[bitcoin-dev] Multiparty signatures
Erik Aronesty
erik at q32.com
Thu Jul 26 02:05:05 UTC 2018
Also we don't need any new opcodes to support this. Done right this could
literally go out into clients immediately.
On Fri, Jul 20, 2018, 4:18 PM Erik Aronesty <erik at q32.com> wrote:
> Sorry there were typos:
>
> - Using MuSig's solution for the blinding factor (e)
> - Using interpolation to enhance MuSig to be M of N instead of M of M
>
> References:
>
> - MuSig
> https://blockstream.com/2018/01/23/musig-key-aggregation-schnorr-signatures.html
> - HomPrf http://crypto.stanford.edu/~dabo/papers/homprf.pdf (sections
> 7.1 and 7.4)
>
> Each party:
>
> 1. Publishes public key G*xi, G*ki, where ki is a random nonce
> 3. Xi = H(G*xi) ... Xi is the parties x coordinate, for the purposes of
> interpolation
> 3. R = G*k = via interpolation of r1=Gk1, r2=Gk2... (see HomPrf)
> 4. L = H(X1,X2,…) (see MuSig)
> 5. X = sum of all H(L,Xi)Xi (see MuSig)
> 6. Computes e = H(R | M | X) .... standard schnorr e... not a share
> 7. Computes si = ki *e+ xi * e ... where si is a "share" of the sig, and
> xi is the private data, and e is the blinding factor
> 8. Publishes (si, e) as the share sig
>
> If an attacker has multiple devices, e is safe, because of the musig
> construction.
>
> But what protects k from the same multiparty birthday attack?
>
> If an attacker has multiple devices, by carefully controlling the
> selection of private keys, the attacker can try to solve
> the polynomial equation to force the selection of a "known k".
>
> A "known k" would allow an attacker to sign messages on his own.
>
> To fix this, we need to somehow "blind k as well".
>
> Does this work?
>
> The revision below seems to solve this problem.
>
> 1. Publishes public key G*xi, G*ki, where ki is a random nonce
> 3. Xi = H(G*xi) ... Xi is the parties x coordinate, for the purposes of
> interpolation
> 3. R = G*k = via interpolation of r1=Gk1, r2=Gk2... (see HomPrf)
> 4. L = H(X1,X2,…) (see MuSig)
> 5. L2 = H2(XN,XN-1,…) (see MuSig... H2 is a "second hash")
> 6. X = sum of all H(L,Xi)Xi (see MuSig)
> 7. Computes e = H(R | M | X) .... standard schnorr e... not a share
> 8. Computes e2 = H(R | M | X2) ... a second blinding factor
> 9. Computes si = ki *e2 + xi * e ... where si is a "share" of the sig, and
> xi is the private data, and e, e2 are blinding factors
> 10. Publishes (si, e, e2) as the share sig
>
> The final signature is computed via interpolation, and e2 is can be
> subtracted to recover a "normal" schnor sig for the set of participants.
>
> Now there's no mechanism for a birthday attack on k.
>
>
>
> On Fri, Jul 20, 2018 at 1:34 PM, Erik Aronesty <erik at q32.com> wrote:
>
>> Hi, thanks for all the help. I'm going to summarize again, and see if
>> we've arrived at the correct solution for an M of N "single sig" extension
>> of MuSig, which I think we have.
>>
>> - Using MuSig's solution for the blinding to solve the Wagner attack
>> - Using interpolation to enhance MuSig to be M of N instead of M of M
>>
>> References:
>>
>> - MuSig
>> https://blockstream.com/2018/01/23/musig-key-aggregation-schnorr-signatures.html
>> - HomPrf http://crypto.stanford.edu/~dabo/papers/homprf.pdf (sections
>> 7.1 and 7.4)
>>
>> Each party:
>>
>> 1. Publishes public key G*xi
>> 3. Xi = H(G*xi) ... Xi is the parties x coordinate, for the purposes of
>> interpolation
>> 3. r = G*x = via interpolation of Gx1, Gx2... (see HomPrf)
>> 4. L = H(X1,X2,…) (see MuSig)
>> 5. X = sum of all H(L,Xi)Xi (see MuSig)
>> 6. Computes e = H(r | M | X) .... standard schnorr e... not a share
>> 7. Computes si = xi - xe ... where si is a "share" of the sig, and xi is
>> the private data
>> 8. Publishes (si, e, G*Xi)
>>
>> Any party can then derive s from m of n shares, by interpolating, not
>> adding.
>>
>>
>>
>>
>
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