57 lines
1.8 KiB
Python
Executable File
57 lines
1.8 KiB
Python
Executable File
#!/usr/bin/env python3
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# Copyright 2010-2021 Google LLC
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# Licensed under the Apache License, Version 2.0 (the "License");
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# you may not use this file except in compliance with the License.
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# You may obtain a copy of the License at
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#
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# http://www.apache.org/licenses/LICENSE-2.0
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#
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# Unless required by applicable law or agreed to in writing, software
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# distributed under the License is distributed on an "AS IS" BASIS,
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# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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# See the License for the specific language governing permissions and
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# limitations under the License.
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"""Magic sequence problem.
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This models aims at building a sequence of numbers such that the number of
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occurrences of i in this sequence is equal to the value of the ith number.
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It uses an aggregated formulation of the count expression called
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distribute().
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Usage: python magic_sequence_distribute.py NUMBER
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"""
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from absl import app
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from absl import flags
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from ortools.constraint_solver import pywrapcp
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FLAGS = flags.FLAGS
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def main(argv):
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# Create the solver.
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solver = pywrapcp.Solver('magic sequence')
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# Create an array of IntVars to hold the answers.
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size = int(argv[1]) if len(argv) > 1 else 100
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all_values = list(range(0, size))
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all_vars = [solver.IntVar(0, size, 'vars_%d' % i) for i in all_values]
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# The number of variables equal to j shall be the value of all_vars[j].
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solver.Add(solver.Distribute(all_vars, all_values, all_vars))
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# The sum of all the values shall be equal to the size.
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# (This constraint is redundant, but speeds up the search.)
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solver.Add(solver.Sum(all_vars) == size)
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solver.NewSearch(
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solver.Phase(all_vars, solver.CHOOSE_FIRST_UNBOUND,
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solver.ASSIGN_MIN_VALUE))
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solver.NextSolution()
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print(all_vars)
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solver.EndSearch()
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if __name__ == '__main__':
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app.run(main)
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