Problem with mathematical operation with findall/3 in a List (Prolog) - prolog

Question

I want to multiply elements in a List with findall/3. Specifically I have two functions double(X,Y) which doubles X and square(X,Y) that returns the squared value of X. My problem is that it the operation works only for the first element of the list.
double(X,Y) :- Y is X*2.
square(X,Y) :- Y is X*X.
map_f(Operation,[H|List],[R|Results]) :-
Predicate=..[Operation,H,R],
call(Predicate),
findall(X,( member(X,List) ), Results).
For example, if I type map_f(double,[3,1,2,6,3,1,6],L). ,
I expect the output: L = [6,2,4,12,6,2,12],
but instead it shows:
?- map_f(double, [3, 1, 2, 6, 3, 1, 6], List).
List = [6, 1, 2, 6, 3, 1, 6]
Yes (0.00s cpu)
Any help will be very appreciated.

Answer 1

If you want to use findall/3, you'd have to write it like this:
?- Xs = [3,1,2,6,3,1,6], findall(Y, ( member(X, Xs), double(X, Y) ), Ys).
Xs = [3, 1, 2, 6, 3, 1, 6],
Ys = [6, 2, 4, 12, 6, 2, 12].
If you really want to pass the predicate as an argument and use =.., the logic is still the same, you'd just have to re-write your definition so that it does the right thing:
map_f(Pred_name, L1, L2) :-
Goal =.. [Pred_name, X, Y],
findall(Y, ( member(X, L1), Goal ), L2).
Then:
?- map_f(double, [3,1,2,6,3,1,6], R).
R = [6, 2, 4, 12, 6, 2, 12].
?- map_f(square, [3,1,2,6,3,1,6], R).
R = [9, 1, 4, 36, 9, 1, 36].
But, instead of:
Goal =.. [Pred_name, Arg1, Arg2], Goal
it is easier to use call/N+1:
call(Pred_name, Arg1, Arg2)
So your definition will become:
map_f(Pred_name, L1, L2) :-
findall(Y, ( member(X, L1), call(Pred_name, X, Y) ), L2).
But really, all of this is completely unnecessary if you only have lists. You can just use maplist/N+1, like that:
?- maplist(double, [3,1,2,6,3,1,6], R).
R = [6, 2, 4, 12, 6, 2, 12].
... which iterates over the lists instead of backtracking over them. You can see a maplist implementation for example here:
https://github.com/SWI-Prolog/swipl-devel/blob/2d20d4e8ac28adfcede7a9bd231ea0d9d12d0bbb/library/apply.pl#L195-L205
If your predicate is a real relation (so if it works both ways), you can also use maplist both ways. findall cannot do that! Here is one silly example:
?- maplist(succ, [1,2,3], R).
R = [2, 3, 4].
?- maplist(succ, R, [1,2,3]).
R = [0, 1, 2].
?- map_f(succ, [1,2,3], R).
R = [2, 3, 4].
?- map_f(succ, R, [1,2,3]).
ERROR: Arguments are not sufficiently instantiated