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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