This exercise is Part 2 of 4 of the Tic Tac Toe exercise series. The other exercises are: Part 1, Part 3, and Part 4.
As you may have guessed, we are trying to build up to a full tic-tac-toe board. However, this is significantly more than half an hour of coding, so we’re doing it in pieces.
Today, we will simply focus on checking whether someone has WON a game of Tic Tac Toe, not worrying about how the moves were made.
If a game of Tic Tac Toe is represented as a list of lists, like so:
game = [[1, 2, 0],
[2, 1, 0],
[2, 1, 1]]where a 0 means an empty square, a 1 means that player 1 put their token in that space, and a 2 means that player 2 put their token in that space.
Your task this week: given a 3 by 3 list of lists that represents a Tic Tac Toe game board, tell me whether anyone has won, and tell me which player won, if any. A Tic Tac Toe win is 3 in a row - either in a row, a column, or a diagonal. Don’t worry about the case where TWO people have won - assume that in every board there will only be one winner.
Here are some more examples to work with:
winner_is_2 = [[2, 2, 0],
[2, 1, 0],
[2, 1, 1]]
winner_is_1 = [[1, 2, 0],
[2, 1, 0],
[2, 1, 1]]
winner_is_also_1 = [[0, 1, 0],
[2, 1, 0],
[2, 1, 1]]
no_winner = [[1, 2, 0],
[2, 1, 0],
[2, 1, 2]]
also_no_winner = [[1, 2, 0],
[2, 1, 0],
[2, 1, 0]]There were many strategies for approaching this problem. The important part is to be careful about checking rows, columns, and diagonals for winners. Things like making sure that a row of 0, 0, 0 doesn’t win with player number 0 is important, and making sure to count the columns and rows at the edges.
A few submitted solutions are below - feel free to comment if you think you have a better one, and I’ll put it up here!
Not using too many functions, but using the numpy library to transpose the game board, turning column-checking into row-checking.
Using set() to simplify the counting.
Happy hacking!