你能用纯序言写 between/3 吗? [英] Can you write between/3 in pure prolog?

问题描述

我一直试图了解如何从 Prolog 谓词中的回溯生成一系列值.内置谓词 between/3 将在回溯时一次生成一个范围内的所有整数，因此如何编写的示例可能对我的任务有所帮助.

I've been trying to understand how to produce a series of values from a Prolog predicate on backtracking. The builtin predicate between/3 will generate all the integers in a range one at a time on backtracking, so an example of how that is written may help me with my task.

我在现有的 Prolog 系统中寻找了一个实现，但是 GNU Prolog 的 between/3 的实现是一个 C 函数，其中的诀窍是它调用了另一个 C 函数Pl_Create_Choice_Point"这允许它在回溯时产生额外的值.

I looked for an implementation in an existing Prolog system, but the implementation of between/3 for GNU Prolog is a C function, and the trick there is that it calls another C function "Pl_Create_Choice_Point" which allows it to produce additional values on backtracking.

推荐答案

bet(N, M, K) :- N =< M, K = N.
bet(N, M, K) :- N < M, N1 is N+1, bet(N1, M, K).

在行动:

$ swipl
?- [bet].
% bet compiled 0.00 sec, 1,064 bytes
true.

?- bet(1,5, K).
K = 1 n
K = 2 n
K = 3 n
K = 4 n
K = 5 n
false.

如果使用cut，可以防止最终搜索失败，并恢复确切的内置 between/3 行为:

If you use a cut, you can prevent the final search failure, and recover the exact builtin between/3 behavior:

bet(N, M, K) :- N < M, K = N.
bet(N, M, K) :- N == M, !, K = N.
bet(N, M, K) :- N < M, N1 is N+1, bet(N1, M, K).

在行动:

?- [bet].
% bet compiled 0.00 sec, 416 bytes
true.

?- between(1,5,K).
K = 1 n
K = 2 n
K = 3 n
K = 4 n
K = 5.

?- [bet].
% bet compiled 0.00 sec, 240 bytes
true.

?- bet(1,5,K).
K = 1 n
K = 2 n
K = 3 n
K = 4 n
K = 5.

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