194 lines
4.8 KiB
Python
194 lines
4.8 KiB
Python
# Copyright 2010 Hakan Kjellerstrand hakank@gmail.com
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#
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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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"""
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KenKen puzzle in Google CP Solver.
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http://en.wikipedia.org/wiki/KenKen
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'''
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KenKen or KEN-KEN is a style of arithmetic and logical puzzle sharing
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several characteristics with sudoku. The name comes from Japanese and
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is translated as 'square wisdom' or 'cleverness squared'.
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...
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The objective is to fill the grid in with the digits 1 through 6 such that:
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* Each row contains exactly one of each digit
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* Each column contains exactly one of each digit
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* Each bold-outlined group of cells is a cage containing digits which
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achieve the specified result using the specified mathematical operation:
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addition (+),
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subtraction (-),
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multiplication (x),
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and division (/).
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(Unlike in Killer sudoku, digits may repeat within a group.)
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...
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More complex KenKen problems are formed using the principles described
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above but omitting the symbols +, -, x and /, thus leaving them as
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yet another unknown to be determined.
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'''
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The solution is:
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5 6 3 4 1 2
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6 1 4 5 2 3
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4 5 2 3 6 1
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3 4 1 2 5 6
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2 3 6 1 4 5
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1 2 5 6 3 4
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This model was created by Hakan Kjellerstrand (hakank@gmail.com)
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Also see my other Google CP Solver models:
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http://www.hakank.org/google_or_tools/
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"""
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import sys
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from ortools.constraint_solver import pywrapcp
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from functools import reduce
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#
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# Ensure that the sum of the segments
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# in cc == res
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#
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def calc(cc, x, res):
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solver = list(x.values())[0].solver()
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if len(cc) == 2:
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# for two operands there may be
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# a lot of variants
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c00, c01 = cc[0]
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c10, c11 = cc[1]
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a = x[c00 - 1, c01 - 1]
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b = x[c10 - 1, c11 - 1]
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r1 = solver.IsEqualCstVar(a + b, res)
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r2 = solver.IsEqualCstVar(a * b, res)
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r3 = solver.IsEqualVar(a * res, b)
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r4 = solver.IsEqualVar(b * res, a)
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r5 = solver.IsEqualCstVar(a - b, res)
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r6 = solver.IsEqualCstVar(b - a, res)
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solver.Add(r1 + r2 + r3 + r4 + r5 + r6 >= 1)
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else:
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# res is either sum or product of the segment
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xx = [x[i[0] - 1, i[1] - 1] for i in cc]
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# Sum
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# # SumEquality don't work:
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# this_sum = solver.SumEquality(xx, res)
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this_sum = solver.IsEqualCstVar(solver.Sum(xx), res)
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# Product
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# # Prod (or MakeProd) don't work:
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# this_prod = solver.IsEqualCstVar(solver.Prod(xx), res)
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this_prod = solver.IsEqualCstVar(reduce(lambda a, b: a * b, xx), res)
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solver.Add(this_sum + this_prod >= 1)
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def main():
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# Create the solver.
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solver = pywrapcp.Solver("KenKen")
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#
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# data
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#
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# size of matrix
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n = 6
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# For a better view of the problem, see
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# http://en.wikipedia.org/wiki/File:KenKenProblem.svg
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# hints
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# [sum, [segments]]
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# Note: 1-based
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problem = [[11, [[1, 1], [2, 1]]], [2, [[1, 2], [1, 3]]],
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[20, [[1, 4], [2, 4]]], [6, [[1, 5], [1, 6], [2, 6], [3, 6]]],
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[3, [[2, 2], [2, 3]]], [3, [[2, 5], [3, 5]]],
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[240, [[3, 1], [3, 2], [4, 1], [4, 2]]], [6, [[3, 3], [3, 4]]],
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[6, [[4, 3], [5, 3]]], [7, [[4, 4], [5, 4], [5, 5]]],
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[30, [[4, 5], [4, 6]]], [6, [[5, 1], [5, 2]]],
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[9, [[5, 6], [6, 6]]], [8, [[6, 1], [6, 2], [6, 3]]],
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[2, [[6, 4], [6, 5]]]]
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num_p = len(problem)
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#
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# variables
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#
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# the set
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x = {}
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for i in range(n):
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for j in range(n):
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x[i, j] = solver.IntVar(1, n, "x[%i,%i]" % (i, j))
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x_flat = [x[i, j] for i in range(n) for j in range(n)]
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#
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# constraints
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#
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# all rows and columns must be unique
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for i in range(n):
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row = [x[i, j] for j in range(n)]
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solver.Add(solver.AllDifferent(row))
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col = [x[j, i] for j in range(n)]
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solver.Add(solver.AllDifferent(col))
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# calculate the segments
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for (res, segment) in problem:
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calc(segment, x, res)
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#
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# search and solution
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#
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db = solver.Phase(x_flat, solver.INT_VAR_DEFAULT, solver.INT_VALUE_DEFAULT)
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solver.NewSearch(db)
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num_solutions = 0
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while solver.NextSolution():
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for i in range(n):
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for j in range(n):
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print(x[i, j].Value(), end=" ")
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print()
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print()
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num_solutions += 1
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solver.EndSearch()
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print()
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print("num_solutions:", num_solutions)
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print("failures:", solver.Failures())
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print("branches:", solver.Branches())
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print("WallTime:", solver.WallTime())
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if __name__ == "__main__":
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main()
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