Is O(N!*N) an acceptable big oh complexity class or do I remove the constant and just say O(N!)?

Question

Is O(N!N) an acceptable big oh complexity class or do I remove the constant and just say O(N!)?

Answer 1

See What is O(log(n!)) and O(n!) and Stirling Approximation, which discusses the relationship between O(n!) and O(n^n). This should help you to determine the appropriate big-O when you multiply those by n.

The additional n in your problem is not a constant, and is not dominated by the n!, so it does not disappear from the function when you convert from the actual value of the function to the Big-O (or Big-Theta) asymptotic complexity class of the function.

For Big-O, it's probably sufficient to say O(n^(n+1)), but that's not sufficient for Big-Theta.

Here's a related problem involving Big-O and factorials: https://math.stackexchange.com/questions/323290/stirlings-approximation

Answer 2

Lets do some algebra ....

  N!      =  N x (N - 1) x (N - 2) x ... 1    // by definition

  N! x N  =  N x (N x (N - 1) x (N - 2) x ... 1)

          =  (N + 1 - 1) x (N x (N - 1) x (N - 2) x ... 1)

          =  ((N + 1) x N x (N - 1) x (N - 2) x ... 1) 
              - (N x (N - 1) x (N - 2) x ... 1)

          =  (N + 1)! - N!

Now, it is "intuitively obvious"¹ that O((N + 1)! - N!) is the same as O((N+1)!) ... and this can be proved from first principles.

Also, it is "intuitively obvious" that O(N!N) is N times as expensive as O(N!) which puts it into a different complexity class. (Just like O(N) and O(N^2) are different complexity classes!)

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But here's the thing. The O(N!) complexity class is clearly problematic for practical computations as N gets large. It is substantially worse than simple exponential.

  N!   ~=  N^N for large N  // by Stirling's Approximation

  N^N   =  e^(NlogN)

  e^(NlogN) >> e^N for large N

So basically, whether you have O(N!) or O(N!N) you have a big problem ... either way.

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^{1 - Statements like "intuitively obvious" are not proper maths. Obviously.}

Answer 3

O(n!*n) is rather high complexity. But since there is multiplication you cannot remove second n. But if there was addiction like o(n!+n) then n could be omitted since for big values of n, n is not significant compared to n!