Power set generated by bits

Question

I have this code which generates power set of an array of size 4 (number is just example, less combinations to write...).

#define ARRAY_SIZE 4


unsigned int i,          


        

Answer 1

Write a function to generate all items of a fixed size. Note that the first such item is 1,..,k while the last is (n-k+1),..,n. This is very simple, you basically have to reimplement basic counting: you increment the last "digit", until you reach n. Then you reset to 1 and continue in the same fashion to its left.

Then just have k run from 1 (0) to n.

Here is one proposal for the internal algorithm. It's rather ugly, though.

void out(std::vector const& item) {
  char del = '{';
  for (auto it = item.begin(); it != item.end(); ++it) {
    std::cout << del << *it;
    del = ',';
  }
  std::cout << "}\n";
}

bool ascending(std::vector const& item) {
  int last = 0;
  for (auto it = item.begin(); it != item.end(); ++it) {
    if (*it <= last)
      return false;
    last = *it;
  }
  return true;
}

void allk(int k, int n) {
  std::vector item;
  item.reserve(k+1);
  for (int i = 1; i <= k; i++)
    item.push_back(i);
  bool valid = true;
  while (valid) {
    out(item);
    do {
      valid = false;
      for (auto it = item.rbegin(); it != item.rend(); ++it) {
        if (*it == n) {
          *it = 1;
        } else {
          ++*it;
          valid = true;
          break;
        }
      }
    } while (valid && !ascending(item));
  }
}