我的IA任务是解决爱因斯坦问题。
我必须使用Prolog中的CSP模型来解决它。我没有得到模型，但只有问题和一些输入数据。我的解决方案必须是通用的，我的意思是，对于一些输入数据，我必须提供一个解决方案。问题的维度是N，例如N可能是5(我们有5栋房子)，但它可能会变化。
我在互联网上找到的许多解决方案都将约束直接放在代码中，但我需要使用输入数据来生成它们。必须使用MAC(保持弧一致性)算法来解决该问题。
我读了很多关于它(爱因斯坦之谜)的书。为了实现这个问题，我需要一个问题的表示。
问题是，我不知道如何在Prolog中准确地表示这个问题(我知道基本的Prolog，没有使用其他库，我们不允许使用clpfd库-Prolog clp求解器)。
我知道我应该从输入(14条线索)中创建约束+来自同一组的所有变量(例如国籍)都应该不同的约束，我可以实现I谓词，比如：
my_all_different(like all_different/1 offered by clpfd).
例如：
Attributes = ['Color', 'Nationality', 'Drink', 'Smoke', 'Pet'].
Values = [['Blue', 'Green', 'Ivory', 'Red', 'Yellow'],
['Englishman', 'Japanese', 'Norwegian', 'Spaniard', 'Ukrainian'],
['Coffee', 'Milk', 'Orange juice', 'Tea', 'Water'],
['Chesterfield', 'Kools', 'Lucky Strike', 'Old Gold', 'Parliament'],
['Dog', 'Fox', 'Horse', 'Snails', 'Zebra']
]).
Statements = 'The Englishman lives in the red house',
'The Spaniard owns the dog',
'Coffee is drunk in the green house',
'The Ukrainian drinks tea',
'The green house is immediately to the right of the ivory house',
'The Old Gold smoker owns snails',
'Kools are smoked in the yellow house',
'Milk is drunk in the middle house',
'The Norwegian lives in the first house',
'The man who smokes Chesterfield lives in the house next to the man with the fox',
'Kools are smoked in the house next to the house where the horse is kept',
'The Lucky Strike smoker drinks orange juice',
'The Japanese smokes Parliament',
'The Norwegian lives next to the blue house'
]).
Question = 'Who owns a zebra'?
现在，我设法解析了这个输入，并获得了一个列表列表：
R = [[red,englishman]
[spaniard,dog]
[green,coffee]
[ukrainian,tea]
[green,ivory,right]
[old,snails]
[yellow,kools]
[milk,middle]
[norwegian,first]
[chesterfield,fox,next]
[kools,horse,next]
[orange,lucky]
[japanese,parliament]
[blue,norwegian,next]].
现在，我想我需要使用这些生成的信息来构建一些约束，根据我所读到的内容，使用二进制约束(我认为用谓词表示)是个好主意，但我也有一些一元约束，那么我应该如何表示约束以包括所有约束呢？
另一个问题是：如何表示变量(我将在其中获得计算数据)，这样我就不需要搜索和修改列表(因为在prolog中，你不能像命令式语言那样修改列表)。
所以我想使用一个变量列表，其中每个变量/元素由一个三元组表示：(var, domain, attrV)，其中var包含变量的当前值，domain是一个列表，比如：[1，2，3，4，..，N]，attrV是相应属性(例如红色)的一个值(N)。一个元素是：(C, [1, 2, 3, 4, 5], red)。
其他问题：我应该如何在prolog中实现MAC算法(使用AC-3算法)，因为我将有一个元组队列，如果不满足约束，这个队列将被修改，这意味着修改变量列表，以及我应该如何修改prolog中的列表。
任何帮助都将不胜感激！
我试图使用您上面给出的链接中的CSP求解器来解决问题的特定版本，但我仍然无法找到解决方案，我想获得解决方案，因为通过这种方式，我将知道如何正确表示通用版本的约束。
添加代码：
% Computational Intelligence: a logical approach.
% Prolog Code.
% A CSP solver using arc consistency (Figure 4.8)
% Copyright (c) 1998, Poole, Mackworth, Goebel and Oxford University Press.
% csp(Domains, Relations) means that each variable has
% an instantiation to one of the values in its Domain
% such that all the Relations are satisfied.
% Domains represented as list of
% [dom(V,[c1,...,cn]),...]
% Relations represented as [rel([X,Y],r(X,Y)),...]
% for some r
csp(Doms,Relns) :-
write('CSP Level'), nl,
ac(Doms,Relns).
% ac(Dom,Relns) is true if the domain constrants
% specified in Dom and the binary relations
% constraints specified in Relns are satisfied.
ac(Doms,Relns) :-
make_arcs(Relns,A),
consistent(Doms,[],A,A),
write('Final Doms '), write(Doms), nl, %test
write('Final Arcs '), write(A), nl. %test
% make_arcs(Relns, Arcs) makes arcs Arcs corresponding to
% relations Relns. There are acrs for each ordering of
% variables in a relations.
make_arcs([],[]).
make_arcs([rel([X,Y],R)|O],
[rel([X,Y],R),rel([Y,X],R)|OA]) :-
make_arcs(O,OA).
% consistent(Doms,CA,TDA,A) is true if
% Doms is a set of domains
% CA is a set of consistent arcs,
% TDA is a list of arcs to do
% A is a list of all arcs
consistent(Doms,CA,TDA,A) :-
consider(Doms,RedDoms,CA,TDA),
write('Consistent Doms '), write(RedDoms), nl, %test
solutions(RedDoms,A),
write('Consistent Doms '), write(RedDoms), nl, %test
write('Consistent Arcs '), write(A), nl. %test
% consider(D0,D1,CA,TDA)
% D0 is the set of inital domains
% D1 is the set of reduced domains
% CA = consistent arcs,
% TDA = to do arcs
consider(D,D,_,[]).
consider(D0,D3,CA,[rel([X,Y],R)|TDA]) :-
choose(dom(XV,DX),D0,D1),X==XV,
% write('D0 '), write(D0),
% write('D1 '), write(D1), nl,
choose(dom(YV,DY),D1,_),Y==YV, !,
prune(X,DX,Y,DY,R,NDX),
( NDX = DX
->
consider(D0,D3,[rel([X,Y],R)|CA],TDA)
; acc_todo(X,Y,CA,CA1,TDA,TDA1),
consider([dom(X,NDX)|D1],D3,
[rel([X,Y],R)|CA1],TDA1)).
% prune(X,DX,Y,DY,R,NDX)
% variable X had domain DX
% variable Y has domain DY
% R is a relation on X and Y
% NDX = {X in DX | exists Y such that R(X,Y) is true}
prune(_,[],_,_,_,[]).
prune(X,[V|XD],Y,YD,R,XD1):-
+ (X=V,member(Y,YD),R),!,
prune(X,XD,Y,YD,R,XD1).
prune(X,[V|XD],Y,YD,R,[V|XD1]):-
prune(X,XD,Y,YD,R,XD1).
% acc_todo(X,Y,CA,CA1,TDA,TDA1)
% given variables X and Y,
% updates consistent arcs from CA to CA1 and
% to do arcs from TDA to TDA1
acc_todo(_,_,[],[],TDA,TDA).
acc_todo(X,Y,[rel([U,V],R)|CA0],
[rel([U,V],R)|CA1],TDA0,TDA1) :-
( X == V
; X == V,
Y == U),
acc_todo(X,Y,CA0,CA1,TDA0,TDA1).
acc_todo(X,Y,[rel([U,V],R)|CA0],
CA1,TDA0,[rel([U,V],R)|TDA1]) :-
X == V,
Y == U,
acc_todo(X,Y,CA0,CA1,TDA0,TDA1).
% solutions(Doms,Arcs) given a reduced set of
% domains, Dome, and arcs Arcs, solves the CSP.
solutions(Doms,_) :-
solve_singletons(Doms),
write('Single Doms '), write(Doms), nl. %test
solutions(Doms,A) :-
my_select(dom(X,[XV1,XV2|XVs]),Doms,ODoms),
split([XV1,XV2|XVs],DX1,DX2),
acc_todo(X,_,A,CA,[],TDA),
( consistent([dom(X,DX1)|ODoms],CA,TDA,A)
; consistent([dom(X,DX2)|ODoms],CA,TDA,A)).
% solve_singletons(Doms) is true if Doms is a
% set of singletom domains, with the variables
% assigned to the unique values in the domain
solve_singletons([]).
solve_singletons([dom(X,[X])|Doms]) :-
solve_singletons(Doms).
% select(E,L,L1) selects the first element of
% L that matches E, with L1 being the remaining
% elements.
my_select(D,Doms,ODoms) :-
select(D,Doms,ODoms), !.
% choose(E,L,L1) chooses an element of
% L that matches E, with L1 being the remaining
% elements.
choose(D,Doms,ODoms) :-
select(D,Doms,ODoms).
% split(L,L1,L2) splits list L into two lists L1 and L2
% with the about same number of elements in each list.
split([],[],[]).
split([A],[A],[]).
split([A,B|R],[A|R1],[B|R2]) :-
split(R,R1,R2).
/* -------------------------------------------------------------------*/
cs1(V, V). %A1 = A2
cs2(V1, V2) :- (V1 is V2 - 1; V2 is V1 - 1). %next
cs3(V1, V2) :- V1 is V2 + 1. %right
zebra(English,Spaniard,Ukrainian,Norwegian,Japanese,
Red,Green,Ivory,Yellow,Blue,
Dog,Snails,Fox,Horse,Zebra,
Coffee,Tea,Milk,Orange_Juice,Water,
Old_Gold,Kools,Chesterfields,Lucky_Strike,Parliaments) :-
csp([dom(English, [1, 2, 3, 4, 5]),
dom(Spaniard, [1, 2, 3, 4, 5]),
dom(Ukrainian, [1, 2, 3, 4, 5]),
dom(Norwegian, [1, 2, 3, 4, 5]),
dom(Japanese, [1, 2, 3, 4, 5]),
dom(Red, [1, 2, 3, 4, 5]),
dom(Green, [1, 2, 3, 4, 5]),
dom(Ivory, [1, 2, 3, 4, 5]),
dom(Yellow, [1, 2, 3, 4, 5]),
dom(Blue, [1, 2, 3, 4, 5]),
dom(Dog, [1, 2, 3, 4, 5]),
dom(Snails, [1, 2, 3, 4, 5]),
dom(Fox, [1, 2, 3, 4, 5]),
dom(Horse, [1, 2, 3, 4, 5]),
dom(Zebra, [1, 2, 3, 4, 5]),
dom(Coffee, [1, 2, 3, 4, 5]),
dom(Tea, [1, 2, 3, 4, 5]),
dom(Milk, [1, 2, 3, 4, 5]),
dom(Orange_Juice, [1, 2, 3, 4, 5]),
dom(Water, [1, 2, 3, 4, 5]),
dom(Old_Gold, [1, 2, 3, 4, 5]),
dom(Kools, [1, 2, 3, 4, 5]),
dom(Chesterfields, [1, 2, 3, 4, 5]),
dom(Lucky_Strike, [1, 2, 3, 4, 5]),
dom(Parliaments, [1, 2, 3, 4, 5])],
[rel([English, Red], cs1(English,Red)),
rel([Spaniard, Dog], cs1(Spaniard,Dog)),
rel([Coffee, Green], cs1(Coffee,Green)),
rel([Ukrainian, Tea], cs1(Ukrainian,Tea)),
rel([Green, Ivory], cs3(Green,Ivory)),
rel([Old_Gold, Snails], cs1(Old_Gold,Snails)),
rel([Kools, Yellow], cs1(Kools,Yellow)),
rel([Milk, Milk], Milk = 3),
rel([Norwegian, Norwegian], Norwegian = 1), %here is the problem
rel([Chesterfields, Fox], cs2(Chesterfields,Fox)),
rel([Kools, Horse], cs2(Kools,Horse)),
rel([Lucky_Strike, Orange_juice], cs1(Lucky_Strike,Orange_juice)),
rel([Japanese, Parliaments], cs1(Japanese,Parliaments)),
rel([Norwegian, Blue], cs2(Norwegian,Blue))]).
我已经做了一些搜索，然后我建议阅读利用样本数据的弧一致性的约束满足
再次编辑到目前为止的努力。遗憾的是，添加最后一个约束会使结果无效。明天我会试着理解的原因
好消息我在下一个/2 中发现了愚蠢的错误
:- include(csp).
next(V1, V2) :-
succ(V1, V2) ; succ(V2, V1).
dom(I, O, D) :-
maplist(dom, I, O),
alldiff(I, [], D).
dom(V, dom(V, [1,2,3,4,5])).
alldiff([], D, D).
alldiff([V|Vs], S, D) :-
maplist(rdiff(V), Vs, Ds),
append(S, Ds, As),
alldiff(Vs, As, D).
rdiff(A, B, D) :- rel(A = B, D).
rel(R, rel([A, B], R)) :- R =.. [_, A, B].
zebra :-
People = [English, Spaniard, Ukrainian, Norwegian, Japanese],
Color = [Red, Green, Ivory, Yellow, Blue],
Pet = [Dog, Snails, Fox, Horse, Zebra],
Drink = [Coffee, Tea, Milk, Orange_Juice, _Water],
Smoke = [Old_Gold, Kools, Chesterfields, Lucky_Strike, Parliaments],
maplist(dom, [People, Color, Pet, Drink, Smoke], DomT, DiffPair),
flatten(DomT, Doms),
maplist(rel,
[English = Red % The Englishman lives in the red house
,Spaniard = Dog % The Spaniard owns the dog
,Ukrainian = Tea % The Ukrainian drinks tea
,Coffee = Green % Coffee is drunk in the green house
,succ(Ivory, Green) % The green house is immediately to the right of the ivory house
,Old_Gold = Snails % The Old Gold smoker owns snails
,Kools = Yellow % Kools are smoked in the yellow house
,Milk = H3 % Milk is drunk in the middle house
,Norwegian = H1 % The Norwegian lives in the first house
,next(Chesterfields, Fox) % The man who smokes Chesterfield lives in the house next to the man with the fox
,next(Kools, Horse) % Kools are smoked in the house next to the house where the horse is kept
,Lucky_Strike = Orange_Juice % The Lucky Strike smoker drinks orange juice
,Japanese = Parliaments % The Japanese smokes Parliament
,next(Norwegian, Blue) % The Norwegian lives next to the blue house
], ConstrS),
flatten([DiffPair, ConstrS], Rels),
csp([dom(H1, [1]), dom(H3, [3])|Doms], Rels),
maplist(writeln,
[people:[English, Spaniard, Ukrainian, Norwegian, Japanese],
color:[Red, Green, Ivory, Yellow, Blue],
pet:[Dog, Snails, Fox, Horse, Zebra],
drink:[Coffee, Tea, Milk, Orange_Juice, _Water],
smoke:[Old_Gold, Kools, Chesterfields, Lucky_Strike, Parliaments]
]).
我已经分离了csp.pl，适用于SWI Prolog。这是
% Computational Intelligence: a logical approach.
% Prolog Code.
% A CSP solver using arc consistency (Figure 4.8)
% Copyright (c) 1998, Poole, Mackworth, Goebel and Oxford University Press.
% csp(Domains, Relations) means that each variable has
% an instantiation to one of the values in its Domain
% such that all the Relations are satisfied.
% Domains represented as list of
% [dom(V,[c1,...,cn]),...]
% Relations represented as [rel([X,Y],r(X,Y)),...]
% for some r
csp(Doms,Relns) :-
ac(Doms,Relns).
% ac(Dom,Relns) is true if the domain constrants
% specified in Dom and the binary relations
% constraints specified in Relns are satisfied.
ac(Doms,Relns) :-
make_arcs(Relns,A),
consistent(Doms,[],A,A).
% make_arcs(Relns, Arcs) makes arcs Arcs corresponding to
% relations Relns. There are acrs for each ordering of
% variables in a relations.
make_arcs([],[]).
make_arcs([rel([X,Y],R)|O],
[rel([X,Y],R),rel([Y,X],R)|OA]) :-
make_arcs(O,OA).
% consistent(Doms,CA,TDA,A) is true if
% Doms is a set of domains
% CA is a set of consistent arcs,
% TDA is a list of arcs to do
% A is a list of all arcs
consistent(Doms,CA,TDA,A) :-
consider(Doms,RedDoms,CA,TDA),
solutions(RedDoms,A).
% consider(D0,D1,CA,TDA)
% D0 is the set of inital domains
% D1 is the set of reduced domains
% CA = consistent arcs,
% TDA = to do arcs
consider(D,D,_,[]).
consider(D0,D3,CA,[rel([X,Y],R)|TDA]) :-
choose(dom(XV,DX),D0,D1),X==XV,
choose(dom(YV,DY),D1,_),Y==YV, !,
prune(X,DX,Y,DY,R,NDX),
( NDX = DX
->
consider(D0,D3,[rel([X,Y],R)|CA],TDA)
; acc_todo(X,Y,CA,CA1,TDA,TDA1),
consider([dom(X,NDX)|D1],D3,
[rel([X,Y],R)|CA1],TDA1)).
% prune(X,DX,Y,DY,R,NDX)
% variable X had domain DX
% variable Y has domain DY
% R is a relation on X and Y
% NDX = {X in DX | exists Y such that R(X,Y) is true}
prune(_,[],_,_,_,[]).
prune(X,[V|XD],Y,YD,R,XD1):-
+ (X=V,member(Y,YD),R),!,
prune(X,XD,Y,YD,R,XD1).
prune(X,[V|XD],Y,YD,R,[V|XD1]):-
prune(X,XD,Y,YD,R,XD1).
% acc_todo(X,Y,CA,CA1,TDA,TDA1)
% given variables X and Y,
% updates consistent arcs from CA to CA1 and
% to do arcs from TDA to TDA1
acc_todo(_,_,[],[],TDA,TDA).
acc_todo(X,Y,[rel([U,V],R)|CA0],
[rel([U,V],R)|CA1],TDA0,TDA1) :-
( X == V
; X == V,
Y == U),
acc_todo(X,Y,CA0,CA1,TDA0,TDA1).
acc_todo(X,Y,[rel([U,V],R)|CA0],
CA1,TDA0,[rel([U,V],R)|TDA1]) :-
X == V,
Y == U,
acc_todo(X,Y,CA0,CA1,TDA0,TDA1).
% solutions(Doms,Arcs) given a reduced set of
% domains, Dome, and arcs Arcs, solves the CSP.
solutions(Doms,_) :-
solve_singletons(Doms).
solutions(Doms,A) :-
select(dom(X,[XV1,XV2|XVs]),Doms,ODoms),
split([XV1,XV2|XVs],DX1,DX2),
acc_todo(X,_,A,CA,[],TDA),
( consistent([dom(X,DX1)|ODoms],CA,TDA,A)
; consistent([dom(X,DX2)|ODoms],CA,TDA,A)).
% solve_singletons(Doms) is true if Doms is a
% set of singletom domains, with the variables
% assigned to the unique values in the domain
solve_singletons([]).
solve_singletons([dom(X,[X])|Doms]) :-
solve_singletons(Doms).
:- redefine_system_predicate(select(_,_,_)).
% select(E,L,L1) selects the first element of
% L that matches E, with L1 being the remaining
% elements.
select(D,Doms,ODoms) :-
% remove(D,Doms,ODoms), !.
system:select(D,Doms,ODoms), !.
% choose(E,L,L1) chooses an element of
% L that matches E, with L1 being the remaining
% elements.
choose(D,Doms,ODoms) :-
% remove(D,Doms,ODoms).
system:select(D,Doms,ODoms).
% split(L,L1,L2) splits list L into two lists L1 and L2
% with the about same number of elements in each list.
split([],[],[]).
split([A],[A],[]).
split([A,B|R],[A|R1],[B|R2]) :-
split(R,R1,R2).
上一次修正后的测试似乎不错/2:
?- zebra.
people:[3,4,2,1,5]
color:[3,5,4,1,2]
pet:[4,3,1,2,5]
drink:[5,2,3,4,1]
smoke:[3,1,2,4,5]
true ;
false.