Power set of an input set as custom collection

Question

I have been reading the Effective Java book and I have stuck with this code I am unable to understand how this code is generating power set.

Code:

public class PowerSet {

    public static final <E> Collection<Set<E>> of(Set<E> s) {
        List<E> src = new ArrayList<>(s);
        if (src.size() >= 30)
            throw new IllegalArgumentException("Set too big " + s);
        return new AbstractList<Set<E>>() {

            @Override
            public int size() {
                return 1 << src.size();
            }

            @Override
            public boolean contains(Object o) {
                return o instanceof Set && src.containsAll((Set) o);
            }

            @Override
            public Set<E> get(int index) {
                Set<E> result = new HashSet<>();
                for (int i = 0; index != 0; i++, index >>= 1)
                    if ((index & 1) == 1)
                        result.add(src.get(i));
                return result;
            }
        };
    }

    public static void main(String[] args) {
        Collection<Set<String>> result = of(Set.of("a", "b", "c"));
        System.out.println(result);
    }
}

Output:

[[], [a], [b], [a, b], [c], [a, c], [b, c], [a, b, c]]

Can someone explain how this code is generating powerset of a given set.

What exactly do you struggle with? Does this help stackoverflow.com/a/24366172/1398418? — Oleg
– Oleg, Commented Jul 24, 2020 at 11:49

tb189 · Accepted Answer · 2020-07-24 11:54:56Z

The code uses the binary representation of the index number as a map of which element of s to include.

For instance, assuming only 3 bits in a number:

index   | a | b | c
--------------------
0 (000) | 0 | 0 | 0 -> take nothing
1 (001) | 0 | 0 | 1 -> take only c
2 (010) | 0 | 1 | 0 -> take only b
3 (011) | 0 | 1 | 1 -> take a and b
4 (100) | 1 | 0 | 0 -> take only a
...

The get method of the generated list follows this logic with the index input given:

index >>= 1 shifts all bits one position to the right with each loop
(index & 1) == 1 checks whether the rightmost bit of index is a 1

The & operator is the binary AND, so 2 & 1 equals binary 010 AND 001, giving 000 (not equal to 1 or 001) and 3 & 1 equals binary 011 AND 001, giving 001 (equal to 1 or 001)

If this evaluates to true, the i-th element is added to the list
This terminates when index == 0, i.e. there are no more bits to shift / elements to add

Example for index = 3:

i | index | (index & 1) == 1 | element added
---------------------------------------------
0 |   011 |             TRUE |   a (0-th element)
1 |   001 |             TRUE |   b (1-th element)
2 |   000 |            FALSE |   - 
(terminates as index == 0)

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