I'm a begginer at Prolog and I need some help with this exercise, this is the knowledge base provided:
album(‘R. Stevie Moore’, ‘Manuscription’).
album(‘Lane Steinberg’, ‘Manuscription’).
album(‘R. Stevie Moore’, ‘The Yung & Moore Show’).
album(‘Yukio Yung’, ‘The Yung & Moore Show’).
album(‘Jessie Evans’, ‘Autonervous’).
album(‘Bettina Koster’, ‘Autonervous’).
album(‘Lucia Pamela’, ‘Walkin on the moon’).
album(‘Shooby Taylor’, ‘The Human Horn’).
album(‘Tiny Tim’, ‘God Bless Tiny Tim’).
album(‘The Legendary Stardust Cowboy’, ‘Rock-It to Stardom’).
vinil(‘Rock-It to Stardom’).
vinil(‘Walking on the Moon’).
cd( ‘God Bless Tiny Tim’).
cd(‘Walking on the Moon’).
cd(‘Autonervous’).
cd(‘Manuscription’).
cassette(‘The Human Horn’).
cassette(‘The Yung & Moore Show’).
mp3(‘Walkin on the Moon’).
I need to make a query that will return to me all the albums that were made by only one musician.
Thanks for your help :)
In the comments you have provided your code album(X,B),album(Y,C), B \= C. Actually, this is not too far away from a correct solution.
A correct solution could be:
one_musician_album(X) :-
album(A, X),
\+ (album(B, X), A \= B).
The meaning of the predicate is: an album X is a "one-musician album" if this album is authored by some musician A and it's not possible to find a different musician B authored the album X.
Test run:
?- one_musician_album(X).
X = 'Walkin on the moon' ;
X = 'The Human Horn' ;
X = 'God Bless Tiny Tim' ;
X = 'Rock-It to Stardom'.
To get all answers you have to type ';' after each answer.
Maybe this is not needed for you, but it's possible to get all answers in a list with findall:
?- findall(X, one_musician_album(X), Albums).
Albums = ['Walkin on the moon', 'The Human Horn', 'God Bless Tiny Tim', 'Rock-It to Stardom'].
Here is a generalization, based on 'all solutions' builtin bagof/3.
Note that
?- bagof(A, album(A,T), As).
T = 'Autonervous',
As = ['Jessie Evans', 'Bettina Koster'] ;
T = 'God Bless Tiny Tim',
As = ['Tiny Tim']
....
then, restricting the authors (As) to be a list of a single element, we get albums with a single author
?- bagof(A, album(A,T), [A]).
A = 'Tiny Tim',
T = 'God Bless Tiny Tim' ;
...
then, we could use findall/3, as in the good answer by Sergey, or, again, bagof/3, with explicit quantification:
?- bagof(T, A^bagof(A, album(A, T), [A]), Ts).
Ts = ['God Bless Tiny Tim', 'Rock-It to Stardom', 'The Human Horn', 'Walkin on the moon'].
Related
I'm writing a "solver" using the SICStus Prolog attributed variables interface:
:- module(attach2, [attach2/2]).
:- use_module(library(atts)).
:- attribute att/2.
attach2(X,Y) :- put_atts(X,att(X,Y)),
put_atts(Y,att(Y,X)).
verify_attribute(_,_,[]).
attribute_goal(V,(true,G,true,G,true)) :- get_atts(V,att(X,Y)), G = attach2(X,Y).
Sample query:
| ?- attach2(X,Y), copy_term(X,Xc,Xcc), copy_term(Y,Yc,Ycc), copy_term(X+Y,XYc,XYcc).
Xcc = attach2:(true,attach2(Xc,_A),true,attach2(Xc,_A),true),
Ycc = attach2:(true,attach2(Yc,_B),true,attach2(Yc,_B),true),
XYc = _C+_D,
XYcc = (attach2:attach2(_C,_D),attach2:attach2(_D,_C)),
attach2(X,Y),
attach2(Y,X) ? ;
no
If I change attribute_goal/2 to ...
attach2(X,Y) :- put_atts(X,att(X,Y)),
put_atts(Y,att(X,Y)).
... the answer get better:
| ?- attach2(X,Y), copy_term(X,Xc,Xcc), copy_term(Y,Yc,Ycc), copy_term(X+Y,XYc,XYcc).
Xcc = attach2:(true,attach2(Xc,_A),true,attach2(Xc,_A),true),
Ycc = attach2:(true,attach2(_B,Yc),true,attach2(_B,Yc),true),
XYc = _C+_D,
XYcc = attach2:attach2(_C,_D),
attach2(X,Y) ? ;
no
My conclusion from this:
copy_term/3 collects goals from attribute_goal/2 and if it encounters more than one attributed variable, then it removes trivial true and duplicate goals.
It this essentially right?
I want to create the Knowledge base equivalent to the statements as below in Prolog.
John likes all kind of food.
Apple and vegetable are food
Anything anyone eats and not killed is food.
Anil eats peanuts and still alive
Harry eats everything that Anil eats.
So this one of the version borrowed from here
Once this file is created I want to find the answer for the query "John likes peanuts." saying likes(john, peanut).
Attempt to write rules in Prolog from my side is as below
alive(anil).
eats(anil,peanut).
food(apple).
food(vegitables).
food(X):-likes(john,X).
eats(X,Y),not(killed(X)):-food(Y).
eats(anil,X):-eats(harry,X).
killed(X):-not(alive(X)).
alive(X):-not(killed(X)).
But I am getting the errors and warnings as below;
ERROR: No permission to modify static procedure `(',')/2'
Warning: Clauses of eats/2 are not together in the source-file
Warning: Current predicate: food/1
Warning: Use ':- discontiguous eats/2.' to suppress this message
Warning: Clauses of alive/1 are not together in the source-file
Warning: Current predicate: killed/1
Warning: Use ':- discontiguous alive/1.' to suppress this message
"John likes all kind of food."
likes(john,X) :- food(X).
"Apple and vegetable are food."
food(apple).
food(X) :- vegetable(X).
"Anything anyone eats and not killed is food."
food(X) :- eats(P,X), alive(P).
"Anil eats peanuts and still alive."
eats(anil,peanuts).
alive(anil).
"Harry eats everything that Anil eats."
eats(harry,X) :- eats(anil,X).
Asking
?- likes(john,X).
produces (in SWI Prolog)
70 ?- likes(john,X).
X = apple ;
ERROR: food/1: Undefined procedure: vegetable/1
Exception: (9) vegetable(_G19387424) ? fail
Exception: (8) food(_G19387424) ? fail
71 ?-
Fixing it with
:- dynamic(vegetable/1).
we get
75 ?- likes(john,X).
X = apple ;
X = peanuts ;
false.
and so, of course,
101 ?- likes(john,peanuts).
true .
/*facts*/
female(Sufia).
female(Bilkis).
female(Nasima).
female(Rabeya).
female(NWife).
female(Eva).
female(Rekha).
female(Shammi).
female(Shanta).
female(TWife).
female(NSWife).
female(Jui).
female(SHWife).
female(TSWife).
male(Anowar).
male(Shahidul).
male(Khazal).
male(Shafik).
male(Nahid).
male(Jahorul).
male(Naim).
male(Tanim).
male(Raiyan).
male(NSon).
male(JHusband).
male(Jubayer).
male(Shoddo).
male(TSon).
male(NSSon).
male(JSon).
male(SHSon).
male(TSSon).
parent(Anowar,Shahidul).
parent(Anowar,Nasima).
parent(Anowar,Shafik).
parent(Shahidul,Nahid).
parent(Shahidul,Eva).
parent(Nasima,Rekha).
parent(Nasima,Shammi).
parent(Nasima,Shanta).
parent(Shafik,Tanim).
parent(Shafik,Raiyan).
parent(Nahid,NSon).
parent(Rekha,Jui).
parent(Rekha,Jubayer).
parent(Shammi,Shoddo).
parent(Tanim,TSon).
parent(NSon,NSSon).
parent(Jui,JSon).
parent(Shoddo,SHSon).
parent(TSon,TSSon).
/*rules*/
mother(X,Y):-parent(X,Y),female(X).
father(X,Y):-parent(X,Y),male(X).
son(X,Y):-parent(X,Y),male(X).
daughter(X,Y) :- female(X),parent(X,Y).
sister(X,Y) :- female(Y),parent(Par,X),parent(Par,Y), X \= Y.
child(X, Y) :-parent(Y, X).
sibling(X,Y):-parent(Z,X),parent(Z,Y),X\=Y.
grandparent(X,Z):-parent(X,Y),parent(Y,Z).
firstCousin(X,Y):-grandparent(Z,X),grandparent(Z,Y),X\=Y.
secondCousin(X,Y):-grandparent(A,X),grandparent(B,Y),sibling(A,B),A\=B.
firstcousinOnceRemoved(X,Y):-child(X,Z),cousin(Z,Y).
firstcousinTwiceRemoved(X,Y):-grandparent(X,Z),cousin(Z,Y).
You need to quote every name, otherwise these symbols are variables, not atoms.
Otherwise, just lowercase the first letter. To illustrate this last option, using my pet project, here is the graph:
as obtained from this source
:- module(answer_so, [answer_so/0]).
:- use_module(genealogy).
answer_so :- genealogy(answer_so, 'Answer SO').
parent_child(P,C) :- parent(P,C).
female(sufia).
female(bilkis).
female(nasima).
female(rabeya).
female(nwife).
female(eva).
female(rekha).
female(shammi).
female(shanta).
female(twife).
female(nswife).
female(jui).
female(shwife).
female(tswife).
male(anowar).
male(shahidul).
male(khazal).
male(shafik).
male(nahid).
male(jahorul).
male(naim).
male(tanim).
male(raiyan).
male(nson).
male(jhusband).
male(jubayer).
male(shoddo).
male(tson).
male(nsson).
male(json).
male(shson).
male(tsson).
parent(anowar,shahidul).
parent(anowar,nasima).
parent(anowar,shafik).
parent(shahidul,nahid).
parent(shahidul,eva).
parent(nasima,rekha).
parent(nasima,shammi).
parent(nasima,shanta).
parent(shafik,tanim).
parent(shafik,raiyan).
parent(nahid,nson).
parent(rekha,jui).
parent(rekha,jubayer).
parent(shammi,shoddo).
parent(tanim,tson).
parent(nson,nsson).
parent(jui,json).
parent(shoddo,shson).
parent(tson,tsson).
Seems there are no marriages...
This is a really quick one, I don't know the correct way to google it!
Say we have an output such it gives:
X = [4|_1218]
what generally has to be changed to get the output just to say 4?
member1(Head, [Head|_]):-!.
member1(Item, [_|Tail]):-
member1(Item, Tail).
findnotin([Head|_], RuledOut, [Head|_]):-
not(member1(Head, RuledOut)).
Edit Completedish program:
find_not_in([], _, _).
find_not_in([Head|Candidates], RuleOut, [Head|Legal]):-
not(member1(Head, RuleOut)),
find_not_in(Candidates, RuleOut, Legal).
find_not_in([_|Candidates], RuleOut, Legal):-
find_not_in(Candidates, RuleOut, Legal).
I have some problem with my code here.
Let's say I have a knowledge base like this:
university('University of Cambridge', 1, 'United Kingdom', 90.3, 92.8, 89.4).
university('University of Oxford', 2, 'United Kingdom', 88.9, 94.8, 88.0).
university('ETH Zurich - Swiss Federal Institute of Technology', 3, 'Switzerland', 86.4, 94.4, 92.2).
university('University of Edinburgh', 4, 'United Kingdom', 83.7, 88.8, 83.6).
Then I call it out like this:
(Ignore the checkC(Country)/checkC(_))
checkC(Country):-
university(U, R, Country, _, _, _),nl,
write('University: '),write(U),nl,
write('Rank: '),write(R),nl,
write('Country: '),write(Country),nl,
fail; true, nl,
write('************'), nl,nl.
checkC(_):-
write('Country not found'),nl,nl.
My question is, why is it if the user inputs a random country name not within the knowledge base, the write('Country not found') won't come out, I found out that it has something to do with the fail; true.
Any help?
While the fail; true trick is nice for the command line, gathering multiple solutions is usually done via the predicates findall, bagof and setof.
I used format for readabilities sake.
checkC(X):-
findall( [U,R,X], university(U,R,X,_,_,_), List),
(List = [] ->
write('Country not found\n\n');
writeCountries(List)).
writeCountries([]).
writeCountries([X|R]):-
writeCountries(R),
format('University: ~w~nRank: ~w~nCountry: ~w~n~n',X).
If you are using swipl, maplist comes in handy:
checkC(X):-
findall( [U,R,X], university(U,R,X,_,_,_), List),
(List = [] ->
write('Country not found\n\n');
maplist(format('University: ~w~nRank: ~w~nCountry: ~w~n~n'),List)).
If you don't have format/2 look if you have writef/2 or printf/2.