How does this code to create a board for a Tic-Tac-Toe game work?

Question

After I couldn’t figure out, how to build a board for a Tic-Tac-Toe game by myself, I randomly discoverd an already finished Tic-Tac-Toe program from another programmer.

What she did was:

board = [' ' for _ in range(9)]

def print_board():
    for row in [board[i*3:(i+1)*3] for i in range(3)]:
        print('| ' + ' | '.join(row) + ' |')

print_board()

Output:

|   |   |   |

|   |   |   |

|   |   |   |

Now I am totally confused with this sector of the code:

board[i*3:(i+1)*3]

What does this part of the code mean and why is it so essential to use it in this case?

If you want the whole context of the program or try it yourself:
https://github.com/kying18/tic-tac-toe/blob/master/game.py

Asked By: d2w3t

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Source

Answer 1

Be aware that this code is more complicated than it needs to be.

An equivalent but simpler version would be

def print_board():
    for i in range(3):
        row = board[i*3:(i+1)*3]
        print('| ' + ' | '.join(row) + ' |')

Answered By: mkrieger1

Answer 2

Link to flawless code

Working board

def printboard(xState, oState):
    # Printing Board By using gameValues 's List
    zero = 'X' if xState[0] else ('O' if oState[0] else 0)
    one = 'X' if xState[1] else ('O' if oState[1] else 1)
    two = 'X' if xState[2] else ('O' if oState[2] else 2)
    three = 'X' if xState[3] else ('O' if oState[3] else 3)
    four = 'X' if xState[4] else ('O' if oState[4] else 4)
    five = 'X' if xState[5] else ('O' if oState[5] else 5)
    six = 'X' if xState[6] else ('O' if oState[6] else 6)
    seven = 'X' if xState[7] else ('O' if oState[7] else 7)
    eight = 'X' if xState[8] else ('O' if oState[8] else 8)
    print(f"{zero} | {one} | {two}")
    print(f"---------")
    print(f"{three} | {four} | {five}")
    print(f"---------")
    print(f"{six} | {seven} | {eight}")

contact if you have any doubts.

Answered By: rishabh11336

Answer 3

You have a board that you want to represent as a 2D object (3×3)

a b c
d e f
g h i

However, you are storing this board as a 1D list

    [a b c d e f g h i]
idx  0 1 2 3 4 5 6 7 8

idx represent the index of each element, so if you want to print the board line by line you need to first print:

a b c  (which corresponds to indices 0,1,2)

then

d e f  (which corresponds to indices 3,4,5)

and finally

g h i  (which corresponds to indices 6,7,8)

As you can see, the start indices will be 0,3,6 and the final indices will be 3,6,9 (since python slicing never includes the last index, and therefore it has to be 1 unit more than the last index you actually want to include)

so we can make a loop that iterates 3 times (we have 3 rows), so the iteration variable will have the values (0,1,2)

Now we have to convert (0,1,2) to the start indices (0,3,6) which is just multiplying by 3, this is where the 3*i comes from

And we have to convert (0,1,2) to the end indices (3,6,9) which is just (i+1)*3

So the slicing for a given loop iteration must be board[i*3:(i+1)*3] which is what you have in the code

Instead of a 1D list, you could have represented the board as a list of lists, and have two indices: one for row and one for column

board = [[a, b, c], 
         [d, e, f], 
         [g, h, i]]

Answered By: Sembei Norimaki

How does this code to create a board for a Tic-Tac-Toe game work?

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