Print list of binary permutations

Question

What I am trying to do is print all the possibilities of a binary number n digits long. In other words, with a 4 digit number:

..etc

To be honest, I have no idea of where to even start with this (other than I figure I'd need to use a loop, and probably an array) so any pointers in the right direction would be appreciated.

I should clarify, all possible numbers within a given range. I.E. all possibilities of a 4 digit binary. Replace 4 with whatever number. It's not homework but I am trying to teach myself java. — Smitty, Dec 11 '11 at 02:14
Generally people who have the motivation to teach themselves a language do not use statements like *"I have no idea of where to even start"*. Get a book, read, try simpler tasks.. — Andrew Thompson, Dec 11 '11 at 02:36
Thanks for your commentary, but I don't see a real need to defend myself on the internet. I was interested in understanding something, therefore I asked a question. — Smitty, Dec 11 '11 at 02:45
No offense, but this questions show lack of understanding how computers do math. — harold, Dec 11 '11 at 10:33

score 16 · Accepted Answer · answered Dec 11 '11 at 02:15

16

Maybe you could use a recursive algorithm:

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("", 4);

That would give you all possible binary numbers with 4 digits.

Hope this helped!

answered Dec 11 '11 at 02:15

eboix

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You wouldn't really be generating numbers, but it would still produce the desired output! – eboix Dec 11 '11 at 02:18

score 7 · Answer 2 · answered Dec 11 '11 at 02:10

7

For an n-bit binary number, there are 2^n "permutations". You just need to loop over the integers from 0 to (1<<n)-1, and convert each one to binary.

answered Dec 11 '11 at 02:10

Oliver Charlesworth

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you mean loop from 0 - (2^n)-1? – ChrisWue Dec 11 '11 at 02:13

score 5 · Answer 3 · edited May 23 '17 at 11:45

5

for(int i=0; i < 128; i++){
  System.out.println(Integer.toBinaryString(i));
}

Adjust the max for as high as you'd like to go.

If you need the padded 0's, there was another question on that just today: Pad a binary String equal to zero ("0") with leading zeros in Java

edited May 23 '17 at 11:45

Community

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1

answered Dec 11 '11 at 02:11

ziesemer

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This won't give the desired output. – Oliver Charlesworth Dec 11 '11 at 02:13
Since it's likely a homework exercise, this is a minor detail that can be remedied by the OP. :-) – ziesemer Dec 11 '11 at 02:14

duffymo · Answer 4 · 2011-12-11T02:26:32.863

It helps to know how many possibilities there are.

2^4 = 16, right?

It'll help to know this as well.

Here's how I'd do it:

/**
 * BinaryDemo
 * @author Michael
 * @since 12/10/11
 */
public class BinaryDemo {

    public static void main(String[] args) {
        if (args.length > 0) {
            int n = Integer.parseInt(args[0]);
            int m = 1;
            for (int i = 1; i <= n; ++i) {
                m *= 2;
            }
            System.out.println("# bits  : " + n);
            System.out.println("# values: " + m);
            String format = "%" + n + "s";
            for (int i = 0; i < m; ++i) {
                System.out.println(String.format(format, Integer.toString(i, 2)));
            }

        } else {
            System.out.println("Usage: BinaryDemo <n>");
        }
    }
}

Thank you for your help, this is very helpful and I'll be digging into this now too :) — Smitty, Dec 11 '11 at 02:43

score 1 · Answer 5 · answered Oct 17 '17 at 21:23

My solutions uses the backtracking algorithm. Just print all combinations using '0' and then '1'. It's not better than accepted answer but just one more way to solve the problem.As an example just call the function like below from main()

ArrayList<Integer> chosen = new ArrayList<>(); 
printAllBinaryDigits(4, chosen);

private static void printAllBinaryDigits(int digits, ArrayList<Integer>chosen) {
            if (digits == 0) {
                System.out.println(chosen);
            } else {
                chosen.add(0);
                printAllBinaryDigits(digits-1, chosen);
                chosen.remove(chosen.size()-1);
                chosen.add(1);
                printAllBinaryDigits(digits-1, chosen);
                chosen.remove(chosen.size()-1);
            }

        }

score 1 · Answer 6 · answered Nov 13 '19 at 08:47

Below is the solution using Recursion as an approach in java

public class NumberOfBinaryPatterns {
    static int[] bitArray = new int[]{0,1};
    public static void main(String args[])
    {   
        System.out.println("Below are the patterns\n");
        int numberOfBits=3; // It can be any value based on the requirement, i.e. 4,5, etc. We can get the input for "numberOfBits" from user on runtime to enhance this further
        drawBinaryPattern(numberOfBits,"");
    }
    private static void drawBinaryPattern(int n,String seed)
    {
        if(n==0){
            System.out.println(seed);
            return;
        }   
        for(int j=0;j<bitArray.length;j++)
        {
            String temp = seed+bitArray[j];
            drawBinaryPattern(n-1,temp);
        }
    }
}

score 0 · Answer 7 · edited Feb 16 '13 at 05:43

To find all possible permutations of a given binary string(pattern) for example

The permutations of 1000 are 1000, 0100, 0010, 0001

void permutation(int no_ones, int no_zeroes, string accum){
    if(no_ones == 0){
        for(int i=0;i<no_zeroes;i++){
            accum += "0";
        }

        cout << accum << endl;
        return;
    }
    else if(no_zeroes == 0){
        for(int j=0;j<no_ones;j++){
            accum += "1";
        }

        cout << accum << endl;
        return;
    }

    permutation (no_ones - 1, no_zeroes, accum + "1");
    permutation (no_ones , no_zeroes - 1, accum + "0");
}

int main(){
    string append = "";

    //finding permutation of 11000   
    permutation(2, 6, append);  //the permutations are 

    //11000
    //10100
    //10010
    //10001
    //01100
    //01010

    cin.get(); 
}

Print list of binary permutations

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