Timeline for How to create a 1-of-2 multisig with P2TR?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| May 23, 2023 at 17:20 | comment | added | Sachin Meier | To be clear, the 0x00 byte indicates an empty vector. It's the length byte, not actual data. so the unlocking part of the stack would be [sig] [] or [] [sig]. And yes, [sig] [sig] should fail. You could easily change the script to accept >= N signatures but why waste data. | |
| May 13, 2023 at 15:44 | comment | added | Peter | @sachin To clarify, the unlocking script for this would be either [00] [sig] or [sig] [00], and the script would fail if [sig0] [sig1] is provided (regardless if one or both signatures were valid). Correct? | |
| Mar 11, 2023 at 17:58 | comment | added | Pieter Wuille | Even then it's better to split the options up into two individual single key options, and either turn both into leaves, or one into a leaf and the other into a key path. But fair enough, in that case it's less clear cut. | |
| Mar 11, 2023 at 15:50 | comment | added | Sachin Meier | Yes this is definitely the better way to go if the only spend option you want is a 1-of-2. I think my answer is still valuable in the case that someone wants a 1-of-2 multisig as only 1 in a set of possible spend conditions. | |
| Mar 8, 2023 at 17:55 | comment | added | Pieter Wuille | That seems a bit silly; you can instead pick one of the keys as internal key, and use the other one in a "<key> OP_CHECKSIG" script. That's strictly cheaper to spend. | |
| Mar 8, 2023 at 17:38 | history | answered | Sachin Meier | CC BY-SA 4.0 |