--- Log opened Sun Jun 25 00:00:40 2023
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12:00 < ademan[m]> probably a dumb question, but how does this work? https://github.com/bitcoin/bips/pull/1458/files#diff-1b8a42137070950593140a8b67adc1df0eb2bddb3ef95f1f1ab6c0c9893872afR96 "Alice performs the tweak with the sum of her input private keys in the following manner[...]" I thought the point was this should work in a coinjoin, so she would not have access to the private keys of all inputs.
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12:18 < ademan[m]> I know it says "her input keys" but in this scheme wouldn't using a subset of input keys force bob to try all combinations of input pubkeys to scan?
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12:29 < Guest85> Hello how is everyone’s day going?
12:30 < Guest85> I have question related to a pending bitcoin deposit if anyone could talk to me about this please
12:30 < Guest85> Or call me on my cell phone I’m at work but need to figure this out please
12:31 < Guest85> I’ll put my cell number
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14:28 < Guest78> Hello
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14:43 < ademan[m]> or is the idea that in a case where alice is paying bob, she would sum her private keys, and in a coinjoin/collaborative case, we need that blind diffie hellman construct that's referenced to produce P_0 ?
15:02 < ademan[m]> ah, after reading the linked https://gist.github.com/RubenSomsen/be7a4760dd4596d06963d67baf140406 I'm assuming P_0 = B + hash(outpoints_hash * sum(a_i) * B || 0) + G is meant to be calculated as P_0 = B + hash(outponts_hash * sum(a_i * B) || 0) + G in the collaborative case?
15:02 < ademan[m]> that might be obvious to a more sophisticated reader but it wasn't obvious to me
15:03 < ademan[m]> (where a_i * B is computed with the linked protocol)
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15:20 < RubenSomsen> ademan[m]: The part of the BIP you are referring to assumes Alice controls all input keys. If she doesn't then yes, you can't just sum privkeys together. Note btw there's also a #silentpayments channel.
15:24 < RubenSomsen> The more complex blind DH steps are only required if you don't control all input keys *and* you want to maintain full privacy against the other input owners. Otherwise you can just calculate a_1*B + a_2*B ... + a_i*B, which is equivalent to sum(a_i)*B.
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17:11 < ademan[m]> thanks! so in the non-blinded way you would be revealing the recipient though, right? (since you need to reveal B to the other participants)
17:23 < ademan[m]> I suppose that still wouldn't reveal which output is going to that recipient though, huh?
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--- Log closed Mon Jun 26 00:00:41 2023