Java split the BitSet

Question

I want to split a bitset in more chunks. The splitting function depends on the cardinality of the bitset, which is the number of bits set to true. For example I have this BitSet with the cardinality 4:

INPUT:  101101

The desired output is the following:

OUTPUT: 100000 - 001000 - 000100 - 000001

Using the Java library called BitSet is there a function or a possible way to achieve that?

Answer 1

The following code applies every possible mask with a single bit set and only keeps the non-zero results:

int[] split(int input) {
    return IntStream.iterate(Integer.reverse(1), mask -> mask >>> 1)
            .limit(Integer.SIZE)
            .map(mask -> input & mask)
            .filter(result -> result != 0)
            .toArray();
}

Answer 2

Yes. Using basic operation AND.

Basicaly:

xxxxxx AND 110000 = xx0000.

Repeat procedure for all subsequences.

Answer 3

You need some loops here:

You should to find the positions of 1 then you can make a loop like this :

then If the position exist in the List then print 1 else print 0

public static void main(String[] args) {
    String Input = "101101";

    //find positions
    List<Integer> listPositivePosition = new ArrayList<>();
    for(int i = 0; i<Input.length(); i++){
        if(Input.charAt(i)=='1'){
            listPositivePosition.add(i);
        }
    }

    for(int i = 0; i<listPositivePosition.size(); i++){
        for(int j = 0; j<Input.length(); j++){
            //If the position exist in the List then print 1 else print 0
            if(j == listPositivePosition.get(i)){
                System.out.print("1");
            }else{
                System.out.print("0");
            }
        }
        System.out.println();
    }

}

Hope this can help you.

Answer 4

Please forgive my method naming convention :)

public static void main(String[] args) {

    final String a = "100100";
    System.out.println(Arrays.toString(foo(a)));
}

private static String[] foo(String a) {
    final long counted = IntStream.range(0, a.length()).filter(i -> a.charAt(i) == '1').count();
    final String[] ret = new String[(int) counted];
    int index = 0;
    for (int i = 0; i < a.length(); i++) {
        if (a.charAt(i) == '1') {
            ret[index] = ret(a, i);
            index++;
        }
    }
    return ret;

}

private static String ret(String a, int i) {
    final StringBuilder sb = new StringBuilder(a.replaceAll(".", "0"));
    sb.setCharAt(i, '1');
    a = sb.toString();
    return a;
}

Answer 5

Pure Java and easy-to-read solution:

List<String> binSplitter(String input) {

    String str = new String(new char[input.length()]).replace("\0", "0");
    List<String> chunks = new ArrayList<>();

    for (int i = 0; i < input.length(); i++) {
        if (input.charAt(i) == '1') {
            chunks.add(str.substring(0, i) + "1" + str.substring(i + 1, input.length()));
        }
    }

    return chunks;
}