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I repeat, you got it very wrong. Transaction number is not correlated with mining energy expenditure. Mining is used to secure the ledger in a way that the sa

by spiorf 9y ago

I repeat, you got it very wrong. Transaction number is not correlated with mining energy expenditure. Mining is used to secure the ledger in a way that the same amount of energy is needed to alter it. The block hash begins with a number of zeros. Try for yourself how many tries it takes to find a string that hashes to a hash beginning with 3 zeros. Bitcoin block hashes begin with 13 or 14 zeros IIRC. This means that you need to try trillions of combinations again if you want to alter a block. And then you must keep finding other hashes with enough zeros fast enough to outcompete the whole network. But the cost of hashing is the same, with 1, 10 or 1000 transactions in a block. And the block time is on average fixed at 10 minutes. You don't hash transactions directly, you hash a block header that keeps only the merkle root hash of the transactions. The merkle root is fixed size.

Retric 9y ago
Hash costs are a large fraction of overall costs, but not 100%. Network bandwidth is not free. The more transactions the more beefy the full nodes must be, worse every node pays the full costs or lose out on some transaction fees. But, it gets worse simply maintaining value takes hashing power to avoid a bad actor double spending and destroying faith in the system. Thus, even as a value store you need to pay for large scale hashing.
pdpi 9y ago
> Transaction number is not correlated with mining energy expenditure. You're only kind of right. The mining algorithm cares not one bit about number of transactions, so higher tx volume has no direct effect on the mining energy expenditure. But it does have a second order effect. More transactions usually entails higher fees (as people compete for block space to put their transactions in). More fees means higher mining rewards, means higher incentive to mine, means more mining activity, and, therefore, higher energy expenditure.
- decentralised 9y ago
  > The mining algorithm cares not one bit about number of transactions Actually it does to some extent because the transaction hashes are used to calculate the block header hash.
  - 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 ago
      I'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 ago
        No, 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
hwillis 9y ago
You can't just scale the number of transactions in a block forever and still have a stable currency. If you have only a few miners working on massive blocks, then they confer very little confidence onto the transactions in the blocks. More transactions, more mining. The whole point of the system is to verify transactions and it stops working if it doesn't do that.
- imnotatwork 9y ago
  > You can't just scale the number of transactions in a block forever and still have a stable currency. If you mean that over time the incentive to centralization become stronger, yes, you are right. There must be competition to enter the blocks, otherwise when mining subsidy ends, there will be no incentive to secure the ledger. If you mean that ten times the transaction have a computational cost 10 times greater (or 5, or 2), you're wrong. > The whole point of the system is to verify transactions and it stops working if it doesn't do that. Plenty of cryptocurrencies are mining tons of empty blocks. on the short term, if there are no transactions, mining continues with the same difficulty. The effect is long term: if noone is using the currency for transaction, it has no value so less and less people mine it. The difficulty drops, and the security drops, pulling value down even more. You are almost right, but it is a very indirect effect, and takes years to manifest itself.

by spiorf 9y ago

Retric 9y ago

Hash costs are a large fraction of overall costs, but not 100%. Network bandwidth is not free. The more transactions the more beefy the full nodes must be, worse every node pays the full costs or lose out on some transaction fees. But, it gets worse simply maintaining value takes hashing power to avoid a bad actor double spending and destroying faith in the system. Thus, even as a value store you need to pay for large scale hashing.

pdpi 9y ago

> Transaction number is not correlated with mining energy expenditure. You're only kind of right. The mining algorithm cares not one bit about number of transactions, so higher tx volume has no direct effect on the mining energy expenditure. But it does have a second order effect. More transactions usually entails higher fees (as people compete for block space to put their transactions in). More fees means higher mining rewards, means higher incentive to mine, means more mining activity, and, therefore, higher energy expenditure.

decentralised 9y ago
> The mining algorithm cares not one bit about number of transactions Actually it does to some extent because the transaction hashes are used to calculate the block header hash.
- 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 ago
    I'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 ago
      No, 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

hwillis 9y ago

You can't just scale the number of transactions in a block forever and still have a stable currency. If you have only a few miners working on massive blocks, then they confer very little confidence onto the transactions in the blocks. More transactions, more mining. The whole point of the system is to verify transactions and it stops working if it doesn't do that.

imnotatwork 9y ago
> You can't just scale the number of transactions in a block forever and still have a stable currency. If you mean that over time the incentive to centralization become stronger, yes, you are right. There must be competition to enter the blocks, otherwise when mining subsidy ends, there will be no incentive to secure the ledger. If you mean that ten times the transaction have a computational cost 10 times greater (or 5, or 2), you're wrong. > The whole point of the system is to verify transactions and it stops working if it doesn't do that. Plenty of cryptocurrencies are mining tons of empty blocks. on the short term, if there are no transactions, mining continues with the same difficulty. The effect is long term: if noone is using the currency for transaction, it has no value so less and less people mine it. The difficulty drops, and the security drops, pulling value down even more. You are almost right, but it is a very indirect effect, and takes years to manifest itself.