My solution to the 10 queens riddle, posed by Rodrigo from Mathspp

queens.py

from collections import namedtuple
from copy import deepcopy

MIN_ROW, MAX_ROW = 0, 9
MIN_COLUMN, MAX_COLUMN = 0, 9

total_successful_combinations = 0
total_failed_combinations = 0

Field = namedtuple('Field', 'row, column')

board = [[" "] * 10 for column in range(10)]


def update_board(last_board, current_field=Field(0, 0)):
    current_board = deepcopy(last_board)

    # try placing a Queen onto the board
    if current_board[current_field.row][current_field.column] == ' ':
        current_board[current_field.row][current_field.column] = 'Q'
    else:
        # must be an 'X' if it's not an empty field
        # print_board(current_board)
        # print("No luck this time!\n")
        global total_failed_combinations 
        total_failed_combinations += 1
        return

    # check whether all Queens have been placed successfully
    if current_field.row == 9:
        print_board(current_board)
        print("***Success!***\n")
        global total_successful_combinations
        total_successful_combinations += 1
        return

    # get rid of options in same row
    for col in range(MIN_COLUMN, MAX_COLUMN + 1):
            if not Field(current_field.row, col) == current_field:
                current_board[current_field.row][col] = 'X'

    # get rid of options in same column
    for row in range(MIN_ROW, MAX_ROW + 1):
        if not Field(row, current_field.column) == current_field:
            current_board[row][current_field.column] = 'X'

    # get rid of options diagonally to the right
    col = current_field.column
    for row in range(current_field.row + 1, MAX_ROW + 1):
        col += 1
        if col < MAX_COLUMN + 1 and not Field(row, col) == current_field:
            current_board[row][col] = 'X'


    # # get rid of options diagonally to the left
    col = current_field.column
    for row in range(current_field.row + 1, MAX_ROW + 1):
        col -= 1
        if col > MIN_COLUMN - 1 and not Field(row, col) == current_field:
            current_board[row][col] = 'X'
        else:
            break

    # go to next line and call update_board recursively:
    next_row = current_field.row + 1
    for col in range(MIN_COLUMN, MAX_COLUMN + 1):
        next_field = Field(next_row, col)
        next_board = current_board
        update_board(next_board, next_field)


def print_board(board):
    for col in board:
        for char in col:
            print(char, end=' ')
        print()


def main(board):
    for col in range(MIN_COLUMN, MAX_COLUMN + 1):
        next_field = Field(0, col)
        next_board = board
        update_board(next_board, next_field)
    print(f'{total_successful_combinations=} and {total_failed_combinations=}')


if __name__ == "__main__":
    main(board)

# total_successful_combinations=724 and total_failed_combinations=312612
# if the riddle is meant to also include all possible variations of different queens being placed in the same fields, then the answer must be multiplied by 10! = 3628800 (permutations without repetition)
# in this case the answer would be 2'627'251'200 possible combinations

# sample output:
"""
X X X X X X X X X Q
X X X X X X X Q X X
X X X X Q X X X X X
X X Q X X X X X X X
Q X X X X X X X X X
X X X X X Q X X X X
X Q X X X X X X X X
X X X X X X X X Q X
X X X X X X Q X X X
X X X Q X X X X X X
***Success!***
"""