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3.66 to think in slowing clock #12

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@soulomoon

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for the first pair procedure, you have a clock tick in normal speed
for the second pair procedure, 2 times slower
3: 4 times slower per tick
n: 2^(n-1) n s p t

then to get to the next pair procedure,
you have to take two step, per step is going to be a tick of a clock in that pair pro's world.
so you get total time of getting to (n, 1)'s location.
(n, 1):2^n - 1

then you consider walk to the other direction,
walk to (n, m) from (n, 1):
you get that it take one step to (n, 2)
and two step each to the next.
then you have every thing you get.

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