I don't know why this isn't working... here's the code.
cameToTheParty(date(15,9,2011), flor).
cameToTheParty(date(22,9,2011), marina).
cameToTheParty(date(15,9,2011), pablo).
cameToTheParty(date(22,9,2011), pablo).
cameToTheParty(date(15,9,2011), leo).
cameToTheParty(date(22,9,2011), flor).
cameToTheParty(date(15,9,2011), fer).
cameToTheParty(date(22,9,2011), mati).
cameToThePartyThatDay(Peoples, Date):-
bagof(X,cameToTheParty(Date,X),Peoples).
When I try
?- cameToThePartyThatDay(People,Day).
it says
People = [flor, pablo, leo, fer],
Day = date(15, 9, 2011) ;
People = [marina, pablo, flor, mati],
Day = date(22, 9, 2011).
But, when I try the following with a variable date's field, or an actual date, like...
member(X,cameToThePartyThatDay(People,date(15,9,2011))).
it just says
false.
The issue is that member is trying to find an element from the list cameToThePartyThatDay(People,date(15,9,2011)), which is not, in fact, a list.
What you want to do is:
cameToThePartyThatDay(People,date(15,9,2011)),
member(X,People).
... so that People is unified with the list of people who came to the party that day, and then member can pull elements from the list of People.
member(X,cameToThePartyThatDay(People,date(15,9,2011)))
is a wrong way to use member/2 because
cameToThePartyThatDay(People,date(15,9,2011))
isn't a list.
A right way could be
cameToThePartyThatDay(People, date(15, 9, 2011)),
member(X, People)
For Prolog, the boldface part in the following expression:
member(X,cameToThePartyThatDay(People,date(15,9,2011))).
Is not a call. In fact predicates are not functions: they do not return anything. According to Prolog the boldface part is a functor.
In order to let it work, you first call cameToThePartyThatDay and then you use People in the member/2 predicate, like:
cameToThePartyThatDay(People,date(15,9,2011)),
member(X,People).
Related
This is my first-day learning prolog and I would like to write a program that decides what shoes I should wear based on the weather and what kind of appointment I have at the office. The main "function" would be declared for example:
"go :- outfit(snow, 15, casual, F1), write(F1)."
Where snow is the weather, 15 is the temperature(not relevant now), and casual is the formality of the appointment. "write(F1)" will display the "output" so the variable F1 needs to be the result of said relation(s). Here are the rules for what shoes to wear:
%Rules for weather
rain(Weather) :- (Weather = 'rain').
snow(Weather) :- (Weather = 'snow').
nice(Weather) :- (Weather = 'nice').
clear(Weather) :- (Weather = 'clear').
%Rules for formality
formal(Appointment) :- (Appointment = 'formal').
semiformal(Appointment) :- (Appointment = 'semiformal').
casual(Appointment) :- (Appointment = 'casual').
%Rules for when to wear a type of footwear
dressShoes(Appointment) :- formal(Appointment).
boots(Appointment, Weather) :- not(formal(Appointment)), (rain(Weather);snow(Weather)).
sneakers(Appointment, Weather) :- not(formal(Appointment)), (nice(Weather);clear(Weather)).
This is where my issue is, I am not sure how to tie the last three relations to a single relation that fills the variable "F1" for my final "outfit" function. I am a C++ guy, so I would like to essentially place a string into F1 like "[sneakers]" or "[boots]" but this is part of my growing pains with prolog. Any help is much appreciated.
Some misunderstandings about Prolog I guess. This kind of rule:
rule(Variable) :- (Variable = 'value').
You don't need to quote 'value', it is already an atom. Whatever book you are reading, look up atoms. It becomes:
rule(Variable) :- (Variable = value).
You don't need the extra parentheses in the rule definition. It becomes:
rule(Variable) :- Variable = value.
You don't need the explicit unification in the body. There is nothing else happening between the head and the body. So you don't need a variable, either. It becomes:
rule(value).
Applying this to your program, I get:
rain(rain).
snow(snow).
nice(nice).
clear(clear).
%Rules for formality
formal(formal).
semiformal(semiformal).
casual(casual).
Those rules say pretty much nothing ;-)
Your example at the very top:
go :- outfit(snow, 15, casual, F1), write(F1).
Does exactly the same as just calling outfit(snow, 15, casual, F1). So what is the purpose of the "go" and the "write"? Skip them I guess.
The logic of your program: can you explain it without code? Your code is so unusual that I have to guess.
If you want to say, "If the appointment is formal, put on dress shoes", you could write it like this:
occasion_shoes(Occasion, Shoes) :-
formality_occasion(Formality, Occasion),
formality_shoes(Formality, Shoes).
formality_occasion(formal, evening_party).
formality_occasion(semi_formal, office).
formality_shoes(formal, dress_shoes).
Do you see what is going on? You match the occasion to the shoes. To do that, you look up the formality of the occasion in the table formality_occasion/2 and then match it to the formality of the shoes in the formality_shoes/2 table.
If you are struggling to model your problem, you can also read up on relational database design.
With the following rules:
test('John', ebola).
test('John', covid).
test('Maria', covid).
How can I create a predicate that would tell me if John or Maria took (both) the Ebola and Covid tests?
I want to do something similar to this (I know it's wrong, just the idea):
tests(Persona, Ebola, Covid) :-
Ebola = test(Persona, ebola),
Covid = test(Persona, covid).
Prolog is relational not functional. test(X, Y) either holds or fails and doesn't return a value like what you thought. Here's what you should have written:
tests(Persona) :-
test(Persona, ebola),
test(Persona, covid).
You can query tests('John') which is true since both test/2 calls succeeds. The query tests('Maria') fails because test('Maria', ebola) fails.
Does it answer your question ?
I'm coding in prolog, but I have a issue. How can I make that the answer appears just once? for example I just want that "X = uni, X = uca, X = unam" but instead it just keep showing me the options repeatedly.
This is some of it:
is(uni, college).
is(uca, college).
is(unan, college).
is(computation, carrer).
In this part I'm assigning available places to the carrer
has(computation, available_places, 200).
and finally assigning the carrer to a college
offers(unan, computation).
offers(uni, computation).
offers(uca, computation).
and I make the query like this:
which(X):- is(X, college), is(Y, carrer), offers(X, Y),has(Y, available_places, Z), Z<300.
but the result as I said at the beggining, show me the names of the college repeadtedly. Any idea how to solve this? D:
In prolog I'm trying to unify every valid pairing of needs with resources
needs([ece2090,1,m,13,16]).
needs([ece3520,1,tu,11,14]).
needs([ece4420,1,w,13,16]).
resources([joel, [ece2090,ece2010,ece3520,ece4420],[[m,13,16]]]).
resources([sam, [ece2010,ece4420],[]]).
resources([pete, [ece3520],[[w,13,16]]]).
using this formula
make_bid([Class,Sect,Day,Ts,Te],[Name,Cap,Unavail],[Class,Sect,Day,Ts,Te,Name,_]) :-
no_conflict_all_unavailable(Day,Ts,Te,Unavail),
course_capable(Class,Cap),
writef('%w %w %w\n',[Class,Sect,Name]),
fail.
and running this test.
test(Listing) :- needs(N), resources(R), make_bid(N,R,Listing).
The point of this part of the program is to pair every class with a teacher that both has the qualifications to teach the class and is not unavailable during that time. It's supposed to give a list.
?- test(Listing).
ece3520 1 joel
ece3520 1 pete
ece4420 1 joel
ece4420 1 sam
false.
When run, the above is generated. This is correct, but it's in a format that's useless to me, since I need it to be a variable of its own to do further computations. Then the solution is to use bagof or findall, right?
So I remove the fail clause from the main part of the program and then change the test to this
test(Bag) :- needs(N), resources(R), bagof(Listing,make_bid(N,R,Listing),Bag).
but it generates this
ece3520 1 joel
Bag = [[ece3520, 1, tu, 11, 14, joel, _G4310]]
If you look closely you'll see that there's no period at the end as well as a lack of a true/false statement. This would lead one to believe it is infinitely looping. This isn't the case however, as the Bag matrix is fully formed and I can simply type "." to end the program (instead of, you know, aborting it).
It only generates the first valid solution. Why is this happening?
You've structured your test predicate so that bagof/3 is called for every instance combination of needs(N) and resources(R) and so it collects each result of make_bid in it's own bagof/3 result:
ece3520 1 joel
Bag = [[ece3520, 1, tu, 11, 14, joel, _G4310]]
The first line is the write that is in make_bid predicate. The second line is the Bag result for the single query to make_bid for one pair of needs/resources. The last argument in the list, _G4310, occurs because your predicate uses _ and it's anonymous (never used/instantiated).
Your current make_bid is designed to write the results in a loop rather than instantiate them in multiple backtracks. So that could be changed to:
make_bid([Class, Sect, Day, Ts, Te], [Name, Cap, Unavail], [Class, Sect, Day, Ts, Te, Name, _]) :-
no_conflict_all_unavailable(Day, Ts, Te, Unavail),
course_capable(Class, Cap).
(NOTE: I'm not sure why you have _ at the end of the 3rd list argument. What does it represent?)
If you want to collect the whole result in one list, then you canb use findall/3:
findall([Class, Sect, Name], (needs(N), resources(R), make_bid(N, R, [Class, Sect, _, _, _, Name, _]), Listings).
This will collect a list of elements that look like, [Class, Sect, Name]. You could use bagof/3 here, but you'd need an existential quantifier for the variables in the make_bid/3 call that you don't want to bind.
If you wanted the entire Listing list, then:
findall(L, (needs(N), resources(R), make_bid(N, R, L)), Listings).
But each element of Listings will be a list whose last element is an anonymous variable, since that's how make_bid/3 is structured.
I'm trying to to get a list of three items with their relevant information based on a collection of information:
product(I):-
I = [_,_,_,_], %Type,Brand,Category,Value
cheaper(item(apple,_,_),item(_,kay,_,_),I),
cheaper(item(bar,_,_,_),item(_,_,fruit,_),I),
member(item(_,kay,_,2),I),
member(item(apple,granny,_,_),I),
member(item(bar,_,chocolate,_),I),
/* Below not given */
member(item(cracker,_,_,_),I),
member(item(_,_,biscuit,_),I),
member(item(_,_,_,4),I),
member(item(_,_,_,5),I).
cheaper(X,Y,H) :- %Used to determine the item values
item(X,_,_,A),
item(Y,_,_,B),
A<B.
When I try running it I encounter an error:
?- product(I).
ERROR: cheaper/3: Undefined procedure: item/4
Exception: (8) item(item(apple, _G3604, _G3605), _G3651, _G3652, _G3653) ?
I understand that item isn't a procedure, however what can I used for checking the value for apple against the value for bar?
First, the obvious note, you're calling cheaper wrong once:
cheaper(item(apple,_,_),item(_,kay,_,_),I),
↑
Only three values, not four.
If item isn't a procedure, you mustn't call it, but use destructuring.
Also you want those items you're checking with cheaper to be part of the list, right? If so, you'll have to check that. And you can use unification to extract the values that you need:
cheaper(X,Y,I) :-
member(X,I),
member(Y,I),
[item(_,_,_,A),item(_,_,_,B)] = [X,Y],
A<B.
Now you'll get some errors regarding not instantiated argument. That's because you are checking not (yet) set variables if they are greater than each other. To avoid this, move the cheaper/3 calls to the end of your clause body:
product(I):-
I = [_,_,_,_], %Type,Brand,Category,Value
member(item(_,kay,_,2),I),
member(item(_,_,_,4),I),
member(item(_,_,_,5),I),
member(item(apple,granny,_,_),I),
member(item(bar,_,chocolate,_),I),
/* Below not given */
member(item(cracker,_,_,_),I),
member(item(_,_,biscuit,_),I),
cheaper(item(apple,_,_,_),item(_,kay,_,_),I), % note the 4th argument
cheaper(item(bar,_,_,_),item(_,_,fruit,_),I).
With this, you'll get one solution and then it fails with an error. This is, because you only give three values for the price slot and you have four items and prolog will check A > 2.
Sorry, in my other answer I didn't look for what the poster was trying to achieve and I think that this is better than a complete reedit. (glorious SO mods let me know if I'm wrong)