3
$\begingroup$

A similar question was asked here: Solving recurrences using substitution method, but I am still somewhat hazy as to how this process works.

Say, for $T(n) = T(\lceil n/5 \rceil + 36) + n \log n$

Statement: we wish to prove that $T(n) \le 5 \ n \log n$ for $n \ge 0$

Base case, $n = 0$ and $T(0) \le 0$

$T(0) = 0$

$T(0) \le 0 \log 0$

Holds, since $0 \log 0 = 0$

Inductive step, for $n > 0$

(Inductive Hypothesis is: $T(n) = 5 \ n \log n$)

$T(n) = T(\lceil n/5 \rceil + 36) + n \log n$

$\le 5 \, (\lceil n/5 \rceil + 36) \log (\lceil n/5 \rceil + 36) + n \log n$ by Ind. Hyp.

$\le 5 \, (n/5 + 36) \log (n/5 + 36) + n \log n$ by property of ceiling

$= (n + 180) \log ((n + 180)/5) + n \log n$ by algebra

This is the point where it gets hazy. According to the link, what I should do is say that for all $n \ge 180$, then

$\le (n + n) \log ((n + n)/5) + n \log n$

$= 2n \log (2n/5) + n \log n$

$= 2n \, [\log n + \log (2/5)] + n \log n$

$= 2n \log n + 2n \log (2/5) + n \log n$

$= 3n \log n + kn$ where $k = \log (2/5) * 2$

$\le 5n \log n$

Does this complete the proof? Or am I in error? I still don't understand why I can just set $n \ge 180$ and therefore get rid of the irritating arithmetic within the log.

$\endgroup$
2
$\begingroup$

The link doesn't say what you say it does. I don't see the number 180 appearing anywhere on that link. And that link doesn't say the purpose is to prove $T(n) \le 5 n \log n$ for all $n \ge 0$; it says to prove that $T(n) = O(n \log n)$. Check the definition of big-O notation; it doesn't require proving $T(n) \le c n \log n$ for all $n \ge 0$; it suffices to prove it for all $n \ge 180$ (or for all $n \ge n_0$ for any other constant $n_0$ of your choice).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.