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divisible_by_9_through_1

<table align="left"> <td> <a href="https://colab.research.google.com/github/google/or-tools/blob/main/examples/notebook/contrib/divisible_by_9_through_1.ipynb">Run in Google Colab</a> </td> <td> <a href="https://github.com/google/or-tools/blob/main/examples/contrib/divisible_by_9_through_1.py">View source on GitHub</a> </td> </table>

First, you must install ortools package in this colab.

python

%pip install ortools

Divisible by 9 through 1 puzzle in Google CP Solver.

From http://msdn.microsoft.com/en-us/vcsharp/ee957404.aspx ' Solving Combinatory Problems with LINQ' ''' Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number must also satisfy these divisibility requirements:

The number should be divisible by 9.
If the rightmost digit is removed, the remaining number should be divisible by 8.
If the rightmost digit of the new number is removed, the remaining number should be divisible by 7.
And so on, until there's only one digit (which will necessarily be divisible by 1). '''

Also, see 'Intel Parallel Studio: Great for Serial Code Too (Episode 1)' http://software.intel.com/en-us/blogs/2009/12/07/intel-parallel-studio-great-for-serial-code-too-episode-1/

This model is however generalized to handle any base, for reasonable limits. The 'reasonable limit' for this model is that base must be between 2..16.

Compare with the following models:

MiniZinc: http://www.hakank.org/minizinc/divisible_by_9_through_1.mzn
Comet : http://www.hakank.org/comet/divisible_by_9_through_1.co
ECLiPSe : http://www.hakank.org/eclipse/divisible_by_9_through_1.ecl
Gecode : http://www.hakank.org/gecode/divisible_by_9_through_1.cpp

This model was created by Hakan Kjellerstrand ([email protected]) Also see my other Google CP Solver models: http://www.hakank.org/google_or_tools/

python

import sys

from ortools.constraint_solver import pywrapcp

#
# Decomposition of modulo constraint
#
# This implementation is based on the ECLiPSe version
# mentioned in
# - A Modulo propagator for ECLiPSE'
#   http://www.hakank.org/constraint_programming_blog/2010/05/a_modulo_propagator_for_eclips.html
# The ECLiPSe source code:
# http://www.hakank.org/eclipse/modulo_propagator.ecl
#


def my_mod(solver, x, y, r):

  if not isinstance(y, int):
    solver.Add(y != 0)

  lbx = x.Min()
  ubx = x.Max()
  ubx_neg = -ubx
  lbx_neg = -lbx
  min_x = min(lbx, ubx_neg)
  max_x = max(ubx, lbx_neg)

  d = solver.IntVar(max(0, min_x), max_x, "d")

  if not isinstance(r, int):
    solver.Add(r >= 0)
    solver.Add(x * r >= 0)

  if not isinstance(r, int) and not isinstance(r, int):
    solver.Add(-abs(y) < r)
    solver.Add(r < abs(y))

  solver.Add(min_x <= d)
  solver.Add(d <= max_x)
  solver.Add(x == y * d + r)


#
# converts a number (s) <-> an array of integers (t) in the specific base.
#
def toNum(solver, t, s, base):
  tlen = len(t)
  solver.Add(
      s == solver.Sum([(base**(tlen - i - 1)) * t[i] for i in range(tlen)]))


def main(base=10):

  # Create the solver.
  solver = pywrapcp.Solver("Divisible by 9 through 1")

  # data
  m = base**(base - 1) - 1
  n = base - 1

  digits_str = "_0123456789ABCDEFGH"

  print("base:", base)

  # declare variables

  # the digits
  x = [solver.IntVar(1, base - 1, "x[%i]" % i) for i in range(n)]

  # the numbers, t[0] contains the answer
  t = [solver.IntVar(0, m, "t[%i]" % i) for i in range(n)]

  #
  # constraints
  #
  solver.Add(solver.AllDifferent(x))

  for i in range(n):
    mm = base - i - 1
    toNum(solver, [x[j] for j in range(mm)], t[i], base)
    my_mod(solver, t[i], mm, 0)

  #
  # solution and search
  #
  solution = solver.Assignment()
  solution.Add(x)
  solution.Add(t)

  db = solver.Phase(x, solver.CHOOSE_FIRST_UNBOUND, solver.ASSIGN_MIN_VALUE)

  solver.NewSearch(db)
  num_solutions = 0
  while solver.NextSolution():
    print("x: ", [x[i].Value() for i in range(n)])
    print("t: ", [t[i].Value() for i in range(n)])
    print("number base 10: %i base %i: %s" % (t[0].Value(), base, "".join(
        [digits_str[x[i].Value() + 1] for i in range(n)])))
    print()
    num_solutions += 1
  solver.EndSearch()

  print("num_solutions:", num_solutions)
  print("failures:", solver.Failures())
  print("branches:", solver.Branches())
  print("WallTime:", solver.WallTime())


base = 10
default_base = 10
max_base = 16
if len(sys.argv) > 1:
  base = int(sys.argv[1])
  if base > max_base:
    print("Sorry, max allowed base is %i. Setting base to %i..." %
          (max_base, default_base))
    base = default_base
main(base)

# for base in range(2, 17):
#     print
#     main(base)