# Tag Info

## Hot answers tagged collision-resistance

9

Well, first of all, you need to be clear about the meanings of various cryptographical primitives. Cryptographic hash function; this is a function that takes an input string, and generates a hash. The idea is that we don't know how to create two input strings with the same hash, and so the hash can be used as a replacement for the original string. Now, ...

7

Preimage resistance is about the most basic property of a hash function which can be thought. It means: For a given $h$ in the output space of the hash function, it is hard to find any message $x$ with $H(x) = h$. (Note that the it is hard here and in the next definitions is not formally defined, but can be formalized by looking at families of hash ...

5

The answer to the original question would have been: Yes, it would be impossibly difficult to exhibit workfactor, salt, password0, password1 such that bcrypt(workfactor, salt, password0) = bcrypt(workfactor, salt, password1); but even if that was feasible, it would not matter much, because in normal use at least one of the password is unknown to an adversary ...

4

Accidental collisions are interesting for certain applications, and one would expect accidental collisions to occur less frequently in a system than malicious collisions. So, if you are not worried about malicious collisions, only accidental, it is easy to compute how many digests you would need to compute before seeing an accidental collision. If the ...

3

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 a user that can see the database of hashes does not see the password directly, but this property depends crucially on the hash being resistant to a pre-image ...

3

A collision attack is the ability to find two inputs that produce the same result, but that result is not known ahead of time. In a typical case (e.g., the attack on MD5) only a relatively small number of specific inputs are known to produce collisions. Collision resistance obviously means that a collision attack is difficult (for some definition of ...

2

No, it's not possible. RSA based signatures can only be generated if you know the private key. At the RSA primitive level, signing is the same as decrypting a message (where the data being decrypted in this instance is a hash of the message). If you change the hash, then verification of that signature would fail. Without knowledge of the corresponding ...

2

Yes, it makes sense to truncate the hash to 128 bits. The security proof actually says that if finding a preimage for F requires effort 2^n, then breaking the Lamport signature scheme with G having k-bit digests requires effort (2^n)/(2k). So strictly speaking, with F truncated to 128 bits and G having 256 bits (2k=512=2^9), you will have 128-9=119 bits of ...

2

Yes, there have been real life inputs with collisions working on the full length of MD5. There is a pair of X.509 certificates that share an MD5 hash. There is also a pair of PostScript documents that are an MD5 collision. There are also two binary strings a mere 6 bits different that are a collision. The whole length of MD5 has been broken. Attacks that ...

2

While the EAX mode permits truncating the tag to any length $\tau$ between 0 and $n$ bits, where $n$ is the block size of the underlying block cipher, this should only be taken as a statement that truncating the tag in EAX mode does not introduce any security issues beyond the obvious (an attacker only needs $2^\tau$ attempts to froge a $\tau$-bit tag by ...

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