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Taking into account this paper I will write here a definition that the authors provide.

$\textbf{Definition:}$ (linear secret sharing scheme). A $(t,n)$ secret sharing scheme is a linear secret sharing scheme when the $n$ shares, $v_1,v_2,...,v_n$ can be presented as in Equation $\ref{5}$

$$(v_1,v_2,...,v_n)=(k_1,k_2,...,k_t)H,\label{5}\tag{5}$$

where $H$ is a public $t × n$ matrix whose any $t × t$ submatrix is not singular. The vector $(k_1,k_2,...,k_n)$ is randomly chosen by the dealer.

According to Definition, we can see that Shamir’s $(t, n)$ secret sharing scheme is a linear scheme. Let

$$f(x)=a_0+a_1x+\cdots+a_{t-1}x^{t-1}, \label{6}\tag{6}$$

The shares $v_i = f(i)$, $i = 1, 2, ..., n$ can be presented as in Equation $\ref{7}$

$$(v_1,v_2,...,v_n)=(a_0,a_1,...,a_{t-1})H,\label{7}\tag{7}$$

How is $\ref{7}$ equivalent to $\ref{6}$? in some definitions it quotes $y_i= f(x_i)$ or $y_i= f(x_i)\bmod{p}$ how do they differ with $\ref{7}$?

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  • $\begingroup$ the secret sharing scheme of Shamir is linear after all? why? $\endgroup$ Jan 17, 2022 at 9:51
  • $\begingroup$ @kelalaka in $(5)$ you can replace index $n$ of $k_n$ with $t$...I don't want to interupt your edit...because you are always helpful $\endgroup$ Jan 17, 2022 at 11:28
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    $\begingroup$ No probs, see my edits and learn :) $\endgroup$
    – kelalaka
    Jan 17, 2022 at 11:43

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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 missed a minus sign in the paper.

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  • $\begingroup$ well the weird is with all these definitions that in some case they write $f(x)=...mod{p}$ in other cases $f(x)=...$ without modulo and in some cases $y_i\equiv_p f(x_i)$...to be quite frank, i can not understand the difference...do you? $\endgroup$ Jan 17, 2022 at 13:35
  • $\begingroup$ in other words the definition says give me the points $(s,a_1,a_2,...a_{t-1})$ recall that $a_0=s$ and i can find a mapping $H(s,a_1,a_2,...a_{t-1})=(v_1,v_2...,v_n)$ such that the pairs $(i,v_i)$ $\forall i \in n$ are points of the polynomial function $H=f(x)=s+\sum_{i=1}^{t-1}a_ix^i$? $\endgroup$ Jan 17, 2022 at 13:51
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    $\begingroup$ @HungerLearn: The math in Shamir's secret sharing is done in a finite field. The integers modulo a prime $p$ form such a finite field, but there are also other types of finite fields. (In particular, any set with $p^n$ elements, where $p$ is a prime and $n$ is a positive integer, can be given multiplication and addition operators that make it a finite field.) The confusion of notation you mention probably reflects that: some authors are assuming a prime-order field and using notation from modular arithmetic, while others just assume a generic field. $\endgroup$ Feb 16, 2022 at 15:04

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