我最近在 Google Play 应用商店中发现了一款名为Cryptogram的小游戏。有几十个类似的应用程序。这个想法是将数字与颜色相匹配，以使所有方程式听起来都正确。
我能够很快地通过手工解决问题 1-8 和问题 10，但问题 9 对我来说更难。
问题 9
经过一段时间的修补和猜测，我放弃了，决定编写一个解决方案。作为本科生，我使用 Prolog/Datalog 完成了一些小任务以及一些Project Euler问题。以前我见过15 行数独求解器，它使用 Prolog 的有限域约束逻辑编程 (clpfd) 库，我决定自己试一试。我正在使用SWI-Prolog。
:- use_module(library(clpfd)).
problem(Colors) :-
Colors = [Pink, Cyan, Yellow, Green, Purple, Red, Brown, White, Lime],
Colors ins 0..9,
all_distinct(Colors),
% The leading digit of a number can't be 0
Pink #\= 0,
Red #\= 0,
White #\= 0,
Green #\= 0,
Lime #\= 0,
Cyan #\= 0,
% I originally tried to write a predicate generalizing numbers and a list of digits
% but got in way over my head with CLPFD.
Number1_1 #= (Pink * 1000) + (Cyan * 100) + (Pink * 10) + Yellow,
Number1_2 #= (Green * 10) + Purple,
Number1_3 #= (Cyan * 100) + (Red * 10) + Purple,
Number2_1 #= (Red * 1000) + (Brown * 100) + (White * 10) + Red,
Number2_2 #= (Lime * 10) + Yellow,
Number2_3 #= (Red * 1000) + (Lime * 100) + (Purple * 10) + Pink,
Number3_1 #= (White * 1000) + (Purple * 100) + (Cyan * 10) + White,
Number3_2 #= (Green * 1000) + (Cyan * 100) + (Yellow * 10) + Purple,
Number3_3 #= (Cyan * 1000) + (Red * 100) + (Yellow * 10) + Red,
% I'm not 100% sure whether to use floored or truncated division here.
% I thought the difference would be a float vs integer output,
% but that doesn't make sense with finite domains.
Number1_1 // Number1_2 #= Number1_3,
Number1_1 rem Number1_2 #= 0,
Number2_3 #= Number2_1 + Number2_2,
Number3_3 #= Number3_1 - Number3_2,
Number3_1 #= Number1_1 - Number2_1,
Number3_2 #= Number1_2 * Number2_2,
Number3_3 #= Number1_3 + Number2_3.
当我在 SWI-Prolog 中运行此查询时的输出让我觉得我误解了 CLPFD 中的一个大概念：
?- problem([Pink, Cyan, Yellow, Green, Purple, Red, Brown, White, Lime]).
Pink in 3..9,
_7756#=Pink+10*Purple+1000*Red+100*Lime,
_7810#=1010*Pink+100*Cyan+Yellow,
all_distinct([Pink, Cyan, Yellow, Green, Purple, Red, Brown, White|...]),
Cyan in 1..7,
_7946#=1000*Cyan+10*Yellow+101*Red,
_7994#=100*Cyan+10*Yellow+1000*Green+Purple,
_8048#=10*Cyan+100*Purple+1001*White,
_8096#=100*Cyan+Purple+10*Red,
Yellow in 0..9,
_8162#=Yellow+10*Lime,
Green in 1..7,
_8216#=10*Green+Purple,
Purple in 0..9,
Red in 1..7,
_8294#=1001*Red+100*Brown+10*White,
Brown in 0..9,
White in 2..8,
Lime in 1..9,
_7756 in 1103..7568,
_8096+_7756#=_7946,
_8294+_8162#=_7756,
_8096 in 110..779,
_7810//_8216#=_8096,
_7810 in 3334..9799,
_8048+_8294#=_7810,
_7810 rem _8216#=0,
_8048 in 2313..8778,
_7946+_7994#=_8048,
_7946 in 1213..7678,
_7994 in 1100..7565,
_8216*_8162#=_7994,
_8216 in 12..79,
_8162 in 14..99,
_8294 in 1021..7486.
我希望颜色列表中的每种颜色都绑定到 0..9 范围内的单个不同整数，但这不是正在发生的事情。你能帮我找到解决这个问题的方法吗？
编辑
所以我选择了一种任意颜色，并开始在约束所说的有效范围内为其分配数字。我在 Cyan 绑定到 1 的情况下运行了这个查询。
?- problem([Pink, 1, Yellow, Green, Purple, Red, Brown, White, Lime]).
false.
这没有任何意义。前面的“输出”说“1..7 中的青色”，我认为这意味着该范围内的任何值都是有效的。但是，如果我为 Cyan 选择另一个任意值：
?- problem([Pink, 2, Yellow, Green, Purple, Red, Brown, White, Lime]).
Pink = 7,
Yellow = 6,
Green = 3,
Purple = 4,
Red = 1,
Brown = 8,
White = 5,
Lime = 9.
我得到了我正在寻找的答案。虽然 Cryptogram 解决了，但我仍然不明白为什么 Prolog 的 CLPFD 库没有完全独立地找到它。
编辑 2
我使用您的建议来清理代码。我还重新引入了将数字与数字联系起来的谓词。此代码块完美运行。
:- use_module(library(clpfd)).
digit_number(0, [], 1).
digit_number(Number, [Digit|Tail], DigitPlace) :-
digit_number(NextNumber, Tail, NextDigitPlace),
DigitPlace #= NextDigitPlace * 10,
PlaceNumber #= Digit * (NextDigitPlace),
Number #= PlaceNumber + NextNumber.
digit_number(Number, ColorList) :-
digit_number(Number, ColorList, _).
problem(Colors) :-
Colors = [Pink, Cyan, Yellow, Green, Purple, Red, Brown, White, Lime],
Colors ins 0..9,
all_distinct(Colors),
digit_number(Number1_1, [Pink, Cyan, Pink, Yellow]),
digit_number(Number1_2, [Green, Purple]),
digit_number(Number1_3, [Cyan, Red, Purple]),
digit_number(Number2_1, [Red, Brown, White, Red]),
digit_number(Number2_2, [Lime, Yellow]),
digit_number(Number2_3, [Red, Lime, Purple, Pink]),
digit_number(Number3_1, [White, Purple, Cyan, White]),
digit_number(Number3_2, [Green, Cyan, Yellow, Purple]),
digit_number(Number3_3, [Cyan, Red, Yellow, Red]),
Number1_1 // Number1_2 #= Number1_3,
Number1_1 rem Number1_2 #= 0,
Number2_1 + Number2_2 #= Number2_3,
Number3_1 - Number3_2 #= Number3_3,
Number1_1 - Number2_1 #= Number3_1,
Number1_2 * Number2_2 #= Number3_2,
Number1_3 + Number2_3 #= Number3_3,
label(Colors).