Re:Memory access is O(N^[1/3])
by dooglius on Wednesday October 08, 2025 @06:48PM
(#45521488)
> if you model memory access as O(N^1/3), linear search worse case is not O(N), it is O(N^4/3)
This would be true if we modeled memory access as Theta(N^{1/3}) but that's not the claim. One can imagine the data organized/prefetched in such a way that a linear access scan is O(1) but a random access is expected Theta(N^{1/3}). You see this same sort of pattern with well-known data structures like a (balanced) binary tree; random access is O(log(n)), but a linear scan is O(n).
This would be true if we modeled memory access as Theta(N^{1/3}) but that's not the claim. One can imagine the data organized/prefetched in such a way that a linear access scan is O(1) but a random access is expected Theta(N^{1/3}). You see this same sort of pattern with well-known data structures like a (balanced) binary tree; random access is O(log(n)), but a linear scan is O(n).