# Tag Info

48

The difference is in the choice of $m_1$. In the first case (second preimage resistance), the attacker is handed a fixed $m_1$ to which he has to find a different $m_2$ with equal hash. In particular, he can't choose $m_1$. In the second case (collision resistance), the attacker can freely choose both messages $m_1$ and $m_2$, with the only requirement that ...

23

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 ...

23

Observation: An individual 1-byte pearson hash behaves like an 8 bit block cipher, encrypting the initial state using the message as key. This means that given a fixed message, each possible initial state produces a different output. This implies that a combined hash will never contain duplicate bytes. Without this property a hash would forget about the ...

20

With the definitions that a function $F$ is collision-resistant when a [computationally bounded] adversary can't [with sizable odds] exhibit any $(a,b)$ with $a\ne b$ and $F(a)=F(b)$; first-preimage-resistant when, given $f$ determined as $F(a)$ for an unknown random $a$, a [computationally bounded] adversary can't [with sizable odds] exhibit any $b$ with $... 18 With all well-regarded hash functions, the bits of the hash all have equal worth: as far as anyone knows (unless they aren't telling), the bits are not correlated. If you take$k$bits of an$n$-bit hash, you get a$k$-bit hash function. Truncating SHA-256 to 255 bits gives you a hash that's almost as good as SHA-256: it has$2^{255}$strength against ... 16 Right now, the best published attack against MD5's preimage resistance (first preimage, actually, but it applies to second preimage resistance as well) finds preimages in cost$2^{123.4}$average cost, which is slightly better than the generic attack (average cost of$2^{128}$), but still way beyond the technologically feasible. The attack rebuilds the ... 13 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 ... 12 Because it is not secure enough. Hash functions rely a lot on diffusion (a single bit change must change half of the other bits) and confusion (the value of a bit should depend on the value of other bits). This is also known as the avalanche effect. Because it lacks a permutation, my first intuition: it lacks diffusion and has weakness to differential ... 11 It is neither pre-image resistant, second pre-image resistant nor collision resistant. It is easy to compute square-roots modulo a prime (assuming, of course, a square root exists, it will half the time). If$p = 3 \bmod 4$, then the simple formula$x^{(p+1)/4} \bmod p$will work; for$p = 1 \bmod 4$, it's a tad more complicated but still sufficiently easy,... 8 I prefer using definitions that explicitly specify who does what. Weak collision resistance: After Bob creates some message x1, it is "computationally infeasible" for an attacker Mallory to compute some other message x2 such that h(x1) == h(x2). Strong collision resistance: It is "computationally infeasible" for an attacker Mallory to find any two messages ... 8 Yes, it has happened. If you look at the SHA3 hash zoo, there are a number of hashes who has the best attack listed as "2nd preimage". One general place this can occur is if you have a hash function with a weak message compression step, but a fairly strong finalization step. Here, we might not be able to generate first preimages (because we don't know what ... 8 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}+ 2^{n-...

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The best way to approach problems like this is to start by assuming that a simple solution exists. That assumption might be wrong, of course, but: since this is a textbook problem, it probably does have a relatively simple solution that you should be able to figure out based on what you've learned, and even if that wasn't the case, you should still at ...

8

It's harder for the designer to make a hash function collision-resistant than second-preimage-resistant, because it's harder for the adversary to find a second preimage than to find a collision.

7

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 ...

7

You are mistaken. A collision attack is where you need to find any $x$ and $y$ such that $hash(x)=hash(y)$. Thus, you have much more freedom in finding the collision. This makes it "easier" for the adversary.

6

Can we exhibit collisions, or second-preimages (with implies the former), for the ChaCha core? No, likely not. The Salsa20 and ChaCha cores both consist of a large number of "quarter-rounds" each of which is invertible and bijective. The only reason neither core is a bijection (and thus can have collisions) is the final addition of the input elements into ...

6

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 ...

6

Take a function $H:\mathbb S\to\{0,1\}^k$ where $\mathbb S$ is a large finite subset of $\{0,1\}^*$, such that $H$ "compress data" [however this is defined], and $H$ is [conjectured] collision-resistant [thus second-preimage-resistant] and first-preimage-resistant; e.g., SHA-512, for $k=512$. Let $«0»$ and $«1»$ be two public distinct elements of $\mathbb S$....

6

Hash + digital signature If the hash is not collision resistant, the attacker can produce two messages having the same hash. They'll request a signature on the first and present the signature on the second, a forgery. When second pre-image resistance is violated, this attack becomes much more severe, since now the attacker doesn't need control over both ...

6

First of all, are my assumptions above correct? What would the complexity of the three attacks be? I am also curious what "unit" a number like $2^{256}$ implies - is it something like floating-point operations that would be needed to run? Almost, this number is the number of times the primitive (in this case SHA-256) must be called to break the ...

6

W-OTS+ is stronger, as it makes weaker assumptions on the hash function. Let us take a rather extreme example, let us consider W-OTS and W-OTS+ based on the MD5 hash function. Now, the proof for W-OTS is quite invalid; it assumes that the hash function is collision resistant, and we know how to generate collisions with MD5. On the other hand, W-OTS+ based ...

6

A secure hash function based on the discrete log problem is $H(x\|y)=x\cdot G+y\cdot P$, where $P$ is a random point for which the discrete log is not known and "$\|$" denotes concatenation; this was shown by Damgård. This can be proven collision resistant under the discrete log assumption (if you can find a collision then this can be used to find the ...

6

Any hash function? Yes, certainly. In fact, most hash functions are not even designed to be resistant to preimage attacks. This includes CRCs and standard checksums like Fletcher. Creating preimages with them is trivial. The oldest popular hash algorithm is MD2 which has a preimage attack with a complexity of 273 and a memory requirement of storing 273 ...

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Apart from the slightly reduced resistances, there is no problem: Resistances for SHA3-512; Pre-image resistance decreased to $2^{511}$ or $2^{504}$, if 1 bit or 1 byte trimmed, respectively. Secondary preimage resistance decreased to o $2^{511}$ or $2^{504}$, if 1 bit or 1 byte trimmed, respectively. Collision resistance decreased to o $\sqrt{2^{511}} = \... 5 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 ... 5 Without the specific reference I can't be sure this is what you are talking about, but generally a "long message" attack is a way to defeat second preimage resistance with less complexity than expected. It uses a time-space tradeoff to find a second preimage with complexity$2^{n/2}$for a$n$-bit hash function (normally you would expect$2^n$). In the ... 5 For a$n$-bit ideal hash: Finding a collision is expected to require a little above$2^{n/2}$hashes, in short because after having computed$m$hashes, there are$m(m-1)/2$opportunities of collisions. For details, see Birthday problem for cryptographic hashing, 101. Finding a preimage (first or second) is expected to require about$2^n\$ hashes, in short ...

4

Preliminary: Almost the same article is available for free without breaking any law, nor downloading 5GB (formatting is shifted by at most one third of a page). It is also (as well as all other articles of IACR crypto conferences from 2000-2011) in the IACR Online Proceedings, specifically in the FSE 2008 section, but then you need to subtract about 223 from ...

4

The answer you are looking for is that a pre-image/2nd pre-image attack on two combined hash functions is at least as difficult as an attack on the stronger of the two hash functions. Take a look at Joux's paper on multi-collisions. There is a section on pre-image attacks for concatenated outputs. (I suggest reading the entire thing. It's one of those rare ...

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