Hi Peter,
> Point is, the attacker is thousands of UTXOs can also DoS rounds by simply
> failing to complete the round. In fact, the double-spend DoS attack requires
> more resources, because for a double-spend to be succesful, BTC has to be
> spent
> on fees.
>
> It's just a fact of life that a
Hi aj,
I think there's another workaround for the x-only issue with
TAPLEAF_UPDATE_VERIFY.
So the opcode will need a function f that ensures that the new internal
key f(P'), where P' = P + X, has even y. You describe what happens for
the canonical choice of
f(P') = if has_even_y(P') then P' else
Half-aggregation has been mentioned several times on this list in various
contexts. To have a solid basis for discussing applications of half-aggregation,
I think it's helpful to have a concrete specification of the scheme and a place
for collecting supplemental information like references to cryp
On Thu, Jul 7, 2022 at 8:29 PM Eric Voskuil wrote:
> Value is subjective, though a constraint of 1tx per 10 minutes seems
> unlikey to create a fee of 5000x that of 5000tx. This is of course why I
> stated my assumption. Yet this simple example should make clear that at
> some point a reduction i
The attacker isn't guaranteed to spend *any* funds to disrupt the protocol
indefinitely, that's the issue here. In this scenario, her input double
spend is at an impractical feerate, and is never included in a block,
sitting at the bottom of the mempool.
The other users' only practical choice is t
On Tue, Jul 05, 2022 at 08:46:51PM +, alicexbt wrote:
> Hi Peter,
>
> > Note that Wasabi already has a DoS attack vector in that a participant can
> > stop
> > participating after the first phase of the round, with the result that the
> > coinjoin fails. Wasabi mitigates that by punishing par
> What do you do if the "first" word (of 12), happens to be the last word in
> the list alphabetically?
>
That couldn't happen. If one word is the very last from the wordlist, it
would end up at the end of your mnemonic once you rearrange your 12 words
alphabetically.
However!
(@vjudeu) Choosing
What do you do if the "first" word (of 12), happens to be the last word in
the list alphabetically? So that seems like a dead end.
Since users are never expected to memorize the "whole list" (of 2048 words)
in any case, it seems that the smarter thing to do (if this "order"
criterion is desirable)
vju...@gazeta.pl, what you describe is not possible without a hard fork,
just like Eric said. There is no atomic way to move Bitcoin off of Bitcoin.
You can use Bitcoin txns, or you can use trust/custody, or you can make a
shitcoin. There is no way to actually divide or transfer sats to another
ne
> Simply fork off an inflation coin and test your theory. I mean, that’s the
> only way it can happen anyway.
That would be an altcoin. But it can be done in a simpler way: we have 21
million coins. It doesn't matter if it is 21 million, if it is 100 million, or
if it is in some normalized rang
@vjudeu
> better to allow transaction joining.. to make fees more smoothly
I'm not familiar with RSK transaction joining. However, I don't think this
addresses the issues Corey brought up - which is that the
appropriate amount of security (ie miner revenue) isn't linked with any
bitcoin market beh
Isn't it enough to just generate a seed in the same way as today, then sort the
words alphabetically, and then use that as a seed? I know, the last word is a
checksum, but there are only 2048 words, so it is not a big deal to get any
checksum we want. If that is insecure, because of lower possib
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