Security of Shamir secret sharing with regards to the threshold Consider Shamir Secret Sharing Scheme (SSSS). This is completely unfounded, but my intuition tells me that the threshold k we set (AKA the number of shares required to construct the original secret) with regards to the total number of shares distributed (n)  has a profound relation with the security of the whole system. 

There are two scenarios I'm imagining where the threshold that would affect the security.


  1) When k = n.
  
  2) When k is dramatically lower than n.


Is any of this true?
Absolutely no information about the secret is gained by having fewer than $k$ shares.  This is true regardless of what $k$ is, and for an adversary with unlimited computational power.  The secret is information theoretically hidden.

In at nutshell the reason for this is that, for every possible value of the secret, there exists a polynomial that interpolates it and the all the shares obtained.

Each share is a point on a degree $k-1$ polynomial $f$.  The secret is also a point on the polynomial, typically it is taken to be the point $(0,S)$ (i.e. $f(0)=S$, so the constant term of the polynomial is the secret $S$).

SSS uses the following fact: for every $k$ points, there exists exactly one polynomial of degree $k-1$ that goes through all these points. This can be shown using Legrange interpolation.

Let's say that you knew $k-1$ shares and were trying to figure out if $S'$ was the secret.  You think to your self, well if $S'$ is the secret, then I must have that $f(0)=S'$.  Then, I would have a total of $k$ points on $f$, namely $k-1$ from the shares I know of and I'm assuming you have the point $(0,S')$.  Such an $f$ always exists and could very well have been the polynomial generated by SSS, and so I cannot rule out $S'$ as the secret.

This is true for every value of $S'$, and so I gain no information about $S$.

Edit: a tangent about the case of $k=n$.  You can actually do something much simpler than SSS.  Randomly sample $n-1$ bit strings $s_1,\cdots,s_{n-1}$.  Then let $s_n=S\oplus s_1\oplus\cdots\oplus s_{n-1}$.  Note that the $s_i$ satisfy $S=s_1\oplus\cdots\oplus s_n$.  If any one share is unknown (WLOG, say $s_n$ is unknown), then $s_n$ is functionally a OTP that protects $S$.