Dorian and Abigail play a game of tic-tac-toe. Dorian plays with crosses (x) and Abigail plays with noughts (o). However, we do not know who amongst Dorian and Abigail started this game.
In the game shown below, six moves have already been made:
o | o |
---+---+---
o | x |
---+---+---
x | x |
Who will win the game?
Let the board be numbered as
1|2|3
4|5|6
7|8|9
The last move for x must be 7. If not then, either {7,5} or {7,8} had two x and Dorian could have picked 3 (resp. 9) to win.
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Let's now assume that Dorian also did the last move to bring the game to its current state. We will work backwards to arrive at an illogical situation.
The board before that move is
o o .
o x .
. x .
We can see that Abigal must haved necessarily picked 1 as her last move as otherwise se could have won by either {1,2,3} or {1,4,7}.
The board before that is hence
. o .
o x .
. x.
Dorian placed an x in either 5 or 8. We don't know which. However what we know is that either of these moves is deeply illogical: Ttey ultimately allow Abigail placing an o 1 and secure victory.
Therefore the board is not the logical conclusion of any game where x played last.
So Abigal must have placed the last move and Dorian will therefore place an x to win in the next round.
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To be honest, I would not be surprised if the other path where O played last would lead to some kind of similar of contradiction but with a quick inspection it seems more plausible.
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Posted by Yannis
on 2022-06-28 13:20:41 |
An attempt
