From mboxrd@z Thu Jan 1 00:00:00 1970 Return-Path: Received: from smtp1.linuxfoundation.org (smtp1.linux-foundation.org [172.17.192.35]) by mail.linuxfoundation.org (Postfix) with ESMTPS id 46950B7C for ; Fri, 24 Feb 2017 22:20:22 +0000 (UTC) X-Greylist: whitelisted by SQLgrey-1.7.6 Received: from mail-io0-f174.google.com (mail-io0-f174.google.com [209.85.223.174]) by smtp1.linuxfoundation.org (Postfix) with ESMTPS id C098E18C for ; Fri, 24 Feb 2017 22:20:20 +0000 (UTC) Received: by mail-io0-f174.google.com with SMTP id l66so2808289ioi.1 for ; Fri, 24 Feb 2017 14:20:20 -0800 (PST) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=bittorrent-com.20150623.gappssmtp.com; s=20150623; h=mime-version:in-reply-to:references:from:date:message-id:subject:to :cc; bh=apRFdcYHU8VoMC5fRKoY9Bz/aL0S1caImkffHdneXmA=; b=P9HiKMVKvbUe6MDFOTyEJbLClhDC1tO6sVNKzQQ2UaHH5Nfl3SBQKijhUrpMn80gtn rRNH8baBwhrCp0o8Pf4fMSgJD9JPA1xGtzfOU0K+koZQC2o4vaiURUfYyGmFKV1ZRLGT 9Xoxuhd8dfs2DqNA0dq2LFkSw/wt+rcXcwWhdXypBWjcdv8xi+ETmSaYXX7+a6O/n9RK 57EMqpxL6elA1+EsFPD1BuwBnc+101DjX0azXQ24stNC7dz0qtuRKoFayS62AbYbohAu QxRKlCz4FHAG0ev6Vz4hP4UqBABB5A7b5VXWw4m5jyeC1Dm8tc/yeqsPvLH3zv2aXBm/ mKZw== X-Google-DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=1e100.net; s=20161025; h=x-gm-message-state:mime-version:in-reply-to:references:from:date :message-id:subject:to:cc; bh=apRFdcYHU8VoMC5fRKoY9Bz/aL0S1caImkffHdneXmA=; b=sp0TSSLXhaNVeYqLaE3wWBA3AZQ7oZeUEEChkL6A9w6OyopjwnsMZhD20J4CZ81C1R qjNOoD1ADF1qWwrcyBFhUY+ohJ/iNw5/mLlO69yPWpu89ngtaoGJnC5ZcSd+gWZFo7Uz hQLDbVLI7us+E8Q95ssAdFa8QH+dDjwyVs8OJWG05OxaEbeeCF8Gn6pIX8AJIQ9YsT5S WSCr2C5IY+8DI2MdpYNB9VPTdjt4XQcFYxWmwaCgmdkvLO/9UofgboLAI+MchzmQft1c PsL02gV8CeiUb0y1sE/+axFpIHVXMzmimhUa1tdFxw8/xDqexNgfWRIX1bMWSCZEG0GX BXHg== X-Gm-Message-State: AMke39mDzUATkS+F3vrMNT8POEs6i7k54A1sTqNQwB4OcEJViw6IRe+JPNsk4osENr+JUHfNKuNOyhwZcpkt6Gwu X-Received: by 10.107.26.205 with SMTP id a196mr4651415ioa.214.1487974820033; Fri, 24 Feb 2017 14:20:20 -0800 (PST) MIME-Version: 1.0 Received: by 10.36.73.150 with HTTP; Fri, 24 Feb 2017 14:20:19 -0800 (PST) In-Reply-To: <20170224043613.GA32502@savin.petertodd.org> References: <20170223235105.GA28497@savin.petertodd.org> <20170224010943.GA29218@savin.petertodd.org> <20170224025811.GA31911@savin.petertodd.org> <20170224031531.GA32118@savin.petertodd.org> <20170224043613.GA32502@savin.petertodd.org> From: Bram Cohen Date: Fri, 24 Feb 2017 14:20:19 -0800 Message-ID: To: Peter Todd Content-Type: multipart/alternative; boundary=001a113fd05695bf9205494e22bc X-Spam-Status: No, score=-2.6 required=5.0 tests=BAYES_00,DKIM_SIGNED, DKIM_VALID,HTML_MESSAGE,RCVD_IN_DNSWL_LOW autolearn=ham version=3.3.1 X-Spam-Checker-Version: SpamAssassin 3.3.1 (2010-03-16) on smtp1.linux-foundation.org Cc: Bitcoin Protocol Discussion Subject: Re: [bitcoin-dev] A Better MMR Definition X-BeenThere: bitcoin-dev@lists.linuxfoundation.org X-Mailman-Version: 2.1.12 Precedence: list List-Id: Bitcoin Protocol Discussion List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , X-List-Received-Date: Fri, 24 Feb 2017 22:20:22 -0000 --001a113fd05695bf9205494e22bc Content-Type: text/plain; charset=UTF-8 So your idea is to cluster entries by entry time because newer things are more likely to leave and updating multiple things near each other is cheaper? That can be done with my tool. Instead of using hashes for the values being stored, you use position entries. The first entry gets a value of all zeros, the next one a one followed by all zeros, then the next two correspond to the first two with the second bit flipped to one, then the next four the first four with the third bit flipped to one, etc. It probably performs a little bit better to do it two bits at a time instead of one so that the entries are 00, 01, 10, 11, 0001, 0010, 0011, 0101, 0110, 0111, 1001, etc. If you were to really use this you'd probably want to to add some optimizations to use the fact that the terminals fit in 64 bits instead of 256, but it mostly works unchanged, and gets whatever benefits there are to this clustering plus the high performance implementation tricks I've built which I keep complaining that nobody's giving feedback on. I'm not sold on this being a win: The empirical access patterns are unknown, it requires an extra cache miss per lookup to find the entry number, it may be that everything is optimized well enough without it for there to be no meaningful gains, and it's a bunch of extra complexity. What should be done is that a plain vanilla UTXO set solution is optimized as well as it can be first, and then the insertion ordering trick is tried as an optimization to see if it's an improvement. Without that baseline there's no meaningful basis for comparison, and I'm quite confident that a naive implementation which just allocates individual nodes will underperform the thing I've come up with, even without adding optimizations related to fitting in 64 bits. On Thu, Feb 23, 2017 at 8:36 PM, Peter Todd wrote: > On Thu, Feb 23, 2017 at 07:32:43PM -0800, Bram Cohen wrote: > > On Thu, Feb 23, 2017 at 7:15 PM, Peter Todd wrote: > > > > > > > > Glad we're on the same page with regard to what's possible in TXO > > > commitments. > > > > > > Secondly, am I correct in saying your UTXO commitments scheme requires > > > random > > > access? While you describe it as a "merkle set", obviously to be > merkelized > > > it'll have to have an ordering of some kind. What do you propose that > > > ordering > > > to be? > > > > > > > The ordering is by the bits in the hash. Technically it's a Patricia > Trie. > > I'm using 'merkle tree' to refer to basically anything with a hash root. > > The hash of what? The values in the set? > > > > Maybe more specifically, what exact values do you propose to be in the > set? > > > > > > > > That is unspecified in the implementation, it just takes a 256 bit value > > which is presumably a hash of something. The intention is to nail down a > > simple format and demonstrate good performance and leave those semantics > to > > a higher layer. The simplest thing would be to hash together the txid and > > output number. > > Ok, so let's assume the values in the set are the unspent outpoints. > > Since we're ordering by the hash of the values in the set, outpoints will > be > distributed uniformly in the set, and thus the access pattern of data in > the > set is uniform. > > Now let's fast-forward 10 years. For the sake of argument, assume that for > every 1 UTXO in the set that corresponds to funds in someone's wallet that > are > likely to be spent, there are 2^12 = 4096 UTXO's that have been permanently > lost (and/or created in spam attacks) and thus will never be spent. > > Since lost UTXO's are *also* uniformly distributed, if I'm processing a new > block that spends 2^12 = 4096 UTXO's, on average for each UTXO spent, I'll > have to update log2(4096) = 12 more digests than I would have had those > "dead" > UTXO's not existed. > > Concretely, imagine our UTXO set had just 8 values in it, and we were > updating > two of them: > > # > / \ > / \ > / \ > / \ > / \ > # # > / \ / \ > / \ / \ > # . . # > / \ / \ / \ / \ > . X . . . . X . > > To mark two coins as spent, we've had to update 5 inner nodes. > > > Now let's look at what happens in an insertion-ordered TXO commitment > scheme. > For sake of argument, let's assume the best possible case, where every UTXO > spent in that same block was recently created. Since the UTXO's are > recently > created, chances are almost every single one of those "dead" UTXO's will > have > been created in the past. Thus, since this is an insertion-ordered data > structure, those UTXO's exist in an older part of the data structure that > our > new block doesn't need to modify at all. > > Concretely, again let's imagine a TXO commitment with 8 values in it, and > two > of them being spent: > > # > / \ > / \ > / \ > / \ > / \ > . # > / \ / \ > / \ / \ > . . . # > / \ / \ / \ / \ > . . . . . . X X > > To mark two coins as spent, we've only had to update 3 inner nodes; while > our > tree is higher with those lost coins, those extra inner nodes are amortised > across all the coins we have to update. > > > The situation gets even better when we look at the *new* UTXO's that our > block > creates. Suppose our UTXO set has size n. To mark a single coin as spent, > we > have to update log2(n) inner nodes. We do get to amortise this a bit at > the top > levels in the tree, but even if we assume the amortisation is totally free, > we're updating at least log2(n) - log2(m) inner nodes "under" the amortised > nodes at the top of the tree for *each* new node. > > Meanwhile with an insertion-ordered TXO commitment, each new UTXO added to > the > data set goes in the same place - the end. So almost none of the existing > data > needs to be touched to add the new UTXOs. Equally, the hashing required > for the > new UTXO's can be done in an incremental fashion that's very L1/L2 cache > friendly. > > > tl;dr: Precisely because access patterns in TXO commitments are *not* > uniform, > I think we'll find that from a L1/L2/etc cache perspective alone, TXO > commitments will result in better performance than UTXO commitments. > > > Now it is true that Bitcoin's current design means we'll need a map of > confirmed outpoints to TXO insertion order indexes. But it's not > particularly > hard to add that "metadata" to transactions on the P2P layer in the same > way > that segwit added witnesses to transactions without modifying how txids > were > calculated; if you only connect to peers who provide you with TXO index > information in blocks and transactions, you don't need to keep that map > yourself. > > Finally, note how this makes transactions *smaller* in many circumstances: > it's > just a 8-byte max index rather than a 40 byte outpoint. > > -- > https://petertodd.org 'peter'[:-1]@petertodd.org > --001a113fd05695bf9205494e22bc Content-Type: text/html; charset=UTF-8 Content-Transfer-Encoding: quoted-printable
So your idea is to cluster entries by entry time because n= ewer things are more likely to leave and updating multiple things near each= other is cheaper?

That can be done with my tool. Instea= d of using hashes for the values being stored, you use position entries. Th= e first entry gets a value of all zeros, the next one a one followed by all= zeros, then the next two correspond to the first two with the second bit f= lipped to one, then the next four the first four with the third bit flipped= to one, etc. It probably performs a little bit better to do it two bits at= a time instead of one so that the entries are 00, 01, 10, 11, 0001, 0010, = 0011, 0101, 0110, 0111, 1001, etc. If you were to really use this you'd= probably want to to add some optimizations to use the fact that the termin= als fit in 64 bits instead of 256, but it mostly works unchanged, and gets = whatever benefits there are to this clustering plus the high performance im= plementation tricks I've built which I keep complaining that nobody'= ;s giving feedback on.

I'm not sold on this be= ing a win: The empirical access patterns are unknown, it requires an extra = cache miss per lookup to find the entry number, it may be that everything i= s optimized well enough without it for there to be no meaningful gains, and= it's a bunch of extra complexity. What should be done is that a plain = vanilla UTXO set solution is optimized as well as it can be first, and then= the insertion ordering trick is tried as an optimization to see if it'= s an improvement. Without that baseline there's no meaningful basis for= comparison, and I'm quite confident that a naive implementation which = just allocates individual nodes will underperform the thing I've come u= p with, even without adding optimizations related to fitting in 64 bits.

On Thu, = Feb 23, 2017 at 8:36 PM, Peter Todd <pete@petertodd.org> wr= ote:
On Thu, Feb 23, 201= 7 at 07:32:43PM -0800, Bram Cohen wrote:
> On Thu, Feb 23, 2017 at 7:15 PM, Peter Todd <pete@petertodd.org> wrote:
>
> >
> > Glad we're on the same page with regard to what's possibl= e in TXO
> > commitments.
> >
> > Secondly, am I correct in saying your UTXO commitments scheme req= uires
> > random
> > access? While you describe it as a "merkle set", obviou= sly to be merkelized
> > it'll have to have an ordering of some kind. What do you prop= ose that
> > ordering
> > to be?
> >
>
> The ordering is by the bits in the hash. Technically it's a Patric= ia Trie.
> I'm using 'merkle tree' to refer to basically anything wit= h a hash root.

The hash of what? The values in the set?

> > Maybe more specifically, what exact values do you propose to be i= n the set?
> >
> >
> That is unspecified in the implementation, it just takes a 256 bit val= ue
> which is presumably a hash of something. The intention is to nail down= a
> simple format and demonstrate good performance and leave those semanti= cs to
> a higher layer. The simplest thing would be to hash together the txid = and
> output number.

Ok, so let's assume the values in the set are the unspent outpoi= nts.

Since we're ordering by the hash of the values in the set, outpoints wi= ll be
distributed uniformly in the set, and thus the access pattern of data in th= e
set is uniform.

Now let's fast-forward 10 years. For the sake of argument, assume that = for
every 1 UTXO in the set that corresponds to funds in someone's wallet t= hat are
likely to be spent, there are 2^12 =3D 4096 UTXO's that have been perma= nently
lost (and/or created in spam attacks) and thus will never be spent.

Since lost UTXO's are *also* uniformly distributed, if I'm processi= ng a new
block that spends 2^12 =3D 4096 UTXO's, on average for each UTXO spent,= I'll
have to update log2(4096) =3D 12 more digests than I would have had those &= quot;dead"
UTXO's not existed.

Concretely, imagine our UTXO set had just 8 values in it, and we were updat= ing
two of them:

=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0#
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 / \
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0/=C2=A0 =C2=A0\
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 /=C2=A0 =C2=A0 =C2=A0\
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0/=C2=A0 =C2=A0 =C2=A0 =C2=A0\
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 /=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0\
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0#=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2= =A0#
=C2=A0 =C2=A0 =C2=A0 =C2=A0 / \=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0/ \
=C2=A0 =C2=A0 =C2=A0 =C2=A0/=C2=A0 =C2=A0\=C2=A0 =C2=A0 =C2=A0 =C2=A0/=C2= =A0 =C2=A0\
=C2=A0 =C2=A0 =C2=A0 #=C2=A0 =C2=A0 =C2=A0.=C2=A0 =C2=A0 =C2=A0.=C2=A0 =C2= =A0 =C2=A0#
=C2=A0 =C2=A0 =C2=A0/ \=C2=A0 =C2=A0/ \=C2=A0 =C2=A0/ \=C2=A0 =C2=A0/ \
=C2=A0 =C2=A0 .=C2=A0 =C2=A0X .=C2=A0 =C2=A0. .=C2=A0 =C2=A0. X=C2=A0 =C2= =A0.

To mark two coins as spent, we've had to update 5 inner nodes.


Now let's look at what happens in an insertion-ordered TXO commitment s= cheme.
For sake of argument, let's assume the best possible case, where every = UTXO
spent in that same block was recently created. Since the UTXO's are rec= ently
created, chances are almost every single one of those "dead" UTXO= 's will have
been created in the past. Thus, since this is an insertion-ordered data
structure, those UTXO's exist in an older part of the data structure th= at our
new block doesn't need to modify at all.

Concretely, again let's imagine a TXO commitment with 8 values in it, a= nd two
of them being spent:

=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0#
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 / \
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0/=C2=A0 =C2=A0\
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 /=C2=A0 =C2=A0 =C2=A0\
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0/=C2=A0 =C2=A0 =C2=A0 =C2=A0\
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 /=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0\
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0.=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2= =A0#
=C2=A0 =C2=A0 =C2=A0 =C2=A0 / \=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0/ \
=C2=A0 =C2=A0 =C2=A0 =C2=A0/=C2=A0 =C2=A0\=C2=A0 =C2=A0 =C2=A0 =C2=A0/=C2= =A0 =C2=A0\
=C2=A0 =C2=A0 =C2=A0 .=C2=A0 =C2=A0 =C2=A0.=C2=A0 =C2=A0 =C2=A0.=C2=A0 =C2= =A0 =C2=A0#
=C2=A0 =C2=A0 =C2=A0/ \=C2=A0 =C2=A0/ \=C2=A0 =C2=A0/ \=C2=A0 =C2=A0/ \
=C2=A0 =C2=A0 .=C2=A0 =C2=A0. .=C2=A0 =C2=A0. .=C2=A0 =C2=A0. X=C2=A0 =C2= =A0X

To mark two coins as spent, we've only had to update 3 inner nodes; whi= le our
tree is higher with those lost coins, those extra inner nodes are amortised=
across all the coins we have to update.


The situation gets even better when we look at the *new* UTXO's that ou= r block
creates. Suppose our UTXO set has size n. To mark a single coin as spent, w= e
have to update log2(n) inner nodes. We do get to amortise this a bit at the= top
levels in the tree, but even if we assume the amortisation is totally free,=
we're updating at least log2(n) - log2(m) inner nodes "under"= the amortised
nodes at the top of the tree for *each* new node.

Meanwhile with an insertion-ordered TXO commitment, each new UTXO added to = the
data set goes in the same place - the end. So almost none of the existing d= ata
needs to be touched to add the new UTXOs. Equally, the hashing required for= the
new UTXO's can be done in an incremental fashion that's very L1/L2 = cache
friendly.


tl;dr: Precisely because access patterns in TXO commitments are *not* unifo= rm,
I think we'll find that from a L1/L2/etc cache perspective alone, TXO commitments will result in better performance than UTXO commitments.


Now it is true that Bitcoin's current design means we'll need a map= of
confirmed outpoints to TXO insertion order indexes. But it's not partic= ularly
hard to add that "metadata" to transactions on the P2P layer in t= he same way
that segwit added witnesses to transactions without modifying how txids wer= e
calculated; if you only connect to peers who provide you with TXO index
information in blocks and transactions, you don't need to keep that map=
yourself.

Finally, note how this makes transactions *smaller* in many circumstances: = it's
just a 8-byte max index rather than a 40 byte outpoint.

--
http= s://petertodd.org 'peter'[:-1]@petertodd.org

--001a113fd05695bf9205494e22bc--