I have a fact set as so...
fav_fruit(male, young, apple).
fav_fruit(female, young, bannana).
fav_fruit(male, old, bannana).
fav_fruit(female, old, apple).
fav_fruit(female, young, apple).
I'm need find out if there is any group of (Gender,Age) where there is more the one favorite fruit (the answer for the facts above would be (female,young.)
What I've been trying to figure out is how to use the aggregate or findall functions in prolog to return some type of list like....
female, young = 2 (apple, bannana)
male, young = 1 (apple)
male, old = 1 (bannana)
female, old = 1 (apple)
...that way I could just check the total for each member and test whether it it greater then 1.
Any ideas would be greatly appreciated.
How about
fav_fruit_class(Gender-Age, List):-
findall(Gender-Age, fav_fruit(Gender, Age, _), LGenderAge),
sort(LGenderAge, SGenderAge),
member(Gender-Age, SGenderAge),
findall(Fruit, fav_fruit(Gender, Age, Fruit), List).
The first findall+sort gets a list of classes (Gender/Age). The second findall gets the list of favorite fruits for each class.
Related
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).
Each list represents a customer and his car brand and colour:
customer([_Name,_Surname,_Car,_Color]).
I would like to know if I can check in more than 2 lists that no customers have the same car of the same color. Because each customer has different name and surname from the others, (meaning that only _Car and _Color can take the same values in lists), I think one way is to check that the lists have not exactly 2 same elements.
An example of lists is:
customer1([john,brown,vw,black]).
customer2([will,smith,audi,green]).
customer3([nick,cave,bmw,blue]).
customer4([jim,beam,bmw,black]).
In this example none has the same brand and the same color, so it's true.
What I imagine is something like:
result:-customer1(X1,Y1,Z1,W1),
customer2(X2,Y2,Z2,W2),
customer3(X3,Y3,Z3,W3),
customer4(X4,Y4,Z4,W4),
(check that no Z's AND W's are the same).
Is this possible?
this query would answer about different persons having the same Car of same Color:
?- customer([Name1,Surname1,Car,Color]), customer([Name2,Surname2,Car,Color]), ( Name1 \= Name2 ; Surname1 \= Surname2).
This will tell you if two lists share the same car & color:
sameCarAndColor( [_,_,Car,Color], [_,_,Car,Color] ).
To check more than 2, just use this on each possible pair of lists.
% A quiz team structure takes the form:
% team(Captain, Vice_captain, Regular_team_members).
% Captain and Vice_captain are player structures;
% Regular_team_members is a list of player structures.
% player structures take the form:
% player(First_name, Surname, details(Speciality,Recent_score)).
team(player(niall,elliott,details(history,11)),
player(michelle,cartwright,details(fashion,19)),
[player(peter,lawlor,details(science,12)),
player(louise,boyle,details(current_affairs,17))]).
I've been given the database above (I haven't copied over all the entries of people as it would be too long).
I've been asked to get the surname of any vice-captain whose team includes a captain or a regular team member whose speciality is science.
I can get the surname of the vice-captains using the code below but I can't return only those teams that include a captain or regular team members whose speciality is science. What would I need to add to do this?
part_two(Surname):-
team(_,player(_,Surname,_),_).
I've also been asked to also get the first name and the surname of any captain whose regular team members number more than one and who all have the same surnames.
This is my attempt so far:
part_three(First_name,Surname):-
team(Captain,_,Regular_team_members),
first_name(Captain,First_name),
surname(Captain,Surname),
Regular_team_members=[_,_|_].
I just need to exclude the details of those captains whose regular team members don't all have the same surname.
part_two(Surname):-
team(Captain, Vice_captain, Regular_team_members),
surname(Vice_captain, Surname),
member(Player, [Captain|Regular_team_members]),
specialty(Player, science).
% 'abstract data structures' accessors
surname(player(_First_name, Surname, _Details), Surname).
specialty(player(_First_name, _Surname, details(Speciality, _Recent_score)), Speciality).
Since you're going anyway to scan the Regular_team_members list, looking for appropriate constraint, you can get a simpler 'program' first 'joining' the Captain to other players.
You could change a little what you have already written as follows:
part_two(Surname):-
team(P,player(_,Surname,_),L),
( P=player(_,_,details(science,_)) -> true ; member(player(_,_,details(science,_)),L) ).
Example:
Database:
team(player(niall,elliott,details(history,11)),
player(michelle,cartwright,details(fashion,19)),
[player(peter,lawlor,details(history,12)),
player(louise,boyle,details(current_affairs,17))]).
team(player(niall1,elliott1,details(science,11)),
player(michelle1,cartwright1,details(fashion,19)),
[player(peter,lawlor,details(history,12)),
player(louise,boyle,details(current_affairs,17))]).
team(player(niall2,elliott2,details(history,11)),
player(michelle2,cartwright2,details(fashion,19)),
[player(peter,lawlor,details(science,12)),
player(louise,boyle,details(current_affairs,17))]).
team(player(niall3,elliott3,details(science,11)),
player(michelle3,cartwright3,details(fashion,19)),
[player(peter,lawlor,details(science,12)),
player(louise,boyle,details(current_affairs,17))]).
Now querying:
?- part_two(X).
X = cartwright1 ;
X = cartwright2 ;
X = cartwright3.
so I have a set of facts:
course(cs61, "intro to cs")
...
course(cs62b, "data structure")
...
grade(adam, cs61, spring11, "A")
...
grade(chrisitian, cs61, fall2010, "A-")
I need to define a predicate good_standing(Student) to determine whether the Student got A in every class he took. I must use 2 different approaches:
use setof/3
use NAF
on the 1st. I have tried to figure out get Result_list: the set of students that got A from every class he took. and then call member(Student, Result_list). But I don't know what to get the Result_list since the syntax is a bit strange to me.
1) For the NAF solution you should keep in mind that the good-standing student is the one with no grades lower than A, i.e., the one such that there is no course he/she took and the grade is lower than A. We also require this person to take at least one course.
good_standing(Student) :- grade(Student, _,_,_),
\+ (grade(Student, Course, _, Grade),
lower(Grade,"A")).
with lower/2 defining relation between the grades:
lower("A-","A").
lower("B","A").
...
The solution above works if students get only one grade per course.
2) For the setof solution you can find all grades per student and check that they are A.
grade(Student, Grade) :- grade(Student, _,_,Grade).
good_standing(Student) :- setof(Grade, grade(Student,Grade), ["A"]).
Alternatively, as suggested by false we can make the existential quantification explicit within setof:
good_standing(Student) :- setof(Grade,
Course^Period^grade(Student,Course,Period,Grade),
["A"]).
I'm implementing a variation on Einstein's Riddle and i'm having some trouble.
When trying to calculate the solution i try this:
solve(Street) :- Street = [_House1,_House2,_House3,_House4,_House5],
%hint one goes here
%hint two goes here
%etc.
I can then ask the solution by typing: solve(Street)
However this comes up as solution:
house(flower, food, pet, sport)
house(flower, food, pet, sport)
house(x , food, pet, sport)
house(flower, food, pet, sport)
house(x, flower, pet, sport)
As you can see there's 2 times x, the rest are all types of foods, flowers, pets and sports.
But every type is unique: if one person likes flower X, noone else can like X.
Now, the reason why my solution gives 2 x's is easy to see: we are given an amount of hints but in all the hints there are only mentioned 4 flowers. So Prolog doesn't know there is another flower, and just uses x twice, just because it's possible and fulfills all the other hints.
What i want to say is that all the types of foods and flowers etc. in Street are unique so he should leave some blank when he used all types already. 3 would look like: house(x , food, pet ,sport) and 5 would look like: house(_, flower, pet, sport).
I also tried adding this to the hints: (let's say "cactus" is one of the flowers not mentioned in the hints)
member(house(cactus,_,_,_), Street)
However then my program doesn't end...
A hint may look like this:
is_neighbour(house(_,_,_,football),house(_,_,fish,_), Street),
with : is_neighbour(A,B,List) giving true when A and B are next to each other in List.
The hint can be translated to: the person who loves football lives next to the person who has fish.
If any more info need to be provided i'm willing to elaborate. :)
To express that no flower is reported twice, and also to make sure that all flowers are bound, you can use the permutation/2 predicate: the list of all flowers should be a permutation of the list of specified flowers. This would read like [untested]
flowers([], []).
flowers([house(Flower,_,_,_)|Street], [Flower|Rest]) :- flowers(Street, Rest).
-- ...
flowers(Street, Flowers),
permutation(Flowers, [kaktus, tulpe, nelke, rose, fingerhut]),
Edit: for 10 flowers, using permutations is probably too slow. An alternative approach is
flower(kaktus).
flower(tulpe).
flower(nelke).
--...
flowers(Street,[F1,F2,F3,F4,F5,F6,F7,F8,F9,F10]),
flower(F1), flower(F2), F1\=F2,
flower(F3), F3\=F1, F3\=F2,
flower(F4), F4\=F1, F4\=F2, F4\=F3,
--...