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Jan 5, 2018 at 19:28 vote accept someone1
Jan 5, 2018 at 11:33 answer added Yuval Filmus timeline score: 0
Jan 4, 2018 at 17:50 comment added someone1 Right. So then what would a good bound be? Both the substitutions I've made probably lead to different bounds as well.
Jan 4, 2018 at 17:38 comment added Raphael No, you have $n^2$ in $f$ but $\log_b a = 1$.
Jan 4, 2018 at 17:05 comment added someone1 Doesn't the accepted version's case for c = -1 work?
Jan 4, 2018 at 16:19 comment added Raphael I don't why it's been accepted, it clearly doesn't apply to the question.
S Jan 4, 2018 at 16:11 history suggested Nehorai Elbaz CC BY-SA 3.0
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Jan 4, 2018 at 16:11 review Suggested edits
S Jan 4, 2018 at 16:11
Jan 4, 2018 at 15:37 comment added someone1 cs.stackexchange.com/questions/86194/… Please see the accepted answer for this question.
Jan 4, 2018 at 15:35 comment added Raphael "The solution shows a solution using master method to prove the answer as O(n)" -- please re-check both the problem statement and the solution, because this seems wrong. Case 1 and 2 certainly don't apply since here, $f \in \omega(n \log^k n)$ for all $k$. I didn't check if case 3 applies, but if it does the answer is not $O(n)$.
Jan 4, 2018 at 15:29 history edited Raphael
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S Jan 4, 2018 at 15:20 history suggested Nehorai Elbaz CC BY-SA 3.0
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Jan 4, 2018 at 15:13 review Suggested edits
S Jan 4, 2018 at 15:20
S Jan 4, 2018 at 15:12 history suggested Nehorai Elbaz CC BY-SA 3.0
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Jan 4, 2018 at 15:10 review Suggested edits
S Jan 4, 2018 at 15:12
Jan 4, 2018 at 14:58 history asked someone1 CC BY-SA 3.0