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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?

Vaibhav
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2 Answers2

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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

  1. Can you apply case 1?0 < c < logb(a) = 1. Is n/logn = O(n^c). No, because n/logn grow infinitely faster than n^c
  2. Can you apply case 2? No. c = 1 You need to find some k > 0 such that n/log n = Theta(n log^k n )
  3. Can you apply case 3 ? 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

  1. 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
UmNyobe
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  • UmNyobe, may be i'm also unclear about the case 1 then. Here's my understanding. Case 1 says that f(n) 'part' of the recurrence must be O(n^(log a base b - e)) where e is a constant > 0 So what's wrong with n/log n = O(n^(1-some c : 0 – Vaibhav Jan 23 '15 at 10:10
  • No `x/logx` grows [infinitely faster](http://www.wolframalpha.com/input/?i=lim++%28x%2Flogx%29%2F%28x%5E0.99999999%29) than `x^c` , 0< c <1 – UmNyobe Jan 23 '15 at 10:26
  • Mine goodness.. I thought the fraction has x divided by a function of x, viz. log x so it has got to be less than x.. So, what about your point # 3? It's not even big-omega (n^c). – Vaibhav Jan 23 '15 at 10:43
  • same as case 1. please do the math. – UmNyobe Jan 23 '15 at 10:51
  • Nope! I didn't get why you say that n/log n grows 'infinitely' faster than n^c.. and then it is not even >= c. n^c.. (big-omega). I understand the explanation of 4T(n/2), n/log n being O(n^2) to fall in case 1 for "exponent strictly bigger than the exponent of the polynomial part of n/log(n)" from Simone's answer.. – Vaibhav Jan 23 '15 at 11:13
  • I guess I get it now.. n/log n is not big-omega(n^c) because that case says f(n) should be big-omega(n^(log a base b + e)) : e > 0. It is because of this e.. right? – Vaibhav Jan 23 '15 at 13:00
  • UmNyobe, can you please confirm that hence, all rec. of the form T(n) = a T(n/a) + n/log n are not solvable by MM.. ? – Vaibhav Jan 23 '15 at 13:16
  • looking the way you stated seems more complicated than anything else. I used a proof by contradiction. n^c, c> 1 is big omega n. if f(n) is big omega n^c, then f is big omega n (big O and omega are transitive). Showing that n/log n is not even big omega of function `n` means it is not going to be big omega n^c, c > 1. You can also take one value of c (like 2) and show that it doesn work. And yes `T(n) = a T(n/a) + n/log n` are not solvable by MM is the conclusion you can have. – UmNyobe Jan 23 '15 at 14:16
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This is because in Q18 we have a = 4 and b = 2, thus we get that n^{log(b,a)} = n^2 which has an exponent strictly bigger than the exponent of the polynomial part of n/log(n).

Simone
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  • Simone, so what is the story between 'strictly' (n^2) and 'seemingly' (n) bigger in contrast to 2T(n/2) (the first rec.)? f(n) = n/log n is O(n) and O(n^2).. – Vaibhav Jan 23 '15 at 10:03