I have the next code in a file called "testing.pl":
fact(1).
fact(2).
fact(3).
funcA(X) :- funcB(X).
funcB(X) :- fact(X).
testing :- funcA(_X).
Then, in SWI-Prolog interpreter I query funcA(X). and the output is:
X = 1 ;
X = 2 ;
X = 3.
But, whenever I query testing. the output is:
true ;
true ;
true.
So, my questions is:
How can I use the conclusion of a rule as a premise of another rule (funcA(X) at the right of testing), but having the same effect as if I query that premise (funcA(X))?
In the example above, I would like to write testing. at some point of my "testing.pl" file and to get the funcA(X) to do the same like when I query with the interpreter, so funcB(X) will check for all values that X can take from fact(N) and return it.
My desire result would be to write testing. and to get on screen:
X = 1 ;
X = 2 ;
X = 3.
Thanks.
You can manually print anything on the terminal, using for example predicates like portray_clause/1, format/2 etc.
The only additional piece you need is a way to force backtracking over all answers. One way to do it is using false/0.
So, in your case, you can write for example:
testing :-
funcA(X),
format("X = ~q ;~n", [X]),
false.
And now invoking testing/0 yields, without any further interaction:
?- testing.
X = 1 ;
X = 2 ;
X = 3 ;
In addition, the predicates now fails, so you also get false/0 if you use it interactively (from the toplevel), but not if you invoke testing from the shell via swipl -g testing ....
Also take a look at the important variable_names/1 option that is available to use the intended variable names.
I lave tailoring this output to your exact use case as an exercise. Ideally, the output should be a proper Prolog term, so that you can actually test it easily by reading it with read/1 and comparing it to a reference result.
Related
Completely new to prolog. Interesting journey so far in trying to change how I think, so appreciate any help here.
I am trying to assert facts for a pre-defined set of names. For example, assume I have a a set of people [alice, bob, ...] in one file. I would like to assert facts about these folks in other files, but want to make sure that these folks exist and that is checked when the facts are loaded/compiled(?).
For example, assume I don't have 'chuck' in the list and I make an assertion:
user: swipl app.pl
?- full_name(chuck, "Charlie Steel").
should result in an error.
What is the best way I can do this?
So, here's the code I came up with:
person(deborah).
person(tony).
read_my_file(Filename) :-
open(Filename, read, In),
read_my_file1(In),
close(In).
read_my_file1(In) :-
read(In, Term),
( Term == end_of_file
-> true
; assert_or_abort(Term),
read_my_file1(In)
).
assert_or_abort(Term) :-
( full_name(Person, Name) = Term
-> ( person(Person)
-> assertz(full_name(Person, Name))
; format(user, '~w is not a person I recognize~n', [Person])
)
; format(user, '~w is not a term I know how to parse~n', [Term])
).
The trick here is using read/2 to obtain a Prolog term from the stream, and then doing some deterministic tests of it, hence the nested conditional structure inside assert_or_abort/1. Supposing you have an input file that looks like this:
full_name(deborah, 'Deborah Ismyname').
full_name(chuck, 'Charlie Steel').
full_name(this, has, too, many, arguments).
squant.
You get this output:
?- read_my_file('foo.txt').
chuck is not a person I recognize
full_name(this,has,too,many,arguments) is not a term I know how to parse
squant is not a term I know how to parse
true.
?- full_name(X,Y).
X = deborah,
Y = 'Deborah Ismyname'.
I got the following event: item(C,X,G,P), (where C is a number for the product,X it's name,G it's price,P it's cost).
When i use the command item(n3001,_,_,P) directly on the prolog console i get as answer
G = 1.25 X = 100 but when i write the equation p3(C)-: item(C,_,_,P). then i consult the text i get as answer yes.
My question clarified is how come the one time i get the value of P which i want and the other time i get whether it's true or false?
There are no return values in Prolog and p3/1 does not constitute a function but a relation. Your definition
p3(C) :-
item(C,_,_,P).
reads: If item(C,_,_,P) succeeds then p3(C) succeeds as well. For the sake of argument, let's assume that your code includes the following fact:
item(n3001,100,1.25,1).
If you query
?- p3(n3001).
Prolog unifies C in the head of your rule with n3001 and then tries your goal item(C,_,_,P) which succeeds. Hence the rule succeeds and Prolog tells you:
?- p3(n3001).
yes
If you want to know the price corresponding to n3001 you have to to define a rule where P appears in the head of the rule as well, e.g.:
p3(C,P) :-
item(C,_,_,P).
If you query that you'll get to see the value of P corresponding to n3001:
?- p3(n3001,P).
P = 1
If you query item/4 directly P appears in the arguments and therefore you get to see a substitution for it that satisfies your query:
?- item(n3001,_,_,P).
P = 1
When I run query human(Who). on the below .pl file
human(ann).
human(george).
human(mike).
I only get back Who = ann .
Instead of
Who = ann ;
Who = george ;
Who = mike.
Am using prolog 6.6.6. How do I get it to show the full list?
The answer you got was the following. Do you note the space before the dot?
Who = ann .
^ SPACE!!!
This space means: The query was aborted. Maybe you typed return. Or maybe you have a somewhat illconfigured terminal.
To better check this, try:
?- X = 1 ; X = 2 ; X = 3.
There you should get all three answers, too. If not, it is definitely your terminal
What you are seeing is the default behaviour of prolog.
The query
?- findall(Object,Goal,List).
Should work for you.
Eg.
findall(X, human(X), L).
It will populate the list with all possible answers.
Trying to create a predicate (timePeriod/2) that calculates the time period between two dates for a specific fact. I've managed to do this by myself, but face issues when 'other answers' exist in the same list (i.e. easier to explain with examples).
I have the following knowledge-base facts;
popStar('Jackson',1987,1991).
popStar('Jackson',1992,1996).
popStar('Michaels',1996,2000).
popStar('Newcastle',2000,2007).
popStar('Bowie',2008,2010).
And the following function, calculates the time between dates for a specific fact (as per below).
Predicate (timePeriod/2) -
timePeriod(PS,X) :-
bagof((Name,Start,End),popStar(Name,Start,End),PSs),X is End-Start+1)
Using Bowie as an example; it returns X=3 (which is correct).
However, when there is repetition in the list, with more than one answer available, the predicate just states 'false'. Using the facts 'Jackson' as an example, I want to be able to calculate both of the time periods for both facts; at the same time.
So, if the predicate would work for both of the Jackson facts, the predicate timePeriod would state X=10.
Would really appreciate if anyone could suggest what to change in order for this to work correctly.
Thanks.
You probably don't quite understand what foreach/3 does. I don't think I fully understand foreach/3 either. I know for sure that it is not the same as say:
for each x in xs:
do foo(x)
Another thing: "tuples" in Prolog are not what you might expect, coming from a language like Python or Haskell. This: (a,b,c) is actually this: ','(a,','(b,c)). Much better is to use a flat term, the generic form would be triple(a,b,c). For a pair, the idiom is First-Second.
So, you can simplify your call to bagof/3 to this:
?- bagof(From-To, pop_star(Name, Start, End), Ts).
Name = 'Bowie',
Ts = [2008-2010] ;
Name = 'Jackson',
Ts = [1987-1991, 1992-1996] ;
Name = 'Michaels',
Ts = [1996-2000] ;
Name = 'Newcastle',
Ts = [2000-2007].
Once you have a list as above, you need to sum the differences, which would be maybe something like:
periods_total(Ps, T) :-
maplist(period_length, Ps, Ls),
sum_list(Ls, T).
period_length(From-To, Length) :-
Length is To - From + 1.
And then you can query like this:
?- bagof(From-To, pop_star('Jackson', From, To), Ps), periods_total(Ps, T).
Ps = [1987-1991, 1992-1996],
T = 10.
?- bagof(From-To, pop_star(Name, From, To), Ps), periods_total(Ps, T).
Name = 'Bowie',
Ps = [2008-2010],
T = 3 ;
Name = 'Jackson',
Ps = [1987-1991, 1992-1996],
T = 10 ;
Name = 'Michaels',
Ps = [1996-2000],
T = 5 ;
Name = 'Newcastle',
Ps = [2000-2007],
T = 8.
SWI-Prolog has a nice library to handle aggregation: it builds upon standard 'all solutions' predicates like findall/3,setof/3,bagof/3, so you should first grasp the basic of these (as Boris explained in his answer). With the library, a single query solves your problem:
timePeriod(PS,X) :-
aggregate(sum(P), B^E^(popStar(PS,B,E),P is E-B+1), X).
Trying to create a predicate (timePeriod/2) that calculates the time period between two dates for a specific fact. I've managed to do this by myself, but face issues when 'other answers' exist in the same list (i.e. easier to explain with examples).
I have the following knowledge-base facts;
popStar('Jackson',1987,1991).
popStar('Jackson',1992,1996).
popStar('Michaels',1996,2000).
popStar('Newcastle',2000,2007).
popStar('Bowie',2008,2010).
And the following function, calculates the time between dates for a specific fact (as per below).
Predicate (timePeriod/2) -
timePeriod(PS,X) :-
bagof((Name,Start,End),popStar(Name,Start,End),PSs),X is End-Start+1)
Using Bowie as an example; it returns X=3 (which is correct).
However, when there is repetition in the list, with more than one answer available, the predicate just states 'false'. Using the facts 'Jackson' as an example, I want to be able to calculate both of the time periods for both facts; at the same time.
So, if the predicate would work for both of the Jackson facts, the predicate timePeriod would state X=10.
Would really appreciate if anyone could suggest what to change in order for this to work correctly.
Thanks.
You probably don't quite understand what foreach/3 does. I don't think I fully understand foreach/3 either. I know for sure that it is not the same as say:
for each x in xs:
do foo(x)
Another thing: "tuples" in Prolog are not what you might expect, coming from a language like Python or Haskell. This: (a,b,c) is actually this: ','(a,','(b,c)). Much better is to use a flat term, the generic form would be triple(a,b,c). For a pair, the idiom is First-Second.
So, you can simplify your call to bagof/3 to this:
?- bagof(From-To, pop_star(Name, Start, End), Ts).
Name = 'Bowie',
Ts = [2008-2010] ;
Name = 'Jackson',
Ts = [1987-1991, 1992-1996] ;
Name = 'Michaels',
Ts = [1996-2000] ;
Name = 'Newcastle',
Ts = [2000-2007].
Once you have a list as above, you need to sum the differences, which would be maybe something like:
periods_total(Ps, T) :-
maplist(period_length, Ps, Ls),
sum_list(Ls, T).
period_length(From-To, Length) :-
Length is To - From + 1.
And then you can query like this:
?- bagof(From-To, pop_star('Jackson', From, To), Ps), periods_total(Ps, T).
Ps = [1987-1991, 1992-1996],
T = 10.
?- bagof(From-To, pop_star(Name, From, To), Ps), periods_total(Ps, T).
Name = 'Bowie',
Ps = [2008-2010],
T = 3 ;
Name = 'Jackson',
Ps = [1987-1991, 1992-1996],
T = 10 ;
Name = 'Michaels',
Ps = [1996-2000],
T = 5 ;
Name = 'Newcastle',
Ps = [2000-2007],
T = 8.
SWI-Prolog has a nice library to handle aggregation: it builds upon standard 'all solutions' predicates like findall/3,setof/3,bagof/3, so you should first grasp the basic of these (as Boris explained in his answer). With the library, a single query solves your problem:
timePeriod(PS,X) :-
aggregate(sum(P), B^E^(popStar(PS,B,E),P is E-B+1), X).