# Difference between revisions of "TADM2E 2.21"

From Algorithm Wiki

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6. True | 6. True | ||

7. False | 7. False | ||

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+ | Edit: I'm not sure why (7) is False. | ||

+ | It's possible that I'm mistaken but I think that: | ||

+ | |||

+ | <math>n^{-1/2} = \dfrac{n}{\sqrt{n}} = \sqrt{n}</math> | ||

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+ | Since a square root of N dominates logn: | ||

+ | |||

+ | <math>\log n = O(\sqrt{n})</math> | ||

+ | |||

+ | Then 7 should be True. |

## Revision as of 18:26, 11 November 2018

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.