I'm interested in formulae made up from lots of conjunctions (part of a larger problem). I want to write a program that takes something like this:
:- get_params(conj(conj(a,b),c),X)
and returns a list of all the parameters of the conjunctions i.e. X=[a,b,c]. At the moment I can do
:- get_params(conj(a,b),X) to get X=[a,b]
using simple Prolog pattern matching but how would you go about doing things such as
:- get_params(conj(conj(a,b),c),X) to get X=[a,b,c]
It seems really simple but I've been struggling all day!
Since you are describing a list, consider using DCG notation:
params(conj(A,B)) --> !, params(A), params(B).
params(X) --> [X].
Example:
?- phrase(params(conj(conj(a,b),c)), Ps).
Ps = [a, b, c].
Assuming that all conj functors are binary:
get_params(X, Y, L) :-
get_params(X, L1),
get_params(Y, L2),
append(L1, L2, L).
get_params(conj(X, Y), L) :-
get_params(X, Y, L), !.
get_params(A, [A]).
Related
I want to add in the DB a constant and a linked variable:
?- assertz(my(x, A))
So that in the future I can define A and get the only one entry. Sth like that:
?- assertz(my(x, A)), ..., A = 2.
?- my(A, B).
A = x,
B = 2.
Can this be done?
As I noted in the comments your idea of a link like a pointer is not the way to approach solving your problem.
A common solution is to walk the tree and construct a new tree as you walk the tree by replacing the leaf of the tree with a new leaf that contains the value from the input tree along with the associated value, what you are thinking should be linked.
Since you are somewhat new to Prolog I will do this with two examples. The first will just walk a tree and only return true when successfully walked. It can be used to understand how to walk a tree and run with gtrace to single step the code to understand it.
The second example will expand on the tree walk and add the type (link as you think) to the leaf item. The the old leaf for something simple like an atom a, will become a new leaf in the tree like (a,atom).
Also this was quickly written as a demonstration only. I am sure it will have problems if pressed into doing anything more than the single example.
:- module(example,
[
example/1
]).
example(walk) :-
Term = term_size(a(1,"Hello",'Atom',1+2,[a,$,T])),
walk(Term).
example(infer_type) :-
Term = term_size(a(1,"Hello",'Atom',1+2,[a,$,T])),
infer_type(Term,Is),
write(Is).
walk([]) :- !.
walk([T]) :- var(T), !.
walk(L) :- is_list(L), !, L = [H|T], walk(H), walk(T).
walk(T) :- compound(T), !, T =.. [_|Args], !, walk(Args).
walk(T) :- integer(T), !.
walk(T) :- var(T), !.
walk(T) :- atomic(T), !.
walk(T) :- T =.. [Arg|Args], !, walk(Arg), walk(Args).
infer_type([],[]) :- !.
infer_type([T],[(T,var)]) :- var(T), !.
infer_type(L,S) :- is_list(L), !, L = [H|T], infer_type(H,I), infer_type(T,Is), S = [I|Is].
infer_type(T,S) :- compound(T), !, T =.. [F|Args], !, infer_type(Args,Is), S =.. [F|Is].
infer_type(T,(T,integer)) :- integer(T), !.
infer_type(T,(T,var)) :- var(T), !.
infer_type(T,(T,atom)) :- atomic(T), !.
infer_type(T,S) :- T =.. [Arg|Args], !, infer_type(Arg,I), infer_type(Args,Is), S =.. [I|Is].
Example run
Note: I know there are warnings; it is a demo not production code.
Welcome to SWI-Prolog (threaded, 64 bits, version 8.5.3)
?- working_directory(_,'C:/Users/Groot').
true.
?- [example].
Warning: c:/users/Groot/example.pl:20:
Warning: Singleton variables: [T]
Warning: c:/users/Groot/example.pl:24:
Warning: Singleton variables: [T]
true.
?- example(walk).
true.
?- example(infer_type).
term_size(a((1,integer),(Hello,atom),(Atom,atom),(1,integer)+(2,integer),[(a,atom),(($),atom),(_25642,var)]))
true.
As an exercise I did not identify the string as a string, the change should be easy.
ass(a).
ass(b).
ass(c).
con(c,r).
arg(A, L) :- forall(member(S, L), (ass(S), \+ con(S,A))).
If I run arg(r, [a,b]) it will work but if I run arg(r,X) it returns: ERROR out of global stack. I would like it to return [a,b]. I understand this is because L is unbounded, but how can I fix this.
In the predicate:
arg(A, L) :- forall(member(S, L), (ass(S), \+ con(S,A))).
May have a limitation in your case as described in the SWI Prolog documentation for forall/2:
If your intent is to create variable bindings, the forall/2 control
structure is inadequate. Possibly you are looking for maplist/2,
findall/3 or foreach/2.
So in this case, you may be better off with:
arg(A, L) :- findall(S, (ass(S), \+ con(S,A)), L).
Which will yield:
?- arg(r, X).
X = [a, b].
?- arg(r, [a,b]).
true.
?-
I have a term which may or may not contain the atom 'this'. The term may also contain variables.
I need to replace 'this' with a variable I. How can I do this?
I tried to do something like this:
term_to_atom((f(a), g(this, b), ...), A),
tokenize_atom(A, L),
replace(this, I, L, L2)
It seemed to work. The problem is, I need to go back to the original term and I can't do it...
SWI-Prolog has atomic_list_concat/2 and atom_to_term/2 which should help you go back to the original term.
main :-
term_to_atom((f(a), g(this, b)), A),
tokenize_atom(A, L),
replace(this, 'I', L, L2),
atomic_list_concat(L2, A2),
atom_to_term(A2, T, []),
writeln(T).
?- main.
f(a),g(_G69,b)
true .
Take a look at this predicate (replace/4):
replace(Term,Term,With,With) :-
!.
replace(Term,Find,Replacement,Result) :-
Term =.. [Functor|Args],
replace_args(Args,Find,Replacement,ReplacedArgs),
Result =.. [Functor|ReplacedArgs].
replace_args([],_,_,[]).
replace_args([Arg|Rest],Find,Replacement,[ReplacedArg|ReplacedRest]) :-
replace(Arg,Find,Replacement,ReplacedArg),
replace_args(Rest,Find,Replacement,ReplacedRest).
An example of what you need:
| ?- replace(f(1,23,h(5,this)),this,Var,Result).
Result = f(1,23,h(5,Var))
yes
I have a problem with predicate which works in that way that it takes list of atoms:
nopolfont([to,jest,tekśćik,'!'],L).
and in result
L = [to,jest,tekscik,'!'].
I have problem with make_swap and swap predicates. So far I have:
k(ś,s).
k(ą,a).
% etc.
swap(X,W) :- name(X,P), k(P,Y), !, name(Y,W).
swap(X,X).
make_swap(A,W)
:- atom(A),!,
name(A,L),
swap(L,NL),
name(W,NL).
nopolfont([],[]).
nopolfont([H|T],[NH|S]) :- make_swap(H,NH), nopolfont(T,S).
Is there any elegant way to do this?
This is also quite elegant:
polish_char_replacer(X, Y) :-
k(X, Y),
!.
polish_char_replacer(X, X).
nopolfont(Atoms1, Atoms2) :-
maplist(replace(polish_char_replacer), Atoms1, Atoms2).
replace(Goal, Atom1, Atom2) :-
atom_chars(Atom1, Chars1),
maplist(Goal, Chars1, Chars2),
atom_chars(Atom2, Chars2).
Probably as elegant as it can get:
k(ś,s).
k(ą,a).
swap(X,W) :- name(P,[X]), k(P,Y), !, name(Y,[W]).
swap(X,X).
list_swap([], []).
list_swap([H|T], [W|S]) :-
swap(H, W),
list_swap(T, S).
atom_swap(A,W) :-
atom(A), !,
name(A, L),
list_swap(L,S),
name(W, S).
nopolfont([],[]).
nopolfont([H|T],[NH|S]) :-
atom_swap(H,NH),
nopolfont(T,S).
Also, obviously define this, to get the expected result, but I assume this is in the % etc
k(ć, c).
I am learning prolog and I am reading a book called Programming Prolog for Artificial Intelligence. As practice I want to learn how to extend one of the examples in this book. Can someone please help?
Say you have these facts:
parent(pam, bob). %pam is a parent of bob
parent(george, bob). %george is a parent of bob
How would I write a prolog predicate that would give me a list of bobs parents? For example:
list_parents(bob, L).
L = [pam, george] ;
L = [george, pam] ;
true.
An all-solutions predicate like findall/3 might do the trick:
list_parents(P, L) :-
findall(Parent, parent(Parent, P), L).
Simply put, findall/3 finds all bindings for Parent in the 'backtrack-able' goal parent(Parent, P), and puts all bindings of Parent into the list L. Note that this won't remove duplicates, but you can do a sort/2 to L before returning it to create a set. Executing this:
?- list_parents(bob, L).
L = [pam, george].
If you don't have findall/3 in your PROLOG implementation, you could do it manually like this:
list_parents(P, L) :-
list_parents(P, [], L).
list_parents(P, Acc, L) :-
parent(Parent, P),
\+ member(Parent, Acc), !,
list_parents(P, [Parent|Acc], L).
list_parents(_, L, L).
This version sends calls to list_parents/2 off to an accumulator-version, list_parents/3. The latter tries to collect Parent bindings also, as long as we haven't seen them before (hence the \+ member check), and returns the list where no new Parent bindings accumulated into the Acc list can be found. Executing this gives us the same result as the first option:
?- list_parents(bob, L).
L = [pam, george].
Try this:
parent(pam, bob). %pam is a parent of bob
parent(george, bob). %george is a parent of bob
list_parents(A, Es, [X|Xs]) :- parent(X, A), \+ member(X, Es), list_parents(A, [X|Es], Xs).
list_parents(A, Es, []).
That was an inefficient method, a better method will need a "solutions" higher-order predicate.
list_parents(X, Ys) :- solutions(parent, [X, W], 1, Ys)