我怎么算原子Prolog？

Question

我该如何解决这个问题？

输入：     count_atoms（T，计数）

输出      count_atoms（A（B，C（d，e）中，f）中，计数）。      数= 4;

我真的不知道......请你能帮帮我吗？

Answer 1

也许基于堆栈的方法可以提供帮助。您可以编写至少四个辅助谓词，如下所示：

% Increases accumulator if T is atomic + call to count_atoms/6
count_atoms(T, Stack, StackSize, Accumulator, Count) :- ...

% Gets the arity of T if T is compound + call to count_atoms/6
count_atoms(T, Stack, StackSize, Accumulator, Count) :- ...

% Gets Nth subterm of T and puts it on the stack + call to count_atoms/6
count_atoms(T, N, Stack, StackSize, Accumulator, Count) :- ...

% Pops element from stack + call to count_atoms/5
count_atoms(T, _, Stack, StackSize, Accumulator, Count) :- ...

但你仍然需要一个count_atoms / 2谓词和一个停止算法并产生结果。

Answer 2

使用SWI-Prolog的库（aggregate）和递归：

count_atoms(T, Count) :-
       atom(T)
    -> Count = 1
    ;  compound(T)
    -> aggregate_all(sum(C), (arg(_, T, A), count_atoms(A, C)), Count)
    ;  Count = 0
    .

试验：

?- count_atoms(a(b,c(1,e),f),Count).
Count = 3.

但我担心这不是你的任务的解决方案。对于更基本的东西，您可以使用compound分解任何=..项并递归参数列表。