How to list the ones of a counting number?

Question

What are the options to quickly list the 1's of a Java BigInteger? We can assume the BigInteger is positive, but it might be a rather large Java BigInteger with sparely distributed 1's. Nevertheless we would like to find them quickly. I would like to have an enumerator or iterator for the bit-positions of the 1's of a Java BigInteger.

Chai T. Rex's answer does what you want, but I am not sure if it is faster than simply testing each bit with `testBit()` in a loop (Henry's comment). You could test both and see which is faster or at least fast enough. If the 1 bits are pretty sparse, the `getLowestSetBit()` -- `clearBit()` method could be faster. — Rudy Velthuis, Feb 04 '18 at 11:35

score 1 · Answer 1 · answered Feb 03 '18 at 12:06

You can use BigInteger's getLowestSetBit and clearBit methods. For example, to print them out:

public static final void printOneIndexes(BigInteger n) {
    while (!n.equals(BigInteger.ZERO)) {
        int i = n.getLowestSetBit();
        System.out.printf(" %d", i);
        n = n.clearBit(i);
    }
}

score 0 · Answer 2 · answered Feb 03 '18 at 09:48

0

Use .toString(radix) method. Like this: bigInteger.toString(2) and then check for ones in a string

answered Feb 03 '18 at 09:48

Robert K.

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Why would you convert to a string, when you can test the bits directly `BigInteger.testBit(int n)`? – Henry Feb 03 '18 at 09:50
@Ridcully the result would be the same as `bigInteger.bitCount()` which might be even easier to read and understand. – Henry Feb 03 '18 at 12:22
@Ridcully because this is not the question. OP wants the bit positions, not the number of them. – Henry Feb 03 '18 at 18:16

Rudy Velthuis · Accepted Answer · 2018-02-04T23:36:24.157

One can use getLowestSetBit(), but tests showed that the following is about 3-4 times as fast:

public static final void printOneIndexes(BigInteger n)
{
    for (int i = 0; i < n.bitLength(); i++)
    {
        if (n.testBit(i))
            System.out.printf(" %d", i);
    }
}

Of course, if you remove the printing, and store the results in a list, it can be much faster still.

I tested it with a BigInteger with only a few bits set, created by:

public static BigInteger makeSparseInt()
{
    BigInteger b = BigInteger.ZERO.setBit(1000000);

    Random r = new Random();

    for (int i = 0; i < 100; i++)
    {
        b = b.setBit(r.nextInt(1000000));
    }
    return b;
}

How to list the ones of a counting number?

3 Answers3