From: Rusty Russell <rusty@rustcorp.com.au>
To: Kalle Rosenbaum <kalle@rosenbaum.se>
In-Reply-To: <CAPswA9yUUUH7XM_wC=+QXnbEhp49hOyFAtfJztq+r2p0rNmeiQ@mail.gmail.com>
References: <871th3t1go.fsf@rustcorp.com.au>
	<CAPswA9yUUUH7XM_wC=+QXnbEhp49hOyFAtfJztq+r2p0rNmeiQ@mail.gmail.com>
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Date: Sat, 27 Jun 2015 14:16:35 +0930
Message-ID: <87a8vlvjv8.fsf@rustcorp.com.au>
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Cc: bitcoin-dev@lists.linuxfoundation.org
Subject: Re: [bitcoin-dev] [RFC] IBLT block testing implementation
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Kalle Rosenbaum <kalle@rosenbaum.se> writes:
> 2015-06-23 7:53 GMT+02:00 Rusty Russell <rusty@rustcorp.com.au>:
>> Hi all,
>>
>>         I've come up with a model for using IBLT to communicate blocks
>> between peers.  The gory details can be found on github: it's a
>> standalone C++ app for testing not integrated with bitcoin.
>>
>>         https://github.com/rustyrussell/bitcoin-iblt
>
> Good to see that you're working on this. Really exciting!
>
> I want to take the opportunity to link to my previous work on IBLTs, for
> those that haven't seen it, where I investigate the behaviour of the IBLT
> when changing different parameters, like cell count, hashFunctionCount etc:
> https://github.com/kallerosenbaum/bitcoin-iblt/wiki

Yep, I stole the hashFunctionCount = 3 straight from there, same
with 64-byte bucket contents.

>>From glancing over your implementation, I see that you don't use a
> keyHashSum, in fact you don't use a key at all, but only a
> concatenatenation of (txid48, fragid, tx-chunk) as value. Here the
> txid48+fragid functions as a kind of keyHashSum. I think this might be a
> very good idea,
>
> If you have a false positive with count == 1, then you would probably
> detect it if fragid is outside reasonable limit from from base_fragid. Did
> you implement your idea to remove all the count==1 fagments in ascending
> order of (fragid-base_fragid)?

Yep!  I keep records of all the 1 and -1 buckets; separate lists
depending on how far they are off the base.  Lists for 0, 1, 2, ... 7,
then powers of 2.  See todo in iblt.cpp.

> Anyhow, I think we should make some more comparable tests, just as you
> proposed last year when I didn't reply, sorry... My code is a more straight
> forward implementation of the IBLT paper (http://arxiv.org/pdf/1101.2245.pdf)
> and encoding blocks is done pretty much as Gavin proposed in his gist. That
> should function as a baseline so that we can validate that your
> optimizations actually work. Please contact me directly if you are
> interested in that, and we'll figure something out.

Absolutely, will do that offline.

> More comments/questions inline:
>
>>
>> Assumptions and details:
>>
>> 1. The base idea comes from Gavin's Block Propagation gist:
>>         https://gist.github.com/gavinandresen/e20c3b5a1d4b97f79ac2
>>
>> 2. It relies on similarity in mempools, with some selection hints.  This
>>    is designed to be as flexible as possible to make fewest assumptions
>>    on tx selection policy.
>>
>> 3. The selection hints are: minimum fee-per-byte (fixed point) and
>>    bitmaps of included-despite-that and rejected-despite-that.  The
>>    former covers things like child-pays-for-parent and the priority
>>    area.  The latter covers other cases like Eligius censoring "spam",
>>    bitcoin version differences, etc.
>>
>
> There is a chance that the bit prefix of the added or removed tx is not
> unique within the receiver's mempool. In that case the receiver can
> probably just use the earliest matching transaction and hope for the best.
> If not -> bad luck. Is that how you do it?

No; they add or remove all matching.  If they add too many, that's the
easy case, of course.  They can't remove too many (since they know that
bit prefix was unique on the other end).

>> 4. Cost to represent these exceptional added or excluded transactions is
>>    (on average) log2(transactions in mempool) bits.
>
> These exceptional tx *could* instead be encoded in the IBLT, just as if
> they were unknown differences. Your bitmaps is probably a more compact
> representation, but it's also more complex.

It's pretty easy to cut the bitmaps to zero and test this (comment them
out in wire_encode.cpp, for example).

But the overhead of IBLT is some factor greater than txsize (need to
measure that factor, too!); whereas these are a log(#txs-in-mempool) bits.

>> 5. At Peiter Wuille's suggestion, it is designed to be reencoded between
>>    nodes.  It seems fast enough for that, and neighboring nodes should
>>    have most similar mempools.
>
> What is the purpose of reencoding when relaying? Is that to improve the
> failure probability as new tx may have arrived in the mempool of the
> intermediary node?

Yep, and estimation ability.  The theory is that adjacent nodes will
have closer mempools, allowing for smaller IBLTs.  The size of mempool
differences for each block can be fed back, so you have an idea of the
likely delta to peers (ie. add that to the actual difference between
your mempool and the new block, to estimate the amount of IBLT
reconstruction required).

>> 6. It performs reasonably well on my 100 block sample in bitcoin-corpus
>>    (chosen to include a mempool spike):
>>
>>    Average block range (bytes):                         7988-999829
>>    Block size mean (bytes):                             401926
>>
>>    Minimal decodable BLT size range (bytes):            314-386361
>>    Minimal decodable BLT size mean (bytes):             13265
>>
>> 7. Actual results will have to be worse than these minima, as peers will
>>    have to oversize the IBLT to have reasonable chance of success.
>>
>
> Yes, I have made some analysis on this here:
> https://github.com/kallerosenbaum/bitcoin-iblt/wiki/FailureProbability. We
> use quite different data structures for encoding blocks in IBLT, but it
> might apply to your implementation as well. An interesting result is that
> the space saving percentage actually increases as blocks grow.

Let's pick a 5% as our failure target (given most nodes will get blocks
from more than 1 peer, and our other estimates of mempool differences
will be conservative).  Seems like 16/16 transactions takes ~400 cells
for recovery, 64/64 takes ~1400, 128/128 takes ~2480, 256/256 says
~4600.

Using that 128/128 => 198k number, and your txs were about 300B, that
implies an overhead of 2.6, right?

We're probably better estimating in terms of "cells recovered" (ie. sum
of cells for txs which we were missing, plus number of txs we added
erroneously).

Cheers,
Rusty.