In tic-tac-toe we have 3X3 matrix. Suppose we play it and the game ends in a tie position. This means neither 'X' nor 'O' won. Now we pad the matrix equally on all its sides to make a bigger 5X5 matrix, the interior 3X3 holding the 'X' and 'O's in tie position). Since, the game has been extended, we can now have at most 5 'X's in an extended row/column, so we give 5 points for that. Similarly, we can also have 3 consecutive 'X's in a diagonal because we already have an interior tie matrix. We give 3 points for that(and so on...4 points for 4 consecutive horizontal/vertical 'X's or 3 points for 3 consecutive horizontal/vertical/diagonal 'X's). We calculate total points of each player till the matrix fills. Note that if there were 3 consecutive 'X's in a row(for which player had got 3 points) and if he puts another 'X' to make it 4 consecutive in a row, he/she will get additional 4 points.
So, can we prove that there will always be at least one winner and the game will not end in a tie in this extended matrix. I am only getting the approach of permutation and combinations but 5X5, it seems difficult on pen and paper. So I am also open to an intuitive solution.
