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07:10 < Hansie> Hi there.  In Elliptic Curve Cryptography, is it possible to verify if a Pedersen commitment `vH + kG` only has an element on `G` and nothing on `H`, thus with `v = 0`, by just evaluating the commitment on face value, without trying to solve for `k`? Commitment bases `G` and `H` are known. (`v` is the value and `k` is the blinding factor.) I know t
07:10 < Hansie> he Pedersen commitment is computationally binding and perfectly hiding.
07:13 < sarang> If you don't want to reveal `k` you can sign with it, since a commitment to zero is a public key
07:13 < Madars_> yes. if dlog holds, then if you know a decommitment of the form cm = v*H, it is computationally infeasible to find another one (as that would break the dlog). but of course for *every* v', there exists a k' s.t. cm = v' * G + k' * H, so "have nothing on H" only has a computational meaning
07:19 < Hansie> Thanks, but I am not sure I understand the answer. Is it feasible to verify in polynomial time if `v=0`?
07:20 < Hansie> ... without knowing `k`?
07:21 < sarang> When you say "verify" do you mean that a prover wishes to prove that her commitment is to `v=0` without revealing the blinder `k`, such that a verifier can be convinced of this without attempting to brute-force blinders?
07:22 < Hansie> Yes, thats it!
07:23 < sarang> In the case, the prover's commitment is of the form `C=kG` and the prover can simply sign an appropriate message with `k`
07:23 < sarang> If the discrete log between the two Pedersen generators is unknown, the prover cannot do this with nonzero value except negligibly
07:25 < sarang> A successful signature (and properly constructed message involving `C`, etc.) convinces the verifier that the commitment was to zero
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07:30 < adiabat> digi_james: thanks, I can answer here, I dunno if it's too noisy / OT as I also made #utreexo (which nobody uses :)
07:30 < adiabat> 1) just the inclusion proofs are enough to perform deletion operations, even if you only have roots
07:31 < Hansie> sarang: Ok, so if we have `C = 0H + (k_1G+k_2G+k_3G)` where 3 parties each hold their `k_n` value private but are not present in this final Tx, would it still be possible? The 3 parties can sign an appropriate message with their respective `k_n` when those commitments are created.
07:31 < adiabat> for addition operations, no proofs are needed, just the roots and the elements to add
07:32 < adiabat> 2 & 3) so far I've done everything in RAM and don't have serialization code for partial / sparse forests.  It seems like it'd work OK on disk but would be slower.
07:35 < sarang> Hansie: I suppose at that point it's basically a multisignature on the combined key... but I know that andytoshi and friends have been finding all the possible ways that multisignatures can go horribly wrong :D
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07:44 < Hansie> sarang: So if we can prove that each individual `C_n=0H+k_nG` is a commitment to zero with an appropriate signature, would it be enough to additionally verify that `C=C_1+C_2+C_3` and in so doing prove that `C = 0H + (k_1G+k_2G+k_3G)`?
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07:47 < sarang> Are the individual commitments `C_n` provided separately to the verifier?
07:47 < Hansie> Yes, the verifier can have access to those commitments.
07:48 < sarang> If each prover signs for their own commitment to zero, then of course the sum is also a commitment to zero
07:48 < sarang> (this of course only holds if each prover has a commitment _to zero_ on their own)
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07:51 < Hansie> Yes, and thank you sarang, this scheme is the answer I was looking for. I basically want to add a bunch of commitments together and prove that the result is a commitment to `0` without the owners of those commitments being present.
07:53 < sarang> Wait, are the individual commitments _not_ to zero?
07:53 < sarang> The individual signers cannot use their commitments as public keys if they are not separately zero-valued
07:54 < Hansie> So `(v_1H+k_1G) + (v_2H+k_2G) - (v3H+k_3G) = 0H + (k_1G+k_2G-k_3G)`
07:54 < sarang> In the case of the equation you listed, individual signatures don't work
07:54 < sarang> because `v_1H+k_1G` is not a public key with known private key unless `v_1 = 0`
07:55 < aj> sarang: you can do schnorr signatures of k_1G, k_2G, k_3G and add them up though
07:55 < sarang> If some outside entity knew the masks, they could sign for the sum/difference... but at that point they could simply brute-force the values and recover all commitments
07:56 < sarang> aj: wouldn't that require appropriate precommitments and such?
07:56 < sarang> Which wouldn't work non-interactively?
07:56 < aj> sarang: i think you just wouldn't be able to tell the difference between "it doesn't add up to 0H after all" and "someone's not following the protocol"?
07:57 < Hansie> I thought the parties could construct their zero commitments with `k_n` and appropriate signatures when they create the actual commitments.
07:58 < sarang> Hansie: I think I'm confused about the structure of your individual commitments... you said "zero commitments" which I assume are not the same as whatever "usual commitments" to non-zero values the signers are working with
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08:00 < Hansie> Yes, someone creates a normal commitment, and at the same time a commitment to zero with the same blinding factor and signature to prove it
08:00 < sarang> If you're including both commitments (to zero and to non-zero) it's trivial to brute-force the value if it's in a limited range
08:00 < sarang> Not sure if that's important to your scheme or not
08:03 < Hansie> Confidentiality is important but can be sacrificed if needed.
08:03 < sarang> You implied that your goal is to make this non-interactive, such that any third party can perform this aggregation-to-zero?
08:05 < Hansie> Yes, that is the goal. The additional metadata can be kept secret by the 3rd party until needed.
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08:05 < Hansie> Actually not any 3rd party, but a semitrusted 3rd party.
08:05 < sarang> How much trust is semitrust? :)
08:07 < Hansie> They will be able to brute force the values before the final transaction is posted, but never the blinding factors.
08:08 < Hansie> Trusted not to reveal that information
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15:34 < roasbeef> https://eprint.iacr.org/2019/768
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16:00 < nsh> "Abstract: We show how to perform a full-threshold n-party actively secure MPC protocol over a subgroup of order p of an elliptic curve group E(K). This is done by utilizing a full-threshold n-party actively secure MPC protocol over Fp in the pre-processing model (such as SPDZ), and then locally mapping the Beaver triples from this protocol into equivalent triples for the elliptic curve. This allows us to transform essentially {\em any} one-party
16:00 < nsh> protocol over an elliptic curve, into an n-party one. As an example we show how to transform the shuffle protocol of Abe into an n-party protocol. This application requires us to also give an MPC protocol to derive the switches in a Waksman network from a generic permutation, which may be of independent interest."
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--- Log closed Thu Jul 04 00:00:00 2019