On 10/17/23, Antoine Riard via bitcoin-dev <bitcoin-dev@lists.linuxfoundation.org> wrote:
Hi Zeeman,
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.
This is not that the HTLC-timeout does not confirm. It is replaced in cycles by C's HTLC-preimage which is still valid after `B->C` HTLC timelock has expired. And this HTLC-preimage is subsequently replaced itself.
See the test here: https://github.com/ariard/bitcoin/commit/19d61fa8cf22a5050b51c4005603f43d72f...
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.
This is correct that A tries to recover the HTLC funds on the `A===B` channel.
However, there is no need to consider the fee rates nor mempool congestion as the exploit lays on the replacement mechanism itself (in simple scenario).
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`.
This is correct that `C` broadcasts an HTLC-success transaction at block height 144.
However `C` broadcasts this high feerate transaction at _every block_ between blocks 100 and 144 to replace B's HTLC-timeout transaction.
Let me also explain to non-Lightning experts why HTLC-timeout is presigned in this case and why `B` cannot feebump it.
Note `B` can feebump the HTLC-timeout for anchor output channels thanks to sighash_single | anyonecanpay on C's signature.
Le mar. 17 oct. 2023 à 11:34, ZmnSCPxj <ZmnSCPxj@protonmail.com> a écrit :
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