In a game of tic-tac-toe, (or Xs and Os, or noughts and crosses), the first player places an $X$ in one of the squares on a $3 \times 3$ board, and then the second player places an $O$, and then the first player places an $X$ again, and so on until either one of the players gets three-in-a-row and wins, or else the squares are all filled and the game ends in a draw.
Tic-tac-toe is a popular game among children, but for most the game becomes uninteresting once they realize that there exists an optimal strategy: If both players follow a simple rule for where to place each piece, essentially constantly blocking each other, the game is guaranteed to end in a draw. We say that the game is "solved": There is an optimal strategy, an *algorithm* if you will, and if both players commit to following it, then the game is decided before it has even begun.
But through the introduction of a little *quantum physics*, the game suddenly becomes a lot more interesting again!
Fear not though: You do not need to understand any quantum physics to understand quantum tic-tac-toe—or TiqTaqToe, as it is also known. Experience has shown time and again that the ruleset is actually quite easy for new players to grasp! But finding a good *strategy* to outsmart your opponent, on the other hand, turns out to be quite challenging—even to those of us who both understand quantum physics, and who have been playing this game for a while!
### 1. Rules
#### 1.1 Setup
Each player is given a set of d4 dice of various colours: The first player ($X$) gets five colours, and the second player ($O$) gets four colours, so that we have nine colours in total—the same as the number of squares on the board.
![[IMG_8650.jpeg|400]]
#### 1.2 Making a Move
For each turn, the player will:
- Place either one or two dice of a single colour that has not yet been used.
- If the player chooses to place one die, then it must be placed in an empty square with the number $4$ up.
- If the player chooses to place two dice (of the same colour), then the dice must be placed in two different squares, at least one of which are empty. The other square could also be empty, or it could contain one of the opponent's dice.
- *You may not place a die in a square that already houses one of your own.*
Let us look at each of the possible moves in turn.
##### The Classical Move
If the first player places a singular die in an empty square, then they have to place the die with the number $4$ pointing up. This shows that "all" of the colour is in that square; in other words, that the probability that the die will lead to an $X$ being drawn in that same square upon observation (which happens towards the end of the game—we will get to that shortly) is $4/4 = 100\%$.
![[IMG_8652 1.jpeg|400]]
##### The Superposition Move
Alternatively, the player may choose to place two dice *of the same colour* in two different squares. Then the player must place the dice with the number $2$ up. This shows that each square houses "half" of the colour. More precisely, it shows that there is a $2/4 = 50\%$ chance that the colour turns into an $X$ in the first square, and a $2/4 = 50\%$ chance that the $X$ instead ends up in the second square.
![[IMG_8653 1.jpeg|400]]
Note that the squares do not have to be adjacent.
![[IMG_8654 1.jpeg|400]]
So one may place either one or two dice of a colour per turn. One may *not* place more than two dice at once, and one may not place dice of a colour that was already used on a previous turn. Furthermore, it is not allowed to place dice with the numbers $1$ or $3$ facing up, even if they would sum to $4$.
##### The Entangling Move
If the first player already put down a singular die, then the second player may "invade" the opponent's square by placing two dice of a colour, one in their square and the other in an empty square. As before, they will place the dice with the number $2$ pointing up.
![[IMG_8655 2.jpeg|400]]
But given that we are interpreting the numbers as probabilities, this gives us something of a strange situation: The board is now telling us that there is a $100\%$ probability for an $X$ to be observed in the middle square, *and* a $50\%$ probability that an $O$ is observed in the middle square. There can't be more than one symbol per square, so this is clearly impossible.
The game solves this conundrum by ruling that when player $2$ "invades" player $1