I am trying to understand this article. Can someone explain to me how the ECDSA adaptor signing is work?

From the article:

1. ECDSA adaptor signing $$s' = (H(m) + R t p)r^{-1}$$

As I understand this is standard formula where $x = t p$ - is a private key for the signature s'. So public key is $P = t p G$

2. Decryption ECDSA adaptor signature: $$s = s' t^{-1} = (H(m) + R t p)(rt)^{-1}$$

I can't understand this. Because it follows that: $$s = s' t^{-1} = (H(m) + R t p)(rt)^{-1} = (H(m) t^{-1} + R p)r^{-1}$$

But this is not standard formula for ECDSA signature (( It should be $$s = (H(m) + R p)r^{-1}$$

I don't think the article is wrong. Most likely I'm missing something. Can anyone help with understanding these formulas?

  • $\begingroup$ I'm confused too. The article claims $R'=Rt=rT$, where $T=tG$, which would only make sense if $R=rG$. But $R$ is not $rG$ and is not a point, $R$ is the x-coord of the point $rG$. $\endgroup$
    – knaccc
    Apr 2, 2022 at 10:50


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