# Tag Info

## Hot answers tagged encryption

9

Your problem is equivalent to solving $m^3 \equiv c \pmod n$. This corresponds to breaking RSA for $e=3$. We know no efficient way for doing that without factoring $n$. If you know the factors of $n$ you can compute the private exponent $d$ as $e \cdot d \equiv 1 \pmod{\varphi(n)}$ using the extended euclidean algorithm and then compute $m$ as $c^d$.

6

Before we start with vectorial Boolean functions, let's recall the definition of the nonlinearity of a Boolean function: $$\mathcal{NL}(f) = \min_{a \in \mathbb{F}_2^n} d_H(f, \ell_a \oplus b),$$ where $\ell_a \oplus b$ represents the affine Boolean function defined by the bitvector $a$: $\ell_a(x) = a \cdot x$ ($\cdot$ is the dot product). The above ...

3

It depends on the application if base 64 is being used to represent keys. Many applications that implement/use cryptography have been originally designed in a time where ASCII based communication was commonplace. If you would directly use BER / DER - a binary encodoing of parameters - then you had a high chance of losing data. For instance, you would not be ...

3

Base-64 is simply a way to represent binary data using the ASCII character set. It's used because a single base64 digit represents a whole number of bits (6 bits), where each decimal digit represents ~3.3 bits, which can make conversion a little tricky. Being able to represent more bits per digit also means its more space efficient than decimal. It takes ...

2

For some modes of operation you can easily show that an involution would be insecure: OFB would be most clearly insecure, since the keystream just repeats the nonce/IV and its corresponding encrypted block. CFB would likewise be insecure, since zero blocks encrypt just like with OFB. This is of more limited advantage to an attacker, but far from secure. ...

1

Shift cipher or ceasar cipher attains perfect secrecy only in the special case with the assumption that $26$ keys are used in equal probability in the shift cipher, and to encrypt each symbol we use a different key which is choosen equiprobably (i.e. perfectly random) from the key space. It is easy to check all keys given a plaintext when the key is fixed ...

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