Difference between revisions of "TADM2E 2.21"
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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: | ||
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| + | <math>n^{-1/2} = \dfrac{n}{\sqrt{n}} = \sqrt{n}</math> | ||
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| + | Since a square root of N dominates logn: | ||
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| + | <math>\log n = O(\sqrt{n})</math> | ||
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| + | 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.
