rich1.0npub1g5zswf6y48f7fy90jf3tlcuwdmjn8znhzaa4vkmtxaeskca8hpss23ms3lhttps://nostr.ae/npub1g5zswf6y48f7fy90jf3tlcuwdmjn8znhzaa4vkmtxaeskca8hpss23ms3lnjumphttps://nostr.ae📅 Original date posted:2020-06-20 📝 Original message: Good morning Dave, > ZmnSCPxj noted that pay-to-preimage doesn't work with PTLCs.[2] I was > hoping one of Bitcoin's several inventive cryptographers would come > along and describe how someone with an adaptor signature could use that > information to create a pubkey that could be put into a transaction with > a second output that OP_RETURN included the serialized adaptor > signature. The pubkey would be designed to be spendable by anyone with > the final signature in a way that revealed the hidden value to the > pubkey's creator, allowing them to resolve the PTLC. But if that's > fundamentally not possible, I think we could advocate for making > pay-to-revealed-adaptor-signature possible using something like > OP_CHECKSIGFROMSTACK.[3] Not a cryptographer, I just play one on the Internet, but maybe the pay-for-signature construction could work...? Assuming a PTLC has a pointlocked branch, which involves signing with MuSig(A, B). A offers to B the amount if B reveals the secret `t` behind `T = t * G`; A knows `T` but not `t`. This is done by B handing over `R[B]` and `s'[B]`: R = R[A] + R[B] + T s'[B] = r[B] + h(MuSig(A, B) | R | m) * b Then A provides its partial signature to B. s[A] = r[A] + h(MuSig(A, B) | R | m) * a B has to complete the signature by: s = s[A] + s'[B] + t Since A knows both `s[A]` and `s'[B]`, once it knows `s`, it can compute `t`. Now, we can massage the equation for `s`: s = r[A] + h(MuSig(A, B) | R | m) * a + r[B] + h(MuSig(A, B) | R | m) * b + t ; multiply both sides by G s * G = r[A] * G + h(MuSig(A, B) | R | m) * a * G + r[B] * G + h(MuSig(A, B) | R | m) * b * G + t * G ; replace with public points s * G = R[A] + h(MuSig(A, B) | R | m) * A + R[B] + h(MuSig(A, B) | R | m) * B + T Note that A can compute `s * G` above, because it generated `R[A]`, was given `R[B]` and `T`, and knows who `A` and `B` are. So what A needs to do is to offer a fund that can only be claimed by leaking knowledge of `s` behind `s * G`. A can do this by creating a new keypair `A[p4s] = a[p4s] * G` and putting a fund into it. Then A generates an `R[A][p4s] = r[A][p4s] * G`, and computes: R[p4s] = R[A][p4s] + s * G s'[A][p4s] = r[A][p4s] + h(A | R[p4s] | m) * a[p4s] The signed message could be a signature to `SIGHASH_NONE`, finally an actual use for that flag. A reveals publicly (in an `OP_RETURN` as you suggest): * `R[A][p4s]` * `s * G` * `s'[A][p4s]` * `A[p4s]` - Already the Schnorr output pubkey. In order to complete the above signature, a third party C has to learn `s` from B. The third party has to scan every onchain 1-of-1 signature for an `s` that matches `s * G`, so there is greater processing (point multiplies are more expensive than hashes, also there are more 1-of-1s). But once learned, the third party can complete the signature and claim the funds. And A then learns `s`, from which it can derive `t`. The third party learns about which channel (i.e. the UTXO that was spent to create the PTLC in the first place), but never learns `t` or `T`, which is a small but nice privacy bonus. Regards, ZmnSCPxj