Tic Tac Toe Game In Python

What we will see through this article is we will find exhaustively about Spasm Tac Toe Game In Python. Spasm Tac-Toe is an extremely straightforward two-player game. So just two players can play simultaneously.

This game is otherwise called Noughts and Cross or Xs and operating system game. One player plays with X, the other player plays with O. In this game we have a board with a 3X3 framework. The quantity of stages can be expanded.

This game is played by two people. In the first place, we draw a board with a 3 × 3 square lattice. On the off chance that the main player chooses ‘X’ and attracts something the square framework, the subsequent player has the chance to attract ‘O’ the accessible spaces.

Essentially, players draw ‘X’ and ‘O’ then again to fill in the spaces until a player prevails with regards to attracting 3 focuses a line on a level plane, in an upward direction, or slantingly. Then, at that point, the player dominates the match, any other way, the game will get a draw when every one of the seats are filled.

How about we see some ongoing interactions outwardly.

Tic Tac Toe Game in Python

Below is the code for the above game :

# Tic-Tac-Toe Program using
# random number in Python

# importing all necessary libraries
import numpy as np
import random
from time import sleep

# Creates an empty board
def create_board():
	return(np.array([[0, 0, 0],
					[0, 0, 0],
					[0, 0, 0]]))

# Check for empty places on board
def possibilities(board):
	l = []

	for i in range(len(board)):
		for j in range(len(board)):

			if board[i][j] == 0:
				l.append((i, j))
	return(l)

# Select a random place for the player
def random_place(board, player):
	selection = possibilities(board)
	current_loc = random.choice(selection)
	board[current_loc] = player
	return(board)

# Checks whether the player has three
# of their marks in a horizontal row
def row_win(board, player):
	for x in range(len(board)):
		win = True

		for y in range(len(board)):
			if board[x, y] != player:
				win = False
				continue

		if win == True:
			return(win)
	return(win)

# Checks whether the player has three
# of their marks in a vertical row
def col_win(board, player):
	for x in range(len(board)):
		win = True

		for y in range(len(board)):
			if board[y][x] != player:
				win = False
				continue

		if win == True:
			return(win)
	return(win)

# Checks whether the player has three
# of their marks in a diagonal row
def diag_win(board, player):
	win = True
	y = 0
	for x in range(len(board)):
		if board[x, x] != player:
			win = False
	if win:
		return win
	win = True
	if win:
		for x in range(len(board)):
			y = len(board) - 1 - x
			if board[x, y] != player:
				win = False
	return win

# Evaluates whether there is
# a winner or a tie
def evaluate(board):
	winner = 0

	for player in [1, 2]:
		if (row_win(board, player) or
			col_win(board,player) or
			diag_win(board,player)):

			winner = player

	if np.all(board != 0) and winner == 0:
		winner = -1
	return winner

# Main function to start the game
def play_game():
	board, winner, counter = create_board(), 0, 1
	print(board)
	sleep(2)

	while winner == 0:
		for player in [1, 2]:
			board = random_place(board, player)
			print("Board after " + str(counter) + " move")
			print(board)
			sleep(2)
			counter += 1
			winner = evaluate(board)
			if winner != 0:
				break
	return(winner)

# Driver Code
print("Winner is: " + str(play_game()))

Program Execution of Tic-Tac-Toe game

Output

The result will print on the screen utilizing the print() capability like displayed beneath:

[[0 0 0]
 [0 0 0]
 [0 0 0]]
Board after 1 move
[[0 0 0]
 [0 0 0]
 [1 0 0]]
Board after 2 move
[[0 0 0]
 [0 2 0]
 [1 0 0]]
Board after 3 move
[[0 1 0]
 [0 2 0]
 [1 0 0]]
Board after 4 move
[[0 1 0]
 [2 2 0]
 [1 0 0]]
Board after 5 move
[[1 1 0]
 [2 2 0]
 [1 0 0]]
Board after 6 move
[[1 1 0]
 [2 2 0]
 [1 2 0]]
Board after 7 move
[[1 1 0]
 [2 2 0]
 [1 2 1]]
Board after 8 move
[[1 1 0]
 [2 2 2]
 [1 2 1]]
Winner is: 2

Final Words

What we realized through this article is Spasm Tac Toe Game In Python. Likewise in the event that you have any questions kindly leave a remark through the remark box. What’s more, we ask that you benefit every individual who imparted this article to your companions.