CODEWITHSUNDEEP
algorithm
2hamed
2hamed

Reputation: 9057

What is the algorithm to produce every k-bit number?

I want an algorithm to do the following: When the input is for example 3 I want every 3-bit number to be produced like the following: 

000
001
010
011
100
101
110
111

EDIT: Thank you for all your answers, But I'd prefer an algorithm that treats '1's and '0's as characters and the whole answer as an string so I could extend the answer for characters as well. Like producing every possible combination of a,b,c with the length of 3.

Upvotes: 0

Views: 286

Answers (5)

Antoine Gersant
Antoine Gersant

Reputation: 536

Here is some pseudo code which should help you :

function listNumbers (bits : Int) : List<String> {
  l = [];
  if (bits == 0) {
    l.append("");        
  } else {
    prev = listNumbers(bits-1);
    for (number in prev) {
      l.append("0" + number);
      l.append("1" + number);
    }
  }
  return l;
}

Upvotes: 1

Asgeir
Asgeir

Reputation: 1112

What you want do do is generate combinations the code in the article shouldn't be too hard to generalize for all characters using modular arithmetic. Or alternately you map characters onto numeric values, calculate the permutations and then map back to characters.

Upvotes: 0

eboix
eboix

Reputation: 5131

Maybe you could use a recursive algorithm. This is written in Java:

public void printBin(String soFar, int iterations) {
    if(iterations == 0) {
        System.out.println(soFar);
    }
    else {
        printBin(soFar + "0", iterations - 1);
        printBin(soFar + "1", iterations - 1);
    }
}

You would execute this like this:

printBin("", 3);

That would give you all possible binary numbers with 3 digits.

Note, however, that with really big amounts of digits, you might get overflows.

Hope this helped!

Upvotes: 0

Jahan Zinedine
Jahan Zinedine

Reputation: 14874

It means every number between zero and 2^n-1 which n is your bit number

Upvotes: 1

Luchian Grigore
Luchian Grigore

Reputation: 258618

A straight-forward algorithm would be:

  • calculate 2^n-1; in your case, 7.

  • for i = 0 : 7 convert i to binary form

  • output binary form

Upvotes: 7

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