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2

NP Complete need not mean safer, since you want strength not only in the worst case (for the attacker) but in all cases. Some cryptosystems based on NP complete problems (most notoriously based on the knapsack problem) have been broken. Having said that, cryposystems based on the Discrete Logarithm problem in prime order groups, as well as those based on ...


2

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 $c_1,c_2,c_3\in GF(9)$ ...


2

Credit to @lulu's comment above. List down the coprimes of $26$ smaller than itself: $1,3,5,7,9,11,15,17,19,21,23,25$. Then calculate the inverse of each one. Here is a piece of C code that you might find useful: int Inverse(int n,int a) { int x1 = 1; int x2 = 0; int y1 = 0; int y2 = 1; int r1 = n; int r2 = a; while (r2 != 0) ...


1

for $a \gt 0$ $$ x^{4a+b} = (x^4+1)x^{4(a-1) +b}-x^{4(a-1) +b}\equiv x^{4(a-1) +b} $$ because characteristic is 2


1

Let $\Pr(X_1=0)$ be $p$, so $\Pr(X_1=1)=1-p$. The event $X_1\oplus X_2=0$ can occur in two possible disjoint ways: (i) $X_1=0$ and $X_2=0$ or (ii) $X_1=1$ and $X_2=1$. The probability of (i) is $(p)(1/2)$, and the probability of (ii) is $(1-p)(1/2)$. If we add them, we get: $$\Pr(X_1\oplus X_2=0)=(p)(1/2)+(1-p)(1/2)=1/2.$$ Since the only possible ...


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Intuitively, ignore the $\pmod {11}$s and you have $(7^A)^B=7^{AB}=(7^B)^A$ The outer $\pmod {11}$ clearly is no trouble-you just have to convince yourself the inner one isn't either.


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$60 = 17\cdot 3 + 9$ $17 = 9\cdot 1 + 8$ $9 = 8\cdot1 + 1$ $1 = (9 - 8) = (9 - (17 - 9)) = (2\cdot 9 - 17) = (2\cdot (60-17\cdot 3) -17) = 2\cdot 60 - 7\cdot 17$ So $17\cdot -7 \mod_{60} = 1$ Thus $53$ is the inverse.


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Take the greatest common divisor of these numbers. To explain this a bit more... Some $\textit{trigrams}$, meaning strings of 3 letters, in the English Language appear more frequently than others on average. For example; \begin{equation} ING \quad THE \quad AND \end{equation} So for the text that has been enchiphered using Vigenere, you would normally ...


1

First, consider the definition of group isomorphism: Given two groups $(G, \otimes)$ and $(H, \odot)$, a group isomorphism from $(G, \otimes)$ to $(H, \odot)$ is a bijective function $f : G \to H$ such that for all $u$ and $v$ in $G$ it holds that $f(u \otimes v) = f(u) \odot f(v)$. Now, consider a homomorphic encryption such as ElGamal ...



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