Generate 10,000 unique 9-digit numbers

Question

public static void main(String[] args) {
    for(int i = 0; i < 10000; i++) {
        long number = (long) Math.floor(Math.random() * 90000000) + 100000000L;
        System.out.println(number);
    }
}

The above code doesn't generate 10,000 random numbers. Why?

You mean, it generates 10,000 random numbers, but they repeat? Or what is the problem? — Thilo, May 20 '14 at 11:32
Random doesn't mean unique: see Birthday paradox http://en.wikipedia.org/wiki/Birthday_problem in your case you can expect about 2*sqrt(90000000) = 2e4 unique numbers — Dmitry Bychenko, May 20 '14 at 11:33
A good example of Random Numbers in a range .. http://stackoverflow.com/questions/363681/generating-random-numbers-in-a-range-with-java — Kenneth Clark, May 20 '14 at 11:37
This question appears to be off-topic because this is not a code review or debugging service. — Raedwald, May 20 '14 at 12:01

Thilo · Answer 1 · 2014-05-20T11:43:09.100

If you don't require uniform distribution, you could start with 100,000,000 and add an increment randomly chosen between [1 and 89,999]. Repeat 10,000 times.

This will give you a non-repeating sequence of numbers that are somewhat randomly generated.

But they will not be uniformly distributed (it is very unlikely to get numbers in the high ranges, like 999,xxx,xxx), and they will be strictly ascending, so you have a good chance to guess the next one if you know the previous one ("good" = 1/89999).

score 0 · Answer 2 · answered May 20 '14 at 11:38

0

Do it this way instead:

public static void main(String[] args) {
    Set<Integer> uniques = new HashSet<>(15000);
    Random random = new Random();
    for (int i = 0; i < 10000;) {
        int number = random.nextInt(Integer.MAX_VALUE);
        if (uniques.add(number)) {
            System.out.println(i + ": " + number);
            i++;
        }
    }
}

answered May 20 '14 at 11:38

Harmlezz

7,972
27
35

maybe pre-size the HashSet. It's going to get big. – Thilo May 20 '14 at 11:40

score 0 · Answer 3 · answered May 20 '14 at 11:40

Using Set, you can ensure uniqueness. Try,

 Set<Long> resSet=new HashSet<>();
 for(int i = 0;  resSet.size()<10000; i++){
     long number = (long) Math.floor(Math.random() * 90000000) + 100000000L;
     resSet.add(number);
 }
 System.out.println(resSet);

score 0 · Answer 4 · edited May 20 '14 at 12:09

0

I had a similar problem and I did the following, generate random number and check it has been generated before, if not, just use and note it.

List<String> gen = new ArrayList<String>();
long number=0;
int i=0;

while(i<10000)
{
   while(!gen.contains(number) || i==0)
      number = (long) Math.floor(Math.random() * 90000000) + 100000000L;
   gen.add(number);
   System.out.println(number);
   i++;
}

I didn't tested the code since I did it in .NET, but I searched for code equivalencies.

Hope it helps!

edited May 20 '14 at 12:09

рüффп

5,172
34
67
113

answered May 20 '14 at 11:47

ZeroWorks

1,618
1
18
22

i did try this and i got 10,000 random numbers,but these numbers are of 8 and 9-digit values(characters).....i want only 9 digit values – user3651409 May 21 '14 at 10:38
I used your algorithm to get number... so try this instead... replace: number = (long) (Math.floor(Math.random() * 999999999D) + 100000000L); Adding 100000000 to random should return a 9-digit value. – ZeroWorks May 21 '14 at 12:10

score 0 · Answer 5 · answered May 20 '14 at 11:51

A minor contribution to @Masud's code:

         Set<Long> resSet = new HashSet<Long>();
         while(true) {
             long number = (long) Math.floor(Math.random() * 90000000) + 100000000L;
             resSet.add(number);
             if(resSet.size() == 10000)
                 break;
         }

This way you can surely get 10000 unique numbers (since the for loop does not guarantee that).

One drawback of this approach is that it "perishes" several numbers (which are the same as previous ones). I don't know if there is some more clever approach to this.

Generate 10,000 unique 9-digit numbers

5 Answers5