Checking if the tictactoe game has ended and returning the winner - numpy

Question

I'm trying to code the function which checks if the tictactoe game has ended and returns the winner if the game has ended. Can you please help me code this part ?
input: s, a 3*3 integer matrix depicting the current state of the game.
if the element of the matrix is equal to 0 means the place is empty, and if the element is equal to 1 means the place is taken by 'X' player and if the element is equal to -1 means that the place is taken by 'O' player.
output:
e: the result, an integer scalar with value 0, 1 or -1.
if e = None, the game has not ended yet.
if e = 0, the game ended with a draw.
if e = 1, X player won the game.
if e = -1, O player won the game.
I tried some user defined functions to work this out but couldn't do it properly. can you please help me code this part ?
[code]
def check_game(s):
e = None
def sum_all_straights(S):
r3 = np.arange(3)
return np.concatenate([s.sum(0),s.sum(1),s[[r3,2-r3],r3].sum(1)])
def winner(S):
ss = sum_all_straights(S)
ssa = np.absolute(ss)
win, = np.where(ssa==3)
if win.size:
e = ss[win[0]]//3
sas = sum_all_straights(np.absolute(S))
if (sas>ssa).all():
e = 0
return e

Answer 1

Here is one approach: We can check for a winner by summing the fields of each straight line. If at least one of the sums is 3 or -3 then player X or O has won.
If there is no winner we check for a draw. The game is drawn if no player can win anymore, i.e. if each straight line has at least one X and one O. We can detect that by first taking the absolute value of the board and then summing, call that "SAS"---this counts the chips on each straight line regardless of who they belong to---and first summing and then take the absolute value, call that "SSA". If both players have chips on a line there will be some cancellation in the sum and SSA will be smaller than SAS. If this is true for each straight line the game is drawn.
Implementation:
import numpy as np
def sum_all_straights(S):
r3 = np.arange(3)
return np.concatenate([S.sum(0),S.sum(1),S[[r3,2-r3],r3].sum(1)])
# ^ ^ ^
# | | |
# | ------------+-----
# | | |
# | ------------ |
# This returns the three vertical sums, | the three horizontal sums and
# the two diagonal sums, using advanced indexing
def winner(S):
ss = sum_all_straights(S)
ssa = np.absolute(ss)
win, = np.where(ssa==3)
if win.size:
return ss[win[0]]//3
sas = sum_all_straights(np.absolute(S))
if (sas>ssa).all():
return 0
examples = b"""
XO. OOO ... XXO
OXO XX. .X. OXX
XXO OXX .O. XOO
"""
examples = np.array(examples.strip().split()).reshape(3,4,1).view("S1").transpose(1,0,2)
examples = 0 + (examples==b'X') - (examples==b'O')
for example in examples:
print(example,winner(example),"\n")
Demo:
[[ 1 -1 0]
[-1 1 -1]
[ 1 1 -1]] None
[[-1 -1 -1]
[ 1 1 0]
[-1 1 1]] -1
[[ 0 0 0]
[ 0 1 0]
[ 0 -1 0]] None
[[ 1 1 -1]
[-1 1 1]
[ 1 -1 -1]] 0