Subject: Re: [bitcoin-dev] A Better MMR Definition

From: Peter Todd <[email-masked]>
Date: Thu, 23 Feb 2017 23:36:13 -0500

On Thu, Feb 23, 2017 at 07:32:43PM -0800, Bram Cohen wrote:
> On Thu, Feb 23, 2017 at 7:15 PM, Peter Todd <[email-masked]> 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