4 ms·
Yes, but the hashing of the merkle tree to obtain the merkle root takes like 10^4 hashes for 10^3 transactions. Compared with 2^80 that is about the number of
by spiorf 9y ago
Yes, but the hashing of the merkle tree to obtain the merkle root takes like 10^4 hashes for 10^3 transactions.
Compared with 2^80 that is about the number of hash operations needed to find a POW block header, it it trascurable.
- decentralised 9y agoI'm not sure about that calculation to be honest. The merkle root is very simple to calculate and verify that's why it's so useful. The POW difficulty is in finding a hash starting with the right number of zeroes and containing the previous block header hash. Did i get this wrong?
- imnotatwork 9y agoNo, you're right. Numbers i said are my guesses, but the order of magnitude should be about that. I don't remember exactly how many hash ops you have to do to put 1000 txes in a merkle tree, but it should be more than 1000 and less than 10000. My point was exactly that. Computational cost of the merkle tree is negligible when confronted with the computational cost of the POW