尝试在Python中使用随机数为数独创建矩阵

Question

I need to create a sudoku template, with it all filled. I need to do with random numbers, so I check if the number already exists in the row and in the column. The problem is that when a number is repeated in the column, it does not try another one, getting an infinite loop. Can Someone help me? Remembering that I am required to do with random numbers.

import random
matrix = [[None for i in range(9)] for j in range(9)]

def criarSdk():
    for l in range(9):
        for c in range(9):
            if matrix[l][c] is None:
                tmp3 = False
                print("linha l={}, coluna c={}".format(l,c))
                while tmp3 is False:
                    ale = random.randint(1, 9)
                    tmp1 = veriLine(matrix,ale, l)
                    tmp2 = veriCol(matrix,ale, c)
                    tmp3 = tmp1 and tmp2
                    if tmp3 is True:
                        matrix[l][c] = ale

def veriLine(vetor, value, line):
    tmp = True
    for c in range(9):
        if value == vetor[line][c]:
            tmp = False

    return tmp

def veriCol(vetor, value, col):
    tmp = True
    for l in range(9):
        if value == vetor[l][col]:
            tmp = False

    return tmp

criarSdk()
for i in range(9):
    print(matrix[i])

Answer 1

Your problem is not with the code, but with your algorithm.

If you fill up a sudoku like this, you can end up in a situation where no digit can be filled in for some space.

Consider this:

1 2 3 4 5 6 7 8 9
2 1 5 3 4 7 8 6 x

You can imagine the first line getting filled with random values (which happen to be in order) and then the values on the second line also getting filled in with random values, but what value to put in x? You can't put in 9 , because it would conflict on the column. But you can't put in anything else either, since everything else is already on the line.

In this case, your code will loop endlessly, trying random numbers, but none of them will ever work.

You need to look into backtracking and come up with a better solution. Note that your requirement for randomly picking a number doesn't have to mean you keep calling random.randint() for a new option, you could also randomly pick from a set of numbers you haven't tried yet.