"""
Calvin puzzle in cpmpy.

From 'An Exercise for the Mind: A 10 by 10 Math Puzzle: 
      A Pattern Recognition Game: Meditation on an Open Maze'
http://www.chycho.com/?q=Puzzle
'''
The Purpose of the Game

To take a 10 by 10 grid, representing 100 squares, and completely fill every square 
based on two types of movements. 
...

Movement Type I)  If the next number in the sequence is going to be placed vertically 
or horizontally, then it must be placed exactly three squares away from the previous 
number (there must be a two square gap between the numbers).

Movement Type II) If the next number in the sequence is going to be placed diagonally, 
then it must be placed exactly two squares away from the previous number (there must 
be a one square gap between the numbers). 
'''


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 *

def calvin_puzzle(n=5,num_sols=1):

  x = intvar(1,n*n,shape=(n,n),name="x")
  x_flat = [x[i,j] for i in range(n) for j in range(n)]

  model = Model(AllDifferent(x_flat))
  model += (x[0,0] == 1)

  for k in range(1,n*n):
    i = intvar(0,n-1,name=f"i_{k}")
    j = intvar(0,n-1,name=f"j_{k}")
    a = intvar(-3,3,name=f"a_{k}")
    b = intvar(-3,3,name=f"b_{k}")

    # Ensure that we are in the square
    model += [i + a >= 0, i+a <= n-1, j+b >= 0, j+b <= n-1]
    
    # model += [a != -1, a != 1, b != -1, b != 1]

    model += [ ((abs(a) == 3) & (b == 0)) |
               ((abs(b) == 3) & (a == 0)) |           
               ((abs(a) == 2) & (abs(b) == 2))
            ]

    # Valid moved
    # model += [ ((a == 3) & (b == 0))  |
    #            ((a == -3) & (b == 0)) |
               
    #            ((b == 3) & (a == 0))  |
    #            ((b == -3) & (a == 0)) |
               
    #            ((a == 2) & (b == 2))  |
    #            ((a == 2) & (b == -2)) |
    #            ((a == -2) & (b == 2)) | 
    #            ((a == -2) & (b == -2))
    #         ]

    # 1) First: fix this k, i.e.
    model += [k == Element(x_flat,(i*n)+j)]
    
    # 2) Then, find the position of the next value, i.e.
    model += [(k+1) == Element(x_flat, ((i+a)*n)+(j+b))]
 
  ss = CPM_ortools(model)
  # ss.ort_solver.parameters.search_branching = ort.PORTFOLIO_SEARCH
  # ss.ort_solver.parameters.cp_model_presolve = False
  ss.ort_solver.parameters.linearization_level = 0
  ss.ort_solver.parameters.cp_model_probing_level = 0

  num_solutions = ss.solveAll(solution_limit=num_sols,display=x)
  print('num_solutions:', num_solutions)
  print("Num conflicts:", ss.ort_solver.NumConflicts())
  print("NumBranches:", ss.ort_solver.NumBranches())
  print("WallTime:", ss.ort_solver.WallTime())


for n in range(2,10+1):
  print("\nn:",n)
  calvin_puzzle(n,1)