Why is Blakley secret sharing in finite space?

The paper says it requires a finite space. I wonder what the reason is, since from my understanding it should be perfectly fine to use the "normal" infinite space.

So why does Blakley use a finite space? Is there a possible attack in infinite space?

• Points in a finite space tend to be more relieable and easy to encode for computers, especially if you have to solve a linear system of equations to find your secret (there are rounding errors if you use floats there). – SEJPM Jan 10 at 10:42
• So just from security perspective it would not matter? – Ceriath Jan 10 at 10:51

So why does Blakley use a finite space?

There are multiple reasons:

• It is really inconvenient to use and represent infinite sets with computers, because the representation takes a potencially infinite amount of storage.
• If you use floats or doubles for representation, then you also have to fight the problem of rounding errors, which are especially notorious in linear algebra when you are solving linear systems of equations, potentially preventing you from learning the accurate secret even if you have all the shares.
• There is very few practical downsides to using finite spaces, especially as you can use finite fields of arbitrary size with exact precision rather easily.

However, I think that for theoretical / mathematical purposes the restriction to finite spaces is not needed though I haven't checked whether this fact is used in the security proof for perfect security somewhere.

The perfect security argument refers to hyperplanes being uniformly random. Please note that:

There is no uniform distribution on an infinite space. Each point or hyperspace has probability zero of being chosen.

You'd need an infinite number of random bits to even specify the angle a hyperplane makes with an axis.