http://www.apache.org/licenses/LICENSE-2.0

%pip install ortools

import sys

from ortools.constraint_solver import pywrapcp


#
# Make a transition (automaton) list of tuples from a
# single pattern, e.g. [3,2,1]
#
def make_transition_tuples(pattern):
  p_len = len(pattern)
  num_states = p_len + sum(pattern)

  tuples = []

  # this is for handling 0-clues. It generates
  # just the minimal state
  if num_states == 0:
    tuples.append((1, 0, 1))
    return (tuples, 1)

  # convert pattern to a 0/1 pattern for easy handling of
  # the states
  tmp = [0]
  c = 0
  for pattern_index in range(p_len):
    tmp.extend([1] * pattern[pattern_index])
    tmp.append(0)

  for i in range(num_states):
    state = i + 1
    if tmp[i] == 0:
      tuples.append((state, 0, state))
      tuples.append((state, 1, state + 1))
    else:
      if i < num_states - 1:
        if tmp[i + 1] == 1:
          tuples.append((state, 1, state + 1))
        else:
          tuples.append((state, 0, state + 1))
  tuples.append((num_states, 0, num_states))
  return (tuples, num_states)


#
# check each rule by creating an automaton and transition constraint.
#
def check_rule(rules, y):
  cleaned_rule = [rules[i] for i in range(len(rules)) if rules[i] > 0]
  (transition_tuples, last_state) = make_transition_tuples(cleaned_rule)

  initial_state = 1
  accepting_states = [last_state]

  solver = y[0].solver()
  solver.Add(
      solver.TransitionConstraint(y, transition_tuples, initial_state,
                                  accepting_states))


def main(rows, row_rule_len, row_rules, cols, col_rule_len, col_rules):

  # Create the solver.
  solver = pywrapcp.Solver('Regular test')

  #
  # variables
  #
  board = {}
  for i in range(rows):
    for j in range(cols):
      board[i, j] = solver.IntVar(0, 1, 'board[%i, %i]' % (i, j))
  board_flat = [board[i, j] for i in range(rows) for j in range(cols)]

  # Flattened board for labeling.
  # This labeling was inspired by a suggestion from
  # Pascal Van Hentenryck about my Comet nonogram model.
  board_label = []
  if rows * row_rule_len < cols * col_rule_len:
    for i in range(rows):
      for j in range(cols):
        board_label.append(board[i, j])
  else:
    for j in range(cols):
      for i in range(rows):
        board_label.append(board[i, j])

  #
  # constraints
  #
  for i in range(rows):
    check_rule(row_rules[i], [board[i, j] for j in range(cols)])

  for j in range(cols):
    check_rule(col_rules[j], [board[i, j] for i in range(rows)])

  #
  # solution and search
  #
  db = solver.Phase(board_label, solver.CHOOSE_FIRST_UNBOUND,
                    solver.ASSIGN_MIN_VALUE)

  print('before solver, wall time = ', solver.WallTime(), 'ms')
  solver.NewSearch(db)

  num_solutions = 0
  while solver.NextSolution():
    print()
    num_solutions += 1
    for i in range(rows):
      row = [board[i, j].Value() for j in range(cols)]
      row_pres = []
      for j in row:
        if j == 1:
          row_pres.append('#')
        else:
          row_pres.append(' ')
      print('  ', ''.join(row_pres))

    print()
    print('  ', '-' * cols)

    if num_solutions >= 2:
      print('2 solutions is enough...')
      break

  solver.EndSearch()
  print()
  print('num_solutions:', num_solutions)
  print('failures:', solver.Failures())
  print('branches:', solver.Branches())
  print('WallTime:', solver.WallTime(), 'ms')


#
# Default problem
#
# From http://twan.home.fmf.nl/blog/haskell/Nonograms.details
# The lambda picture
#
rows = 12
row_rule_len = 3
row_rules = [[0, 0, 2], [0, 1, 2], [0, 1, 1], [0, 0, 2], [0, 0, 1], [0, 0, 3],
             [0, 0, 3], [0, 2, 2], [0, 2, 1], [2, 2, 1], [0, 2, 3], [0, 2, 2]]

cols = 10
col_rule_len = 2
col_rules = [[2, 1], [1, 3], [2, 4], [3, 4], [0, 4], [0, 3], [0, 3], [0, 3],
             [0, 2], [0, 2]]

if len(sys.argv) > 1:
  file = sys.argv[1]
  exec(compile(open(file).read(), file, 'exec'))
main(rows, row_rule_len, row_rules, cols, col_rule_len, col_rules)