New TGFN episode is out, in which nostr:nprofile1qythwumn8ghj7un9d3shjtnswf5k6ctv9ehx2ap0qyvhwumn8ghj7un9d3shjtnndehhyapwwdhkx6tpdshsz9thwden5te0wfjkccte9ejxzmt4wvhxjme0qyt8wumn8ghj7ur4wfcxcetjv4kxz7fwvdhk6tcpzdmhxue69uhk7enxvd5xz6tw9ec82c30qqsyy2wzruqsr27rhfzjx0shd6t4l20xwxa33fnj900hwf46y4z9l7ge022gg and nostr:nprofile1qythwumn8ghj7un9d3shjtnwdaehgu3wvfskuep0qyvhwumn8ghj7un9d3shjtnndehhyapwwdhkx6tpdshsz8thwden5te0dehhxarj9ekh2arfdeuhwctvd3jhgtnrdakj7qg7waehxw309ahx7um5wgkhqatz9emk2mrvdaexgetj9ehx2ap0qyghwumn8ghj7mn0wd68ytnhd9hx2tcqyzqh9wfq2framl5ek7pnypuz6qcjlgc95xvlk2lg50n9v03qkncwy68zk7f tell us about FROSTR — stick around past the end credits for a bonus segment, in which they trash nostr:nprofile1qyd8wumn8ghj7mr0vd4kymmc9enxjct5dfskvtnrdakj7qg6waehxw309ac8junpd45kgtnxd9shg6npvchxxmmd9uq3kamnwvaz7tmjv4kxz7fwwajhxar9wfhxyarr9e3k7mf0qqsrhuxx8l9ex335q7he0f09aej04zpazpl0ne2cgukyawd24mayt8gu390qy's promenade project.
Discussion
Here i was, thinking it was an editting mistake
Forgot the link: https://fountain.fm/episode/gWYJ5PgxwzdnBjuNuzyc
I tried to zap here, as I'm having issues with fountain lately. I was told... Recipient lack receiving capacity?
That rings true, I'll look into it
It's a bit of a shame that no one in the pod explained exactly how the 2-of-3 shamir secret sharing works, since it's so simple!
So, here is my contribution:
First, let's call S our secret. S is a number or can be represented as such.
We choose a random line that pass through the point (0,S). The lines with this property are the following:
y = a * x + S (where a is a coefficient)
So we choose a at random, and we have generated a line that pass through our secret S. If you know the line, you know the secret by simply substituting x = 0 into the equation.
Now, we want to split this knowledge into 3 pieces. The way to do it is to choose 3 more points on the line, different from (0,S). For example we choose:
P1 = (1, a+S)
P2 = (2, 2a+S)
P3 = (3, 3a+S)
(check that these points are on the line)
Now, how many lines go through two distinct points?
That's easy, only one! So if we know any 2 of these 3 points, we can reconstruct S.
Let's imagine we know P1 and P2.
Then, we can find S this way
2*P1 - P2 = 2*(1, a+S) - (2, 2a+S) = (2, 2a+2S) - (2, 2a+S)
= (2-2, 2a+2S-2a-S) = (0,S)
ta daaaa.
We have our secret back!
This is how a 2-of-3 SSS works. To increase the second number, you simply generate more points (P4, P5, ...). To increase the first number, you need to increase the degree of the polynomial.
wtf is a polynomial? The line we used:
y = a * x + S
is a polynomial, of the smallest degree, which is 1. The degree of a polynomial is the highest exponent of x.
This is a parabola for our secret.
y = a1 * x^2 + a2 * x + S
Note that there are two random coefficient now. With this parabola we can create 3-of-whatever SSS schemes for our secret S. This means we'll need at least 3 points to reconstruct it.
The number of points you need to reconstruct the secret is always the degree of the polynomial + 1. But why? This is left as an exercise for the reader.
