Score:0

How to prove that an elliptic curve point is smaller or greater than half of the curve's order?

cn flag

Is it possible to tell if a point on an elliptic curve is less than half of the curve's order?

If I have a point $ = [a]$ on a curve with prime order q, is there an efficient way to know that $a < q/2$?

I understand that range proofs would work for this, but is there a quicker way? Specifically, I am working with secp256k1, but any advice is greatly appreciated.

kelalaka avatar
in flag
Are you looking for this [Proving that two points on elliptic curve are within range](https://crypto.stackexchange.com/q/66058/18298) Especially [Bullet Proofs](https://eprint.iacr.org/2017/1066.pdf)
Score:2
ru flag

If there were a polynomial time solution then this would provide a polynomial time solution to the elliptic curve discrete logarithm problem. We strongly believe this not to be the case.

To see the reduction to the elliptic curve discrete logarithm, suppose that I have a point $P_0=[x_0]G$ where the order of $G$ is $q$ and I wish to know $x_0$. I calculate $2^{-1}\pmod q$, calculate $[2^{-1}]P_0$ and run my magic algorithm. If the algorithm says that there exists an $a$ with $0<a<q/2$, then I know that $x_0$ is even other wise I know that is is odd. Writing $b_0$ for the low bit of $x$, I write $x_1=(x_0-b_0)/2$ and compute $P_1=[2^{-1}](P_0-[b]G)=[x_1]G$. I can now repeat the process to recover the low bit $b_1$ of $x_1$ and so on, terminating when $P_n=G$. This will take at most $\log_2 q$ steps.

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