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To: Bastien TEINTURIER <bastien@acinq.fr>
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Cc: Bitcoin Protocol Discussion <bitcoin-dev@lists.linuxfoundation.org>,
 lightning-dev <lightning-dev@lists.linuxfoundation.org>
Subject: Re: [bitcoin-dev] [Lightning-dev] RBF Pinning with Counterparties
	and Competing Interest
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X-List-Received-Date: Mon, 22 Jun 2020 08:15:45 -0000

Good morning Bastien,

> Thanks for the detailed write-up on how it affects incentives and central=
ization,
> these are good points. I need to spend more time thinking about them.
>
> > This is one reason I suggested using independent pay-to-preimage
> > transactions[1]
>
> While this works as a technical solution, I think it has some incentives =
issues too.
> In this attack, I believe the miners that hide the preimage tx in their m=
empool have
> to be accomplice with the attacker, otherwise they would share that tx wi=
th some of
> their peers, and some non-miner nodes would get that preimage tx and be a=
ble to
> gossip them off-chain (and even relay them to other mempools).

I believe this is technically possible with current mempool rules, without =
miners cooperating with the attacker.

Basically, the attacker releases two transactions with near-equal fees, so =
that neither can RBF the other.
It releases the preimage tx near miners, and the timelock tx near non-miner=
s.

Nodes at the boundaries between those that receive the preimage tx and the =
timelock tx will receive both.
However, they will receive one or the other first.
Which one they receive first will be what they keep, and they will reject t=
he other (and *not* propagate the other), because the difference in fees is=
 not enough to get past the RBF rules (which requires not just a feerate in=
crease, but also an increase in absolute fee, of at least the minimum relay=
 feerate times transaction size).

Because they reject the other tx, they do not propagate the other tx, so th=
e boundary between the two txes is inviolate, neither can get past that bou=
ndary, this occurs even if everyone is running 100% unmodified Bitcoin Core=
 code.

I am not a mempool expert and my understanding may be incorrect.

Regards,
ZmnSCPxj