/* 

  Honey division puzzle in Picat.

  From Martin Chlond Integer Programming Puzzles:
  http://www.chlond.demon.co.uk/puzzles/puzzles1.html, puzzle nr. 6.
  Description  : Honey division puzzle
  Source       : H E Dudeney - Amusements in Mathematics
  """
  6. Once upon a time there was an aged merchant of Baghdad who was much respected by all 
  who knew him. He had three sons, and it was a rule of his life to treat them all exactly 
  alike. Whenever one received a present, the other two were each given one of equal 
  value. One day this worthy man fell sick and died, bequeathing all his possessions to his three 
  sons in equal shares.

  The only difficulty that arose was over the stock of honey. There were exactly twenty-one 
  barrels. The old man had left instructions that not only should every son receive an equal 
  quantity of honey, but should receive exactly the same number of barrels, and that no honey 
  should be transferred from barrel to barrel on account of the waste involved. Now, as seven 
  of these barrels were full of honey, seven were half full, and seven were empty, this was found 
  to be quite a puzzle, especially as each brother objected to taking more than four barrels 
  of the same description - full, half full, or empty.

  Can you show how they succeeded in making a correct division of the property? 
  (Dudeney)
  """

  This model was inspired by the XPress Mosel model created by Martin Chlond.
  http://www.chlond.demon.co.uk/puzzles/sol1s6.html


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

*/


% import util.
import cp.


main => go.


go =>
  Son = 3,
  Cap = 3,
  S = 1..Son,
  C = 1..Cap,

  % Howfull = [1, 0.5, 0]
  HowFull = [2, 1, 0], % multiplies with 2 for the integer version

  X = new_array(Son,Cap),
  X :: 1..4,

  SumX #= sum([X[I,J] : I in S, J in C]),

  %  each son gets 7 barrels
  foreach(I in S)
      sum([X[I,J] : J in C]) #= 7
  end,

  %  each son gets 3.5 units
  foreach(I in S)
     % multiplies with 2 for the integer version
     sum([HowFull[J]*X[I,J] : J in C]) #= 7
  end,

  %  use 7 of each barrel capacity
  foreach(J in C)
     sum([X[I,J] : I in S]) #= 7
  end,

  solve([ff], X.to_list() ++ [SumX]),

  writeln(sumX=SumX),
  foreach(Row in X) writeln(Row.to_list()) end,
  
  nl.