问题是
布朗、克拉克、琼斯和史密斯是四个重要的公民，他们作为建筑师、银行家、医生和律师为社区服务，尽管不一定分别。布朗比琼斯更保守，但比史密斯更自由，比比他年长的男人更优秀，收入也比比克拉克年轻的男人高
收入超过建筑师的银行家既不是最年轻的，也不是最年长的。
医生比律师更穷，没有建筑师那么保守
不出所料，最年长的男人是最保守的，收入最高，最年轻的男人是最好的高尔夫球手。
每个人的职业是什么？
我写过
jobs(L) :- L = [[brown,_,_,_,_,_],
[clark,_,_,_,_,_],
[jones,_,_,_,_,_],
[smith,_,_,_,_,_]],
% [name,job,conservative,golf,income,age]
% conserative: 1 = least conservative, 4 = most conservative
% golf: 1 = worst golfer, 4 = best golfer
% income: 1 = least income, 4 = highest income
% age: 1 = youngest, 4 = oldest
% Brown is more conservative than Jones. Brown is less conservative than Smith.
member([brown,_,C1,_,_,_],L),
member([jones,_,C2,_,_,_],L),
C1 > C2,
member([smith,_,C3,_,_,_],L),
C1 < C3,
% Brown is a better golfer than those older than him.
member([brown,_,_,G1,_,A1],L),
member([_,_,_,G2,_,A2],L),
G1 > G2,
A2 > A1,
% Brown has a higher income than those younger than Clark.
member([brown,_,_,_,I1,_],L),
member([clark,_,_,_,_,A3],L),
member([_,_,_,_,I2,A4],L),
I1 > I2,
A3 > A4,
% Banker has a higher income than architect. Banker is neither youngest nor oldest.
member([_,banker_,_,I3,A5],L),
member([_,architect,_,_,I4,_],L),
I3 > I4,
(A5 = 2;A5 = 3),
% Doctor is a worse golfer than lawyer. Doctor is less conservative than architect.
member([_,doctor,C4,G3,_,_],L),
member([_,lawyer,_,G4,_,_],L),
member([_,architect,C5,_,_,_],L),
G3 < G4,
C4 < C5,
% Oldest is most conservative and has highest income.
member([_,_,4,_,4,4],L),
% Youngest is the best golfer.
member([_,_,_,4,_,1],L).
当我问它时
?- jobs(L).
我得到
ERROR: >/2: Arguments are not sufficiently instantiated
我不确定错误意味着什么，我相信我已经翻译了所有线索。
如果您只使用有限域约束而不是较低级别的算术，则代码完全按预期工作。例如，使用 (#>)/2 而不是 (>)/2 。
在使用约束使其超出此实例化错误后，您将注意到，您的代码有一个拼写错误：banker_ 。此外，您没有正确表述含义，因此您的谓词将产生false。
下面是代码的略微修改版本，更改为使用有限域约束并更正提到的两个错误：
:- use_module(library(clpfd)).
older_worse_golfer([], _, _).
older_worse_golfer([[_,_,_,G,_,A]|Rest], G0, A0) :-
A #> A0 #==> G #< G0,
older_worse_golfer(Rest, G0, A0).
younger_higher_income([], _, _).
younger_higher_income([[_,_,_,_,I,A]|Rest], I0, A0) :-
A #< A0 #==> I0 #> I,
younger_higher_income(Rest, I0, A0).
man_profession_rest([M,P|Rest], M-P, Rest).
jobs(Ls, Vs) :-
Ls = [[brown,_,_,_,_,_],
[clark,_,_,_,_,_],
[jones,_,_,_,_,_],
[smith,_,_,_,_,_]],
maplist(man_profession_rest, Ls, _, Rests),
append(Rests, Vs),
Vs ins 1..4,
% [name,job,conservative,golf,income,age]
% conserative: 1 = least conservative, 4 = most conservative
% golf: 1 = worst golfer, 4 = best golfer
% income: 1 = least income, 4 = highest income
% age: 1 = youngest, 4 = oldest
% Oldest is most conservative and has highest income.
member([_,_,4,_,4,4], Ls),
% Brown is more conservative than Jones. Brown is less
% conservative than Smith.
memberchk([brown,_,C1,_,_,_], Ls),
memberchk([jones,_,C2,_,_,_], Ls),
memberchk([smith,_,C3,_,_,_], Ls),
C1 #> C2,
C1 #< C3,
% Brown is a better golfer than those older than him.
memberchk([brown,_,_,G1,_,A1], Ls),
older_worse_golfer(Ls, G1, A1),
% IMPLIED: Brown is not the oldest
A1 #< 4,
% Brown has a higher income than those younger than Clark.
memberchk([brown,_,_,_,I1,_], Ls),
memberchk([clark,_,_,_,_,A3], Ls),
younger_higher_income(Ls, I1, A3),
% IMPLIED: Clark is not the youngest
A3 #> 1,
% Banker has a higher income than architect. Banker is neither
% youngest nor oldest.
I3 #> I4,
A5 in 2..3,
member([_,banker,_,_,I3,A5], Ls),
member([_,architect,_,_,I4,_], Ls),
% Doctor is a worse golfer than lawyer. Doctor is less
% conservative than architect.
member([_,doctor,C4,G3,_,_], Ls),
member([_,lawyer,_,G4,_,_], Ls),
member([_,architect,C5,_,_,_], Ls),
G3 #< G4,
C4 #< C5,
% Youngest is the best golfer.
member([_,_,_,4,_,1], Ls).
您可以使用label/1来搜索具体的解决方案。正如您在以下查询中看到的那样，有一个关于专业的独特解决方案：
?- time(setof(MP, Ls^Vs^Rs^(jobs(Ls, Vs),
label(Vs),
maplist(man_profession_rest, Ls, MP, Rs)), MP)).
这会产生：
% 124,485 inferences, 0.041 CPU in 0.042 seconds (97% CPU, 3043643 Lips)
MP = [[brown-banker, clark-doctor, jones-architect, smith-lawyer]].
这甚至不需要所有收入水平等都不同。如果需要，可以通过添加以下内容轻松表达此约束：
transpose(Rests, RestsT), maplist(all_different, RestsT)
到这个公式。
（继续由CapelliC开辟的道路...）从域中进行选择并（更好的是，同时）应用规则通常是解决此类难题的方式。尽快仔细测试，尽快消除错误的选择——但不要早点。
我们无法在算术上比较未知值，这就是错误的含义：>比较其参数实例化的两个已知算术值。但是，如果Prolog逻辑变量尚未实例化，则意味着其值仍然未知。
逻辑编程（CLP）中，我们可以预先注册这些约束，但不能在vanilla Prolog中注册。尽管许多现代Prolog都有CLP包或谓词可用。SWI Prolog也有。但是在原版Prolog代码中，我们必须小心。
mselect([A|As],S,Z):- select(A,S,S1), mselect(As,S1,Z).
mselect([],Z,Z). % (* instantiate a domain by selecting from it *)
puzzle(L):- % (* [_,_,Conserv,Golf,Income,Age] *)
L = [ [brown,_,C1,G1,I1,A1],
[clark,_,C2,_ ,I2,A2],
[jones,_,C3,_ ,I3,A3],
[smith,_,C4,_ ,I4,A4] ],
L1 = [[_,_,4,_,4,4], [_,_,_,4,_,1]], % (* 6,7 - oldest, youngest *)
mselect( L1, L, L2), % (* L2: neither youngest nor oldest *)
mselect( [A3,A4], [1,2,3,4], [A2,A1]), A2 > 1, % (* 3b. 1 < A2 < A1 *)
select( C2, [1,2,3,4], [C3,C1,C4]), % (* 1. C3 < C1 < C4 *)
select( [_, banker, _ ,GB,IB,_ ], L2, [P3] ),
mselect( [ [_, archct, CA,GA,IA,_ ], % (* second view into the same matrix *)
[_, doctor, CD,GD,ID,_ ] ], [P3|L1],
[ [_, lawyer, _ ,GL,IL,_ ] ] ),
CD < CA, % (* 5b. *)
mselect( [ID,IL], [1,2,3,4], [IA,IB]), % (* 4a. IA < IB *)
mselect( [GA,GB], [1,2,3,4], [GD,GL]), % (* 5a. GD < GL *)
% (* 2. ( X in L : A1 < AX ) => G1 > GX *)
% (* 3. ( Y in L : AY < A2 ) => I1 > IY ... so, not(A1<A2)! i.e. % 3b. 1 < A2 < A1 *)
forall( (member(X,L), last(X,AX), AX>A1), (nth1(4,X,GX), G1>GX) ),
forall( (member(Y,L), last(Y,AY), A2>AY), (nth1(5,Y,IY), I1>IY) ).
测试： ([_,_,Conserv,Golf,Income,Age])
7 ?- time(( puzzle(_X), maplist(writeln,_X),nl, false; true )).
[brown,banker,3,3,3,3]
[clark,doctor,1,1,1,2]
[jones,archct,2,4,2,1]
[smith,lawyer,4,2,4,4]
[brown,banker,3,3,3,3]
[clark,doctor,1,1,2,2]
[jones,archct,2,4,1,1]
[smith,lawyer,4,2,4,4]
[brown,banker,3,3,2,3]
[clark,doctor,1,1,3,2]
[jones,archct,2,4,1,1]
[smith,lawyer,4,2,4,4]
% (* 2,299 inferences, 0.000 CPU in 0.120 seconds (0% CPU, Infinite Lips) *)
true.
根据谜题问题的提出方式，这实际上是一种解决方案。
这是我对这个问题的回答：
puzzle(Puzzle) :-
Names = [brown,clark,jones,smith],
permute(Names,Conservatives),
% Brown is more conservative than Jones.
ismore(brown,jones,Conservatives),
% Brown is less conservative than Smith.
isless(brown,smith,Conservatives),
permute(Names,Golfs),
permute(Names,Ages),
% Brown is a better golfer than those older than him.
worsethans(brown,Golfs,WorseAtGolfThanBrown),
betterthans(brown,Ages,OlderThanBrown),
members(OlderThanBrown,WorseAtGolfThanBrown),
permute(Names,Incomes),
% Brown has a higher income than those younger than Clark.
worsethans(brown,Incomes,WorseIncomeThanBrown),
worsethans(clark,Ages,YoungerThanClark),
members(YoungerThanClark,WorseIncomeThanBrown),
permute([banker,architect,lawyer,doctor],Jobs),
% Banker has a higher income than architect.
lookup(banker,Jobs,Names,Banker),
lookup(architect,Jobs,Names,Architect),
ismore(Banker,Architect,Incomes),
% Banker is neither youngest nor oldest.
([_,Banker,_,_]=Ages;[_,_,Banker,_]=Ages),
% Doctor is a worse golfer than lawyer.
lookup(doctor,Jobs,Names,Doctor),
lookup(lawyer,Jobs,Names,Lawyer),
ismore(Lawyer,Doctor,Golfs),
% Doctor is less conservative than architect.
ismore(Architect,Doctor,Conservatives),
% Oldest is most conservative and has highest income.
[Oldest,_,_,_]=Ages,
[Oldest,_,_,_]=Conservatives,
[Oldest,_,_,_]=Incomes,
% Youngest is the best golfer.
[_,_,_,Youngest]=Ages,
[Youngest,_,_,_]=Golfs,
Puzzle = [Names,Jobs,c(Conservatives),g(Golfs),i(Incomes),a(Ages)].
它需要以下支持谓词：
ismore(X,Y,Zs) :-
append(Xs,[Y|_],Zs),
member(X,Xs).
isless(X,Y,Zs) :-
append(_,[Y|Xs],Zs),
member(X,Xs).
betterthans(X,Ys,Zs) :-
append(Zs,[X|_],Ys).
worsethans(X,Ys,Zs) :-
append(_,[X|Zs],Ys).
%lookup(X,Xs,Ys,Y)
lookup(X,[X|_],[Y|_],Y).
lookup(X,[_|Xs],[_|Ys],Y) :-
lookup(X,Xs,Ys,Y).
members([], _).
members([X|Xs], Ys) :-
member(X, Ys),
members(Xs, Ys).
select([X|Xs], X, Xs).
select([X|Xs], Y, [X|Ys]) :- select(Xs, Y, Ys).
permute([], []).
permute(Xs, [X|Zs]) :-
select(Xs, X, Ys),
permute(Ys, Zs).
现在，我唯一的问题是这给了我不止一个答案。除非我的逻辑错误，否则这就是我得到的：
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,brown,clark,jones]),g([clark,brown,smith,jones]),i([smith,brown,clark,jones]),a([smith,brown,jones,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,brown,clark,jones]),g([clark,brown,smith,jones]),i([smith,brown,jones,clark]),a([smith,brown,jones,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,brown,clark,jones]),g([clark,brown,smith,jones]),i([smith,jones,brown,clark]),a([smith,brown,jones,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,brown,clark,jones]),g([clark,brown,smith,jones]),i([smith,brown,clark,jones]),a([smith,jones,brown,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,brown,clark,jones]),g([clark,brown,smith,jones]),i([smith,brown,jones,clark]),a([smith,jones,brown,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,brown,clark,jones]),g([clark,brown,smith,jones]),i([smith,jones,brown,clark]),a([smith,jones,brown,clark])]
[[brown,clark,jones,smith],[banker,doctor,architect,lawyer],c([smith,brown,jones,clark]),g([jones,brown,smith,clark]),i([smith,brown,clark,jones]),a([smith,brown,clark,jones])]
[[brown,clark,jones,smith],[banker,doctor,architect,lawyer],c([smith,brown,jones,clark]),g([jones,brown,smith,clark]),i([smith,brown,jones,clark]),a([smith,brown,clark,jones])]
[[brown,clark,jones,smith],[banker,doctor,architect,lawyer],c([smith,brown,jones,clark]),g([jones,brown,smith,clark]),i([smith,clark,brown,jones]),a([smith,brown,clark,jones])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,clark,brown,jones]),g([clark,brown,smith,jones]),i([smith,brown,clark,jones]),a([smith,brown,jones,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,clark,brown,jones]),g([clark,brown,smith,jones]),i([smith,brown,jones,clark]),a([smith,brown,jones,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,clark,brown,jones]),g([clark,brown,smith,jones]),i([smith,jones,brown,clark]),a([smith,brown,jones,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,clark,brown,jones]),g([clark,brown,smith,jones]),i([smith,brown,clark,jones]),a([smith,jones,brown,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,clark,brown,jones]),g([clark,brown,smith,jones]),i([smith,brown,jones,clark]),a([smith,jones,brown,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,clark,brown,jones]),g([clark,brown,smith,jones]),i([smith,jones,brown,clark]),a([smith,jones,brown,clark])]
如果我也说克拉克的收入大于布朗，我可以把它限制在一个单一的解决方案中。
谁能确认我的答案是否正确，是否应该有更多的限制？
在使用
变量之前，您需要将变量绑定到域，最简单的方法是排列/2：
L = [ [brown,J1,C1,G1,I1,A1],
[clark,J2,C2,G2,I2,A2],
[jones,J3,C3,G3,I3,A3],
[smith,J4,C4,G4,I4,A4]],
permutation([1,2,3,4], [C1,C2,C3,C4]),
permutation([1,2,3,4], [I1,I2,I3,I4]),
permutation([1,2,3,4], [A1,A2,A3,A4]),
permutation([1,2,3,4], [G1,G2,G3,G4]),
permutation([banker,archit,doctor,lawyer], [J1,J2,J3,J4]),
现在可以使用规则
% Brown is more conservative than Jones. Brown is less conservative than Smith.
member([brown,_,CB,GB,IB,AB],L),
member([jones,_,CJ,_,_,_],L),
CB > CJ,
member([smith,_,CS,_,_,_],L),
CB < CS,
效率方面，当您选择（通过成员）命名成员时，一次"获取"所有相关变量（稍后使用棕色属性）。还要注意，引用不同的选择变量 J1、C1 等可能会导致不需要的绑定。
一个难以表达的规则是
% Brown is a better golfer than those older than him.
member([_,_,_,GO1,_,AO1],L),
(AO1 > AB, GB > GO1 ; AO1 < AB),
member([_,_,_,GO2,_,AO2],L),
(AO2 > AB, GB > GO2 ; AO2 < AB),
member([_,_,_,GO3,_,AO3],L),
(AO3 > AB, GB > GO3 ; AO3 < AB),
vardiff(GO1,GO2,GO3),
vardiff(AO1,AO2,AO3), % bug: AO1 was GO1
其中 vardiff/3 是一个简单的便利：
vardiff(A,B,C) :- A=B,A=C,B=C.
当然，如果你的Prolog有可用的，CLP（FD）是一个更好的选择。
我的解决方案基于CapelliC所说的
% [name,job,conservative,golf,income,age]
% conserative: 1 = least conservative, 4 = most conservative
% golf: 1 = worst golfer, 4 = best golfer
% income: 1 = lowest income, 4 = highest income
% age: 1 = youngest, 4 = oldest
jobs(L) :- L =
[[brown,J1,C1,G1,I1,A1],
[clark,J2,C2,G2,I2,A2],
[jones,J3,C3,G3,I3,A3],
[smith,J4,C4,G4,I4,A4]],
permutation([1,2,3,4], [C1,C2,C3,C4]),
permutation([1,2,3,4], [I1,I2,I3,I4]),
permutation([1,2,3,4], [A1,A2,A3,A4]),
permutation([1,2,3,4], [G1,G2,G3,G4]),
permutation([banker,architect,doctor,lawyer], [J1,J2,J3,J4]),
% Brown is more conservative than Jones. Brown is less conservative than Smith.
member([brown,_,CB,GB,IB,AB],L),
member([jones,_,CJ,_,_,_],L),
member([smith,_,CS,_,_,_],L),
CB > CJ,
CB < CS,
% Brown is a better golfer than those older than him.
member([_,_,_,G01,_,A01],L),
(A01 > AB, GB > G01 ; A01 < AB),
member([_,_,_,G02,_,A02],L),
(A02 > AB, GB > G02 ; A02 < AB),
member([_,_,_,G03,_,A03],L),
(A03 > AB, GB > G03 ; A03 < AB),
vardiff(G01,G02,G03),
vardiff(G01,A02,A03),
% Brown has a higher income than those younger than Clark.
member([clark,_,_,_,_,AC],L),
member([_,_,_,_,I01,A04],L),
(A04 < AC, IB > I01; AC < A04),
% Banker has a higher income than architect. Banker is neither youngest nor oldest.
member([_,banker,_,_,IBa,ABa],L),
member([_,architect,CAr,_,IAr,_],L),
IBa > IAr,
(ABa = 1, ABa = 4),
% Doctor is a worse golfer than lawyer. Doctor is less conservative than architect.
member([_,doctor,CDo,GDo,_,_],L),
member([_,lawyer,_,GLa,_,_],L),
GDo < GLa,
CDo < CAr,
% Oldest is most conservative and has highest income.
member([_,_,4,_,4,4],L),
% Youngest is the best golfer.
member([_,_,_,4,_,1],L).
vardiff(A,B,C) :- A=B, A=C, B=C.
我得到
3 ?- jobs(L).
L = [[brown,architect,2,4,1,1],[clark,banker,3,1,2,2],[jones,doctor,1,2,3,3],[smith,lawyer,4,3,4,4]] ;
L = [[brown,architect,2,4,1,1],[clark,banker,3,2,2,2],[jones,doctor,1,1,3,3],[smith,lawyer,4,3,4,4]] ;
L = [[brown,architect,2,4,1,1],[clark,banker,3,3,2,2],[jones,doctor,1,1,3,3],[smith,lawyer,4,2,4,4]] ;
L = [[brown,architect,2,4,1,1],[clark,banker,3,1,2,3],[jones,doctor,1,2,3,2],[smith,lawyer,4,3,4,4]] ;
L = [[brown,architect,2,4,1,1],[clark,banker,3,2,2,3],[jones,doctor,1,1,3,2],[smith,lawyer,4,3,4,4]] ;
L = [[brown,architect,2,4,1,1],[clark,banker,3,3,2,3],[jones,doctor,1,1,3,2],[smith,lawyer,4,2,4,4]] ;
L = [[brown,architect,2,4,1,1],[clark,banker,3,1,3,2],[jones,doctor,1,2,2,3],[smith,lawyer,4,3,4,4]] ;
L = [[brown,architect,2,4,1,1],[clark,banker,3,2,3,2],[jones,doctor,1,1,2,3],[smith,lawyer,4,3,4,4]] ;
L = [[brown,architect,2,4,1,1],[clark,banker,3,3,3,2],[jones,doctor,1,1,2,3],[smith,lawyer,4,2,4,4]] ;
L = [[brown,architect,2,4,1,1],[clark,banker,3,1,3,3],[jones,doctor,1,2,2,2],[smith,lawyer,4,3,4,4]] ;
L = [[brown,architect,2,4,1,1],[clark,banker,3,2,3,3],[jones,doctor,1,1,2,2],[smith,lawyer,4,3,4,4]] ;
L = [[brown,architect,2,4,1,1],[clark,banker,3,3,3,3],[jones,doctor,1,1,2,2],[smith,lawyer,4,2,4,4]] ;
重复的多个答案，所以我删除了其中一些。
所有的线索都得到了满足，除了克拉克是第二老的，例如
L = [[brown,architect,2,4,1,1],[clark,banker,3,1,2,3],[jones,doctor,1,2,3,2],[smith,lawyer,4,3,4,4]] ;
此线索被违反
% Brown has a higher income than those younger than Clark.
对于我所有的答案，布朗是最年轻的，所以像这样的线索
% Brown is a better golfer than those older than him.
% Brown has a higher income than those younger than Clark.
% Youngest is the best golfer.
似乎有点毫无意义...