1
$\begingroup$

I am doing a small project using elliptic curve in cryptography. My doubt is, can I directly convert a projective to a Jacobian coordinate system without using the affine conversion in elliptic curve cryptography? Actually, I am stuck at this point…

$\endgroup$
1
$\begingroup$

If the question is how to take a point in projective coordinates (a triplet $(A,B,C)$, where the corresponding affine coordinate is $(AC^{-1}, BC^{-1})$), and you want to convert it into the same point in Jacobian coordinates $(D, E, F)$ (where the affine coordinate is $(DF^{-2}, EF^{-3})$), I believe this works:

$$D = AC$$

$$E = BC^2$$

$$F = C$$

We can easily see that $DF^{-2} = AC^{-1}$ and $EF^{-3} = BC^{-1}$, hence $(D,E,F)$ refers to the same point as the original, and this conversion takes only three multiplies.

$\endgroup$
  • $\begingroup$ Thank you @poncho. To reverse the operation will the below equations work? $$ A = DF $$ $$ B = E $$ $$ C = F^{3} $$ $\endgroup$ – vijita Jun 30 '14 at 8:39
  • $\begingroup$ @vijita: that reversed operation also preserves $DF^{-2}=AC^{-1}$ and $EF^{-3} = BC^{-1}$, and so, yes, those equations also "works". $\endgroup$ – poncho Jun 30 '14 at 12:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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