# Tag Info

14

The example is using a shorthand notation for the rotors that somewhat obscures the way they actually work. For example, the first rotor in your example, BDFHJLCPRTXVZNYEIWGAKMUSQO, actually applies the following permutation of the alphabet: ABCDEFGHIJKLMNOPQRSTUVWXYZ ↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓ BDFHJLCPRTXVZNYEIWGAKMUSQO Applying this rotor in the ...

7

The symbol of the circle with the + in it is one of many symbols for exclusive-or. XOR, EOR, EXOR, ⊻, ⊕, ↮, and ≢. Binary OR is true when either input is true; binary XOR is true when exactly one input is true. If both inputs are true, the XOR result is false. One property of this is that if either input bit flips, the output bit will also flip. That's sort ...

6

This example is correct. The inversed versions are the inverse permutation; that is, if the forward direction is the permutation $P$, then the inverse permutation $P^{-1}$ has the property that $P^{-1}(P(X)) = X$ for all $X$. That is, if $X$ is a plaintext letter, and we run it through in the forward direction (giving us $P(X)$), and then run it through in ...

4

That definition is a standard definition which defines encryption as a function $E$. That function takes two inputs, a $\kappa$ bit key and a $n$ bit message. Hence it is defined over the cartesian product - denoted as $\times$ - over these two sets, i.e. all bitstring of length $\kappa$ and $n$ respectively. It maps - denoted as $\rightarrow$ - to an $n$ ...

3

As one of the authors of the paper, let me give you an answer. The operation $F$ is indeed applied to both $x$ and $x'$. By stating that $\oplus$ is invariant under rotation, we mean that if you first rotate $x$ and $x'$ and take the difference with $\oplus$, you get the same result as if you first take the difference with $\oplus$ and then rotate the ...

2

It's exclusive or (or XOR), corresponds to $x \oplus y := x+y \pmod 2$ for single bits, i.e., scalars. It is sometimes used for binary vectors as well, whereby two bitvectors of length $n$ $$\mathbf{x}=(x_1,\ldots,x_n)$$ and $$\mathbf{y}=(y_1,\ldots,y_n)$$ result in $$\mathbf{x}\oplus \mathbf{y}=(x_1\oplus y_1,\ldots,x_n \oplus y_n)$$ i.e., a bitwise ...

2

It just means vectors of dimension $n$, where each entry is in $\mathbb{Z}_q$. I.e., $(x_1, \ldots, x_n) \in \mathbb{Z}_q^n$ means that $x_i \in \mathbb{Z}_q$ for all $i = 1, \ldots, n$.

1

It contains the equivalence classes represented by $\{1,\ldots,6\}$. To elaborate on Nova's comment below, $\mathbb{Z}_p$ is the collection of congruence classes when you divide by $p$. When $p=7$, two integers represent the same congruence class when they have the same remainder when divided by $7$. As you probably know, the only possible remainders are ...

1

It means $Q$ divides $P-1$. In other words, $P-1$ is a multiple of $Q$.

1

The explanation of why the security parameter $k$ is given in the unary form of $1^k$ as mentioned at the second footnote at page 366 of Foundations of Cryptography Volume 2 is to allow a smooth transition to fully non-uniform formulations...Specifically $1^n$ indicates that the $n^{th}$ circuit is to be used.

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