CODEWITHSUNDEEP
javarandombinary
willkara
willkara

Reputation: 177

Trying to randomly generate a binary number from 000 to 110

I have code that randomly generates a binary number from 000 to 111 but I'm having trouble having it create a number from 000 to just 110. I know that I can somehow rerun the code everything it comes out with 111 but I can't seem to figure out the way to have it do that.

public String binNumber() {
        StringBuilder storage = new StringBuilder();
        int i = 0;
        while (i < 3) {
            int binny = this.giveMeBinary();
            storage.append(String.valueOf(binny));

            i++;
        }


        return storage.toString();
    }

public int giveMeBinary() {
        Random rg = new Random();
        int bin = rg.nextInt(2);
        return bin;

    }

Upvotes: 0

Views: 12100

Answers (5)

Sidharth Taneja
Sidharth Taneja

Reputation: 564

I created a method which can return a random binary number below 128 (value 128 can be alter in the code)-

    public static void RandomBinary(int len){
    if(len<128) {
        Random random = new Random();
        boolean res;
        String randomBinary = "";
        for (int i = 0; i < len - 1; i++) {
            res = random.nextBoolean();
            if (res)
                randomBinary = randomBinary.concat("1");
            else
                randomBinary = randomBinary.concat("0");
        }
        System.out.print(randomBinary);
    }else {
        System.out.print("KeyGen maximum length allowed is 128. Please enter value below 128");
    }
}

Upvotes: 0

nolt2232
nolt2232

Reputation: 2644

If you want to re-roll, try this:

String num = "";
while (true)
{
      num = binNumber();
      if (!num.equals("111"))
      {
          break;
      }
}
return num;

Upvotes: 1

chubbsondubs
chubbsondubs

Reputation: 38852

Something like:

Random random = new Random();

int nextNumber = random.nextInt( 7 );
System.out.println( Integer.toBinaryString( nextNumber ) );

Remember your numbers you create in a computer are always in binary. You just need to print it out differently. So even though you've given it a decimal number that number is converted by the compiler to binary. 7 == 0x111 when its compiled.

Upvotes: 1

Adam Mihalcin
Adam Mihalcin

Reputation: 14478

The better way to do this is to generate a random number from 0 to 6 inclusive, and then convert to a string

public String binNumber() {
    Random rg = new Random();
    int n = rg.nextInt(7);
    return Integer.toBinaryString(n);
}

Upvotes: 8

Jeremy
Jeremy

Reputation: 5433

"Rerolling" if you get 111 is definitely a bad idea; that's potentially an algorithm with infinite running time, although in practice it would do quite well.

Why work in base 2? Work in base 10, easily get a random number between 0 and 6, and then translate back to binary.

Upvotes: 3

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