Creating a power set of a Sequence

Power set is easy to generate if one is familiar with bits. For the set of N elements, there will be 2^N subsets which will go to power set (including empty set and initial set). So each element will be either IN or OUT (1 or 0 in other words).

Taking this into consideration, it is easy to represent subsets of the set as bit masks. Then enumerating through all possible bit masks, it is possible construct the whole power sets. In order to do this we need to examine each bit in bit mask and take element of input set if there is 1 in that place. Below is example for string (collection of chars) as input. It can be easily rewritten to work for collection of any type values.

private static List<string> PowerSet(string input)
{
    int n = input.Length;
    // Power set contains 2^N subsets.
    int powerSetCount = 1 << n;
    var ans = new List<string>();

    for (int setMask = 0; setMask < powerSetCount; setMask++)
    {
        var s = new StringBuilder();
        for (int i = 0; i < n; i++)
        {
            // Checking whether i'th element of input collection should go to the current subset.
            if ((setMask & (1 << i)) > 0)
            {
                s.Append(input[i]);
            }
        }
        ans.Add(s.ToString());
    }

    return ans;
}

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Example

Suppose you have string "xyz" as input, it contains 3 elements, than there will be 2^3 == 8 elements in power set. If you will be iterating from 0 to 7 you will get the following table. Columns: (10-base integer; bits representation (2-base); subset of initial set).

0   000    ...
1   001    ..z
2   010    .y.
3   011    .yz
4   100    x..
5   101    x.z
6   110    xy.
7   111    xyz

You can notice that third column contains all subsets of initial string "xyz"

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Another approach (twice faster) and generic implementation

Inspired by Eric’s idea, I have implemented another variant of this algorithm (without bits now). Also I made it generic. I believe this code is near to fastest of what can be written for Power Set calculation. Its complexity is the same as for bits approach O(n * 2^n), but for this approach constant is halved.

public static T[][] FastPowerSet<T>(T[] seq)
{
    var powerSet = new T[1 << seq.Length][];
    powerSet[0] = new T[0]; // starting only with empty set

    for (int i = 0; i < seq.Length; i++)
    {
        var cur = seq[i];
        int count = 1 << i; // doubling list each time
        for (int j = 0; j < count; j++)
        {
            var source = powerSet[j];
            var destination = powerSet[count + j] = new T[source.Length + 1];
            for (int q = 0; q < source.Length; q++)
                destination[q] = source[q];
            destination[source.Length] = cur;
        }
    }
    return powerSet;
}