In this tutorial we are going to see how we can implement the tic tac toe game in Python. We can either make use of random numbers for the computer move or we can develop a simple algorithm which will play the role of a computer.

在本教程中，我们将了解如何在Python中实现井字游戏。 我们可以利用随机数来移动计算机，也可以开发一种简单的算法来扮演计算机的角色。

Let us first see the representation of our board. For the purpose of this tutorial every place in the board is given a number. The board will look like this.

首先让我们看一下董事会的代表情况。 出于本教程的目的，董事会中的每个位置都有一个编号。 董事会将看起来像这样。

So when we say the 9th position is being assigned an X it means the last box of the board will be assigned an X.

因此，当我们说第9个位置被分配为X时，这意味着木板的最后一个框将被分配为X。

Using Random Numbers:

使用随机数：

Here we can keep a list of blank spaces on the board. Everytime when it’s computer’s turn we will generate a random number between [0,size_of_list-1] and computer will play that the move on the number present on that index. Here is the code-snippet for the computer move using a random number generator.

在这里，我们可以在板上保留空白列表。 每次轮到计算机时，我们都会在[0，size_of_list-1]之间生成一个随机数，然后计算机将对该索引上存在的数字进行移动。 这是使用随机数生成器进行计算机移动的代码片段。

def computer_move():move=-1empty_place = []# If I can win, others don't matter.for i in range(1,10):if board[i-1] == i-1 :empty_place.append(i)if len(empty_place) > 0 :idx = random.randint(0,len(empty_place)-1)move = empty_place[idx]return make_move(board,computer,move)return (False,False)

Here empty_place will store all the positions on the board which are empty and then we will take a random positions from the list and make sure that the computer’s move.

在这里empty_place将存储板上所有空的位置，然后我们从列表中随机抽取一个位置并确保计算机移动。

Developing Computer Algorithm: 

开发计算机算法：

The algorithm will consist of five steps:

该算法将包括五个步骤：

1. First we will see if there is a move that computer can make which will win the game. If there is so, take that move. Otherwise, go to step 2.

   首先，我们将看看计算机是否可以采取行动来赢得比赛。 如果是这样，那就采取行动。 否则，请执行步骤2。

2. See if there is a move that player can make which will cause computer to lose game. If there is so, move there to block the player. Otherwise, go to step 3.

   查看玩家是否可以采取行动，这将导致计算机输掉游戏。 如果有，请移动到那里以阻止播放器。 否则，请转到步骤3。

3. Check if any of the corner spaces (spaces 1, 3, 7, or 9) are free. If so, move there. If no corner piece is free, then go to step 4.

   检查是否有任何角落空间(空间1、3、7或9)可用。 如果是这样，请移到那里。 如果没有角码可用，则转到步骤4。

4. Check if the center is free or not. If so, move there. If it isn’t, then go to step 5.

   检查中心是否免费。 如果是这样，请移到那里。 如果不是，请转到步骤5。

5. Move on any of the side pieces (spaces 2, 4, 6, or 8). Now there are no more steps, because if execution reaches step 5 the side spaces are the only spaces left.

   在任何侧边上移动(空格2、4、6或8)。 现在没有更多的步骤了，因为如果执行到达步骤5，则仅剩下边距。

Our simple algorithm is now ready. Let us first complete other requirements and then we will see the code for this algorithm.

我们的简单算法现已准备就绪。 让我们首先完成其他要求，然后我们将看到该算法的代码。

We will need to display the board after every turn.

我们将需要在每转之后显示板。

For the board we will take a list of size 9 which will contain the position at the start and later they will be changed to X and O.

对于板，我们将采取大小9将包含在起始位置的列表，之后他们将被更改为 X 和 O。

At the start of the game, we first need to decide player will get X or O and whose turn will be first. This can be either done automatically using random number or we can take input from user. For this tutorial we are doing it using random numbers.

在游戏开始时，我们首先需要确定玩家将获得X或O，并且其回合将是第一位。 这可以使用随机数自动完成，也可以从用户那里获取输入。 在本教程中，我们使用随机数进行操作。

We will then create a new board (i.e A list having values from 1 to 9).

然后，我们将创建一个新的面板(即，一个值从1到9的列表)。

To check if move is valid: 

要检查移动是否有效：

Everytime we are making a move we should keep in mind that the position should not be already occupied (If a position contain the X or O then its occupied otherwise it will contain position number).

每次我们移动时，都应记住该位置不应已经被占用(如果一个位置包含X或O，则该位置将被占用，否则将包含位置编号)。

To check if player has won: 

要检查玩家是否赢了：

To check if a player won we have to check 2 diagonal entries, 3 rows and 3 columns. We will make a list of all the winning combinations and go through each after every turn of player or computer to check if they won.

要检查玩家是否获胜，我们必须检查2条对角线，3行和3列。 我们将列出所有获胜组合的清单，并在每次转动玩家或计算机后逐一检查它们是否获胜。

The winning combinations are as below:

获奖组合如下：

Row-wise: (0,1,2), (3,4,5), (6,7,8)

逐行：(0,1,2)，(3,4,5)，(6,7,8)

Column-wise: (0,3,6), (1,4,7), (2,5,8)

列方式：(0,3,6)，(1、4、7)，(2、5、8)

Diagonal-wise: (0,4,8), (2,4,6)

对角方向：(0,4,8)，(2,4,6)

To check next move of computer:

要检查计算机的下一步动作：

If in a turn computer cannot win and we aren’t blocking the players pieces then we have to decide what move computer will play (i.e. If we are at step 3 of algorithm). Now according to the preference of our algorithm we should first check corner places, then center and then the rest. So we will create a list which will store these steps in the correct order and we will go through this list and make our move.

如果转牌时计算机不能赢，并且我们没有挡住玩家的棋子，那么我们必须决定计算机将要进行的移动(即，如果我们处于算法的第3步)。 现在，根据我们算法的偏好，我们应该首先检查拐角位置，然后检查中心，然后检查其余部分。 因此，我们将创建一个列表，该列表将以正确的顺序存储这些步骤，然后我们将遍历该列表并继续进行操作。

Corner: (1,3,7,9)

角球：(1,3,7,9)

Center: (5)

中心：(5)

Others: (2,4,6,8).

其他：(2,4,6,8)。

Now let us see how the code looks :

现在让我们看一下代码的样子：

Python Tic Tac Toe游戏代码 (Python Tic Tac Toe Game Code)

import random
import sysboard=[i for i in range(0,9)]
player, computer = '',''moves=((1,7,3,9),(5,),(2,4,6,8))winners=((0,1,2),(3,4,5),(6,7,8),(0,3,6),(1,4,7),(2,5,8),(0,4,8),(2,4,6))
# Table
tab=range(1,10)# Function to print the board
def print_board():x=1counter = 0for i in range(0,9):if board[i] == 'X' or board[i] == 'O' :print(board[i],end = ' ')else :print(" ",end = ' ')counter+=1if counter%3 ==0 :print("",end = '\n----------\n')else :print("| ",end='')# This function will decide player will play with X or O
def select_char():if random.randint(0,1) == 0:return ('X','O')return ('O','X')# Function to decide a valid move.
def can_move(brd, player, move):if move in tab and brd[move-1] == move-1:return Truereturn False# Function to check if win condition is satisfied.
def can_win(brd, player, move):places=[]x=0for i in brd:if i == player: places.append(x);x+=1win=Truefor tup in winners:win=Truefor ix in tup:if brd[ix] != player:win=Falsebreakif win == True:breakreturn windef make_move(brd, player, move, undo=False):if can_move(brd, player, move):brd[move-1] = playerwin=can_win(brd, player, move)if undo:brd[move-1] = move-1return (True, win)return (False, False)def random_move():move=-1empty_place = []# If I can win, others don't matter.for i in range(1,10):if board[i-1] == i-1 :empty_place.append(i)if len(empty_place) > 0 :idx = random.randint(0,len(empty_place)-1)move = empty_place[idx]return make_move(board,computer,move)return (False,False)# AI goes here
def computer_move():move=-1# If the computer can win in this turn, it will take the turn.for i in range(1,10):if make_move(board, computer, i, True)[1]:move=ibreakif move == -1:# If player can win next turn then we have to block that.for i in range(1,10):if make_move(board, player, i, True)[1]:move=ibreakif move == -1:# Otherwise, try to take one of the desired places.for tup in moves:for mv in tup:if move == -1 and can_move(board, computer, mv):move=mvbreakreturn make_move(board, computer, move)def is_full():for i in range(0,9):if board[i] == i:return Truereturn Falseplayer, computer = select_char()
print('Player is [%s] and computer is [%s]' % (player, computer))turn = random.randint(0,1)if turn == 0 :print("Player will play first.")
else :print("Computer will play first.")print_board()computer_move()print()result="It's a tie !!"while is_full():print_board()print('Make your move  [1-9] : ', end='')move = int(input())moved, won = make_move(board, player, move)if not moved:print(' >> Invalid number ! Try again !')continueif won:result=' You won !!'breakelif computer_move()[1]:result=' You lose !!'break;print_board()
print(result)

Output:

输出：

Player is [X] and computer is [O] Computer will play first. | | ———- | | ———- | | ———-

播放器为[X]，计算机为[O]。 计算机将首先播放。 | | ———- | | ———- | | ————

O | | ———- | | ———- | | ———- Make your move [1-9] : 2 O | X | ———- | | ———- O | | ———- Make your move [1-9] : 4 O | X | O ———- X | | ———- O | | ———- Make your move [1-9] : 5 O | X | O ———- X | X | O ———- O | | ———- Make your move [1-9] : 8 O | X | O ———- X | X | O ———- O | X | ———- You won !!

O | | ———- | | ———- | | ———- 行动[1-9]：2 O | X | ———- | | ———- O | | ———- 行动[1-9]：4 O | X | O ———- X | | ———- O | | ———- 行动[1-9]：5 O | X | O ———- X | X | O ———- O | | ———- 行动[1-9]：8 O | X | O ———- X | X | O ———- O | X | ------- 你赢了！

In the replace computer_move with random_move in the function calling then the computer will take random moves instead of using the algorithm as discussed before.

在函数调用中用random_move替换computer_move的情况下，计算机将采取随机移动的方式，而不是使用前面讨论的算法。

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翻译自: https://www.thecrazyprogrammer.com/2019/11/python-tic-tac-toe-game.html