"""
KenKen puzzle in cpmpy.

http://en.wikipedia.org/wiki/KenKen
'''
KenKen or KEN-KEN is a style of arithmetic and logical puzzle sharing
several characteristics with sudoku. The name comes from Japanese and
is translated as 'square wisdom' or 'cleverness squared'.
...
The objective is to fill the grid in with the digits 1 through 6 such that:

  * Each row contains exactly one of each digit
  * Each column contains exactly one of each digit
  * Each bold-outlined group of cells is a cage containing digits which
    achieve the specified result using the specified mathematical operation:
      addition (+),
      subtraction (-),
      multiplication (x),
      and division (/).
      (Unlike in Killer sudoku, digits may repeat within a group.)

...
More complex KenKen problems are formed using the principles described
above but omitting the symbols +, -, x and /, thus leaving them as
yet another unknown to be determined.
'''

The solution is:

  5 6 3 4 1 2
  6 1 4 5 2 3
  4 5 2 3 6 1
  3 4 1 2 5 6
  2 3 6 1 4 5
  1 2 5 6 3 4


This cpmpy model was written by Hakan Kjellerstrand (hakank@gmail.com)
See also my cpmpy page: http://hakank.org/cpmpy/
  
"""
from cpmpy import *
import cpmpy.solvers
import numpy as np
from cpmpy_hakank import *



#
# Ensure that the sum of the segments
# in cc == res
#
def calc(cc, x, res):

  constraints = []
  if len(cc) == 2:

    # for two operands there may be
    # a lot of variants
    c00, c01 = cc[0]
    c10, c11 = cc[1]
    a = x[c00-1,c01-1]
    b = x[c10-1,c11-1]
    constraints += [(a+b==res) |
                    (a*b==res) |
                    (a*res==b) |
                    (b*res==a) |
                    (a-b==res) |
                    (b-a==res)
      ]

  else:
    # res is either sum or product of the segment
    xx = [x[i[0]-1, i[1]-1] for i in cc]
    constraints += [(sum(xx) == res) |
                    # (reduce(lambda a, b: a * b, xx) == res)
                    # (prod(xx,res)
                    (prod1(xx) == res)
                    ]

  return constraints

def kenken2():

  model = Model()

  # data

  # size of matrix
  n = 6

  # For a better view of the problem, see
  #  http://en.wikipedia.org/wiki/File:KenKenProblem.svg

  # hints
  #    [sum, [segments]]
  # Note: 1-based
  problem = [[11, [[1, 1], [2, 1]]], [2, [[1, 2], [1, 3]]],
             [20, [[1, 4], [2, 4]]], [6, [[1, 5], [1, 6], [2, 6], [3, 6]]],
             [3, [[2, 2], [2, 3]]], [3, [[2, 5], [3, 5]]],
             [240, [[3, 1], [3, 2], [4, 1], [4, 2]]], [6, [[3, 3], [3, 4]]],
             [6, [[4, 3], [5, 3]]], [7, [[4, 4], [5, 4], [5, 5]]],
             [30, [[4, 5], [4, 6]]], [6, [[5, 1], [5, 2]]],
             [9, [[5, 6], [6, 6]]], [8, [[6, 1], [6, 2], [6, 3]]],
             [2, [[6, 4], [6, 5]]]]

  num_p = len(problem)

  #
  # variables
  #

  # the set
  x = intvar(1,n,shape=(n,n),name="x")

  #
  # constraints
  #

  # all rows and columns must be unique
  model += [AllDifferent(row) for row in x]
  model += [AllDifferent(col) for col in x.transpose()]

  # calculate the segments
  for (res, segment) in problem:
    model += [calc(segment, x, res)]

  def print_sol():
    for i in range(n):
      for j in range(n):
        print(x[i, j].value(), end=" ")
      print()
    print()
    

  ss = CPM_ortools(model)
  num_solutions = ss.solveAll(solution_limit=0,display=print_sol)
  print("num_solutions:", num_solutions)


kenken2()