# Tag Info

1

In general, you never want to use CRC/weak checksum for any computations on secret material (like keys). CRC is a linear function and by showing CRC of a key, you reveal a lot of equations that hold among the key bits. This is equivalent to showing the same number of bits of the key as the length of the checksum. The proper way of doing it has been ...

3

If $f:\{0,1\}^m\rightarrow \{0,1\}^n$ with $n\geq m,$ then of course there are, the set of one-to-one functions, but such a function is not a cryptographic hash function, since it lacks the compression property. If $n<m,$ (or more generally if $|X|>|Y|$ for $f:X\rightarrow Y$), collisions will happen.

4

When designing security for a physical safe, one of the critical specifications is how long will the safe resist attack, this tells you how quickly you must detect and respond to an attack on the safe. Yes, however there's a key difference between physical safes and cryptography. With a physical safe, the attackers must be present on site (if they ...

0

Collision resistance relates to how difficult it is to find a collision, not the total lack of collisions. In other words, a hash function $H(x)$ is collision resistant if and only if all attackers given reasonable computational resources (probabilistic polynomial algorithms) have only a negligible chance of finding two messages, $m_0$ and $m_1$ such that ...

0

By "one way function" do you mean preimage resistant, or do you mean that the function doesn't ever reveal the input? Assume H(x) is a collision-resistant function. Let L(x) = the last 256 bits of x. Then, let G(x) = H(x) || L(x) That is, G(x) is the concatenation of a collision-resistant hash of x, and the last 256 bits of x. Now, over all ...

0

A Cryptographic hash function as described in the literature has 3 criteria: Preimage resistance: Given $H,y$, it is "hard" to find an $x$ with $H(x)=y$ Second Preimage resistance: Given $H,x,$ it is hard to find $x'\neq x$ with $H(x')=H(x)$ Collision resistance. It is hard to find 2 $x,y$ with $H(x)=H(y)$ The very definition used (the first 1 and the ...

2

Thomas Pornin already explained why such a thing is not usually possible, but I would like to quote a graphic from Rogaway and Shrimpton's "Cryptographic Hash-Function Basics: Definitions, Implications, and Separations for Preimage Resistance, Second-Preimage Resistance, and Collision Resistance" (pdf): The dotted arrow from Collision resistance to ...

7

It is going to be pretty hard to achieve collision resistance without one-wayness. Indeed, negation of one-wayness means that for a given output, you can find a corresponding input. So a collision is easily obtained by simply choosing a random input m, hashing it into output x, then finding a preimage m' for the obtained output x. The only way for such a ...

0

You'd still need to iterate over all the possible passwords to do this, using a brute force or dictionary attack (i.e. a pre-image attack with regards to the hash). That is, unless the hash algorithm used isn't one way; i.e. you can retrieve more information about the passwords. A 32 bit hash is not common, so this could be possible. Fortunately most ...

3

Per my comment, I'd like to suggest a definition for "non-iterative hash function", and propose some constructions that fit the definition. I will also suggest an alternate name (though it may not help much with searching for papers on the topic). Let $\mathcal{M}$ be the message space of a hash function, e.g. $\mathcal{M}=\{0,1\}^{*<\ell}$, the set of ...

5

If $H(x) = x$, $x$ is a fixed point. If for a value the output of the function is the same as the input, it is called a fixed point. A length extension attack is unrelated to the concept of fixed points. There is a good question about understanding length extension attacks here.

Top 50 recent answers are included