如何在所有参数模式的后继算术中实现阶乘序列? [英] How to implement the factorial sequence in successor arithmetics for all argument modes?

问题描述

下面的 Prolog 程序定义了一个谓词 fact/2，用于在后续算术中计算整数的阶乘:

事实(0，s(0)).事实(s(X)，Y):-事实(X，Z)，产品(s(X)，Z，Y).产品(0，_，0).产品(s(U)，V，W):-总和(V，X，W)，产品(V，U，X).总和(0，Y，Y).总和(s(X)，Y，s(Z)):-总和(X，Y，Z).

它适用于这种参数模式下的查询:

?- 事实(s(0), s(0)).真的;错误的.

它也适用于这种参数模式下的查询:

?- 事实(s(0), Y).Y = s(0);错误的.

它也适用于这种参数模式下的查询:

?- 事实(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查询会耗尽资源:

?- 事实(X, s(0)).X = 0;X = s(0);超出堆栈限制 (0.2Gb)堆栈大小:本地:4Kb，全局:0.2Gb，跟踪:0Kb筹码深度:2,503,730，最后跟注:100%，选择点数:13在:[2,503,730] sum('<garbage_collected>'，_1328，_1330)[38] prod('<garbage_collected>', <compound s/1>, '<garbage_collected>')[33] 事实('<garbage_collected>'，<compound s/1>)[32] 事实('<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?

解决方案

第一个问题一定是为什么?failure-slice 有助于理解问题:

<上一页>事实(0，s(0)):- 错误.fact(s(X), Y) :- fact(X, Z), false, prod(s(X), Z, Y).

仅当给出第一个参数时，此片段才会终止.如果不是，则无法防止不终止，因为 Y 在可见部分不受任何限制.所以我们必须改变那部分.一个简单的方法是观察第二个参数不断增加.其实增长的挺快的，不过为了终止，一个就够了:

fact2(N, F) :-事实2(N，F，F).事实2(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 证明.

fact2(A,B)terminates_if b(A);b(B).％ 最佳.找到循环:[fact2(s(_),s(_))].NTI 耗时 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| 成比例的时间空间！那不是感叹号，而是阶乘符号……

The following Prolog program defines a predicate fact/2 for computing the factorial of an integer in successor arithmetics:

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).

It works with queries in this argument mode:

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

It also works with queries in this argument mode:

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

It also works with queries in this argument mode:

?- 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))))))
;  …

But it exhausts resources with queries in this argument mode:

?- 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>')

How to implement the factorial sequence in successor arithmetics for all argument modes?

解决方案

The first question must be why? A failure-slice helps to understand the problem:

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).

And, should I add, this can be even proved.

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

But, there is a caveat...

If only F is known, the program will now require temporally space proprotional to |F|! That is not an exclamation point but a factorial sign...

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