Variation of Tic-Tac-Toe where players have numbers 1 to 9 for their moves

n3t.py

# board is an array with indices in following arrangement
#
# 1 | 5 | 2
# ---------
# 6 | 0 | 7
# ---------
# 3 | 8 | 4

win_indices = [
    (1,5,2), (6,0,7), (3,8,4), #left to right
    (1,6,3), (5,0,8), (2,7,4), #top to bottom
    (1,0,4), (2,0,3)           #diagonal
]

# Player A uses the numbers 1,2,...,9
# Player B uses the numbers -1,-2,...,-9
# 0 means the field is not taken

# returns 1 for player A, -1 for player B, 0 for no winner
def winner(board):
    for a,b,c in win_indices:
        if board[a] > 0 and board[b] > 0 and board[c] > 0:
            return 1
        if board[a] < 0 and board[b] < 0 and board[c] < 0:
            return -1
    return 0

# enumerate valid moves for player A (player=1) or player B (player=-1)
def moves(player, board):
    player_numbers = range(9,0,-1) if player > 0 else range(-9,0)
    numbers_left = set(player_numbers)
    numbers_left.difference_update(board)
    for i in range(9):
        number = abs(board[i])
        for x in player_numbers:
            if abs(x) > number and x in numbers_left:
                yield i,x

def evaluate(player, board):
    winning_player = winner(board)
    if winning_player != 0: #we already have a winner
        return winning_player
    
    best = -player
    for i,x in moves(player, board):
        old, board[i] = board[i], x #update board
        best_opponent = evaluate(-player, board)
        board[i] = old #reset - avoids slow copy of array
        if best_opponent == 0: #player can force a draw
            best = 0
        if best_opponent == player: #player can force a win
            best = player
            break
    return best

print(evaluate(-1, [9] + [0]*8))