make a line. the y-intercept is the secret. straight lines need two points to derive. a simple curve needs three. etc.
the number of points you need is the n in the n/m shares.
if you have 1 n for a straight line, you can’t derive it without 1 more point. if you have 2 n for a curve, you can’t derive it without 1 more point.
from gfshare man page:
What Shamir’s method relies on is the creation of a random polynomial, the sampling of various coordinates along the curve of the polynomial and then the interpolation of those points in order to re-calculate the y-intercept of the polynomial in order to reconstruct the secret. Consider the formula for a straight line: Y = Mx + C. This formula (given values for M and C) uniquely defines a line in two dimensions and such a formula is a polynomial of order 1. Any line in two dimensions can also be uniquely defined by stating any two points along the line. The number of points required to uniquely define a polynomial is thus one higher than the order of the polynomial. So a line needs two points where a quadratic curve needs three, a cubic curve four, etc.
When we create a N-of-M share, we encode the secret as the y-intercept of a polynomial of order N-1 since such a polynomial needs N points to uniquely define it. Let us consider the situation where N is 2: We need a polynomial of order 1 (a straight line). Let us also consider the secret to be 9, giving the formula for our polynomial as: Y = Ax + 9. We now pick a random coefficient for the graph, we’ll use 3 in this example. This yields the final polynomial: Sx = 3x + 9. Thus the share of the secret at point x is easily calculated. We want some number of shares to give out to our secret-keepers; let’s choose three as this number. We now need to select three points on the graph for our shares. For simplicity’s sake, let us choose 1, 2 and 3. This makes our shares have the values 12, 15 and 18. No single share gives away any information whatsoever about the value of the coefficient A and thus no single share can be used to reconstruct the secret.
Now, consider the shares as coordinates (1, 12) (2, 15) and (3, 18) - again, no single share is of any use, but any two of the shares uniquely define a line in two-dimensional space. Let us consider the use of the second and third shares. They give us the pair of simulaneous equations: 15 = 2M + S and 18 = 3M + S. We can trivially solve these equations for A and S and thus recover our secret of 9.