I am using SWI-Prolog and am confused why the option library would be written to give the following outputs:
?- option(a(A), [a=1, a=2, a(3)]).
A = 3.
?- option(b(B), [b=1, b=2]).
B = 1.
I would expect A=1 ... Looking through the option library code though, this result is clearly intended (git link), but why is this not a bug?
option(Opt, Options) :- % make option processing stead-fast
arg(1, Opt, OptVal),
nonvar(OptVal), !,
functor(Opt, OptName, 1),
functor(Gen, OptName, 1),
option(Gen, Options),
Opt = Gen.
option(Opt, Options) :-
get_option(Opt, Options), !.
get_option(Opt, Options) :-
memberchk(Opt, Options), !.
get_option(Opt, Options) :-
functor(Opt, OptName, 1),
arg(1, Opt, OptVal),
memberchk(OptName=OptVal, Options), !.
Since memberchk/2 (which is semi-deterministic, i.e., it succeeds at most once) is used in the code you quote, non-determinism (A=1 ; A = 2 ; etc.) seems explicitly not intended. If anything, contradictory options should maybe raise a domain error, no?
The documentation says the Name = Value syntax is deprecated in favor of the Name(Value) syntax.
It seems reasonable this would be represented in the library(option) code by checking first for the preferred form.
Related
another way to ask the question is:
How I can list all the properties of an atom?
For example:
movie(agora).
director(agora, 'Alejandro Amenabar')
duration(agora, '2h').
so, I will like to receive all the predicates that has agora for argument. In this case it will be: movie, director, duration, with the other parameters ('Alejandro Amenabar', '2h').
I found: this, and this questions, but I couldn't understand well.
I want to have the value of false in the "variable Answer" if PersonInvited doesn't like something about the movie.
My query will be:
answer(Answer, PersonInvited, PersonWhoMadeInvitation, Movie)
Answer: I don't like this director
answer(false, PersonInvited, PersonWhoMadeInvitation, Movie):-
director(Movie, DirectorName),not(like(PersonInvited,DirectorName)).
The same thing will happen with any property like genre, for example.
Answer: I don't like this genre
answer(false, PersonInvited, PersonWhoMadeInvitation, Movie):-
genre(Movie, Genre), not(like(PersonInvited,Genre)).
So, I want to generalize this situation, instead of writing repeatedly every feature of every object.
I found two solutions the 2nd is cleaner from my point of view, but they are different.
Parameters:
PredName: Name of the predicate.
Arity: The Arity of the Predicate.
ParamValue: If I want to filter by one specific parameter.
PosParam: Which is the position of the parameter in the predicate.
ListParam: All the value of the posibles values parameters (mustbe a Variable all the time).
Solution 1:
filter_predicate(PredName, Arity, ParamValue,PosParam, ListParam):-
current_predicate(PredName/Arity),
Arity >= PosParam,
nth(PosParam, ListParam, ParamValue),
append([PredName], ListParam, PredList),
GlobalArity is Arity + 1,
length(PredList, GlobalArity),
Predicate =.. PredList,
Predicate.
Query
filter_predicate(PredName, Arity, agora, 1, Pm).
Output
Arity = 2
Pm = [agora,'Alejandro Amenabar']
PredName = director ?
yes
Solution2:
filter_predicate(PredName, Arity, ParamList):-
current_predicate(PredName/Arity),
append([PredName], ParamList, PredList),
GlobalArity is Arity + 1,
length(PredList, GlobalArity),
Predicate =.. PredList,
Predicate.
Query 1:
filter_predicate(PredName, Arity, [agora, X]).
Output
Arity = 2
PredName = director
X = 'Alejandro Amenabar' ?
Query 2:
filter_predicate(PredName, Arity, [X, 'Alejandro Amenabar']).
Output
Arity = 2
PredName = director
X = agora ?
here is my attempt, using SWI-Prolog
?- current_predicate(so:F/N), N>0, length(As,N), Head =.. [F|As], clause(so:Head,Body), As=[A|_], A==agora.
note that I coded into a module called so the facts, so I qualify with the module name the relevant calls. Such builtins (clause/2 and current_predicate/1) are ISO compliant, while modules (in SWI-prolog) are not. So I'm not sure about portability, etc...
clause/2 it's a builtin that allows for easy writing metainterprets. See the link for an awesome introduction to this Prolog historical 'point of strength'.
The 2 last calls (I mean, As=[A|_], A==agora) avoid matching clauses having a variable as first argument.
Using reading lines into lists with prolog
All your predicates are in a file 'my_file.pl'.
e.g. my_file.pl contains:
movie(agora).
director(agora, 'Alejandro Amenabar').
duration(agora, '2h').
You can use:
getLines(File,L):-
setup_call_cleanup(
open(File, read, In),
readData(In, L),
close(In)
).
readData(In, L):-
read_term(In, H, []),
( H == end_of_file
-> L = []
; L = [H|T],
readData(In,T)
).
pred_arg_file(Pred,Argue,File):-
getLines(File,L),
member(M,L),
M=..List,
member(Argue,List),
List=[Pred|_].
Then you can query:
?-pred_arg_file(Pred,agora,'my_file.pl').
Pred = movie ;
Pred = director ;
Pred = duration ;
false
or
?- findall(Pred,pred_arg_file(Pred,agora,'my_file.pl'),Preds).
Preds = [movie,director,duration].
If you want to return the properties, return the whole List not just the head.
pred_arg_file(List,Argue,File):-
getLines(File,L),
member(M,L),
M=..List,
member(Argue,List).
From my understanding you should change your data representation so that you can query the relations.As other answers have pointed out, So use triples, you can easily write code to change all your relations into this form as a one off. You then need to work out what the best way to store likes or dislikes are. This will effect how negation works. In this example:
relation(starwars,is,movie).
relation(lucas, directs,starwars).
relation(agora, is,movie).
relation('Alejandro Amenabar', directs, agora).
relation(agora, duration, '2h').
like(ma,'Alejandro Amenabar').
like(ma,movie).
like(ma,'2h').
ma_does_not_want_to_go(Film):-
relation(Film,is,movie),
relation(Film,_,Test), \+like(ma,Test).
ma_does_not_want_to_go(Film):-
relation(Film,is,movie),
relation(Test,_,Film), \+like(ma,Test).
ma_wants_to_go(Film):-
relation(Film,is,movie),
\+ma_does_not_want_to_go(Film).
sa_invites_ma(Film,true):-
ma_wants_to_go(Film).
sa_invites_ma(Film,false):-
ma_does_not_want_to_go(Film).
A draft of a solution using Logtalk with GNU Prolog as the backend compiler:
% a movie protocol
:- protocol(movie).
:- public([
director/1,
duration/1,
genre/1
]).
:- end_protocol.
% a real movie
:- object('Agora',
implements(movie)).
director('Alejandro Amenabar').
duration(120).
genre(drama).
:- end_object.
% another real movie
:- object('The Terminator',
implements(movie)).
director('James Cameron').
duration(112).
genre(syfy).
:- end_object.
% a prototype person
:- object(person).
:- public([
likes_director/1,
likes_genre/1
]).
:- public(likes/1).
likes(Movie) :-
conforms_to_protocol(Movie, movie),
( Movie::genre(Genre),
::likes_genre(Genre) ->
true
; Movie::director(Director),
::likes_director(Director) ->
true
; fail
).
:- end_object.
% a real person
:- object(mauricio,
extends(person)).
likes_director('Ridlye Scott').
likes_genre(drama).
likes_genre(syfy).
:- end_object.
Some sample queries:
$ gplgt
...
| ?- {movies}.
...
(5 ms) yes
| ?- mauricio::likes('Agora').
true ?
yes
| ?- mauricio::likes(Movie).
Movie = 'Agora' ? ;
Movie = 'The Terminator' ? ;
no
| ?- 'The Terminator'::director(Director).
Director = 'James Cameron'
yes
The code can be improved in several ways but it should be enough to give you a clear idea to evaluate this solution.
If I understood your question properly I propose the follow:
What if you change your schema or following this idea you can make a method that simulate the same thing.
class(movie, agora).
property(director, agora, 'Alejandro Amenabar').
property(duration, agora, '2h').
If do you want the types of agora, the query will be:
class(Type, agora)
If you want all the properties of agora, that will be:
property( PropertyName, agora, Value).
Given following facts:
route(TubeLine, ListOfStations).
route(green, [a,b,c,d,e,f]).
route(blue, [g,b,c,h,i,j]).
...
I am required to find all the pairs of tube Lines that do not have any stations in common, producing the following:
| ?- disjointed_lines(Ls).
Ls = [(yellow,blue),(yellow,green),(yellow,red),(yellow,silver)] ? ;
no
I came up with the below answer, however it does not only give me incorrect answer, but it also does not apply my X^ condition - i.e. it still prints results per member of Stations lists separately:
disjointed_lines(Ls) :-
route(W, Stations1),
route(Z, Stations2),
setof(
(W,Z),X^
(member(X, Stations1),nonmember(X, Stations2)),
Ls).
This is the output that the definition produces:
| ?- disjointed_lines(L).
L = [(green,green)] ? ;
L = [(green,blue)] ? ;
L = [(green,silver)] ? ;
...
I believe that my logic relating to membership is incorrect, however I cannot figure out what is wrong. Can anyone see where am I failing?
I also read Learn Prolog Now chapter 11 on results gathering as suggested here, however it seems that I am still unable to use the ^ operator correctly. Any help would be appreciated!
UPDATE:
As suggested by user CapelliC, I changed the code into the following:
disjointed_lines(Ls) :-
setof(
(W,Z),(Stations1, Stations2)^
((route(W, Stations1),
route(Z, Stations2),notMembers(Stations1,Stations2))),
Ls).
notMembers([],_).
notMembers([H|T],L):- notMembers(T,L), nonmember(H,L).
The following, however, gives me duplicates of (X,Y) and (Y,X), but the next step will be to remove those in a separate rule. Thank you for the help!
I think you should put route/2 calls inside setof' goal, and express disjointness more clearly, so you can test it separately. About the ^ operator, it requests a variable to be universally quantified in goal scope. Maybe a concise explanation like that found at bagof/3 manual page will help...
disjointed_lines(Ls) :-
setof((W,Z), Stations1^Stations2^(
route(W, Stations1),
route(Z, Stations2),
disjoint(Stations1, Stations2)
), Ls).
disjoint(Stations1, Stations2) :-
... % could be easy as intersection(Stations1, Stations2, [])
% or something more efficient: early fail at first shared 'station'
setof/3 is easier to use if you create an auxiliary predicate that expresses the relationship you are interested in:
disjoint_routes(W, Z) :-
route(W, Stations1),
route(Z, Stations2),
disjoint(Stations1, Stations2).
With this, the definition of disjointed_lines/1 becomes shorter and simpler and no longer needs any ^ operators:
disjointed_lines(Ls) :-
setof((W, Z), disjoint_routes(W, Z), Ls).
The variables you don't want in the result of setof/3 are automatically hidden inside the auxiliary predicate definition.
I have to write a predicate to do work like following:
?- cat(north,south,X).
X = northsouth
?- cat(alley,'91',Y).
X = alley91
?-cat(7,uthah,H).
Bad Input
H = H
Please Help..
atom_concat_redefined(A1, A2, A3) :-
( nonvar(A1) -> atom_chars(A1, Chs1) ; true ),
( nonvar(A2) -> atom_chars(A2, Chs2) ; true ),
( nonvar(A1), nonvar(A2) -> true ; atom_chars(A3, Chs3) ),
append(Chs1, Chs2, Chs3),
atom_chars(A1, Chs1),
atom_chars(A2, Chs2),
atom_chars(A3, Chs3).
This definition produces the same errors in a standard conforming implementation like SICStus or GNU - there should be no other differences, apart from performance. To compare the errors use the goal:
?- catch(atom_concat_redefined(A,B,abc+1), error(E,_), true).
E = type_error(atom,abc+1).
Note the underscore in error(E,_), which hides the implementation defined differences. Implementations provide additional information in this argument, in particular, they would reveal that atom_chars/2 or atom_concat/3 produced the error.
atom_codes/2 it's the ISO approved predicate to convert between an atom and a list of codes. When you have 2 lists corresponding to first two arguments, append/3 (alas, not ISO approved, but AFAIK available in every Prolog), will get the list corresponding to third argument, then, convert that list to atom...
Note that, while append/3 is a 'pure' Prolog predicate, and can work with any instantiation pattern, atom_codes/2 requires at least one of it's argument instantiated. Here is a SWI-Prolog implementation of cat/3, 'working' a bit more generally. I hope it will inspire you to read more about Prolog...
ac(X,Xs) :- when((ground(X);ground(Xs)), atom_codes(X,Xs)).
cat(X,Y,Z) :- maplist(ac, [X,Y,Z],[Xs,Ys,Zs]), append(Xs,Ys,Zs).
edit
as noted by #false I was wrong about append/3. Now I'll try to understand better what append/3 does... wow, a so simple predicate, so behaviour rich!
I would like to determine in Prolog the type of a string of characters, if it is alphabetic, alphanumeric or numeric.
For example:
"I use this page" alphabetic
"0c0d24e" alphanumeric
How can i do?
the predicate available is char_type/2, or better, code_type/2.
To apply to each code in string, use maplist/2. The only problem it's the wrong arguments order of code_type. Then a service predicate is needed (or download lambda, if you're using SWI-Prolog, with ?- pack_install(lambda).).
Without lambda:
code_type_(X,Y) :- code_type(Y,X).
?- maplist(code_type_(alpha), "abc").
true.
With lambda:
?- [library(lambda)].
?- maplist(\C^code_type(C,alpha), "abc").
true.
edit after comments, it's apparent that more flexible parsing is required. A DCG it's the recommended way to go: library(dcg/basics) offers some prebuilt 'categorizer', and highlights the proper way to write your own, combining with code_type: for instance, here is a recently added rule:
%% prolog_var_name(-Name:atom)// is semidet.
%
% Matches a Prolog variable name. Primarily intended to deal with
% quasi quotations that embed Prolog variables.
prolog_var_name(Name) -->
[C0], { code_type(C0, prolog_var_start) }, !,
prolog_id_cont(CL),
{ atom_codes(Name, [C0|CL]) }.
prolog_id_cont([H|T]) -->
[H], { code_type(H, prolog_identifier_continue) }, !,
prolog_id_cont(T).
prolog_id_cont([]) --> "".
see how code_type/2 is used to qualify single characters...
more edit - note: untested
qualify_atom(Atom, Type) :-
atom_codes(Atom, Codes),
qualify_codes(Codes, Type).
qualify_codes(Codes, Type) :-
( maplist(code_type_(alnum), Codes)
-> Type = alnum
; maplist(code_type_(alpha), Codes)
-> Type = alpha
; Type = unknown
).
then, to work on a list
?- maplist(qualify_atom, Atoms, Types).
edit
An update of this answer is mandatory: since library(yall) has been released in SWI-Prolog, and is autoloaded, we can now write:
?- maplist([C]>>code_type(C,alpha), `abc`).
Also, note the change in literal representation: double quotes in SWI-Prolog ver.7+ don't represent anymore a list of character codes.
At some point of my program I have an atom formed by what previously were also atoms, and I want to remove the character spaces within it so that later I can use without any problem:
term_to_atom(Result, 'second2(second2),region(ºMediterranean Sea),months(no_value),third3(third3),recog(ºNew Type),distance(no_value)').
and obtain this
Result = (second2(second2), region(ºMediterraneanSea), months(no_value), third3(third3), recog(ºNewType), distance(no_value))
or also the original would work
Result = (second2(second2), region(ºMediterranean Sea), months(no_value), third3(third3), recog(ºNew Type), distance(no_value))
because if I don't delete those character spaces then term_to_atom will complain about it. How can I solve it?
You can use this procedure:
strip_spaces(S, NoSpaces) :-
atom_codes(S, Cs), delete(Cs, 0' , X), atom_codes(NoSpaces, X).
but delete/3 is deprecated. Another possibility is
strip_spaces(S, NoSpaces) :-
atomic_list_concat(L, ' ', S),
atomic_list_concat(L, NoSpaces).
Either of these will 'eat' each space, but from your problem description, in comments you exchanged with gusbro, this doesn't seems to me the right way to go. Changing the literals seems at DB interface could ask for trouble later.
Parsing your input to a list, instead of a conjunction, can be done with DCGs:
:- [library(http/dcg_basics)].
parse_result(X, R) :-
atom_codes(X, Cs),
phrase(parse_list(R), Cs).
parse_list([T|Ts]) -->
parse_term(T), (",", parse_list(Ts) ; {Ts=[]}).
parse_term(T) -->
string(F), "(", string(Arg), ")",
{atom_codes(Fa,F), atom_codes(Arga,Arg), T =.. [Fa,Arga]}.