% Solution to Expert Systems 1995 Final Exam

% Note: the final was more difficult than I intended, I apologize for any
%  unusual discomfort this may have caused.

% This file can be read by the Prolog interpretor. 

%
% Problem 1.
%

% The data immediately below is not part of the solution, it is there
% so the solution can be conveniently run.

:-[-'/home/ee4785/ee4785/shellOAVbc.pro'].

start:-wipe,deduce(partsCheck,X),sOut<<"Value of partsCheck is "<<X<<nl.

data(part1,isa,part,init).
data(part1,color,red,init).
data(part1,texture,soft,init).
data(part1,shape,round,init).

data(part2,isa,part,init).
data(part2,color,blue,init).
data(part2,texture,rough,init).
data(part2,shape,round,init).

data(part3,isa,part,init).
data(part3,color,green,init).
data(part3,shape,oval,init).


% Exactly one of the two lines below should be commented out.

% data(part3,texture,smooth,init).  % Comment out for partsCheck=true
data(part3,texture,rough,init). % Comment out for partsCheck=false


% The solution starts below:

r1:if not(O:.color=red) 
    or 
     O:.texture=soft or O:.shape=round
   then O:.cond1.

r2:if not(O:.color=blue) 
    or
     O:.texture=T 
    and thereExist(A:.isa=part, not(A=O) and A:.texture=T)
   then O:.cond2.

r3:if forAll(O:.isa=part, true, O:.cond1 and O:.cond2 ) then partsCheck.


%
% Problem 2.
%

trans(M,T):-transp(M,H,H,T).

transp([[F|M]|EM],Left,[M|H],[[F|Right]|More]):-transp(EM,Left,H,[Right|More]).

transp([],L,[],[[]|T]):-transp(L,H,H,T).

transp([[]|_],[],[],[]).

%
% Problem 3.
%

% See file fin95solp3.pro


% 
% Problem 4a.
% 

% Assumption: lowest-numbered rule used after priority considered.
% 
% Step    Database        Agenda
% """"""""""""""""""""""""""""""
% 0       start		<r1>
% 1	start a b c	<r2> r3 r6
% 2	start a b d	r6 <r4>
% 3	start a b c d	r6 <r3>
% 4	start b c d e	<r5>
% 5	start b c d e f
% 
% rx indicates an activation on the agenda.
% <rx> indicates the activation on the agenda which will fire.
% 
% 
% Problem 4b.
% 
% The correct answer could have been written in computer-engineering or
% logic notation. (The answer below is written using "and" and "or" so
% that it can be used with a text device.)
% 
% r1:  not(a or b) or c
% r2:  not(a and b) or c
% r3:  c or e
% 
% Problem 4C.
% 
% Using rules and shell, e would be proven true.  Using logic (as shown
% in 4b), c could either be true or false when both a and b are
% false. Therefore, e could be either true or false.
%