3 added 62 characters in body

Here is a general trick that you can use to make yourself look like you have an amazing memory.

Start with a finite abelian group $(G,+)$ in which you are comfortable doing arithmetic. Be sure to know the sum $$g^* = \sum_{g \in G} g.$$

Take a set $S$ of $|G|$ physical objects with an easily computable set isomorphism $$\varphi : S \longrightarrow G.$$ Allow your audience to remove one random element from $a \in S$ and then shuffle $S$ without telling you which one it what $a$ is. [Shuffling means we need $G$ to be abelian.]

Now inform your audience that you are going to look briefly at each remaining element of $S$ and remember exactly which elements you saw, and determine by process of elimination which element of $S$ was removed.

Now glace through all the remaining elements of $S$ one by one and keep a "running total" to compute $$\varphi(a) = g^* - \sum_{s \in S-{a}} \varphi(s).$$

Finally apply $\varphi^{-1}$ and obtain $a.$

Note that $\varphi$ is not "canonical" in the sense there are definitely choices to be made. On the other hand in should be "natural" in the sense that you should be very comfortable saying $s = \varphi(s).$

The prototypical example is to take $G$ to be $\Bbb Z / 13 \Bbb Z \times V_4,$ $S$ to be a standard deck of 52 cards, and $\varphi(s)$ to be $( \text{rank}(s) , \text{suit}(s) )$.

2 deleted 1 characters in body

Here is a general trick that you can used use to make yourself look like you have an amazing memory.

Start with a finite abelian group $(G,+)$ in which you are comfortable doing arithmetic. Be sure to know the sum $$g^* = \sum_{g \in G} g.$$

Take a set $S$ of $|G|$ physical objects with an easily computable set isomorphism $$\varphi : S \longrightarrow G.$$ Allow your audience to remove one random element from $a \in S$ without telling you which one it is.

Now inform your audience that you are going to look briefly at each remaining element of $S$ and remember exactly which elements you saw, and determine by process of elimination which element of $S$ was removed.

Now glace through all the remaining elements of $S$ one by one and keep a "running total" to compute $$\varphi(a) = g^* - \sum_{s \in S-{a}} \varphi(s).$$

Finally apply $\varphi^{-1}$ and obtain $a.$

Note that $\varphi$ is not "canonical" in the sense there are definitely choices to be made. On the other hand in should be "natural" in the sense that you should be very comfortable saying $s = \varphi(s).$

The prototypical example is to take $G$ to be $\Bbb Z / 13 \Bbb Z \times V_4,$ $S$ to be a standard deck of 52 cards, and $\varphi(s)$ to be $( \text{rank}(s) , \text{suit}(s) )$.

Here is a general trick that you can used to make yourself look like you have an amazing memory.

Start with a finite abelian group $(G,+)$ in which you are comfortable doing arithmetic. Be sure to know the sum $$g^* = \sum_{g \in G} g.$$

Take a set $S$ of $|G|$ physical objects with an easily computable set isomorphism $$\varphi : S \longrightarrow G.$$ Allow your audience to remove one random element from $a \in S$ without telling you which one it is.

Now inform your audience that you are going to look briefly at each remaining element of $S$ and remember exactly which elements you saw, and determine by process of elimination which element of $S$ was removed.

Now glace through all the remaining elements of $S$ one by one and keep a "running total" to compute $$\varphi(a) = g^* - \sum_{s \in S-{a}} \varphi(s).$$

Finally apply $\varphi^{-1}$ and obtain $a.$

Note that $\varphi$ is not "canonical" in the sense there are definitely choices to be made. On the other hand in should be "natural" in the sense that you should be very comfortable saying $s = \varphi(s).$

The prototypical example is to take $G$ to be $\Bbb Z / 13 \Bbb Z \times V_4,$ $S$ to be a standard deck of 52 cards, and $\varphi(s)$ to be $( \text{rank}(s) , \text{suit}(s) )$.