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Yes, this is correct! Congrats
The challenge going forward will be characterizing the function you introduced, .
Just as a quick note, I would have written it like this, but your introduction of D(a,b) is also correct.
pull down to refresh
Yes, this is correct! Congrats
The challenge going forward will be characterizing the function you introduced, .
Just as a quick note, I would have written it like this, but your introduction of D(a,b) is also correct.
Thanks, I feel I understand much better what you meant with V(x) and P(x), then.
Adding to above reasoning, I replace the flow cost with the corrected value: u(x)=−hAc
Yet, I am still not sure how to exactly compose the value of Alice finding the next block. It seems to me that it is dependent on whether Alice will publish once she gets ahead. From what I know this would depend on Alice’s hashrate. I therefore introduce another term, a binary decision function that returns 0 or 1 if Alice would publish the state based on a, b, and x: D(a, b, x):
rV(0,0)=−hAc+λA{D(1,0,x)[P(1,0)−V(0,0)]+[1−D(1,0,x)][V(1,0)−V(0,0)]}
If she publishes, she is one block ahead and gets one block reward:
[P(1,0)−V(0,0)]=R
rV(0,0)=−hAc+λA{D(1,0,x)R+[1−D(1,0,x)][V(1,0)−V(0,0)]}
If she does not publish, she gains the reward of mining with a one-block lead
V(1, 0), which in turn would either resolve to Bob catching up or Alice pulling ahead further, but is left as a black box because it was given in the problem statement.