The following program noMetagolR is given in:
http://www.doc.ic.ac.uk/~shm/Papers/metagol_gram.pdf page 33.
parse(S,G1,G2) :- parse(s(0),S,[],G1,G2).
parse(Q,X,X,G1,G2) :- abduce(acceptor(Q),G1,G2).
parse(Q,[C|X],Y,G1,G2) :- Skolem(P), abduce(delta1(Q,C,P),G1,G3), parse(P,X,Y,G3,G2).
abduce(X,G,G) :- member(X,G).
abduce(X,G,[X|G]) :- not(member(X,G)).
Skolem(s(0)). Skolem(s(1)). ...
An example query is :
parse([],[],G1), parse([0],G1,G2), parse([0,0],G2,G3), parse([1,1],G3,G4), parse([0,0,0],G4,G5), parse([0,1,1],G5,G6), parse([1,0,1],G6,G),not(parse([1],G,G)), not(parse([0,1],G,G)).
The answer substitutions should return a learnt grammar for parity.
The program is said to run in Yap. I normally use SWI-prolog. Either way,
what do I do to make them understand Skolem/1 ? Presumbly this means that Skolem is a variable? I thought maybe using =.. but this does not work.
Also how many Skolem/1 facts are needed?
in SWI-Prolog, you can place in your source the directive
:- set_prolog_flag(allow_variable_name_as_functor,true).
see current_prolog_flag/2
example on REPL:
1 ?- set_prolog_flag(allow_variable_name_as_functor,true).
true.
2 ?- assert(X(1)).
true.
3 ?- X(Y).
Y = 1.