# Tag Info

10

A cryptographic hash function $f : \{0,1\}^{*} \to \{0,1\}^n$ has three properties: (1) preimage resistance, (2) second-preimage resistance, and (3) collision resistance. Even further, these properties form a hierarchy where each property implies the one before it, i.e., a collision-resistant function is also second-preimage resistant, and a second-preimage ...

5

Consider the function $H$ transforming a message $m$ to the SHA-512 hash of the first 1024 bits of $m$ (right-padded with $1024-n$ zero bits if the bit length $n$ of $m$ is less than 1024). $H$ is first-preimage resistant, but not second-preimage resistant: once you have a first preimage $m_1$, it is trivial to get another $m_2$ with the same hash (e.g. ...

4

You are not asking for a collision but for a preimage. Collision attack: the attacker computes two messages m and m', distinct from each other, such that m and m' hash to the same value. Preimage: the attacker is given a goal (a hash value h) and finds a message m which hashes to h. MD5 is weak for collisions, but not for preimages: no attack method is ...

4

Let me try to elaborate on their proof. Suppose you had a hash function $H$ that was second-preimage resistant but not first-preimage resistant. By showing that this leads to a contradiction, we will be showing that with second-preimage resistance, you must have first-preimage resistance. Namely, we will show that the lack of first-preimage resistance is ...

3

Given message $A$, you have to find message $B$, such that the first 64 bits (say, MSB) of their hashes collide: $$MSB_{64}(H(A)) = MSB_{64}(H(B))$$ This problem is called Second Preimage Search for the function $MSB_{64}(H)$, or Partial Second Preimage Search for the hash function $H$ alone. When $H$ is the full round SHA-1, there is no result, ...

3

Pre-image resistant but not 2nd pre-image resistant? describes the relationship between the three basic hash function security notions: Collision Resistance, Second Preimage Resistance and Preimage Resistance. In short, Collision Resistance implies Second Preimage Resistance (but not vice-versa) - there is a good diagram on page 4 of RogawayShrimpton04 that ...

3

In their paper Second Preimages on $n$-Bit Hash Functions for Much Less than $2^n$ Work, Kelsey and Schneier provide: a second preimage attack on all $n$-bit iterated hash functions with Damgard-Merkle strengthening and $n$-bit intermediate states, allowing a second preimage to be found for a $2^k$-message-block message with about \$k\times2^{n/2+1}+ ...

3

Actually, to the best of our knowledge, it's computationally infeasible. By the terminology what we use when we discuss cryptographical hash functions, you're not asking for a hash collision (which is "find two different messages that hash to the same value"), but instead you're asking for a hash preimage (which is "for this hash value, find a message that ...

3

As far as I am aware, there are no practical known second pre-image attacks on MD5, under the conditions you listed. However: if the attacker can control any part of the original, I would worry about using MD5 in this setting. Its security in this setting may be fragile and there may well be cleverer attacks than anything currently in the literature. I ...

3

I have ask as part of my answer, "What problem are you trying to solve?" Do you want a secure unkeyed hash function? If you do, then there are plenty of them around. Even some of the ones that are broken for some uses might be okay for yours (SHA-1 springs to mind -- note the discussion above on HMAC and how broken a hash function has to be). But really, ...

3

We know no practically feasible way to do what you ask for, except if the hash X=acf3602b5eb9a2db3e365d3043682faf or the content of the file wczasp.rb was prepared specially to make that possible. Assuming that the content of file wczasp.rb is arbitrary, what is asked would be a preimage attack. This is further sub-classified as first preimage if only the ...

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