Prolog Logic/Einstein Puzzle

Question

The problem is

Brown, Clark, Jones and Smith are four substantial citizens who serve the community as architect, banker, doctor and lawyer, though not necessarily respe

Answer 1

you need to bind your variables to a domain before to use them, the easiest way is permutation/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]),

now the rules can be used

    % 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,

efficiency wise, when you select (via member) a named member, 'fetch' all related variables at once (brown attributes' are used later). Also beware that referencing in different selection variables J1,C1, etc could lead to unwanted binding.

A rule difficult to express is

    % 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

where vardiff/3 is a simple convenience:

vardiff(A,B,C) :- A\=B,A\=C,B\=C.

Of course, if your Prolog has available, CLP(FD) is a much better choice.