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# Copyright 2018, Gurobi Optimization, LLC

# Sudoku example.

# The Sudoku board is a 9x9 grid, which is further divided into a 3x3 grid
# of 3x3 grids.  Each cell in the grid must take a value from 0 to 9.
# No two grid cells in the same row, column, or 3x3 subgrid may take the
# same value.
# In the MIP formulation, binary variables x[i,j,v] indicate whether
# cell <i,j> takes value 'v'.  The constraints are as follows:
#   1. Each cell must take exactly one value (sum_v x[i,j,v] = 1)
#   2. Each value is used exactly once per row (sum_i x[i,j,v] = 1)
#   3. Each value is used exactly once per column (sum_j x[i,j,v] = 1)
#   4. Each value is used exactly once per 3x3 subgrid (sum_grid x[i,j,v] = 1)
# Input datasets for this example can be found in examples/data/sudoku*.

import sys
import math
from gurobipy import *

if len(sys.argv) < 2:
    print('Usage: filename')

f = open(sys.argv[1])

grid =

n = len(grid[0])
s = int(math.sqrt(n))

# Create our 3-D array of model variables

model = Model('sudoku')

vars = model.addVars(n,n,n, vtype=GRB.BINARY, name='G')

# Fix variables associated with cells whose values are pre-specified

for i in range(n):
    for j in range(n):
        if grid[i][j] != '.':
            v = int(grid[i][j]) - 1
            vars[i,j,v].LB = 1

# Each cell must take one value

model.addConstrs((vars.sum(i,j,'*') == 1
                 for i in range(n)
                 for j in range(n)), name='V')

# Each value appears once per row

model.addConstrs((vars.sum(i,'*',v) == 1
                 for i in range(n)
                 for v in range(n)), name='R')

# Each value appears once per column

model.addConstrs((vars.sum('*',j,v) == 1
                 for j in range(n)
                 for v in range(n)), name='C')

# Each value appears once per subgrid

    quicksum(vars[i,j,v] for i in range(i0*s, (i0+1)*s)
                         for j in range(j0*s, (j0+1)*s)) == 1
             for v in range(n)
             for i0 in range(s)
             for j0 in range(s)), name='Sub')




# Retrieve optimization result

solution = model.getAttr('X', vars)

for i in range(n):
    sol = ''
    for j in range(n):
        for v in range(n):
            if solution[i,j,v] > 0.5:
                sol += str(v+1)

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