Project: SOPS (Provisioning / Security & Compliance)
File: README.md

Forked from We want to split a secret into parts. Any two points on the cartesian plane define a line. Three points define a parabola. Four points define a cubic curve, and so on. In general, `n` points define an function of degree `(n - 1)`. If our secret was somehow an function of degree `(n - 1)`, we could just compute `n` different points of that function and give `n` different people one point each. In order to recover the secret, then we'd need all the `n` points. If we wanted to, we could compute more than n points, but even then still only `n` points out of our whole set of computed points would be required to recover the function. A concrete example: our secret is the function `y = 2x + 1`. This function is of degree 1, so we need at least 2 points to define it. For example, let's set `x = 1`, `x = 2` and `x = 3`. From this follows that `y = 3`, `y = 5` and `y = 7`, respectively. Now, with the information that our secret is of degree 1, we can use any 2 of the 3 points we computed to recover our original function. For example, let's use the points `x = 1; y = 3` and `x = 2; y = 5`. We know that first degree functions are lines, defined by their slope and their intersection point with the y axis. We can easily compute the slope given our two points: it's the change in `y` divided by the change in `x`: `(5 - 3)/(2 - 1) = 2`. Now, knowing the slope we can compute the intersection point with the `y` axis by "working our way back". We know that at `x = 1`, `y` equals `3`, so naturally because the slope is `2`, at `x = 0`, `y` must be `1`. The method we've used for this isn't very general: it only works for polynomials of degree 1. Lagrange interpolation is a more general way that lets us obtain the function of degree `(n - 1)` that passes through `n` arbitrary points. Understanding how to perform Lagrange interpolation isn't really necessary to understand Shamir's Secret Sharing: it's enough to know that there's only one function of degree `(n - 1)` that passes through `n` given points and that computing this function given the points is computationally efficient. But for those interested, here's an explanation: Let's say our points are `(x0, y0),...,(xj, yj),...,(x(n-1), y(n-1))`. Then, the Lagrange polynomial `L(x)`, the polynomial we're looking for, is defined as follows: `L(x) = sum from j=0 to j=(n-1) of {yj * lj(x)}` and `lj(x) = product from m=0 to m=(n-1) except when m=j of {(x - xm)/(xj - xm)}` A concrete example, with 3 points: ``` x0 = 1 y0 = 1 x1 = 2 y1 = 4 x2 = 3 y2 = 9 ``` Let's apply the formula: ``` L(x) = y0 * l0(x) + y1 * l1(x) + y2 * l2(x) ``` Substitute `y_j` for the actual value: ``` L(x) = 1 * l_0(x) + 4 * l_1(x) + 9 * l_2(x) ``` Replace `l_j(x)`: ``` l_0(x) = (x - 2)/(1 - 2) * (x - 3)/(1 - 3) = 0.5x^2 -

+ 3 l_1(x) = (x - 1)/(2 - 1) * (x - 3)/(2 - 3) = - x^2 + 4x - 3 l_2(x) = (x - 1)/(3 - 1) * (x - 2)/(3 - 2) = 0.5x^2 - 1.5x + 1 ``` ``` L(x) = 1 * ( 0.5x^2 - 2.5x + 3) + 4 * ( -x^2 + 4x - 3) + 9 * ( 0.5x^2 - 1.5x + 1) ``` ``` L(x) = ( 0.5x^2 - 2.5x + 3) + ( -4x^2 + 16x - 12) + ( 4.5x^2 - 13.5x + 9) = x^2 + 0x + 0 = x^2 ``` So the polynomial we were looking for is `y = x^2`. So we have the ability of splitting a function into parts, but in the context of computing we generally want to split a number, not a function. For this, let's define a function of degree `threshold`. `threshold` is the amount of parts we want to require in order to recover the secret. Let's set the parameter of degree zero to our secret `S` and make the rest of the parameters random: `y = ax^(threshold) + bx^(threshold-1) + ... + zx^1 + S` With `a, b, ...` random. Then, we want to generate our parts. For this, we evaluate our function at as many points as we want parts. For example, say our secret is 123, we want 5 parts and a threshold of 2. Because the threshold is 2, we're going to need a polynomial of degree 2: `y = ax^2 + bx + 123` We randomly set `a = 7` and `b = 1`: `y = 7x^2 + x + 123` Because we want 5 parts, we need to compute 5 points: ``` x = 0 -> y = 123 # woops! This is the secret itself. Let's not use that one. x = 1 -> y = 131 x = 2 -> y = 153 x = 3 -> y = 189 x = 4 -> y = 239 x = 5 -> y = 303 ``` And that's it. Each of the computed points is one part of the secret. Now that we have our parts, we have to define a way to recover them. Using the example from the previous section, we only need any two points out of the five we created to recover the secret, because we set the threshold to two. So with any two of the five points we created, we can recover the original polynomial, and because the secret is the free term in the polynomial, we can recover the secret. In the previous examples we've only used integers, and this unfortunately has a flaw. First of all, it's impossible to uniformly sample integers to get random coefficients for our generated polynomial. Additionally, if we don't operate in a finite field, information about the secret is leaked for every part someone recovers. For these reasons, Vault's implementation of Shamir's Secret Sharing uses finite field arithmetic, specifically in GF(2^8), with 229 as the generator. GF(2^8) has 256 elements, so using this we can only split one byte at a time. This is not a problem, though, as we can just split each byte in our secret independently. This implementation uses tables to speed up the execution of finite field arithmetic.