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comment A rudimentary challenge–response authentication based on dot products in $\mathrm{GF}(2)^k$
Hint: $p\cdot c_1\oplus p\cdot c_2 = p\cdot (c_1\oplus c_2)$ and so the eavesdropper knows the answer to the challenge $c_1\oplus c_2$ without needing to know $p$.
Sep
22
awarded  Yearling
Jun
19
answered LFSR and Markov chain question
May
21
revised Shamir Secret sharing - Can share generator keep x values secret?
rolled back to a previous revision
May
21
reviewed Reject suggested edit on Shamir Secret sharing - Can share generator keep x values secret?
May
21
reviewed Reject suggested edit on Shamir Secret sharing - Can share generator keep x values secret?
May
21
awarded  Cleanup
May
21
awarded  Custodian
May
21
revised Shamir Secret sharing - Can share generator keep x values secret?
rolled back to a previous revision
May
21
reviewed Reject suggested edit on Shamir Secret sharing - Can share generator keep x values secret?
May
21
answered Shamir Secret sharing - Can share generator keep x values secret?
May
17
answered Berlekamp-Massey to construct minimal LFSR
May
17
comment Berlekamp-Massey to construct minimal LFSR
Perhaps the explanation in this answer might help, even though it uses slightly different notation.
Mar
7
revised How to generate a random integer in interval $[1, 2^n-1]$ from random integer in interval $[0, 2^n-1]$?
added 1 characters in body
Feb
16
answered Do deterministic secret sharing schemes exist?
Nov
26
comment Upper bound Linear Feedback Shift Register
What does a linear feedback shift register with $R$ registers mean? In general, the period of a polynomial $g(x)$ is the least integer $e$ such that $g(x)$ divides $x^e -1$. If $g(x)$ is the feedback polynomial of an LFSR, then $e$ is the upper bound you seek. But determining $e$ for a given $g(x)$ is by no means a trivial task.
Nov
11
awarded  Necromancer
Oct
1
comment RS Erasure Coding and Shamir's Secret Sharing
Yes, with $3$-out-of-$4$ secret sharing, you cannot recover the secret from $4$ shares if one share is corrupt but you don't know which one. (If you do know which share is corrupt, just ignore it and recover using the known $3$ good ones). With $2$-out-of-$4$ sharing, you can recover the secret if all $4$ shares are present as long as (i) at most one share is bad (but you don't know which one) or (ii) one, or even two shares, are known to be bad and you know which ones. As before, ignore the known bad shares and reconstruct from the known good ones (at least two are available).
Oct
1
comment RS Erasure Coding and Shamir's Secret Sharing
@mikeazo How many uncorrupted shares do you need for reconstruction? If $2$ good shares are required to reconstruct, then you use a $[4,2]$ Reed-Solomon code. It can reconstruct the secret if any two good shares are available. It can reconstruct if two good shares are available and a third "My flash drive got laundered" share is present (just ignore the clean guy's share!). And it can reconstruct the secret if all $4$ parties are present and (possibly) one of the shares is bad, but it is not known which one is bad. (If you know which share is bad, just ignore it and use 2 good shares).
Sep
29
comment Why does Shamir's Secret Sharing Scheme need a finite field?
OK, I edited my answer to remove the references to the Fundamental Theorem of Algebra. But isn't it true that if a degree-$n$ polynomial in $\mathbb F_q[x]$ does not have $n$ roots in the finite field $\mathbb F_q$, then all its $n$ roots lie in a finite extension field $\mathbb F_{q^m}$ where $m$ is some finite integer? Or are you saying that the roots lie in no finite extension of $F_q$, and we must always go to nonfinite extensions?