-2
$\begingroup$

In Elliptic Curve Digital Signature Algorithm (ECDSA) I often see 2 different written equations of it:

  1. Elliptic curve point multiplication by a scalar, $Q_{A}=d_{A}\times G$, source
  2. Modular exponentiation, $y:=g^{x}\mod p$, source

Both methods work other than one is multiplication the other is exponent. Are there any advantage doing one or another?

$\endgroup$
1
  • $\begingroup$ Did you read the Wikipages properlty? $\endgroup$
    – kelalaka
    Dec 18, 2019 at 14:40

1 Answer 1

3
$\begingroup$

You are referring to two different protocols.

The second source is linked to the DSA (Digital Signature algorithm). This uses modular exponentiation in a group of prime order over the integers.

The first one is a version of the DSA over Elliptic curves, namely ECDSA (Elliptic Curve Digital Signature Algorithm).

They basically work the same. You have a group, with a defined operation, and you work on elements of this group with scalars. The difference between classical DSA and ECDSA is that the first one has multiplication as a group law (hence $g \times g \times \ldots \times g = g^k$ where k is the number of multiplications) whereas the second one has addition law (for a point $P$ of the curve, $P+P+\ldots+P = kP$)

$\endgroup$
1
  • 1
    $\begingroup$ Yes. Further, some authors note the group used in ECDSA multiplicatively, like $h=g^k$ (without modulo), or $Q=k⋅G$, or $Q=k\,G$, or $Q=kG$ (as this answer does in its last equation), or $Q=[k]G$, where the first source in the question uses $Q=k\times G$. In all cases, that denotes obtaining the result of (virtually) combining $k$ copies of $g$ or $G$ using $k-1$ group operations (sometime with the convention that $k=0$ yields the group's neutral as result, or/and a negative $k$ is first made positive by reducing $k$ modulo the order). $\endgroup$
    – fgrieu
    Dec 18, 2019 at 17:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.