Delete vowels in a list - prolog

Question

Write a program that deletes vowels (String, NoVowelsString) that deletes all vowels from a given string.
So far I've got the condition vowel(X):- member(X,[a,e,i,o,u]). Then I thought of the one that deletes all the elements from the other list:
delete2([],L1,L1).
delete2([H|T],L1,L3) :-
delete2(H,L1,R2),
delete2(T,R2,L3).
So having these two I thought that I could put a condition to those elements being deleted that they have to be a member of [a,e,i,o,u]. Though I still haven't got anywhere.

Answer 1

The following is based on the reification of term equality/inequality.
First, we first define list_memberd_t/3, which behaves just like the memberd_truth/3 but has a different argument order:
list_memberd_t([] ,_,false).
list_memberd_t([Y|Ys],X,Truth) :-
if_(X=Y, Truth=true, list_memberd_t(Ys,X,Truth)).
list_memberd_truth(Xs,X,Truth) :- list_memberd_t(Xs,X,Truth).
For the sake of brevity, let's define memberd_t/3 based on list_memberd_t/3:
memberd_t(X,Xs,Truth) :- list_memberd_t(Xs,X,Truth).
As a parallel to library(apply), let's define tinclude/3:
:- meta_predicate tinclude(2,?,?).
tinclude(P_2,Xs,Zs) :-
list_tinclude_list(Xs,P_2,Zs).
list_tinclude_list([], _P_2,[]).
list_tinclude_list([E|Es],P_2,Fs0) :-
if_(call(P_2,E), Fs0 = [E|Fs], Fs0 = Fs),
list_tinclude_list(Es,P_2,Fs).
tfilter/3 is another name for tinclude/3:
tfilter(P_2,As,Bs) :-
tinclude(P_2,As,Bs).
Next, we define the meta-predicate texclude/3, the opposite of tinclude/3:
:- meta_predicate texclude(2,?,?).
texclude(P_2,Xs,Zs) :-
list_texclude_list(Xs,P_2,Zs).
list_texclude_list([],_,[]).
list_texclude_list([E|Es],P_2,Fs0) :-
if_(call(P_2,E), Fs0 = Fs, Fs0 = [E|Fs]),
list_texclude_list(Es,P_2,Fs).
Now let's use them together!
?- texclude(list_memberd_truth([a,e,i,o,u]),
[d,e,l,e,t,e,' ',v,o,w,e,l,s,' ',i,n,' ',a,' ',l,i,s,t], Filtered).
Filtered = [d, l, t, ' ',v, w, l,s,' ', n,' ', ' ',l, s,t].
Edit
As an alternative to using above texclude/3, let's use tinclude/3 with an auxiliary predicate not/3 to flip the truth value:
:- meta_predicate not(2,?,?).
not(P_2,X,Truth) :-
call(P_2,X,Truth0),
truth_flipped(Truth0,Truth).
truth_flipped(true,false).
truth_flipped(false,true).
Sample query:
?- tinclude(not(list_memberd_truth([a,e,i,o,u])),
[d,e,l,e,t,e,' ',v,o,w,e,l,s,' ',i,n,' ',a,' ',l,i,s,t], Filtered).
Filtered = [d, l, t, ' ',v, w, l,s,' ', n,' ', ' ',l, s,t].

Answer 2

here a solution using DCG. Note how the 'output' is obtained (no arguments passing, only difference lists)
novowels --> ("a";"e";"i";"o";"u"), !, novowels.
% or ..
% novowels --> [C], {memberchk(C, "aeiou")}, !, novowels.
novowels, [C] --> [C], !, novowels.
novowels --> [].
I must confess the second cut doesn't like me, but seems required.
test:
?- phrase(novowels, "abcdefghilmnopq", L),format('~s',[L]).
bcdfghlmnpq
L = [98, 99, 100, 102, 103, 104, 108, 109, 110|...].
edit About the second cut, it seems required by 'left hand' notation: if I code with argument, without cut, I get a correct parsing:
novowels(Cs) --> ("a";"e";"i";"o";"u"), !, novowels(Cs).
% novowels(Cs) --> [C], {memberchk(C, "aeiou")}, !, novowels(Cs).
novowels([C|Cs]) --> [C], novowels(Cs).
novowels([]) --> [].
test:
?- phrase(novowels(L), "abcdefghilmnopq"),format('~s',[L]).
bcdfghlmnpq
L = [98, 99, 100, 102, 103, 104, 108, 109, 110|...] ;
false.
I wonder if this is a bug of the DCG translator, or (more probably) my fault...

Answer 3

Here is the code
deleteV([H|T],R):-member(H,[a,e,i,o,u]),deleteV(T,R),!.
deleteV([H|T],[H|R]):-deleteV(T,R),!.
deleteV([],[]).
What it does?
First it question itself?It's the head a vowel
Yes->We ignore it.
No->We need it.
If it finds an empty list, it constructs the result list, and when returning from backtracking it appends the consonats in front.
This code was tested in SWIProlog.