Using binary counting to count all subsets of an array - c++

Question

So if I am given an array such as
a = {1, 2, 3}
We know that the given subarrays (non contiguous included) are (this represents the power set)
{1} {2} {3} {1,2,3} {1,2} {1,3} {2,3}
I also know that these subsets can be represented by counting in binary from
000 -> 111 (0 to 7), where each 1 bit means we 'use' this value from the array
e.g. 001 corresponds to the subset {3}
I know that this method can somehow be used to generate all subsets, but im not really sure how this can be implemented in c++
So basically what I am asking is how can (if it can) binary counting be used to generate power sets?
Any other methods for generating a power set are also much appreciated!

Answer 1

For your example with 3 set elements you can just do this:
for (s = 1; s <= 7; ++s)
{
// ...
}
Here's a demo program:
#include <iostream>
int main()
{
const int num_elems = 3; // number of set elements
const int elems[num_elems] = { 1, 2, 3 }; // mapping of set element positions to values
for (int s = 1; s < (1 << num_elems); ++s) // iterate through all non-null sets
{
// print the set
std::cout << "{";
for (int e = 0; e < num_elems; ++e) // for each set element
{
if (s & (1 << e)) // test for membership of set
{
std::cout << " " << elems[e];
}
}
std::cout << " }" << std::endl;
}
return 0;
}
Compile and test:
$ g++ -Wall sets.cpp && ./a.out
{ 1 }
{ 2 }
{ 1 2 }
{ 3 }
{ 1 3 }
{ 2 3 }
{ 1 2 3 }
Note that it's a common convention to make the least significant bit correspond to the first set element.
Note also that we are omitting the null set, s = 0, as you don't seem to want to include this.
If you need to work with sets larger than 64 elements (i.e. uint64_t) then you'll need a better approach - you can either expand the above method to use multiple integer elements, or use std::bitset or std::vector<bool>, or use something like #Yochai's answer (using std::next_permutation).

Answer 2

Actually creating the sets is pretty easy - just use bitwise operations >>= and & to test a bit at a time. Assuming input vector/array a[] known to have 3 elements and therefore produce a 7 vector output:
std::vector<std::vector<T>> v(7);
for (int n = 1; n <= 7; ++n) // each output set...
for (int i = 0, j = n; j; j >>= 1, ++i) // i moves through a[i],
// j helps extract bits in n
if (j & 1)
v[n-1].push_back(a[i]);

Answer 3

For compile time size, you may use bitset, something like:
template <std::size_t N>
bool increase(std::bitset<N>& bs)
{
for (std::size_t i = 0; i != bs.size(); ++i) {
if (bs.flip(i).test(i) == true) {
return true;
}
}
return false; // overflow
}
template <typename T, std::size_t N>
void display(const std::array<T, N>& a, const std::bitset<N>& bs)
{
std::cout << '{';
const char* sep = "";
for (std::size_t i = 0; i != bs.size(); ++i) {
if (bs.test(i)) {
std::cout << sep << a[i];
sep = ", ";
}
}
std::cout << '}' << std::endl;
}
template <typename T, std::size_t N>
void display_all_subsets(const std::array<T, N>& a)
{
std::bitset<N> bs;
do {
display(a, bs);
} while (increase(bs));
}
Live example