I am trying to add one item to the end of a list in prolog, but it keeps on failing.
insertAtEnd(X,[ ],[X]).
insertAtEnd(X,[H|T],[H|Z]) :- insertAtEnd(X,T,Z).
letters([a,b,c]).
I do not understand why this below does not work.
insertAtEnd(d,letters(Stored),letters(Stored)).
I am also attempting to store this list in the variable Stored throughout, but I am not sure if the above is correct way to proceed.
you can use append
and put your item as second list
like this:
insertAtEnd(X,Y,Z) :- append(Y,[X],Z).
Prolog implements a relational computation model, and variables can only be instantiated, not assigned. Try
?- letters(Stored),
insertAtEnd(d, Stored, Updated),
write(Updated).
Related
I'm pretty new to prolog and while learning it I stumbled upon a problem that I'm having. I basically have a few database facts. They are:
book(year(1937), title([of,mice,and,men]), rating_out_of_ten(9)).
book(year(2008), title([the,hunger,games]), rating_out_of_ten(8)).
I'm am trying to query the books that have the word "of" in the title. This is what I attempted:
book(year, title, rating_out_of_ten), member(of, title).
It returns false when I do this. Can someone please help. Thanks in advance.
So I'm new to the prolog too but I think I know the solution:
In your query you are actually using the wrong predicate, it should be book, instead of movie
The query should look like this:
book(X,title(Y),Z), member(of,Y)
Arguments have to be capitalized (they have to be variables), however to reach elements of the list you have to use list's name and then a variable, like this:
title(Y)
so you can use Y in the application of the member predicate.
Remember that variable names must start with a capital letter (or underscore). Also, are you really querying on movie when your facts are book?
You could try:
?- book(year(Y), title(T), rating_out_of_ten(R)), member(of, T).
I am very new to Prolog and trying to learn.
For my program, I would like to have the user provide pairs of strings which are "types of".
For example, user provides at command line the strings "john" and "man". These atoms would be made to be equal, i.e. john(man).
At next prompt, then user provides "man" and "tall", again program asserts these are valid, man(tall).
Then the user could query the program and ask "Is john tall?". Or in Prolog: john(tall) becomes true by transitive property.
I have been able to parse the strings from the user's input and assign them to variables Subject and Object.
I tried a clause (where Subject and Object are different strings):
attribute(Subject, Object) :-
assert(term_to_atom(_ , Subject),
term_to_atom(_ , Object)).
I want to assert the facts that Subject and Object are valid pair. If the user asserts it, then they belong to together. How do I force this equality of the pairs?
What's the best way to go about this?
Questions of this sort have been asked a lot recently (I guess your professors all share notes or something) so a browse through recent history might have been productive for you. This one comes to mind, for instance.
Your code is pretty wide of the mark. This is what you're trying to do:
attribute(Subject, Object) :-
Fact =.. [Object, Subject],
assertz(Fact).
Using it works like this:
?- attribute(man, tall).
true.
?- tall(X).
X = man.
So, here's what you should notice about this code:
We're using =../2, the "univ" operator, to build structures from lists. This is the only way to create a fact from some atoms.
I've swapped subject and object, because doing it the other way is almost certainly not what you want.
The predicate you want is assertz/1 or asserta/1, not assert/2. The a and z on the end just tells Prolog whether you want the fact at the beginning or end of the database.
Based on looking at your code, I think you have a lot of baggage you need to shed to become productive with Prolog.
Prolog predicates do not return values. So assert(term_to_atom(... wasn't even on the right track, because you seemed to think that term_to_atom would "return" a value and it would get substituted into the assert call like in a functional or imperative language. Prolog just plain works completely differently from that.
I'm not sure why you have an empty variable in your term_to_atom predicates. I think you did that to satisfy the predicate's arity, but this predicate is pretty useless unless you have one ground term and one variable.
There is an assert/2, but it doesn't do what you want. It should be clear why assert normally only takes one argument.
Prolog facts should look like property(subject...). It is not easy to construct facts and then query them, which is what you'd have to do using man(tall). What you want to say is that there is a property, being tall, and man satisfies it.
I would strongly recommend you back up and go through some basic Prolog tutorials at this point. If you try to press forward you're only going to get more lost.
Edit: In response to your comment, I'm not sure how general you want to go. In the basic case where you're dealing with a 4-item list with [is,a] in the middle, this is sufficient:
build_fact([Subject,is,a,Object], is_a(Subject, Object)).
If you want to isolate the first and last and create the fact, you have to use univ again:
build_fact([Subject|Rest], Fact) :-
append(PredicateAtoms, [Object], Rest),
atomic_list_concat(PredicateAtoms, '_', Predicate),
Fact =.. [Predicate, Subject, Object].
Not sure if you want to live with the articles ("a", "the") that will wind up on the end though:
?- build_fact([john,could,be,a,man], Fact).
Fact = could_be_a(john, man)
Don't do variable fact heads. Prolog works best when the set of term names is fixed. Instead, make a generic place for storing properties using predefined, static term name, e.g.:
is_a(john, man).
property(man, tall).
property(john, thin).
(think SQL tables in a normal form). Then you can use simple assertz/1 to update the database:
add_property(X, Y) :- assertz(property(X, Y)).
i know there is a build-in function findall/3 in prolog,
and im trying to find the total numbers of hours(Thrs) and store them in a list, then sum the list up. but it doesnt work for me. here is my code:
totalLecHrs(LN,THrs) :-
lecturer(LN,LId),
findall(Thrs, lectureSegmentHrs(CC,LId,B,E,THrs),L),
sumList(L,Thrs).
could you tell me what's wrong with it? thanks a lot.
You need to use a "dummy" variable for Hours in the findall/3 subgoal. What you wrote uses THrs both as the return value for sumList/2 and as the variable to be listed in L by findall/3. Use X as the first argument of findall and in the corresponding subgoal lectureSegmentHrs/5 as the last argument.
It looks like the problem is that you're using the same variable (Thrs) twice for different things. However it's hard to tell as you've also used different capitalisation in different places. Change the findall line so that the initial variable has the same capitalisation in the lectureSegmentHrs call. Then use a different variable completely to get the final output value (ie the one that appears in sumList and in the return slot of the entire predicate).
You need to use a different variable because Prolog does not support variable reassignment. In a logical language, the notion of reassigning a variable is inherently impossible. Something like the following may seem sensible...
...
X = 10,
X = 11,
...
But you have to remember that , in Prolog is the conjunction operator. You're effectively telling Prolog to find a solution to your problem where X is both 10 and 11 at the same time. So it's obviously going to tell you that that can't be done.
Instead you have to just make up new variable names as you go along. Sometimes this does get a bit annoying but it's just goes with the territory of a logical languages.
I have these two lists =
fruits([banana, apple, mangoes, pears]).
foodILike([hamburgers, banana, shakes, fries]).
I want to write a prolog predicate that will return true as soon as it sees 1 items in the foodsILike list in the fruits list. How can I go about doing so?
First, for the plain answer:
fruitsILike(F) :-
fruits(Fs)
member(F, Fs),
foodILike(Ls),
member(F, Ls).
You could avoid the membership check by flattening the fruits and foods lists:
fruit(banana).
fruit(apple).
...
foodILike(hamburger).
foodILike(banana).
...
fruitsILike(F) :-
fruit(F),
foodILike(F).
That said, you seem to try and solve problems in Prolog using imperative idioms, and that won't work. First, predicates do not return anything. When calling a predicate, Prolog unifies its arguments with valid values according to the facts and rules in the program. Therefore, the "returned value" are the assignments to unbound variables. Second, Prolog does not do something "as soon as". It iterates over all possible solutions. You get the first solution, then the second solution, and so on.
member can 1) individually generate all the members of a given list and/or 2) give a yes/no answer as to whether a particular element is in a particular list. I believe you want to use the first form on fruits to generate each of the elements of fruit, and the second form on foodILike to see if any of those is present.
I'm working on a college assignment where I must verify if a certain clause (as a fact or as a rule) exists in the current clause database.
The idea is to use a rule whose head is verify(+name, +arguments). This rule should be true if in the database exists another rule whose head is name(arguments)
Any help would be greatly appreciated...
Using call/1 is not a good idea because call/1 actually calls the goal, but you just want to find out if the fact/rule exists, and you don't want to wait after a long calculation that the call might trigger, and you don't want to have something printed on the screen if the called rule in turn calls e.g. writeln/1. In addition, you would want verify/2 to succeed even if the call failed (but the fact/rule is otherwise there).
As a solution, SWI-Prolog offers callable/1
callable(+Term)
True if Term is bound to an atom or a compound term,
so it can be handed without type-error to call/1, functor/3 and =../2.
Here are two version of verify/2, one using call/1 and the other using callable/1.
verify1(Name, Arguments) :-
Term =.. [Name | Arguments],
call(Term).
verify2(Name, Arguments) :-
Term =.. [Name | Arguments],
callable(Term).
father(abraham, isaac) :-
writeln('hello').
father(abraham, adam) :-
fail.
Are you familiar with the concept of unification? What you have to do is: just call a predicate that looks like the one you're trying to find.
So, say in your database is:
father(abraham,isaac).
Now you want to call something like:
verify(father,[abraham,isaac]).
Your predicate body will then have to contain a mechanism of calling father(abraham,isaac). which should then return true. Calling father(abraham,adam) should fail.
You will need two predicates for this: =../2 and call/2. If you are using SWI-Prolog, call help(=..). and help(call) from the interpreter's command line to access the documentation.
I hope I didn't spoil the assignment for you. You still have to find out what to do with partially instantiated predicates (so, say something like verify(father,[abraham,X]). on your own, but it shouldn't be hard from here.
Good luck.