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1

Based on a quick Google search, I assume you're reading the paper "Modified Koblitz Encoding Method for ECC" by Kodali and Sarma (Int.J.Rec.Tr.Eng.Tech., vol. 8, no. 1, Jan 2013). The fact that the paper doesn't actually cite any source for the encoding scheme, or even cite anything by Koblitz, makes me a bit skeptical of the paper's quality. For that ...


2

Yes, it is equally as difficult; if we assign: $$g' = ag$$ $$a' = a^{-1}$$ $$b' = b$$ Then the restatement of your problem is: given $g' = ag$, $a'g' = g$ and $b'g' = abg$, compute $a'b'g' = bg$, which is exactly the ECDH problem. Now, this assumes that $a$ has an inverse; this is not a problem if the curve order is a prime, and is easy to work around if ...


3

After a multiplication you have a number with $2 \cdot 255$ bits. Since $2^{255} = 19 \pmod q$, you can take the upper half, multiply it by 19 and add it to the lower half. This gives you an equivalent number smaller than $20 \cdot 2^{255}$. Repeat this to get a number that's smaller than $2 \cdot q$. Now check if the value is greater or equal to $q$ and ...


1

No. One of the most important principles of cryptography is that knowing the encryption scheme cannot help someone attempting to decrypt the material without the key. The encryption used seems to be a reasonably well-written implementation of several standard algorithms, for which no practical attacks are known. Finding a way to crack these would be a major ...


3

I know that k is what is used as the secret key and that $Q=(Q_x,Q_y)$ is the public key. But playing around on these two pages and seeing the contents of this file have letft me more confused. What is the importance of $P$ and how does it differ from $Q$? Don't get too caught up with variable names. People usually use $P$ and $Q$ to represent points in ...


1

This web page use unusual variable names, which do not match the names you are used to see in other documents in the calculator the curve equation is (Y^2 == X^3 + AX + B) mod p (many documents use a and b in lowercase) in the calculator n is any number (many documents use k, while n is usually the reserved name for the order of a point G chosen as group ...


1

The answer to the question actually goes back to the basics of cryptography. Many people confuse the two fundamental uses of cryptology: Privacy Authentication The first item, privacy, is accomplished by encryption. Encryption, however, does not guarantee that the secret message actually came from its supposed source. It does not even guarantee that the ...


3

The encryption part of NaCl is older. I think NaCl itself still doesn't have official signature support. NaCl's box uses montgomery form public keys together with the montgomery ladder. This ladder only returns the x-coordinate of the result and thus is not compatible with most signature algorithms. Ed25519 on the other hand uses (twisted) edwards form, ...


0

You need to look at the constructors and see which values correspond to which parameters. The ECCurveFp reference is here Note that $P$ is given in decimal $P = \texttt{6277101735386680763835789423207666416083908700390324961279}$ $a = \texttt{fffffffffffffffffffffffffffffffefffffffffffffffc}$ $b = ...


2

In short: the question does not explain well the notion of asymmetry in ECC; and the exposition is not how Elliptic Curve Cryptography works. A reasoning sidestepping the notion of Discrete Logarithm Problem over a finite group can not really explain asymmetry as meant in ECC. Asymmetry is in the knowledge Alice and Bob have about the key, not asymmetry of ...


5

Update: My previous answer, although technically true, didn't answer your question. The issue is that the strong-DDH is not hard when using pairing groups, so my answer was merely stating that your problem is at least as hard as an easy problem (duh!) After some thought, I realized your problem is at least as hard as the 2-weak Bilinear Diffie-Hellman ...


2

In fact you don't need algebra to see this. Only use the geometrical interpretation of the addition.


1

Given that $N$ and $P$ are prime, one obvious way to do this is to select a random value $g$ from $[1, N-1]$, and compute $g^{(N-1)/3} \bmod N$; assuming that $N \equiv 1 \pmod{3}$, this resulting value will either be 1, the displayed value of $\lambda$, or $N-\lambda-1$ (with equal probabilities of each). If $N \not\equiv 1 \pmod{3}$, then the only modular ...



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