Learn Prolog Now! DCG Practice Example

Answer 1

How about something like:

n(L,N) --> n(L,N,0).

n(_,N,N) --> [], !.
n(L,N,K) --> L, {K1 is K + 1}, n(L, N, K1).

abbccd(N,M) -->
    {M1 is 2*M},
    n("a",N),
    n("b",M1),
    n("c",M1),
    n("d",N).

gen :-
    forall((
           between(1,4,N),
        between(1,4,M),
        phrase(abbccd(N,M),S),
        string_to_atom(S,A)
           ),
           writeln(A)).

executing:

 ?- gen.
abbccd
abbbbccccd
abbbbbbccccccd
abbbbbbbbccccccccd
aabbccdd
aabbbbccccdd
aabbbbbbccccccdd
aabbbbbbbbccccccccdd
aaabbccddd
aaabbbbccccddd
aaabbbbbbccccccddd
aaabbbbbbbbccccccccddd
aaaabbccdddd
aaaabbbbccccdddd
aaaabbbbbbccccccdddd
aaaabbbbbbbbccccccccdddd
true.

Answer 2

I believe I figured it out...

s --> x.
s --> a,d.
s --> a,s,d.

x --> [].
x --> b,b,x,c,c.

a --> [a].
b --> [b].
c --> [c].
d --> [d].

?- s([],[]).
Yes

?- s([a,b,c,c,d],[]).
No

?- s([a,a,a,b,b,c,c,d,d,d],[]).
Yes

It's amusing to look at the solution and think, "I racked my brain over that?" But I guess that's half the fun of learning something new, especially when it's something like logic programing coming from an imperative programming background.

Answer 3

@Timothy, your answer works, but it generates duplicates:

?- length(S,_), s(S,[]).
S = [] ;
S = [a, d] ;
S = [a, d] ;            % XXX
S = [b, b, c, c] ;
S = [a, a, d, d] ;
S = [a, a, d, d] ;      % XXX

This can be fixed by removing one clause, leaving the DCG:

s --> x.
s --> a,s,d.

x --> [].
x --> b,b,x,c,c.

% a, b, c, d the same

This generates:

?- length(S,_), s(S,[]).
S = [] ;
S = [a, d] ;
S = [b, b, c, c] ;
S = [a, a, d, d] ;
S = [a, b, b, c, c, d] ;
S = [a, a, a, d, d, d] ;
S = [b, b, b, b, c, c, c, c] ;
S = [a, a, b, b, c, c, d, d] ;
S = [a, a, a, a, d, d, d, d] ;