python - Tic Tac Toe how to check win-CodePudding

I am currently learning to create tic tac toe game using python. I need to compare the board with the list to know whether the player win or not by checking whether the board meet the win condition or not. But I don't know how to do it. Can anyone help?

board = {
1: 'X', 2: 'X', 3: 'X',
4: 'X', 5: 'O', 6: 'O',
7: 'O', 8: 'X', 9: 'O'
}


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

CodePudding user response：

If you note 'X' as 0 and 'O' as 1, the board can be represented as a 9-bit number, for example, for your board:

board = {
1: 'X', 2: 'X', 3: 'X',
4: 'X', 5: 'O', 6: 'O',
7: 'O', 8: 'X', 9: 'O'
}

is equivalent to:

board = 0b000011101  # 29

In a similar manner, the winning combinations can also be represented as 9-bit number:

winCombinations = np.array([
  0b111000000,  # 448
  0b000111000,  # 56
  0b000000111,  # 7
  0b100010001,  # 273
  0b001010100,  # 84
  0b100100100,  # 292
  0b010010010,  # 146
  0b001001001   # 73
])

Using this method, you can check if a board has a winning combination via bitwise operation:

if np.any(board & winCombinations == winCombinations):
    print('O won')
elif np.any(~c & winCombinations == winCombinations):
    print('X won')
else:
    print('Draw')

In case you want to be able to check non-complete boards as well, you can use 18 bits to describe a board (first 9 bits for 'X' and other 9 bits for 'O'). Using this method, the board representation will be:

board = 0b111100010000011101  # 246813

Then the checking will be:

if np.any(board & winCombinations == winCombinations):
    print('O won')
elif np.any(board >> 9 & winCombinations == winCombinations):
    print('X won')
else:
    print('No winner')

CodePudding user response：

A very pythonic way to state the condition when a player wins with like this

def isWinner(board, player):
    return any(
        all(board[cell] == player for cell in line) 
      for line in winCombinations)

Then you use it like this

isWinner(board, 'O'), isWinner(board, 'X')

CodePudding user response：

I don't want to give you the exact answer because you probably want to learn it yourself. So I'll just give you an easy way to solve it !

You should iterate thought your winning combinations (assuming you gathered all the possible winning combinations) and for each winning combination : check if the 3 cell belong to the SAME player ! If they do, you know that this player won !

But be careful, you need to handle empty cells ! Because you don't want your function to return a winner because, for a given winning combinaison, your 3 cells are empty.

CodePudding user response：

This function should do it:

def winner(board, winning_combinations):
  for combination in winning_combinations:
    if board[combination[0]] == board[combination[1]] == board[combination[2]]:
      return board[combination[0]]
  return None

With your inputs it returns 'X'