SIZ TUTORIAL | Creation Of Tic-Tac-Toe Game With Python in 15 lines | 20% Beneficiary to @sIz-official

Introduction

Hello Guys Assalam-o-Alaikum,
Hope you all guys are fine in this Covid circumstances, Today I will tell you about very interesting
game which everybody has at least one time played in their life its name is Tic-tac-Toe game but the twist is that we will code this game in python programming lets start coding and its interpretation.

Achievements	Links
Achievement # 1	Link
Achievement # 2	Link

Step#1

First we are taking input from user to input elements of row 1 to row 3. Note that input must be "X" and "O" remember it's case sensitive.

pattern=[]
for i in range(1,4):
    print('Enter state',i,'for row',str(i),'separated by commas:',end=' ')
    i=input()
    pattern.append(i.split(","))

Step#2:It will print elements("O" or "X") in a matrix form which I will show you in output. It follows the complete pattern as play tic tac toe game on paper.

for row in pattern:
    print('|',end='\t')
    for state in row:
        print(state,end='\t')
    print('|')

Step#3

This is our final step it will check whether our rows elements are same, diagonally all elements are same or columns wise elements match. if "X" win it will display result in the end "X" or if "O" wins it will show result "O"

if (pattern[0][0] == pattern[0][1] == pattern[0][2]) and (pattern[0][0]=='O'or'X'):print(pattern[0][0])
elif (pattern[1][0] == pattern[1][1] == pattern[1][2]) and (pattern[1][0]=='O'or'X'):print(pattern[1][0])
elif (pattern[2][0] == pattern[2][1] == pattern[2][2]) and (pattern[2][0]=='O'or'X'):print(pattern[2][0])
elif (pattern[0][0] == pattern[1][0] == pattern[2][0]) and (pattern[0][0]=='O'or'X'):print(pattern[0][0])
elif (pattern[0][1] == pattern[1][1] == pattern[2][1]) and (pattern[0][1]=='O'or'X'):print(pattern[0][1])
elif (pattern[0][2] == pattern[1][2] == pattern[2][2]) and (pattern[0][2]=='O'or'X'):print(pattern[0][2])
elif (pattern[0][0] == pattern[1][1] == pattern[2][2]) and (pattern[0][0]=='O'or'X'):print(pattern[0][0])
elif (pattern[0][2] == pattern[1][1] == pattern[2][0]) and (pattern[0][2]=='O'or'X'):print(pattern[0][2])
else:
    print(' ')

Output

Enter state 1 for row 1 separated by commas: O,X,X
Enter state 2 for row 2 separated by commas: X,X,O
Enter state 3 for row 3 separated by commas: O,X,O
| O X X |
| X X O |
| O X O |
X

Conclusion

It is a quite interesting let you guys run code if you face any problem you can ask in comment.