Quantum Tic Tac Toe is a variant of Tic-Tac-Toe with "entangled" states (there are even iPhone and Android apps).

In essence, a classic naught or cross isn't placed until the 'quantum state' of the board is measured and forms a closed loop.

Here the pieces X3, X7 and O8 are measured when placing the two O8 pieces are placed and forms a closed loop:

enter image description here

Upon measurement, the player placing the X marks, picks whether O8 collapses onto square 3 or square 5. After picking, and resolving entanglement, the board looks like this (O2 and O4 collapse in 7, and 8 respectively. X5 to square 4.)

enter image description here

Is there a winning strategy for either player? If so, what is it? If not, how can one guarantee a tie?

  • 1
    @thesunneversets - one of the reasons for posting this was wanting to spread the knowledge - I just found it yesterday and had the same thought as you.
    – ripper234
    Apr 7, 2012 at 8:21

1 Answer 1


In Solving Quantum Tic-Tac-Toe, Ishizeki and Matsuura use a computer to search the game space to find solutions, but don't specifically speak to strategy. However, there are a few strategies we can glean from their results:

  1. Go for a half-point victory, not a full point victory. In their search they found the first player, X, cannot guarantee a win by being the only player to get three in a row. But a win can be guaranteed where X and O both have 3 in a row, but X has the lower maximal subscript.
  2. As player X, make your first marks split into opposite corners. If X plays perfectly, a half-point win can be guaranteed by playing the first pair of marks, X1, in opposite corners.
  3. Don't play as O, the second player. Obviously this is not a fair strategy or one you're likely be able to follow in a real game. But since X can guarantee a half-point victory, it follows O can't guarantee a victory.

Since there is a guaranteed path to victory for X, the obvious strategy is to, after playing the first pair in opposite corners, always respond to a move by O by playing such that you are still guaranteed victory (or for O, always play the move that gives you the most options for victory, and hope X makes a mistake). Unfortunately the authors didn't include all of the paths in their search, so it would be necessary to reproduce their results to apply this strategy.

However, I have tried to glean some strategies from the one game they did include in the paper (see Figure 5. I've broken this out into each move in the image below):

  1. Avoid entanglements. By not creating a cycle you avoid giving your opponent the advantage of choosing where your marks are fixed. Of course, if you can make a cycle such that either choice is a win for you, like the placement of X9, then you should do so. See both the resolution of X9 and the alternate resolution.
  2. Play at least one mark of a pair in an empty square. This increases the chances of having the lowest maximum subscript in a tie breaker, as on each play you add another square where you have the mark with the lowest subscript. X does this up until the last move.
  3. Play to increase the number of probable wins. For example on Move 3, after playing X1 in opposite corners, play one X3 in the middle and the other in a corner. When one X1 collapses, the other X1 that is now fixed will be in line with either X3. The same is true for X3 in relation to X1
  4. Don't play on your own marks. X never does this until the final move when there is no other option.
  5. Don't let your last mark be part of your three in a row. Of course your last move has the highest subscript, and can make you lose the tie-breaker. Making this actually happen may require some additional planning.

Breakdown of Figure 5 game

  • 1
    X could force a win on move 7 by resolving O6 the other way and then playing X7 on top of X1, which can only resolve as a diagonal win for X. This is taking advantage of the general tic-tac-toe rule that the middle square is the most important
    – Chris Dodd
    Jul 15, 2013 at 22:32
  • @ChrisDodd that solution gives up X's chance of getting the full point, the solution posted does just as well with the half point forced win but also has a chance to win the full point.
    – GDanger
    May 28, 2015 at 23:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .