#simple big o question

19 messages · Page 1 of 1 (latest)

raven violetBOT
sterile cobalt
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yes

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yes

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yes

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General rule (for some constant $m > 1$): $\log n < \sqrt[m]{n} < n < n^m < m^n < n!$

echo heronBOT
sterile cobalt
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Always ignore lesser terms

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That's the magic of Big O. You only care about the dominating factor

sterile cobalt
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yes

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yes

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hmmmm

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good question

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I think it becomes O(n!)

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let me review my book

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it gives formal definition of big O

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It would be O(n! n)

raven violetBOT
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Solved

Post marked as solved by @uncut sequoia.

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