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!

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    It's the commitment to zero z_j, from section 6.2.2
    – koe
    Commented Jun 23, 2020 at 15:47
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    Yeah, the commitment has three components (plus the message and key prefixing) c_pi = H(m, (K_pi), [alpha_k G], [alpha_k H_p(K_pi)], (C_pi - C'_pi), [alpha_z G]). r_k_pi = alpha_k - c_pi*k_pi and r_z = alpha_z - c_pi*z. You can find discussion of the commitment to zero in section 5.4
    – koe
    Commented Jun 23, 2020 at 16:55
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    Thanks a lot @koe!! Your point is right. BTW, I think c_pi is called challenge not commitment. :)
    – Mooooo
    Commented Jun 25, 2020 at 8:46
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    @Mooooo I'm not quite sure why you keep referring to this as a "commitment-ring"? The ring being signed is a ring of keys, not commitments. Even the commitment to zero is really just another key. Anyway, looks like you've found what you were after.
    – jtgrassie
    Commented Jun 26, 2020 at 2:32
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    @jtgrassie MLSAG is a bit like two knotted bracelets glued together and tied at the same point, so in that sense one bracelet is the set of one-time address keys, and the other bracelet is the corresponding set of commitment to zero keys. I believe Mooooo is using 'ring' loosely to mean what I mean by bracelet. The entire set of keys is technically the 'ring', and there is no technical term for the ring's layers.
    – koe
    Commented Jun 26, 2020 at 19:37

1 Answer 1


#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'.

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