Prolog: ERROR: >/2: Arguments are not sufficiently instantiated
Prolog: ERROR: >/2: Arguments are not sufficiently instantiated
我的 post 末尾的代码应该可以回答以下难题：
Brown, Clark, Jones and Smith are 4 substantial citizens who serve
their community as achitect, banker, doctor and lawyer, though not
necessarily respectively. Brown, who is more conservative 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.
The banker, who earns more than the architect, is neither the youngest nor the oldest.
The doctor, who is a poorer golfer than the lawyer, is less
conservative than the architect.
As might be expected, the oldest man is the most conservative and has the largest income, and the youngest man is the best golfer.
What is each man's profession?
当我尝试启动代码时，出现错误：
ERROR: >/2: Arguments are not sufficiently instantiated
密码是：
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% 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
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
内置谓词 (>)/2 是所谓的 moded 谓词。这意味着您只能在非常特殊的情况下使用它。这不是一个真正的关系，你可以在所有方向上使用。
要更普遍地推理整数，请使用您的 Prolog 系统的 CLP(FD) 约束 代替。
例如，在 GNU Prolog 中，只需 将 (>)/2 替换为 (#>)/2 即可更明确地对整数进行推理。
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)).
到你的程序。
如果你这样做，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 没有错误:
?- solutions(L).
false.
但是，如您所见，您的公式目前过于具体：没有找到一个解决方案。
考虑提交一个单独的问题，看看如何调试此类问题。
我的 post 末尾的代码应该可以回答以下难题：
Brown, Clark, Jones and Smith are 4 substantial citizens who serve their community as achitect, banker, doctor and lawyer, though not necessarily respectively. Brown, who is more conservative 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. The banker, who earns more than the architect, is neither the youngest nor the oldest.
The doctor, who is a poorer golfer than the lawyer, is less conservative than the architect. As might be expected, the oldest man is the most conservative and has the largest income, and the youngest man is the best golfer. What is each man's profession?
当我尝试启动代码时，出现错误：
ERROR: >/2: Arguments are not sufficiently instantiated
密码是：
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% 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
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
内置谓词 (>)/2 是所谓的 moded 谓词。这意味着您只能在非常特殊的情况下使用它。这不是一个真正的关系，你可以在所有方向上使用。
要更普遍地推理整数，请使用您的 Prolog 系统的 CLP(FD) 约束 代替。
例如，在 GNU Prolog 中，只需 将 (>)/2 替换为 (#>)/2 即可更明确地对整数进行推理。
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)).
到你的程序。
如果你这样做，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 没有错误:
?- solutions(L). false.
但是，如您所见，您的公式目前过于具体：没有找到一个解决方案。
考虑提交一个单独的问题，看看如何调试此类问题。