How to calculate big-theta

Question

Can some one provide me a real time example for how to calculate big theta.

Is big theta some thing like average case, (min-max)/2?

I mean (minimum time - big O)/2

Please correct me if I am wrong, thanks

Answer 1

Big-theta notation represents the following rule:

For any two functions f(n) , g(n) , if f(n)/g(n) and g(n)/f(n) are both bounded as n grows to infinity, then f = Θ(g) and g = Θ(f) . In that case, g is both an upper bound and a lower bound on the growth of f .

Here's an example algorithm:

def find-minimum(List) 
  min = +∞
  foreach value in List 
    min = value if min > value
  return min

We wish to evaluate the cost function c(n) where n is the size of the input list. This algorithm will perform one comparison for every item in the list, so c(n) = n .

c(n)/n = 1 which remains bounded as n goes to infinity, so c(n) grows no faster than n . This is what is meant by big-O notation c(n) = O(n) . Conversely, n/C(n) = 1 also remains bounded, so c(n) grows no slower than n . Since it grows neither slower nor faster, it must grow at the same speed. This is what is meant by theta notation c(n) = Θ(n) .

Note that c(n)/n² is also bounded, so c(n) = O(n²) as well — big-O notation is merely an upper bound on the complexity, so any O(n) function is also O(n²) , O(n³) ...

However, since n²/c(n) = n is not bounded, then c(n) ≠ Θ(n²) . This is the interesting property of big-theta notation: it's both an upper bound and a lower bound on the complexity.

Answer 2

Big theta is a tight bound, for a function T(n) : if: Omega(f(n))<=T(n)<=O(f(n)) , then Theta(f(n)) is the tight bound for T(n).

In other words Theta(f(n)) 'describes' a function T(n), if both O [big O] and Omega, 'describe' the same T, with the same f.

for example, a quicksort [with correct median choices], always takes at most O(nlogn), at at least Omega(nlogn), so quicksort [with good median choices] is Theta(nlogn)

EDIT: added discussion in comments:
Searching an array is still Theta(n). the Theta function does not indicate worst/best case, but the behavior of the desired case. ie, searching for an array, T(n)=number of ops for worst case. in here, obviously T(n)<=O(n) , but also T(n)>=n/2 , because at worst case you need to iterate the whole array, so T(n)>=Omega(n) and therefore Theta(n) is asymptotic bound.

Answer 3

From http://en.wikipedia.org/wiki/Big_O_notation#Related_asymptotic_notations , we learn that "Big O" denotes an upper bound, whereas "Big Theta" denotes an upper and lower bound, ie in the limit as n goes to infinity:

f(n) = O(g(n))      -->    |f(n)| < k.g(n)

f(n) = Theta(g(n))  -->    k1.g(n) < f(n) < k2.g(n)

So you cannot infer Big Theta from Big O.

Answer 4

ig-Theta (Θ) notation provides an asymptotic upper and lower bound on the growth rate of an algorithm's running time. To calculate the big-Theta notation of a function, you need to find two non-negative functions, f(n) and g(n), such that:

There exist positive constants c1, c2 and n0 such that 0 <= c1 * g(n) <= f(n) <= c2 * g(n) for all n >= n0. f(n) and g(n) have the same asymptotic growth rate. The big-Theta notation for the function f(n) is then written as Θ(g(n)). The purpose of this notation is to provide a rough estimate of the running time, ignoring lower order terms and constant factors.

For example, consider the function f(n) = 2n^2 + 3n + 1. To calculate its big-Theta notation, we can choose g(n) = n^2. Then, we can find c1 and c2 such that 0 <= c1 * n^2 <= 2n^2 + 3n + 1 <= c2 * n^2 for all n >= n0. For example, c1 = 1/2 and c2 = 2. So, f(n) = Θ(n^2).