- #1

- 16

- 0

Hey,

Is nlogn (or more specifically nlog[base2]n) the same as: n multiplied by logn?

Thanks

Is nlogn (or more specifically nlog[base2]n) the same as: n multiplied by logn?

Thanks

- Thread starter pjhphysics
- Start date

- #1

- 16

- 0

Hey,

Is nlogn (or more specifically nlog[base2]n) the same as: n multiplied by logn?

Thanks

Is nlogn (or more specifically nlog[base2]n) the same as: n multiplied by logn?

Thanks

- #2

- 811

- 6

I am guessing you're talking about Big Oh notation. An interesting thing about Big Oh is that it doesn't matter what base log you're referring to. Given two bases, log_a(x) and log_b(x) will always be proportional to each other for all x. Big Oh notation ignores scalar differences between functions, so O(log_a(x)) = O(log_b(x)).

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