Henno Brandsma
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So in your cipher text, the characters $c_1, c_7, c_{13}, \ldots, c_{1+6i}, \ldots$ should roughly obey the frequency statistics of a permuted or shifted standard English frequency. In the most ...

Alongside the other arguments: signing (with RSA) means exponentiating it modulo $N$ (with power $d$, the secret exponent). So anything you sign that way must be of bitsize smaller than $N$, and ...

It's base64 encoded binary data, after decoding it (e.g. using the base64 command line tool in Linux or MacOS) we see it consists of 544 bytes of random looking data. As this is a multiple of 16, we ...

Because (if $C=P^e$ where $P$ is the plain text): $2^e C= (2P)^c \pmod{N}$ so multiplying the ciphertext by $2^e$ modifies the plain text by a multiplication of $2$, which is a bitshift to the left of ...

The '0b' representation removes leading 0's. The actual value (as a byte) of l (lowercase letter L) as a binary 8bit byte is 01101100 and likewise for the others (they are lower ASCII, so the highest ...

It's a translation of $ed \equiv 1 \pmod{\phi(N)}$, namely that $e$ and $d$ are each other's inverse modulo $\phi(N)$, which says that $ed-1$ must be an integer multiple of $\phi(N)$ (recall that $a \... View answer 1 votes DH in the classic form can be vulnerable to MITM, but the form used in TLS (SSL is deprecated, and hardly used any more, TLS is the more secure successor) is that the server sends its DH-contribution (... View answer Accepted answer 2 votes In short: Trace back where bits come from in every byte in the output state (8 bytes) of the SP_network. In one round, after we have xored the round key bits, we have 8 bytes (a 64 bit word, with ... View answer 1 votes The key is 4 long, so should be a$2 \times 2$matrix. In members of$\mathbb{Z}_{26}$CFDG becomes$2, 5, 3, 6$in the usual A becomes$0$, Z becomes 25 encoding. After some experimenting I found ... View answer Accepted answer 1 votes More standard is that we multiply a column with a plain text vector to get the cipher text vector so the known plain text equation (the first two pairs) yield $$KP=C = K\begin{bmatrix}5 & 8\\17 &... View answer Accepted answer 2 votes These modular equations are not uniquely solvable:$$\begin{bmatrix}7&2\\ 10& 20\end{bmatrix}, \begin{bmatrix}7&2\\ 23& 7\end{bmatrix}, \begin{bmatrix}20&15\\ 10& 20\end{... View answer 1 votes "KEYWORD" is a weird format for a Hill cipher, aren't you confused with the Playfair cipher? There you work with a 5x5 matrix where I and J are often conflated into I (as 26 is one too big) and the ... View answer 1 votes It's indeed standard to add pre-agreed padding characters at the end to make the plain text a multiple of$n$when we use an$n \times n$encryption matrix. So your encoded plain text could be $$\... View answer 1 votes A zero-knowledge proof for the secret key is a probabilistic challenge-response algorithm where the prover has probability 1 to get it right knowing the secret key, and some chance p < 1 if he ... View answer 0 votes You should consider the field \textrm{GF}(2^8) as a "given", a way to multiply and add bytes (the addition is just XOR, the multiply could be given as a table of 256 by 256 entries; you'll see we ... View answer 1 votes It might help to describe the padding arithmetically: first shift p (the pin) 2048 bits to the left, so compute 2^{2048}\cdot p. To replace the trailing 0's by 1's we add 2^{2048}-1. So ... View answer 0 votes It must be sent with the message in order to be decipherable to the legitimate decipherer. The message sent has the IV as the first part. The shared key K is known, so the "packet key" K' is \... View answer 1 votes This is a toy system that is used in didactic contexts as an introduction to RSA, in a way. We have some prime p (we are going to do the power operations modulo this prime). If m is the message (... View answer 3 votes It looks like the library is treating the string as the key to Blowfish, which has a veriable key size; the way the keysetup is done (with a cyclical use of the key bytes, see more details on the ... View answer 1 votes Because \mathbb{Z}/{p\mathbb{Z}} when used for El-Gamal is used in its multiplicative form: we have a generator g such that all powers of g cycle through \{1,2,\ldots,p-1\} (so 0 is excluded)... View answer 3 votes The padding used for RSA is not the PKCS #5/#7 padding (as you seem to suggest in your own answer), but the Wikipedia entry seems to refer to PKCS #1 v1.5 (RFC2313)) which uses a padding 00 || BT || ... View answer 3 votes As @kelalaka said, the answer is still valid as described in this question, so a PBKDF2-SHA512 hash with a large salt and iteration count in the 100000's, usually. View answer Accepted answer 2 votes This part of the standard (at the end) explains why the length of the authenticated data is at the end, namely to ensure a unique input to the MAC for all pair of ciphertext and authenticated data: ... View answer 1 votes In a typical hybrid system (PGP, S-MIME etc.), a message contains both an encrypted key k_s for the symmetrical system, encrypted by the recipient’s public key. A new k_s is generated per message.... View answer Accepted answer 1 votes You apply the extended Euclidean algorithm to e and \phi(N) (which have to have gcd equal to 1) and we get x,y \in \mathbb{Z} such that$$xe + y\phi(N) = 1$$The$x$(taken modulo$\phi(N)$, ... View answer 0 votes You're just taking two succesive operations, namely to the power$k_1$and then to the power$k_2$in the ring of integers modulo$n$, in both cases. You are just confused by the different notations ... View answer 1 votes It's a standard fact in abstract algebra: The statement is just$(g^{k_1})^{k_2} =(g^{k_2})^{k_1}$in the ring$R=\mathbb{Z}/n\mathbb{Z}$, with$k_1, k_2 \in \mathbb{N}$. This statement is valid in ... View answer 2 votes The$S\$-boxes are used in the round function of a Feistel cipher, so they will be exactly the same for encryption and decryption. We're not using a SP-network here like AES, where we need the ...