Mixed Integer Programming QuickStart Sample (IronPython)

import numerics

from System import Array

# The linear programming classes reside in their own namespace.
from Extreme.Mathematics.Optimization import *
# Vectors and matrices are in the Extreme.Mathematics namespace
from Extreme.Mathematics import *

# Illustrates solving mixed integer programming problems
# using the classes in the Extreme.Mathematics.Optimization
# namespace of the Extreme Optimization Numerical Libraries for .NET.

# In this QuickStart sample, we'll use the Mixed Integer
# programming capabilities to solve Sudoku puzzles.
# The rules of Sudoku will be4 expressed in terms of
# linear constraints on binary variables.

# First, create an empty linear program.
lp = LinearProgram()

# Create an array of binary variables that indicate whether
# the cell at a specific row and column contain a specific digit.
# - The first index corresponds to the row.
# - The second index corresponds to the column.
# - The third index corresponds to the digit.
variables = Array.CreateInstance(LinearProgramVariable, 9, 9, 9)

# Create a binary variable for each digit in each row and column.
# The AddBinaryVariable method creates a variable that can have values of 0 or 1.
for row in range(9):
    for column in range(9):
        for digit in range(9):
            variables[row, column, digit] = lp.AddBinaryVariable("x{0}{1}{2}".format(row, column, digit), 0.0)

# To add integer variables, you can use the AddIntegerVariable method.
# To add real variables, you can use the AddVariable method.

# Now add constraints that represent the rules of Sudoku.

# There are 4 rules in Sudoku. They are all of the kind
# where only one of a certain set of combinations 
# of (row, column, digit) can occur at the same time.
# We can express this by stating that the sum of the corresponding
# binary variables must be one.

# AddConstraints is a helper function defined below.
# For each combination of the first two arguments, 
# it builds a constraint by iterating over the third argument.
coefficients = Vector([ 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0 ])
def AddConstraints(lp, variable):
    for i in range(9):
        for j in range(9):
            variables = Array.CreateInstance(LinearProgramVariable, 9)
            for k in range(9):
                variables[k] = variable(i, j, k)
            lp.AddLinearConstraint(variables, coefficients, ConstraintType.Equal, 1.0)

# Rule 1: each posiion contains exactly one digit
AddConstraints(lp, lambda row, column, digit: variables[row, column, digit])
# Rule 2: each digit appears once in each row
AddConstraints(lp, lambda row, digit, column: variables[row, column, digit])
# Rule 3: each digit appears once in each column
AddConstraints(lp, lambda column, digit, row: variables[row, column, digit])
# Rule 4: each digit appears exactly once in each block
AddConstraints(lp, lambda block, digit, index: \
    variables[3 * (block % 3) + (index % 3), 3 * (block / 3) + (index / 3), digit])

# We represent the board with a 9x9 sparse matrix.
# The nonzero entries correspond to the numbers
# already on the board.

# Let's see if we can solve "the world's hardest Sudoku" puzzle:
# http:#www.mirror.co.uk/fun-games/sudoku/2010/08/19/world-s-hardest-sudoku-can-you-solve-dr-arto-inkala-s-puzzle-115875-22496946/
rows = Array[int]([ 0,0,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,8,8 ])
columns = Array[int]([ 2,3,0,7,1,4,6,0,5,6,1,4,8,2,3,7,1,3,8,2,7,5,6 ])
digits = Array[float]([ 5,3,8,2,7,1,5,4,5,3,1,7,6,3,2,8,6,5,9,4,3,9,7 ])
board = Matrix.CreateSparse(9, 9, rows, columns, digits)

# Now fix the variables for the for the digits that are already on the board.
# We do this by setting the lower bound equal to the upper bound:
for triplet in board.NonzeroComponents:
    variables[triplet.Row, triplet.Column, triplet.Value - 1].LowerBound = 1.0

# Solve the linear program.
solution = lp.Solve()

# Scan the variables and print the digit if the value is 1.
for row in range(9):
    for column in range(9):
        theDigit = 0
        for digit in range(9):
            if variables[row, column, digit].Value == 1.0:
                theDigit = digit + 1
                break
        print theDigit.ToString() if theDigit > 0 else ".",
    print