Is n to the power n (i-e n^n) a polynomial? Can T(n) = 2T(n/2) + n^n be solved using the master method?
It's not only not polynomial, it's also worse than factiorial. O(n^n) dominates O(n!). Also in the masters method f(n) must be polynomial, so you can not use it.
Related
Why Big-O notation can not compare algorithms in the same complexity class. Please explain, I can not find any detailed explanation.
So, O(n^2) says that this algorithm requires less or equal number of operations to perform. So, when you have algorithm A which requires f(n) = 1000n^2 + 2000n + 3000 operations and algorithm B which requires g(n) = n^2 + 10^20 operations. They're both O(n^2)
For small n the first algorithm will perform better than the second one. And for big ns second algorithm looks better since it has 1 * n^2, but first has 1000 * n^2.
Also, h(n) = n is also O(n^2) and k(n) = 5 is O(n^2). So, I can say that k(n) is better than h(n) because I know how these functions look like.
Consider the case when I don't know how functions k(n) and h(n) look like. The only thing I'm given is k(n) ~ O(n^2), h(n) ~ O(n^2). Can I say which function is better? No.
Summary
You can't say which function is better because Big O notation stays for less or equal. And following is true
O(1) is O(n^2)
O(n) is O(n^2)
How to compare functions?
There is Big Omega notation which stays for greater or equal, for example f(n) = n^2 + n + 1, this function is Omega(n^2) and Omega(n) and Omega(1). When function has complexity equal to some asymptotic, Big Theta is used, so for f(n) described above we can say that:
f(n) is O(n^3)
f(n) is O(n^2)
f(n) is Omega(n^2)
f(n) is Omega(n)
f(n) is Theta(n^2) // this is only one way we can describe f(n) using theta notation
So, to compare asymptotics of functions you need to use Theta instead of Big O or Omega.
T(n) ={ 2T(n/2) + n^2 when n is even and T(n) = 2T(n/2) + n^3 when n is odd
I solved this separately and i am getting the solution as theta(n^2) if n is even and theta(n^3) if n is odd from case 3 of master's theorem. But i am not supposed to solve this problem separately.
How to solve a recurrence relation like this together?
T(n) ={ 2T(n/2) + n^2 when n is even and T(n) = 2T(n/2) + n^3 when n is odd
Is it solvable by master's theorem or master's theorem does not apply?
Kindly help me with this.
Suppose n = 2^k for some integer k, so n equals to 100...00. Then you can apply master method the even part of the recurrence. and obtain theta(n^2).
Now suppose there is also 1 not in the most significant bit, e.g. 100100..00. So, you will have at least one level in your recursion-tree all nodes of which add up to n^3 * constant, and by this you obtain theta(n^3).
Thus, the answer is theta(n^2) if n is a power of two and theta(n^3) otherwise. But if we first encounter odd n and it is equal to a base case then it might not be cubic.
After some chatting with kelalaka it came to me that if first 1 is k-th from the right in n then if k > (2/3)(1/lg 2)lg n, we don't care any more about (n/2^k)^3. It is still O(n^2).
So the answer to this question What is the difference between Θ(n) and O(n)?
states that "Basically when we say an algorithm is of O(n), it's also O(n2), O(n1000000), O(2n), ... but a Θ(n) algorithm is not Θ(n2)."
I understand Big O to represent upper bound or worst case with that I don't understand how O(n) is also O(n2) and the other cases worse than O(n).
Perhaps I have some fundamental misunderstandings. Please help me understand this as I have been struggling for a while.
Thanks.
It's helpful to think of what big-Oh means: if a function is O(n), then c*n, where c is some positive number, is the upper-bound. If c*n is an upper-bound, it's clear that for integers, c*n^2 would also be an upper-bound. Also c*n^3, c*n^4, c*n^1000, etc.
The below graph shows the growth of functions, which are upper bounds of the function "to the right" of it; i.e., it grows faster on smaller n.
Suppose the running time of your algorithm is T(n) = 3n + 6 (i.e., an arbitrary polynomial of order 1).
It's true that T(n) = O(n) because 3n + 6 < 4n for all n > 5 (to use the definition of big-oh notation). It's also true that T(n) = O(n^2) because 3n + 6 < n^2 for all n > 5 (to use the defintion again).
It's also true that T(n) = Θ(n) because, in addition to the proof that it was O(n), it is true that 3n + 6 > n for all n > 1. However, you cannot prove that 3n + 6 > c n^2 for any value of c for arbitrarily large n. (Proof sketch: lim (cn^2 - 3n - 6) > 0 as n -> infinity).
I understand Big O to represent upper bound or worst case with that I don't understand how O(n) is also O(n2) and the other cases worse than O(n).
Intuitively, an "upper bound of x" means that something will always be less than or equal to x. If something is less than or equal to x, it is also less than or equal to x^2 and x^1000, for large enough values of x. So x^2 and x^1000 can also be upper bounds.
This is what Big-oh represents: upper bounds.
When we say that f(n) = O(g(n)), we mean only that for all sufficiently large n, there exists a constant c such that f(n) <= cg(n). Note that if f(n) = O(g(n)), we can always choose a function h(n) bigger than g(n) and since g(n) is eventually less than h(n), we have f(n) <= cg(n) <= ch(n), so f(n) = O(h(n)) as well.
Note that the O bound is not tight. The theta bound is the intersection of O(g(n)) and Omega(g(n)), where Omega gives the lower bound (it's like O, the upper bound, but bounds from below instead). If f(n) is bounded below by g(n), and h(n) is bigger than g(n), then if follows that f(n) is not (necessarily) bounded below by h(n).
The question is :
T(n) = √2*T(n/2) + log n
I'm not sure whether the master theorem works here, and kinda stuck.
This looks more like the Akra-Bazzi theorem: http://en.wikipedia.org/wiki/Akra%E2%80%93Bazzi_method#The_formula with k=1, h=0, g(n)=log n, a=(2)^{1/2}, b=1/2. In that case, p=1/2 and you need to evaluate the integral \int_1^x log(u)/u^{3/2} du. You can use integration by parts, or a symbolic integrator. Wolfram Alpha tells me the indefinite integral is -2(log u + 2)/u^{1/2} + C, so the definite integral is 4 - 2(log x + 2)/x^{1/2}. Adding 1 and multiplying by x^{1/2}, we get T(x) = \Theta(5x^{1/2} - 2 log x - 4).
Master theorem have only constrains on your a and b which holds for your case. The fact that a is irrational and you have log(n) as your f(n) has no relation to it.
So in your case your c = log2(sqrt(2)) = 1/2. Since n^c grows faster than your log(n), the complexity of the recursion is O(sqrt(n)).
P.S. solution of Danyal is wrong as the complexity is not nlogn and the solution of Edward Doolittle is correct, also it is an overkill in this simple case.
As per master theorem, f(n) should be polynomial but here
f(n) = logn
which is not a polynomial so it can not be solved by master theorem as per rules. I read somewhere about the fourth case as well. I must mention that as well.
It is also discussed here:
Master's theorem with f(n)=log n
However, there is a limited "fourth case" for the master theorem, which allows it to apply to polylogarithmic functions.
If
f(n) = O(nlogba logk n), then T(n) = O(nlogba log k+1 n).
In other words, suppose you have T(n) = 2T (n/2) + n log n. f(n) isn't a polynomial, but f(n)=n log n, and k = 1. Therefore, T(n) = O(n log2 n)
See this handout for more information: http://cse.unl.edu/~choueiry/S06-235/files/MasterTheorem-HandoutNoNotes.pdf
I came across this problem on asymptotic complexity of a function:
The complexities of 3 functions are as follows:
f(n) = O(n)
g(n) = Big-Omega(n)
h(n) = Theta(n)
So what will be the asymptotic complexity of the resultant function [f(n).g(n)] + h(n)
I can figure out the answer to this will be Big-Omega(n) by elementary hit and trial. For example if I say f(n) = n and g(n) = n and h(n) = n. So we can say f(n) is O(n) and g(n) is Big-Omega(n) and h(n) is Theta(n). Now f(n).g(n) is n2 and this will be Big-Omega(n) but not O(n). Now adding this to h(n) is n2+n. Which also is Big-Omega(n) but not Theta(n).
But I'm not able to figure out a proper logical or mathematical proof to this. Can someone please help me out with this?
Here's an attempt at a logical explanation:
f(n) = O(n) means that f's running time is at most linear (may be constant time).
h(n) = Theta(n) means that h's running time running is linear.
g(n) = Big-Omega(n) means that g's running time is atleast linear (may be polynomial, exponential... we don't know).
Now let's analyse the best case: f(n) is constant time, g(n) is linear, h(n) is linear. what can we say about the function f(n)*g(n)+h(n)? that it's also linear.
What can we say about the worst case? nothing, as we have no clue about the behaviour of g(n) in the worst case.
So we can conclude that f(n)*g(n)+h(n) = Big-Omega(n) as this function is linear at the best case.