I'm new to prolog, but basically, I want a program to iterate through the predicates(not sure if I'm using that term correctly) and then come to a final predicate which can process the input and provide one of two answers. The ask and the maplist were my tries at iterating through the program the way a program like Java would.
(Also, sidenote, but is there any way to have the user input Yes and No instead of true. and false.?)
Here's how my code looks currently:
ask(happy,X).
ask(lonely,X).
ask(education,X).
ask(institution,X).
ask(home,X).
ask(parents,X).
ask(social_life,X).
ask(hobbies,X).
ask(opinion,X).
ask(relationship,X).
ask(finances,X).
ask(future,X).
ask(freedom,X).
ask(feelings,X).
maplist(ask(_), Xs).
Xs= [happy(X),lonely(X),education(X),institution(X), home(X),
parents(X), social_life(X), hobbies(X), opinion(X), relationship(X),
finances(X), future(X), freedom(X),feelings(X)].
happy(X):-
write("Are you happy?"),nl,
read(X).
lonely(X):-
write("Are you lonely?"),nl,
read(X).
Maybe this can serve as inspiration
main(AnswersOut) :-
Ls = [happy,lonely,education],
ask_next_question(Ls,[],AnswersOut).
ask_next_question([L|Ls],Answers,AnswersOut) :-
format("Question about: ~w\n",[L]),
read_line_to_string(user_input,Str),
format("You said: ~w\n",[Str]),
ask_next_question(Ls,[L-Str|Answers],AnswersOut).
ask_next_question([],A,A).
Then you can collect answers into a list of pairs for further processing.
Note the use of read_line_to_string/2 which doesn't read a term (necessarily terminated with .) as read/2 does, but an arbitrary String terminated by newline.
Run it:
?- main(A).
Question about: happy
|: Well....
You said: Well....
Question about: lonely
|: Yes
You said: Yes
Question about: education
|: Too high
You said: Too high
A = [education-"Too high", lonely-"Yes", happy-"Well...."].
Related
I am quite new to Prolog and have had some trouble understanding it.
I have some facts named 'problem' I wish to first print out these facts to the user and then ask them to input a value, this value is then read and used later.
From my understanding thus far, it would be best to use a forall to print out these facts and then use read to read the value inputted, but I am having some issue implementing this. Here is what I have so far, any explanation would be appreciated
My question: How do I read in the input from the user regarding the problem and apply that into a variable for later use?
tellMeYourProblem:-
forall(problem(P),
writeln(P)),
answer = read(X),
problem('1').
problem('2').
problem('3').
problem('4').
problem('5').
problem('6').
problem('7').
problem('8').
problem('9').
problem('10').
Note: This answer uses SWI-Prolog.
How do I read in the input from the user regarding the problem?
You are doing that already with read(X), however read/1 reads terms (terms end with periods) and you probably want to read characters. If you are using SWI-Prolog take a look at Primitive character I/O for reading characters and Predicates that operate on strings for reading strings.
How do I apply that into a variable for later use?
When doing basic I/O with a user at a text level, a REPL is a good way to start. Adding a REPL is a bit more complicated so I will give you the code.
tellMeYourProblem :-
output_problems,
read_input.
output_problems :-
forall(problem(P),
writeln(P)).
read_input :-
repeat,
read_string(user_input, "\n", "\r\t ", _, Line),
process_input(Line).
process_input(Line) :-
string(Line),
atom_number(Line, N),
integer(N),
do_something_with(Line),
fail.
process_input("quit") :-
write('Finished'), nl,
!, true.
do_something_with(X) :-
writeln(X).
problem('1').
problem('2').
problem('3').
problem('4').
problem('5').
problem('6').
problem('7').
problem('8').
problem('9').
problem('10').
Also with Prolog, the style is to use snake casing so tellMeYourProblem should be tell_me_your_problem.
Normally in Prolog a REPL is done with ->/2, (Read Input till quit statement Prolog) , but I changed this to add more guard statements so that the exit condition would work, e.g.
string(Line),
atom_number(Line, N),
integer(N)
or putting the guard in the head, e.g.
process_input("quit")
When doing I/O to a screen and keyboard, the thought is to use stdIn and stdOut but for the keyboard SWI-Prolog uses user_input instead. See: Input and output
After all of the boiler plate code for the REPL is the next part you seek which is to do something with the input value, in this case just print it out.
do_something_with(X) :-
writeln(X).
The easiest to write out the facts of problem/1,
is to use the builtin listing/[0,1]. This builtin
accepts a so called predicate indicator. You can
write out the facts via:
?- listing(problem/1).
The predicate is supported by many Prolog systems
such as GNU Prolog, etc.. For how to read input see
for example the post by Guy Coder.
I'm new to Prolog and having a bit of difficulty. I have:
man(ken).
man(tom).
woman(juli).
father(ken, tom).
father(ken, juli).
male(A) :- man(A).
brother(A,B) :- male(A), father(C,A), father(C,B), (A \= B).
I know the male/man is redundant, but it's part of the assignment. Anyway, when I try something like:
|?- brother(tom, juli).
I get "no" as the response. I'm sure I've made a stupid, simple mistake, but my lack of understanding is making it very hard to find. Can anyone see what my problem is?
When you enter:
|?- brother(tom, juli).
You'll see a response something like this (SWI Prolog):
true ? ;
no
| ?-
So it first responds with "true" (gives a match) and then, after you enter ; to show more solutions, it says "no" to indicate there are no further solutions. Some prolog interpreters may say "no" or "false" in this case, with the same meaning. This response from the prolog interpreters initially throws many a new prolog user.
You could, alternatively, press "enter" which just means you're done and don't want to see any further solutions:
true ?
yes
| ?-
Then you get "yes".
How would I state things "in general" about the facts? Suppose I need to state "everyone likes the person who likes him/her", and I have a list of people who may or may not like each other.
This is what I tried so far, but it's sure not the way to do it:
likes(dana, cody).
hates(bess, dana).
hates(cody, abby).
likes(first(Girl, OtherGirl), first(OtherGirl, Girl)).
hates(Girl, OtherGirl):- \+ likes(Girl, OtherGirl).
because this won't even compile.
everybody([dana, cody, bess, abby]).
likes_reflexive(dana, cody).
hates(bess, dana).
hates(cody, abby).
likes_reflexive(X, Y):- likes(X, Y), likes(Y, X).
hates(Girl, OtherGirl):- \+ likes(Girl, OtherGirl).
%% likes_reflikes_reflexive(X, Y):- likes(X, Y), likes(Y, X).
%% user:6: warning: discontiguous predicate likes_reflexive/2 - clause ignored
%% hates(Girhates(Girl, OtherGirl):- \+ likes(Girl, OtherGirl).
%% user:8: warning: discontiguous predicate hates/2 - clause ignored
Unfortunately I don't understand what the warnings say. Hope it makes my intention more clear. I.e. by stating one fact, I also want to state the other related fact.
If you want to change your knowledge base dynamically, you can use asserts. If you want to modify existing predicate, you should define it as dynamic, e.g. :- dynamic(likes/2).. If predicate is undefined, you can omit it.
add_mutual_likes(X, Y) :- asserta(likes(X, Y)), asserta(likes(Y, X)).
:- initialization(add_mutual_likes(dana, cody)).
initialization/1 calls add_mutual_likes(data, cody) goal when file is loaded. add_mutual_likes/2 adds two facts to a database. asserta/1 converts it's argument into a clause and adds it to a database.
| ?- [my].
yes
| ?- listing(likes/2).
% file: user_input
likes(cody, dana).
likes(dana, cody).
yes
| ?- likes(cody, dana).
yes
| ?- likes(dana, cody).
yes
| ?- add_mutual_likes(oleg, semen).
yes
| ?- listing(likes/2).
% file: user_input
likes(semen, oleg).
likes(oleg, semen).
likes(cody, data).
likes(data, cody).
yes
I use gprolog.
Let's start with the warnings. They are merely "style" suggestions. They are telling you that all the definitions for likes and hates should be together. Trust me if you have a big Prolog program it becomes a nightmare to go around tour code to get the full definition of your predicate. It would be like writing half a function in C++ and finish it in another file.
Now, you want to say "everyone likes the person who likes him/her". I'm not sure why you are using that function "first" in the code. This would be sufficient:
likes(dana, cody).
likes(Girl, OtherGirl) :- likes(OtherGirl, Girl).
The second clause reads "Girl likes OtherGirl if OtherGirl likes Girl. This won't work.
If you ask your program "is it true that cody likes dana"
? likes(cody, dana)
Prolog will reason like this:
The answer is yes if dana likes cody (using the second clause).
Yes! Because dana likes cody (using the first clause).
This is not enough to make it a correct program. Since we are in Prolog you can say: "give me another solution" (usually by entering ";" in the prompt).
Prolog will think "I only used the first clause, I haven't tried the second".
The answer is Yes also if dana likes cody (using the second clause).
The answer is Yes according to the second clause, if cody likes dana.
But that's our initial query. Prolog will give you the same answer again and again, looping forever if you asked for all the solutions.
You can do two things here. The first is telling Prolog that one solution is enough. You do this adding a "!" (that basically says, clear all the open branches left to explore).
likes(dana, cody) :- !.
likes(Girl, OtherGirl) :- likes(OtherGirl, Girl).
Another alternative is to "stratify the program".
direct_likes(dana, cody).
likes(Girl, OtherGirl) :- direct_likes(OtherGirl, Girl), !.
likes(Girl, OtherGirl) :- direct_likes(Girl, OtherGirl).
What you want is a fact where Prolog does not care about the order of arguments. Alas, something like that does not exist. What you can do instead is define facts where the implied meaning is that it is valid for all argument orders (like_each in the example below). But of course, you cannot use these facts in that way. Instead, you define the actual predicate to try (hence the or ;) all possible argument orders.
Thus, the solution is:
%% bi-directional like
like_each(dana, cody).
likes(A, B) :- like_each(A, B); like_each(B, A).
%% optional: one-directional like
% likes(cody, sarah).
Also, be careful with
hates(Girl, OtherGirl):- \+ likes(Girl, OtherGirl).
If both variables are unbound (e.g., ?- hates(A,B)), it will always fail. This happens because Prolog first tries to find a match for likes, which always succeeds for two variables, and then negates the result. Thus, you cannot use hates to find all pairs who don't like each other.
I am learning Prolog for an university exam using SWI Prolog and I have some question about this simple program that implement the different predicate that say TRUE if two element are different (if they do not match) and say FALSE if they match.
This is the code:
different(X,X) :- !,
fail.
diferent(_,_).
The problem is that if I try to execute the following query in the Prolog shell I always obtain FALSE:
[debug] 10 ?- different(a,b).
false.
[debug] 11 ?- different(a,a).
false.
As you can see the first query have to say TRUE because a don't match with b
Why?
change diferent(_,_) to different(_,_). Ie it is a spelling error.
Your second predicate is not being examined as it does not match your query.
The program should be
different(X,X) :- !,fail.
different(_,_).
This "exercise" is pointless. Use prolog-dif!
IMO the only reasonable definition of different/2 is:
different(A,B) :- dif(A,B).
If you are using emacs for editing your prolog files (If not I strongly recommend you switching to it) I suggest you using hi-lock-mode. I will highlight in a pattern-like fashion all matches in the file whenever the cursor is over a word. It can save you hours of pain when you develop bigger projects.
I am reading http://cs.union.edu/~striegnk/learn-prolog-now/html/node3.html#subsec.l1.kb1,
but I am having trouble running the following predicate:
SICStus 4.0.1 (x86-win32-nt-4): Tue May 15 21:17:49 WEST 2007
| ?- woman(mia).
! Existence error in user:woman/1
! procedure user:woman/1 does not exist
! goal: user:woman(mia)
| ?-
If, on the other hand, I write it to a file and run consult the file, it seems to work fine...
Am I only allowed to define predicates in a file having later to consult them? Can't I just do it in the editor itself?
It's a little annoying to make predicates in the repl. You could do
| ?- ['user'].
woman(mia).
^D
ie consult user input, or
| ?- assertz(woman(mia)).
assert it. Both awkward IMO -- there might be a better way, though, I just don't know it. In general it is easier to use a script.
You should enter woman(mia). into a file to assert it as a fact. If you write it into the interpreter, it's taken as a query, not a fact.
From the SWI Prolog FAQ:
Terms that you enter at the toplevel are processes as queries, while
terms that appear in a file that is loaded into Prolog is processed as
a set of rules and facts. If a text reads as below, this is a rule.
carnivore(X) :- animal(X), eats_meat(X).
Trying to enter this at the toplevel results in the error below. Why?
Because a rule is a term :-(Head, Body), and because the toplevel
interprets terms as queries. There is no predicate with the name :-
and two arguments.
?- carnivore(X) :- animal(X), eats_meat(X). ERROR: Undefined
procedure: (:-)/2 ERROR: Rules must be loaded from a file ERROR:
See FAQ at http://www.swi-prolog.org/FAQ/ToplevelMode.txt
Isn't this stupid? Well, no. Suppose we have a term
eats_meat(rataplan). If this appears in a file, it states the fact
that rataplan eats meat. If it appears at the toplevel, it asks Prolog
to try proving whether rataplan eats meat.
If a text reads
:- use_module(library(clpfd)).
This is a directive. Directives are similar to queries, but instead of
asking the toplevel to do something, they ask the compiler to do
something. Like rules and facts, such terms belong in files.
Instead of writing to a file you can also use assert in the toplevel (as explained later in the FAQ as well).