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all_interval

examples/notebook/contrib/all_interval.ipynb

2016-064.1 KB
Original Source
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all_interval

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

First, you must install ortools package in this colab.

python
%pip install ortools

All interval problem in Google CP Solver.

CSPLib problem number 7 http://www.cs.st-andrews.ac.uk/~ianm/CSPLib/prob/prob007/index.html ''' Given the twelve standard pitch-classes (c, c , d, ...), represented by numbers 0,1,...,11, find a series in which each pitch-class occurs exactly once and in which the musical intervals between neighbouring notes cover the full set of intervals from the minor second (1 semitone) to the major seventh (11 semitones). That is, for each of the intervals, there is a pair of neigbhouring pitch-classes in the series, between which this interval appears. The problem of finding such a series can be easily formulated as an instance of a more general arithmetic problem on Z_n, the set of integer residues modulo n. Given n in N, find a vector s = (s_1, ..., s_n), such that (i) s is a permutation of Z_n = {0,1,...,n-1}; and (ii) the interval vector v = (|s_2-s_1|, |s_3-s_2|, ... |s_n-s_{n-1}|) is a permutation of Z_n-{0} = {1,2,...,n-1}. A vector v satisfying these conditions is called an all-interval series of size n; the problem of finding such a series is the all-interval series problem of size n. We may also be interested in finding all possible series of a given size. '''

Compare with the following models:

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


def main(n=12):

  # Create the solver.
  solver = pywrapcp.Solver("All interval")

  #
  # data
  #
  print("n:", n)

  #
  # declare variables
  #
  x = [solver.IntVar(1, n, "x[%i]" % i) for i in range(n)]
  diffs = [solver.IntVar(1, n - 1, "diffs[%i]" % i) for i in range(n - 1)]

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

  for k in range(n - 1):
    solver.Add(diffs[k] == abs(x[k + 1] - x[k]))

  # symmetry breaking
  solver.Add(x[0] < x[n - 1])
  solver.Add(diffs[0] < diffs[1])

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

  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("diffs:", [diffs[i].Value() for i in range(n - 1)])
    num_solutions += 1
    print()

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


n = 12
if len(sys.argv) > 1:
  n = int(sys.argv[1])
main(n)