## Level 4 – Expert: TiqTaqToe with Bases In Level 3, [[TiqTaqToe with Interference]], we encountered a new kind of superposition: Now, a single piece could be in a superposition, not in space, but over the *symbols* $X$ and $O$. It is this capability that we now extend, as we allow players to create two different *kinds* of superpositions over symbols. This is where the blue numbers enter the stage. The red numbers represented "minus the opponent's symbol", such that a red number belonging to $X$ would represent "$-Oquot;. The blue numbers, meanwhile, represent "minus the *player's* symbol", so that a blue number belonging to $X$ will represent "$-Xquot;. But don't worry, you won't be using them to cancel yourself! The two different kinds of symbol-superpositions will differ in the relative sign of the dice—$X$ plus $O$ versus $X$ minus $O$—and yet they will behave identically when observed in the standard manner. However, when making a *skew* observation—a choice now available to the initiating player, as explained in detail below—there will be a crucial difference![^5] ### Handing out the Pieces Now, each player should have access to all their dice, so $40$ d4s and one d6 each. Hence, there is no more need for a "bank" aside from the observation token, which should as before be placed between the players with the "XO" side up until the observation phase, (at which point it is given to the initiating player). ### The Spinning Move This move places a classical piece belonging to the opponent into a symbol-superposition, while also placing another piece in a symbol-superposition in a different square. Much like the cancelling move from the previous level, it is performed like an entangling move, except that one of the two $2$s is now *blue*. ![[IMG_0056.jpeg|400]] Like the entangling move, both squares will house a symbol after the observation phase, but *unlike* the entangling move, the two squares are now *independent*. This means that each square will be rolled for independently in the observation phase, so that you can end up with $X$s or $O$s in both squares—an impossibility with the entangling move. So while the *average* outcome might be the same, performing the spinning move comes with a greater risk, but also a potentially greater reward! The name of the move is inspired by the common visage of a spinning coin, often used to illustrate how a qubit—the fundamental building block of quantum computers—can be "both $0$ and $1$ at the same time". Exchange $0$ and $1$ for $X$ and $O$, and this is exactly the kind of superposition we are creating here! The symbol superpositions in the two squares are not *equal*, however: The right square houses the superposition state "half $X$ plus half $Oquot;, while the middle square houses the state "half $X$ *minus* half $Oquot;. We will refer to the first kind of state as a "plus state" and the latter kind as "minus state".[^1] If we observe these states as before, however, this minus sign makes no difference: The result is still 50/50 $X$ or $O$ in each square independently. The difference comes with the option of *skew* observation, introduced below. ### The Half-spinning Move This move is performed exactly like the half-entangling move, except that one of the two $2$s is now blue. ![[IMG_0057.jpeg|400]] However, the resulting state can be hard to interpret at first glance: What does it *mean* for a state to "halfway" be in a symbol-superposition? The answer is that this is now a more complicated state describing two particles in a superposition over both symbols and squares. We will sidestep the need for a clearer intuitition by, in the observation phase, checking for empty squares first. Then, what you end up with is equivalent to either a spinning move or a superposition move, depending on the roll. (If a half-spinning move was performed on *both* halves, it might be easiest to first roll for which half of the original superposition the yellow particle is found to be in, after which two squares will be in a normal superposition, and two squares will be in spin-superpositions.) ### The Cancelling Moves These moves function as before, except that—for reasons that will become clear shortly—Player 2 should always place one red and one *blue* $2$. (Player 1 places one red and one blank $2$, as before.) ![[IMG_0061.jpeg|400]] ### The Observation Phase New to this level is that whoever initiates the observation phase, meaning whoever spent the last open square, gets to choose between two *modes*, or *bases*, in which to observe: Standard, or *skew*. Let's observe the following board. ![[IMG_0087.jpeg|400]] Before placing any symbols, we have to figure out which squares will end up empty. Start by removing the dice from any cancelled squares. ![[IMG_0085.jpeg|400]] Now, one square that might end up empty is the bottom square. Player 1 points to it and asks, "is this square mine?" and rolls. They get a $4$ – the answer is yes! But not so fast—we are not allowed to draw in any symbols yet! For now, instead move the die to a $4$, as if a classical move had been made in the bottom square, and remove the pink dice from the right side of the board. ![[IMG_0084.jpeg|400]] Now, Player 2 has two grey dice in the bottom right and top right squares, one of which has blue numbering: "minus half $Oquot;. The minus sign makes no difference here, however, since any state can be multiplied by minus one without changing it, and so this is just a normal superposition state. Player 2 points to the top-right square and asks, "is this square mine?" They roll a $2$ – the answer is yes, but again we can't draw in any symbols yet, and so we just make it a blank $4$ for now. ![[IMG_0082.jpeg|400]] We are left with two empty squares, and seven squares that we know will house a symbol in the end: Two with classical states, three with symbol superpositions, and two that are entangled.[^2] At this stage, Player $2$, who put down the last dice and is thus the initiating player (as shown by them possessing the observation token), makes their choice between an observation in the *standard* basis, or the *skew* basis. #### Standard Basis Observation These are the kind of observations we have been performing so far. Now, as before, start by replacing classical pieces with their respective symbols. ![[IMG_0088.jpeg|400]] Let us resolve the entangled squares next. Player 1 points to the bottom-left square and asks, "is this square mine?" They roll a $3$, so the answer is no—and, since the squares are entangled, this means that since the purple $X$ piece goes in the top-left square, the blue $O$ piece goes in the bottom-left. ![[IMG_0089.jpeg|400]] The remaining three squares are all independent of each other, so we resolve them one at a time. Player 2 points to the leftmost square and asks, "is this square mine?" They roll a $1$; that's a *no*, so the square goes to Player 1. ![[IMG_0090.jpeg|400]] Next, Player 1 points to the top square and asks, "is this square mine?", and roll a $6$. Lucky! ![[IMG_0091.jpeg|400]] Finally, Player 2 rolls for the rightmost square, and rolls a $4$ – also a yes. ![[IMG_0092.jpeg|400]] The game is undecided, and there are still empty squares left, and so it continues. Now, if player 2 were next to go, they would be guaranteed victory. Do you see how?[^3] However, Player 2 initiated the observation phase, and so the turn goes to Player 1. The best they can do here is a 50/50 chance of victory for either player. Do you see how?[^4] #### Skew Basis Observation In short, observing in the skew basis *switches the behaviour of classical and spinning pieces*: Plus states now behave like $X$ states, and minus states behave $O$ states—and vice versa: $X$ states behave like plus states, and $O$ states like minus states, meaning they behave as if they were in a symbol-superposition! We start again at the following board. ![[IMG_0093.jpeg|400]] Player 2 chooses to observe in the skew basis, and flips the observation token so that the "+-" side faces up. Start by replacing any (*unentangled!*) plus states with the symbol $X$, and any minus states with the symbol $O$. ![[IMG_0094.jpeg|400]] Now you see why Player 2 should always place blue-numbered dice as part of a cancelling move: Otherwise, they would essentially give a piece to Player 1 for free! (At least in the skew basis.) As mentioned, the plus/minus states and the X/O states have switched roles. This means that it is now the *classical* states that need to be rolled for! Player 1 starts by pointing to the bottom square, and asks, "is this mine?" They roll a $5$ – the answer is *no*. Next, Player 2 asks the same question for the top-right square, and rolls a $2$ – a yes! ![[IMG_0096.jpeg|400]] All that remain are the two entangled squares. How is entanglement affected by switching from the standard to the skew basis? The surprising answer turns out to be: Not at all! In other words, entanglement is *basis-independent*: Whether two squares are connected or not does not depend on how you choose to look at them! Therefore, we roll just as we normally would. Player 1 points to the top left square and asks, "is this square mine?" They roll a $4$—the answer is yes—and, thanks to the entanglement, this means that the bottom left square is an $O$. ![[IMG_0097.jpeg|400]] It is Player 1's turn. However, their situation is dire—it appears the Quantum Gods were not on their side this round! %% This is because whichever square player 2 enda up capturing—and they *will* get one of them!—they will get, not one, but *two* three-in-a-rows. Player 1, meanwhile, will get none. The risk of rolling independent squares paid off: Player 2 wins 0-2! %% #### What about Half-cancelling? The half-cancelling move can be tricky to reason about, so let us deal with it separately. Let's say Player 2 performed the move, such that we have the following state. (Recall that Player 2 should place a blue $2$ as part of any cancelling move.) ![[IMG_0100.jpeg|400]] As before, we start by rolling to see whether the half-cancelling succeeded; if it didn't, we place the opponent's piece in a classical state in the square that was attempted cancelled. ![[IMG_0101.jpeg|400]] Otherwise, the square is empty. ![[IMG_0103.jpeg|400]] Next, we have to find out if the rightmost square is empty. Recall that, in a standard basis measurent, there are three possible outcomes, with probabilities one-fifth and two-fifths respectively. The two outcomes that yield a symbol in the bottom square both also yield an empty square in the rightmost square, and so this square is empty with three-fifths probability. Player 2 rolls a d6, after stating the outcomes: "On a $1$ or a $2$, the rightmost square contains a classical $X$, and on a $3$, $4$, or $5$, the rightmost square is empty (and on a $6$ we reroll)." They roll a $5$: It is empty. ![[IMG_0104.jpeg|400]] We now know that there *is* a piece in the bottom square, but the probability for each possible outcome is not equal. At this point the initating player chooses a basis. Let us look at each choice separately. ##### Standard Basis Observation We have two possible outcomes, $X$ or $O$, where the second outcome has double the probability of the first. We see that we have a total probability weight of $1+2=3$; then, the first outcome ($X$) must have probability one third, and the second outcome ($O$) must have probability two thirds. This is easily rolled for using a d6; Player 2 states the possible outcomes: "On a $1$ or a $2$, the square is yours, and otherwise it is mine." They roll a $4$ – it is their's! ![[IMG_0105.jpeg|400]] ##### Skew Basis Observation In the skew basis, this state is a non-equal superposition of a plus state and a minus state, so that we again have two possible outcomes: $X$ or $O$. What is the probability for each? Well, like for the half-cancelling square, this calculation again involves some non-trivial square roots and minus signs. If you're curious, you can find the full derivation in [[The Physics of TiqTaqToe]], but the end result is a surprising one: It turns out that this state, when observed in the skew basis, will yield an $O$ $97 \%$ of the time, and an $X$ only $3\%$ of the time! This means that, just like for the half-cancelled square, we can check for an $O$ using both dice: If player 2 rolls two $1$s, the square goes to the opponent, and otherwise it is their's. They roll, and... ![[IMG_0106.jpeg|400]] If the square did not involve a blue die, as would be the case if it were Player 1 who performed the cancelling move, then the probabilities would be reversed: $97\%$ for $X$, and $3\%$ for $O$. ![[IMG_0107.jpeg|400]] ![[IMG_0114.jpeg|400]] And so find that the half-cancelling move, just like the (full-)cancelling move, has much more power when observed in the skew basis than in the standard basis! What if Player 2 had half-cancelled both halves of Player 1's superposition? ![[IMG_0109.jpeg|400]] Well, first check that the cancelling succeeded by rolling two d6 and checking for double ones. Assuming they did, you then end up with the following state. ![[IMG_0111.jpeg|400]] There are four possible outcomes: An $X$ or an $O$, to the left or to the right. Note, though, that the state to the left is identical to the state to the right. This means that where the piece ends up must be of equal probability, i.e. 50/50. ***DOUBLE CHECK THIS!!*** As mentioned previously, the initating player gets to wait on deciding the basis in which to observe the board until all empty squares have been identified, so first, check where the piece will end up by pointing to (for instance) the bottom square and asking, "is the piece here?" ![[IMG_0113.jpeg|400]] Now, the situation is exactly what we had above, and can be dealt with accordingly: $3\%$ chance for an $X$, and $97\%$ chance for an $O$! ### Balance and Strategy Previously, whoever did *not* initiate the observation phase gained a possible advantage, because they would then be first to go after the observation phase was done, provided no winner was declared. Now, the pendulum swings in the other direction, as whoever places last gets to choose which *kind* of observation to perform—and even gets to make this decision *after* seeing where the pieces end up! What's more, it turns out that the second player can always *ensure* that they go last, and thus control the observation! (We leave it to you to figure out how.) Does this mean we now have a *second*-player advantage? Well, actially implementing that strategy usually involves playing a lot of superposition moves, foregoing the more powerful entangling, cancelling, and spinning moves in the process. What's more, this also allows Player 1 to place one more piece on the board than Player 2, increasing their stake on the board. And so we find that two natural strategies present themselves: Sacrifice everything for a chance to control the observation phase, or let go of that hope and instead work towards a position that is as advantageous as possible in both bases simultaneously. We leave it for you to decide which strategy works best—or to come up with your own! ### Levelling Up With one more level to go, we are finally glimpsing the top of the TiqTaqToe mountain. In the final level, [[Fully Quantum TiqTaqToe]], we will *no longer replace observed pieces with symbols*. Instead, we simply let them return to a classical state, as if the player had just played a classical move in that square. Thus, the piece can be interacted with again and again, before *and* after each observation phase! Additionally, players are now allowed to switch basis *on every turn*. Thus, squares that would not be interactable due to the "max two dice per square" rule, can now be entangled, cancelled, or spun by simply switching the basis first! All this basis flipping needs to be kept track of though, and this is where the observation token really comes in handy. So, when you feel ready, take the final step, and graduate to TiqTaqToe Master in **Level 5: [[Fully Quantum TiqTaqToe]]!** [^1]: You may wonder whether there is any difference between the states "half $X$ minus half $Oquot; and "half $O$ minus half $Xquot;. The answer is no: Only the *relative* sign between the symbols are relevant here. This is because you can multiply any quantum state with $-1$ (and in general, any complex phase) without changing its observable properties. Hence, we simply refer to both cases as a "minus" state. [^2]: NB: Entanglement can be easy to miss here! Here's the rule: Any two partner squares (meaning squares with same-coloured dice) that do *not* include a blue (or red) number are entangled; otherwise, the squares are independent. Remember to always check for this before rolling! [^3]: Right: They simply make a superposition move, filling both squares. Since there are no more empty squares, the game then goes straight to the observation phase again, and regardless of where the $O$ piece ends up, they will have a three-in-a-row! [^4]: If $X$ places a classical piece in the centre, then $O$ will likely entangle with it. Then, regardless of where the pieces land, $O$ will have a three-in-a-row, while $X$ only gets a three-in-a-row in the middle. Thus this would give Player 2 a $50\%$ chance of winning, and otherwise the game is a tie. Not ideal for Player 1! If Player 1 instead plays a *superposition* move, the game will go straight to the observation phase. Then, if their piece lands in the middle, Player 1 wins! Otherwise, the game will continue with Player 2 taking the middle and winning. Not an optimal outcome, but definitely better than the alternative! [^5]: The name *skew basis* is inspired by the example of polarization filters, as explained in [[The Physics of TiqTaqToe]]. More common names are "the orthogonal basis", "the Hadamard basis", or even just "the +/- basis".