How to create a random 16 digits number, with specific first digits?

Question

I want to create a randomly generated 16 digit-number in java.But there is a catch I need the first two digits to be "52". For example, 5289-7894-2435-1967. I was thinking of using a random generator and create a 14 digit number and then add an integer 5200 0000 0000 0000. I tried looking for similar problems and can't really find something useful. I'm not familiar with the math method,maybe it could solve the problem for me.

One google search. The rest should be easy with this. http://stackoverflow.com/questions/3709521/how-do-i-generate-a-random-n-digit-integer-in-java-using-the-biginteger-class — austin wernli, Jan 08 '15 at 15:45
Yes,I've made that thought as well, but I don't know how to do it. This was my problem to begin with. — Marios Ath, Jan 08 '15 at 15:45
You should probably look at these two : http://stackoverflow.com/questions/4655931/12-digit-unique-random-number-generation-in-java http://stackoverflow.com/questions/5328822/generating-10-digits-unique-random-number-in-java — user1676389, Jan 08 '15 at 15:49
Your idea, to randomly create a 14-digit number and then add 5200000000000000 to it (or convert it to a string and prepend `"52"`) makes sense. Do you care about whether every possible 14-digit number could be generated, or is it acceptable if your random number generator misses some possibilities? Do you care about how uniform the distribution is? — ajb, Jan 08 '15 at 15:50

Nic · Accepted Answer · 2015-01-12T19:26:08.543

First, you need to generate a random 14-digit number, like you've done:

long first14 = (long) (Math.random() * 100000000000000L);

Then you add the 52 at the beginning.

long number = 5200000000000000L + first14;

One other way that would work equally well, and will save memory, since Math.random() creates an internal Random object:

//Declare this before you need to use it
java.util.Random rng = new java.util.Random(); //Provide a seed if you want the same ones every time
...
//Then, when you need a number:
long first14 = (rng.nextLong() % 100000000000000L) + 5200000000000000L;
//Or, to mimic the Math.random() option
long first14 = (rng.nextDouble() * 100000000000000L) + 5200000000000000L;

Note that nextLong() % n will not provide a perfectly random distribution, unlike Math.random(). However, if you're just generating test data and it doesn't have to be cryptographically secure, it works as well. It's up to you which one to use.

sol4me · Answer 2 · 2015-01-08T16:19:19.177

3

You can generate 14 random digits and then append at the beginning "52". E.g.

public class Tes {

    public static void main(String[] args) {
        System.out.println(generateRandom(52));
    }

    public static long generateRandom(int prefix) {
        Random rand = new Random();

        long x = (long)(rand.nextDouble()*100000000000000L);

        String s = String.valueOf(prefix) + String.format("%014d", x);
        return Long.valueOf(s);
    }
}

edited Jan 08 '15 at 16:19

answered Jan 08 '15 at 15:47

sol4me

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You could also add the `52` by adding `5200000000000000` (plus or minus a few zeros) to the `x`, if you don't want to use `String` processes to add numbers. It feels cleaner to me. – Nic Jan 08 '15 at 15:48
Actually, the code you've posted, that uses the `String` to add a prefix, is wrong. It will fail if `x` is less than 10^13. You'd need something like `String.format("%014d", x)`. – ajb Jan 08 '15 at 16:15
Are the digits of `(long)(rand.nextDouble()*100000000000000L)` uniformly distributed? Would probably be better to use this: http://stackoverflow.com/a/2546186/1178016 – xehpuk Jan 08 '15 at 16:33

weston · Answer 3 · 2015-01-08T16:12:58.240

3

Random rand = new Random();
String yourValue = String.format((Locale)null, //don't want any thousand separators
                        "52%02d-%04d-%04d-%04d",
                        rand.nextInt(100),
                        rand.nextInt(10000),
                        rand.nextInt(10000),
                        rand.nextInt(10000));

edited Jan 08 '15 at 16:12

answered Jan 08 '15 at 16:02

weston

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score 0 · Answer 4 · answered Jan 08 '15 at 15:46

0

Create a 14 random digit number. with Math.random
Concatenate to it at the begining the "52" String.
Convert the string with Integer.parseInt(String) method

answered Jan 08 '15 at 15:46

Jordi Castilla

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Unfortunately you haven't explained how to do the first step. This is not trivial, since `Random.nextLong()` doesn't allow you specify a bound, unlike `Random.nextInt(bound)`. – ajb Jan 08 '15 at 15:47

How to create a random 16 digits number, with specific first digits?

4 Answers4