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#!/usr/bin/env python3
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# Copyright 2010-2025 Google LLC
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# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
#
# http://www.apache.org/licenses/LICENSE-2.0
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#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
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"""Magic sequence problem.
This models aims at building a sequence of numbers such that the number of
occurrences of i in this sequence is equal to the value of the ith number.
It uses an aggregated formulation of the count expression called
distribute().
Usage: python magic_sequence_distribute.py NUMBER
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"""
from absl import app
from absl import flags
from ortools.constraint_solver import pywrapcp
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FLAGS = flags.FLAGS
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def main(argv):
# 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.
size = int(argv[1]) if len(argv) > 1 else 100
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].
solver.Add(solver.Distribute(all_vars, all_values, all_vars))
# The sum of all the values shall be equal to the size.
# (This constraint is redundant, but speeds up the search.)
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, solver.ASSIGN_MIN_VALUE)
)
solver.NextSolution()
print(all_vars)
solver.EndSearch()
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if __name__ == "__main__":
app.run(main)