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7

One line: worst means any and average means random. Lattice-based cryptosystem Let me restate. Fix security parameter n. What the reduction shows is the existence of a solver for the lattice problem on input any n-dimensional lattce using the adversary breaking a lattice-based cryptosystem with the security parameter n on the average case. Since we can ...


6

It is equivalent to the computational Diffie-Hellman problem; if you can one of the two problems, you can solve the other (with a polynomial number of queries to the oracle which solves the other). If you can solve the Diffie-Hellman problem, you can solve your problem: this can be seen by first noting that, with a Diffie-Hellman solver, given $g^b$, you ...


6

There is no direct inference from $P = NP$ or $P \neq NP$ to security or insecurity of any particular encryption algorithm. As far as practical consequences are concerned, the "$P = NP$" problem is severely overhyped. If $P = NP$ then any problem for which a solution can be verified in polynomial time can also be solved in polynomial time. "Polynomial time" ...


6

A fast 64-bit hash cannot be completely secure, since a $2^{32}$ brute force collision search is completely doable, and even a $2^{64}$ preimage attack could be feasible. As a MAC used for hash table keying, that doesn't really matter (unless you leak the key). Finding just a few collisions isn't a problem and gathering statistics for an attack would ...


5

Actually, it's not true that public key encryption is based on Discrete Log; the ones in common use (DH, ECDH, ECDSA) are (and even RSA can be viewed as "based on Discrete Log", at least from the standpoint of "if you can solve the Discrete Log modulo a composite, you can break RSA"). However, we do have a number of public key systems (NTRU, McEliece) which ...


5

The probability of someone 'getting lucky' with a guess at a key for a decent cryptosystem is crazily low, but yes: it is possible. However, there are methods that can 'survive' even this. For example, consider the one time pad. In this system the key and plaintext are xor'd together to form the ciphertext, and to decrypt you xor the ciphertext and key. ...


5

Intuition The intuition behind the proof is that a valid ciphertext is correctly generated and, thus, an adversary should query to the random oracles to generate random strings in the ciphertext. In addition, notice that the hash value on an unqueried string is undetermined (to the adversary) in the random oracle model. Therefore, the chance to construct a ...


5

This scheme is insecure, as anyone with the public key can generate a forgery of an arbitrary message. To do this, the forger would take the message $M$, the public key $y$, pick an arbitrary $z$, and compute $r = y^{-H(M)} g^{z} \bmod p$ and output $(r,z)$


4

Yes, it is a secure PRF, as long as the attacker can make at most $2^n-1$ queries. The key lemma is purely information-theoretic in nature. From this lemma, you can prove your result using standard methods. Lemma. Let $x=(x_1,\dots,x_q),y=(y_1,\dots,y_q)$ are any sequence of $q$ $n$-bit values, where $q<2^n$ and the $x_i$'s are all distinct. Then ...


4

Theorem. (Barring any errors, which are certainly possible): Let $A$ be any adversary attacking $F$. As long as $A$ makes fewer than $2^n$ queries, there is an adversary $B[A]$ attacking $f$, making at most twice the queries of $A$ and running in (roughly) the same time such that $ \mathsf{Adv}^{\mathsf{prf}}_F(A) = \mathsf{Adv}^{\mathsf{prf}}_{f}(B[A]). $ ...


4

Not if the attacker is given enough time to ask for the values $F(k,x)$ for all $2^n$ values of $x$. We have $\Sigma_x F(k,x) = 0$ which is distinguishable from random.


4

First, on the difference between perfect security and semantic security. Both definitions concern confidentiality, so let us first define what confidentiality means. Note first that an adversary as some a priori knowledge of the message. We can capture that by e.g. having the adversary choose two messages and then flipping a fair coin to decide which one to ...


4

The equation $t'=t+n\cdot t_c$ is an estimation to put an upper limit on $t'$. It might be possible that an attacker $A'$ can use a different, more efficient algorithm. But since the attack will work with using $A$, there exists an attacker $A'$ with at most $t'$. This means it's actually not an equation, but an inequality $t' \leq t + n \cdot t_c$. And then ...


4

As correctly pointed out by Ricky Demer it is not necessarily true. However, this implication does not hold for very specific cases. In the case of random oracle gates the existence of the RO changes the functionality of the "scheme", since with RO there are RO-Gates and without there aren't. In most cases the existence of the RO does not affect the ...


4

From the key creators point of view, notice that:$g_1\bmod p=1$ and $g_2\bmod q=1$. Which means that $(g_1-1)\cdot(g_2-1)\bmod N=0$. Which implies that $g_1-1$ and $g_2-1$ share a common divisor with $p$ and $q$. To obtain $p$ we simply take $\gcd(g_1-1,N)$. To obtain $q$ we simply take $\gcd(g_2-1,N)$, or $N/p$. Because we have been able to factor the ...


3

As @xagawa mentioned in his comment, it depends on what you mean by "controled". In the case of using a programmable random oracle, yes, the reduction (in particular the simulation of the challenger of the real game) decides about what to return as answer to an oracle query. Thereby, the reduction has to guarantee that the "programmed" answers are ...


3

AES is not a one-way permutation; it is a permutation, for sure, but whoever can apply it can also apply its inverse. Crudely said, the AES decryption key is identical to the AES encryption key. A one-way permutation would be like a hash function: everybody can compute it in one direction, with no secret value, but nobody knows how to do it in the other ...


3

If they don't trust the server they sure shouldn't send any money. The "trusted" third party is used to solve the problem of participants who don't trust each other. So by definition, your problem can only be somewhat mitigated, not solved completely. I'm not sure what you mean by "provably fair". If the server can't prove he cannot cheat, it's not provably ...


3

By the definition given there, in a modular protocol design, each method or scheme used in the protocol (such as the use of nonces or the application of a cryptographic schemes) has a clear goal which it is proven to achieve, and sub-protocols can be replaced without re-proving the security of the remaining protocol steps. That's also the meaning ...


3

Reductionist security In a reductionist security proof for some cryptographic protocol $\Pi$ to some alleged hard problem $P$ means, that we can build an algorithm $\cal B$ for solving $P$ if we have access to a hypothetical algorithm $\cal A$ that efficiently breaks the security definition for the protocol $\Pi$. In general, showing a polynomial time ...


3

The standard definition of existential forgery allows the adversary to ask and obtain the signature of any message she wants, and claim success if she can exhibit (with sizable odds) any acceptable (message, signature) pair, for any message for which she did not ask signature. Update: There is also strong existential unforgeability, where the adversary ...


3

I see no reason to expect $x-1+x^N \bmod N^2$ to be indistinguishable from $r$, at least not based upon the assumption you give. The map $f(x)= x^N \bmod N^2$ is a very different map from the map $g(x) = x-1+x^N \bmod N^2$. The range of $f$ is a subgroup of size $\varphi(N)$; that's not true of $g$ (for instance, the range of $g$ can potentially be the ...


3

HMAC-SHA-256 is sufficient for up to 256 bit security. Confer e.g. NIST SP 800-107. This recommendation is based on the premise that collision attacks are infeasible against common uses of HMAC, and that you consequently only have to worry about primary pre-image attacks that attempt to recover the secret key (and use this for forging subsequent messages). ...


3

Answering my own question since there haven't been any response. In fact there is no flaw in the definition. Nor in my attack per se! However, the entire point of this definition is to rule out non-stateful schemes (and stateful schemes that doesn't incorporate the state in a sufficient manner) as being secure. In particular, when the adversary asks for ...


3

In general this is not the case. Consider a PRF F which ignores the first bit of the key. Then you can distinguish F'(k,.)=F(.,k) from a random function (and thus break its security as a PRF) by querying it on inputs x and x', where with x' we denote x with the first bit flipped. Note that the outputs of F'(k,.) will collide on inputs x and x' as ...


3

A viable answer to the question, as posed in the title, is: No If a stream cipher involves a limited number of messages (say 10) of limited length (say 1000 characters each) enciphered with differently keyed streams, then surely there is not sufficient information available for breaking the cipher? As you would expect, this depends completely on the ...


3

Such reductions I know are the reductions in hardcore predicates/functions from computational assumptions, say, from the OWP/OWF/RSA/DCR/CDH/DBDH assumptions, the reductions in (provably-secure) PRGs/PRFs from computational assumptions, say, from the OWP/OWF/RSA/DCR/CDH/DBDH assumptions, and the reductions in LPN/LWE. Re: Can you be more specific? ...


3

This may just be a matter of terminology. A claim that an algorithm is "secure" is meaningless without qualifying/quantifying what it is secure against. Conventionally, the security/strength of cryptographic primitives is described and analysed in terms of computational and memory cost (i.e. secure against an attacker capable of performing a certain number ...


3

NTRU, as it was originally introduced, is based on what people now call the "NTRU assumption", which really just assumes that it is difficult to break NTRU. Annoyingly circular, but that's how it is. Similar to how breaking RSA is not provably as hard as factoring (technically it is based on the "RSA assumption"), but people still think it is hard, NTRU ...


2

Yes, you have shown that these two problems are equivalent by showing reductions in both directions (which is the way to address such problems). Btw., this result is well known which holds of the order of $G$ is prime.



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