First of All Theory
Big O = Upper Limit O(n)
Theta = Order Function - theta(n)
Omega = Q-Notation(Lower Limit) Q(n)
In many Blogs & Books How this Statement is emphasised is Like
"This is Big O(n^3)" etc.
and people often Confuse like weather
O(n) == theta(n) == Q(n)
But What Worth keeping in mind is They Are Just Mathematical Function With Names O, Theta & Omega
so they have same General Formula of Polynomial,
Let,
f(n) = 2n4 + 100n2 + 10n + 50 then,
g(n) = n4, So g(n) is Function which Take function as Input and returns Variable with Biggerst Power,
Same f(n) & g(n) for Below all explainations
Big O(n4) = 3n4, Because 3n4 > 2n4
3n4 is value of Big O(n4) Just like f(x) = 3x
n4 is playing a role of x here so,
Replacing n4 with x'so, Big O(x') = 2x', Now we both are happy General Concept is
So 0 = f(n) = O(x')
O(x') = cg(n) = 3n4
Putting Value,
0 = 2n4 + 100n2 + 10n + 50 = 3n4
3n4 is our Upper Bound
Theta(n4) = cg(n) = 2n4 Because 2n4 = Our Example f(n)
2n4 is Value of Theta(n4)
so, 0 = cg(n) = f(n)
0 = 2n4 = 2n4 + 100n2 + 10n + 50
2n4 is our Lower Bound
This is Calculated to find out that weather lower Bound is similar to Upper bound,
Case 1). Upper Bound is Similar to Lower Bound
if Upper Bound is Similar to Lower Bound, The Average Case is Similar
Example, 2n4 = f(x) = 2n4,
Then Omega(n) = 2n4
Case 2). if Upper Bound is not Similar to Lower Bound
in this case, Omega(n) is Not fixed but Omega(n) is the set of functions with the same order of growth as g(n).
Example 2n4 = f(x) = 3n4, This is Our Default Case,
Then, Omega(n) = c'n4, is a set of functions with 2 = c' = 3
Hope This Explained!!