Reputation: 11
Using Prolog, I first created two facts called grade and food: The first fact is grade(X,Y) where X is the student (rob or matt) and Y is the grade level (freshman or sophomore). The second fact is food(X,Y) where X is the student (rob or matt) and Y is the food (pizza, burger, pasta, wrap).
I created a rule called preference(X,Y), where X is the student (rob or matt) and Y is the students' preference.
I want to enter preference(rob,X).
in the GNU Prolog and have it return:
sophomore, pizza, burger.
However, it keeps returning: sophomore, pizza, pizza.
How do I fix this problem? I've spent hours looking into this. Thanks
This is the code I have:
grade(rob, sophomore).
grade(matt, freshman).
food(rob, pizza).
food(rob, burger).
food(matt, pasta).
food(matt, wrap).
preference(X,Y):-
grade(X,A),
food(X,B),
food(X,C),
Y = (A, B, C).
Upvotes: 1
Views: 154
Reputation: 2436
The way you have defined your facts is nice. The way you query it is not conventional. Here is how I would do it. The "preference" rule is simpler:
grade(rob, sophomore).
grade(matt, freshman).
food(rob, pizza).
food(rob, burger).
food(matt, pasta).
food(matt, wrap).
preference(X, A, Y):-
grade(X, A),
food(X, Y).
You conventionally query the database and get all solutions with backtracking:
?- preference(rob, Grade, Food).
Grade = sophomore,
Food = pizza ;
Grade = sophomore,
Food = burger.
If you want to collect the foods, you can use bagof/setof, like this:
?- bagof(Food, preference(rob, Grade, Food), Foods).
Grade = sophomore,
Foods = [pizza, burger].
What if you want to query all freshmen?
?- bagof(Food, preference(Person, freshman, Food), Foods).
Person = matt,
Foods = [pasta, wrap].
Upvotes: 2
Reputation: 3746
You need to state that the value of B
and C
are different; there are multiple ways to do that, for the simplicity I go with \==/2
(documentation):
preference(X,Y):-
grade(X,A),
food(X,B),
food(X,C),
B\==C,
Y = (A, B, C).
Gives the output
| ?- preference(X,Y).
X = rob
Y = (sophomore,pizza,burger) ? ;
X = rob
Y = (sophomore,burger,pizza) ? ;
X = matt
Y = (freshman,pasta,wrap) ? ;
X = matt
Y = (freshman,wrap,pasta) ? ;
no
If you don't want to have the basically doubled entries you can go with the (in this case lexical) "less than" @</2
:
preference(X,Y):-
grade(X,A),
food(X,B),
food(X,C),
B @< C,
Y = (A, B, C).
| ?- preference(X,Y).
X = rob
Y = (sophomore,burger,pizza) ? ;
X = matt
Y = (freshman,pasta,wrap) ? ;
no
Upvotes: 1
Reputation: 2516
I may be wrong, but I suspect this may be a misunderstanding of prolog in general in addition to a non-intuitive REPL. Prolog doesn't really "return" a value, it just tries to match the variables to values that make your predicates true, and I would be willing to bet you're hitting enter after you see the first result.
The way preference
is currently written B
and C
will match any two foods that rob
is associated with. This could be pizza, pizza
or pizza, burger
or burger, pizza
, or so on. It does not check whether B
and C
are equal. When I run preference(rob,X).
prolog does not only give me the first result UNLESS I hit enter.
| ?- preference(rob,X).
X = (sophomore,pizza,pizza) ? ?
Action (; for next solution, a for all solutions, RET to stop) ?
If you hit a
(or spam ;
a few times) prolog will give you the rest of the results.
| ?- preference(rob,X).
X = (sophomore,pizza,pizza) ? a
X = (sophomore,pizza,burger)
X = (sophomore,burger,pizza)
X = (sophomore,burger,burger)
yes
| ?-
I think that all you really need to get all of a person's preferences is just food
unless you specifically need them in a tuple or list which will take some slightly more complicated logic (let me know in a comment if that's what you're looking for)
| ?- food(rob, X).
X = pizza ? a
X = burger
yes
| ?-
Upvotes: 0