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7

Your problem seems to be at least as hard as the 2-weak Bilinear Diffie-Hellman Inversion Problem (2-wBDHI problem): Given $g, g^x, g^{x^2}, g^y \in \mathbb G$, and $T \in \mathbb G_T$ to determine whether or not $T = e(g,g)^{x^3 y}$. Proof: We first need to define an equivalent version of your problem, where we take some generator $h$ so $g = h^b$. ...


7

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 ...


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 ...


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 $(r,z)$


5

My experience of such proofs is that often you're proving a much stronger statement: in any execution of Game i+1, either everything is distributed identically to Game i or else F must have happened. In other words, the conditional distribution of game i+1 conditioned on F not happening is identical to the distribution of Game i. For example, one of ...


5

As fgrieu notes in the comments, the protocol might not even work reliably in the absence of adversaries: if the tag fails to receive the reader's reply of "True", the keys will get out of sync. (If that happens, the tag will just retry the next exchange with the old key, so this could be fixed by having the reader remember one more "subkey". But the ...


5

You can generate a random string $s_1$ as long as the plaintext. Then XOR this value with the plaintext generating $s_2$. Now encrypt both parts using $\mathrm{Enc}_1$ and $\mathrm{Enc}_2$. You need to decrypt both to XOR the two parts together again. This is similar to secret sharing where you need two parts of a key to decrypt. If $\mathrm{Gen}_1$ and ...


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 ...


4

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 ...


3

The reason NIST chose one algorithm out of the five AES finalists, even though all of them were pretty well-respected (and some were, at the time, considered likely to be more secure then Rijndael) is because NIST is a standards body, and the whole point of the AES project was to find a standard algorithm. The issue with approving lots of algorithms is that ...


3

You have to worry not just about a pair of blinding values being equal, but more complex relationships between them. Thus, finding a proof of security for this approach looks non-trivial to me. Let me elaborate. Suppose $R_j$ is the $j$th blinding variable you use. If $R_i = R_j$, that's a problem, but as you say, that can be made very unlikely. ...


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

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

However I think that the requirement on $\Pr[F]$ is way too strong: one can prove security with much more frequent faults. Maybe this short paper by Alexander W. Dent could be of interest: A Note On Game-Hopping Proofs? In this paper he introduces a fourth kind of game hop, namely transitions based on large failure events, which seems to be exactly what ...


3

Ok, here's a version that definitely works. Three points: If we know $S_2$ is close to $1/2$, then the statement that $S_1$ and $S_2$ are close is equivalent to saying that $S_1$ is close to $1/2$. So I'll prove the latter version. The way to deal with the edge cases, I think, is to demand Pr$[S_2|F] = 1/2$. This is the case whenever the game (not the ...


3

For any $n \in \mathbf{N}$, let $X_n$ be a random variable which always equals $n$, and $Y_n$ be a random variable which equals $n$ or $n+1$ each with probability $1/2$. Then the probability ensembles $X = \{X_n\}_{n\in \mathbf{N}}$ and $Y = \{Y_n\}_{n\in \mathbf{N}}$ are not computationally indisinguishable. A possible distinguisher is an algorithm $D$ ...


3

I would think these numbers would have been put on the google search engine, and yield (probably) many hits. This assumption is wrong. Certificate serial numbers are not indexed by common search engines, nor are they typically posted to any HTML site. Frankly, I'm not sure why you would assume they'd be indexed. The Wordpress certificate is used for ...


2

The problem that arises in the security proof is that the adversary who may win the real game with some probability may however cause the simulation to nearly always abort (by issuing a private key query that requires the simulation to abort) and the probability of an abort may be different for different sets of private key queries. So, the problem is that ...


2

Let $q$ be given by $\:$ for all $n$, $\: q(n) = 1 \;\;$. $\;\;\;\;\;$ For every $P^*\hspace{-0.05 in}$, every $\: x\in L_R \:$, $\frac{p-\kappa(|x|)}{q(|x|)} = \frac{p-\kappa(|x|)}1 = \:p\hspace{-0.04 in}-\hspace{-0.04 in}\kappa(|x|) \: \leq \: p\hspace{-0.04 in}-\hspace{-0.04 in}0 \: = \: p \: \leq \: 1 \;\;$. For every $P^*\hspace{-0.05 in}$, every $\: ...


2

A classic example of this sort of thing would be the Goldreich-Levin hard-core bit for an arbitrary one-way function. The proof of security for the Goldreich-Levin construction involves showing that if Mallory can predict this single-bit value (even with probability slightly better than $1/2$), then it is possible to invert the one-way function. Thus, this ...


2

Not always, it depends on the particular encryption scheme. Strictly speaking, the proofs only say that breaking indistinguishability is equivalent to breaking the hardness assumption they are based on. There are some cryptosystems, like Rabin's, where the security of the key is equivalent to the security of the ciphertexts, i.e. factoring <=> key ...


2

Mostly. The two problems are actually more closely equivalent in a gap model than in a non-gap model. Square-DH clearly reduces to CDH either way, but CDH reduces to two calls to Square-DH (you have 3, but you can use $(u-v)^2$ to make it 2). This is fine if the Square-DH adversary is always right, but maybe the adversary only solves the Square-DH problem ...


2

I am not aware of any work that proposes a Gap problem related to LWE. The reason is probably that LWE is an average-case problem specifically designed for the use in crypto. However, there are the related worst-case problems, e.g. the shortest vector problem (SVP), that come with a Gap version. So, you might want to have a look at GapSVP and GapCVP.


2

It is called circular security. It is problematic because it is not captured by the usual security definitions. I.e., even if an encryption scheme is proven secure by some regular definition, it is usually not a given to be circular secure. To see why consider, for example, the usual definition of semantic security for a public-key encryption scheme $\Pi = ...


2

For CPA security it is actually enough that the first scheme, i.e., $\pi = (gen, enc, dec)$ is CPA secure. Lets define the CPA game of a general scheme $\pi = (Gen, Enc, Dec)$ against an adversary $A$ as follows: We sample $(pk, sk) \leftarrow Gen(1^\lambda)$, and send $pk$ to $A$. $A$ outputs messages $m_0$ and $m_1$. We sample $b \leftarrow \{0,1\}$ (a ...


2

In the standard definition of security for public key encryption schemes there exists only one public key. Therefore the decryption oracle will always decrypt with the secret key that corresponds to the public key given to the attacker. It does not matter how the $c_1$ in your question is computed, it can be computed using the real public key, a different ...


2

What you present is a generalized version of the so called fixed-argument pairing inversion (FAPI) problem. The FAPI problem is given an element $z\in G_T$ and an element $h\in G$ to compute $f\in G$ such that $e(h,f)=z$. Note, that FAPI is implied by the computational Diffie Hellman problem: Given $(g,g^a,g^b)\in G^3$, call the FAPI oracle with $z\gets ...


2

Assuming that the probability distributions of $\pi_{k_1}$ and $\pi_{k_2}$ are both uniform (that is, each permutation can take on any particular setting with probability $1/n!$), then no, adversary does not have enough information to learn anything about the original positions. This remains true even if we assume the adversary can perform unbounded ...


2

Hint: you can notice that $n! > 2^n$ (except for very small $n$).



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