# Difference between revisions of "TADM2E 2.42"

From Algorithm Wiki

(Recovering wiki) |
Manish.khkr (Talk | contribs) |
||

Line 2: | Line 2: | ||

The paper mentioned is "S. Skiena. Encroaching lists as a measure of presortedness. BIT, 28:775-784, 1988." | The paper mentioned is "S. Skiena. Encroaching lists as a measure of presortedness. BIT, 28:775-784, 1988." | ||

+ | |||

+ | |||

+ | '''Other solution :''' | ||

+ | |||

+ | <math> O(nlog(√n)) </math> and <math> O(nlog(n)) </math> belongs to same class of function with respect to Big O notation. | ||

+ | There is no difference between them other than a constant factor. | ||

+ | |||

+ | |||

+ | <math>\lim_{x\to\infty} (nlog(√n)) / (nlog(n))</math> | ||

+ | = <math>\lim_{x\to\infty} 1/2*(nlog(n))/(nlog(n)) </math> | ||

+ | = <math>1/2</math> |

## Latest revision as of 17:22, 17 December 2019

Change the assumptions of the proof.

The paper mentioned is "S. Skiena. Encroaching lists as a measure of presortedness. BIT, 28:775-784, 1988."

**Other solution :**

$ O(nlog(√n)) $ and $ O(nlog(n)) $ belongs to same class of function with respect to Big O notation. There is no difference between them other than a constant factor.

$ \lim_{x\to\infty} (nlog(√n)) / (nlog(n)) $
= $ \lim_{x\to\infty} 1/2*(nlog(n))/(nlog(n)) $
= $ 1/2 $