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Your solution to part 1) is ok. A zero $a$ of $x^3-x^2+1$ will, indeed, give you a normal basis $\mathcal{B}=\{a,a^9, a^{81}=a^3\}$. I'm relatively sure that in part 2) you are asked to do the following. Let $$z=c_1a+c_2a^9+c_3a^{81}$$ be an arbitrary element of $GF(3^6)$ - written using the basis $\mathcal{B}$, so the coefficients $c_1,c_2,c_3\in GF(9)$ ...

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There are different ways to find out. For such a big number we need a method without having to factorize it. One way to start, is to prove that N is composite. Therefore we choose a strong primality-test like the one of Miller–Rabin,Solovay–Strassen or the newer APRCL-test. If this test returns false, the compositeness of N is proven. Next choose a test ...

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If you are using cyclic notation, $\pi=(1234)$. And $\pi^{-1}=(4321)$, which is just $\pi$ written backwards. But in the matrix notation that you are using, you form the inverse by exchanging the top and bottom rows. So $$\pi^{-1} = \begin{pmatrix}2&3&4&1\\1&2&3&4\end{pmatrix}$$ which, after rearranging the columns, is the same as ...

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No. The inverse permutation is the one that puts everything back where it came from ($\pi(1) = 2$, so $\pi^{-1}(2) = 1$, and so on), which means that you get the inverse by swapping the two rows: $$\pi^{-1} = \begin{pmatrix}2&3&4&1\\1&2&3&4\end{pmatrix}$$ Usually one also sorts the columns for easy readability, so that we get $$... 1 The girth for directed graphs is usually defined as the directed girth, that is, the minimum length of a directed cycle (or \infty if no directed cycles exist). For instance, see Jørgen Bang-Jensen and Gregory Gutin, Digraphs: Theory, Algorithms and Applications (Second Edition), Springer Monographs in Mathematics. (The definition is given in chapter 1. ... 1 This is a simple RSA problem: d is the inverse of e modulo \phi(n), where n = pq So apply the extended Euclidean algorithm (see here) to e and \phi(n). You get integers k,l \in \mathbb{Z} with ke + l\phi(n) = 1. Then k is easily seen to be the required inverse, i.e. d (just take the previous equation modulo \phi(n) where the right ... 1 Yes, if there is no known plaintext, and just a single cipher text m^\ast, you do need k even if p is known to everyone. In this case the plaintexts are numbers of a special form (because they're built up from smaller alphabet numbers in your case), so there might be a weakness there, and there are some other issues, but essentially yes. But as said, ... 1 The rule is already given here: Let \begin{matrix}\blacksquare\square\\\square\blacksquare\end{matrix} be pattern 1 and \begin{matrix}\square\blacksquare\\\blacksquare\square\end{matrix} be pattern 2. Notice that they add up to a black square, and when added to themselves, they create a half-black square. If the source pixel is black: For half of ... 1 The correctness of the cryptosystem (in the sense that decryption is inverse to encryption) depends on$$m^{k\phi(n)+1}\equiv m\pmod n$$for all m. Euler's theorem says that this is the case when m and n are coprime -- which is not enough here -- but if n is known to be square-free it is actually the case for all m. Constructing n as the ... 1 Let's see what the right question is. The basic proposition is this. Suppose, given y, b and prime p, you want to find x such that b^x \equiv y \mod p, where b is a primitive root mod p. Suppose p-1 = cd with c and d coprime. Then y^c \equiv b^{xc} \equiv (b^c)^x \mod p, where b^c is a d'th root of 1 mod p. This will ... 1 Chinese remainder theorem says: if a and b are co-prime and pa + qb = 1 and x\equiv n(\mod a)\\ x\equiv m(\mod b) then x\equiv pam+qbn (\mod ab) 19*64=1216\\ 27*19-8*64 = 1\\ x\equiv 40(\mod 64)\\ x\equiv 13(\mod 19)\\ x\equiv 13*(-8)*64+40*27*19 (\mod 1216)\\  1 Let$$a=q_an+r_ab=q_bn+r_b$$for quotients q_a,q_b and remainders 0\le r_a,r_b<n of a,b modulo n. Then$$\begin{align} a+b&=(q_a+q_b)n+(r_a+r_b)\\ &=\left(q_a+q_b+\delta\right)n +\left(r_a+r_b-\delta n\right) \end{align}$$for$$\delta=\left\lfloor\frac{r_a+r_b}{n}\right\rfloor where $\lfloor x\rfloor$ is the greatest integer (less ...

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