TADM2E 2.21

From Algorithm Wiki
Revision as of 04:52, 28 November 2018 by Kelvinjhwong (Talk | contribs)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search

First we should see the dominance relations given in the book as it will help in determining the complexity for cases involving square root of n.

1. True 2. False, sqrt(n) dominates logn. 3. True 4. False 5. True 6. True 7. False


Edit: I'm not sure why (7) is False. It's possible that I'm mistaken but I think that:

$ n^{-1/2} = \dfrac{n}{\sqrt{n}} = \sqrt{n} $

Since a square root of N dominates logn:

$ \log n = O(\sqrt{n}) $

Then 7 should be True.

Edit on Edit: The mistake here is that $ n^{-1/2} = \dfrac{1}{\sqrt{n}} $ and not $ \sqrt{n} $. Since $ \dfrac{1}{\sqrt{n}} $ grows slowly compared to $ \log n $, the answer is False.

(Addendum on Edit on Edit: $ \dfrac{1}{\sqrt{n}} $ doesn't grow at all, it tends to zero as n goes to positive infinity)