human
I found an interesting problem on https://www.codewars.com/kata/561bed6a31daa8df7400000e:
Based on the well known ['Eight Queens' problem](https://en.wikipedia.org/wiki/Eight_queens_puzzle).

#### Summary
Your challenge is to place N queens on a chess board such that none of the queens are attacking each other.

#### Details
A standard 8x8 chess board has its rows (aka ranks) labelled 1-8 from bottom to top, and its columns (aka files) labelled a-h from left to right. Any square can be denoted by a combination of letter and number. For example, the top left square is a8. A queen may attack any square on its rank or file. It may also attack any square diagonally.

In this kata, chessboard may have any size from 1 to 10. Columns are marked with letters a-j, and rows with single digit from 1 to 0 (first row is marked with 1, ninth row is marked with 9, and tenth row is marked with 0, so `a1` denotes bottom left field of the chessboard, and `j0` - top right field on a 10x10 board).

You will be given the position of one queen on the board (e.g. `'c3'`). Your challenge is to compute the position of the remaining queens and return the full solution (including the original input) as a comma-separated string (e.g. `'d8,a7,e6,h5,b4,g3,c2,f1'`). Fields do not have to be sorted in any particular way, but you have to adhere to above format (lowercase letters, no spaces, fields separated with commas).

Note that there may be multiple valid solutions for a given starting position. Your function only needs to produce (any) one of these. All given starting positions produce valid solutions.

If you wish to write more tests, the validity of a solution may be checked using preloaded function (see sample tests).

Hint: check the wiki article for the various algorithms that are applicable to this problem.

----
Next steps:
 - [N queens problem (with no mandatory queen position)](https://www.codewars.com/kata/52cdc1b015db27c484000031)
 - [N queens problem (with one mandatory queen position) - challenge version](https://www.codewars.com/kata/5985ea20695be6079e000003)

I tried it in Python, but could not do it. Can you solve it?

 ... Write me a Python code to compute the position of the remaining queens and return the full solution (including the original input) as a comma-separated string (e?
assistant
#Here is Python code to compute the position of the remaining queens and return the full solution (including the original input) as a comma-separated string (e.

MOVES = ( (1,1), (-1,1), (1,-1), (-1,-1) )

def find_empty(row, X):
    for y,spot in enumerate(row):
        if spot == 0:
            yield (X,y)

def populating_board(x,y, board):
    board[x][y] += 1

    pos = [(x,y)]
    for i in range(len(board)):
        if (x,y) != (i,y):
            board[i][y] += 1
            pos.append((i,y))
        
        if (x,y) != (x,i):
            board[x][i] +=1
            pos.append((x,i))

    for dx,dy in MOVES:
        c = 0
        while True:
            c += 1
            xx = x+c*dx
            yy = y+c*dy

            if not (0<=xx<len(board) and 0<=yy<len(board)):
                break

            if (xx,yy) != (x,y):
                board[xx][yy] += 1
                pos.append((xx,yy))
    
    return pos

def queens(position, size):
    board = [[0]*size for i in range(size)]
    y,x = ord(position[0])-97, size - (size==10 and not int(position[1]) and 10 or int(position[1]))
    initial_X = x

    #populating board with not possible Q positions
    populating_board(x,y,board)
    queens = [position]
    
    def dfs(counter, next_X):
        nonlocal board
        if counter == size:
            return True
        

        while next_X > 0 and 0 not in board[next_X]:
            next_X -= 1
            
        while next_X < size-1 and 0 not in board[next_X]:
            next_X += 1

        for x,y in find_empty(board[next_X], next_X):
            queens.append(chr(y+97)+str(size -(size==10 and not x and 10 or x)))
            to_restore = populating_board(x,y,board)
            
            if dfs(counter+1, x):
                return True

            queens.pop()
            for pos_x, pos_y in to_restore:
                board[pos_x][pos_y] -= 1
    
    if dfs(1, x):
        return ','.join(queens)