How can I convert uppercase input to the lowercase? - prolog

I have a predicate that takes two arguments in which the first argument can be a compound one and the second argument is always B. I have also defined some new operators like + and &.
pvc(A, B) :- somestuff(A, B).
Here the user may type something like pvc((x+y)&(A+b), B).
As a beginner in Prolog, what I want to do is to convert the compound A to all lowercase and call somestuff with new AN. So it shall be somestuff((x+y)&(a+b), B).
I tried something like pvc(A, B) :- downcase_atom(A,AN),somestuff(AN, B). But it doesn't seem like to be the correct way. I will appreciate any help.

So, you will need to induct on the structure of your thing, and the easiest way to do this is with the univ operator =../2. First handle your base case, which you did:
downcase_compound(A, B) :- atom(A), downcase_atom(A, B).
Now you will take the structure apart and induct on it using univ:
downcase_compound(A, B) :-
compound(A),
A =.. [Functor|Args],
downcase_compound(Functor, DowncaseFunctor),
maplist(downcase_compound, Args, DowncaseArgs),
B =.. [DowncaseFunctor|DowncaseArgs].
The trick here is simply breaking your compound down into bits you can use, and then recursively calling downcase_compound/2 on those bits. See it in action:
?- downcase_compound((x+y,('A'+b)), X).
X = (x+y, a+b)

Related

How to create anonymous variables from a string?

I have defined a predicate find_word/2 that when given a list of letters (with some letters possibly ungrounded), produces possible words that match the pattern given in the list. This is something like a hangman solver.
word('entity', n, 11).
word('physical entity', n, 1).
word('abstraction', n, 0).
% ... and 200,000 more entries ...
% Example: find_word([_,o,u,n,t,r,y], X) -> X = country
find_word(LetterList, Word) :-
word(Word, _, _),
atom_chars(Word, LetterList).
The code above works as intended. The challenge is that I receive hangman problems from outside the Prolog system as a string (e.g. app_e), where the underscores in the string represent the missing letters to be found by the prolog program above. i.e. I need to convert the app_e string into a list that can be fed into find_word/2.
On my first attempt, I used atom_chars\2:
?- atom_chars(app_e, L), find_word(L, Word).
Unfortunately, this does not work as hoped because atom_chars(app_e, L) -> L = [a, p, p, '_', e]. i.e. the '_' isn't a wildcard.
In summary, given a string app_e, how do I transform it into a list that can be fed into find_word\2 to achieve the same effect as find_word([a,p,p,_,e], Word).?
I think atom_chars/2 is working as intended here, you just need a little cleanup step to finish turning your input into the desired form, which I think you can do quite straightforwardly like so:
charvar('_', _).
charvar(C, C) :- C \= '_'.
Usage looks like this:
?- maplist(charvar, [a,p,p,'_',e], X).
X = [a, p, p, _3398, e] .
Don't worry about the fact that this variable is not rendered as an underscore; your own probably wouldn't be either:
?- X=[_].
X = [_3450].

Prolog - Domain error: 'acyclic_term ' expected

What I have to do is, write a predicate Multiplication/3, whose first argument is an integer, second argument is a list, and the third argument is the result of multiplying the integer with the list, for example:
?-Multiplication(3,[2,7,4],Result).
should return
Result = [6,21,12].
Here's my code:
Multiplication(X,[],Result).
Multiplication(X,[Head|Tail],Result) :-
Y is X*Head,
append([Result], [Y], L),
append([],L,Result), // HERE
Multiplication(X,Tail,Result).
And I get the following error:
Domain error: 'acyclic_term ' expected, found '#(lists:append([],S_1,S_1),[S_1=[S_1,1]])'
on the second append call.
If anyone knows why I receive the error, how to fix it or another way to solve this, I'm open to ideas.
Your two goals append([Result], [Y], L), append([],L,Result) are exactly the same as:
L = [Result,Y], L = Result.
or even simpler:
L = [L,Y]
which would result either in silent failure or an infinite term. Instead, your Prolog produces an error, so that you can correct your program.
In your original code:
Multiplication(X,[Head|Tail],Result) :-
Y is X*Head,
append([Result], [Y], L),
append([],L,Result), // HERE
Multiplication(X,Tail,Result).
You're getting a "cycle" because you're appending Result to something to get L, then appending something to L to get Result. That's not good. You also have a capitalized predicate name, which is a syntax error. (I assume that, since you ran your code, it wasn't capitalized in the original version.)
You're new proposed solution is overly complicated. Why do you need the 4th argument? Also, your base case for return (which is return(X, [], Result) doesn't make sense, as it has to singleton variables. The use of append/3 is overkill since recursion handles the iteration through the list elements for you.
Starting from the top, you have a common pattern in Prolog where you want to run a query on corresponding elements of two or more lists. A simple recursive solution would look something like this:
multiplication(_, [], []). % Multiplying anything by the empty list is the empty list
multiplication(M, [X|Xs], [XX|XXs]) :-
XX is M * X,
multiplication(M, Xs, XXs).
Another way to implement this kind of pattern in Prolog is with maplist/3. You can first define the query on corresponding elements:
multiply(X, Y, Product) :- Product is X * Y.
Then use maplist/3:
multiplication(M, List, Result) :-
maplist(multiply(M), List, Result).
Maplist will do a call(multiply(M), ...) on each corresponding pair of elements of List and Result.
I edited the code and came up with this:
multiplication(X,[],Result,Result).
multiplication(X,[Head|Tail],List,Result) :-
Y is X*Head,
append(List, [Y], L),
multiplication(X,Tail,L,Result).
return(X,[],Result).
return(X,L,Result) :-
multiplication(X,L,_,Result).
and the query:
return(2,[1,2],Result).
After the first run, it seems to return Result as it should be, but it runs forever.

Prolog program to return atoms in a proposition formula

I am a newbie to prolog and am trying to write a program which returns the atoms in a well formed propositional formula. For instance the query ats(and(q, imp(or(p, q), neg(p))), As). should return [p,q] for As. Below is my code which returns the formula as As. I dont know what to do to split the single F in ats in the F1 and F2 in wff so wff/2 never gets called. Please I need help to proceed from here. Thanks.
CODE
logical_atom( A ) :-
atom( A ),
atom_codes( A, [AH|_] ),
AH >= 97,
AH =< 122.
wff(A):- ground(A),
logical_atom(A).
wff(neg(A)) :- ground(A),wff(A).
wff(or(F1,F2)) :-
wff(F1),
wff(F2).
wff(and(F1,F2)) :-
wff(F1),
wff(F2).
wff(imp(F1,F2)) :-
wff(F1),
wff(F2).
ats(F, As):- wff(F), setof(F, logical_atom(F), As).
First, consider using a cleaner representation: Currently, you cannot distinguish atoms by a common functor. So, wrap them for example in a(Atom).
Second, use a DCG to describe the relation between a well-formed formula and the list of its atoms, like in:
wff_atoms(a(A)) --> [A].
wff_atoms(neg(F)) --> wff_atoms(F).
wff_atoms(or(F1,F2)) --> wff_atoms(F1), wff_atoms(F2).
wff_atoms(and(F1,F2)) --> wff_atoms(F1), wff_atoms(F2).
wff_atoms(imp(F1,F2)) --> wff_atoms(F1), wff_atoms(F2).
Example query and its result:
?- phrase(wff_atoms(and(a(q), imp(or(a(p), a(q)), neg(a(p))))), As).
As = [q, p, q, p].
This should do what you want. It extracts the unique set of atoms found in any arbitrary prolog term.
I'll leave it up to you, though, to determine what constitutes a "well formed propositional formula", as you put it in your problem statement (You might want to take a look at DCG's for parsing and validation).
The bulk of the work is done by this "worker predicate". It simply extracts, one at a time via backtracking, any atoms found in the parse tree and discards anything else:
expression_atom( [T|_] , T ) :- % Case #1: head of list is an ordinary atom
atom(T) , % - verify that the head of the list is an atom.
T \= [] % - and not an empty list
. %
expression_atom( [T|_] , A ) :- % Case #2: head of listl is a compound term
compound(T) , % - verify that the head of the list is a compound term
T =.. [_|Ts] , % - decompose it, discarding the functor and keeping the arguments
expression_atom(Ts,A) % - recurse down on the term's arguments
. %
expression_atom( [_|Ts] , A ) :- % Finally, on backtracking,
expression_atom(Ts,A) % - we simply discard the head and recurse down on the tail
. %
Then, at the top level, we have this simple predicate that accepts any [compound] prolog term and extracts the unique set of atoms found within by the worker predicate via setof/3:
expression_atoms( T , As ) :- % To get the set of unique atoms in an arbitrary term,
compound(T) , % - ensure that's its a compound term,
T =.. [_|Ts] , % - decompose it, discarding the functor and keeping the arguments
setof(A,expression_atom(Ts,A),As) % - invoke the worker predicate via setof/3
. % Easy!
I'd approach this problem using the "univ" operator =../2 and explicit recursion. Note that this solution will not generate and is not "logically correct" in that it will not process a structure with holes generously, so it will produce different results if conditions are reordered. Please see #mat's comments below.
I'm using cuts instead of if statements for personal aesthetics; you would certainly find better performance with a large explicit conditional tree. I'm not sure you'd want a predicate such as this to generate in the first place.
Univ is handy because it lets you treat Prolog terms similarly to how you would treat a complex s-expression in Lisp: it converts terms to lists of atoms. This lets you traverse Prolog terms as lists, which is handy if you aren't sure exactly what you'll be processing. It saves me from having to look for your boolean operators explicitly.
atoms_of_prop(Prop, Atoms) :-
% discard the head of the term ('and', 'imp', etc.)
Prop =.. [_|PropItems],
collect_atoms(PropItems, AtomsUnsorted),
% sorting makes the list unique in Prolog
sort(AtomsUnsorted, Atoms).
The helper predicate collect_atoms/2 processes lists of terms (univ only dismantles the outermost layer) and is mutually recursive with atoms_of_prop/2 when it finds terms. If it finds atoms, it just adds them to the result.
% base case
collect_atoms([], []).
% handle atoms
collect_atoms([A|Ps], [A|Rest]) :-
% you could replace the next test with logical_atom/1
atom(A), !,
collect_atoms(Ps, Rest).
% handle terms
collect_atoms([P|Ps], Rest) :-
compound(P), !, % compound/1 tests for terms
atoms_of_prop(P, PAtoms),
collect_atoms(Ps, PsAtoms),
append(PAtoms, PsAtoms, Rest).
% ignore everything else
collect_atoms([_|Ps], Rest) :- atoms_of_prop(Ps, Rest).
This works for your example as-is:
?- atoms_of_prop(ats(and(q, imp(or(p, q), neg(p))), As), Atoms).
Atoms = [p, q].

Partial Dictionary/Record Unification?

I understand that some Prologs support dictionary-like associative data structures out of the box. For the implementations that do, do they support some notion of partial unification with another structure that doesn't actually contain all of the keys?
For example, in the syntax of core.logic/miniKanren:
(run* [q]
(== {:foo 1 :bar 2} (partial-map :foo q)))
This would return a single result where q is bound to 1.
Do Prologs give this operation or this partial structure a name?
In general one works around the poor selection of fundamental data types in Prolog the standard way: by adding libraries and using interfaces. SWI-Prolog, for example, comes with the assoc library that implements an AVL tree-based association data structure. (As an aside, balanced trees are more common in functional and logic programming than hash tables because it's easier to create "persistent" data structures on trees than hash tables—persistent in the FP sense of sharing internal structure.)
Using this library looks something like this:
?- [library(assoc)].
% library(assoc) compiled into assoc 0.00 sec, 97 clauses
true.
?- empty_assoc(Assoc).
Assoc = t.
?- empty_assoc(Assoc), get_assoc(test, Assoc, V).
false.
?- empty_assoc(Assoc), put_assoc(test, Assoc, foo, Assoc2).
Assoc = t,
Assoc2 = t(test, foo, -, t, t).
?- empty_assoc(Assoc),
put_assoc(test, Assoc, foo, Assoc2),
get_assoc(test, Assoc2, Value).
Assoc = t,
Assoc2 = t(test, foo, -, t, t),
Value = foo.
Once you have something that gives you an interface like this, you can define all kinds of logical relations on top of it. Once you have logical relations, Prolog's normal unification machinery will take care of the rest—no special support for this or that data type is required. Based on your requirements, I think what you want is like a subset relation, except checking both that all of one association are in the other and they all have the same value. I guess that would look something like this:
association_subset(Left, Right) :-
forall(gen_assoc(Assoc, Left, Value), get_assoc(Assoc, Right, Value)).
This predicate will only be true if the Left association is a subset of the Right association, as defined above. We can test it and see if it's doing what we want:
simple(Assoc) :-
empty_assoc(Empty),
put_assoc(foo, Empty, foo_test, V1),
put_assoc(bar, V1, bar_test, Assoc).
complex(Assoc) :-
simple(Assoc1),
put_assoc(baz, Assoc1, bazzle, Assoc).
unrelated(Assoc) :-
empty_assoc(Empty),
put_assoc(baz, Empty, bazzle, Assoc).
...
?- simple(X), complex(Y), association_subset(X, Y).
X = t(foo, foo_test, <, t(bar, bar_test, -, t, t), t),
Y = t(baz, bazzle, -, t(bar, bar_test, -, t, t), t(foo, foo_test, -, t, t)).
?- simple(X), simple(Y), association_subset(X, Y).
X = Y, Y = t(foo, foo_test, <, t(bar, bar_test, -, t, t), t).
?- simple(X), unrelated(Y), association_subset(X, Y).
false.
?- complex(X), simple(Y), association_subset(X, Y).
false.
We can translate this to your exact question like so:
left(Assoc) :-
empty_assoc(Empty),
put_assoc(foo, Empty, 1, Assoc).
right(Assoc) :-
left(Assoc1),
put_assoc(bar, Assoc1, 2, Assoc).
?- left(L), right(R), association_subset(L, R), get_assoc(foo, L, Q).
L = t(foo, 1, -, t, t),
R = t(foo, 1, <, t(bar, 2, -, t, t), t),
Q = 1.
I realize that this answer doesn't really answer the question you asked, but I hope it answers the question beneath the question. In other words, there doesn't need to be special support for these data structures—the above predicate could be defined over association lists as well, you can see that all you'd need is the usual ways of making empty associations, adding, testing for, and generating the keys/values of the association and the underlying data structure becomes irrelevant. No special support is necessary either data-structure-wise or unification-wise. Special syntax would certainly make it nicer to look at! But it isn't necessary to get the behavior you desire.
Some Prolog systems such as Eclipse have a record notation. This can be used
when you know in advance the possible keys of your map. But it needs
a type declaration. The record notation is also found in Prolog decendant
languages such as Erlang.
The idea is very simple. You first declare
a record type (some syntax invented here):
:- rectype T{K1,...,Kn}.
Now you can use inside your Prolog program
records, just write (again some syntax invented here):
... T{F1 = V1, .., Fn = Vm} ...
At compile type the record will be converted into a compound
and can then easily be used in normal unification. The conversion
reorders the key value pairs according to the record type
declaration, then drops the keys and uses the positions only.
Unused positions are replaced by annonymous variables or
by default values if the record type declaration also
covers this.
... T(W1, ..., Wn) ...
Your example would work as follows:
:- rectype myrec{foo, bar}
?- myrec{foo=1,bar=2} = myrec{foo=q}
The latter query would be internally executed as:
?- myrec(1,2) = myrec(q,_).
For more details how Eclipse does it, see for example here:
http://www.eclipseclp.org/doc/bips/kernel/syntax/struct-1.html
For dynamic maps where the keyset is not static you can implement
dynamic data structures as the other post about SWI-Prolog AVL
trees shows. Or ask your Prolog system for a handle to a specific
data structure. Implement these with a FFI (Foreign Function Interface)
or access these which are already bundled with the Prolog system.
Eclipse for example bundles a couple, see "Description" section
in the below article:
http://www.eclipseclp.org/doc/bips/kernel/record/index.html
Bye
It is not that clear to me what you actually want, (you have removed the hashing aspect), but maybe you rather want feature terms or feature structures?
They are popular to linguists and have been part of Life.
It is possible to implement them with the help of attributed variables, but so far, I have not seen much of a demand for them.
You can also simulate them a bit clumsily with syntactic unification. It is clumsy because you need to represent each feature with a separate argument (you can do this slightly better, but it is also more complex then). So if your program contains n features, a feature structure will contain n different arguments most of which will never be touched. However, unification will then work directly as intended.

ERROR: "Out of global stack" when processing Prolog list of pairs

In SWI-Prolog, I have a list whose elements are pairs of the form Key-ValuesList. For instance, one such list may look like:
[1-[a,b],2-[],3-[c]]
I would like to transform this list into a nested list of pairs of the form Key-[Value], where Value is an element in ValuesList. The above example would be transformed into:
[[1-[a],2-[],3-[c]], [1-[b],2-[],3-[c]]]
My current solution is the following:
% all_pairs_lists(+InputList, -OutputLists).
all_pairs_lists([], [[]]).
all_pairs_lists([Key-[]|Values], CP) :-
!,
findall([Key-[]|R], (all_pairs_lists(Values,RCP), member(R,RCP)), CP).
all_pairs_lists([Key-Value|Values], CP) :-
findall([Key-[V]|R], (all_pairs_lists(Values,RCP), member(V,Value), member(R,RCP)), CP).
Using this predicate, a call of the form
all_pairs_lists([1-[a,b],2-[],3-[c]],OutputLists).
Binds the variable OutputLists to the desired result mentioned above. While it appears correct, this implementation causes an "Out of global stack" error when InputList has very long lists as values.
Is there a less stack consuming approach to doing this? It would seem like quite a common operation for this type of data structure.
Well, to sum it up, you're doing it wrong.
In Prolog, when we want to express a relation instead of a function (several results possible instead of one), we don't use findall/3 and member/2 directly. We rather state what the relation is and then maybe once it's done if we need a list of results we use findall/3.
Here what it means is that we want to express the following relation:
Take a list of Key-Values and return a list of Key-[Value] where Value is a member of the Values list.
We could do so as follows:
% The base case: handle the empty list
a_pair_list([], []).
% The case where the Values list is empty, then the resulting [Value] is []
a_pair_list([Key-[]|List], [Key-[]|Result]) :-
a_pair_list(List, Result).
% The case where the Values list is not empty, then Value is a member of Values.
a_pair_list([Key-[Not|Empty]|List], [Key-[Value]|Result]) :-
member(Value, [Not|Empty]),
a_pair_list(List, Result).
Once this relation is expressed, we can already obtain all the info we wish:
?- a_pair_list([1-[a, b], 2-[], 3-[c]], Result).
Result = [1-[a], 2-[], 3-[c]] ;
Result = [1-[b], 2-[], 3-[c]] ;
false.
The desired list is now just a fairly straight-forward findall/3 call away:
all_pairs_lists(Input, Output) :-
findall(Result, a_pair_list(Input, Result), Output).
The important thing to remember is that it's way better to stay away from extra logical stuff: !/0, findall/3, etc... because it's often leading to less general programs and/or less correct ones. Here since we can express the relation stated above in a pure and clean way, we should. This way we can limit the annoying use of findall/3 at the strict minimum.
As #Mog already explained clearly what the problem could be, here a version (ab)using of the basic 'functional' builtin for list handling:
all_pairs_lists(I, O) :-
findall(U, maplist(pairs_lists, I, U), O).
pairs_lists(K-[], K-[]) :- !.
pairs_lists(K-L, K-[R]) :- member(R, L).
test:
?- all_pairs_lists([1-[a,b],2-[],3-[c]],OutputLists).
OutputLists = [[1-[a], 2-[], 3-[c]], [1-[b], 2-[], 3-[c]]].

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