why doesnt it work?

Question

board = [
[7,8,0,4,0,0,1,2,0],
[6,0,0,0,7,5,0,0,9],
[0,0,0,6,0,1,0,7,8],
[0,0,7,0,4,0,2,6,0],
[0,0,1,0,5,0,9,3,0],
[9,0,4,0,6,0,0,0,5],
[0,7,0,3,0,0,0,1,2],
[1,2,0,0,0,7,4,0,0],
[0,4,9,2,0,6,0,0,7]
]
def solve(bo):
for column in bo:
for n, i in enumerate(column):
if i == 0:
#Find the box that the epty space is in
pos_x = (column.index(i) // 3) * 3
pos_y = (bo.index(column) // 3) * 3
#insert numbers inside empty space
for k in range(1, 10):
if validation(k, column, bo, pos_x, pos_y) == True:
column[n] = k

return print_board(bo)

def validation(num, col, game_board, box_x, box_y):
if num in col:
return False
else:
box_nums = []
for x in range(0, 3):
box_nums.append(game_board[box_y][box_x * 3:(box_x * 3) + 3])
box_y += 1
if num in box_nums[0] or box_nums[1] or box_nums[2]:
box_nums.clear()
return False
return True

def print_board(the_board):
for k in the_board:
print(k)

print(solve(board))

----------------------------------
line 22 never works ps:the code is still not finished

Answer 1

Not sure which line is number 22, but I'm pretty sure the line

if num in box_nums[0] or box_nums[1] or box_nums[2]:

doesn't do what you intended. If you want to check if num is in box_nums[0] or num is in box_nums[1] or num is in box_nums[2], you have to express it just like that.

if num in box_nums[0] or num in box_nums[1] or num in box_nums[2]:
...

Answer 2

Kirill Vidov This is how I would rewrite your solve() function:

def solve(bo):
for n, row in enumerate(bo):
for m, cell in enumerate(row):
if cell == 0:
pos_x = n - n%3
pos_y = m - m%3
for k in range(1, 10):
if validation(...):
row[m] = k

I realise you said the code is still not finished, but I don't think you have done enough in the validation function. You have checked if the number appears in the same row (you have used the word column, but it is actually the row), and if it appears in the same 3x3 block, but not if it appears in the same column.

Feel free to take a look at a checker I wrote a while back. Note, this is obviously just a checker, not a solver. https://code.sololearn.com/c455G0Sdzi3i/?ref=app

Answer 3

Great! 👍

Answer 4

Jhonie San Juan dont worry about it, Russ solved my problem, but thanks

Answer 5

Russ greatly appreciated, thank you once again

Answer 6

Yes, it currently only checks rows, not columns.

Answer 7

Russ my bad, i got confused with rows and columns

Answer 8

Russ omg that worked

Answer 9

Russ thank you so much

Answer 10

Russ is there a way to make it shorter?

Answer 11

import numpy

board = [
[7, 8, 0, 4, 0, 0, 1, 2, 0],
[6, 0, 0, 0, 7, 5, 0, 0, 9],
[0, 0, 0, 6, 0, 1, 0, 7, 8],
[0, 0, 7, 0, 4, 0, 2, 6, 0],
[0, 0, 1, 0, 5, 0, 9, 3, 0],
[9, 0, 4, 0, 6, 0, 0, 0, 5],
[0, 7, 0, 3, 0, 0, 0, 1, 2],
[1, 2, 0, 0, 0, 7, 4, 0, 0],
[0, 4, 9, 2, 0, 6, 0, 0, 7]
]

def check(y, x, n):
global board
for i in range(0, 9):
if board[y][i] == n:
return False
for i in range(0, 9):
if board[y][i] == n:
return False
xO = (x // 3) * 3
yO = (y // 3) * 3
for i in range(0, 3):
for j in range(0, 3):
if board[yO + i][xO + i] == n:
return False
return True

def solve():
global board
for y in range(9):
for x in range(9):
if board[y][x] == 0:
for n in range(1, 10):
if check(y, x, n):
board[y][x] = n
solve()
board[y][x] = 0
return
print(numpy.matrix(board))
exit(0)

solve()

Answer 12

Jhonie San Juan nice one, but I want to make an original solver

Answer 13

Ah, okie nice. You want this to be dynamic. I actually got that from youtube somewhere in the past so i just searched it again. I will read that again so i may understand what you want. Thanks

Answer 14

Russ for now I only checked if it appears in the same line horizontally, I need to check vertically

Answer 15

you need to import "numpy" for that to work