Question

[_, [ X , _ ],_] will match a list like [d, [X,a], s]. Is there a way to match it to any pattern where there is one or more anonymous variables? ie. [[X,a],s] and [[d,a],[p,z], [X,b]] would match?

I am trying to write a program to count the elements in a list ie. [a,a,a,b,a,b] => [[a,4],[b,2]] but I am stuck:

listcount(L, N) :-  listcountA(LS, [], N).
listcountA([X|Tail], [? [X, B], ?], N) :- B is B+1, listcountA(Tail, [? [X,B] ?], N).
listcountA([X|Tail], AL, N) :- listcountA(Tail, [[X,0]|AL], N).

Thanks.

Answer 1

A variable match a term, and the anonimus variable is not exception. A list is just syntax sugar for a binary relation, between head and tail. So a variable can match the list, the head, or the tail, but not an unspecified sequence.

Some note I hope will help you:

listcount(L, N) :- listcountA(LS, [], N).

In Prolog, predicates are identified by name and num.of.arguments, so called functor and arity. So usually ‘service’ predicates with added arguments keep the same name.

listcountA([X|Tail], [? [X, B], ?], N) :- B is B+1, listcountA(Tail, [? [X,B] ?], N).

B is B+1 will never succeed, you must use a new variable. And there is no way to match inside a list, using a ‘wildcard’, as you seem to do. Instead write a predicate to find and update the counter.

A final note: usually pairs of elements are denoted using a binary relation, conveniently some (arbitrary) operator. For instance, most used is the dash.

So I would write

listcount(L, Counters) :-
    listcount(L, [], Counters).

listcount([X | Tail], Counted, Counters) :-
    update(X, Counted, Updated),
    !, listcount(Tail, Updated, Counters).
listcount([], Counters, Counters).

update(X, [X - C | R], [X - S | R]) :-
    S is C + 1.
update(X, [H | T], [H | R]) :-
    update(X, T, R).
update(X, [], [X - 1]).  % X just inserted

update/3 can be simplified using some library predicate, ‘moving inside’ the recursion. For instance, using select/3:

listcount([X | Tail], Counted, Counters) :-
    ( select(X - C, Counted, Without)
    ->  S is C + 1
    ;   S = 1, Without = Counted
    ),
    listcount(Tail, [X - S | Without], Counters).
listcount([], Counters, Counters).