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2

It all comes down to your threat model, right? Just because an implementation is done in hardware does not mean that power and fault attacks must be considered. If I host the hardware in my secure facility with armed guards at the door, but the hardware is connected to a machine which is connected to the internet, I might feel that it is okay to not be ...


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What's the meaning of 160 bit Curve in Elliptic curve? It's the size of the field, as the other answers explain. It affects the size of keys and signatures (which can be equal to the number or e.g. double it), as well as security and performance. Why they don't say 100 bit curve? 100-bit curves are too small. It's also a convention. Other ...


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It refers to the bit length of the order of the elliptic curve group (number of points on the curve). The security of a curve is equal to the square root of its order. So we pick values that are twice as long as the security level we want (160-bit for 80-bit security, 256-bit for 128-bit security, etc) We don't use 100-bit curves because we prefer security ...


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An elliptic curve to be used in cryptography is defined over a finite field $\mathbb{F}_p$ (but also can be a binary polynomial finite field $\mathbb{F}_{2^m}$). The bit length of this $p$, or the $m$ in the case of the binary polinomials, is what is used later to describe the elliptic curve in terms of size.


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When you go from Affine to Jacobian, $X$ and $Y$ stay the same, and $Z$ is equal to $1$ Affine -> Jacobian: $(X',Y',Z') = (X,Y,1)$ Jacobian -> Affine: $(X',Y') = (\frac{X}{Z^2}, \frac{Y}{Z^3} )$


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The sect curves are curves over a binary field. From SEC 2: Recommended Elliptic Curve Domain Parameters (chapter 3): The example elliptic curve domain parameters over $\mathbb{F}_{2^m}$ have been given nicknames to enable them to be easily identified. The nicknames were chosen as follows. Each name begins with sec to denote ‘Standards for Efficient ...


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Basically you are talking about “superencipherment”. This has a long history in cyptography, but it eats up space like crazy. Your observation of PK cyptography being vulnerable is true. Solving 'Prime' would bring the house down.


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They rely on problems not so different as you might think. They are based either in the factoring problem or in the discrete logarithm problem, which have a deep connection between each other. Once you have an algorithm that can efficiently solve one, you most likely would be able to adapt it to reproduce an answer for the other in polynomial time. Thus ...


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, ...


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I don't blame whomever down-voted this, I realize now it was a stupid question. In order to do this, first convert the Edwards curve to Montgomery form, then to Weierstrass form, per this thread. But, as the above thread explains, as well as CodesInChaos' comment, this will result in poor performance. Thanks to CodesInChaos for pointing me in the right ...


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If an attacker can choose the points $P_i$, than this system is not semantically secure. For example, they may choose $P_2=2P_1$, and the corresponding encryption $Q_2$ would be equal to $2Q_1$. If the points are chosen at random, this system is semantically secure if decisional Diffie-Hellman assumption holds for the curve. This assumption is presumed to ...



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