I want to use atom_chars/2 on the expression of 3+4, but I get
ERROR: atom_chars/2: Type error: 'atom' expected, found '3+4' (a compound).
I'm thinking that if I can add " " on both sides of the compound, it would work, e.g.
atom_chars("3+4", Result).
but I don't know how I can do that, or is there other approaches to do this?
Please give me some advice.
EDIT: What I mean is that the input has to be 3+4, instead of '3+4', so what I want to do is to write a predicate before the atom_chars/2 to convert 3+4 to '3+4'.
For instance: for compound2atom(X,Y),
-?compound2atom(3+4,Y).
Y='3+4'.
If you are using SWI-Prolog, there is with_output_to/2 or format/3:
?- with_output_to(atom(A), write(3+4)).
A = '3+4'.
?- with_output_to(chars(C), write(3+4)).
C = ['3', +, '4'].
?- format(atom(A), "~w", [3+4]).
A = '3+4'.
?- format(chars(C), "~w", [3+4]).
C = ['3', +, '4'].
But if you look hard enough you should be able to find some predicate that does that, for example term_to_atom/2.
My personal preference leans towards format/3.
Related
Given following facts:
route(TubeLine, ListOfStations).
route(green, [a,b,c,d,e,f]).
route(blue, [g,b,c,h,i,j]).
...
I am required to find all the pairs of tube Lines that do not have any stations in common, producing the following:
| ?- disjointed_lines(Ls).
Ls = [(yellow,blue),(yellow,green),(yellow,red),(yellow,silver)] ? ;
no
I came up with the below answer, however it does not only give me incorrect answer, but it also does not apply my X^ condition - i.e. it still prints results per member of Stations lists separately:
disjointed_lines(Ls) :-
route(W, Stations1),
route(Z, Stations2),
setof(
(W,Z),X^
(member(X, Stations1),nonmember(X, Stations2)),
Ls).
This is the output that the definition produces:
| ?- disjointed_lines(L).
L = [(green,green)] ? ;
L = [(green,blue)] ? ;
L = [(green,silver)] ? ;
...
I believe that my logic relating to membership is incorrect, however I cannot figure out what is wrong. Can anyone see where am I failing?
I also read Learn Prolog Now chapter 11 on results gathering as suggested here, however it seems that I am still unable to use the ^ operator correctly. Any help would be appreciated!
UPDATE:
As suggested by user CapelliC, I changed the code into the following:
disjointed_lines(Ls) :-
setof(
(W,Z),(Stations1, Stations2)^
((route(W, Stations1),
route(Z, Stations2),notMembers(Stations1,Stations2))),
Ls).
notMembers([],_).
notMembers([H|T],L):- notMembers(T,L), nonmember(H,L).
The following, however, gives me duplicates of (X,Y) and (Y,X), but the next step will be to remove those in a separate rule. Thank you for the help!
I think you should put route/2 calls inside setof' goal, and express disjointness more clearly, so you can test it separately. About the ^ operator, it requests a variable to be universally quantified in goal scope. Maybe a concise explanation like that found at bagof/3 manual page will help...
disjointed_lines(Ls) :-
setof((W,Z), Stations1^Stations2^(
route(W, Stations1),
route(Z, Stations2),
disjoint(Stations1, Stations2)
), Ls).
disjoint(Stations1, Stations2) :-
... % could be easy as intersection(Stations1, Stations2, [])
% or something more efficient: early fail at first shared 'station'
setof/3 is easier to use if you create an auxiliary predicate that expresses the relationship you are interested in:
disjoint_routes(W, Z) :-
route(W, Stations1),
route(Z, Stations2),
disjoint(Stations1, Stations2).
With this, the definition of disjointed_lines/1 becomes shorter and simpler and no longer needs any ^ operators:
disjointed_lines(Ls) :-
setof((W, Z), disjoint_routes(W, Z), Ls).
The variables you don't want in the result of setof/3 are automatically hidden inside the auxiliary predicate definition.
I have to write a predicate to do work like following:
?- cat(north,south,X).
X = northsouth
?- cat(alley,'91',Y).
X = alley91
?-cat(7,uthah,H).
Bad Input
H = H
Please Help..
atom_concat_redefined(A1, A2, A3) :-
( nonvar(A1) -> atom_chars(A1, Chs1) ; true ),
( nonvar(A2) -> atom_chars(A2, Chs2) ; true ),
( nonvar(A1), nonvar(A2) -> true ; atom_chars(A3, Chs3) ),
append(Chs1, Chs2, Chs3),
atom_chars(A1, Chs1),
atom_chars(A2, Chs2),
atom_chars(A3, Chs3).
This definition produces the same errors in a standard conforming implementation like SICStus or GNU - there should be no other differences, apart from performance. To compare the errors use the goal:
?- catch(atom_concat_redefined(A,B,abc+1), error(E,_), true).
E = type_error(atom,abc+1).
Note the underscore in error(E,_), which hides the implementation defined differences. Implementations provide additional information in this argument, in particular, they would reveal that atom_chars/2 or atom_concat/3 produced the error.
atom_codes/2 it's the ISO approved predicate to convert between an atom and a list of codes. When you have 2 lists corresponding to first two arguments, append/3 (alas, not ISO approved, but AFAIK available in every Prolog), will get the list corresponding to third argument, then, convert that list to atom...
Note that, while append/3 is a 'pure' Prolog predicate, and can work with any instantiation pattern, atom_codes/2 requires at least one of it's argument instantiated. Here is a SWI-Prolog implementation of cat/3, 'working' a bit more generally. I hope it will inspire you to read more about Prolog...
ac(X,Xs) :- when((ground(X);ground(Xs)), atom_codes(X,Xs)).
cat(X,Y,Z) :- maplist(ac, [X,Y,Z],[Xs,Ys,Zs]), append(Xs,Ys,Zs).
edit
as noted by #false I was wrong about append/3. Now I'll try to understand better what append/3 does... wow, a so simple predicate, so behaviour rich!
I'm currently trying to to interpret user-entered strings via Prolog. I'm using code I've found on the internet, which converts a string into a list of atoms.
"Men are stupid." => [men,are,stupid,'.'] % Example
From this I would like to create a rule, which then can be used in the Prolog command-line.
% everyone is a keyword for a rule. If the list doesn't contain 'everyone'
% it's a fact.
% [men,are,stupid]
% should become ...
stupid(men).
% [everyone,who,is,stupid,is,tall]
% should become ...
tall(X) :- stupid(X).
% [everyone,who,is,not,tall,is,green]
% should become ...
green(X) :- not(tall(X)).
% Therefore, this query should return true/yes:
?- green(women).
true.
I don't need anything super fancy for this as my input will always follow a couple of rules and therefore just needs to be analyzed according to these rules.
I've been thinking about this for probably an hour now, but didn't come to anything even considerable, so I can't provide you with what I've tried so far. Can anyone push me into the right direction?
Consider using a DCG. For example:
list_clause(List, Clause) :-
phrase(clause_(Clause), List).
clause_(Fact) --> [X,are,Y], { Fact =.. [Y,X] }.
clause_(Head :- Body) --> [everyone,who,is,B,is,A],
{ Head =.. [A,X], Body =.. [B,X] }.
Examples:
?- list_clause([men,are,stupid], Clause).
Clause = stupid(men).
?- list_clause([everyone,who,is,stupid,is,tall], Clause).
Clause = tall(_G2763):-stupid(_G2763).
I leave the remaining example as an easy exercise.
You can use assertz/1 to assert such clauses dynamically:
?- List = <your list>, list_clause(List, Clause), assertz(Clause).
First of all, you could already during the tokenization step make terms instead of lists, and even directly assert rules into the database. Let's take the "men are stupid" example.
You want to write down something like:
?- assert_rule_from_sentence("Men are stupid.").
and end up with a rule of the form stupid(men).
assert_rule_from_sentence(Sentence) :-
phrase(sentence_to_database, Sentence).
sentence_to_database -->
subject(Subject), " ",
"are", " ",
object(Object), " ",
{ Rule =.. [Object, Subject],
assertz(Rule)
}.
(let's assume you know how to write the DCGs for subject and object)
This is it! Of course, your sentence_to_database//0 will need to have more clauses, or use helper clauses and predicates, but this is at least a start.
As #mat says, it is cleaner to first tokenize and then deal with the tokenized sentence. But then, it would go something like this:
tokenize_sentence(be(Subject, Object)) -->
subject(Subject), space,
be, !,
object(Object), end.
(now you also need to probably define what a space and an end of sentence is...)
be -->
"is".
be -->
"are".
assert_tokenized(be(Subject, Object)) :-
Fact =.. [Object, Subject],
assertz(Fact).
The main reason for doing it this way is that you know during the tokenization what sort of sentence you have: subject - verb - object, or subject - modifier - object - modifier etc, and you can use this information to write your assert_tokenized/1 in a more explicit way.
Definite Clause Grammars are Prolog's go-to tool for translating from strings (such as your English sentences) to Prolog terms (such as the Prolog clauses you want to generate), or the other way around. Here are two introductions I'd recommend:
http://www.learnprolognow.org/lpnpage.php?pagetype=html&pageid=lpn-htmlse29
http://www.pathwayslms.com/swipltuts/dcg/
I'd like to know why I get an error with my SWI Prolog when I try to do this:
(signal(X) = signal(Y)) :- (terminal(X), terminal(Y), connected(X,Y)).
terminal(X) :- ((signal(X) = 1);(signal(X) = 0)).
I get the following error
Error: trabalho.pro:13: No permission to modify static procedure
'(=)/2'
It doesn't recognize the "=" in the first line, but the second one "compiles". I guess it only accepts the "=" after the :- ? Why?
Will I need to create a predicate like: "equal(x,y) :- (x = y)" for this?
Diedre - there are no 'functions' in Prolog. There are predicates. The usual pattern
goes
name(list of args to be unified) :- body of predicate .
Usually you'd want the thing on the left side of the :- operator to be a predicate
name. when you write
(signal(X) = signal(Y))
= is an operator, so you get
'='(signal(X), signal(Y))
But (we assume, it's not clear what you're doing here) that you don't really want to change equals.
Since '=' is already in the standard library, you can't redefine it (and wouldn't want to)
What you probably want is
equal_signal(X, Y) :- ... bunch of stuff... .
or
equal_signal(signal(X), signal(Y)) :- ... bunch of stuff ... .
This seems like a conceptual error problem. You need to have a conversation with somebody who understands it. I might humbly suggest you pop onto ##prolog on freenode.net or
some similar forum and get somebody to explain it.
Because = is a predefined predicate. What you actually write is (the grounding of terms using the Martelli-Montanari algorithm):
=(signal(X),signal(Y)) :- Foo.
You use predicates like functions in Prolog.
You can define something like:
terminal(X) :- signal(X,1);signal(X,0).
where signal/2 is a predicate that contains a key/value pair.
And:
equal_signal(X,Y) :- terminal(X),terminal(Y),connected(X,Y).
I am writing a little Prolog program, which is expected to do something as following:
?-input([allan,is,a,name]).
true.
?-input([Is,allan,a,name]).
true.
and here is my code:
% Simple answering agent
input(Text) :-
phrase(sentence(S), Text),
perform(S).
sentence(statement(S)) --> statement(S).
sentence(query(Q)) --> query(Q).
statement(Statement) -->
[Thing, 'is', 'a', Category],
{ Statement =.. [Category, Thing]}.
query(Query) -->
['Is', Thing, 'a', Category],
{ Query =.. [Category, Thing]}.
perform(statement(S)) :- asserta(S).
perform(query(Q)) :- Q.
the input([allan,is,a,name]). part seems working fine, but there is a trouble with the query part, for which if I type in input([Is,allan,a,name])., it prints
Is = 'Is'
Can someone please take a look at this problem for me, thank you.
Well, the problem is that Is is a variable and thus prolog instantiates it (with 'Is'). It would be good practise to ensure that all the members of the list are atoms but for a quick fix you could just do:
query(Query) -->
[_, Thing, 'a', Category],
{ Query =.. [Category, Thing]}.
this way, Is won't get instantiated and prolog will just say true. The only problem is that a statement could be interpreted as a query:
9 ?- input([allan, is, a, name]).
true ;
false.
10 ?- input([is, is, a, name]).
true .
11 ?- input([allan, is, a, name]).
true ;
true.
which can be fixed with some cuts (or be saying that Thing should be different than 'is' - if that's acceptable)
Edit: for a more general solution: it really depends on what kind of sentences you want to parse and what compromises the user can make. For example, it might be possible to ask him to give you words that are prolog atoms; if a word that starts with an upper case letter is requested he will have to use ''. Otherwise, it will be better to just give them in a string/atom ('Is allan a name' or "Is allan a name"). It's easy to separate it to individual atoms: use atomic_list_concat/3. For what is allan you still don't need to do something special; it's a 3 word sentence while the rest were 4 so you can separate it immediately.