# Tag Info

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When you are trying to do Attribute Based Encryption schemes, the collection of points has a nice property such that if you scale the x-axis by a scalar, the polynomial still passes through the same y-intercept. So, you can use this scheme to represent a hash for point combinations; and give different users an incompatible set of points (to help in ...

3

This is about encoding a quantity. It can represent anything you want. This encoding would be part of a publicly known protocol. If you are in $\mathbb{F}_p$ the quantity $s=a_0$ can represent one out of any $p$ quantities. Standard ways of encoding text include the ASCII code (look it up) for which $p\geq 256$ is sufficient. If you use $\mathbb{F}_p$ you ...

0

A secret sharing scheme is not a classical encryption scheme. So I don't think one can cast it in any meaningful way into that framework. As for your other questions if you are operating in the finite field with $p$ elements where $p$ is a prime all computations are modulo $p$. That $\equiv_p$ notation is a horrible notation but I presume it stands for ...

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Well, one can assign shares as $v_i=f(x_i)$ or $v_i=f(i)$ as long as the $x_i$ are distinct it will work. The authors chose to use $v_i=f(i)$. The observation that Shamir secret sharing is linear follows directly by using the definition of matrix multiplication. There is a typo in the paper though, the matrix entry quoted should be $h_{i,j}=j^{i-1}$ and they ...

1

Clarification The way I understood your question was: Participants will collaborate in sets $(P_1, P_2, \ldots)$ of $t+1$ participants each, and reconstruct the secret. They will keep doing this, until every participant has learned the secret (at least once) The question then is to find bounds for the number of required distinct sets $P_i$. In words: "...

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Although I am not sure for this and this is not a complete asnwer and I hope if anybody sees what I am writing he/she could verify it, If the players are $n$, the $(k,n)$ secret sharing scheme means that I will split $s$ in $n$ parts, such that the polynomial $f$ is a polynomial function of degree $t$ that takes as input $s$ and the $t$ and with the process $... 2 Short answer: No A permutation polynomial is a polynomial$f:\mathbb{Z}_p\rightarrow\mathbb{Z}_p$which is a bijection meaning the list$[f(x): x \in \mathbb{Z}_p]$is a permutation of the elements of the field. Example: For example$f(x)=x^3$gives the list$[0,1,3,2,4]$as$x$ranges over$\mathbb{Z}_5$. But these polynomials leak information since if you ... 1 Is this scheme some well known scheme? It appears to be the well known$(n,n)$secret sharing scheme, using a group operation (note: you said a finite field; however since you never use the multiplication operation, it works just as well over any finite group [1]). That is:$n-1$of the secrets are random group elements$r_i$The last group element is$r_{...

2

Is the $t$ out of $n$, namely $(t,n)$, threshold in the secret sharing scheme related to the entropy of the random variable that is shared according to the scheme? No. You can use a $(t,n)$ sharing scheme (for any $t>0$) to share a value that has one bit of entropy - e.g. everyone knows that it is either a 0 or a 1. And, even in that case, with $t-1$ ...

0

Have you looked at Shamir secret sharing? For your case, it seems like all $K$ players are required to reconstruct $y$. I think this is true because if a single player $j$ decides not to share their value $a_jx_j$, then the players would add up their values and get: $$\sum_{i\neq j,i=1}^K a_ix_i = y - a_jx_j$$ Since $a_jx_j$ is (hopefully) uniformly random, ...

2

Provided that Shamir's secret sharing is implemented correctly, the attacker gains no advantage from knowing up to $k-1$ shares, and the size of the field does not matter. Yes, the attacker can carry out a brute force guessing attack to (possibly, given enough time) find the password. But they can do that even if the password isn't shared, and knowing up to ...

2

Let the Shamir's Secret Sharing (SSS) is constructed from the finite field $K = \mathbb F_{p^m}$, i.e. $K$ is a finite field extension with $p^m$ elements, $order(K) = p^m$. When an attacker accesses the $k-1$ of the $k$ shares of SSS, the remaining all values from the $K$ have the same probability to be a candidate of the last share. This is due to the ...

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Assume that $p$ and $q$ are different primes (if $q$ divides $p$, for instance, the problem is trivial). Modulus conversion is usually not a simple task in the context of secret-sharing. The most common use-case for this type of primitives is, for example, taking a bit $b\in\{0,1\}$ that is secret-shared over a large prime $p$ as $b = a+b\bmod p$, and ...

3

Let $E$ be a secure elliptic curve, and $G$ is the basepoint with order $\ell$. $(n,n)$ scheme: $n$ key holders each randomly pick a private key $x_i$, compute a public key $X_i = [x_i]G$ and a hash commitment $H(X_i)$. The key holders first exchange hash commitments, and only when all hash commitments have been shared do they then share their public keys ...

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