same value for seed used to create java Random on two machines

Question

If I use the same seed value for Random in a java program and run this on two different machines ,will I get the same set of numbers?

for example

    long seed = 123L;//may be taken from some database or something
    java.util.Random ran = new java.util.Random(seed);
    int ret = 0;
    for (int i= 0; i< 10; i++){
        ret = ran.nextInt(1000);
        System.out.println("ret="+ret);
    }

I always get

ret=782
ret=450
ret=176
ret=789
ret=795
ret=657
ret=834
ret=837
ret=585
ret=453

If I run this multiple times on my computer,I would get the same set of numbers.. but suppose someone manages to get the secret seed value I used(by guessing or from the secret location where it was stored) and run this code on his machine,will he get the same set of numbers?

Yes. If you want a secure way of generating random numbers, then use SecureRandom. — Boris the Spider
– Boris the Spider, Commented Jun 23, 2013 at 12:19

Stuart Golodetz · Accepted Answer · 2013-06-23 12:25:42Z

Yes, the contract specifying the way in which random numbers are generated is the same in both cases, so they'll produce the same sequence of numbers if given the same seed. Implementations of Random must use prescribed algorithms in order to ensure that this is the case. A more precise way of putting it (from the relevant documentation) is:

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. In order to guarantee this property, particular algorithms are specified for the class Random. Java implementations must use all the algorithms shown here for the class Random, for the sake of absolute portability of Java code. However, subclasses of class Random are permitted to use other algorithms, so long as they adhere to the general contracts for all the methods.

If you're relying on this behavior make sure that you test across JVM versions and platforms. Write once, debug everywhere.
@Jason: Certainly good advice. Writing automatically-run tests to check this kind of thing might not be a bad idea either. That way, you're slightly more likely to find out about invalid assumptions you've made when building on a new platform.

built1n · Accepted Answer · 2013-06-23 12:28:51Z

The JRE should generate the same sequence of random numbers, given that they use same seed, and algorithm are used. The Java documentation on Random says:

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.

The algorithms are used by all JRE implementations, too:

In order to guarantee this property, particular algorithms are specified for the class Random. Java implementations must use all the algorithms shown here for the class Random, for the sake of absolute portability of Java code. However, subclasses of class Random are permitted to use other algorithms, so long as they adhere to the general contracts for all the methods.

Simiil · Accepted Answer · 2013-06-23 12:21:46Z

2

Yes, thats the point.

For example: In Minecraft you can obtain level seeds to initialize the Random Generator, and everyboy with this seed will get the same map.

If you read the JavaDoc, you'll see that next(int bits) (and nextInt() is just next(32)) will update the seed to (seed * 0x5DEECE66DL + 0xBL) & ((1L << 48) - 1) and return (int)(seed >>> (48 - bits)). Thats always the same on any computer for the same seed.

answered Jun 23, 2013 at 12:21

Simiil

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Comments

Juned Ahsan · Accepted Answer · 2013-06-23 12:23:01Z

Answer is yes because the way Random.nextInt() is impelmented. It simply uses the seed and does some calcuaiton to generate the numbers. The code does not use any mahcine specific params for the generation. Here is the code of nextInt():

 public int nextInt(int n) {
     if (n<=0)
        throw new IllegalArgumentException("n must be positive");

     if ((n & -n) == n)  // i.e., n is a power of 2
         return (int)((n * (long)next(31)) >> 31);

     int bits, val;
     do {
         bits = next(31);
         val = bits % n;
     } while(bits - val + (n-1) < 0);
     return val;
 }

Read more about Random nextInt method here:

http://docs.oracle.com/javase/6/docs/api/java/util/Random.html#nextInt(int)

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