For finding all solutions to a goal, look at bagof or findall. Or am I missing something?
Like this:
?- findall(Moves, plan(State, Goal, _, Moves), AllMoves).
The whole idea of these predicates is that you say which arguments you want to collect and get a list of all possible instantiations under that predicate. In this sense you normally have a "return" value (an argument that gets instantiated with the result) that you can then look at or print, instead of printing it explicitly in the predicate that finds solutions.
A simplistic example:
foo(1). foo(2). foo(3). foo(4). foo(5). foo(6).
bar(R) :- foo(A), A mod 2 =:= 0.
findall(R, bar(R), Even).
Now to recursion: how does it work? You cannot share variables between different clauses of the same predicate. For example, this is wrong:
baz(0, B).
baz(X, B) :- X0 is X - 1, B1 is B + 1, baz(X0, B1).
because B is a singleton variable in the first clause of baz
. Instead, you can do:
baz(0, B, B).
baz(X, B, Result) :- X0 is X - 1, B1 is B + 1, baz(X0, B1, Result).
which you can now call:
?- baz(10, 2, Result).
Result = 12
but you will still run into problems after the first answer.
You get the single correct plan probably because the first clause of plan
does not meet the requirements of subsetB
, and you get to the second clause. There, you make a Moves that has a free variable at its Tail, but this is not a problem yet. The problem is, however, that when you find your first solution (all in the second plan
clause, recursively), Moves is now bound to a list of actions, and instead of starting to look for a new solution, you get into the second clause again by backtracking, with the already filled in Moves, which probably messes up the rest of the algorithm.
To make it correct, you probably need to make sure that when your plan
backtracks, it starts to look for a new solution, with a clean Moves. You can start by instantiating Moves to an empty list and collecting results in an accumulator, as shown in the simplistic baz
predicate above.