At the Landauer kT limit, you need 1014 kWh to perform your 1035 FLOPs. That’s 10,000x the yearly US electricity production. You’d need a quantum computer or a Dyson sphere to solve that problem.
Assuming 10^6 bit erasures per FLOP (as you did; which source are you using?), one only needs 8.06*10^13 kWh (= 2.9*10^(-21)*10^(35+6)/(3.6*10^6)), i.e. 2.83 (= 8.06*10^13/(2.85*10^13)) times global electricity generation in 2022, or 18.7 (= 8.06*10^13/(4.30*10^12)) times the one generated in the United States.
At the Landauer kT limit, you need 1014 kWh to perform your 1035 FLOPs. That’s 10,000x the yearly US electricity production. You’d need a quantum computer or a Dyson sphere to solve that problem.
Either that, or magic. :D
Hi there,
Assuming 10^6 bit erasures per FLOP (as you did; which source are you using?), one only needs 8.06*10^13 kWh (= 2.9*10^(-21)*10^(35+6)/(3.6*10^6)), i.e. 2.83 (= 8.06*10^13/(2.85*10^13)) times global electricity generation in 2022, or 18.7 (= 8.06*10^13/(4.30*10^12)) times the one generated in the United States.