I am trying to solve the following recurrence:
$T(n) = 3T(n^{\frac{2}{3}}) + \log n$
but am not sure how to do so, since Master theorem does not apply. I tried to draw the recursion tree as follows:
but am not sure where to go from there, such as trying to figure out the height of the tree or the number of nodes in the last layer. Any guidance on how to find the overall big theta of the recurrence would be appreciated.
As you expand the formula, we will have:
T(n) = 3 log(n^{2/3}) + 3^2 log(n^((2/3)^2)) + ... + 3^k log(n^((2/3)^k)) + log(n)
In the above equation, k is the height of the tree. If we suppose n = 2 ^ ((3/2)^k), finally we will have 2 in n^((2/3)^k). Hence, k = log_{3/2)(log(n)). Also, we know that log(n^a) = a log(n):
T(n) = 2 log(n) + 2^2 log(n) + ... + 2^k log(n) + log(n) =
log(n) (1 + 2 + 2^2 + ... + 2^k) =
(2^(k+1) - 1) log(n)
Hence, as 2^k = O(log^2(n)), T(n) = O(log^2(n) * log(n)) = \O(log^3(n)).
Related
Can anybody please explain the time complexity of T(n)=2T(n/4)+O(1) using recurrence tree? I saw somewhere it says O(n^1/2).
Just expand the equation for some iteration, and use the mathematical induction to prove the observed pattern:
T(n) = 2T(n/4) + 1 = 2(2T(n/4^2) + 1) + 1 = 2^2 T(n/4^2) + 2 + 1
Hence:
T(n) = 1 + 2 + 2^2 + ... + 2^k = 2^(k+1) - 1 \in O(2^(k+1))
What is k? from the expansion 4^k = n. So, k = 1/2 log(n). Thus, T(n) \in O(2^(1/2 log(n) + 1)) = O(sqrt(n)). Note that 2^log(n) = n.
A divide and conquer algorithm solves a problem of size n by dividing it into 2
subproblems, each of size n-1, and takes O(n) time to combine their solutions. What is the runtime of this algorithm?
I'm not quite sure how to structure this recurrence relation and determine what the runtime is. Is the following relation correct?
T(n) = 2T(n-1) + O(n)
How can I get the runtime from this, if so?
Thank you so much!
Yes, your recurrence relation correctly describes your problem. To make things concrete, let's say the recurrence relation is: T(n) = 2T(n-1) + n (that is +n rather than +O(n).
Then, telescoping the recurrence relation (and assuming T(0) = 0).
T(n) = n + 2(n-1) + 4(n-2) + 8(n-3) + ... + 2^n(n-n)
= (1 + 2 + 4 + ... + 2^n)n - (0*2^0 + 1*2^1 + ... + n*2^n)
= n*(2^(n+1)-1) - 2(n*2^n-2^n+1)
= 2^(n+1) - n - 2
Checking this is correct:
2T(n-1) + n
= 2(2^n - (n-1) - 2) + n
= (2^(n+1) - 2n + 2 - 4) + n
= 2^(n+1) - n - 2
= T(n)
How do I determine the running time (in terms of Big-Theta) for the algorithm of input size n that satisfies recurrence relation T(n) = T(n-1)+n where n >= 1 and with initial condition T(1) = 1?
Edit: I was practicing a past exam paper. Got stuck on this question. Need guidance
Look at it this way: T(n) = T(n-1) + n = T(n-2) + (n-1) + n = T(n-3) + (n-2) + (n-1) + n. Which means if n >= 1 then you will get something like T(n) = 1 + 2 + 3 + ... + n. If you work out the pattern of this series you will see that (n+1)n/2. Therefore, Ө(n^2)
I thought it would be something like this...
T(n) = 2T(n-1) + O(n)
= 2(2T(n-2)+(n-1)) + (n)
= 2(2(2T(n-3)+(n-2))+(n-1))+(n)
= 8T(n-3) + 4(n-2) + 2(n-1) + n
Which ends up being something like the summation of 2i * (n-i), and my book says this ends up being O(2n). Could anybody explain this to me? I don't understand why it's 2n and not just O(n) as the (n-i) will continue n times.
This recurrence has already been solved on Math Stack Exchange. As I solve this recurrence, I get:
T(n) = n + 2(T(n-1))
= n + 2(n - 1 + 2T(n-2)) = 3n - 2 + 2^2(T(n-2))
= 3n - 2 + 4(n - 2 + 2(T(n-3))) = 7n - 10 + 2^3(T(n-3))
= 7n - 10 + 8(n - 3 + 2(T(n-4))) = 15n - 34 + 2^4(T(n-4))
= (2^4 - 1)n - 34 + 2^4(T(n-4))
...and so on.
Effectively the recurrence boils down to:
T(n) = (2n+1) * T(1) − n − 2
See the Math Stack Exchange link for how we arrive at this solution. Taking T(1) to be constant, the dominating factor in the above recurrence is (2(n + 1)).
Therefore, the rate of growth of given recurrence is O(2n).
I have this recurrence:
T(n)= 2T(n/2) + (n-1)
My try is as follow:
the tree is like this:
T(n) = 2T(n/2) + (n-1)
T(n/2) = 2T(n/4) + ((n/2)-1)
T(n/4) = 2T(n/8) + ((n/4)-1)
...
the hight of the tree : (n/(2h))-1 = 1 ⇒ h = lg n - 1 = lg n - lg 2
the cost of the last level : 2h = 2lg n - lg 2 = (1/2) n
the cost of all levels until level h-1 : Σi=0,...,lg(2n) n - (2i-1), which is a geometric series and equals (1/2)((1/2)n-1)
So, T(n) = Θ(n lg n)
my question is: Is that right?
No, it isn't. You have the cost of the last level wrong, so what you derived from that is also wrong.
(I'm assuming you want to find the complexity yourself, so no more hints unless you ask.)
Edit: Some hints, as requested
To find the complexity, one usually helpful method is to recursively apply the equation and insert the result into the first,
T(n) = 2*T(n/2) + (n-1)
= 2*(2*T(n/4) + (n/2-1)) + (n-1)
= 4*T(n/4) + (n-2) + (n-1)
= 4*T(n/4) + 2*n - 3
= 4*(2*T(n/8) + (n/4-1)) + 2*n - 3
= ...
That often leads to a closed formula you can prove via induction (you don't need to carry out the proof if you have enough experience, then you see the correctness without writing down the proof).
Spoiler: You can look up the complexity in almost any resource dealing with the Master Theorem.
This can be easily solved with Masters theorem.
You have a=2, b=2, f(n) = n - 1 = O(n) and therefore c = log2(2) = 1. This falls into the first case of Master's theorem, which means that the complexity is O(n^c) = O(n)