Thursday, June 19, 2014



If S is a set, then P(S), the 'powerset' of S is the set of all subsets of S including the empty set and S itself. If S has cardinality N then the cardinality of P(S) is 2N (why?). That is, there are 2N subsets associated with S.

Here's a function to compute P(S) in OCaml. It's an instance of the 'divide and conquer' strategy of problem solving.

  let rec sets l =
    match l with
    | [] -> [[]]
    | x :: xs -> let l = sets xs in 
                   l @ ( (fun y -> x :: y) l)

This program translates to C++ naturally and with brevity thanks to C++ 11 lambdas.

#include <boost/utility.hpp>

#include <set>
#include <iterator>
#include <algorithm>

template <class I, class D>
D sets (I begin, I end, D dst)
  typedef typename std::iterator_traits<I>::value_type value_type;
  typedef std::set<value_type> set_t;

  if (begin == end)
      *dst++ = set_t (); //the empty set
      std::set<set_t> l;
      std::set<set_t>::iterator back=l.end ();
      sets (boost::next (begin), end, std::inserter (l, back));

      std::transform (l.begin (), l.end (), dst, 
        [](set_t const& s) -> set_t const& { return s; });
      std::transform (l.begin (), l.end (), dst, 
        [=](set_t s) -> set_t { s.insert (*begin); return s; });

  return dst;