I have this Prolog code that returns: [[vincent,vincent],[vincent,marcellus],[marcellus,vincent],[marcellus,marcellus],[pumpkin,pumpkin],[honey_bunny,honey_bunny]].
:- initialization main.
loves(vincent, mia).
loves(marcellus, mia).
loves(pumpkin, honey_bunny).
loves(honey_bunny, pumpkin).
jealous(X, Y) :-
loves(X, Z),
loves(Y, Z).
main :-
findall([X, Y], jealous(X, Y), L),
write(L),
halt.
How to get the only results when X != Y?
I tried the following code to get the same results as before.
jealous(X, Y) :-
X \== Y,
loves(X, Z),
loves(Y, Z).
With \=, I got [].
How to get only [vincent,marcellus] as a result?
The order of the goals in your attempted solution is wrong. When called with two distinct variables, the (\==)/2 standard predicate always succeed. The solution is to call the predicate only when its arguments are instantiated:
jealous(X, Y) :-
loves(X, Z),
loves(Y, Z),
X \== Y.
With this fix, your query now returns:
?- findall([X, Y], jealous(X, Y), L).
L = [[vincent, marcellus], [marcellus, vincent]].
So, no one is jealous of himself anymore. But you still get a redundant solution. We can modify the jealous/2 predicate to sort the names in the returned solutions. For example:
jealous(X, Y) :-
loves(X0, Z),
loves(Y0, Z),
X0 \== Y0,
( X0 #< Y0 ->
X = X0, Y = Y0
; X = Y0, Y = X0
).
Now, by using setof/3 instead of findall/3, we get:
?- setof([X, Y], jealous(X, Y), L).
L = [[marcellus, vincent]].
One final observation. A list is a poor solution for representing a pair. The traditional way is to use either X-Y or (X, Y).
Whenever possible, use dif/2 instead of (\==)/2.
dif/2 will help you write logically sound programs.
For details, look at prolog-dif!
Related
I am new to prolog and currently stuck trying to understand how to implement this.
I need a predicate to find the second and the second last elements of a list using recursion, so for example:
second_secondLast([1,2], X, Y). must return X=2, Y=1.
second_secondLast([1,2,3], X, Y). must return X=2, Y=2.
second_secondLast([1], X, Y). must print 'Error' and return false.
First, I have the error-checking clauses:
second_secondLast([], X, Y) :- print("Error"), !, fail.
second_secondLast([_], X, Y) :- print("Error"), !, fail.
Next, I tried something like this:
second_secondLast([Y,X],X,Y) :- !.
second_secondLast(L, X, Y) :-
second(L,X),
secondLast(L,Y).
second([_,S|_], X) :- X = S.
secondLast([P,_], Y) :- Y = P.
secondLast([F|R], Y) :- secondLast(R, Y).
However, the output using [1,2,3] is X=Y, Y=2.
I'm not sure if it is possible to force the output to be X=2 instead, or if there is a better method to do this.
First of all, the output X=Y, Y=2. has nothing to do with your program, it is an idiosyncracy of swipl (and maybe other interactive environments for Prolog implementations).
I think, your program looks fine, but you are asking for possible improvements.
second([_,S|_], S). is a more elegant version of your second([_,S|_], X) :- X = S..
Likewise, secondLast([P,_], P). is more elegant than your secondLast([P,_], Y) :- Y = P..
I would also prefer secondLast([_|R], Y) :- secondLast(R, Y). to your
secondLast([F|R], Y) :- secondLast(R, Y)..
Your error-checking clauses look fine to me.
You could also get rid of the predicate second and alter the definition of second_secondLast by using
second_secondLast([H,X|T], X, Y):-
secondLast([H,X|T], Y).
instead of your
second_secondLast(L, X, Y) :-
second(L,X),
secondLast(L,Y).
That change would also make it a bit more efficient.
Another possibility is to use
second_secondLast(L, X, Y):-
L= [_,X|_],
secondLast(L, Y).
Then you could also get rid of the predicate secondLast and alter the above clause to
second_secondLast(L, X, Y):-
L= [_,X|_],
append(_, [Y,_], L).
.
There is always a ton of possibilities...
I'm trying to rule to check if someone likes a friend other than the one stated in the argument. So for example,
likes(Alice,Bob).
likes(Bob,Alice).
likes(Alice,Jeff).
likes(Jeff,Alice).
I'm trying to create a rule friends(X,Y) that if both of them like each other, we look for another pair that either X or Y is linked to. Any help?
The first solution returns the first person found that one of the two people likes, in case these 2 people like each other.
likes(alice,bob).
likes(bob,alice).
likes(alice,jeff).
likes(jeff,alice).
friends_aux(X, Y, Z) :- likes(Y, Z), Z \= X.
friends_aux(X, Y, Z) :- likes(X, Z), Z \= Y.
friends(X, Y, Z) :- likes(X, Y), likes(Y, X), friends_aux(X, Y, Z), !.
And the second solution returns true if any of the two people have any other couple.
friends2_aux(X, Y) :- likes(Y, Z), Z \= X.
friends2_aux(X, Y) :- likes(X, Z), Z \= Y.
friends2(X, Y) :- likes(X, Y), likes(Y, X), friends2_aux(X, Y), !.
I attached an image of the results.
Results
Hopefully this is what you wanted.
I need a predicate elements(X,L) where X is an operation like neg X, X and Y and returns a list L with all the elements from the operation, like [X,Y].
I already have these:
elements(neg X, [X]).
elements(X and Y, [X,Y]).
elements(X or Y, [X,Y]).
elements(X imp Y, [X,Y]).
But I don´t know how to make it work with complex operations like X imp (Y or Z).
I'm not great at picking operator precedences but here's what I have for your input:
:- op(100, fx, neg).
:- op(200, xfy, and).
:- op(200, xfy, or).
:- op(300, xfy, imp).
elements(Term, Variables) :- term_variables(Term, Variables).
This seems to do what you specified:
?- elements(X imp (Y and Z), Q).
Q = [X, Y, Z].
Is there more to your problem than this?
sibling(X, Y):- father(Z, X), father(Z, Y), not (X=Y).
sister(X, Y):- father(Z, X), father(Z, Y), female(X).
brother(X, Y):- father(Z, X), father(Z, Y), male(X).
i'm having a bit problem with using the not function. i've tried not X=Y. but to no avail, the sibling rule still produce error.
if i were to delete the not x=y, the output will be a bit kind of "ugly".
how should i write the not function?
The ISO predicate implementing not provable is called (\+)/1.
However, as #coder explains in the comments, it is much better to use dif/2 to express that two terms are different.
dif/2 is a pure predicate that works correctly in all directions, also if its arguments are not yet instantiated.
For example, with (\+)/1, we get:
?- \+ (X = Y ).
false.
No X and Y exist that satisfy this goal, right? Wrong:
?- X = a, Y = b, \+ (X = Y ).
X = a,
Y = b.
In contrast, with dif/2:
?- dif(X, Y).
dif(X, Y).
and in particular:
?- X = a, Y = b, dif(X, Y).
X = a,
Y = b.
See prolog-dif for more information. dif/2 is with us since the very first Prolog system. I strongly recommend you use it.
SWI Prolog has no notoperator. it can be used as a regular compound term, e.i. not(X).
It must be no space between functor and open parenthesis:
foo( argument list ).
This is the cause of the error.
SWI Prolog suggests ISO-standard replacement for not/1: (\+)/1
I've got these definitions:
memberx(X, [X|_]).
memberx(X, [_|T]) :- memberx(X, T).
intersectionx([], _, []).
intersectionx([H|T], Y, [_|Z]) :- memberx(H, Y), !, intersectionx(T, Y, Z).
intersectionx([_|T], Y, Z) :- intersectionx(T, Y, Z).
I get the following result:
?- intersectionx([1], [1], Z).
Z = [_G305].
Why doesn't it result in Z = [1]??
Z = [_G305].
means that this answer is true for all terms. That is, it is not
only true for Z = [1] - as you expect, but it is also true for Z = [2].
Clearly, that is not what you expected.
So where is the error? A simple way to detect it is to watch out for anonymous
variables denoted _.
Consider:
intersectionx([H|T], Y, [_|Z]) :- memberx(H, Y), !, intersectionx(T, Y, Z).
^^^
What you have written means that the intersection of a list starting with
H and another list will be (provided the goals on the right hand side
are all true) a list starting with anything... Replace anything by that H!