Jim Posen <jim.po...@gmail.com> writes:
> I'm still not following why this doesn't accumulate.
> In the example route, let's look at it from the point of view of C. C sees
> the following regardless of whether D or E or someone behind E is the last
> hop in the route:
> B -> HTLC(expire = X + delta) -> C -> HTLC(expire = X) -> D
> So D is not required to reveal the preimage before time X, and in the case
> of an on-chain settle, C needs to be able to redeem the HTLC output through
> the timeout clause before time X + delta. C can't redeem the HTLC (with
> sufficient confirmations) at least until the settlement transaction is
> confirmed. So it seems to me that regardless of the overall route and the
> maximum CSV on it, the delta for the C hop has to be greater than the CSV
> delay on the update transaction. And that this must be true at every hop
> for the same reason.

That'd be a purely reactionary behavior, i.e., chosing the delta in such
a way that I can both settle the channel and have enough time to react
to turn around and reveal the preimage. So with the assumptions we had
before (CSV = 144 and CLTV delta = 144) you'd have an effective delta of
288 on each hop, yes. That's basically the case in which each channel
reacts serially.

You can trivially parallelize these closures by looking ahead and
noticing that each hop really just cares about its own closure deadline,
i.e., each node just cares to close 288 blocks before the CLTV expires,
not that its delta w.r.t. to the downstream channel is that far in the
future. So all we care about is that once we are due to give the
upstream hop the preimage we've already closed the downstream channel
and can now read the HTLC preimage from that channel.

The CSV timeout isn't part of the delta on each hop, but we need to
implement the deadline computation as:

CLTV - CLTV delta - CSV

instead of LN-penaltiy's

CLTV - CLTV delta
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