Subsets in Prolog

Solution 1:

Here goes an implementation:

subset([], []).
subset([E|Tail], [E|NTail]):-
  subset(Tail, NTail).
subset([_|Tail], NTail):-
  subset(Tail, NTail).

It will generate all the subsets, though not in the order shown on your example.

As per commenter request here goes an explanation:

The first clause is the base case. It states that the empty list is a subset of the empty list.

The second and third clauses deal with recursion. The second clause states that if two lists have the same Head and the tail of the right list is a subset of the tail of the left list, then the right list is a subset of the left list.

The third clause states that if we skip the head of the left list, and the right list is a subset of the tail of the left list, then the right list is a subset of the left list.

The procedure shown above generates ordered sets. For unordered sets you might use permutation/3:

unordered_subset(Set, SubSet):-
  length(Set, LSet),
  between(0,LSet, LSubSet),
  length(NSubSet, LSubSet),
  permutation(SubSet, NSubSet),
  subset(Set, NSubSet).

Solution 2:

On http://www.probp.com/publib/listut.html you will find an implementation of a predicate called subseq0 that does what you want to:

subseq0(List, List).
subseq0(List, Rest) :-
   subseq1(List, Rest).

subseq1([_|Tail], Rest) :-
   subseq0(Tail, Rest).
subseq1([Head|Tail], [Head|Rest]) :-
   subseq1(Tail, Rest).

A short explanation: subseq0(X, Y) checks whether Y is a subset subsequence of X, while subseq1(X, Y) checks whether Y is a proper subset subsequence of X.

Since the default representation of a set is a list with unique elements, you can use it to get all subsets as in the following example:

?- subseq0([1,2,3], X).
X = [1, 2, 3] ;
X = [2, 3] ;
X = [3] ;
X = [] ;
X = [2] ;
X = [1, 3] ;
X = [1] ;
X = [1, 2] ;
false.