When you play quantum tic-tac-toe using dice (see [[TiqTaqToe with d4 dice]]), it is possible to include an extra move that allows the player to *cancel* part of the opponents state, and ensure that a square instead resolves to empty. In this variant of the game, you are now given the ability to cancel the opponent's pieces, and thus "empty" the square in which the opponent had placed their die! This leads to a longer game, since you will end up with more empty squares after observing the board, and opens for new and exciting strategies to outsmarting your opponent. In the following, let the blue player be $X$, and the orange player be $O$. Recall that the number on the die represents the probability (amplitude-squared) that a piece is observed in a given square: If the number on the die is $a$, then the probability of observing it in the given square is $a/4$. We expand the dice set with one extra die per colour, with the numbers painted the colour of the opponent: ![[IMG_9103.jpeg|400]] Now, the coloured-in dice represent a piece of the opponent with a *phase* of $-1$: If the blue dice normally represent $X$, a blue die with red numbers represent $-O$, and if the orange dice normally represent $O$, an orange die with green numbers represent $-X$. The rules of the game are exactly like before, except with the addition of two moves. Neither move adds a piece to the board, but rather *alters the probabilities of a piece that the opponent already placed*. ##### Alternative Implementation If you have a TiqTaqToe dice set without the fifth, coloured-in dice per colour, you may still implement the new moves by letting a die placed like a right-side-up triangle $\triangle$ represent the usual die placements (e.g. $X$), and a die placed like an up-side-down triangle $\nabla$ represent a cancelling die (e.g. $-O$). Just remember that at most one die per colour may be cancelling! Player $1$ performing a cancelling move to free the middle square will then look like this: ![[IMG_8300.jpg|400]] ### 1. New Moves #### 1.1 The Cancelling Move The cancelling move functions superficially just like a normal entangling move, except that one of the two dice placed is cancelling. At most one of the placed dice may be cancelling, but you may choose which one—that's the square that will end up empty! Let's say player $1$ starts by performing a classical move on the middle square. ![[IMG_9104.jpeg|400]] then $X$ can cancel by playing an entangling move with one half $X$ and one half $-O$: ![[IMG_9106.jpeg|400]] Observing this board would yield an empty middle square, and either an $X$ or an $O$ with equal probability in the square to the right: ![[IMG_9107.jpeg||400]] Note how there are an equal number of pieces on the board in the end, regardless of whether the rightmost square ends up resolving to an $X$ or an $O$: We haven't *added* a piece to the board, only *altered* the state of the one that was already there! #### 1.2 The Half-cancelling move If player $1$ started by performing a superposition move ![[IMG_9113.jpeg|400]] then player $2$ can as usual entangle with one part of the superposition, except they are now free to let one of the two dice be cancelling. Let's say player $2$ wants to free up the middle square. ![[IMG_9110.jpeg|400]] Then with overwhelming probability, we get the following situation: ![[IMG_9112.jpeg|400]] Remember that the half-cancelling move, like the cancelling move, does not place an extra piece on the board, but rather alters the state of an already existing one. This gives us three possible, mutually exclusive outcomes: an $X$ in the leftmost square, an $X$ in the rightmost square, or an $O$ in the rightmost square. Furthermore, we see that the probability that we get an $X$ in the rightmost square should be half that of each of the other outcomes. The result is that the dice resolve to an $X$ in the leftmost square with probability $2/5$, an $O$ in the rightmost square with probability $2/5$, and an $X$ in the rightmost square with probability $1/5$. But let us rewind a bit: I said the middle square ends up empty with *overwhelming* probability—but not an exact one! Looking at the board, one might imagine that since $1/4 - 2/4 = -1/4$, we should get something like a quarter probability that the piece in question ends up resolving to an $X$ in the middle square would be something like a quarter. In quantum mechanics, however, we work with the *square roots* of the probabilities, rather than the probabilities themselves, and performing the calculation shows that there is only a $\approx 3.3\%$ chance of the piece ending up in the cancelled square! ##### Rolling the Half-cancelling Move ###### Using a d100 die How do you roll for these probabilities using dice? If you have access to a d100 die, which is usually implemented by rolling a d10 die twice, once for the tenths place and once for the ones place (where a double $0$ outcome is interpreted as 100), you can decide the outcome in one go by rolling and comparing the result to the following table (where the probability of failure has been rounded down to $3\%$): |Roll |Outcome | |:---:|:--------------------------------------:| |1-19 |Opponent's piece in entangled square | |21-59|Opponent's piece in non-entangled square| |61-99|Your piece in entangled square | |20, 60, 100|Opponent's piece in *cancelled* square| ###### Using the d4 dice If you do not have access to a d100 die, I recommend starting by checking whether cancellation failed. To do this, you may implement a d32 die using three d4 dice in the following way (the probability of failure is then instead rounded down to $1/32 \approx 3.1\%$): Pick up two of the opponent's dice of the same colour, and one of your own, for example the cancelling die. If both the opponent's dice land on a $4$ *and* your own die lands on a $1$ or a $2$, then cancelling failed, and the opponent's piece ends up in the square you attempted to cancel. In case it helps remember the rule for rolling, you may imagine that the two dice of the opponent stood together to defeat yours against overwhelming odds. Assuming cancelling didn't fail, you next have to figure out which of the three outcomes the piece will resolve to. Recall that each outcome then has probability either $2/5$ or $1/5$. If you had a d5 die, resolving would be as easy as (for example) mapping the first outcome (opponent's piece in entangled square) to the number $1$, the second outcome (opponent's piece in non-entangled square) to the numbers $2$ and $3$, and the last outcome (your piece in entangled square) to the numbers $4$ and $5$—and then roll it! If you have any d6 dice lying around, the easiest way to implement of a d5 die is simply to roll a d6 die and reroll whenever the outcome is $6$. Otherwise, we can implement a d8 die using two d4 dice in the following way: Roll two different d4s, designating one as control and the other as outcome. If the number on the first one is $1$ or $2$, the second die gives the outcome. If the number on the first die is $3$ or $4$, add $4$ to the outcome of the other. Then to roll a d5, simply reroll whenever the result is $6$ or higher! If *both* halves are cancelled, using two cancelling moves, the probability that each cancelling failed is still 3%, and so the above rule can be used to check for each. If cancelling succeeded, the remaining outcomes have probabilities 1/6 and 2/6 respectively, and so can be rolled using a d6 as above. ### 3. How Does This Connect to Nature? Phases are what sets quantum mechanics apart from just normal probability theory. One feature of phases is that they may *cancel* each other. This is the same phenomenon which makes water be still after a wave peak collides with an equally-sized wave bottom: they *cancel* each other. Noise-cancelling headphones operate via the same principle, only on *sound* waves rather than water waves! ![[Pasted image 20241104020002.png|400]] ### 4. Calculating the Probabilities #### 4.1 The Cancelling Move Say player $1$ performed a classical move and placed a $4$-die in square one. You, player $2$, want to cancel it, and so play a cancelling move on squares $1$ and $2$, with the cancelling die in square one. ![[IMG_9106.jpeg|400]] Let the state be described by $\ket{ab} = \ket{a} \ket{b}$, where $a \in \{E,X,O\}$ describes the state of square one and $b \in \{E,X,O\}$ describes the state of square two. Then the move has the following effect: $\ket{ab} = \ket{XE} \longrightarrow \frac{1}{\sqrt{2}} (\frac{\ket{X} - \ket{X}}{\sqrt{2}}\ket{E} + \ket{E}\frac{\ket{X} + \ket{O}}{\sqrt{2}})= \frac{\ket{EX} + \ket{EO}}{2}\, .$ If we calculate the probability of each outcome (remembering that probabilities are obtained by squaring the amplitudes), we get that each has probability $(1/2)^2 = 1/4$. Since we only have two outcomes, this means that the total probability is only $50\%$. We therefore have to *renormalize* the state, i.e. multiply all amplitudes by the same constant chosen so that the sum of the probabilities becomes $100\%$ again. In this case, this renormalization constant is simply $r = 2$. The state we end up with is then: $\ket{ab} = \frac{\ket{EX} + \ket{EO}}{2} \longrightarrow \frac{\ket{EX} + \ket{EO}}{\sqrt{2}}\, ,$ and we see that we have $(1/\sqrt{2})^2 = 1/2 = 50\%$ probability of each outcome, as expected. #### 4.2 The Half-cancelling Move This time, say player $1$ performed a superposition move, and placed a $2$-die on square one and a $2$-die on square two. ![[IMG_9113.jpeg|400]] Let $\ket{abc} = \ket{a}\ket{b}\ket{c}$ describe the state of square one, two and three. Then we have the following starting state: $\ket{abc} = \frac{1}{\sqrt{2}}(\ket{XEE} + \ket{EXE})\, .$ You want to cancel square two, and so you play a half-cancelling move by entangling in square two and three, but leaving a cancelling die in square two. ![[IMG_9110.jpeg|400]] Reading off the board, we get that the move then has the following effect: $\ket{abc} \longrightarrow \frac{1}{\sqrt{2}} \ket{XEE} + \frac{1}{2} \ket{EXE} - \frac{1}{\sqrt{2}} \ket{EXE} + \frac{1}{2} \ket{EEX} + \frac{1}{\sqrt{2}} \ket{EEO}\, ,$ where each term on the right corresponds to each die in the picture when read from left to right. Using that $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$, we can contract the matching terms to get: $\ket{abc} = \frac{1}{\sqrt{2}} \ket{XEE} + \frac{1-\sqrt{2}}{2} \ket{EXE} + \frac{1}{2} \ket{EEX} + \frac{1}{\sqrt{2}} \ket{EEO}\, .$ Evidently, this state is in need of renormalization, as seen by the first and last term already summing up to $1$ after squaring. The sum of probabilities is currently: $(\frac{1}{\sqrt{2}})^2 + (\frac{1-\sqrt{2}}{2})^2 + (\frac{1}{2})^2 + (\frac{1}{\sqrt{2}})^2 = \frac{1}{2} + \frac{(1-\sqrt{2})^2}{4} + \frac{1}{4} + \frac{1}{2}$ $= \frac{1}{4}(5 + (1 - \sqrt{2})^2) = \frac{1}{4}(5 + 1 - 2\sqrt{2} + 2) = \frac{1}{4}(8 - 2\sqrt{2}) = \frac{4 - \sqrt{2}}{2} \approx 1.29\, .$ In order to renormalize the state $\ket{abc}$, we then have to divide each of the amplitudes by the square root of this number, i.e. multiply the whole state with the renormalization constant $r = \sqrt{\frac{2}{4-\sqrt{2}}} \approx 0.88\, .$ After multiplying, the final state is then: $\ket{abc} = \sqrt{\frac{2}{4-\sqrt{2}}}\frac{1}{\sqrt{2}} \ket{XEE} + \sqrt{\frac{2}{4-\sqrt{2}}}\frac{1-\sqrt{2}}{2} \ket{EXE} + \sqrt{\frac{2}{4-\sqrt{2}}}\frac{1}{2} \ket{EEX} + \sqrt{\frac{2}{4-\sqrt{2}}}\frac{1}{\sqrt{2}} \ket{EEO}\, ,$ which we can simplify to $\ket{abc} = \frac{1}{\sqrt{4-\sqrt{2}}} \ket{XEE} + \frac{\sqrt{2} - 2}{2\sqrt{4 - \sqrt{2}}} \ket{EXE} + \frac{1}{\sqrt{8-\sqrt{8}}} \ket{EEX} + \sqrt{\frac{1}{4-\sqrt{2}}} \ket{EEO}\, .$ From this, the probabilities can now easily be obtained by just reading off the state and squaring the amplitudes. |Outcome |Amplitude squared|Probability| |:---:|:--------------------------:|:-----------:| |$\ket{XEE}, \ket{EEO}$|$(\frac{1}{\sqrt{4-\sqrt{2}}})^2 = \frac{1}{4-\sqrt{2}} \approx 0.387$|$\approx 39\%$| |$\ket{EEX}$|$(\frac{1}{\sqrt{8-\sqrt{8}}})^2 = \frac{1}{8-\sqrt{8}} \approx 0.193$|$\approx 19\%$| |$\ket{EXE}$|$(\frac{\sqrt{2} - 2}{2\sqrt{4 - \sqrt{2}}})^2 = \frac{3 - \sqrt{8}}{8 - \sqrt{8}} \approx 0.033$|$\approx 3\%$ Note how the probabilities still add up to $100\%$, even after rounding! In the case where both sides are cancelled, the calculation follows exactly as above, except now the renormalization constant is $r = (\frac{3}{2} + \frac{(1-\sqrt{2})^2}{2})^{-1/2} = \frac{1}{\sqrt{3-\sqrt{2}}}$. Assuming $O$ cancels the third and fourth squares, the resulting probabilities are: |Outcome |Amplitude squared|Probability| |:---:|:--------------------------:|:-----------:| |$\ket{OEEE}, \ket{EOEE}$|$\frac{1}{2(3 - \sqrt{2})} \approx 0.315$|$\approx 32\%$ |$\ket{XEEE}, \ket{EXEE}$|$\frac{1}{4(3 - \sqrt{2})} \approx 0.158$|$\approx 16\%$| |$\ket{EEXE}, \ket{EEEX}$|$\frac{3-2\sqrt{2}}{4(3 - \sqrt{2})} \approx 0.027$|$\approx 3\%$| Thankfully, we still get that the cancellations fail with roughly 3% probability, so we can use the same rule as above for rolling them. Assuming cancelling succeeded, after renormalization the remainding probabilities are one and two sixths respectively, and so can be rolled using a d6 as above. #### Phase moves - *Cancelling* - Entangling with a classical state while cancelling out one of the squares. - *Half-cancelling* - Entangling with half of a superposition while cancelling one of the squares, albeit imperfectly. - Board renormalizes. - *Spreading* - Cancelling half of the opponent's superposition state. - After cancelling, the result is that half the opponent's state is spread to two fourths. - *Cancelling* - Entangling with a classical state while cancelling in *both* squares.e