# Tag Info

4

As otus suggests in the comments, it's better to first calculate the frequency of each letter in the decrypted message, and then compare the frequency distribution to what would be expected for English text. For the comparison, you can use chi-squared ($\chi^2$) testing. (Actually, for just comparing the likelihoods of different decryptions, you don't even ...

3

The requirement was introduced in IUT Recommendation X.509 (November 1993), informative appendix D.5.2: It must be ensured that e > log2(n). If not, then the simple operation of taking the integer eth root of a ciphertext block will disclose the plaintext. This advice was removed in the 2000 edition of the standard. It is arguably misguided, and at the ...

3

This specific hash function is weak; it appears that what this hash function does is pad out the string to be hashed into a 32 byte string, and then take the 8 4-byte substrings, and maps each substring individually into an individual byte. This immediately makes it trivial to find a preimage; start with a random 31 byte preimage (there appears to be a bug ...

2

As CodesInChaos notes in the comments, having more ciphertext–plaintext pairs doesn't help with brute force guessing attacks. Well, that is, except for the minor issue of unicity. Basically, to narrow the results of your brute force attack down to a single key, you do need to have enough ciphertext–plaintext pairs that the length of the known plaintext ...

2

Yes and yes, as mentioned in the comments. It is worth noting that Bitcoin wallets use a scheme similar to this in BIP32, a method of creating n various EC keypairs from a single seed deterministically: https://github.com/bitcoin/bips/blob/master/bip-0032.mediawiki

2

Yes, this is exactly what KDFs and PRFs are designed for. That is, no reasonably efficient attacker will be able to tell if you used an actual random key or something generated from the KDF/PRF. This is of course assuming that your initial seed/master secret was of sufficient entropy, and the way you derive the various values are not done in a silly way. ...

1

If you are able to compute $m^1 \pmod{N}$, then you have (obviously) recovered the message $m$. So, you should be able to use the extended Euclidean algorithm to express $m^1 \pmod{N}$ in terms of $c_A$ and $c_B$. Hint: It will involve exponentiations and multiplications.

1

So the idea is to use the IV of MD5 as a key to create a MAC. Like CodesInChaos mentions in a comment, it would be pretty much equivalent to using $H(k||m)$, if your IV is randomly chosen. By only using it on fixed length messages you avoid the length extension attack, but that is not the only attack on hash constructions that try to create a MAC. In this ...

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