Currently trying to create a library recommendation system in prolog for a college assignment and being quite new to prolog I'm quite lost and was wondering if I could have some of this explained to me in fine detail.
Here are my facts and rules currently:
book(after_dark, haruki_murakami,fiction,182).
book(python, charlie, revision, 560).
book(nt_bible, sams, reference, 480).
book(monty_python, cleese, comedy, 300).
buildLibrary(Lib) :- findall(book(Title, Author, Genre, Size), book(Title, Author,
Genre, Size), Lib).
holiday(B,L) :- //this should take the list formed in buildLibrary along with a variable that represents a book
//and is true if and only if its genre is comedy or fiction and less than 400 pages
Expected input:
buildLibrary(L)
holidays(book(after_dark,haruki_murakami,fiction,182),L)
Ideally this should return true as it meets the requirements outlined
How do I go about setting the rule for holiday? Once I know how to do this I feel like I can get it working, I've tried multiple things and they've all returned errors in SWL prolog, once again thanks for any help!
If I correctly understand your implementation then you could do something like:
holiday(B,L) :- buildLibrary(Lib), check(L,B).
check(book(X, Y, Genre, Size),[book(X, Y, Genre, Size)|_]):-
(Genre = comedy ;Genre = fiction), Size < 400.
check(B,[book(_, _, Genre, Size)|T]):-
dif(Genre,comedy),dif(Genre,fiction), check(B,T).
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.
So I've this problem that is bugging me a lot:
We've the fact student(sID,LR,[courses]).
In which sID is the student ID, LR to determine if he is a left or right-handed person, and [courses] is an array of courses.
Next, we've the predicate listStudents(C,L) which C represents one Course and L the List. The predicate wants to return a List of Students that are in C Course.
Example:
student(23,1,[math101,phy203]).
student(24,0,[math101,phy203,art]).
student(25,1,[phy203,art]).
?- listStudents(math101,L)
L=[23,24]
Now, I've tried for several things like:
listStudents(C,L):- student(C,_,Y), member(C,Y)
and I realized later it's wrong as it returns me
L=23
L=24
Afterwards I was reading about findall() and made this
listStudents(C,L):- findall(S, study(C,S), L).
study(C,S):- estudiante(S,_,Y), member(C,Y).
which makes a return that I did not expect
L=[23,23,24,24,24]
I have been working with Prolog since today, and wanted to create a simple test case. The basic idea was to have multiple sports defined, and it looks as follows:
soccer :- category(ball_sport),
check(has_11_players_per_team),
check(large_ball),
check(use_feet).
tennis :- category(ball_sport),
...
category(ball_sport) :-
check(has_a_ball).
Now I wanted to create a testcase, to see if both sports are of the ball_sport category, but have no idea to check these sports against eachother .. I thought it would be something like the code below, but it's obvious not. Is there an easy way to check these predicate categories? Thanks
both_ballsports(sport_one, sport_two) :-
has_category(sport_one, ball_sport),
has_category_sport_two, ball_sport).
It seems that first of all, you want to declaratively state attributes of a sport.
For example:
sport_attributes(soccer, [ball_sport,players(22),ball(large),use(feet)]).
sport_attributes(tennis, [ball_sport,players(2),players(4),ball(small),use(racket)]).
Note that I am relating sports to attributes. For comparison, the goals of the form check(X) you use above all seem to lack a critical argument, namely the actual sport for which they hold (or not). For example, the goal check(use_feet) either holds or not, but there is no means to qualify a unary predicate of this kind and state different facts for different sports.
Note the naming convention: We describe what each argument means, separated by underscores.
With this representation, both_ballsports/2 could look like this:
both_ballsports(Sport1, Sport2) :-
ballsport(Sport1),
ballsport(Sport2).
ballsport(Sport) :-
sport_attributes(Sport, As),
member(ball(_), As).
Sample query and answer:
?- both_ballsports(Sport1, Sport2).
Sport1 = Sport2, Sport2 = soccer ;
Sport1 = soccer,
Sport2 = tennis ;
Sport1 = tennis,
Sport2 = soccer ;
Sport1 = Sport2, Sport2 = tennis ;
false.
This can be used in all directions!
I'm super new to Prolog, like, my professor assigned us a program and just asked us to watch a couple youtube videos. No lecture.
So anyway, here's the issue:
I'm supposed to create a pharmacist software that looks up drug interactions.
When I enter a specific drug, then Drug-variable, and Interaction-variable, I get the first drug and interaction in the list (of like, 100 drugs that interact with temazepam):
?- interacts(temazepam,Drug,Interaction).
Drug = thalidomide,
Interaction = neutropenia .
Part 1) How can I get every drug and its interaction from, say, temazepam?
Partial program listed below [because I have 1609 lines of drug interactions listed]:
interacts(X,Y,Interaction):-
Drug(X),
Drug(Y),
Interaction.
Interaction:-
Drug(X),
Drug(Y).
interacts(gatifloxacin,zolpidem,attempted_suicide).
interacts(gatifloxacin,zolpidem,insomnia).
interacts(gatifloxacin,warfarin,cardiac_decompensation).
interacts(gatifloxacin,isosorbide-5-mononitrate,arteriosclerotic_heart_disease).
interacts(gatifloxacin,rosiglitazone,hyperglycaemia).
interacts(gatifloxacin,bortezomib,hyperglycaemia).
interacts(gatifloxacin,mometasone,asthma).
interacts(gatifloxacin,cisplatin,hyperglycaemia).
interacts(gatifloxacin,cisplatin,bone_marrow_failure).
interacts(gatifloxacin,montelukast,difficulty_breathing).
interacts(temazepam,thalidomide,neutropenia).
interacts(temazepam,thalidomide,thrombocytopenia).
interacts(temazepam,timolol,drowsiness).
interacts(temazepam,tizanidine,acid_reflux).
interacts(temazepam,tizanidine,heart_attack).
interacts(temazepam,tolterodine,amnesia).
Part 2) I need to be able to list an interaction and get back every drug that caused it.
I guess just the side-effect then all drug interactions listed would be better than listing drug1+sideEffect = drug2.
Example:
?- interacts(Drug,Drug,amnesia).
Part 3) I should be able to enter a single drug, and get everything with interactions and everything with no interactions.
Example:
?- interacts(valacyclovir,Drug,Interaction).
Drug = zolpidem,
Interaction = anxiety
But for everything
Excuse me for the edits!
Thanks so much in advance!
You can use the built-in predicate findall/3 for that:
drug_allinteractions(Drug,AI) :-
findall((D,I),interacts(Drug,D,I),AI).
The only goal of drug_allinteractions/2 is using findall/3 to query interacts/3 and put its second and third argument into the list AI as a tuple (D,I). Example query: Wich interacting drugs with what interaction-effect are known for gatifloxacin?:
?- drug_allinteractions(gatifloxacin,L).
L = [(zolpidem,attempted_suicide),(zolpidem,insomnia),(warfarin,cardiac_decompensation),(isosorbide-5-mononitrate,arteriosclerotic_heart_disease),(rosiglitazone,hyperglycaemia),(bortezomib,hyperglycaemia),(mometasone,asthma),(cisplatin,hyperglycaemia),(cisplatin,bone_marrow_failure),(montelukast,difficulty_breathing)]
Alternatively, if you just want to query interacts/3 and not write a program:
?- findall((D,I),interacts(gatifloxacin,D,I),AI).
AI = [(zolpidem,attempted_suicide),(zolpidem,insomnia),(warfarin,cardiac_decompensation),(isosorbide-5-mononitrate,arteriosclerotic_heart_disease),(rosiglitazone,hyperglycaemia),(bortezomib,hyperglycaemia),(mometasone,asthma),(cisplatin,hyperglycaemia),(cisplatin,bone_marrow_failure),(montelukast,difficulty_breathing)]
As for your added part 2): You can use findall on your original query:
?- findall((D1,D2),interacts(D1,D2,amnesia),AI).
AI = [(temazepam,tolterodine)]
Note, that unlike in your example I wrote two different variables D1 and D2 for the drugs, otherwise you are asking which drug has the interaction-effect amnesia with itself.
Considering your added part 3) I'm not entirely sure what you want. Your query reads: "Show me all drugs that interact with valacyclovir plus the associated effect". That is basically the same as your very first query, just for a different drug. You can query for all drugs in the relation interacts/3 interactively without showing the interacting drugs and the effects by:
?- interacts(D,_,_).
D = gatifloxacin ? ;
...
Or query for an entire list without duplicates by using setof/3:
?- setof(D1,D2^I^interacts(D1,D2,I),AI).
AI = [gatifloxacin,temazepam]
If you, however, try to find a list of drugs that are not interacting with a given drug, you can write a predicate, say drug_noninteractingdrug/2...
:- use_module(library(lists)).
drug_noninteractingdrug(D,NID) :-
dif(D,NID), % D and NID are different
setof(D2,D^interacts(D,D2,_),L), % L ... all drugs interacting with D
interacts(NID,_,_), % NID ... drug...
\+member(NID,L). % ... that is not in L
... and query this using setof/3:
?- setof(NID,drug_noninteractingdrug(gatifloxacin,NID),NIDs).
NIDs = [temazepam]
With your given minimal example this query of course only yields one drug. Note that you need to include library(lists) for the predicate member/2.
I'm developing a Guess Who? game using Prolog. The mechanics of the game are very easy. A player (in this case, the human) chooses one person among many possible ones, and the other player (the computer) starts asking yes/no questions about some attributes of the person. Eventually, with the given answers, there will only be one possible person.
So far, I've been able to make a set of rules and predicates so that the computer can guess the person based on the questions that have been asked so far. I have a set of suspects - those are the people that, with the available clues, could fit.
suspect('Person 1') :- eyes(blue) , age(old) , gender(male).
The predicates for the attributes are defined so that they will be true either if the question regarding that attribute has not been asked yet, or if the question has been asked and the answer matches the attribute of the suspect.
gender(X) :- not(asked_gender) ; value_of(gender, X).
That way, if two suspects share the same eyes and age and different gender, as long as the gender remains unasked, both of them will be plausible suspects.
However, the hard part now is to automate the process of asking those questions. Basically, I'm looking forward to a solution where Prolog were able to get the possible values for the attributes from the suspects' predicates, instead of listing theme somewhere else. I'm pretty sure there must be a way of doing this, given prolog is able to use the program's code as data itself.
How could I do that?
This works in SWI-Prolog :
:- dynamic value/1.
suspect('Person 1') :- eyes(blue) , age(old) , gender(male).
suspect('Person 2') :- eyes(green) , age(young) , gender(male).
suspect('Person 3') :- eyes(brown) , age(young) , gender(male).
fetch(Criterion, Value) :-
retractall(value(_)),
assert(value([])),
forall(clause(suspect(_), Body),
check(Body, Criterion)),
retract(value(Value)).
check((F, T), Criterion) :-
F =..[Criterion, V1],
retract(value(V2)),
( member(V1, V2) -> V3 = V2; V3 = [V1 | V2]),
assert(value(V3)).
check(T, Criterion).
check((_F, T), Criterion) :-
check(T, Criterion).
check((F), Criterion) :-
F =..[Criterion, V1],
retract(value(V2)),
( member(V1, V2) -> V3 = V2; V3 = [V1 | V2]),
assert(value(V3)).
check((_F), _Criterion).
For example :
?- fetch(gender, Value).
Value = [male].
?- fetch(eyes, Value).
Value = [brown,green,blue].
Well, I would imagine a construction like this:
go :-
findall(People,suspect(People),SuspectList),
length(SuspectList,1),
member(Perb,SuspectList),
write('It is '),write(Perb),write('!!'),nl,!.
go :-
askQuestion,
go.
Where in askQuestion/0 you'd ask questions with the read/1 predicate and assert/1 the answers.
This is where you could try and make it 'intelligent' or you could just iterate over the different questions.