You're using bytewise xor, and three plaintexts $x,y,z$ with a small alphabet (A-Z, which I assume means ASCII values 0x41-0x5a). Then knowing $a \oplus x$, at a certain index, we know all possible ...

The whole point of public key cryptography is that there is a secret key, which, if the system is to be any good, cannot be easily computed from the public key, but is known to the legitimate owner of ...

You are right about the interpretation of the power 10: it's a tenfold iteration. So we apply the function 10 times, starting with $x$, feeding the output as input for the next step. So C-like (I ...

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 an element of the field of all integers modulo $p$, and these are represented by the numbers $0,\ldots, p-1$. And $g$ will be one of them.

First of all, a more usual padding scheme would add 5 times the same byte 0x05 (in your example) so the check not just removes 5 bytes, but also checks that the 4 bytes before it have the same value. ...

To address the other issue (with the CFB-1, CFB-8, registers, etc.): Note that in the picture we encrypt whole blocks of the previous ciphertext, and xor the result with the next plaintext block, to ...

Consider Shor's algorithm, which solves RSA in polynomial time on a quantum computer and the fact that there is publicly available software that could factor a 512 bit modulus on a modern PC in a few ...

The Playfair cipher has a key consisting of a square of $5 \times 5$ letters (usually the J is not used, or I/J are considered one letter). Filling the square can be done in $25!$ ways (pick a ...

After base64 decoding we get (hex) 5d f8 be 87 82 2b ea 5e f3 5c 23 fe 37 81 0f bd which has a size of two blocks. Of your small word-list, only marketing has so many letters that it needs two ...

The key is not stored with the encrypted data, the encrypted key is. This is part of the header. In short, when the volume is created a random key (the master key) is generated and this random key is ...

There is only one, it's declared in <openssl/evp.h>. This post from 2009 names it as yet to be documented, and this seems to be still true. It does have a man page on my Linux systems. BTW recent ...

You should think of Rijndael's S-box as a function that maps bytes to bytes, where a byte (octet) is considered to be a member of a finite field of size $2^8$ (with xor as addition). It's not seen as ...

I assume that $m = pq$ (85) is the modulus, and the number you computed (64) is $\phi(m)$. Now we must choose the exponents ($i$ for public, $j$ for secret? , weird notation) in such a way that $ij = ... View answer 2 votes IMHO it's just a warning to the reader that this is not a standard hash-based design like BSD-crypt or PBKDF2, which are traditional choices. They use the Salsa20/8 Core mixing function because its ... View answer 1 votes You are looking at the SHA-1 specification, not the SHA-256 one. You are quoting pages 15,16 in the PDF you linked to. The remarks Thomas made also apply to SHA-256 though: the message block of 16 32-... View answer Accepted answer 3 votes Keeping the (initial) IV secret only adds security for the first block of your cipher text. The IV for the second block is by definition the first cipher text block, and so on. So if we would have the ... View answer 2 votes Reducing mod 3 means indeed reducing each coefficient mod 3, and again we choose the representatives symmetrically around 0, so each coefficient becomes -1,0 or 1 (instead of, which is also possible, ... View answer 4 votes Yes, there is. There is a polynomial algorithm for solving general polynomial equations, like$x^n - a \mod{p}$in a finite field. And for simple powers, its even easier. So I wouldn't use this for a ... View answer 3 votes The format is not an official standard, but it is pretty straight forward. There are indeed 4 fields, in order$p$,$q$,$d$and$u$(I'm not quite sure what the function of the$u$is?, it's not the ... View answer Accepted answer 2 votes The randomness is not enough for IND-CCA-2. If we get a message (so$L$bits of data plus the$y$to reconstruct the random seed), we can modify it, say flip the first bit, and ask the decryption ... View answer Accepted answer 1 votes CN is common name, OU is orginizational unit, O = organization. So we could have CN="John Johnson", OU="Sales", O="Acme Corp." or something like that. http://docs.oracle.com/javase/1.4.2/docs/... View answer Accepted answer 5 votes Let Alice have a key pair (PubKeyA, PrivKeyA), where the first is public, the second private, and similarly for Bob (PubKeyB, PrivKeyB). Alice and Bob know each other's public keys in a reliable way (... View answer 9 votes Some additions to the other answer: any given letter can only correspond to a fairly limited number of ciphertext letters: only the ones in the same column or row, and never to itself. So a highly ... View answer 1 votes Once you know (you didn't question that) that$a = 3$, then knowing 4 goes to 17, using$f(x) = 3x + b$, we just substitute$x = 4$to get$3 * 4 + b = 12 + b = 17 \mod 26$, where we can easily see ... View answer 6 votes Often the hash (iterated and salted mostly) of a password is saved in a database, instead of the password. If a user logs in, the hash is computed and compared against the stored hash value. This way ... View answer 4 votes I'll leave the probability to someone else, but to ensure the non-smoothness of$p-1$most libraries (including OpenSSL) will generate a prime$p'$such that$2p' + 1$is also prime, and then ... View answer 11 votes As the other poster rightly pointed out, it's a Playfair cipher. Even without the known plaintext, the program "playn" here will give the right text in less than a second. (you can compile it ... View answer 5 votes Suppose that we have Eve, that knows what$e$is going to be, and does not need to know the prover's private key$s$, just the public one$v$. She then sends$g^k \cdot v^{-e}\$ as her first "move", ...