Earlier quoted context omitted.
Ohh you're teasing me.
Here's the tease: https://www.youtube.com/watch?v=zMfdef-nYGY
Thanks a ton. IIRC GA aren't that expensive. . might be fun to spend some money on this.
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For example, consider the use case of Rosetta that is shown on the page. In Prolog, we can express it as follows: First, let us define suitable operators, so that we do not need so many parentheses:
:- op(200, fy, ∧).
:- op(200, fy, ∨).
:- op(200, fy, ¬).
With these definitions in place, we can implement the SAT solver that is shown on the page as follows: sat(F) :- interpret(F, t).
interpret(∧Ls0, T) :-
maplist(interpret, Ls0, Ls),
apply(and, Ls, T).
interpret(∨Ls0, T) :-
maplist(interpret, Ls0, Ls),
apply(or, Ls, T).
interpret(¬Expr, T) :-
interpret(Expr, T0),
not(T0, T).
interpret(t, t).
interpret(f, f).
interpret(v(f), f).
interpret(v(t), t).
apply(and, Ls, T) :- foldl(and, Ls, t, T).
apply(or, Ls, T) :- foldl(or, Ls, f, T).
and(t, t, t). and(t, f, f). and(f, t, f). and(f, f, f).
or(t, t, t). or(t, f, t). or(f, t, t). or(f, f, f).
not(f, t). not(t, f).
Thus, we can post: ?- sat(∧[v(R),v(O),∨[v(S),v(E),¬t], v(T), ¬v(E)]).
R = O, O = S, S = T, T = t,
E = f ;
false.
On backtracking, all satisfying assignments of variables to truth values are reported. In this concrete case, there is only a single solution.Interestingly, the Prolog version can not only be used to find satisfying assignments, but also to generate whole formulas that evaluate to true:
?- sat(F).
F = ∧[] ;
F = ∧[∧[]] ;
F = ∧[∧[], ∧[]] ;
F = ∧[∧[], ∧[], ∧[]] .
However, this particular enumeration is called unfair, since there are true formulas that will never be generated.