4 ms·
I like this general approach, but presumably you don't want the point where 50% of the sequences have never been transmitted (on average, assuming randomly dist
by x1798DE 9y ago
I like this general approach, but presumably you don't want the point where 50% of the sequences have never been transmitted (on average, assuming randomly distributed data), you want the first point where at least 1 hasn't been transmitted, so you should be solving for the point where the probability of a sequence of that length not appearing times the number of sequences of that length is equal to 1.
That said it's a logarithmic scale so the length is probably not that much shorter than the one he came up with.
- mirimir 9y agoHow can you ever do better than a probabilistic estimate for the shortest sequence length where one has never been transmitted?
- deleted 9y ago[deleted]
- nkurz 9y agothe length is probably not that much shorter than the one he came up with What's your intuition for what "not that much shorter" means here? I'm not seeing the math clearly, but it seems like solving for 1/2 vs solving for 1/(2^many) would make a big difference when many is large double digits. I'm not sure how big "big" is, but I'd guess it would be large enough to dwarf the chosen two decimal places and 2-bit difference over a 5 year projection. Another factor that would move things in the same direction (of a shorter string being the correct answer) is that if we are presuming random bits, I think we can multiply the total number bits transmitted by the length of the bit string. This would be to account for the fact that the match doesn't have to start on any particular boundary, rather the target can be "swept through" the entire corpus bit-by-bit.