Quote from: larry_vw_1955 on Today at 07:02:03 AM

Quote from: Epictetus on Today at 06:42:18 AM

Please explain how a lower complexity reduces the amount of energy used when miners are encouraged to increase energy usage in order to be one of the first N2 miners. Whether it takes 1000000 tries on average or 10¹⁸ tries on average, miners are going to use as much energy as they can in order to be first.

Quote

Yes but Round one is only let's say 2min for the sake of argument. Once this 2min is over Only N2 miners will advance to Round 2 to work on the second puzzle. Let's assume Round 2 calculation time is 6min. We have N1 miners working for 2min and N2 miners working for (2+6)=8min. In Bitcoin we have all N1 working all 8min. We save energy on Round 2. N1-N2 miners are IDLE during 6min of Round 2. This is where we save energy. Now I do understand that miners can try to game the system and rent their machines to the winners of Round 2. But I don't know how practical it can be to do so.

Doesn't this scheme compromise the security of the bitcoin network. I mean, the security aspect of it is based on the total hashrate at every moment in time. if at any moment it decreases...then there's a potential weakness.

I don't think so. We can find an optimal value of N2 that does not undermine the security of Bitcoin. Intuitively an upper bound of such value is O(SquareRoot N1). Assuming that the winner of Second Round is among N2 List of N2 miners instead of only N2 first miners. An optimal value of N2 is could be in O(Log(N1)) based on a similar calculation here by EPFL team of Distributed algo: https://arxiv.org/abs/1908.01738