> Already, we have a factor of O(log(n)) here. Doesn’t that mean that O(log(n)) ... | Hacker News

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		layer8 9 days ago \| parent \| context \| favorite \| on: Faster Than Dijkstra? > Already, we have a factor of O(log(n)) here. Doesn’t that mean that O(log(n)) is really O(log²(n))?

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thfuran 9 days ago [–]

You have to define what n is.

layer8 8 days ago | [–]

It’s clear from the parent comment that the number of bits needed to represent the input is meant here.

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