How does one construct a 27x27 tabula recta for 2-of-3 secret sharing? In the answer to this post at security.stackexchange, someone describes how, using pencil and paper, one can do 2-of-3 secret sharing with a tabula recta.  But I don't understand how he created his table, and although I can see some sort of pattern I wouldn't be able to reconstruct it in a pinch. 

Another question I have is, is this sharing scheme information-theoretically secure?  Does any individual share, taken alone, provide any information about the secret?
We start by coding a secret $k\in\{0,1,2\}$ as three shares $x$, $y$, $z$ by


choosing share $x$ uniformly at random in $\in\{0,1,2\}$
choosing share $y$ as $(x-k)\bmod 3$
choosing share $z$ as $(y-k)\bmod 3$


One share gives no clue about $k$, but we can recover $k$ from any two shares:


from $x$ and $y$ as: $k=(x-y)\bmod 3$
from $y$ and $z$ as: $k=(y-z)\bmod 3$
from $z$ and $x$ as: $k=(z-x)\bmod 3$


Note: per standard notation of modular arithmetic, for $n&gt;0$, the quantity $i\bmod n$ is defined as the $j$ with $0\le j&lt;n$ such that $i-j$ is a multiple of $n$.

The table we need to carry this has $k=(x-y)\bmod 3$ at column $x$ and lines $y$, that is:

    |  0  1  2
----+----------
- 0 |  0  1  2
- 1 |  2  0  1
- 2 |  1  2  0




Now, if the secret consists of two digits each in $\{0,1,2\}$, we can express it as three shares of two  digits each in $\{0,1,2\}$, by using the above method for each digit of the secret, yielding the corresponding digit for each of the three shares.

We can simplify doing that by building the following table, where each digit of the result is obtained by applying the previous table to the corresponding digit of each input:

     |  00  01  02  10  11  12  20  21  22
-----+-------------------------------------
- 00 |  00  01  02  10  11  12  20  21  22
- 01 |  02  00  01  12  10  11  22  20  21
- 02 |  01  02  00  11  12  10  21  22  20
- 10 |  20  21  22  00  01  02  10  11  12
- 11 |  22  20  21  02  00  01  12  10  11
- 12 |  21  22  20  01  02  00  11  12  10
- 20 |  10  11  12  20  21  22  00  01  02
- 21 |  12  10  11  22  20  21  02  00  01
- 22 |  11  12  10  21  22  20  01  02  00


We can then replace two-digit pairs by $3^2=9$ letters A..I (in row and column indexes, and the table itself), and we get:

     |  A   B   C   D   E   F   G   H   I
-----+-------------------------------------
- A  |  A   B   C   D   E   F   G   H   I
- B  |  C   A   B   F   D   E   I   G   H
- C  |  B   C   A   E   F   D   H   I   G
- D  |  G   H   I   A   B   C   D   E   F
- E  |  I   G   H   C   A   B   F   D   E
- F  |  H   I   G   B   C   A   E   F   D
- G  |  D   E   F   G   H   I   A   B   C
- H  |  F   D   E   I   G   H   C   A   B
- I  |  E   F   D   H   I   G   B   C   A


We can do the same thing for secrets consisting of three digits, then replace three-digit triples by $3^3=27$ signs: letters A..Z and . . That's how the table in my answer was constructed.

The mapping of $3^3=27$ three-digit triples to letters is arbitrary. We would obtain a table that works just as well for any mapping. We can additionally reorder lines and columns in a convenient order, e.g. alphabetical. That allows to generate different easy-to-use tables that work just as well.

We can also extend to $3^4=81$ symbols, allowing uppercase and lowercase, digits, and 19 typographical symbols (but if the symbols are ultimately printed, the font must allow recognizing between O and 0, l and 1 ).