Using std::bitset to count subsets of a finite set - c++

Question

I would like to check how many subsets S of the set [1, ..., 15]
there are so that it is impossible to choose two elements from S
such that their sum is a multiple of 3.
The algorithm to check this is as follows: there is a natural bijection between
the subsets of [1, ..., 15] and the strings of length 15 with two characters (assume
the two characters are '0' and '1' to fix a convention), where the character '0' in position i means that the integer i is not in the subset, while the character '1' in position i means that the integer i is in the subset.
For example, the string "111001000000000" is associated to the subset {1, 2, 3, 6}. This subset does not fulfill the constraint described above.
I wrote a C++ code to generate all such strings, convert them to a vector of
ints between 1 and 15, and check for all couples in this set if there is one
whose sum is a multiple of 3.
This is the code:
#include <algorithm>
#include <bitset>
#include <cmath>
#include <iostream>
#include <vector>
bool check(const std::vector<int>& dset) {
if (dset.size() == 1) {
if (dset[0] % 3 == 0) { return false; }
}
for (size_t i = 0; i < dset.size() - 1; ++i) {
auto a = dset[i];
for (size_t j = i + 1; j < dset.size(); ++j) {
auto b = dset[j];
if ((a + b) % 3 == 0) { return false; }
}
}
return true;
}
int main() {
const int N = 15; // We consider subsets of [1, ..., N].
int approved = 1; // We automatically approve the empty set.
std::bitset<N> set;
for (int n = 1; n < std::pow(2, N); ++n) {
set = std::bitset<N>(n);
std::vector<int> dset(set.count());
size_t j = 0;
for (int i = 1; i <= N; ++i) {
if (set[i - 1]) {
dset[j++] = i;
}
}
// Sweep through all couples in dset.
if (check(dset)) {
++approved;
}
}
std::cout << approved << " out of " << std::pow(2, N) << std::endl;
}
The problem is that my code returns 373, which is the wrong answer (the correct one should be 378).
I guess I am doing something wrong here, but I cannot find the error in my code.

Answer 1

In check() function you don't need to check for dset size being 1, since it is taken care below in the for loop from index i = 0 till valid size. Remove below if statement which should result valid number whose sum is multiple of 3:
if (dset.size() == 1) {
if (dset[0] % 3 == 0) { return false; }
}