Small fixes #18
Small fixes #18
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| @@ -1566,9 +1566,9 @@ | |||
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| But for the case $600 = 10 \times 10 \times 6$ - we first ``chunk'' in 10s, then | |||
| again in 10s, leaving only 6 components for the final step. That | |||
| requires revealing $2\times 10-1 = 19$ commitments at each of the two reducing | |||
| requires revealing $2\times (10 - 2) = 18$ commitments at each of the two reducing | |||
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I guess we can clarify this and the next bit of arithmetic via #19 but also 2 x (10-2) is not 18 :)
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My true (in)ability is exposed!
| @@ -1863,7 +1863,7 @@ | |||
| receive back a challenge $x$, both sides recalculate $C'$, continue until a | |||
| final step (each step a halving and a new $L, R$), and in the last step reveal | |||
| scalars for the now single values $a, b$, and the Verifier makes the final | |||
| check that $C^{*} = a^{*}b^{*}G + a^{*}G_1 + b^{*}H_1$, where * indicates the $\log_2n$-th transformed values. | |||
| check that $C^{*} = (a^{*} \cdot b^{*})G + a^{*}G_1 + b^{*}H_1$, where * indicates the $\log_2n$-th transformed values. | |||
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I believe these are now single values not vectors (as per previous line), so we don't want a dot product here. Admittedly it does look really crappy using a * superscript here, though.
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You're right. When I was first reading this, I thought the protocol could end on vectors of length 2, for which you would need the dot product. I think the number of messages stays the same, but I think I invented that out of thin air.
| @@ -2143,7 +2143,7 @@ | |||
| can see that this requires: | |||
| \begin{align*} | |||
| & \textbf{H}' = \textbf{y}^{-n}\textbf{H} \\ | |||
| & P = A + xS -zG + \left(z\textbf{y}^n + z^2\textbf{2}^n\right)\textbf{H}' \\ | |||
| & P = A + xS -z\textbf{G} + \left(z\textbf{y}^n + z^2\textbf{2}^n\right)\textbf{H}' \\ | |||
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z is committed to via a single generator G. Whereas the vector (bolded)G is used as a shorthand. See eqns (2), (3), (4) earlier on.
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I think you might be confusing the inner product proof (which has zG) with the range proof (which has z\vec{G}). We need z\vec{G} to extract a_L and a_R (see "Then consider coefficients of G:").
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Thanks for this review work :) It's been many years so forgive me if I struggle sometimes to answer/address points here and there :) |
These are small errors / oddities that I noticed while going through the document. Great work, by the way.