使用 Prolog 解决脑筋急转弯（Master Mind）

Posted 2023-05-08

【中文标题】使用 Prolog 解决脑筋急转弯（Master Mind）

【英文标题】：Using Prolog to solve a brain teaser (Master Mind)

【发布时间】：2020-07-31 05:30:51

【问题描述】：

工作的朋友与我们的 whatsapp 小组分享了这个：

这把锁有一个 3 位数的密码。 只用这些提示你能猜到吗？

我们使用类似于真值表的方法解决了这个问题。不过我很好奇，这在 Prolog 中如何解决？

【问题讨论】：

它在 ECLiPSe、Picat、MiniZinz 或 Potassco 中看起来如何？

肯定是Master Mind。 1977 年，Donald Knuth 证明了密码破解者可以在五步或更少步内解决该模式，使用的算法逐渐减少可能的模式数量。 Donald 再次出击。

这不是斑马拼图，但我会让标签留在那儿。

也很感兴趣：Mastermind is NP-Complete - 2005-12-13

【参考方案1】：

check 谓词的简单编码：

check( Solution, Guess, NValues, NPlaces ) :-
    Solution = [A,B,C],
    Guess   = [X,Y,Z],
    findall( t, (member(E, Guess), member(E, Solution)), Values ),
    length( Values, NValues ),
    ( A=X -> V1    is 1    ; V1      is 0  ),
    ( B=Y -> V2     is 1+V1 ; V2      is V1 ),
    ( C=Z -> NPlaces is 1+V2 ; NPlaces is V2 ).

然后简单地转录线索，不涉及创造力：

puzzle( [A,B,C] ):-
    findall( X, between(0,9,X), XS ),
    select(A,XS,RA), select(B,RA,RB), member(C,RB),
    /* "291": one digit is right and in its place
       "245": one digit is right but in the wrong place
       "463": two digits are right but both are in the wrong place
       "578": all digits are wrong
       "569": one digit is right but in the wrong place */
    check( [A,B,C], [2,9,1], 1, 1 ),
    check( [A,B,C], [2,4,5], 1, 0 ),
    check( [A,B,C], [4,6,3], 2, 0 ),
    check( [A,B,C], [5,7,8], 0, 0 ),
    check( [A,B,C], [5,6,9], 1, 0 ).

运行它：

23 ?- time( puzzle(X) ).
/* 13,931 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) */
X = [3, 9, 4] ;
/* 20,671 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) */
false.

【讨论】：

确实不错。

通常我会尝试变得聪明并通过穿插检查来最小化生成，但是您预先选择了 all，我也这样做了。:) :) 它确实允许对于这个简单的代码。

所以，我为重新表述的问题重新编码，也使用计数，得出了一种相邻的做事方式。是的

【参考方案2】：

这是一种“生成，然后测试”方法。另一种方法是使用 CLP(FD)。

% This anchors the values of A,B,C to the digits

base([A,B,C])  :- member(A,[0,1,2,3,4,5,6,7,8,9]),
                  member(B,[0,1,2,3,4,5,6,7,8,9]),
                  member(C,[0,1,2,3,4,5,6,7,8,9]).

% "291": one digit is right and in its place
% "245": one digit is right but in the wrong place
% "463": two digits are right but both are in the wrong place
% "578": all digits are wrong
% "569": one digit is right but in the wrong place

clue1([A,B,C]) :- A=2 ; B=9; C=1.
clue2([A,B,C]) :- member(2,[B,C]); member(4,[A,C]); member(5,[A,B]).
clue3([A,B,C]) :- permutation([_,6,3], [A,B,C]), [A,B,C]\=[_,6,3].
clue3([A,B,C]) :- permutation([4,_,3], [A,B,C]), [A,B,C]\=[4,_,3].
clue3([A,B,C]) :- permutation([4,6,_], [A,B,C]), [A,B,C]\=[4,6,_].
clue4([A,B,C]) :- A\=5 , B\=7 , C\=8.
clue5([A,B,C]) :- member(5,[B,C]); member(6,[A,C]); member(9,[A,B]).

solution(L)    :- base(L),clue1(L),clue2(L),clue3(L),clue4(L),clue5(L).

准备好了！

?- setof(L,solution(L),Solutions).
Solutions = [[3, 9, 4], [4, 9, 6], [6, 9, 4]].

上面的尝试是错误的，因为...

实际的问题陈述比起初怀疑的要清晰。

这样说是正确的：

"291": one digit is right and in its place
       (and of the other digits, none appears)
"245": one digit is right but in the wrong place
       (and of the other digits, none appears)
"463": two digits are right but both are in the wrong place
       (and the third digit does not appear)
"578": all digits are wrong
       (none of the digits appears in any solution)
"569": one digit is right but in the wrong place
       (and of the other digits, none appears)

这会导致新代码执行显式命中计数，因为通过成员资格检查使上述显式变得乏味。

这最终与 Will Ness 的解决方案相同，只是编码有所不同。

出现另一个问题：在计算“错误位置的值”时，必须计算可能的配对，即丢弃已用于计数的配对元素。另请参阅：Master Mind Rule ambiguity。像我一样使用member/2 不会那样做，必须使用selectchk/3 删除匹配的元素并继续使用减少的列表。下面的代码已相应修复。错误版本在此示例中有效，因为问题仅在错误位置出现重复数字时才会出现。

:- use_module(library(clpfd)).

% This anchors the values of A,B,C to the digits

base([A,B,C])  :- member(A,[0,1,2,3,4,5,6,7,8,9]),
                  member(B,[0,1,2,3,4,5,6,7,8,9]),
                  member(C,[0,1,2,3,4,5,6,7,8,9]).

% "291": one digit is right and in its place
%        (and of the other digits, none appears)
% "245": one digit is right but in the wrong place
%        (and of the other digits, none appears)
% "463": two digits are right but both are in the wrong place
%        (and the third digit does not appear)
% "578": all digits are wrong
%        (== none of them appears in the solution)
% "569": one digit is right but in the wrong place
%        (and of the other digits, none appears)

% Compare guess against clue and:
%
% - Count the number of digits that are "on the right place"
%   and discard them, keeping the part of the guess and clue as
%   "rest" for the next step.
% - Count the number of digits that are "on the wrong place"
%   and discard any pairings found, which is done with 
%   selectchk/3. If one uses member/2 as opposed to 
%   selectchk/2, the "wrong place counting" is, well, wrong.

% Note: - Decisions (guards and subsequent commits) made explicit
%         Usual style would be to share variables in the head instead,
%         then have a "green" or "red" cut as first occurence in the body.
%       - Incrementing the counter is done "early" by a constraint "#="
%         instead of on return by an effective increment,
%         because I feel like it (but is this worse efficiency-wise?)
%       - Explicit repetiton of "selectchk/3" before the green cut,
%         because I want the Cut to stay Green (Could the compiler 
%         optimized this away and insert a Red Cut in the preceding
%         clause? Probably not because Prolog does not carry enough
%         information for it to do so)

right_place_counting([],[],0,[],[]).

right_place_counting([G|Gs],[C|Cs],CountOut,Grest,Crest) :-
   G=C,
   !,
   CountOut#=CountMed+1,
   right_place_counting(Gs,Cs,CountMed,Grest,Crest).

right_place_counting([G|Gs],[C|Cs],CountOut,[G|Grest],[C|Crest]) :-
   G\=C,
   !,
   right_place_counting(Gs,Cs,CountOut,Grest,Crest).

% ---

wrong_place_counting([],_,0).

wrong_place_counting([G|Gs],Cs,CountOut) :-
    selectchk(G,Cs,CsRest),
    !,
    CountOut#=CountMed+1,
    wrong_place_counting(Gs,CsRest,CountMed).

wrong_place_counting([G|Gs],Cs,CountOut) :-
    \+selectchk(G,Cs,_),
    !,
    wrong_place_counting(Gs,Cs,CountOut).

% ---

counting(Guess,Clue,RightPlaceCount,WrongPlaceCount) :-
   right_place_counting(Guess,Clue,RightPlaceCount,Grest,Crest),
   wrong_place_counting(Grest,Crest,WrongPlaceCount).


clue1(Guess) :- counting(Guess,[2,9,1],1,0).
clue2(Guess) :- counting(Guess,[2,4,5],0,1).
clue3(Guess) :- counting(Guess,[4,6,3],0,2).
clue4(Guess) :- counting(Guess,[5,7,8],0,0).
clue5(Guess) :- counting(Guess,[5,6,9],0,1).

solution(L)  :- base(L),clue1(L),clue2(L),clue3(L),clue4(L),clue5(L).

确实

?- solution(L).
L = [3, 9, 4] ;
false.

【讨论】：

嗨@david，感谢您对它的破解。我认为只有 394 是正确答案。根据规则 5... 数字 9 和 6 不能同时出现在答案中

我认为线索 3 的编码不太正确。 （我认为这暗示只有两个数字是正确的，即不是全部三个）

嗨@DavidTonhofer，[4, 9, 6] 或 [6, 9, 4] 不能作为答案，因为线索 5 说 *一位数是正确的......”。因此这意味着并且 9 和 6 不能同时是答案的一部分。它必须包含 9 或 6。

@DavidTonhofer rep = 7,777 嘿嘿。 :)

@WillNess 是时候……解开第七个封印了！