Monero Asked on August 24, 2021
Refer to ZtM2 chp3.5, we can see that in step 5
rπ = α − cπ * kπ (mod l)
It’s fine for public-key-ring part because there is private key kπ.
But what if it is commitment-ring? How to do with no kπ?
Please help to answer it. Thanks!
#Thanks for the reminder from @Koe.
Refer to ZtM2 chp 5.4...
The kπ here for "commitment-ring" should be xj-xj'
So that the challenge is
cπ+1 = Hn(m, Kπ, [α1G], [α1*Hp(K)], commitment_to_zero, [α2G])
Where commitment_to_zero can be derived fromxj-xj'
.
Correct answer by Mooooo on August 24, 2021
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