# Tag Info

## Hot answers tagged elliptic-curves

13

I'd say that the whole argument hinges around a "secret attack" that possibly the NSA may know of, enabling them to break some instances of elliptic curves that the rest of the World considers as safe, because the secret attack is, well, secret. This yields to the only possible answer to your question: since secret attacks are secret, they are not known to ...

9

ECDSA is a digial signature algorithm ECIES is an Intergrated Encryption scheme ECDH is a key secure key exchange algorithm. First you should understand what are the purpose of these algorithms. Digital signature algorithms are used to authenticate a digital content.A valid digital signature gives a recipient reason to believe that the message was created ...

8

The idea of "safe curve" is somewhat overrated. What you really want is a safe implementation which won't leak secret information when employed in some practical context. Leakage may occur in a variety of ways; some examples include timing attacks and implementation behaviour when encountering anomalous input. This is not an exhaustive list, and, depending ...

7

It happens that a line usually (not always) cuts three points in a elliptic curve by the Bezout theorem. This is the case for the points and the curve you are asking for. So the sum of two points are defined like the inverse of the third point intercepted by the line that cut $P$ and $Q$ (let's name it $R$). So we need to find $-R$ because $P+Q=-R$ by ...

7

The tangent line at point $P(x_1,y_1)$ is: $\frac{\partial f}{\partial x}(x-x_1)+\frac{\partial f}{\partial y}(y-y_1)$ Solving differentials, we obtain: $\frac{\partial f}{\partial x}=3x^2+2x$ $\frac{\partial f}{\partial y}=-2y-1$ For $x_1=0$ and $y_1=0$ we have: $\frac{\partial f}{\partial x}=3x^2+2x = 0$ $\frac{\partial f}{\partial y}=-2y-1 = -1$ ...

7

ElGamal appears to be used instead of Diffie-Hellman (or IES) in OpenPGP mostly because when that format was put together, there were some unresolved intellectual property issues surrounding both RSA and Diffie-Hellman, while ElGamal was unproblematic. This trend for ElGamal seems to stick around, mostly by force of habit, e.g. when switching to ...

7

There are actually only 5 unique $x$-coordinates one needs to be concerned about: $(0, \ldots)$ $(1, \ldots)$ $(-1, \ldots)$ $(x_1, \ldots)$ $(x_2, \ldots)$, where $$\begin{eqnarray} x_1 =& 393823572354896145817230607815530211125 \\ & 29911719440698176882885853963445705823 \end{eqnarray}$$ and $$\begin{eqnarray} x_2 =& ... 6 This has been basically asked already: Should we trust the NIST recommended ECC parameters? History Once it was found that NSA allegedly had inserted backdoor to a cryptographic standard, people started thinking what standard it was. The most common guess is that the Dual EC DRBG is the backdoored standard. However, some amount of (possibly justified) ... 6 It is all pairings... this is a rather complex matter. I recommend reading Ben Lynn's PhD dissertation; it is about as nice an introductory text on pairings as you can get. The definition is rather mind-twisting: You first define divisors, which are rather formal objects. It is the free group of the curve points: for each curve point P, you define a ... 6 CodesInChaos has it correct, however since you're just learning, I think I'll lay it out rather explicitly. When we have an abstract group, there are two ways of expressing operations in the group. One way is writing the operation as if it were the multiplication operation, for example, if we apply the operation to elements A and B and the result is ... 6 BouncyCastle has a really bad ECC implementation. It uses affine coordinates which incur a huge performance hit (factor 20 or so) since it computes a field inversion after every single step. Good implementations use Jacobi coordinates (or a similar approach) where denominators are kept and there is only one field inversion at the end. It's also potentially ... 6 Over large characteristic fields, I am not aware of any "point generation method" that can be computed faster than a base field exponentiation, and I would be very surprised if such a thing existed even if you do not require constant running time. So your best bet in general is probably Icart's function (I'd pick that one over Elligator if I didn't need ... 6 Actually, it is not possible to uniquely recover the public key from an ECDSA signature (r,s). This remains true even if we also assume you know the curve, the hash function used, and you also have the message that was signed. However, with the signature and the message that was signed, and the knowledge of the curve, it is possible to generate two ... 6 I don't know about computing things in parallel, so I will ignore that part of the question. First, please note that the encryption algorithm is rarely the the weak point of the security. It is far more likely that you will have problems with the implementation, some spyware installed on your computer, a weak password (If you use qwerty as your password, ... 5 There are some known groups in which computational Diffie-Hellman assumption is equivalent to discrete logarithm problem. Besides, It has been shown that the equivalence holds "when a small amount of extra information depending on the group order is provided". Furthermore, those extra informations has been computed for certain elliptic curve groups used in ... 5 It mainly depends on how the algorithm was selected. If it was selected by a public competition like for AES, then it is likely to be secure. If it was forced in by the NSA such as Dual-EC random number generator, then you may have some doubts. Other questions you may want to ask yourself are: Is this an "original" algorithm or was the problem that it ... 5 Your second equation seems a bit off. In the curve of equation:$$ y^2 + cy = x^3 + ax + b $$in a binary field \mathbb{F}_{2^m}, to add point P_1 = (x_1,y_1) to point P_2 = (x_2,y_2), resulting in point P_3 = (x_3,y_3), then the two equations are: \begin{eqnarray*} x_3 &=& \lambda^2 + x_1 + x_2\\ y_3 &=& \lambda (x_1+x_3) + y_1 + c ... 5 The implications are that someone screwed up some calculation. By Lagrange's theorem, the co-factor must be an integer. 5 The problem doesn't lie with curves in Weierstrass form necessarily, but with naive implementations of elliptic curve arithmetic on such curves. Basically, if you implement an ECC scheme (ECDH, ECDSA or whatever) on a smart card using a curve in Weierstrass form in the most straightforward way possible (by writing a simple double-and-add loop for ... 5 The bad news is that projective coordinates do not work with Pollard's Rho like you want it to. Rho needs an unambiguous point representation to find meaningful collisions, and in projective coordinates each point can have up to p-1 valid distinct representations. The good news is that, sticking to affine coordinates, you can avoid most of the cost of the ... 5 Now to calculate Q this will take a lot of time since it means I will need to perform point addition an insane number of times unless I'm not understanding something about it. You're missing a point; elliptic curve point addition is associative; that is, for any three points A, B, C, we have:$$(A + B) + C = A + (B+C) Now, why is this a big deal? ...

4

Given a EC public key, can a different, but plausible and functional private key be derived to match the public key? No, a public key will correspond to only one private key (with one minor exception, which I will explain below). With Elliptic Curve systems, the private key is an integer $d$ between 1 and $q$ (the order the generator point $G$), and ...

4

The arithmetic done during a point addition is done using the addition and multiplication operations in the field; when you are using a prime field, that is equivalent to doing addition and multiplication modulo the prime (23 in this case). Now, because we're doing the arithmetic modulo 23, we get to notice that $147 \equiv 9 \bmod 23$, and so those refer ...

4

Actually, the problem is that the above quote uses the term "discrete log" in a way that's different from what you're thinking of. When someone uses the term "discrete log", they can mean two things: A discrete log in the group $Z^*_p$; that is, given $p$, $g$ and $g^x \bmod p$, recover $x$ A discrete log in some other group; that is, given a group $G$, a ...

4

Actually, it is possible to define RSA in such a way that the RSA ciphertexts are indistinguishable from random bit strings of the same length. The method is quite simple: When you select the RSA key, you deliberately pick a modulus that is just under a power of 256; for example, if you are generating a 2048 bit key, you select a modulus between $2^{2048} ... 4 You are using the wrong value as the modulus; you ought to be using the value$r$(which is also listed in the document).$p$is the characteristic of the field that the elliptic curve you're using is defined on. In this case, we're not interested in that; instead what we're interested in is the order of the curve, that is, that value$r$such that$rP = ...

4

The curve equation $Y^2=X^3+AX+B$ is traditional because it greatly simplifies a lot of theory. I like to use it for teaching. But all of these curve equations are in a sense equivalent, and for any smooth cubic curve, you can usually find an isomorphic curve of desired form. However, a long time ago, people realized that different curve equations have ...

4

The number of points on the curve $|E({\mathbb F}_p)|$ is defined as $|E({\mathbb F}_p)|=p+1-t$ where $t$ is the so called trace of Frobenius. Using Hasse's theorem one can bound $t$ as $|t| \leq 2\sqrt p$, which gives you an estimation for the number of points for $E({\mathbb F}_p)$. Now you could use a naive algorithm and simply run through all elements ...

4

Type-1 (symmetric pairings) are dead for curves over fields of small characteristic. Over prime fields of large prime characteristic they are not really dead, but as they only offer small embedding degrees ($k=2$), they are not really attractive from a performance point of view. You have to choose very large curves (which makes the curve arithmetic slow) ...

4

Designing such signature schemes from scratch without having strong experience is very likely to fail and very dangerous (see the tons of bad papers out there being accepted to "dubious" conferences and journals). Your proposed scheme Your verification relation is to check if: $sP - Q + R \stackrel{?}{=} zP + mP$ where $Q$ is the public key of the signer ...

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