How can I check that a particular constant exists in Prolog? - prolog

Say I have the following:
person(james).
person(pete).
father(james, pete).
Is there a way to validate that both the arguments to father have been defined i.e to stop a typo such as father(jmes, pete).

There is an easy way of checking these kinds of errors using Prolog: You can call rules at initialization time, i.e. one that checks that your fathers relations is properly defined. This can be useful to catch those typo errors.
:- use_module(library(error)).
check_fathers :-
% Take an element of the father relation
(father(P, _); father(_, P)),
% Proceed, if P is not a person
\+ person(P),
% Throw an error
syntax_error(father_is_no_person(P)).
check_fathers.
:- check_fathers.
person(james).
person(pete).
father(jame, pete). % TYPO! jame instead of james
Then, the program will produce the following output:
ERROR: Syntax error: father_is_no_person(jame)
Warning: <filename>:<line>:
Warning: Goal (directive) failed: user:check_fathers

Your assumptions are wrong in this sense:
father(james, pete). is a fact. It is something that you tell the Prolog processor is uneniably true (similar to a row in a table in a relational database).
So is father(jmes, pete).. Another fact. You are stating that this is so.
There is nothing to check, really.
On the other hand, if father(jmes, pete) appears in a a goal position, the Prolog processor will say "no, I have no indiciation that this is true":
father(james, pete). % that's a fat fact!
?- father(jmes, pete). % that's a query
false. % and the answer is "no, there is no evidence of that"
What you you would like to have is to have something like Java enum types: have the compiler make sure that a keyword comes indeed from a selected set of allowed keywords.
Prolog has has only basic typing, so this cannot be done directly.
But you could do this:
% if X is an unbound variable or a member of the indicated list,
% we are good
allowed_father(X) :-
(var(X);memberchk(X,[james,vader])),!.
% otherwise, we don't just fail, we actually throw an exception
allowed_father(X) :-
throw(error(unknown_father(X))).
% ---
% if X is an unbound variable or a member of the indicated list,
% we are good
allowed_son(X) :-
(var(X);memberchk(X,[pete,luke])),!.
% otherwise, we don't just fail, we actually throw an exception
allowed_father(X) :-
throw(error(unknown_son(X))).
and then you can call the above predicates whenever you need to perform a check:
query_anout(Father) :-
allowed_father(Father),
...
But it's rather awkward. Plus the thrown exception is not ISO-standard and the resulting error message may be confusing (I always rage against ISO standard exceptions which are doubleplusungood awkward)

Related

How to check whether some variable returns something in predicate

Lets assume I have facts as follows:
airport(nyc,'newyork').
I want want to display a message if the user inputs an airport that doesn't exist.
My Attempt:
isAirport(Air) :-
airport(Air,T),
(var(T) -> true
;
write('Airport not found'),
fail
).
However, this doesn't seem to work.
First let's see what happens if you query a conjunction (the , operator) first:
?- airport(nyc, _), write('found!').
found!
true.
?- airport(abc, _), write('found!').
false.
This means, isAirport(abc) directly fails after trying airport(abc,_) without the rest of your predicate being evaluated. In many cases, you can therefore get by without an explicit if-then-else construct and just write something of the form
predicate(X) :-
first_condition(X),
second_condition(X).
and it will only succeed if both conditions are fulfilled for X.
In case you really want to create some user interface, this is a bit more tricky, because I/O is inherently non-logical, in particular when there is backtracking involved. We usually call a program which behaves like we would expect from a logical formula pure and when it contains non-logical constructs like I/O or the cut operator ! are called impure.
Unfortunately, the if-then-else construct (-> and ;) and negation (\+) are implemented via cut and therefore impure as well. Luckily, most of the time people want a conditional, a pure disjunction is sufficient:
case(1,a).
case(2,b).
We have an automatic branching from the execution mechanism of Prolog:
?- case(X,Y).
X = 1,
Y = a ;
X = 2,
Y = b.
But sometimes we really want to do something that needs the impure constructs, like user input. Then the easiest way to keep the nice logical properties of our program is to separate the task into pure and impure ones:
main :-
uinput(Data),
pure_predicate(Data, Result),
write(Result).
After we have done all the impure parts, Data is unified with the user data we wanted. Let's have a look at the implementation of uinput/1:
uinput(data(Airport,D-M-Y)) :-
format('~nAirport? '),
read(Airport),
( ground(Airport), airport(Airport, _) )
->
(
format('~nDay? '),
read(D),
format('~nMonth? '),
read(M),
format('~nYear? '),
read(Y),
( ground(D-M-Y), isDate(D-M-Y) )
->
true
;
throw(failure('unknown date'))
)
;
throw(failure('unknown airport'))
.
We successively read terms from the input and throw an exception if we can't handle it. For the if-then-else construct to work, we need to take special care. If we compare the two queries:
?- between(1,3,X), write(X).
1
X = 1 ;
2
X = 2 ;
3
X = 3.
and
?- between(1,3,X) -> write(X); false.
1
X = 1.
you can see that the if-then-else is losing solutions. This means we need to make sure that our condition is deterministic. Asking for a user input term to be ground is already a good idea, because without variables, there is only one solution term. Still, a call to one of the data-predicates airport/1 and isDate/1 might generate the same term multiple times or not terminate at all. In this particular case, we just need to make sure that each airport has a unique shortcut name, we can also generate dates without repetition:
airport(nyc, 'New York City').
airport(wdc, 'Washington DC').
isDate(X-Y-Z) :-
between(1,31,X),
between(1,12,Y),
between(1970,2100,Z).
Another trick in the implementation of uinput is that we just succeed with true when we have validated everything. The only effect of is now that Data is instantiated with whatever the user entered.
If we give a dummy implementation of the actual implementation, we can already try the implementation oursevles:
pure_predicate(_Data, Result) :-
% here goes the actual stuff
Result='we have found something awesome'.
On the prompt we can use the pure predicate without trouble:
?- pure_predicate(someinputdata,Y).
Y = 'we have computed something awesome'.
On the other hand, we can also use the full predicate as follows:
?- main(_).
Airport? wdc.
Day? |: 1.
Month? |: 2.
Year? |: 2000.
we have found something awesome
true.
Since we are using read, we have to input prolog terms and terminate with a dot ., but everything worked as expected.
In case the user input fails, we get:
?- main(_).
Airport? bla(X).
ERROR: Unhandled exception: failure('unknown airport')
Please note that we only went through this trouble to actually fail early and give a user message in that case. For the actual computation, this is completely unneccessary.
In the code below you are making false assumption that T will remain unbound in case if airport will not be found in database:
airport(Air, T)
What happens actually is that call to airport(Air, T) will make isAirport(Air) to fail immediately and your var(T) and other parts will not be executed at all.
Try this code instead:
isAirport(Air) :-
airport(Air, _T), ! ; write('Airport not found'), fail.

On ways to work around an unexpected "Undefined procedure" error

NB: Just to be clear, my motivation for the question below is to learn my way around Prolog and SWI-Prolog, not to get past a particular error message. In fact, I already know one way to get past this error. My question asks about whether several other alternatives are also possible.
An exercise in my Prolog textbook asks one to describe the outcome one should expect from several queries, assuming one has consulted the following knowledgebase beforehand:
x(a).
z(b).
z(c).
z(d).
w(V) :- x(V).
w(V) :- y(V).
w(V) :- z(V).
On SWI-Prolog, at least, most of these queries fail, because SWI-Prolog intreprets y as undefined.
From the solutions to the exercises at the end of the book I can tell that this is not the authors' intended outcome. Maybe there's a Prolog implementation for which the exercise would turn as the solution presents it.
Be that as it may, I'd like to learn about good ways to work around the problem.
Consider, for example, the query w(x).. the book's solution claims that the query w(x). should evaluate to false.. In fact, this is what happens:
?- w(x).
ERROR: w/1: Undefined procedure: y/1
Exception: (7) y(x) ?
(At this point, SWI-Prolog is expecting me to enter some letter indicating how to respond to the exception. More about this later.)
I'm looking for ways to either turn the interaction above to
?- w(x).
false.
?-
...or at least for a suitable <ONE-LETTER RESPONSE> I could give to SWI-Prolog so that it arrives at the conclusion false. IOW, so that
?- w(x).
ERROR: w/1: Undefined procedure: y/1
Exception: (7) y(x) ? <ONE-LETTER RESPONSE>
false.
?-
I know of at least one answer to my question, namely simply to delete or comment out the line:
w(V) :- y(V).
I would like to know of other possible solutions, such as, for example, the "suitable " I alluded to earlier.
Another possibility would be some SWI-Prolog global configuration that would result in the above interaction to change to
?- w(x).
false.
?-
A third possibility would be to "define" y in some minimal way. The only way I can come up with is by adding the fact
y(dummy).
to the knowledgebase. Is there a more minimal way to define y, one that does not require introducing an extraneous constant into the domain of discourse?
(This is not specific to SWI)
The first Prolog systems back in the 1970s actually behaved in the way you describe. Soon it became apparent that this is a frequent source of errors. Simple misspellings remained undetected for too long. Current implementations produce a clean existence error. This is standard behaviour since 1995.
However, you can go back into the olden tymes with the ISO Prolog flag unknown which has three values error (default), fail, and warning.
?- inex.
ERROR: Undefined procedure: inex/0 (DWIM could not correct goal)
?- set_prolog_flag(unknown, fail).
Warning: Using a non-error value for unknown in the global module
Warning: causes most of the development environment to stop working.
Warning: Please use :- dynamic or limit usage of unknown to a module.
Warning: See http://www.swi-prolog.org/howto/database.html
true.
?- inex.
false.
?- set_prolog_flag(unknown, warning).
Warning: Using a non-error value for unknown in the global module
Warning: causes most of the development environment to stop working.
Warning: Please use :- dynamic or limit usage of unknown to a module.
Warning: See http://www.swi-prolog.org/howto/database.html
true.
?- inex.
Warning: toplevel: Undefined procedure: inex/0 (DWIM could not correct goal)
false.
As you can read above, SWI proposes to use a dynamic declaration in stead - which in turn has its own problems... It is much better to declare instead:
:- discontiguous(y/1).
An undefined procedure error raises an exception so if you want the exception to be raised because you don't want to change y/1 predicate (delete or define it) you need to catch the exception and then return false like this:
x(a).
z(b).
z(c).
z(d).
w(V) :- x(V).
w(V) :- catch(y(V), error(Err,_Context),my_handler(Err)).
w(V) :- z(V).
my_handler(Err):- write(Err),fail.
Example:
?- w(x).
existence_error(procedure,y/1)
false.

turning off Redefined static procedure in prolog

anyone of you could tell me how to turn off "Redefined static procedure" warnings?
I red online documentation of swi-prolog and i found this predicate no_style_check(ultimate) that in principle should turn off these warnings, but when i execute this predicate
main:-
no_style_check(singleton),
no_style_check(discontiguous),
no_style_check(multiple),
require,
test_all.
i received this error
ERROR: Domain error: style_name' expected, foundmultiple'
Anyone knows an alternative way to do this or could tell me why i receive this error ?
Thanks in advance!
Prolog is a pretty loosey-goosey language, so by default it warns you when you do certain things that are not wrong per se, but tend to be a good indication that you've made a typo.
Now, suppose you write something like this:
myfoo(3, 3).
myfoo(N, M) :- M is N*4+1.
Then from the prompt you write this:
?- asserta(myfoo(7,9)).
ERROR: asserta/1: No permission to modify static procedure `myfoo/2'
ERROR: Defined at user://1:9
What's happening here is that you haven't told Prolog that it's OK for you to modify myfoo/2 so it is stopping you. The trick is to add a declaration:
:- dynamic myfoo/2.
myfoo(3, 3).
myfoo(N, M) :- M is N*4+1.
Now it will let you modify it just fine:
?- asserta(myfoo(7,9)).
true.
Now suppose you have three modules and they each advertise themselves by defining some predicate. For instance, you might have three files.
foo.pl
can_haz(foo).
bar.pl
can_haz(bar).
When you load them both you're going to get a warning:
?- [foo].
true.
?- [bar].
Warning: /home/fox/HOME/Projects/bar.pl:1:
Redefined static procedure can_haz/1
Previously defined at /home/fox/HOME/Projects/foo.pl:1
true.
And notice this:
?- can_haz(X).
X = bar.
The foo solution is gone.
The trick here is to tell Prolog that clauses of this predicate may be defined in different files. The trick is multifile:
foo.pl
:- multifile can_haz/1.
can_haz(foo).
bar.pl
:- multifile can_haz/1.
can_haz(bar).
In use:
?- [foo].
true.
?- [bar].
true.
?- can_haz(X).
X = foo ;
X = bar.
:- discontiguous does the same thing as multifile except in a single file; so you define clauses of the same predicate in different places in one file.
Again, singleton warnings are a completely different beast and I would absolutely not modify the warnings on them, they're too useful in debugging.

singleton variables in prolog

I was testing my new version of SWI prolog and keep coming across the error :singleton variable.
Example:
member(X,[X|T]).
member(X,[X|T]) :- member(X,T).
finds the member of a list such as :
member(yolands,[yolanda,tim])
X = yes
but instead I get a singleton variables error for X and T
if I do the following:
member(X,[X|_]).
member(X,[_|T]) :- member(X,T).
It works but looks ugly!
Can anyone explain why single variables ar enot allowed and if this ANSI standard?
Singleton variables are useless in Prolog, and are easily introduced by editing typos.
The warning is welcome to me, as it allows to easily spot such frequent cause of error.
Being a warning, you can run code containing singletons, but any value these eventually will assume will be lost.
I don't think that ISO standard (never heard about ANSI) forbids such variables.
You could rewrite your example in this way
member(X, [Y|T]) :- X = Y ; member(X, T).
and then forget about the singleton.
You have a bug here:
member(X,[X|T]) :- member(X,T).
What you're actually saying (as opposed to what you think you're saying) is that member/2 holds if X is at the head of the list and present in the tail of the list. This predicate will only ever be true for the first N copies of the same thing at the beginning of a list, so it's a very strange thing to say!
?- member(X, [a,a,c]).
X = a ;
X = a ;
false.
?- member(X, [b,a,a]).
X = b ;
false.
Now, you could correct the bug and still have a singleton warning by doing something like this:
member(X, [Y|T]) :- member(X, T).
But this is neither as good as the conventional definition with two heads or #CapelliC's version (+1) with an explicit OR. I think you should wait until you understand Prolog a little better before putting much stock in your sense of Prolog code aesthetics. If you stick with it for a while you'll come to appreciate this warning as well as the use of anonymous variables.
What makes singleton variables useless in Prolog is that they're named but nothing is known about them and they have no effect on the rest of the computation. The underscore highlights that absolutely anything could go in there without affecting the meaning. What makes
member(X, [X|T]).
true is that the X is position 1 is the same as the X at the head of the list in position 2. Lists must either be empty or have a head and a tail, but what's in the tail is not relevant here, what matters is that X is also the head. The T could be the rest of the list, or it could be an improper list, or it could be a breadbox or lightning or the smell of the air on a spring day. It has no bearing on the truth of member(X, [X|T]).
The singleton warning tells you "you've reserved a name for something here, but you never call anything by that name." The first thing I do when I get this message and it isn't an obvious typo is replace the name with _ and see if my code still makes sense. If it doesn't, I have a logic error. If it does, it was probably unnecessary.
You can read about it on the official page of SWI-Prolog FAQ
The most common cases this warning appears are:
Spelling mistakes in variables
Forget to use/bind a variable
SWI suggest some ways to ignore it:
Use anonymous variable named _ for this purpose.
Use your variable starting with _ (like _T, _X), to avoid warning and document what you ignore.
If you are aware of what you are doing, you can use :- style_check(-singleton). and all warnings should go away.

Defining a rule that the user cannot query

How do I define a rule that the user cannot query?
I only want the program itself to call this rule through another rule.
Ex:
rule1():- rule2().
rule2():- 1<5.
?-rule1().
true
?-rule2().
(I don't know what the answer will be, I just want this query to fail!)
Use a Logtalk object to encapsulate your predicates. Only the predicates that you declare public can be called (from outside the object). Prolog modules don't prevent calling any predicate as using explcit qualification bypasses the list of explicitly exported predicates.
A simple example:
:- object(rules).
:- public(rule1/1).
rule1(X) :-
rule2(X).
rule2(X) :-
X < 5.
:- end_object.
After compiling and loading the object above:
?- rules::rule1(3).
true.
?- rules::rule2(3).
error(existence_error(predicate_declaration,rule2(3)),rules::rule2(3),user)
If you edit the object code and explicitly declare rule2/1 as private you would get instead the error:
?- rules::rule2(3).
error(permission_error(access,private_predicate,rule2(3)),rules::rule2(3),user)
More information and plenty of examples at http://logtalk.org/
First, some notes:
I think you mean "predicate" instead of "rule". A predicate is a name/k thing such as help/0 (and help/1 is another) and can have multiple clauses, among them facts and rules, e.g. length([], 0). (a fact) and length([H|T], L) :- ... . (a rule) are two clauses of one predicate length/2.
Do not use empty parenthesis for predicates with no arguments – in SWI-Prolog at least, this will not work at all. Just use predicate2 instead of predicate2() in all places.
If you try to call an undefined predicate, SWI-Prolog will say ERROR: toplevel: Undefined procedure: predicate2/0 (DWIM could not correct goal) and Sicstus-Prolog will say {EXISTENCE ERROR: predicate2: procedure user:predicate2/0 does not exist}
Now, to the answer. Two ideas come to my mind.
(1) This is a hack, but you could assert the predicate(s) every time you need them and retract them immediately afterwards:
predicate1 :-
assert(predicate2), predicate2, retractall(predicate2).
If you want a body and arguments for predicate2, do assert(predicate2(argument1, argument2) :- (clause1, clause2, clause3)).
(2) Another way to achieve this would be to introduce an extra argument for the predicate which you do not want to be called by the user and use it for an identification that the user cannot possibly provide, but which you can provide from your calling predicate. This might be a large constant number which looks random, or even a sentence. This even enables you to output a custom error message in case the wrong identification was provided.
Example:
predicate1 :-
predicate2("Identification: 2349860293587").
predicate2(Identification) :-
Identification = "Identification: 2349860293587",
1 < 5.
predicate2(Identification) :- Identification \= "Identification: 2349860293587",
write("Error: this procedure cannot be called by the user. Use predicate1/0 instead."),
fail.
I don't use the equivalent predicate2("Identification: 2349860293587") for the first clause of predicate2/0, because I'm not sure where the head of the clause might appear in Prolog messages and you don't want that. I use a fail in the end of the second clause just so that Prolog prints false instead of true after the error message. And finally, I have no idea how to prevent the user from looking up the source code with listing(predicate2) so that will still make it possible to simply look up the correct identification code if s/he really wants to. If it's just to keep the user from doing accidental harm, it should however suffice as a protection.
This reminds me to facility found in Java. There one can query the
curent call stack, and use this to regulate permissions of calling
a method. Translated to Prolog we find in the old DEC-10 Prolog the
following predicate:
ancestors(L)
Unifies L with a list of ancestor goals for the current clause.
The list starts with the parent goal and ends with the most recent
ancestor coming from a call in a compiled clause. The list is printed
using print and each entry is preceded by the invocation number in
parentheses followed by the depth number (as would be given in a
trace message). If the invocation does not have a number (this will
occur if Debug Mode was not switched on until further into the execution)
then this is marked by "-". Not available for compiled code.
Since the top level is usually a compiled predicate prolog/0, this could be
used to write a predicate that inspects its own call stack, and then decides
whether it wants to go into service or not.
rule2 :- ancestors(L), length(L,N), N<2, !, write('Don't call me'), fail.
rule2 :- 1<5.
In modern Prologs we don't find so often the ancestors/1 predicate anymore.
But it can be simulated along the following lines. Just throw an error, and
in case that the error is adorned with a stack trace, you get all you need:
ancestors(L) :- catch(sys_throw_error(ignore),error(ignore,L),true).
But beware stack eliminiation optimization might reduce the stack and thus
the list returned by ancestors/1.
Best Regards
P.S.: Stack elimination optimization is already explained here:
[4] Warren, D.H.D. (1983): An Abstract Prolog Instruction Set, Technical Note 309, SRI International, October, 1983
A discussion for Jekejeke Prolog is found here:
http://www.jekejeke.ch/idatab/doclet/prod/en/docs/10_pro08/13_press/03_bench/05_optimizations/03_stack.html

Resources