I recently stumbled upon a resource where the 2T(n/2) + n/log n type of recurrences were declared unsolvable by MM.
I accepted it as a lemma, until today, when another resource proved to be a contradiction (in some sense).
As per the resource (link below): Q7 and Q18 in it are the rec. 1 and 2 respectively in the question whereby, the answer to Q7 says it can't be solved by giving the reason 'Polynomial difference b/w f(n) and n^(log a base b)'. On the contrary, answer 18 solves the second recurrence (in the question here) using case 1.
http://www.csd.uwo.ca/~moreno/CS433-CS9624/Resources/master.pdf
Can somebody please clear the confusion?
If you try to apply the master theorem to
T(n) = 2T(n/2) + n/log n
You consider a = 2, b = 2 which means logb(a) = 1
0 < c < logb(a) = 1. Is n/logn = O(n^c). No, because n/logn grow infinitely faster than n^cc = 1 You need to find some k > 0 such that n/log n = Theta(n log^k n )c > 1, is n/logn = Big Omega(n^c) ? No because it is not even Big Omega(n)If you try to apply the master theorem to
T(n) = 4T(n/2) + n/log n
You consider a = 4, b = 2 which means logb(a) = 2
Can you apply case 1? c < logb(a) = 2. is n/logn = O(n^0) or n/logn = O(n^1). Yes indeed n/logn = O(n). Thus we have
T(n) = Theta(n^2)
note: Explanation about 0 < c <1, case 1
The case 1 is more about analytics.
f(x) = x/log(x) , g(x) = x^c , 0< c < 1
f(x) is O(g(x)) if f(x) < M g(x) after some x0, for some M finite, so
f(x) is O(g(x)) if f(x)/g(x) < M cause we know they are positive
This isnt true here We pose y = log x
f2(y) = e^y/y , g2(y) = e^cy , 0< c < 1
f2(y)/g2(y) = (e^y/y) / (e^cy) = e^(1-c)y / y , 0< c < 1
lim inf f2(y)/g2(y) = inf
lim inf f(x)/g(x) = inf