年龄词prob|em在prolog中

我知道:
Nick = 2*Kate
(Kate + Lea)= 2*Nick
(Chris + Nick)= 2*(Kate + Lea)
Ashley = 24
Ashley + Kate + Lea = 2*(Chris + Nick)

取消一些后,我发现:
Ashley = 12*Kate => Kate = 2
Lea = 3*Kate => Lea = 6
Nick = 2*Kate => Nick = 4
Chris = 6*Kate => Chris = 12

我试过这个:

:- use_module(library(clpfd)).

ages(As) :-
   As = [Ashley, Lea, Nick, Chris, Kate],
   Ashley #= 24,
   12 * Kate #= Ashley,
   Lea #= 3*Kate,
   Nick #= 2*Kate,
   Chris #= 6*Kate.

输出:

Ashley = 24,
Chris = 12,
Kate = 2,
Lea = 6,
Nick = 4

上面的解决方案工作正常,但老师告诉我,我不能使用取消.我有点困惑,因为没有取消它不起作用.

:- use_module(library(clpfd)).

ages(As) :-
   As = [Ashley, Lea, Nick, Chris, Kate],
   Ashley #= 24,
   Lea + Kate #= 2 * Nick,
   Chris + Nick #= 2 * (Kate + Lea),
   Nick #= 2 * Kate,
   Ashley + Kate + Lea #= 2*(Nick + Chris).

输出:

Ashley = 24,
Lea+Kate+24#=2*Nick+2*Chris,
Chris+Nick#=2*Lea+2*Kate,
Lea+Kate#=2*Nick,
2*Kate#=Nick

有任何想法吗？

如果age/1按原样调用,则会得到剩余目标(有关详细信息,请参阅CLP(FD)文档)作为答案:

?- ages(As).
As = [24, _G1642, _G1645, _G1648, _G1651],
_G1642+ -2*_G1645+ -2*_G1648+_G1651#= -24,
-2*_G1642+_G1645+_G1648+ -2*_G1651#=0,
_G1642+ -2*_G1645+_G1651#=0,
2*_G1651#=_G1645.

为了获得实际数字,您必须限制数字的范围并随后标记.由于已知没有人达到200年,因此0至200的年龄限制似乎是合理的.然后您的查询提供实际数字作为答案:

?- ages(As), As ins 0..200, label(As).
As = [24, 6, 4, 12, 2].

这解决了您的直接问题.但是,我想补充两点.首先,它产生更清晰可读的代码,以将域限制和标记部分与描述实际关系的谓词分开.其次,它有助于结果的可读性在列表中具有成对的名称年龄.为了说明这些要点,我们定义一个谓词ages2/1来调用核心关系ages2_/2,然后限制和标记年龄.核心关系将年龄与名称分开,并在第二个列表中提供所述年龄,毕竟这些是您要标记的变量.这允许您在调用谓词的单个列表中具有任意复杂的术语ages2/1.随后您可以像在中一样发布约束ages/1.将这些想法放在一起可能看起来像这样:

ages2(NAs) :-
   ages2_(NAs,As),
   As ins 0..200,
   label(As).

ages2_(NAs,As) :-
   NAs = [ashley-Ashley, lea-Lea, nick-Nick, chris-Chris, kate-Kate],
   As = [Ashley, Lea, Nick, Chris, Kate],
   Ashley #= 24,
   Lea + Kate #= 2 * Nick,
   Chris + Nick #= 2 * (Kate + Lea),
   Nick #= 2 * Kate,
   Ashley + Kate + Lea #= 2*(Nick + Chris).

?- ages2(As).
As = [ashley-24, lea-6, nick-4, chris-12, kate-2].