}

Another approach is to use a ByteBuffer that wraps a 4-byte array. You can then generate a random four bytes (by calling nextBytes(byte[])), zero out the sign bit, and then read the value as an int. I don't believe this offers any advantage over the above, but I thought I'd just throw it out there. It's basically the same as my first solution (that masks with Integer.MAX_VALUE).

In an earlier version of this answer, I suggested using:

int s = rng.nextInt(Integer.MAX_VALUE);

However, according to the docs this will generate integers in the range 0 (inclusive) to Integer.MAX_VALUE (exclusive). In other words, it won't generate the value Integer.MAX_VALUE. In addition, it turns out that next(int) is always going to be faster than nextInt(int).

edited Oct 27, 2017 at 19:24

answered Apr 29, 2011 at 1:52

Ted Hopp

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9 Comments

jjb Over a year ago

Random.next() is protected, though, so you can't call it directly. You can subclass Random and expose something like nextPositiveInt() which returns next(31) pretty easily.

Ciro Santilli OurBigBook.com Over a year ago

Integer.SIZE - 1 would be slightly nicer.

Ted Hopp Over a year ago

@CiroSantilli六四事件法轮功纳米比亚胡海峰 - Yes, that would make it clearer where the value came from. I'll update my answer accordingly.

Felizett Over a year ago

Isn't (-1) >>> 1 just Integer.MAX_VALUE? I could just use rng.nextInt() & Integer.MAX_VALUE, or am I missing something?

Ted Hopp Over a year ago

@aldok - Correct. As I explained in my answer, this is a good thing because it will mean that 0 will occur (roughly) as often as any positive integer value. (This is true whether using secure random or java.util.Random.)

|

Josh KG · Accepted Answer · 2016-04-27 16:02:27Z

20

Since there is an equal chance of positive or negative numbers why not just:

Math.abs(rand.nextInt())

Nice and easy!

edited Apr 27, 2016 at 16:02

Josh KG

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answered Apr 27, 2016 at 15:14

kaifong

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2 Comments

Ted Hopp Over a year ago

This won't work. From the docs for Math.abs(int): "Note that if the argument is equal to the value of Integer.MIN_VALUE, the most negative representable int value, the result is that same value, which is negative."

kaifong Over a year ago

Then they should have called the method Math.absExceptIfTheArgumentIsEqualToIntegerMinValueInWhichCaseGoodBloodyLuckToYou(int);

martin clayton · Accepted Answer · 2011-04-29 01:49:40Z

11

Negative numbers are allowed - maybe you've read of the similar Random method nextInt( int ) which does limit the returned values to be zero or greater.

answered Apr 29, 2011 at 1:49

martin clayton

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1 Comment

Ciro Santilli OurBigBook.com Over a year ago

As mentioned by Ted, nextInt(Integer.MAXVALUE) leaves out Integer.MAXVALUE, so it is somewhat as good as the obvious Math.abs which gets one value wrong (Integer.MAXVALUE).

thomas · Accepted Answer · 2011-04-29 01:55:23Z

1

Check out the documentation for java.util.Random:

http://download.oracle.com/javase/6/docs/api/java/util/Random.html

Are you you trying to get random numbers from 0 to 28? If so, you need to use nextInt(int) as mentioned previously. The seed has no bearing on the range of possible outputs or their relative likelihoods.

answered Apr 29, 2011 at 1:55

thomas

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Comments

debracey · Accepted Answer · 2011-04-29 01:52:13Z

0

Per the documentation http://download.oracle.com/javase/6/docs/api/java/util/Random.html#nextInt():

Returns the next pseudorandom, uniformly distributed int value from this random number generator's sequence. The general contract of nextInt is that one int value is pseudorandomly generated and returned. All 2^32 possible int values are produced with (approximately) equal probability.

Just multiply by -1 if the value is negative

answered Apr 29, 2011 at 1:52

debracey

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1 Comment

Ted Hopp Over a year ago

Multiplying by -1 is not a good idea. First, it doesn't work: the negation of Integer.MIN_VALUE is Integer.MIN_VALUE again (due to overflow), so you can't get rid of all negative numbers that way. Even if it did work, the result would be a non-uniform distribution: zero would have half the probability of any positive integer.

luisZavaleta · Accepted Answer · 2011-04-29 01:54:14Z

0

You also can use Math.random() which returns values between 0 and 1

answered Apr 29, 2011 at 1:54

luisZavaleta

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nbk · Accepted Answer · 2019-11-10 19:11:13Z

0

int randomNumber = new Random().newInt(10);

This will return a random number between 0 and 9.

edited Nov 10, 2019 at 19:11

nbk

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answered Nov 9, 2019 at 23:17

Gayathri Krishnan

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Khang Hoang Nguyen · Accepted Answer · 2019-12-10 00:21:33Z

0

If you happen to work with numbers that have the possibility of having a negative value you can turn it into a positive value using a conditional declaration automatically by multiplying the value to a negative one. You can also turn a positive into a negative value using this same method.

The examples are below.

int a = -5;
a = a < 0? ( a == -2147483648? 0 : -a ) 
           : a;

int b = -2147483648;
b = b < 0? ( b == -2147483648? 0 : -b ) 
           : b;

int c = 12345;
c = c < 0? ( c == -2147483648? 0 : -c ) 
           : c;

edited Dec 10, 2019 at 0:21

answered Mar 19, 2017 at 11:21

Khang Hoang Nguyen

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5 Comments

Ted Hopp Over a year ago

Unfortunately, this doesn't work for Integer.MIN_VALUE because Integer.MIN_VALUE == -Integer.MIN_VALUE due to overflow.

Khang Hoang Nguyen Over a year ago

To fix the integer min value problem or max value problem, if you do not mind the number being one number off you can do this, b = b < 0? b == Integer.MIN_VALUE? (b + 1) * -1 : b * -1 : b;

Khang Hoang Nguyen Over a year ago

If you need the number to be precise, you need a second line to correctly correct the overflow problem. The solution above is only for when you working with a random number and do not mind if the number can be one digit different.

Ted Hopp Over a year ago

For OP's case (where 0 is also an allowed value along with all positive integers) the best way to deal with this is to map Integer.MIN_VALUE to 0 and to negate all other negative numbers. That way, all non-negative numbers have equal probability of being generated (assuming the underlying generator generates all integers with equal probability). Every positive value can be generated in two ways: either by being produced directly by the PRNG or as the negation of a negative number other than Integer.MIN_VALUE. Zero can also be generated two ways: directly and from Integer.MIN_VALUE.

Khang Hoang Nguyen Over a year ago

I do agree with Ted Hopp comment from above, instead of (b + 1) * -1 just change that to 0.

Jeppz · Accepted Answer · 2021-02-19 06:25:33Z

0

int s = rng.nextInt(bound); //bound 29 in this case

This will have a bound of 0 to 29.

edited Feb 19, 2021 at 6:25

Jeppz

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answered Jul 29, 2014 at 1:55

Nawaz

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