That awkward moment you realize the person your hugging is conceal carrying

That’s interesting … how many sats a day on average!? Are you doing that!?

Interesting. Though, I need more sodium than most since I sweat a ton at work.

Oof. I just had the weirdest cramp ever:under my jaw.

I really do think I'm deficient in potassium and magnesium. (No, I'm not going to a fiat doctor to get that checked.)

Can anyone recommend a potassium supplement that's not full of crud? No, I don't want to eat a banana since that's off the table until next year at the earliest.

#cramps #potassium #magnesium #asknostr

Some people are so toxic they'll see us saying gm to each other and call it right wing extremism

Unless I ever qualify for free electricity, they both loose money, just my nodes loose less.

I’m thinking of getting solar panels so I can loose even more money by getting free electricity when it’s not raining for around £20K 😂

Yes

But at least my node earns more than my Bitaxe miner 😂

A train of thought coming from: "we can't do ZKPs using sigma protocols or bulletproofs on chain, because we only have ECC operations against one generator (G)":

Imagine one of the very simplest things you can do with 2 generators: proof of discrete log equivalence or DLEQ for short. This is basically the simplest possible generalization of Schnorr; one secret, two bases. This requires two ephemeral commitments (the nonce k mult. the 2 generators) and only one scalar response (one "signature", if you like).

Of course, you can't verify such a thing on chain; you can't check sJ = R_2 + hash(..)Q where Q = xJ and P = xG. You can only verify the first version, w.r.t. G: sG = R + eP, which is ofc standard Schnorr, here bip340.

But .. is there a way adaptors could help? Because an adaptor is something you verify offchain, and so it isn't restricted to one base/ generator.

Imagine your goal is "unlock a payment only if P and Q have the same DL wrt G, J" (yes I know, bizarre, but I'm fumbling towards the first sentence, here).

Obviously if we're including "verify something offchain before taking an action", we'll find it sanest to stick to a 2 party protocol first. Let's see:

1. Alice sends Bob P, Q, R1, R2 (having calculated from a random k, R1 = kG, R2 =kJ, and already knowing x_a s.t. Q=x_a J and P=x_a G); her private key is x_a

2. Alice pays into a 2-2 multisig with Bob, where her key is P, and the aggregate key is Pagg, with U1. When Alice signs it (not yet), the signature will look like:

s_a = k1 + H(R_agg + T, Pagg, txmsg_U1)x_a

Notice the +T which will be for an adaptor that Alice adds. This process is already well understood for Schnorr multisig. I am glossing over Musig related details, claiming they don't change the following.

3. Alice provides an adaptor over J for ``spending'' U1. Before writing this out, consider: we want the s-value to be the same over J as it is over G. This is problematic, because the s value comes from a structure that is basically : s = k + e*x, where e is the hash, and as per bip340, that hash *must* be exactly: H(R, P, m) where P is the onchain key, R is the onchain nonce and m is the actual transaction message. So if we do things naively, i.e. make an adaptor over the J base like:

s'J = R2 + H(R2+T, Q, message)Q

.. then we're going to fail to create the desired scenario: that the real signature 's' is the same between the J and G bases. In order to solve this we have to play a trick. We don't change the nonce or public key fields in the hash at all, i.e. we do:

s' = k2 + H(R_agg + T, Pagg, txmsg_U1)x

and have Bob verify:

s'J =?= R2 + H(R_agg + T, Pagg, txmsg_U1)Q

Superficially that is unsound; the reason is that such signatures can be forged if the corresponding nonce point R2 is not fixed into the hash/FS challenge, which it appears it isn't (also not "key-prefixing" by including the public key, here Q, in the same challenge is, at the least, problematic). However this can be solved, albeit in an unorthodox way: include both R2 and Q, which are curve points, as data inside the bitcoin transaction! (spending U1). It will just be "dead" data (e.g. OP_DROP) because you won't be able to sign against them; they use the wrong basepoint, but it is still easy).

4. Bob verifies this adaptor using (to repeat): s'J =?= R2 + H(R_agg + T, Pagg, txmsg_U1)Q.

5. Now Bob knows that if he sees the valid s for this equation, he gets the private key t of T w.r.t. J. Importantly, he also knows that if that one s corresponds to the s that is needed to spend U1, then Q and P have the discrete log equivalence property we are proving. But simply convincing Bob of a fact like that is uninteresting in itself; you can do that by simply directly sending him ZKPs of various types. So, :

6. Alice also provides offchain a DLEQ proof that T = tJ and T2 = tG

7. Alice makes another adaptor on a payment of another utxo U2, using the same t: s'_2 = k_3 + H(R_3 + T2, P2, m)P2.

8. Bob now knows that if U1 is spent, it reveals s, proves that DL Q/J = P/G, and that he can spend U2 using s_2 = s'_2 + t, having extracted t from s - s', and knowing that that t value corresponds to T2.

This could be seen as "just an adaptor based atomic swap with a bunch of extra steps", but I'd call it more "atomic swap gated by DLEQ", i.e. it will fail unless Q/J = P/G holds.

NO ONE KNOWS WHAT BITCOIN PRICE WILL DO

Cleanse yourself from doom and hopium, ignore the chart bros, everything comes at its own time.