math checks out. log2(8 billion) ~= 32.9
It's a little more complicated than that. You have to be summing everyone who is still tied to all the previous tracks. It needs to be a geometric sum formula.
It'll just be one fewer junctions. 2^n is always one more than the sum of 2^1+...2^(n-1)
I think you have to include 2^0 for that to be true?
e.g 2^0 = 1, 2^1 = 2 2^0 + 2^1 = 1 + 2 = 3, 2^2 = 4 … 7, 8 15,16 31, 32 etc.
math checks out. log2(8 billion) ~= 32.9
It's a little more complicated than that. You have to be summing everyone who is still tied to all the previous tracks. It needs to be a geometric sum formula.
It'll just be one fewer junctions. 2^n is always one more than the sum of 2^1+...2^(n-1)
I think you have to include 2^0 for that to be true?
e.g 2^0 = 1, 2^1 = 2 2^0 + 2^1 = 1 + 2 = 3, 2^2 = 4 … 7, 8 15,16 31, 32 etc.