How to code a prolog program that has same amount of elements

Question

I want to code in prolog, a program that returns true if the list has the exact same amount of a and b or if the list is empty. For Example :{} is true, {aaababbb} is true, {abba} is also true. But {aabbb} is false, {a} is false, {b} is false.

Here is what I tried to do :

language([]).

langage(L) :- langage(L, []).

langage([a, b | A], A).

langage([a|A], B) :- langage(A, C), C = [b|B].

To do this, you would need to [count the number of occurrences](https://stackoverflow.com/questions/9088062/count-the-number-of-occurrences-of-a-number-in-a-list) of `a` and `b` in the list. — Anderson Green, Dec 11 '21 at 21:40

score 1 · Accepted Answer · answered Dec 12 '21 at 16:51

This is a trivial modification to my previous answer, in which the Bal variable keeps the number of as and bs the same.

ab_list2(ABs) :-
    length(ABs, Len),
    ab_list2_(Len, 0, [], ABs).

% 0 characters left to add
% 0 is the balance of as and bs, i.e. same amount of each
ab_list2_(0, 0, ABs, ABs).

ab_list2_(CharsToAdd, Bal, SoFar, ABs) :-
    succ(CharsToAdd0, CharsToAdd),
    add_char(Char, Inc),
    Bal1 is Bal + Inc,
    % Ensure that the balance can be zero for the complete list
    CharsToAdd0 >= abs(Bal1),
    ab_list2_(CharsToAdd0, Bal1, [Char|SoFar], ABs).


add_char(b, -1).
add_char(a, 1).

Result in swi-prolog:

?- time(findnsols(10, L, ab_list2(L), Ls)).
% 387 inferences, 0.000 CPU in 0.000 seconds (100% CPU, 1297961 Lips)
Ls = [[], [a, b], [b, a], [a, a, b, b], [a, b, a, b], [b, a, a, b], [a, b, b, a], [b, a, b, a], [b, b, a, a], [a, a, a, b, b, b]]

How to code a prolog program that has same amount of elements

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