Nobody cares to answer my very important long question so I keep it short.
% https://stackoverflow.com/questions/29230608/swi-prolog-print-multiple-variables
pretty_row(Ls) :- maplist(format, Ls).
uci(w-k, 'K').
uci(b-q, 'q').
ls([w-k, b-q]).
?- ls(L), pretty_row(L).
% Expected output
Kq
% Prolog
pretty_row(Ls) :- maplist(format, Ls).
uci(w-k, 'K').
uci(b-q, 'q').
ls([w-k, b-q]).
ucis([], []).
ucis([P|Rest], [Z|ZRest]) :- uci(P, Z), ucis(Rest, ZRest).
I am not sure if there is a built in predicate that does what ucis do.
Related
ass(a).
ass(b).
ass(c).
con(c,r).
arg(A, L) :- forall(member(S, L), (ass(S), \+ con(S,A))).
If I run arg(r, [a,b]) it will work but if I run arg(r,X) it returns: ERROR out of global stack. I would like it to return [a,b]. I understand this is because L is unbounded, but how can I fix this.
In the predicate:
arg(A, L) :- forall(member(S, L), (ass(S), \+ con(S,A))).
May have a limitation in your case as described in the SWI Prolog documentation for forall/2:
If your intent is to create variable bindings, the forall/2 control
structure is inadequate. Possibly you are looking for maplist/2,
findall/3 or foreach/2.
So in this case, you may be better off with:
arg(A, L) :- findall(S, (ass(S), \+ con(S,A)), L).
Which will yield:
?- arg(r, X).
X = [a, b].
?- arg(r, [a,b]).
true.
?-
Let's say there is a simple database of people in Prolog
person(john).
person(mary).
person(john).
person(susan).
I need to match the entires exactly once:
john-mary, john-john, john-susan, mary-john, mary-susan, john-susan
I tried coming up with something like this:
match:- person(X),!,person(Y), write(X),write(-), write(Y),nl.
run:- person(X), match(X), fail.
But it's matching many times, and matches a person to him/herself, which shouldn't be.
Basically, what I need is to iterate over all Xs and make Prolog to look strictly "below" for Ys.
A quick solution would be to number your people:
person(1, john).
person(2, mary).
person(3, john).
person(4, susan).
Then you could match people like this:
match(X-Y) :-
person(I, X), person(J, Y), I < J.
Since you have two john entries, I'm not sure any other solution is going to work. Normally you could fake an ordering using #>/2 but that would require your atoms to be unique, and since they aren't, it would prevent the john-john solution.
Edit: Since we're willing to use findall/3 to materialize the database of people, we can treat this as a list problem and find a functional solution. Let's get all the combinations in a list:
combinations([X|Rest], X, Y) :- member(Y, Rest).
combinations([_|Rest], X, Y) :- combinations(Rest, X, Y).
With this predicate in hand, we can find the solution:
combined_folks(People) :-
findall(P, person(P), Persons),
findall(X-Y, combinations(Persons, X, Y), People).
?- combined_folks(X).
X = [john-mary, john-john, john-susan, mary-john, mary-susan, john-susan].
That actually turned out to be pretty clean!
person(john).
person(mary).
person(john).
person(susan).
match :- findall(P,person(P),People), match_all(People).
match_all([_]) :- !.
match_all([P|People]) :- match_2(P,People), match_all(People).
match_2(_,[]) :- !.
match_2(P1,[P2|People]) :- format('~a-~a~n',[P1,P2]), match_2(P1,People).
?- match.
I'm new to prolog and want to save all queries in a file instead of typing them by hand.
I have these facts in facts.pl:
likes(wallace, cheese).
likes(grommit, cheese).
likes(wendolene, sheep).
friend(X, Y) :- \+(X = Y), likes(X, Z), likes(Y, Z).
After reading the answer of this question,
I come up with the following code queries.pl:
main :-
write(likes(wallace, cheese)),
halt.
:- initialization(['facts.pl']).
:- initialization(main).
Here I want to examine if likes(wallace, cheese) holds,
what I expected is outputing something like yes or no but the actual output is likes(wallace, cheese)
I've googled a lot and attempted
X = likes(wallace, cheese), write(X).
X is likes(wallace, cheese), write(X).
X := likes(wallace, cheese), write(X).
but none of them works.
It might be a really easy question for you, but I have no idea about how to get things right.
BTW, I'm using GNU Prolog 1.4.1
I think you need a way to 'tag' each query: here a simple way
query(likes(wallace, cheese)).
query(likes(mickey, whisky)).
% service predicates, check the library and use that if available
forall(X,Y) :- \+ (X, \+ Y).
writeln(T) :- write(T), nl.
main :-
forall(query(Q), (Q -> writeln(yes:Q) ; writeln(no:Q))),
halt.
I have a problem with predicate which works in that way that it takes list of atoms:
nopolfont([to,jest,tekśćik,'!'],L).
and in result
L = [to,jest,tekscik,'!'].
I have problem with make_swap and swap predicates. So far I have:
k(ś,s).
k(ą,a).
% etc.
swap(X,W) :- name(X,P), k(P,Y), !, name(Y,W).
swap(X,X).
make_swap(A,W)
:- atom(A),!,
name(A,L),
swap(L,NL),
name(W,NL).
nopolfont([],[]).
nopolfont([H|T],[NH|S]) :- make_swap(H,NH), nopolfont(T,S).
Is there any elegant way to do this?
This is also quite elegant:
polish_char_replacer(X, Y) :-
k(X, Y),
!.
polish_char_replacer(X, X).
nopolfont(Atoms1, Atoms2) :-
maplist(replace(polish_char_replacer), Atoms1, Atoms2).
replace(Goal, Atom1, Atom2) :-
atom_chars(Atom1, Chars1),
maplist(Goal, Chars1, Chars2),
atom_chars(Atom2, Chars2).
Probably as elegant as it can get:
k(ś,s).
k(ą,a).
swap(X,W) :- name(P,[X]), k(P,Y), !, name(Y,[W]).
swap(X,X).
list_swap([], []).
list_swap([H|T], [W|S]) :-
swap(H, W),
list_swap(T, S).
atom_swap(A,W) :-
atom(A), !,
name(A, L),
list_swap(L,S),
name(W, S).
nopolfont([],[]).
nopolfont([H|T],[NH|S]) :-
atom_swap(H,NH),
nopolfont(T,S).
Also, obviously define this, to get the expected result, but I assume this is in the % etc
k(ć, c).
I'm interested in formulae made up from lots of conjunctions (part of a larger problem). I want to write a program that takes something like this:
:- get_params(conj(conj(a,b),c),X)
and returns a list of all the parameters of the conjunctions i.e. X=[a,b,c]. At the moment I can do
:- get_params(conj(a,b),X) to get X=[a,b]
using simple Prolog pattern matching but how would you go about doing things such as
:- get_params(conj(conj(a,b),c),X) to get X=[a,b,c]
It seems really simple but I've been struggling all day!
Since you are describing a list, consider using DCG notation:
params(conj(A,B)) --> !, params(A), params(B).
params(X) --> [X].
Example:
?- phrase(params(conj(conj(a,b),c)), Ps).
Ps = [a, b, c].
Assuming that all conj functors are binary:
get_params(X, Y, L) :-
get_params(X, L1),
get_params(Y, L2),
append(L1, L2, L).
get_params(conj(X, Y), L) :-
get_params(X, Y, L), !.
get_params(A, [A]).