I am learning about Masters Theorem and solving recurrence relations using it, but I can only find questions like:
T(n) = 16T(n/2) + 1
On google, where Masters Theorem is clearly applicable. I just came across a question:
T(n) = 2^(n/2) T(n/2) + 2^m
Can I apply Masters Theorem in such a case?
Related
My solution:
I am trying to find running time of T(n) = T(n/2) + T(n/4) + Θ(n) but I have different answers using recursion tree and master theorem. Was I wrong from the recursion tree diagram part or did I use wrong asymptotic notation?
You can't use the master theorem (not match to any cases). You should use the Akra-Bazzi theorem (a general case of the master theorem).
Master theorem can be used to solve recurrence relations like
T(n)= aT(n/b)+f(n).
So, if f(n)=O(n) or if f(n)=cn are both the values same?
can I use master theorem for f(n)=cn also?
Asumming that c is a constant and that I understand your question correctly, the solution will be the same for both f(n) = O(n) and f(n) = cn, since cn = O(n) and thus the Master theorem can be applied to solve the recurrance.
If I understood the question correctly, f(n)=cn (where c is a constant) is in O(n); the master theorem can be applied.
I am quite frustrated over this.
In CLRS 3rd edition, page 95 (chapter 4.5), it mentions that recurrences like
T(n) = 2T(n/2) + n lg n
cannot be solved with the Master Theorem because the difference
f(n)/n^(log_b(a)) = (n lg n)/n^1 = lg n
is not polynomial.
But then I come across pages like this this where, at the bottom of the page, it mentions the exact same recurrence and says that it CAN be solved with the Master Theorem because it falls into an "extended case 2" even though the difference is non-polynomial. It becomes n lg^2 n (incrementing the log factor on f(n) by one).
Then I come across pages like this where in example (e) seems like a clear application of Extended Case 2 (the recurrence is T(n) = 4T(n/2) + n^2 lg n), but then the solution is not n^2 log^2 n, but rather n^2 log n! Am I wrong or is the paper wrong?
Can anyone please clear up the contradictions and make it very clear exactly when Master Theorem can be used and when it cannot? When does the polynomial-difference check matter, and when does it not? Is the extended case 2 usable, or does it actually violate something?
EDIT:
I tried solving recurrence (e) directly from the second paper and I get:
T(n) = n^2 lg^2(n)/2 + n^2 lg(n)/2
Is this not big theta n^2 lg^2 n?
The book states that it cannot be solved using Case 3:
even though it appears to have the proper form: ... You might mistakenly think that case 3 should apply
However, this recurrence formula can be solved using master theorem, case 2.
T(n) = 2T(n/2) + nlgn:
We define:
a = 2, b = 2, f(n) = nlgn
Using Master theorem case 2:
c = log_2(2) = 1
k = 1
And f(n) is indeed in Theta(nlogn).
So, all conditions to master theorem case 2 apply, and we can deduce:
T(n) is in Theta(n^c * log(n)^(k+1)) = Theta(n*log(n)^2)
Long story short, Master theorem have 3 cases. Each case have it's prerequisites to be applied. Case 3 have more complicated prerequisites, because it also requires convergence.
Since the prerequisites for case 3 does not apply for this formula, you cannot use case 3. However, the prerequisites of case 2 - do apply, and you can use it.
I am trying to apply the Master's Theorem to a recurrence of this type:
T(n) = T(n/2) + 2^n
However, f(n) = 2^n doesn't seem to fit any of the three cases described in the master's theorem, which all seem to have base n instead of base 2. How can I solve a recurrence of this type, could anyone please help ? Thanks.
If none of the cases of the theorem applies, then the theorem won't solve your recurrence. It can't solve every single recurrence out there.
To address your issue: what you get by repeatedly substituting the recursive case is T(n) = 2^n + 2^(n/2) + 2^(n/4) + ... + 2, and since there are log n many terms to add up, you end up with something below 2^(n+1), so in total you're in Θ(2^n).
We can take log on both sides and solve. It will fall in case 3 of Master's theorem.
How do you go about finding the asymptotic complexity based off a running time? For example:
If the run time of a recursive algorithm is given as
T(n) = 2 T(n/2) + O(n)
considering the Master Theorem, what is its asymptotic complexity?
Could someone explain the steps on how to figure this out?
For the Master Theorem, there are 3 different cases to solve.
First step is to understand which case applies.
For questions involving Master Theorem, our general formula is T(n) = aT(n/b) + f(n)
So first thing to do is compare n^(logb a) with f(n).
Simply whichever is bigger that's the complexity of it(these are Cases 1 and 3) and if they are equal then you multiply the result with lgn, such as if you have a case like T(n) = 16 T(n/4) + O(n^2) then
n^(logb a) would make n^2 as well as f(n) = n^2 so the answer would be Θ(n^(2)*lgn), which is by using the case #2 of the Master Theorem.