First of all, if the answer to my question is here or here
I couldn't find it so please don't kill me.
I want to write a Prolog predicate which returns a list of this form:
(list[elem, elem2], list[elem3, elem4], list[elem5, elem6]).
now I can think of several ways to return a list of this form:
([elem, elem2], [elem3, elem4],[elem5, elem6]).
but how do I make the word "list" appear there as well? what is it even? a fact? another predicate?
Thanks in advance!
You can create a compound term using the standard =../3 built-in predicate. For example:
| ?- Term =.. [list, [1,2,3]].
Term = list([1,2,3])
yes
But note that the syntax that you're trying to use, list[elem5, elem6], is not valid. Are you trying to mimic an array representation? If so, maybe use instead list(elem5, elem6)? For example:
| ?- Term =.. [list, elem5, elem6].
Term = list(elem5, elem6)
yes
Related
let say i have the following facts :
book(65).
own(named('Peter'), 65).
now got the query as a list of clauses :
[what(A), own(named('Peter'), A)]
or
[who(X), book(A), own(X, A)] .
how do I make a rule that accept this list and return the result. Keep in mind that the question could be Why,When,Who...
I went the usual way :
query_lst([]).
%% query_lst([what(Q)|T], Q) :- query_lst(T).
query_lst([H|T]) :- write('?- '),writeln(H),
call(H), query_lst(T).
but this does not allow binding of Q in wh(Q) to the answer which could be in any of the facts that are called by call()
Additional complication I did not forsee is that the query :
(what(A), own(named('Peter'), A).
would fail, because there is no what(X), fact in the DB.
I have to just bind somehow the variable A /that is in what()/ to query_lst(Goals,A) and of course remove what(X) from the list /which i can do with select/3 /
any idea how to bind list-Wh-var to query_lst result ?
my current solution (assumes Q is first element):
query_lst([G|Gs],Res) :- G =.. [Q,Res], member(Q,[what,why,who,when]), lst2conj(Gs,Conj), call(Conj).
Simply convert the list of goals into a conjunction and call it:
list_to_conjunction([], true).
list_to_conjunction([Goal| Goals], Conjunction) :-
list_to_conjunction(Goals, Goal, Conjunction).
list_to_conjunction([], Conjunction, Conjunction).
list_to_conjunction([Next| Goals], Goal, (Goal,Conjunction)) :-
list_to_conjunction(Goals, Next, Conjunction).
Then:
query_list(Goals) :-
list_to_conjunction(Goals, Conjunction),
call(Conjunction).
You got an answer, but it was an answer to your question, not to what you really wanted. Also, you edited your question after you accepted that answer, which isn't very helpful. Typically it's better to open a new question when you have... a new question.
Here is an answer to what you seem to want, which is not exactly what you asked. You have lists of the form [WhPart | Rest] where the WhPart is a wh-word with a variable, and the Rest is a list of goals. You want to execute these goals and get the variable in the wh-term bound.
The good news is that, since the variable in the wh-word also occurs in the goals, it will be bound if you execute them. No extra work is needed. Executing the goals is enough. If the wh-part is really at the start of the list, you can do the whole thing like this:
query([_WhPart | Body]) :-
call_body(Body).
call_body([]).
call_body([Goal | Goals]) :-
call(Goal),
call_body(Goals).
For example:
?- query([who(X), book(A), own(X, A)]).
X = named('Peter'),
A = 65.
?- query([what(A), own(named('Peter'), A)]).
A = 65.
As you can see, there is no need to convert the query to a conjunctive goal: Executing the queries in sequence is exactly the same as executing their conjunction.
Also, it doesn't actually matter which wh-word is used; the only thing that really matters is the variable contained within the term. For this reason the above version does no checking at all, and the _WhPart could be anything. If you want to check that it is a valid term, you can do the following:
query([WhPart | Body]) :-
wh(WhPart),
call_body(Body).
wh(who(_X)).
wh(what(_X)).
wh(when(_X)).
This buys you some "type checking":
?- query([foo(A), own(named('Peter'), A)]).
false.
But not a lot, since you don't know if the wh-word actually fits what is being asked:
?- query([when(A), own(named('Peter'), A)]).
A = 65.
How can I make prolog answer No , if search_answer didn't find an answer , and
Yes with L = [Answer]
search_answer : predicate that returns a list or let's A as a free variable.
found_list(L) :-
search_answer(L).
For example , when asked found_list(L) , although search_answer didn't find an answer , Prolog still answers Yes. I print L , and it is equal to _496 , a free variable.
Given the above piece of code, found_list answers L = [...] if search_answer found a list , else returns L = L, while I want to answer no
I tried the following , but doesn't work
found_list(L) :-
search_answer(L) , is_list(L).
liar_detector is my search_answer predicate , with L = answer
and liars is my found_answer
In found_list(L) :- search_answer(A). both L and A are singleton. I assume you saw the warning. You need to fix that for this predicate to make sense. Do you really want L = [Answer] or L = Answer?
You can achieve what you're after this way:
found_list(A) :-
search_answer(A).
This will fail (result in "no") if search_answer(A) doesn't succeed, and your result will be A if it does succeed.
If you want the result as an answer within a list, you can do this:
found_list([A]) :-
search_answer(A).
I'm not sure what the value is of either of these. The first found_list/1 is just a simple wrapper on search_answer/1 without any additional logic. The second simply makes a single-element list out of the search_answer/1 result. If A is already a list, you don't need to put it inside of brackets ([...]), otherwise you just get a list within a list. I suspect you are really trying to do something else but haven't explained.
In response to the updated question, the following code should work if A is a simple unbound term:
found_list(A) :-
search_answer(A),
is_list(A).
However, is_list/1 will succeed if its argument has a list structure even though it's elements may be unbound:
| ?- X = [_], is_list(X).
X = [_]
yes
| ?-
So, for example, if search_answer(A) succeeds with A = [_], then found_list(A) will suceed with A = [_].
ground/1 can be useful here since:
| ?- ground(X).
no
| ?- ground([_|_]).
no
| ?- ground([a,b]).
yes
| ?-
Thus, the following solution should work:
found_list(A) :-
search_answer(A),
ground(A).
If your intention is not to backtrack to search_answer(A) if A is not ground, but just fail, you could implement found_list/1 as:
found_list(A) :-
search_answer(A),
( ground(A) -> true ; !, false ).
I think, though, there may be a more fundamental issue with the code, as it shouldn't have a behavior that you feel compelled to work around like this.
I googled this but cant find the answer, so here you go:
I have this function in prolog:
ing(Lis) :- findall(I,( recipe2(_,ingredients(I,_)) ),Lis).
This function search and returns me a list of lists like this:
L = [['wheat flour', egg, salt], ['wheat flour', cheese, olives, tomato, salt, basil], ['wheat flour', potatoes, salt], [milk, egg, sugar]].
I want to unify that list of lists in only one list, so i can get out duplicates. I know i have to use recursion, but thats all i know.
Thanks in advance.
You may simply modify the predicate like such:
ing(Lis) :-
setof(E, X^Y^I^( recipe2(X, ingredients(I,Y)), member(E, I) ), Lis).
member/2 is a built-in predicate that unifies the first argument with an element of a list in the second argument. It is non-deterministic.
The use of X^Y^I^ are existential quantifiers to ensure that you only get your results in one solution. It essentially says,
There exists an X, Y, and I for any element E that is a part of an
ingredient list, (I).
Using setof/3 also ensures that any solution you get will be a collection of unique elements.
Documentation (SWI-Prolog) for member/2 and setof/3
I am new on prolog. Can you help me to solve this problem, please.
exam(math, paul).
exam(phys, paul).
exam(cmpe, sofia).
exam(bio, george).
I want to implement the predicate otherExam(L, N). L is a lesson, and N is the list of all lessons (except L) haved by student of L.
otherExam(math,X). returns [phys]
otherExam(cmpe,X). returns []
otherExam(chem,X). returns false (no such lesson)
otherExam(math,[phys]) returns true
otherExam(X,[phys]). returns math
I haved stucked with this problem. If you help me, I will be very happy :))
check findall/3 and select/3
findall/3 is used to get all the possible results from a query
for example, findall(X,ancestor(X,paul),L) will find all X such as X is an ancestor of paul and will put them in the list L
I'm new to prolog and I just can't figure this out.
I'm trying to build a simple program that receives a list of predicates, searches for a specific predicate in the list, and applies a function to that predicate's parameters.
Something along these lines:
?- program([pred1(a,b,p), pred2(d,b,p), pred2 (a,c,p)]).
program (list1) :-
search(pred2(X,Y,p),list1).
doSomething (X,Y) % with the X and Y returned from search function, both of them.
Basically, I want to use all values that would return from an objective search(pred2(X,Y,p),list1) and use them on another function.
Okay, I tried some stuff in prolog and came to this:
member(X, [X | _]).
member(X, [_ | R]) :- member(X, R).
prog(L1,Out) :- member(pred2(X,Y), L1).
?- prog ([(pred1(a,b),pred2(c,b),pred2(d,a)],Out).
It gives true 2 times as it is supposed to, but I wanted to get Out = [c,b] and Out = [d,a]. How I can achieve this?
Regarding Oak's answer: I get that it isn't a procedural language but I can't figure out how to access values and use them in prolog. Your example wasn't that helpful.
For starters, I would avoiding calling these things "functions". Prolog is not a procedural language, and rules / predicates are not functions.
Basically, when you use a rule you're really asking Prolog, "give me all the values which will satisfy this rule". The rule, by itself, does not return anything.
So say you had the following in a procedural language:
f(g(3))
How would you do it in Prolog? You would need to write some predicate f(X,Y) and some predicate g(X,Y), and then you would need to use the query f(3,Y), g(Y,Z) - which means to ask Prolog to find you values for Y and Z which will satisfy this. Z is what you're interested in.
the best way to approach these filter & project requirements in prolog in my opinion is to write your filter expression such that it takes one argument and succeeds if the input argument passes the filter -
iseven(Num) :- 0 is Num % 2 .
Then write the the projection code as taking one argument that is the input, and one that is the output -
triple(NumIn, NumOut) :- NumOut is NumIn * 3 .
Then tie them together -
triple_evens(NumIn, NumOut) :- iseven(NumIn), triple(NumIn, NumOut).
Then to run this on every member of a list, we should use findall -
triple_evens_in_list(Lin, Lout) :-
findall(Num, ( member(NumL, Lin),
triple_evens(NumL, Num)
), LOut).
This could be generalized to take as arguments the name of the filter & map predicates of course. And it could be compressed down to one stmt too in the form -
findall(Num, ( member(M, List), 0 is M % 2, Num is M * 3 ), ListOut).