布朗，克拉克，琼斯和史密斯是4名服务的重要公民 他们的社区是建筑师，银行家，医生和律师，但不是 必然分别。布朗比谁更保守 琼斯，但比史密斯更自由，是一个比男人更好的高尔夫球手 比他年轻，收入比男人多 比克拉克更老。 这位银行家的收入高于建筑师，既不是最年轻也不是最年长的。
比律师更差的高尔夫球手的医生更少 比建筑师保守。 正如可以预料的那样，最年长的人是最保守的，收入最多的人，最年轻的男人是最好的高尔夫球手。 每个人的职业是什么？
当我尝试启动代码时，我收到错误：
ERROR: >/2: Arguments are not sufficiently instantiated
代码是：
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% We represent each "person" with a six-tuple of the form
%
% [ name , profession , age , income , politics , golf ranking ]
%
% where name is either brown, clark, jones, or smith
% profession is either banker, lawyer, doctor, or architect
% age is a range 1 .. 4, with 1 being the youngest and 4 the oldest
% income is a range 1 .. 4, with 1 being the least and 4 the most
% politics is a range 1 .. 4, with 1 being conservative, 4 liberal
% golf ranking is a range 1 .. 4, 1 for the best rank, 4 for the worst
%
solutions(L) :- L = [ [brown, _, _, _, _, _], [clark, _, _, _, _, _],
[jones, _, _, _, _, _], [smith, _, _, _, _, _] ],
clue1(L),
clue2(L),
clue3(L),
clue4(L),
constrained_profession(L),
constrained_age(L),
constrained_income(L),
constrained_politics(L),
constrained_golf_rank(L).
%
% clue #1
% brown, who is more conservateive than jones but
% more liberal than smith, is a better golfer than
% the men who are younger than he is and has a larger
% income than the men who are older than clark
%
clue1(L) :- member(P1,L), member(P2,L), member(P3,L),
P1 = [brown, _, A1, _, L1, G1],
P2 = [jones, _, _, _, L2, _],
P3 = [smith, _, _, _, L3, _],
liberaler( P2, P1 ),
liberaler( P1, P3 ),
not( clue1_helper_a(L) ),
not( clue1_helper_b(L) ).
% for all men younger than brown he is a better golfer ===>
% it is not the case that there exists a man younger than brown
% such that brown is not a better golfer than him.
% The "is not the case" is taken care of in clue1.
clue1_helper_a(L) :- member(P1,L), P1 = [brown, _, A1, _, L1, G1],
member(PU,L), PU = [_, _, AU, _, _, GU],
younger(PU,P1),
not(golfier(P1, PU)).
% for all men older than clark, brown makes more money than they do ===>
% it is not the case that there exists a man older than clark such that
% brown does not make more money than him.
% The "is not the case" is taken care of in clue1.
clue1_helper_b(L) :- member(P1,L), P1 = [brown, _, _, _, _, _],
member(P2,L), P2 = [clark, _, _, _, _, _],
member(PU,L), PU = [_, _, _, _, _, _],
younger(P2,PU),
not(richer(P1, PU)).
%
% clue #2
% the banker, who earns more than the archiect, is
% neither the youngest nor the oldest
%
clue2(L) :- member(P1,L), member(P2,L),
P1 = [_, banker, A1, I1, _, _],
P2 = [_, architect, _, I2, _, _],
richer(P1,P2),
not( A1 = 1 ),
not( A1 = 4 ).
%
% clue #3
% the doctor, who is a pooer golfer than the lawyer, is
% less conservative than the architect.
%
clue3(L) :- member(P1, L), member(P2, L), member(P3,L),
P1 = [_,doctor, _, _, L1, G1],
P2 = [_,lawyer, _, _, _, G2],
P3 = [_,architect, _, _, L3, _],
golfier(P2,P1),
liberaler(P1,P3).
%
% clue #4
% as might be expected, the oldest man is the most
% conservative and has the largest income, and the
% youngest man is the best golfer.
clue4(L) :- member(P1,L), member(P2,L),
P1 = [_, _, 4, 4, 1, _],
P2 = [_, _, 1, _, _, 1].
%
% relations
%
younger(X,Y) :- X = [_, _, AX, _, _, _], Y = [_, _, AY, _, _, _], AX < AY.
liberaler(X,Y) :- X = [_, _, _, _, LX, _], Y = [_, _, _, _, LY, _], LX > LY.
golfier(X,Y) :- X = [_, _, _, _, _, GX], Y = [_, _, _, _, _, GY], GX < GY.
richer(X,Y) :- X = [_, _, _, IX, _, _], Y = [_, _, _, IY, _, _], IX > IY.
%
% constraints
%
constrained_profession(L) :-
member(P1,L), member(P2,L), member(P3,L), member(P4,L),
P1 = [_, banker, _, _, _, _],
P2 = [_, lawyer, _, _, _, _],
P3 = [_, doctor, _, _, _, _],
P4 = [_, architect, _, _, _, _].
constrained_age(L) :-
member(P1,L), member(P2,L), member(P3,L), member(P4,L),
P1 = [_, _, 1, _, _, _],
P2 = [_, _, 2, _, _, _],
P3 = [_, _, 3, _, _, _],
P4 = [_, _, 4, _, _, _].
constrained_income(L) :-
member(P1,L), member(P2,L), member(P3,L), member(P4,L),
P1 = [_, _, _, 1, _, _],
P2 = [_, _, _, 2, _, _],
P3 = [_, _, _, 3, _, _],
P4 = [_, _, _, 4, _, _].
constrained_politics(L) :-
member(P1,L), member(P2,L), member(P3,L), member(P4,L),
P1 = [_, _, _, _, 1, _],
P2 = [_, _, _, _, 2, _],
P3 = [_, _, _, _, 3, _],
P4 = [_, _, _, _, 4, _].
constrained_golf_rank(L) :-
member(P1,L), member(P2,L), member(P3,L), member(P4,L),
P1 = [_, _, _, _, _, 1],
P2 = [_, _, _, _, _, 2],
P3 = [_, _, _, _, _, 3],
P4 = [_, _, _, _, _, 4].
% end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
答案 0 :(得分：3)
内置谓词(>)/2是一个所谓的模式谓词。这意味着您只能在非常特殊的情况下使用它。你可以在各个方向使用它并不是真正的关系。
为了更一般地推理整数，请使用Prolog系统的 CLP（FD）约束 而不是。
例如，在GNU Prolog中，只需将(>)/2替换为(#>)/2 就可以更加声明地推理整数。
在其他Prolog系统中，您当前可能需要导入专用库。例如，在SICStus Prolog中，您需要添加指令：
:- use_module(library(clpfd)).
到您的计划。
如果执行此操作，然后使用以下稍微修改过的谓词定义：
younger(X,Y) :- X = [_, _, AX, _, _, _], Y = [_, _, AY, _, _, _], AX #< AY. liberaler(X,Y) :- X = [_, _, _, _, LX, _], Y = [_, _, _, _, LY, _], LX #> LY. golfier(X,Y) :- X = [_, _, _, _, _, GX], Y = [_, _, _, _, _, GY], GX #< GY. richer(X,Y) :- X = [_, _, _, IX, _, _], Y = [_, _, _, IY, _, _], IX #> IY.
示例查询运行且没有错误：
?- solutions(L). false.
但是，如您所见，您的配方目前太具体：找不到单一的解决方案。
考虑提交一个单独的问题，看看如何调试此类问题。