In this post, a couple comments suggested that a tabula recta could be used to perform Shamir Secret Sharing. I'm specifically interested in this statement:
you could just have a table of all 100 inverses mod 101
If this simply means a 10x10 table, that's no problem. That's smaller than a regular tabula recta. But even if it's 100x100, I am still interested in how this would work.
Taking the question as literally as possible, I have provided my pencil and paper tables for a scheme among three parties with a polynomial of degree 2 and a field of $\mathbb{Z}_5$ to keep things feasible for a pencil and paper example. Given the tables, in order to reconstruct any secret look up the table to be used by the share from party 1 (table numbers are denoted in a box to the upper left of each table), the row to be used by the share from party 2 and the column to be used by the share for party 3.
I've provided an example polynomial $4x^2+3x+1$ which evaluates correctly to 1, the secret, at $x=0$, 3 when $x=1$, 3 when $x=2$, and finally 1 when $x=3$. Then, to reveal the secret, take these shares and look them up as described. I've boxed in red the correct value from following the described procedure. This set of tables should work for any polynomial valid in the field described and for shares based on players associated with the same values for $x$.
$\mathbb{Z}_5$">
