I am trying to check if a string is a working call that can be executed. To do this I parse the string, get the first word, and if it matches a database of predefined functions, it should succeed.
Q has the string, A will be used later, not now. Example of string is: append a and b.
is_uni(Q, A):-
split_string(Q, " ", ",", [X|Y]),
uni_db(Z),
member(X, Z).
uni_db([
append,
member,
append1
]).
You need to use atom_codes/2 predicate to convert stings to atoms, in example you need to convert "append" to append in order to work.
is_uni(Q,A):-
split_string(Q, " ", ",", [X|Y]),
atom_codes(W,X),
uni_db(Z),
member(W, Z).
Example:
?- is_uni("append a and b",A).
true ;
false.
You are confusing strings with atoms.
"append" and 'append', a.k.a. append, are different. You can use atom_string/2 to convert between them:
..., atom_string(A, X), ...
You are re-implementing built-in features.
Why store commands in a list and iterate with member/2? Just define some facts:
uni_db(append).
uni_db(member).
uni_db(append1).
And then, you simply have to check whether uni_db(A). This is better supported by an implementation and more likely to be done efficiently.
Related
I want to parse a logical expression using DCG in Prolog.
The logical terms are represented as lists e.g. ['x','&&','y'] for x ∧ y the result should be the parse tree and(X,Y) (were X and Y are unassigned Prolog variables).
I implemented it and everything works as expected but I have one problem:
I can't figure out how to parse the variable 'x' and 'y' to get real Prolog variables X and Y for the later assignment of truth values.
I tried the following rule variations:
v(X) --> [X].:
This doesn't work of course, it only returns and('x','y').
But can I maybe uniformly replace the logical variables in this term with Prolog variables? I know of the predicate term_to_atom (which is proposed as a solution for a similar problem) but I don't think it can be used here to achieve the desired result.
v(Y) --> [X], {nonvar(Y)}.:
This does return an unbound variable but of course a new one every time even if the logical variable ('x','y',...) was already in the term so
['X','&&','X'] gets evaluated to and(X,Y) which is not the desired result, either.
Is there any elegant or idiomatic solution to this problem?
Many thanks in advance!
EDIT:
The background to this question is that I'm trying to implement the DPLL-algorithm in Prolog. I thought it would by clever to directly parse the logical term to a Prolog-term to make easy use of the Prolog backtracking facility:
Input: some logical term, e.g T = [x,'&&',y]
Term after parsing: [G_123,'&&',G_456] (now featuring "real" Prolog variables)
Assign a value from { boolean(t), boolean(f) } to the first unbound variable in T.
simplify the term.
... repeat or backtrack until a assignment v is found so that v(T) = t or the search space is depleted.
I'm pretty new to Prolog and honestly couldn't figure out a better approach. I'm very interested in better alternatives! (So I'm kinda half-shure that this is what I want ;-) and thank you very much for your support so far ...)
You want to associate ground terms like x (no need to write 'x') with uninstantiated variables. Certainly that does not constitute a pure relation. So it is not that clear to me that you actually want this.
And where do you get the list [x, &&, x] in the first place? You probably have some kind of tokenizer. If possible, try to associate variable names to variables prior to the actual parsing. If you insist to perform that association during parsing you will have to thread a pair of variables throughout your entire grammar. That is, instead of a clean grammar like
power(P) --> factor(F), power_r(F, P).
you will now have to write
power(P, D0,D) --> factor(F, D0,D1), power_r(F, P, D1,D).
% ^^^^ ^^^^^ ^^^^
since you are introducing context into an otherwise context free grammar.
When parsing Prolog text, the same problem occurs. The association between a variable name and a concrete variable is already established during tokenizing. The actual parser does not have to deal with it.
There are essentially two ways to perform this during tokenization:
1mo collect all occurrences Name=Variable in a list and unify them later:
v(N-V, [N-V|D],D) --> [N], {maybesometest(N)}.
unify_nvs(NVs) :-
keysort(NVs, NVs2),
uniq(NVs2).
uniq([]).
uniq([NV|NVs]) :-
head_eq(NVs, NV).
uniq(NVs).
head_eq([], _).
head_eq([N-V|_],N-V).
head_eq([N1-_|_],N2-_) :-
dif(N1,N2).
2do use some explicit dictionary to merge them early on.
Somewhat related is this question.
Not sure if you really want to do what you asked. You might do it by keeping a list of variable associations so that you would know when to reuse a variable and when to use a fresh one.
This is an example of a greedy descent parser which would parse expressions with && and ||:
parse(Exp, Bindings, NBindings)-->
parseLeaf(LExp, Bindings, MBindings),
parse_cont(Exp, LExp, MBindings, NBindings).
parse_cont(Exp, LExp, Bindings, NBindings)-->
parse_op(Op, LExp, RExp),
{!},
parseLeaf(RExp, Bindings, MBindings),
parse_cont(Exp, Op, MBindings, NBindings).
parse_cont(Exp, Exp, Bindings, Bindings)-->[].
parse_op(and(LExp, RExp), LExp, RExp)--> ['&&'].
parse_op(or(LExp, RExp), LExp, RExp)--> ['||'].
parseLeaf(Y, Bindings, NBindings)-->
[X],
{
(member(bind(X, Var), Bindings)-> Y-NBindings=Var-Bindings ; Y-NBindings=Var-[bind(X, Var)|Bindings])
}.
It parses the expression and returns also the variable bindings.
Sample outputs:
?- phrase(parse(Exp, [], Bindings), ['x', '&&', 'y']).
Exp = and(_G683, _G696),
Bindings = [bind(y, _G696), bind(x, _G683)].
?- phrase(parse(Exp, [], Bindings), ['x', '&&', 'x']).
Exp = and(_G683, _G683),
Bindings = [bind(x, _G683)].
?- phrase(parse(Exp, [], Bindings), ['x', '&&', 'y', '&&', 'x', '||', 'z']).
Exp = or(and(and(_G839, _G852), _G839), _G879),
Bindings = [bind(z, _G879), bind(y, _G852), bind(x, _G839)].
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 would like to determine in Prolog the type of a string of characters, if it is alphabetic, alphanumeric or numeric.
For example:
"I use this page" alphabetic
"0c0d24e" alphanumeric
How can i do?
the predicate available is char_type/2, or better, code_type/2.
To apply to each code in string, use maplist/2. The only problem it's the wrong arguments order of code_type. Then a service predicate is needed (or download lambda, if you're using SWI-Prolog, with ?- pack_install(lambda).).
Without lambda:
code_type_(X,Y) :- code_type(Y,X).
?- maplist(code_type_(alpha), "abc").
true.
With lambda:
?- [library(lambda)].
?- maplist(\C^code_type(C,alpha), "abc").
true.
edit after comments, it's apparent that more flexible parsing is required. A DCG it's the recommended way to go: library(dcg/basics) offers some prebuilt 'categorizer', and highlights the proper way to write your own, combining with code_type: for instance, here is a recently added rule:
%% prolog_var_name(-Name:atom)// is semidet.
%
% Matches a Prolog variable name. Primarily intended to deal with
% quasi quotations that embed Prolog variables.
prolog_var_name(Name) -->
[C0], { code_type(C0, prolog_var_start) }, !,
prolog_id_cont(CL),
{ atom_codes(Name, [C0|CL]) }.
prolog_id_cont([H|T]) -->
[H], { code_type(H, prolog_identifier_continue) }, !,
prolog_id_cont(T).
prolog_id_cont([]) --> "".
see how code_type/2 is used to qualify single characters...
more edit - note: untested
qualify_atom(Atom, Type) :-
atom_codes(Atom, Codes),
qualify_codes(Codes, Type).
qualify_codes(Codes, Type) :-
( maplist(code_type_(alnum), Codes)
-> Type = alnum
; maplist(code_type_(alpha), Codes)
-> Type = alpha
; Type = unknown
).
then, to work on a list
?- maplist(qualify_atom, Atoms, Types).
edit
An update of this answer is mandatory: since library(yall) has been released in SWI-Prolog, and is autoloaded, we can now write:
?- maplist([C]>>code_type(C,alpha), `abc`).
Also, note the change in literal representation: double quotes in SWI-Prolog ver.7+ don't represent anymore a list of character codes.
I have a list L =[a+b,b+c] and I want to convert it to a string and print the output a+bb+c.
Can you please help me convert this list to a string? I tried using atomic_list_concat in SWI-Prolog but it gave a type error for a+b.
In SWI-Prolog:
?- with_output_to(atom(Atom), maplist(write, [a+b, b+c])).
Atom = 'a+bb+c'.
You can replace write with a call to a custom predicate if you need more control over how your terms (e.g. a+b) are written.
The members of your list are compound terms, so you have to make them atomic before calling atomic_list_concat:
custom_print(L) :-
maplist(term_to_atom, L, L1),
atomic_list_concat(L1, L2),
print(L2).
go :- match(Mn,Fn),
write('--Matching Result--'),
nl,
write(Mn),
write(' match with '),
write(Fn),
match(Mn1,Fn1).
person(may,female,25,blue).
person(rose,female,20,blue).
person(hock,male,30,blue).
person(ali,male,24,blue).
match(Mn,Fn):-person(Fn,'female',Fage,Fatt),
person(Mn,'male',Mage,Matt),
Mage>=Fage,
Fatt=Matt.
Hi,this is my code...but it's only can show the 1 output...but there are 3 pair of matching in match(X,Y).how to show them all in my go function.
Thank you
You get all your matches if you force backtracking, usually by entering ; (e.g. in SWI Prolog). But you also see that you are getting unnecessary outputs true. This is because the last clause in go is match(Mn1,Fn1). This clause succeeds three times and binds the variables Mn1,Fn1 but then only true is output, because you do not write() after that clause. The fourth time match(Mn1,Fn1) fails and by backtracking you come back to the first clause match(Mn,Fn) that matches, the match is output, etc.
You surely do not want to have this behavior. You should remove the last clause match(Mn1,Fn1) in go. Now by pressing ; you get the 3 matches without any output true in between.
But what you likely want is that the program does the backtracking. To achieve this, you just need to force backtracking by adding false as the last clause. To get proper formatting of the output, use the following program. The last clause go2. is added to get true at the very end.
go2 :- write('--Matching Result--'), nl,
match(Mn,Fn),
write(Mn), write(' match with '), write(Fn), nl,
fail.
go2.
This technique is called failure driven loop.
If you have any predicate that has multiple results and want to to find all of them, you should use findall/3
For example, in your case, you could do something like:
findall([X,Y], match(X,Y),L).
L will be a list that will contain all the X,Y that satisfy match(X,Y) in the format [X,Y].
for example, assuming that:
match(m1,f1).
match(m2,f2).
the result will be L = [ [m1,f1], [m2,f2] ]
note that you can define the format as you wish, for example you could write:
findall(pair(X,Y), match(X,Y), L).
L = [ pair(m1,f1), pair(m2,f2) ]
findall( X, match(X,Y), L).
L = [ m1, m2]
findall( 42, match(X,Y), L).
L = [42, 42]
then you have to recurse on the list to print them.
However, if you wish to find one result, run some code and then continue you could use forall/2:
forall(match(X,Y), my_print(X,Y).
Prolog is a lazy language. Which means that it will stop once it has found a condition that made your problem true. This will be the very first match alone.
IF your code is working (I haven't tried it), then you should try and run the match-statement like this in your prolog inspector: match(X,Y)
The prolog inspector will return all states and print them for you.