prolog - 如何在所有参数模式的后继算术中实现阶乘序列？

Question

以下 Prolog 程序定义了一个谓词 fact/2 用于计算后继算术中整数的阶乘:

fact(0, s(0)).
fact(s(X), Y) :-
  fact(X, Z),
  prod(s(X), Z, Y).

prod(0, _, 0).
prod(s(U), V, W) :-
  sum(V, X, W),
  prod(V, U, X).

sum(0, Y, Y).
sum(s(X), Y, s(Z)) :-
  sum(X, Y, Z).

它适用于这种参数模式下的查询:

?- fact(s(0), s(0)).
   true
;  false.

它也适用于这种参数模式下的查询:

?- fact(s(0), Y).
   Y = s(0)
;  false.

它也适用于这种参数模式下的查询:

?- fact(X, Y).
   X = 0, Y = s(0)
;  X = Y, Y = s(0)
;  X = Y, Y = s(s(0))
;  X = s(s(s(0))), Y = s(s(s(s(s(s(0))))))
;  …

但是在这种参数模式下它会耗尽资源:

?- fact(X, s(0)).
   X = 0
;  X = s(0)
;
Stack limit (0.2Gb) exceeded
  Stack sizes: local: 4Kb, global: 0.2Gb, trail: 0Kb
  Stack depth: 2,503,730, last-call: 100%, Choice points: 13
  In:
    [2,503,730] sum('<garbage_collected>', _1328, _1330)
    [38] prod('<garbage_collected>', <compound s/1>, '<garbage_collected>')
    [33] fact('<garbage_collected>', <compound s/1>)
    [32] fact('<garbage_collected>', <compound s/1>)
    [31] swish_trace:swish_call('<garbage_collected>')

如何在所有参数模式的后继算术中实现阶乘序列？

Answer 1

第一个问题一定是为什么？ failure-slice有助于理解问题:

fact(0, s(0)) :- false.fact(s(X), Y) :- fact(X, Z), false, prod(s(X), Z, Y).

This fragment alone terminates only if the first argument is given. If it is not, then there is no way to prevent non-termination, as Y is not restricted in any way in the visible part. So we have to change that part. A simple way is to observe that the second argument continually increases. In fact it grows quite fast, but for the sake of termination, one is enough:

fact2(N, F) :-
   fact2(N, F, F).

fact2(0, s(0), _).
fact2(s(X), Y, s(B)) :- fact2(X, Z, B), prod(s(X), Z, Y).

而且，我应该补充一下，这甚至可以是 proved .

fact2(A,B)terminates_if b(A);b(B).
    % optimal. loops found: [fact2(s(_),s(_))]. NTI took    0ms,73i,73i

但是，有一个警告......

如果只知道 F，程序现在将需要与 |F| 成正比的时间空间!这不是感叹号而是阶乘符号...