Difference between solving T(n) = 2T(n/2) + n/log n and T(n) = 4T(n/2) + n/log n using Master Method

Question

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?

Solution 1

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

Solution 2

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).

Comments

[Vaibhav]

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]

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<c<1)) ? So, may be not case 3 but can apply case 1 to both..

[UmNyobe]

No x/logx grows infinitely faster than x^c , 0< c <1

[Vaibhav]

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).

[UmNyobe]

same as case 1. please do the math.

[Vaibhav]

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]

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]

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.. ?

[UmNyobe]

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.