I would like to write a predicate different_from(Xs,X) which is a check that succeeds if and only if X is different from all the elements of the list Xs.
So the query
different_from([3,2,5],4)
should succeed but this following query should fail:
different_from([3,2,5],2)
Since the predicate is a check it should not instantiate either of its arguments.
thank you in advance for any help you can provide.
This is a very simple rule consisting of two clauses:
The first clause says that different_from is successful when the list is empty,
The second clause says that different_from is successful when the first element of the list does not match the element being searched, and also when different_from for the tail is successful as well.
Here is the same thing written in Prolog syntax:
different_from([], _).
different_from([H|T], E) :- E \= H, different_from(T, E).
Related
I want to write the rule unique(List) that checks if all elements in List are unique. I'm not allowed to use member, but I am allowed to use 'not'. Therefore I wrote the rule 'member' myself.
I wrote this:
member(Element, [Element|_]).
member(Element, [_|List]) :-
member(Element, List).
unique([H|T]) :-
not(in_list(H, T)),
unique(T).
The member-rule is working, it checks if Element is a member of List. But the unique- rule doesn't work. For the unique-rule is was expecting it would check if H was in T and then do the same for the Header of the Tail and so on. The 'not' makes from a false statement a True-output.
When I run this rule with query ?-unique([1,2,3,4,5]) it gives False. So what's my mistake?
I just discoverd the solutions a few moments later.
I added this above:
unique([]).
unique([_,[]]).
I have database like this:
movie(matrix,wachowski,thriller).
movie(terminator, cameron, thriller).
movie(Gladiator, scott, costume).
movie(star wars, lucas, fantasy).
movie(star trek, abrams, fantasy).
And I want to know who direct fantasy film except Abrams.
I suppose I need to use 'not' predicate, but I don't know exactly how it works.
?- movie(X,not(abrams),fantasy).
But unfortunately it doesn't work.
One more query is what kind of films is not a thriller:
?- movie(X,_,not(thriller)).
Still not working.
Next problem is I need to use predicate direct(Director, listsOfMovie) based on bagof.
?- direct(Director, listsOfMovie) :- bagof(Director,movie(Director,listsOfMovie,_), listsOfMovie).
Still without success :(
Anyone can help?
Use of not
You can't use Prolog predicates like functions. not/1 is a predicate which accepts a query as an argument. So this isn't doing what you think:
movie(X,not(abrams),fantasy).
This is querying movie with a second argument of not(abrams). You don't have any facts or predicates that match movie(_, not(_), _) so it will always fail.
If you want to know which films were not thrillers, you might render it:
movie(X, _, Type),
Type \= thriller.`
Using not, it might be:
not( movie(X, _, thriller) ).
If you wanted the syntax of movie(_, not(_), _) to work, you could write a predicate for it:
movie( Name, not(Director), Type ) :-
movie(Name, D, Type),
D \= Director.
Now we have either a fact or a predicate head that matches the form, movie(_, not(_), _), and then the query, movie(X, not(abrams), Y) would work. But it's not normally done this way.
Using bagof/3
Let's look at your use of bagof. In the simplest case, bagof is supposed to take three arguments:
bagof(X, {query involving X}, ListOfSatisfingXs)
So bagof will run the {query involving X} generating each X that makes it true, creating ListOfSatisfingXs, a unique, sorted list of such instantiations of X. In other words, ListOfSatisfingXs is the unique, sorted values of X that make {query involving X} succeed.
In your case, you've gotten the arguments to bagof a bit mixed up:
direct(Director, listsOfMovie) :-
bagof(Director, movie(Director, listsOfMovie, _), listsOfMovie).
Here, you're reusing your Director argument as your bagof argument, which is not good (since it's not intended). Since you're looking for a list of movies, the first argument should represent the movie. Your query to movie is using listsOfMovie, your intended target argument to hold the list result, which it shouldn't. And finally, listsOfMovie is an atom, not a variable, since it doesn't start with a capital letter.
The corrected version would be:
director_movies(Director, ListOfMovies) :-
bagof(Movie, movie(Director, Movie, _), ListOfMovies).
Here, the bagof is getting the *Unique, sorted list of Movie values such that movie(Director, Movie, _) is true and providing that resulting list in ListOfMovies.
Hi I have to solve a problem in Prolog, that sounds like this: deletes all the sublists of a list that are increasing. For example the list [1,[2],[3,4],6] becomes [1,6].
So far I have tried this but it's not working. Any help please ?
domains
el=integer
list=el*
element=integer;list
lista=element*
goal
elim([1,[2],[3],4)],L),
write(L).
predicates
elim(lista,lista)
is_increasing(lista)
is_list(lista)
clauses
is_increasing([A,B|T]) :-
B>A,
is_increasing([B|T]).
is_list([_|_]).
is_list([]).
elim([],[]).
elim([E|Es],[E|Ts]) :-
is_list(E),
is_increasing(E),
elim(Es, Ts).
attempt to modularize your code: first write an is_increasing/1. Since it appears that a list of 1 element is increasing, you can do as simply as
is_increasing([A,B|T]) :- B > A, is_increasing([B|T]).
is_increasing([_]).
then you can use it to discard elements while copying. Beware to check that an element is a list before calling. Here is a possible definition
is_list([_|_]).
is_list([]).
edit
there is a bad declaration, as advised by mbratch
element=i(integer);l(list)
should be
element=integer;list
Also, you forgot is_increasing([_])., and anyway you're not using at all is_list or is_increasing.
The rule eliminating sublists of course should read
elim([E|Es], Ts) :- is_list(E), is_increasing(E), elim(Es, Ts).
just add the base case and a copy. i.e. elim is a 3 clauses predicate...
edit apart the rule above, you need only a base case
elim([],[]).
and a copy
elim([E|Es],[E|Ts]) :- elim(Es, Ts).
just try to understand why the order of rules is also important in Prolog...
I'm trying to write a query that will make sure an element is present in a list of lists, I tried this implementation:
membernested(E,[H|T]):-member(E,H).
membernested(E,[H|T]):-membernested(E,[T]).
but Prolog won't answer the query, any thoughts?
Change you second clause to:
membernested(E,[H|T]) :- membernested(E,T).
The tail of the list [H|T] is T, not [T]. There's no need to enclose it in another list.
I got stuck to implement a logic. At some instance in my program I have a list say named as List.
The length of this List is variable and I don't know in advance. Now I have to pass this list in a functor to create a fact and I am unable to implement it. For eg:
if List is [first] then it should add the fact functor(first).
if List is [first,second] then it should add the fact functor(first,second).
if List is [first,second,third] then it should add the fact functor(first,second,third).
and so on...
I was trying by =.. but here I am unable to map that variable length constraint. For fixed length I am able to perform but I don't know in advance that how many elements will be there in list.
Any suggestions to implement this logic. Thanks.
I don't quite understand your problem with =.. but this worked for me:
assert_list(List) :-
Term =.. [my_functor|List],
assert(Term).
Note that I use my_functor instead of simply functor because functor/3 is a built-in predicate so you cannot assert ternary functor facts (functor(first, second, third)).
Calling it:
?- assert_list([first,second,third]).
true.
Checking that it works:
?- listing(my_functor).
:- dynamic user:my_functor/3.
user:my_functor(first, second, third).
true.
Note that technically, the different n-ary my_functor/n predicates are not the same predicates. You must use different queries in your program for each n. To circumvent this, you could simply assert the list as one and only argument of my_functor:
?- List = [first, second, third],
assert(my_functor(List)).
true.
?- listing(my_functor).
:- dynamic user:my_functor/3.
user:my_functor([first, second, third]).
true.
My SWI-Prolog version is 5.7.5.