how to solve the error in this prolog code - prolog

What is wrong in this code
frequency(f1).
frequency(f2).
frequency(f3).
allocation(z1,z2,z3,z4,z5,z6,z7,z8) :-
frequency(z1), frequency(z2), frequency(z3), frequency(z4), frequency(z5), frequency(z6), frequency(z7), frequency(z8),
z1\==z2, z1\==z3, z2\==z4, z2\==z3, z3\==z4, z3\==z6, z4\==z5, z4\==z6, z4\==z7, z6\==z7, z6\==z8.
?- allocation(f1,f2,f3,f1,f2,f3,f2,f3).
syntax error: . or operator expected after expression


The following code highlighting should make it clear why the goal allocation(f1,f2,f3,f1,f2,f3,f2,f3) cannot succeed:
allocation(Z1,Z2,Z3,Z4,Z5,Z6,Z7,Z8) :-
dif(Z1,Z2), dif(Z1,Z3),
dif(Z2,Z3), dif(Z2,Z4),
dif(Z3,Z4), dif(Z3,Z6),
dif(Z4,Z5), dif(Z4,Z6), dif(Z4,Z7),
dif(Z6,Z7), dif(Z6,Z8),
maplist(frequency, [Z1,Z2,Z3,Z4,Z5,Z6,Z7,Z8]).
Note the use of dif/2 instead of (\==)/2 in the code above?
prolog-dif preserves logical-purity, for easier debugging!

Related

How to repair error in Einstein Zebra in prolog >

I have a problem with zebra puzzle.
rightOf(A,B,(B,A,_,_,_)).
rightOf(A,B,(_,B,A,_,_)).
rightOf(A,B,(_,_,B,A,_)).
rightOf(A,B,(_,_,_,B,A)).
middleHouse(A,(_,_,A,_,_)).
firstHouse(A,(A,_,_,_,_)).
nextTo(A,B,(A,B,_,_,_)).
nextTo(A,B,(_,A,B,_,_)).
nextTo(A,B,(_,_,A,B,_)).
nextTo(A,B,(_,_,_,A,B)).
nextTo(A,B,(B,A,_,_,_)).
nextTo(A,B,(_,B,A,_,_)).
nextTo(A,B,(_,_,B,A,_)).
nextTo(A,B,(_,_,_,B,A)).
house[Nationality,Pet,Sport,Drinks,Colour].
zebra_owner(Owner) :-
houses(Houses),
exists(house(Owner,zebra,_,_,_), Houses).
length(5,Houses),
exists(house(british,_,_,_,red),Houses),
exists(house(spanish,dog,_,_,_),Houses),
exists(house(_,_,_,coffe,green),Houses),
exists(house(ukrainian,_,_,tea,_),Houses),
rightOf(house(_,_,_,_,green),house(_,_,_,_,ivory),Houses),
exists(house(_,snail,tennis,_,_),Houses),
exists(house(_,_,chess,_,yellow),Houses),
middleHouse(house(_,_,_,mlik,_),Houses),
firstHouse(house(norwegian,_,_,_,_),Houses),
nextTo(house(_,_,rugby,_,_),house(_,fox,_,_,_),Houses),
nextTo(house(_,_,chess,_,_),house(_,horse,_,_,_),Houses),
exists(house(_,_,volleyball,orangejuice,_),Houses),
exists(house(japanese,_,_,_,_),Houses),
nextTo(house(norwegian,_,_,_,_),house(_,_,_,_,blue),Houses),
nextTo(house(_,_,_,tea,_),house(_,_,_,milo,_),Houses),
exists(house(zebra_owner,zebra,_,_,_),Houses).
Under
house[Nationality,Pet,Smokes,Drinks,Colour].
I see the error: Singleton variables: house[Nationality,Pet,Smokes,Drinks,Colour].
And when I write ?- zebra_owner(Owner)
I see the error: procedure 'houses(A)' does not exist. Reachable from:zebra_owner(A).`

You cannot write
house[Nationality,Pet,Sport,Drinks,Colour].
that's non-existent syntax.
To express that predicate house/1 is valid for list [Nationality,Pet,Sport,Drinks,Colour], write:
house([Nationality,Pet,Sport,Drinks,Colour]).
You also must pay attention to predicate naming: houses/1 is not house/1. Predicate houses/1 is defined nowhere (the error message is wrong, it should really say predicate houses/1 is undeclared).

Prolog syntax error: illegal start of term

So I'm trying to test my the predicate reduce/4 defined as following:
:- op(20,xfy,?=).
reduce(R,X ?=Y,P,Q) :-
R=rename,
regle(X ?=Y,R),
length(P,N),
trouve(Y,P,N,M),
remove(N,M,P,Q).
but i keep getting this error:
?- reduce(rename,X ?= Y,[?=(X,Y),?=(f(a),g(Z)),?=(X,a)],Q].
ERROR: Syntax error: Illegal start of term
ERROR: reduce(rename,X ?= Y,[?=(X,Y),?=(f(a),g(Z)),?=(X,a)],
ERROR: ** here **
ERROR: Q] .
I'm new to Prolog so maybe it is something stupid.

You use ] instead of ) to close the bracket of reduce
reduce(rename,X ?= Y,[?=(X,Y),?=(f(a),g(Z)),?=(X,a)],Q].
% ^open ^close
You can fix this with:
?- reduce(rename,X ?= Y,[?=(X,Y),?=(f(a),g(Z)),?=(X,a)],Q).
I cannot validate if this solves your problem, because you did not provide a definition for regle/2, etc. But the fact that I get a semantic error, seems to solve the syntax error nevertheless.

I am not sure as to why 'find_identity(X)' is returning as false [Code Below]

Snippets of my code are shown below. After doing a lot of cumbersome tracing, I have reached a point where I think it's the external modules which are the source of my problem, as the findall in links_of_actor never gets called, but links_of_actor itself gets called. I would be very grateful to anyone who can help and can clarify/add more of my code to this question if needed.
find_identity(A) :-
all_actor_links(ALs),
find_identity(ALs,[],A).
find_identity([a(A,_)],_,A).
find_identity(ALs,Ms,A) :-
agent_ask_oracle(oscar,o(1),link,L),
( memberchk(L,Ms) -> find_identity(ALs,Ms,A)
; otherwise -> purge_actors(ALs,L,NewALs),
find_identity(NewALs,[L|Ms],A)
).
links_of_actor(A,Ls) :-
actor(A),
wp(A,WT),
findall(L,wt_link(WT,L),Ls1),
findall(L,link(L),Ls2),
intersection(Ls1,Ls2,Ls).
actor_links(A,a(A,Ls)) :-
links_of_actor(A,Ls).
all_actor_links(ALs) :-
findall(A,actor(A),As),
maplist(actor_links,As,ALs).
----------------------------------------------------- Support Functions ---------------------------------------------------
% wp(Q,WT) <- issue query Q to Wikipedia and return the page in wikitext format
wp(Q,WT):-
wp_cache(Q,WT),!.
wp(Q,WT):-
wp_query2URL(Q,URL),
http_get(URL,R,[]),
atom_json_term(R,RR,[]),
wt_get(RR,WT0),
( atomic_list_concat1(_L,'#REDIRECT',WT0) -> wt_link(WT0,QQ),wp(QQ,WT)
; otherwise -> WT=WT0
),
assert(wp_cache(Q,WT)).
------------------------------------------------------------ Edit ----------------------------------------------------------------
After using the guitracer that is available with prolog, I have found that the program fails at http_get(URL,R,[]) in the wp(Q,WT) predicate. However, I am still not sure as to why this isn't working - Could it be possibly because of my internet?
To clarify, the predicate actor is defined in my file: actor('Billy Bob Thornton'). etc. and so are the links : link('Barack Obama')..

Using these definitions, the error can be localized by adding a # in front of those goals that have to succeed.
...
wp(Q,WT):-
#wp_query2URL(Q,URL),
#http_get(URL,R,[]),
#atom_json_term(R,RR,[]),
#wt_get(RR,WT0),
...

Well, did you try to see if the URL you gave to http_open is actually valid? You can always test from the top-level if you can http_open an URL:
?- use_module(library(http/http_client)).
true.
?- http_open("stackoverflow.com", R, []).
R = % a lot of stuff

prolog get syntax error when increase stack size

Trying to solve the puzzle task with prolog and got some problems.
1002 Stack Overflow. Re-configure with Setup if necessary.
So, I've tried to increase stack size in setup and run program again.
But it causes the other error: Syntax error on line...
The error line is line with operator "not" in predicate.
Here is my code:
domains
age = integer
childname,ffood,fear = string
child = child(childname,age,ffood,fear)
children = child*
predicates
solve
name(child,childname)
fear(child,fear)
age(child,age)
ffood(child,ffood)
keys(children)
solution(children)
elder(child)
member(children,child)
structure(children)
clauses
member([X|_],X).
member([_|SP],X):-member(SP,X).
name(child(A,_,_,_),A).
age(child(_,A,_,_),A).
ffood(child(_,_,A,_),A).
fear(child(_,_,_,A),A).
structure([child("Dima",_,_,_),child("Kate",_,_,_),child("Misha",_,_,_),child("Sveta",_,_,_),child("Ura",_,_,_)]).
elder(child(_,A,_,_)):-A=7;A=8.
solve:-structure(Children),keys(Children),solution(Children).
keys(Struct):-
member(Struct,child(_,4,_,_)),
member(Struct,child(_,5,_,_)),
member(Struct,child(_,6,_,_),
member(Struct,child(_,7,_,_),
member(Struct,child(_,8,_,_)),
member(Struct,child(_,_,"Banana",_),
member(Struct,child(_,_,"Icecream",_),
member(Struct,child(_,_,"Pizza",_),
member(Struct,child(_,_,"Pasta",_),
member(Struct,child(_,_,"Chocolate",_),
member(Struct,child(_,_,_,"Thunderstorm"),
member(Struct,child(_,_,_,"Spiders"),
member(Struct,child(_,_,_,"Ghosts"),
member(Struct,child(_,_,_,"Dogs"),
member(Struct,child(_,_,_,"Darkness"),
member(Struct,Child1),
name(Child1,"Kate"),
elder(Child1),
not(fear(Child1,"Darkness")),
not(ffood(Child1,"Chocolate")),
member(Struct,Child2),
name(Child2,"Sveta"),
elder(Child2),
not(fear(Child2,"Darkness")),
not(ffood(Child2,"Chocolate")),
ffood(Child2,"Pizza"),
not(fear(Child2,"Spiders")),
member(Struct,Child3),
age(Child3,5),
fear(Child3,"Ghosts"),
member(Struct,Child4),
age(Child4,6),
fear(Child4,"Thunderstorm"),
not(ffood(Child4,"Chocolate")),
not(ffood(Child4,"Pasta")),
member(Struct,Child5),
age(Child5,4),
ffood(Child5,"Banana"),
member(Struct,Child6),
age(Child6,8),
not(fear(Child6,"Dogs")),
member(Struct,Child7),
name(Child7,"Dima"),
not(age(Child7,5)),
not(fear(Child7,"Darkness")),
not(fear(Child7,"Spiders")),
not(ffood(Child7,"Banana")),
member(Struct,Child8),
name(Child8,"Misha"),
not(fear(Child8,"Darkness")),
not(fear(Child8,"Spiders")),
not(ffood(Child8,"Banana")).
solution (Children):-
write ("Solve:"), write (Children).
goal
solve.
Found strange this prolog behavior... maybe somebody had the same problem?

I tried to reformat your code with SWI-Prolog:
keys(Struct):-
member(Struct,child(_,4,_,_)),
member(Struct,child(_,5,_,_)),
member(Struct,child(_,6,_,_),
member(Struct,child(_,7,_,_),
member(Struct,child(_,8,_,_)),
member(Struct,child(_,_,"Banana",_),
...
seems you're missing some parenthesis...
after the obvious correction, I get
?- solve.
Solve:[child(Dima,6,Icecream,Thunderstorm),child(Kate,8,Pasta,Spiders),child(Misha,5,Chocolate,Ghosts),child(Sveta,7,Pizza,Dogs),child(Ura,4,Banana,Darkness)]
true .

Simple error for someone who knows prolog. There is no fullstop to be causing the error

likes(a,b).
likes(b,d).
likes(c,e).
likes(d,f).
likes(e,h).
likes_trans(X,Y) :- likes(X,Z),likes_trans(Z,Y).
likes_trans(X,Y) :- likes(X,Y).
listfriends(X,R) :- findall(Y,likes_trans(X,Y),R).
likes_both(X,Y,R) := listfriends(X,P), listfriends(Y,S), member(R,P), member(S,P).
Why does the last line 'likes_both' give me the error (when loading the file)
Full stop in clause-body? Cannot redefine
Any suggestions much appreciated, I know I'm missing something fundamental here.

It looks like the := on the last line should be replaced with :-
likes_both(X,Y,R) :- listfriends(X,P), listfriends(Y,S), member(R,P), member(S,P).
/* Here ---------^^ */

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