Re: [bitcoin-dev] MATT: [demo] Optimistic execution of arbitrary programs

["Johan Torås Halseth" <johanth@gmail.com>]

Hi, aj. Thanks for taking a look!

> "O(n log n)" sounds wrong? Isn't it O(P + log(N)) where P is the size
> of the program, and N is the number of steps (rounded up to a power of 2)?

Thanks, you are right. That's a typo, it should indeed be O(log n). n
being the number of steps in the program. I think P doesn't matter
here, as we never put the whole program on-chain, just break it down
into n steps.

> > node = h( start_pc|start_i|start_x|end_pc|end_i|end_x|h( h(sub_node1)|h(sub_node2) )
> But I don't think that works -- I think you want to know h(sub_node1)
> and h(sub_node2) directly, so that you can compare them to the results
> you get if you run the computation, and choose the one that's incorrect.

This denotes only how to create the commitment. When we traverse the
tree, the node scripts enforce that h(sub_n
ode{1,2}) that is consistent with the commitment is in the witness,
achieving exactly what you suggest.

> I'm not seeing what forces the prover to come up with a balanced state
> tree

To achieve this the participants agree up front (when the contract is
created) what is the exact length of the trace (or equivalent the
depth of the tree). If the actual execution is shorter, we fill the
rest with no-ops.

This means that we know the moment the challenge protocol starts the
transactions that are going to be played (kinda like a CTV tree), so
if one of the participants creates a trace from a non-balanced state
tree, it will be rejected by the script at that level. It is indeed
important that the state tree is built in a deterministic way.

> There seems to be an error in the "what this would look like for 4 state
> transitions" diagram -- the second node should read "0|0|2 -> 0|1|4"

Yes, fixed! Thanks :)

- Johan


On Mon, Oct 2, 2023 at 5:10 PM Anthony Towns <aj@erisian.com.au> wrote:
>
> On Fri, Sep 29, 2023 at 03:14:25PM +0200, Johan Torås Halseth via bitcoin-dev wrote:
> > TLDR; Using the proposed opcode OP_CHECKCONTRACTVERIFY and OP_CAT, we
> > show to trace execution of the program `multiply` [1] and challenge
> > this computation in O(n logn) on-chain transactions:
>
> "O(n log n)" sounds wrong? Isn't it O(P + log(N)) where P is the size
> of the program, and N is the number of steps (rounded up to a power of 2)?
>
> You say:
>
> > node = h( start_pc|start_i|start_x|end_pc|end_i|end_x|h( h(sub_node1)|h(sub_node2) )
>
> But I don't think that works -- I think you want to know h(sub_node1)
> and h(sub_node2) directly, so that you can compare them to the results
> you get if you run the computation, and choose the one that's incorrect.
> Otherwise you've got a 50/50 chance of choosing the subnode that's
> actually correct, and you'll only be able to prove a mistake with
> 1/2**N odds?
>
> Not a big change, it just becomes 32B longer (and drops some h()s):
>
>   node = start_pc|start_i|start_x|end_pc|end_i|end_x|h(sub_node1)|h(sub_node2)
>   leaf = start_pc|start_i|start_x|end_pc|end_i|end_x|null
>
> I'm not seeing what forces the prover to come up with a balanced state
> tree -- if they don't have to have a balanced tree, then I think there
> are many possible trees for the same execution trace, and again it would
> become easy to hide an error somewhere the challenger can't find. Adding a
> "start_stepcount" and "end_stepcount" would probably remedy that?
>
> There seems to be an error in the "what this would look like for 4 state
> transitions" diagram -- the second node should read "0|0|2 -> 0|1|4"
> (combining its two children), not "0|0|2 -> 1|0|2" matching its left
> child.
>
> I'm presuming that the counterparty verifies they know the program (ie,
> all the leaves in the contract taptree) before agreeing to the contract
> in the first place. I think that's fine.
>
> Cheers,
> aj
>