Date: Fri, 16 Apr 2021 03:47:45 +0000
To: Lloyd Fournier <lloyd.fourn@gmail.com>,
 Bitcoin Protocol Discussion <bitcoin-dev@lists.linuxfoundation.org>
From: ZmnSCPxj <ZmnSCPxj@protonmail.com>
Reply-To: ZmnSCPxj <ZmnSCPxj@protonmail.com>
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Subject: Re: [bitcoin-dev] PSA: Taproot loss of quantum protections
Precedence: list

Good morning LL,

> On Tue, 16 Mar 2021 at 11:25, David A. Harding via bitcoin-dev <bitcoin-d=
ev@lists.linuxfoundation.org> wrote:
>
> > I curious about whether anyone informed about ECC and QC
> > knows how to create output scripts with lower difficulty that could be
> > used to measure the progress of QC-based EC key cracking.=C2=A0 E.g.,
> > NUMS-based ECDSA- or taproot-compatible scripts with a security strengt=
h
> > equivalent to 80, 96, and 112 bit security.
>
> Hi Dave,
>
> This is actually relatively easy if you are willing to use a trusted setu=
p. The trusted party takes a secp256k1 secret key and verifiably encrypt it=
 under a NUMS public key from the weaker group. Therefore if you can crack =
the weaker group's public key you get the secp256k1 secret key. Camenisch-D=
amgard[1] cut-and-choose verifiable encryption works here.
> People then pay the secp256k1 public key funds to create the bounty. As l=
ong as the trusted party deletes the secret key afterwards the scheme is se=
cure.
>
> Splitting the trusted setup among several parties where only one of them =
needs to be honest looks doable but would take some engineering and analysi=
s work.

To simplify this, perhaps `OP_CHECKMULTISIG` is sufficient?
Simply have the N parties generate individual private keys, encrypt each of=
 them with the NUMS pubkey from the weaker group, then pay out to an N-of-N=
 `OP_CHECKMULTISIG` address of all the participants.
Then a single honest participant is enough to ensure security of the bounty=
.

Knowing the privkey from the weaker groups would then be enough to extract =
all of the SECP256K1 privkeys that would unlock the funds in Bitcoin.

This should reduce the need for analysis and engineering.

Regards,
ZmnSCPxj