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3

To quote yyyyyyy from the comments: The $_R$ has nothing to do with the field — it is associated to $\in$! To quote your first link: "For a set $S$, by $a\in_R S$, we mean that $a$ is randomly chosen from $S$." and to quote SEJPM from the comments: If $p\in \mathbb P$ (with $\mathbb P$ being the set of all primes) then the notations ...


2

Use the number of occurrences of each letter as the unknowns you're solving for. This gives you 32 linear equations to solve for 16 unknowns. Use standard approaches for solving systems of linear equations. Finally use the fact that the letters are ordered alphabetically to reconstruct the $x_i$ values.


3

You cannot uniquely invert it. Your hash will have the form $$y_1 \quad y_2 \quad \dots \quad y_m$$ with $y_i=\sum_{k=1}^m x_k^{i-1}.$ First look at the special case $m=2$. $y_1$ will always be $y_1=m=2,\,$ the other equation is $y_2=x_1 + x_2$. You cannot uniquely determine both $x_1,x_2$. In the general case you have $y_1=m$ and $m-1$ equations for the ...


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I think you could calculate it manually or write code to solve this. You need to calculate X values from starting with end. For example : $X_{32}$ = 84 = $X_{32}^{31}$ $mod_{117} $ When you get $X_{32}$ value then you can calculate $X_{31}$: $X_{31}$ = 62 = $X_{31}^{30}$+$X_{32}^{30}$ $mod_{117} $ The only unknown value is $X_{31}$ here. You need ...


2

How does value of $p$ comes to $2^{512}$ for security complexity of $2^{72}$? Personally, I couldn't connfirm this number, but rather landed at $2^{66}$. If you don't know how to find this number, just plug the $2^{512}$ into the complexity equation given and take the logarithm to the base two for better readability. Note that Schnorr's complexity ...


2

The term "finite field cryptography" exists to distinguish from group-based cryptography. It is true that every field contains two groups, but a group is not necessarily part of a field. Every prime number $p$ has a finite field $\mathbb{Z}_p$, and these prime fields are used in cryptosystems such as RSA and DSA. The Diffie-Hellman key exchange operates ...


4

To answer this question, we need to understand what is meant by: the field $\operatorname{GF}(2^8)$ generated by the irreducible polynomial $m(x) = (\mathtt{0x11B})$. and I'll try a gentle, progressive introduction to that. One simple way to see this field is that as the set of the 256 octets $\{\mathtt{0x00},\mathtt{0x01},\dots,\mathtt{0xFF}\}$; with ...


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"Finite field cryptography" is fancy language for group-based cryptography done over the integers modulo a prime (instantiating a field) to distinguish this more "classic" approach from the new fancier elliptic curve cryptography. Example: Finite Field Diffie-Hellman: Diffie-Hellman done in $\mathbb F_p$.


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The Extended Euclidean Algorithm produces the greatest common divisor and the Bézout coefficients for two integers $a$ and $b$. If $gcd(a,b) = 1$ and $a<b$, then the Bézout coefficients also give you the multiplicative inverse of $a$ in $\mathbb{Z}/b\mathbb{Z}$. Because the operations used in the Extended Euclidean Algorithm are defined on polynomials, ...



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