解决此Prolog Caliban问题时的实例化问题

时间:2019-04-16 03:11:44

标签: prolog zebra-puzzle

我正在尝试使用Prolog解决以下Caliban问题:

布朗,克拉克,琼斯和史密斯是为他们服务的4位重要公民     社区为行为学家,银行家,医生和律师,尽管不一定     分别。       布朗比琼斯更保守,但比史密斯更宽松,       比比他年轻并且拥有       比克拉克(Clark)年长的男人收入更高。       比建筑师收入更高的银行家也不是最小的       也不是最古老的。       高尔夫球手比律师差的医生不那么保守       比建筑师。       不出所料,最年长的男人最保守,       收入最高,最年轻的人是最好的高尔夫球手。     每个男人的职业是什么?

我在网上找到了此代码,并试图自己运行它:

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% 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].

但在与?-solutions(L)一起运行时,请继续获取>/2: Arguments are not sufficiently instantiated

有人对此问题有解决方案吗?

0 个答案:

没有答案