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Added organize_day.py marathon2.py olympic.py lectures.py p_median.py mr_smith.py

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hakank
2010-10-26 05:25:08 +00:00
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# Copyright 2010 Hakan Kjellerstrand hakank@bonetmail.com
#
# 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
#
# 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.
"""
Lectures problem in Google CP Solver.
Biggs: Discrete Mathematics (2nd ed), page 187.
'''
Suppose we wish to schedule six one-hour lectures, v1, v2, v3, v4, v5, v6.
Among the the potential audience there are people who wish to hear both
- v1 and v2
- v1 and v4
- v3 and v5
- v2 and v6
- v4 and v5
- v5 and v6
- v1 and v6
How many hours are necessary in order that the lectures can be given
without clashes?
'''
Compare with the following models:
* MiniZinc: http://www.hakank.org/minizinc/lectures.mzn
* SICstus: http://hakank.org/sicstus/lectures.pl
* ECLiPSe: http://hakank.org/eclipse/lectures.ecl
* Gecode: http://hakank.org/gecode/lectures.cpp
This model was created by Hakan Kjellerstrand (hakank@bonetmail.com)
Also see my other Google CP Solver models: http://www.hakank.org/google_or_tools/
"""
import sys
from constraint_solver import pywrapcp
def main():
# Create the solver.
solver = pywrapcp.Solver('Lectures')
#
# data
#
#
# The schedule requirements:
# lecture a cannot be held at the same time as b
# Note: 1-based
g = [
[1, 2],
[1, 4],
[3, 5],
[2, 6],
[4, 5],
[5, 6],
[1, 6]
]
# number of nodes
n = 6
# number of edges
edges = len(g)
#
# declare variables
#
v = [solver.IntVar(0, n-1, 'v[%i]' % i) for i in range(n)]
# maximum color, to minimize
# Note: since Python is 0-based, the
# number of colors is +1
max_c = solver.IntVar(0, n-1, 'max_c')
#
# constraints
#
solver.Add(max_c == solver.Max(v))
# ensure that there are no clashes
# also, adjust to 0-base
for i in range(edges):
solver.Add(v[g[i][0]-1] != v[g[i][1]-1])
# symmetry breaking:
# - v0 has the color 0,
# - v1 has either color 0 or 1
solver.Add(v[0] == 0)
solver.Add(v[1] <= 1)
# objective
objective = solver.Minimize(max_c, 1)
#
# solution and search
#
db = solver.Phase(v,
solver.CHOOSE_MIN_SIZE_LOWEST_MIN,
solver.ASSIGN_CENTER_VALUE)
solver.NewSearch(db, [objective])
num_solutions = 0
while solver.NextSolution():
num_solutions += 1
print 'max_c:', max_c.Value()+1, 'colors'
print 'v:', [v[i].Value() for i in range(n)]
print
print 'num_solutions:', num_solutions
print 'failures:', solver.failures()
print 'branches:', solver.branches()
print 'wall_time:', solver.wall_time(), 'ms'
if __name__ == '__main__':
main()

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# Copyright 2010 Hakan Kjellerstrand hakank@bonetmail.com
#
# 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
#
# 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.
"""
Marathon puzzle in Google CP Solver.
From Xpress example
http://www.dashoptimization.com/home/cgi-bin/example.pl?id=mosel_puzzle_5_3
'''
Dominique, Ignace, Naren, Olivier, Philippe, and Pascal
have arrived as the first six at the Paris marathon.
Reconstruct their arrival order from the following
information:
a) Olivier has not arrived last
b) Dominique, Pascal and Ignace have arrived before Naren
and Olivier
c) Dominique who was third last year has improved this year.
d) Philippe is among the first four.
e) Ignace has arrived neither in second nor third position.
f) Pascal has beaten Naren by three positions.
g) Neither Ignace nor Dominique are on the fourth position.
(c) 2002 Dash Associates
author: S. Heipcke, Mar. 2002
'''
Compare with the following models:
* MiniZinc: http://www.hakank.org/minizinc/marathon2.mzn
* SICStus Prolog: http://www.hakank.org/sicstus/marathon2.pl
* ECLiPSe: http://hakank.org/eclipse/marathon2.ecl
* Gecode: http://hakank.org/gecode/marathon2.cpp
This model was created by Hakan Kjellerstrand (hakank@bonetmail.com)
Also see my other Google CP Solver models: http://www.hakank.org/google_or_tools/
"""
import sys
from constraint_solver import pywrapcp
def main():
# Create the solver.
solver = pywrapcp.Solver('Marathon')
#
# data
#
n = 6
runners_str = ['Dominique', 'Ignace', 'Naren',
'Olivier', 'Philippe', 'Pascal']
#
# declare variables
#
runners = [solver.IntVar(1, n, 'runners[%i]' % i) for i in range(n)]
Dominique, Ignace, Naren, Olivier, Philippe, Pascal = runners
#
# constraints
#
solver.Add(solver.AllDifferent(runners, True))
# a: Olivier not last
solver.Add(Olivier != n)
# b: Dominique, Pascal and Ignace before Naren and Olivier
solver.Add(Dominique < Naren)
solver.Add(Dominique < Olivier)
solver.Add(Pascal < Naren)
solver.Add(Pascal < Olivier)
solver.Add(Ignace < Naren)
solver.Add(Ignace < Olivier)
# c: Dominique better than third
solver.Add(Dominique < 3)
# d: Philippe is among the first four
solver.Add(Philippe <= 4)
# e: Ignace neither second nor third
solver.Add(Ignace != 2)
solver.Add(Ignace != 3)
# f: Pascal three places earlier than Naren
solver.Add(Pascal + 3 == Naren)
# g: Neither Ignace nor Dominique on fourth position
solver.Add(Ignace != 4)
solver.Add(Dominique != 4)
#
# solution and search
#
db = solver.Phase(runners,
solver.CHOOSE_MIN_SIZE_LOWEST_MIN,
solver.ASSIGN_CENTER_VALUE)
solver.NewSearch(db)
num_solutions = 0
while solver.NextSolution():
num_solutions += 1
runners_val = [runners[i].Value() for i in range(n)]
print 'runners:', runners_val
print "Places:"
for i in range(1, n+1):
for j in range(n):
if runners_val[j] == i:
print "%i: %s" % (i, runners_str[j])
print
print 'num_solutions:', num_solutions
print 'failures:', solver.failures()
print 'branches:', solver.branches()
print 'wall_time:', solver.wall_time(), 'ms'
if __name__ == '__main__':
main()

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# Copyright 2010 Hakan Kjellerstrand hakank@bonetmail.com
#
# 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
#
# 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.
"""
Mr Smith in Google CP Solver.
From an IF Prolog example (http://www.ifcomputer.de/)
'''
The Smith family and their three children want to pay a visit but they
do not all have the time to do so. Following are few hints who will go
and who will not:
o If Mr Smith comes, his wife will come too.
o At least one of their two sons Matt and John will come.
o Either Mrs Smith or Tim will come, but not both.
o Either Tim and John will come, or neither will come.
o If Matt comes, then John and his father will
also come.
'''
The answer should be:
Mr_Smith_comes = 0
Mrs_Smith_comes = 0
Matt_comes = 0
John_comes = 1
Tim_comes = 1
Compare with the following models:
* ECLiPSe: http://www.hakank.org/eclipse/mr_smith.ecl
* SICStus Prolog: http://www.hakank.org/sicstus/mr_smith.pl
* Gecode: http://www.hakank.org/gecode/mr_smith.cpp
* MiniZinc: http://www.hakank.org/minizinc/mr_smith.mzn
This model was created by Hakan Kjellerstrand (hakank@bonetmail.com)
Also see my other Google CP Solver models: http://www.hakank.org/google_or_tools/
"""
import sys
from constraint_solver import pywrapcp
def main():
# Create the solver.
solver = pywrapcp.Solver('Mr Smith problem')
#
# data
#
n = 5
#
# declare variables
#
x = [solver.IntVar(0, 1, 'x[%i]' % i) for i in range(n)]
Mr_Smith, Mrs_Smith, Matt, John, Tim = x
#
# constraints
#
#
# I've kept the MiniZinc constraints for clarity
# and debugging.
#
# If Mr Smith comes then his wife will come too.
# (Mr_Smith -> Mrs_Smith)
solver.Add(Mr_Smith-Mrs_Smith <= 0)
# At least one of their two sons Matt and John will come.
# (Matt \/ John)
solver.Add(Matt+John >= 1)
# Either Mrs Smith or Tim will come but not both.
# bool2int(Mrs_Smith) + bool2int(Tim) = 1 /\
# (Mrs_Smith xor Tim)
solver.Add(Mrs_Smith + Tim == 1)
# Either Tim and John will come or neither will come.
# (Tim = John)
solver.Add(Tim == John)
# If Matt comes /\ then John and his father will also come.
# (Matt -> (John /\ Mr_Smith))
solver.Add(Matt - (John*Mr_Smith) <= 0)
#
# solution and search
#
db = solver.Phase(x,
solver.INT_VAR_DEFAULT,
solver.INT_VALUE_DEFAULT)
solver.NewSearch(db)
num_solutions = 0
while solver.NextSolution():
num_solutions += 1
print 'x:', [x[i].Value() for i in range(n)]
print
print 'num_solutions:', num_solutions
print 'failures:', solver.failures()
print 'branches:', solver.branches()
print 'wall_time:', solver.wall_time(), 'ms'
if __name__ == '__main__':
main()

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# Copyright 2010 Hakan Kjellerstrand hakank@bonetmail.com
#
# 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
#
# 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.
"""
Olympic puzzle in Google CP Solver.
Benchmark for Prolog (BProlog)
'''
File : olympic.pl
Author : Neng-Fa ZHOU
Date : 1993
Purpose: solve a puzzle taken from Olympic Arithmetic Contest
Given ten variables with the following configuration:
X7 X8 X9 X10
X4 X5 X6
X2 X3
X1
We already know that X1 is equal to 3 and want to assign each variable
with a different integer from {1,2,...,10} such that for any three
variables
Xi Xj
Xk
the following constraint is satisfied:
|Xi-Xj| = Xk
'''
Compare with the following models:
* MiniZinc: http://www.hakank.org/minizinc/olympic.mzn
* SICStus Prolog: http://www.hakank.org/sicstus/olympic.pl
* ECLiPSe: http://hakank.org/eclipse/olympic.ecl
* Gecode: http://hakank.org/gecode/olympic.cpp
This model was created by Hakan Kjellerstrand (hakank@bonetmail.com)
Also see my other Google CP Solver models: http://www.hakank.org/google_or_tools/
"""
import sys
from constraint_solver import pywrapcp
def minus(solver, x, y, z):
solver.Add(z == abs(x - y))
def main():
# Create the solver.
solver = pywrapcp.Solver('Olympic')
#
# data
#
n = 10
#
# declare variables
#
Vars = [solver.IntVar(1, n, 'Vars[%i]' % i) for i in range(n)]
X1,X2,X3,X4,X5,X6,X7,X8,X9,X10 = Vars
#
# constraints
#
solver.Add(solver.AllDifferent(Vars, True))
solver.Add(X1 == 3)
minus(solver, X2, X3, X1)
minus(solver, X4, X5, X2)
minus(solver, X5, X6, X3)
minus(solver, X7, X8, X4)
minus(solver, X8, X9, X5)
minus(solver, X9, X10, X6)
#
# solution and search
#
db = solver.Phase(Vars,
solver.INT_VAR_SIMPLE,
solver.INT_VALUE_DEFAULT)
solver.NewSearch(db)
num_solutions = 0
while solver.NextSolution():
num_solutions += 1
print 'Vars:', [Vars[i].Value() for i in range(n)]
print
print 'num_solutions:', num_solutions
print 'failures:', solver.failures()
print 'branches:', solver.branches()
print 'wall_time:', solver.wall_time(), 'ms'
if __name__ == '__main__':
main()

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# Copyright 2010 Hakan Kjellerstrand hakank@bonetmail.com
#
# 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
#
# 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.
"""
Organizing a day in Google CP Solver.
Simple scheduling problem.
Problem formulation from ECLiPSe:
Slides on (Finite Domain) Constraint Logic Programming, page 38f
http://eclipse-clp.org/reports/eclipse.ppt
Compare with the following models:
* MiniZinc: http://www.hakank.org/minizinc/organize_day.mzn
* Comet: http://www.hakank.org/comet/organize_day.co
* Gecode: http://hakank.org/gecode/organize_day.cpp
This model was created by Hakan Kjellerstrand (hakank@bonetmail.com)
Also see my other Google CP Solver models: http://www.hakank.org/google_or_tools/
"""
import sys
from constraint_solver import pywrapcp
#
# No overlapping of tasks s1 and s2
#
def no_overlap(solver, s1, d1, s2, d2):
b1 = solver.MakeIsLessOrEqualVar(s1 + d1, s2) # s1 + d1 <= s2
b2 = solver.MakeIsLessOrEqualVar(s2 + d2, s1) # s2 + d2 <= s1
solver.Add(b1 + b2 >= 1)
def main():
# Create the solver.
solver = pywrapcp.Solver('Organizing a day')
#
# data
#
n = 4
tasks = range(n)
work, mail, shop, bank = tasks
durations = [4,1,2,1]
# task [i,0] must be finished before task [i,1]
before_tasks = [
[bank, shop],
[mail, work]
]
# the valid times of the day
begin = 9
end = 17
#
# declare variables
#
begins = [solver.IntVar(begin, end, 'begins[%i]% % i') for i in tasks]
ends = [solver.IntVar(begin, end, 'ends[%i]% % i') for i in tasks]
#
# constraints
#
for i in tasks:
solver.Add(ends[i] == begins[i] + durations[i])
for i in tasks:
for j in tasks:
if i < j:
no_overlap(solver,
begins[i], durations[i],
begins[j], durations[j])
# specific constraints
for (before, after) in before_tasks:
solver.Add(ends[before] <= begins[after])
solver.Add(begins[work] >= 11)
#
# solution and search
#
db = solver.Phase(begins + ends,
solver.INT_VAR_DEFAULT,
solver.INT_VALUE_DEFAULT)
solver.NewSearch(db)
num_solutions = 0
while solver.NextSolution():
num_solutions += 1
print 'begins:', [begins[i].Value() for i in tasks]
print 'ends:', [ends[i].Value() for i in tasks]
print
print 'num_solutions:', num_solutions
print 'failures:', solver.failures()
print 'branches:', solver.branches()
print 'wall_time:', solver.wall_time(), 'ms'
if __name__ == '__main__':
main()

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# Copyright 2010 Hakan Kjellerstrand hakank@bonetmail.com
#
# 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
#
# 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.
"""
P-median problem in Google CP Solver.
Model and data from the OPL Manual, which describes the problem:
'''
The P-Median problem is a well known problem in Operations Research.
The problem can be stated very simply, like this: given a set of customers
with known amounts of demand, a set of candidate locations for warehouses,
and the distance between each pair of customer-warehouse, choose P
warehouses to open that minimize the demand-weighted distance of serving
all customers from those P warehouses.
'''
Compare with the following models:
* MiniZinc: http://hakank.org/minizinc/p_median.mzn
* Comet: http://hakank.org/comet/p_median.co
This model was created by Hakan Kjellerstrand (hakank@bonetmail.com)
Also see my other Google CP Solver models: http://www.hakank.org/google_or_tools/
"""
import sys
from constraint_solver import pywrapcp
def main():
# Create the solver.
solver = pywrapcp.Solver('P-median problem')
#
# data
#
p = 2
num_customers = 4
customers = range(num_customers)
Albert, Bob, Chris, Daniel = customers
num_warehouses = 3
warehouses = range(num_warehouses)
Santa_Clara, San_Jose, Berkeley = warehouses
demand = [100,80,80,70]
distance = [
[ 2, 10, 50],
[ 2, 10, 52],
[50, 60, 3],
[40, 60, 1]
]
#
# declare variables
#
open = [solver.IntVar(warehouses, 'open[%i]% % i')
for w in warehouses]
ship = {}
for c in customers:
for w in warehouses:
ship[c,w] = solver.IntVar(0, 1,'ship[%i,%i]' % (c,w))
ship_flat = [ship[c,w]
for c in customers
for w in warehouses]
z = solver.IntVar(0, 1000, 'z')
#
# constraints
#
z_sum = solver.Sum([demand[c]*distance[c][w]*ship[c,w]
for c in customers
for w in warehouses])
solver.Add(z == z_sum)
for c in customers:
s = solver.Sum([ship[c,w]
for w in warehouses])
solver.Add(s == 1)
solver.Add(solver.Sum(open) == p)
for c in customers:
for w in warehouses:
solver.Add(ship[c,w] <= open[w])
# objective
objective = solver.Minimize(z, 1)
#
# solution and search
#
db = solver.Phase(open + ship_flat,
solver.INT_VAR_DEFAULT,
solver.INT_VALUE_DEFAULT)
solver.NewSearch(db, [objective])
num_solutions = 0
while solver.NextSolution():
num_solutions += 1
print "z:", z.Value()
print 'open:', [open[w].Value() for w in warehouses]
for c in customers:
for w in warehouses:
print ship[c,w].Value(),
print
print
print 'num_solutions:', num_solutions
print 'failures:', solver.failures()
print 'branches:', solver.branches()
print 'wall_time:', solver.wall_time(), 'ms'
if __name__ == '__main__':
main()