Efficient Reorgs on Cryptonets

Every PoW driven cryptonet has a state. The state of Bitcoin (and forks) is the particular set of Unspent Transaction Outputs (UTXOs) at the time - essentially the set of all Bitcoin able to be spent.

When a new block arrives, the usual process to update the state is simple:

Start with S[n,0] (state at block n)
Apply the first transaction from the new block (B[0]) to S

S[n,k] + B[k] -> S[n,k+1] for all k in B

S[n+1,0] = S[n,max(k)+1]

However, what happens when a new block arrives causing a reorganisation of the main chain?

.       3a← 4a    <-- 3a and 4a are not in the main chain currently
1 ← 2 ← 3 ← 4     <-- 3 and 4 are in the main chain

> 5a arrives, causing the reorg:

1 ← 2 ← 3a← 4a← 5a   <-- New main chain
        3 ← 4        <-- Old main chain, 3 and 4 no longer in the main chain

In this case block #2 was the lowest common ancestor (a pivot point)
of the two competing chains 3a->5a and 3->4.

The problem of reorgs

Let’s presume the distance from the lowest common ancestor (LCA) and the new head is n.

Bitcoin et al solve the issue by stepping backwards through time.

Since Bitcoin transactions spend outputs, and outputs may be spent only once, playing the blockchain backwards is trivial:

for each transaction:
	remove it's outputs from the list of UTXOs.
	add the outputs it spends to the list of UTXOs.

And bam! You can then play time forward from the LCA to calculate the new state. How nice.

What happens, though, when we move to a cryptonet that only operates on balances and doesn’t use the input/output system of Bitcoin?

Well, provided we’re recording every transaction it’s quite simple. A transaction moving X coins from A to B results in A-=X and B+=X. That is trivial to reverse. However, the caveat is that we must record every transaction. Once we start including complex mechanisms within the protocol that produce transactions that are not recorded but simply implied, we can no longer play time ‘backwards’ as S[m] depends on S[m-1] and without knowing S[m-1] to calculate the implied transactions, we can’t play time backwards. Of course, if we know S[m-1] we don’t need to do any of this anyway, so we’re sort of stuck. Examples of this sort of mechanism can be found in the way contracts create transactions in Ethereum and the market evaluation in Marketcoin.

Remembering S[m-1] is easy but what if the reorg is of length 2, or 3, or 10? We can’t just remember all the states.

So, we can see that we have a problem.

Efficiently remembering states

The intuitive solution (to me, at least) is to know some but not all states at strategic intervals between the genesis block and the current head. When a reorg of length n occurs, the network has already committed to evaluating n new states. I define ‘efficient’ here to mean evaluating no more than 2n new states (in the worst case). Unfortunately, this means we’ll need to remember about 2*log(2,h) states, where h is the height of the chain head. All the UTXOs in Bitcoin take up a few hundred meg of RAM, so for 500,000 blocks we’re looking at no more than 40 states, but that’s still ~10 GB of space (by Bitcoin’s standards) which isn’t ideal. It’s unlikely that we’ll see long reorganisations, but we’d still be storing half of the figures mentioned above, which, while better, isn’t perfect.

One solution may be to record the net implied change of state as the last transaction, but that solution might be more painful than the cure, and requires introducing extra complexity into the network architecture, which I’m against, so we won’t consider this option here.

In addition to the above constraint on ‘efficient’, we also require that for each block building on the main chain we should only have to calculate one new state (the updated current state). This implies that when we step through the blockchain, we only ever forget cached states, with the exception of the new state produced by the next block.


Current head is of height n.

A[n] = {cached states at height n}

Block n+1 arrives:

assert A[n] is a superset of {all a in A[n+1] s.t. a is not of height n+1}

Thus A[n+1] can be described as the set of some or all of the states in A[n] and the state at n+1, and therefore our collection of states does not requrie regeneration on each new block.

I propose a solution below that has a number of desired properties:

  • A reorg of length n requires computing no more than 2n states
  • Space efficient: k states saved where ld(h) <= k <= 2*ld(h)
  • Incremental: only one new state has to be calculated for each new block
Initial conditions:
- Reorg length: n
- Current height: h >= 3
- i = 0; i < h

2^k < h - i <= 2^(k+1) is always the case for some k
if h-i == 2: set k to 1. (it would otherwise be 0)

After finding k, and while h-i > 1:
1. Cache states at height i + 2^k and i + 2^(k-1).
2. i += 2^(k-1)

and in python: (testing all combinations up to 2^13)

import math

h = 3
states = set([1,2])

while h <= 2**13:
	newStates = set()
	# find largest k s.t. 2^k < h
	i = 0
	while h-i >= 2:
		k = math.log(h-i)//math.log(2)
		i += int(2**(k-1))
	ts1 = set(states) # temp set for testing superset requirement
	ts1.add(h) # add the current state (instead of removing it from newStates)
	assert ts1 >= newStates # ts1 is a superset of newStates
	l = list(newStates) # temp list just to print
	print(h, math.log(h)//math.log(2)+1, len(l), l)
	states = newStates
	h += 1

Because of the ~log(n) space requirement a very fast block time is not a major concern. A chain with a target time of 1 minute requires about 1.5x the storage capacity of an equivelant chain with a target time of 10 minutes in the first year, and this ratio rapidly approaches 1 in the following years.

That said, after the first year with a 1 minute block time, we’d be storing around 30 states. If we ignored all states more than 2000 blocks deep (a day and a bit) we’re still storing more than 15, which isn’t a particularly great optimisation. (When we have events like the Fork of March 2013 we would like clients to adjust quickly and efficiently).

I have some ideas about state-deltas to try and solve this issue (which is ungood, but not doubleplusungood) but that can wait for a future post.