On 10/17/23, ZmnSCPxj via bitcoin-dev
<bitcoin-dev@lists.linuxfoundation.org> wrote:
Good morning Antoine et al.,
Let me try to rephrase the core of the attack.
There exists these nodes on the LN (letters `A`, `B`, and `C` are nodes,
`==` are channels):
A ===== B ===== C
`A` routes `A->B->C`.
The timelocks, for example, could be:
A->B timeelock = 144
B->C timelock = 100
The above satisfies the LN BOLT requirements, as long as `B` has a
`cltv_expiry_delta` of 44 or lower.
After `B` forwards the HTLC `B->C`, C suddenly goes offline, and all the
signed transactions --- commitment transaction and HTLC-timeout transactions
--- are "stuck" at the feerate at the time.
At block height 100, `B` notices the `B->C` HTLC timelock is expired without
`C` having claimed it, so `B` forces the `B====C` channel onchain.
However, onchain feerates have risen and the commitment transaction and
HTLC-timeout transaction do not confirm.
In the mean time, `A` is still online with `B` and updates the onchain fees
of the `A====B` channel pre-signed transactions (commitment tx and
HTLC-timeout tx) to the latest.
At block height 144, `B` is still not able to claim the `A->B` HTLC, so `A`
drops the `A====B` channel onchain.
As the fees are up-to-date, this confirms immediately and `A` is able to
recover the HTLC funds.
However, the feerates of the `B====C` pre-signed transactions remain at the
old, uncompetitive feerates.
At this point, `C` broadcasts an HTLC-success transaction with high feerates
that CPFPs the commitment tx.
However, it replaces the HTLC-timeout transaction, which is at the old, low
feerate.
`C` is thus able to get the value of the HTLC, but `B` is now no longer able
to use the knowledge of the preimage, as its own incoming HTLC was already
confirmed as claimed by `A`.
Is the above restatement accurate?
----
Let me also explain to non-Lightning experts why HTLC-timeout is presigned
in this case and why `B` cannot feebump it.
In the Poon-Dryja mechanism, the HTLCs are "infected" by the Poon-Dryja
penalty case, and are not plain HTLCs.
A plain HTLC offerred by `B` to `C` would look like this:
(B && OP_CLTV) || (C && OP_HASH160)
However, on the commitment transaction held by `B`, it would be infected by
the penalty case in this way:
(B && C && OP_CLTV) || (C && OP_HASH160) || (C && revocation)
There are two changes:
* The addition of a revocation branch `C && revocation`.
* The branch claimable by `B` in the "plain" HTLC (`B && OP_CLTV`) also
includes `C`.
These are necessary in case `B` tries to cheat and this HTLC is on an old,
revoked transaction.
If the revoked transaction is *really* old, the `OP_CLTV` would already
impose a timelock far in the past.
This means that a plain `B && OP_CLTV` branch can be claimed by `B` if it
retained this very old revoked transaction.
To prevent that, `C` is added to the `B && OP_CLTV` branch.
We also introduce an HTLC-timeout transaction, which spends the `B && C &&
OP_CLTV` branch, and outputs to:
(B && OP_CSV) || (C && revocation)
Thus, even if `B` held onto a very old revoked commitment transaction and
attempts to spend the timelock branch (because the `OP_CLTV` is for an old
blockheight), it still has to contend with a new output with a *relative*
timelock.
Unfortunately, this means that the HTLC-timeout transaction is pre-signed,
and has a specific feerate.
In order to change the feerate, both `B` and `C` have to agree to re-sign
the HTLC-timeout transaction at the higher feerate.
However, the HTLC-success transaction in this case spends the plain `(C &&
OP_HASH160)` branch, which only involves `C`.
This allows `C` to feebump the HTLC-success transaction arbitrarily even if
`B` does not cooperate.
While anchor outputs can be added to the HTLC-timeout transaction as well,
`C` has a greater advantage here due to being able to RBF the HTLC-timeout
out of the way (1 transaction), while `B` has to get both HTLC-timeout and a
CPFP-RBF of the anchor output of the HTLC-timeout transaction (2
transactions).
`C` thus requires a smaller fee to achieve a particular feerate due to
having to push a smaller number of bytes compared to `B`.
Regards,
ZmnSCPxj
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