I am trying to solve the following Caliban problem using Prolog:
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?
I found this code online and am trying to run it myself:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% 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].
but keep getting >/2: Arguments are not sufficiently instantiated when running with ?-solutions(L).
Anyone have a solution to this issue?
