Difference between revisions of "2.47"
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Revision as of 19:45, 10 September 2020
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 :
[math]\displaystyle{ O(nlog(√n)) }[/math] and [math]\displaystyle{ 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]\displaystyle{ \lim_{x\to\infty} (nlog(√n)) / (nlog(n)) }[/math]
= [math]\displaystyle{ \lim_{x\to\infty} 1/2*(nlog(n))/(nlog(n)) }[/math]
= [math]\displaystyle{ 1/2 }[/math]
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