# Tag Info

## Hot answers tagged lwe

11

They're actually sampling $5n$ elements from $\Psi_{16}$. Perhaps Protocol 2 on page 5 shows this most clearly, where $\textbf{s}, \textbf{e} \stackrel{\$}{\leftarrow} \Psi_{16}^n$and$\textbf{s}', \textbf{e}', \textbf{e}'' \stackrel{\$}{\leftarrow} \Psi_{16}^n$ are sampled (on line 3 on Alice' side, and line 1 on Bob's). This probably also answers part of ...

9

Yes, LPN is (essentially by definition) equivalent to the hardness of decoding a random linear code over $\mathbb{F}_2$. No, there is no known reductions between LPN and LWE. It is usually believed that LPN is (in some sense) "harder to break" than LWE, simply because we know much less attacks on LPN. It seems to have less structure that could be exploited ...

9

From Status Report on the Second Round of the NIST Post-Quantum Cryptography Standardization Process 3.12 NewHope NewHope is a KEM based on the presumed hardness of the RLWE problem. At its core is Regev’s original idea for public-key encryption from plain LWE but specialized to a power-of-2 cyclotomic ring structure, enabling smaller ciphertext and key ...

7

Let $R$ be the ring $\mathbb{Z}_p[X]/(X^n+1)$, where $n$ is a power of 2. The Ring-LWE assumption says that for any randomly chosen, fixed $s\in R$, the distribution of $$((a_1,a_1s+e_1),\ldots,(a_k,a_ks+e_k))$$ is indistinguishable from the distribution $$((a_1,u_1),\ldots,(a_k,u_k)),$$ where $a_i,u_i$ are uniformly random in $R$ and $e_i$ are chosen from ...

7

We don’t always use power-of-two cyclotomics for RLWE. Many cryptosystems use other cyclotomics, or subfields thereof, or even other fields altogether. For example, many FHE schemes use non-two-power cyclotomics for “packing” and SIMD operations on plaintexts. However, it is simplest to properly define and use RLWE over two-power cyclotomics, in large part ...

7

Firstly, why is this true? This is easy to see, if we consider the finite field as a finite group with the addition operation (and ignore the multiplication operation) If we consider the value $X - Y$, where $X, Y$ are uniformly and independently distributed elements of the group, then there are $n^2$ equiprobable $X, Y$ pairs possible (each with a ...

6

The previous (galvatron's) answer gave two good reasons why Ring-LWE is more efficient. They explain why the running time of LWE schemes is worse than of Ring-LWE ones. (The n^2 vs n equations is more of an efficiency issue because in both cases, this part of the key is expanded from a 256-bit seed using an XOF. So it just takes a lot longer to expand an ...

6

There is no known reduction from LWE to MLWE (or to RLWE). That is, it could be that both MLWE and RLWE are broken, yet LWE is secure. However, this seems highly unlikely. To support the security of LWE, we have reductions showing that breaking the average-case hardness of LWE requires breaking the worst-case hardness of some lattice problems - which would ...

5

I'm assuming $n$ is a power of $2$ and that $q$ is an odd prime larger than $n$. I'm discarding the trivial case $s_1 = s_2$. If you consider everything $\mod q$, then it is most likely over the choice of $a$ that there exists $s_1 \neq s_2$ such that $\|a s_1 - a s_2\| = \sqrt{n}$. Indeed, $a$ is invertible in $R_q$ with probability about $1 - n/q$. Take $... 5 As mentioned in the comments, there is still a lot of active research being done in the area of algorithms for solving Learning With Errors, as clearly it's an important topic for estimating the security of LWE-based cryptography. Therefore it's probably already not a good idea to rely on papers from more than 5 years ago if you want your parameter choices ... 5 Denote by$X$the random variable which is the sum over all$S$. As mentioned, this is a Gaussian of standard deviation at most$\sqrt{m}r$with$r = \alpha q$. Hence, by properties of the (sub-)Gaussian distribution you have that $$\operatorname{Pr}\left[|X| > t\right]\leq 2\exp\left(\frac{-\pi t^2}{r^2m}\right)$$ so, for$t = \frac{q}{2}$you have $$... 5 The probability of error is negligible "as a function of n", meaning that the probability of error will decrease (quickly) as n grows. Increasing n should solve your issue. 5 There is a simple trick (known in the LWE literature as the Hermite normal form of the problem) that takes an existing LPN problem and transforms it into a problem in which the secret has the same Bernoulli distribution as the error. This trick is proved in Lemma 2 of Applebaum et al. for a more general case, or on the (Ring-)LPN attacks of Kirchner (Section ... 5 It's worth mentioning that the conditions needed for f(X_0, X_1) to be uniformly random based off the distributions of X_0, X_1 are quite mild usually. In particular what you need is: X_0 and X_1 to be independent At least one of X_0, X_1 to be uniformly random (say that it is X_0) f(\cdot, X_1) : G\to G to be a bijection [1] for each choice ... 5 One main difference is that in Ring-LWE, the ring R is the full ring of integers \mathcal{O}_K of a number field K, whereas in Poly-LWE it is of the form R=\mathbb{Z}[x]/f(x) for some irreducible f(x); this ring is (isomorphic to) an order of the number field K=\mathbb{Q}(x)/f(x), but may not be the full ring of integers. Another important ... 4 I believe it is also used in other lattice based schemes that use standard LWE. For example, the Frodo paper. They used a seed_A and a Gen(\cdot) function to compute A. Then Alice sends seed_A instead of A for the actually exchange. Gen(\cdot) is a prior-agreed pseudorandom function that extracts and extends the seed. 4 The number of rows in matrix \textbf{A} shows the number of LWE samples and m=1 means the adversary has access only to one sample. This would be a naive adversary model and would not be suitable for assessing the true hardness of the Problem. To consider a more powerful adversary model, a (seemingly) stronger definition is to assume that the adversary ... 4 This lemma is used to conclude that a sample from \mathcal{D}_{\mathbb{Z}^n,\alpha q} is (with overwhelming probability) less than or equal to \alpha q \sqrt{n}. Now, because all values \textbf{s}_{\textbf{A}},\textbf{s}_{\textbf{B}},\textbf{e}'_{\textbf{A}},\textbf{e}'_{\textbf{B}} are sampled from \mathcal{D}_{\mathbb{Z}^n,\alpha q} , so \textbf{... 4 I asked the question to the author directly. To answer the first question, authors of NewHope estimate their security very conservatively, whereas the estimator takes many other things into account. Specifically, here is the answer I got from the author. Hi, Id' suggest to use their claimed security. To explain the differences: The estimate you ... 4 δ_0: the root Hermite factor required β: the BKZ block size d: the dimension of the lattice being reduced m: the number of LWE samples used 4 If the adversary is a classical algorithm, then the answer to your question is not known. But if the adversary is a quantum algorithm that can query the oracle in superposition, then the answer is yes: by making queries to the oracle on certain (efficiently produceable) quantum states, it can recover a Type-I trapdoor for A. For classical algorithms, the ... 4 Out of curiosity, what is the current state of the art on the sampling over D_{\mathbb{Z},\alpha q} This is a fairly involved question to answer. There are a number of competing ways to sample it, which you can roughly divide into: Techniques that work for any probability distribution Techniques that are specific to the discrete Gaussian Table 1 of [... 4 The output of bootstrapping has relatively small noise because it starts from an encryption (of the secret key) that has very small noise, and performs some homomorphic operations on it. These operations increase the noise somewhat, but it starts out so small that the result still doesn’t have much noise. It doesn’t really matter that the ciphertext being ... 4 Given (A, Ax + e) and (A, x^tA+e'), you can do (at least) one potentially interesting thing to solve LWE. Namely, compute the sample$$(A+ A^t, Ax + e + (x^tA+e')^t) = (A+A^t, (A + A^t)x + e + {e'}^t)$$so you can reduce LWE to a case of LWE where the random matrix$A + A^t$is symmetric. It isn't really clear to me that this helps you cryptanalytically -... 4 You can. There is a certain caveat that should be mentioned here --- the LWE problems hardness is controlled (in part) by the size of the modulus$q$. Two important parameter regimes are$q$being polynomially large in the security parameter, and super-polynomially large. Smaller modulus is better for both efficiency and security. I think only recently we ... 3 The LWE assumption tells you that$(\mathbf{a},\mathbf{a}\mathbf{s}+e)$looks random for a hidden random vector$\mathbf{s}$, a random vector$\mathbf{a}$and small error$e$. Invoking the LWE assumption k times, you get that$(\mathbf{a},\mathbf{a}\mathbf{s}_1+e_1,\mathbf{a}\mathbf{s}_2+e_2,\ldots,\mathbf{a}\mathbf{s}_k+e_k)$is indistinguishable from$(\...

3

It is easy to reduce this problem to LWE, since adding any element to a uniformly random value gives a uniformly random distribution. For example, here, $(A',b')$ is distributed the same as if $A'$ was drawn uniformly randomly, and $R$ set as $A'-A$. You're now in the setting of the LWE assumption and can replace $b'$ by a uniformly random value without ...

Only top voted, non community-wiki answers of a minimum length are eligible