# How is Blakley's Secret Sharing Scheme perfect?

Currently, I am trying to understand Blakley's secret sharing scheme. However, there seem to be multiple descriptions of it, which are very different from another, but everyone states that the scheme would be perfectly secure.

From my understanding it works like this:

• a point $$S$$ in $$t$$-dimensional space is created with one coordinate being secret and the others randomly selected
• $$n$$ hyperplanes through the point $$S$$ are created and distributed
• to reconstruct the secret, $$t$$ hyperplanes are required, which will intersect at the point $$S$$.

However, from my understanding, if someone would possess one or more hyperplanes, they would know that $$S$$ lies somewhere on that hyperplane (or the intersection of multiple hyperplanes)

So my question is;

• how is Blakley's scheme considered perfectly secure when an insider knows that the secret lies on his plane?
• If I give you a line, can you find my secret point on it? Jan 8 '19 at 12:31
• since a line is infinite, i can't. But from my understanding perfect means, that no information at all is revealed about the secret, while in this case it would reveal that it is somewhere on that line Jan 8 '19 at 12:38
• OK, better formulation: If I give you a line, can you find the x-coordinate of my secret point on it?
– SEJPM
Jan 8 '19 at 12:41
• Defn: A secret sharing scheme is called perfect if the probability of deception has the same value for all illegal constellations of participants. and see page 21 Jan 8 '19 at 12:44

Since the secret is only one coordinate (e.g. $$x$$) the information that the secret is on a specific infinite line is as much information as having none at all, since guessing $$x$$ of a point on $$y = 5*x$$ is worth as much as guessing $$x$$ of a point on $$y = z * x$$
• And you can use the link I provided to make it more formal. as $r \leq (t-1)$ Jan 8 '19 at 12:52