Prolog逻辑/爱因斯坦之谜 [英] Prolog Logic/Einstein Puzzle

问题描述

问题是

布朗，克拉克，琼斯和史密斯是四个重要公民，虽然不一定分别担任建筑师，银行家，医生和律师，但为社区服务. 布朗比琼斯更保守，但比史密斯更宽松.比年龄更大的布朗，他的高尔夫球手更好，而且比克拉克的年轻人，布朗的收入更高.

Brown, Clark, Jones and Smith are four substantial citizens who serve the community as architect, 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 older than he is and has a larger income than the men who are younger than Clark

比建筑师收入更高的银行家，既不是最年轻也不是最年长的.

The banker who earns more than the architect, is neither the youngest or the oldest.

比高尔夫球员更穷的高尔夫球手的医生，不如建筑师那么保守

The doctor, who is a poorer golfer than the lawyer, is less conservative than the architect

可以预见，年龄最大的人是最保守的人，收入最高，而年龄最小的人是最好的高尔夫球手

As might be expected, the oldest man is the most conservative and has the largest income, and the youngest man is the best golfer

每个人的职业是什么?

我写了

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).

当我问到

?- jobs(L).

我知道

ERROR: >/2: Arguments are not sufficiently instantiated

我不确定错误是什么意思，我相信我已经翻译了所有线索.

I'm not sure what the error means, I believe I've translated all the clues.

推荐答案

(继续CapelliC开拓的道路...)从域中选择并(最好是 while )应用规则这种困惑的方式.尽快进行仔细测试，以尽早消除错误的选择-但不要很快.

(continuing the trail blazed by CapelliC...) Selecting from domains and (better yet, while) applying the rules is usually the way to go in such puzzles. Carefully testing as soon as possible, to eliminate wrong choices as soon as possible - but not sooner.

我们无法对未知值进行算术比较，这就是错误的含义:>比较其参数实例化到的两个已知算术值.但是，如果尚未实例化Prolog逻辑变量，则意味着其值仍然未知.

We can't arithmetically compare unknown values, this is what the error means: > compares two known arithmetical values to which its arguments are instantiated. But if a Prolog logical variable is not yet instantiated it means that its value is still unknown.

在约束逻辑编程(CLP)中，我们可以预先注册此类约束，但不能在原始Prolog中注册.尽管许多现代的Prolog中都有CLP软件包或谓词. SWI Prolog也有.但是在原始的Prolog代码中，我们必须小心.

In constraint logical programming (CLP) we can register such constraints upfront, - but not in vanilla Prolog. Though many a modern Prolog has CLP packages or predicates available in them. SWI Prolog has it too. But in vanilla Prolog code, we must be careful.

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.

这实际上是一个解决方案，具体方法取决于提出难题的方式.

This is actually one solution, according to the way the puzzle question is asked.

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