/* 

  Analogy in Picat.

  Evans' Analogy program.

  This Picat version is based on the Prolog code from
  http://www.ee.pdx.edu/~mperkows/ANALOGY/009.moreProlog-reasoning-by-analogy.ppt
  """
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  %%% analogy.pro          Dave Reed          1/23/02
  %%%
  %%% A program based on Evans' analogy reasoner.
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  """

  This Picat model was created by Hakan Kjellerstrand, hakank@gmail.com
  See also my Picat page: http://www.hakank.org/picat/

*/
import util.

main => go.

go ?=>
  member(Q,[q1,q2,q3,q4]),
  nl,
  println(q=Q),
  analogy(Q, _Answer),
  % write(Answer), nl,
  fail,
  nl.

go => true.

% A small square inside a large triangle (A)
% is to
% A small triangle inside a large square (B)
% as
% A small circle inside a large square   (C)
% is to
% A ....  (the solution)
%
question(q1,A,B,C,L) =>
  A = $inside((small,square), (large, triangle)),
  B = $inside((small, triangle), (large, square)),
  C = $inside((small, circle), (large, square)),
  L = $[inside((small, circle), (large, triangle)),
       inside((small, square), (large, circle)),
       inside((small, triangle), (large, square))].

question(q2, A,B,C,L) =>
  A = $inside((small, circle), (large, square)),
  B = $inside((small, square), (large, circle)),
  C = $above((small, triangle), (large, triangle)),
  L = $[above((small, circle), (large, circle)),
        inside((small, triangle), (large, triangle)),
        above((large, triangle), (small, triangle))].

question(q3,A,B,C,L) =>
  A = $above((small, square), (large, circle)),
  B = $above((large, square), (small, circle)),
  C = $above((small, circle), (large, triangle)),
  L = $[above((large, circle), (small, triangle)),
        inside((small, circle), (large, triangle)),
        above((large, triangle), (small, square))].

% This has an ambiguity: see comment below.
question(q4,A,B,C,L) =>
  A = $above((small, square), (large, square)),
  B = $above((large, square), (small, square)),
  C = $above((large, circle), (small, triangle)),
  L = $[above((small, circle), (large, triangle)),
        inside((small, circle), (large, triangle)),
        above((small, triangle), (large, circle))].

analogy(Question, Solution) => 
  question(Question, F1, F2, F3, Answers),
  transform(Rule, F1, F2), % F1 is to F2
  transform(Rule, F3, Solution), % as F3 is to Solution
  member(Solution, Answers),
  print_solution(F1,F2,F3,Solution).

print_solution(F1,F2,F3,Solution) => 
  printf("%w\nis to\n%w\nas\n%w\nis to\n%w\n",convert(F1),convert(F2),convert(F3),convert(Solution)),
  nl.

convert(S) = to_fstring("a %w %w a %w", S[1].sc(),S.name,S[2].sc()).

% replace comma with space
sc(S) = S.to_string().replace(',',' ').

transform(Type,A,B) ?=>
  Type = invertPosition,
  A = $inside((small,Figure1), (large, Figure2)),
  B = $inside((small,Figure2), (large, Figure1)).

transform(Type,A,B) ?=>
  Type = invertPosition,
  A = $above((Size1, Figure1), (Size2, Figure2)),
  B = $above((Size2, Figure2), (Size1, Figure1)).

transform(Type,A,B) ?=>
  Type = invertSizes,
  A = $inside((small, Figure1), (large, Figure2)),
  B = $inside((small, Figure2), (large, Figure1)).

transform(Type,A,B) => 
  Type = invertSizes,
  A = $above((Size1, Figure1), (Size2, Figure2)),
  B = $above((Size2, Figure1), (Size1, Figure2)).
   
   

/*
?- analogy(q1, Answer).

Answer = inside(small sized square, large sized circle) ; 
Answer = inside(small sized square, large sized circle) ; 

No

?- analogy(q2, Answer).

Answer = above(large sized triangle, small sized triangle) ;
Answer = above(large sized triangle, small sized triangle) ;

No

?- analogy(q3, Answer).

Answer = above(large sized circle, small sized triangle) ;

No

It is possible for 2 different transformations to produce different 
answers must either refine the transformations (difficult)
use heuristics to pick most likely answer (approach taken by Green)
?- analogy(q4, Answer).

Answer = above(small sized triangle, large sized circle) ;
Answer = above(small sized circle, large sized triangle) ;

No


*/