Re: [ANN][2POW] Luck - A new consensus algorithm to eliminate large mining pools
I'm not sure to fully understand your idea.
For the 4th part in your whitepaper, the difficulty function f(l) must be sensitive enough to l to satisfy E2(m)<E1(m), which means that f(max) must be much larger than f(min). How is f(l) designed?
I really don't understand the selfish mining attack in your whitepaper, and I don’t understand how the 1/m^2 was derived?
For the 4th part in your whitepaper, the difficulty function f(l) must be sensitive enough to l to satisfy E2(m)<E1(m), which means that f(max) must be much larger than f(min). How is f(l) designed?
I really don't understand the selfish mining attack in your whitepaper, and I don’t understand how the 1/m^2 was derived?
I think it can be understood like this.
when two miners do not cooperate in mining, their luck l1 will be better, and the mining difficulty f(l1) will be lower;
when they cooperate, their luck l2 will be worse, The mining difficulty f(l2) is higher.
Although the power of their cooperation is *2, when f(l2)>2*f(l1), it can satisfy E2<E1.
From another perspective, the variance of cooperation is greater than that of non-cooperation. one-cpu-one-vote is the result.
Thank you very much. You are right. From the perspective of minimizing risk, the expected output of cooperation is low and the variance is large. Rational miners will choose to mine independently.
