Java generating random sequence of numbers with random access to the sequence elements

Question

In my program I need to produce many large sequences of random integers.

I know that I can just use the Random object to create the sequence, store that sequence in a list or an array and then query it for the i'th element when it needed but as the sequence can be very long (1M elements) and I have many different sequences I wish to know if I can store only the seed of each sequence and do something like:

public static int getIthNumber(int seed, int i){
    Random r = new Random(seed);
    for (int j=0; j< i-1; ++j) r.nextInt();
    return r.nextInt();
}

but without the overhead of the loop..

Peter Lawrey · Accepted Answer · 2012-10-10 22:16:19Z

3

If you need random access you can avoid using a proper sequence.

public static int getIthNumber(int seed, int i){
    return new Random(seed * 10123457689L + i * 101111).nextInt();
}

answered Oct 10, 2012 at 22:16

Peter Lawrey

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3 Comments

Peter Lawrey Over a year ago

they are big primes from primes.utm.edu/curios Pick any you find amusing. ;) The product of the two prime you use must be larger than the longest possible sequence otherwise the sequences will over lap. Note: Random only uses 48-bit of the seed and will repeat after 2^^48 values.

bennyl Over a year ago

Thanks again, can you please explain or point me to an explanation about the motivation to select two prime numbers p1 and p2 so that p1*p2 > maxseqlen as you proposed.

Peter Lawrey Over a year ago

If you have two small primes e.g. 31 and 3, it means that after 93 (31 * 3) values from one it will have the same sequence as another. (There will also be another seed which is the same after 2 * 93 ... n * 93. In the extreme case if you just add seed + i seeds 1,2,3,4 will be same after 3,2,1,0 values.

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