The Merkle root of the transactions is calculated by building a binary tree from the transactions as leaves where each parent node is the hash of its two children. As the coinbase transaction is the first transaction in the block, you can store the hashing partners on the coinbase transaction's Merkle branch and only repeat all the hashes to which the coinbase transaction contributes. This reduces the required number of hashes to calculate the new Merkle root for n transactions from `O(n)`

to `O(log(n))`

given that no other transactions in the block were changed.

There is no way around repeating the whole left flank of hashes in the binary tree, since changing the extranonce in the coinbase transaction changes the outcome of the bottom left hash combining coinbase transaction and first regular transaction.

If one has only one change in the Merkle tree the complexity to compute the hash is

`O(lg(n))`

because other branches are hashes are reused. No further speedup is possible (except Asicboost) as even one bit change in coinbase transaction causes avalon effect. – MCCCS – 2020-06-18T15:07:01.097Thanks! This answers the question. I take it that the coinbase Tx must go into the start then – Wil – 2020-06-18T17:20:30.230

The extra nonce is indeed additional space in coinbase. I don't see any reason why you cannot precompute all the extranonce fields. So, even the

`log(n)`

time could be amortized – sanket1729 – 2020-06-19T07:54:20.350I think you can get ln(n)/ln(2) with precomputation. But it is less if the coinbase tx didn’t go into the bottom. And I guess it needs to go into the bottom? – Wil – 2020-06-20T09:00:06.340

Each transactions is a leaf at the bottom level of the Merkle tree. – Murch – 2020-07-18T17:20:08.783