尝试用 cvxpy 解决数独问题

Question

我正在尝试使用 cvxpy 优化包解决数独问题。我对优化和 cvxpy 都很陌生。

约束是：

所有值都在 1 到 9 之间
所有行的总和 = 45
所有列的总和 = 45
所有平方和 = 45
给定的数字（我正在尝试解决 17 条线索的数独问题）。

所以，这是我的代码：

from cvxpy import *
import numpy as np

x = Variable((9, 9), integer=True)
obj = Minimize(sum(x)) #whatever, if the constrains are fulfilled it will be fine 
const = [x >= 1, #all values should be >= 1
         x <= 9, #all values should be <= 9
         sum(x, axis=0) == 45,  # sum of all rows should be 45
         sum(x, axis=1) == 45,  # sum of all cols should be 45
         sum(x[0:3, 0:3]) == 45, sum(x[0:3, 3:6]) == 45, #sum of all squares should be 45
         sum(x[0:3, 6:9]) == 45,
         sum(x[3:6, 0:3]) == 45, sum(x[3:6, 3:6]) == 45,
         sum(x[3:6, 6:9]) == 45,
         sum(x[6:9, 0:3]) == 45, sum(x[6:9, 3:6]) == 45,
         sum(x[6:9, 6:9]) == 45,
         x[0, 7] == 7, #the values themselves
         x[0, 8] == 1,
         x[1, 1] == 6,
         x[1, 4] == 3,
         x[2, 4] == 2,
         x[3, 0] == 7,
         x[3, 4] == 6,
         x[3, 6] == 3,
         x[4, 0] == 4,
         x[4, 6] == 2,
         x[5, 0] == 1,
         x[5, 3] == 4,
         x[6, 3] == 7,
         x[6, 5] == 5,
         x[6, 7] == 8,
         x[7, 1] == 2,
         x[8, 3] == 1]

prob = Problem(objective=obj, constraints=const)
prob.solve()

我从cvxpy得到的是

prob.status
Out[2]: 'infeasible_inaccurate'

这肯定是一个有效的数独，我检查了数百万次。

为什么我没有得到答案？

如有任何帮助，我们将不胜感激！

Answer 1

文档 (http://cvxr.com/cvx/doc/solver.html) 表明根本原因可能是一个接近可行的解决方案，但由于浮点误差超出公差而看起来不可行，这与所有等式约束相符。

不过，退一步说，这组约束不足以指定有效的数独解决方案。更好的公式是让 0-1 个变量由（行、列、大方块、数字）和约束索引，对于每个 row/column/square、row/digit、column/digit、square/digit组合，匹配变量之和等于1.

Answer 2

这是您默认使用的 ECOS_BB 问题。它不是一个可靠的整数规划求解器，我建议不要使用它。

其他建议：不要使用import *。最好使用 import cvxpy as cp 以避免与其他同名函数混淆。另外，顺便说一句，这里不需要numpy。

下面的脚本returns一个GUROBI的可行方案（没有GUROBI许可证也可以使用GLPK）：

import cvxpy as cp

x = cp.Variable((9, 9), integer=True)

# whatever, if the constrains are fulfilled it will be fine
objective = cp.Minimize(cp.sum(x))
constraints = [x >= 1,  # all values should be >= 1
               x <= 9,  # all values should be <= 9
               cp.sum(x, axis=0) == 45,  # sum of all rows should be 45
               cp.sum(x, axis=1) == 45,  # sum of all cols should be 45
               # sum of all squares should be 45
               cp.sum(x[0:3, 0:3]) == 45, cp.sum(x[0:3, 3:6]) == 45,
               cp.sum(x[0:3, 6:9]) == 45,
               cp.sum(x[3:6, 0:3]) == 45, cp.sum(x[3:6, 3:6]) == 45,
               cp.sum(x[3:6, 6:9]) == 45,
               cp.sum(x[6:9, 0:3]) == 45, cp.sum(x[6:9, 3:6]) == 45,
               cp.sum(x[6:9, 6:9]) == 45,
               x[0, 7] == 7,  # the values themselves
               x[0, 8] == 1,
               x[1, 1] == 6,
               x[1, 4] == 3,
               x[2, 4] == 2,
               x[3, 0] == 7,
               x[3, 4] == 6,
               x[3, 6] == 3,
               x[4, 0] == 4,
               x[4, 6] == 2,
               x[5, 0] == 1,
               x[5, 3] == 4,
               x[6, 3] == 7,
               x[6, 5] == 5,
               x[6, 7] == 8,
               x[7, 1] == 2,
               x[8, 3] == 1]

prob = cp.Problem(objective, constraints)
prob.solve(solver=cp.GUROBI)

print(x.value)

这就是输出

In [2]: run sudoku.py
[[1. 6. 1. 4. 7. 9. 9. 7. 1.]
 [6. 6. 1. 1. 3. 9. 9. 9. 1.]
 [8. 7. 9. 1. 2. 9. 1. 7. 1.]
 [7. 7. 1. 9. 6. 1. 3. 2. 9.]
 [4. 9. 5. 9. 5. 1. 2. 1. 9.]
 [1. 2. 9. 4. 9. 1. 9. 1. 9.]
 [8. 1. 1. 7. 8. 5. 2. 8. 5.]
 [9. 2. 9. 9. 4. 1. 1. 1. 9.]
 [1. 5. 9. 1. 1. 9. 9. 9. 1.]]

Answer 3

Sum == 45 不是一个好的约束，因为它不能防止重复值。您不需要 numpy 或 cvxpy 来解决这个问题，因为一个简单的回溯算法就可以解决它。

count=0

def findnext():
    global x
    for row in range(0,9):
        for col in range(0, 9):
            if x[row][col] == 0:
                return (row, col)
    return (-1,-1)

def colok(row, col):
    global x
    for c in range(0,9):
        if c != col and x[row][c] == x[row][col]:
            # print "x[%d][%d] == x[%d][%d]" % (row,c,row,col)
            return False
    return True

def rowok(row, col):
    global x
    for r in range(0,9):
        if r != row and x[r][col] == x[row][col]:
            return False
    return True

def sqok(row, col):
    global x
    sqy = int(row/3)*3
    sqx = int(col/3)*3
    for r in range(0,3):
        for c in range(0, 3):
            if (sqx+c != col or sqy+r != row) and x[sqy+r][sqx+c] == x[row][col]:
                print "x[%d][%d] == x[%d][%d] = %d" % (row,c,row,col,x[row][col])
                return False

    return True

def backtrack():
    global x
    global count
    (row, col) = findnext()
    if row < 0:
        return True
    for v in range(1,10):
        x[row][col] = v
        if rowok(row, col) and colok(row,col) and sqok(row,col):
            count += 1
            if backtrack():
                return True
    x[row][col] = 0
    return False

结果：

ending after calling backtrack 98248 times
2 8 3 6 5 4 9 7 1  
5 6 1 9 3 7 4 2 8  
9 7 4 8 2 1 5 6 3  
7 5 8 2 6 9 3 1 4  
4 3 6 5 1 8 2 9 7  
1 9 2 4 7 3 8 5 6  
3 1 9 7 4 5 6 8 2  
8 2 7 3 9 6 1 4 5  
6 4 5 1 8 2 7 3 9

尝试用 cvxpy 解决数独问题

Trying to solve Sudoku with cvxpy

python

algorithm

optimization

sudoku

cvxpy