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】：
标签： prolog zebra-puzzle
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】：
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]].
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】：