On Wed, Nov 06, 2013 at 01:06:47PM -0500, Christophe Biocca wrote:
> I might try building this sometime soon. I think it may also serve an
> educational purpose when trying to understand the whole network's behaviour.
> What level of accuracy are we looking for though? Obviously we need to
> fully emulate the steps of the network protocol, and we need to be able to
> specify time taken for transmission/processing for each node. Do we care
> about the actual contents of the messages (to be able to simulate double
> spend attempts, invalid transactions and blocks, SPV node communication),
> and their validation (actual signatures and proof of work)?
> I imagine the latter is pretty useless, beyond specifying that the
> signature/proof of work is valid/invalid.
> If we could build up a set of experiments we'd like to run on it, it would
> help clarify what's needed.
> Off the top of my head:
> - Peter Todd's miner strategy of sending blocks to only 51% of the
> hashpower.

Speaking of, I hadn't gotten around to doing up the math behind that
strategy properly; turns out 51% I was overly optimistic and the actual
threshold is 29.3%

Suppose I find a block. I have Q hashing power, and the rest of the
network 1-Q. Should I tell the rest of the network, or withhold that
block and hope I find a second one?

Now in a purely inflation subsidy environment, where I don't care about
the other miners success, of course I should publish. However, if my
goals are to find *more* blocks than the other miners for whatever
reason, maybe because transaction fees matter or I'm trying to get
nLockTime'd announce/commit fee sacrifices, it gets more complicated.

There are three possible outcomes:

1) I find the next block, probability Q
2) They find the next block, probability 1-Q
2.1) I find the next block, probability Q, or (1-Q)*Q in total.
2.2) They find the next block, probability (1-Q)^2 in total.

Note how only in the last option do I lose. So how much hashing power do
I need before it is just as likely that the other miners will find two
blocks before I find either one block, or two blocks? Easy enough:

Q + (1-Q)*Q = (1-Q)^2 -> Q^2 - Q + 1/2 -> Q = (1 - \sqrt(2))/2

Q ~= 29.2%

So basically, if I'm trying to beat other miners, once I have >29.3% of
the hashing power I have no incentive to publish the blocks I mine!

But hang on, does it matter if I'm the one who actually has that hashing
power? What if I just make sure that only >29.3% of the hashing power
has that block? If my goal is to make sure that someone does useless
work, and/or they are working on a lower height block than me, then no,
I don't care, which means my original "send blocks to >51% of the
hashing power" analysis was actually wrong, and the strategy is even
more crazy: "send blocks to >29.3% of the hashing power" (!)

Lets suppose I know that I'm two blocks ahead:

1) I find the next block: Q                    (3:0)
2) They find the next block: (1-Q)             (2:1)
2.1) I find the next block: (1-Q)*Q            (3:1)
2.2) They find the next block: (1-Q)^2         (2:2)
2.2.1) I find the next block: (1-Q)^2 * Q      (3:2)
2.2.2) They find the next block: (1-Q)^3       (2:3)

At what hashing power should I release my blocks? So remember, I win
this round on outcomes 1, 2.1, 2.2.1 and they only win on 2.2.2:

Q + (1-Q)*Q + (1-Q)^2*Q = (1-Q)^3 -> Q = 1 - 2^-3

Q ~= 20.6%

Interesting... so as I get further ahead, or to be exact the group of
miners who have a given block gets further ahead, I need less hashing
power for my incentives to be to *not* publish the block I just found.
Conversely this means I should try to make my blocks propagate to less
of the hashing power, by whatever means necessary.

Now remember, none of the above strategy requires me to have a special
low-latency network or anything fancy. I don't even have to have a lot
of hashing power - the strategy still works if I'm, say, a 5% pool. It
just means I don't have the incentives people thought I did to propagate
my blocks widely.

The other nasty thing about this, is suppose I'm a miner and recently
got a block from another miner: should I forward that block, or not
bother? Well, it depends: if I have no idea how much of the hashing
power has that block, I should forward the block. But again, if my goal
is to be most likely to get the next block, I should only forward in
such a way that >30% of the hashing power has the block.

This means that if I have some information about what % already has that
block, I have less incentive to forward! For instance, suppose that
every major miner has been publishing their node addresses in their
blocks - I'll have a pretty good idea of who probably has that most
recent block, so I can easily make a well-optimized decision not to
forward. Similarly because the 30% hashing power figure is the
*integral* of time * hashes/second, if miners are forwarding
near-target-headers, I might as well wait a few seconds and see if I see
any near-target-headers; if I do for this block then I have evidence
that hashing power does have it, and I shouldn't forward.

So yeah, we're fucked and have got to fix this awful incentive structure
somehow before the inflation subsidy gets any smaller. Also, raising the
blocksize, especially by just removing the limit, is utter madness given
it can be used to slow down block propagation selectively, so the
hashing power that gets a given block is limited repeatably to the same

P.S: If any large pools want to try this stuff out, give me a shout. You
have my PGP key - confidentiality assured.

P.P.S: If you're mining on a pool with more than, like, 1% hashing
power, do the math on varience... Seriously, stop it and go mine on a
smaller pool, or better yet, p2pool.


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