Time & Space Complexity Study Notes

• Example-1:

         Let f(n) = n² + n + 5. Then f(n) is O(n²), and f(n) is O(n³), but f(n) is not O(n).

• Example-2: 

         Let f(n) =  3ⁿ . Then f(n) is O(4ⁿ) but not f(n) is not O(2ⁿ)

O(1) < O(log n) < O(n) < O(n²) < O(2ⁿ)

Omega (Ω): 

• If we write f(n) = Ω(g(n)), then there exists a function f(n) such that ∀ n ≥ n_0,  f(n) ≥ cg(n) with any constant c and a positive integer n₀. Or
• f(n) = Ω(g(n)) means we can say Function g(n) is an asymptotic lower bound for f(n).
• Example-1: Let f (n) = 2n² + 4n + 10. Then f (n) is Ω(n²)

• If we write f(n) = θ(g(n)), then there exists a function f(n) such that ∀ n ≥ n_0,  c₁g(n) ≤ f(n) ≤ c₂g(n) with a positive integer n₀, any positive constants c₁ and c₂. Or
• f(n) = θ(g(n)) means we can say Function g(n) is an asymptotically tight bound for f(n).

• Example-1:

                  Let f (n) = 2n² + 4n + 10. Then f (n) is θ(n²)

• If we write f(n) = o(g(n), then there exists a function such that f(n) < c g(n) with any positive constant c and a positive integer n₀. Or
• f(n) = o(g(n) means we can say Function g(n) is an asymptotically tight upper bound of f(n).
• Example:  n^1.99 = o(n²)

• If we write f(n) = ω(g(n)), then these exists a function such that f(n) > cg(n) with any positive constant c and a positive integer n₀. Or
• f(n) = ω(g(n)) means we can say g(n) is asymptotically tight lower bound of f(n).
• Example:  n^2.00001 = ω(n²) and n² ≠ ω(n²)

3.Transitive Property

IV. Analysis of Algorithms

• Algorithm can be classified by the amount of time they need to complete compared to their input size.
• The analysis of an algorithm focuses on the complexity of algorithm which depends on time or space.

• Worst case time complexity: It is the function defined by the maximum amount of time needed by an algorithm for an input of size n.
• Average case time complexity: The average-case running time of an algorithm is an estimate of the running time for an "average" input. Computation of average-case running time entails knowing all possible input sequences, the probability distribution of occurrence of these sequences, and the running times for the individual sequences.
• Amortized Running Time: It is the time required to perform a sequence of (related) operations is averaged over all the operations performed.  Amortized analysis guarantees the average performance of each operation in the worst case.
• Best case time complexity: It is the minimum amount of time that an algorithm requires for an input of size n.  

T(N) = 2T(N/2)+bN     Here T(N/2) is substituted with 2T((N/2)/2)+b(N/2)

T(N) = 4T(N/4)+2bN (Similarly substitute for T(N/4)T(N) = 8T(N/8)+3bNAfter (i-1) substititions,T(N) = 2iT(N/2i)+ibN

Using recursion method, n element problem can be further divided into two or more sub problems. In the following 

3. Recursion Tree Method:

figure, given problem with n elements is divided into two equal sub problems of size n/2. For each level of the tree the number of elements is N. When the tree is split so evenly the sizes of all the nodes on each level. Maximum depth of tree is logN (number of levels).

Recurrence relation T(n) = 2 T(n/2) + O(n) =  Θ(Nlog(N)).

Example-1: Find the Time complexity for T(N) = 4T(N/2)+N.

Let T(N) = aT(N/b)+f(N).

Then a=4, b=2, and f(N)=N

N^log_ba = N^log₂4 = N².

f(N) is smaller than N^log_ba

Using case 1 of master theorem with ε=1.

T(N) = Θ(N^log_ba) = Θ(N²).

Example-2:  Find the Time complexity for  T(N) = 2T(N/2) = Nlog(N)

a=2, b=2, and f(N)=Nlog(N)

Using case 2 of the master theorem with k=1.

T(N) = Θ(N(logN)²).

Example-3:  Find the Time complexity for T(N) = T(N/4) + 2N

a=1, b=4, and f(N)=2N

log_ba = log₄1 = 0

N^log_b = N⁰ = 1

Using case 3: f(N) is Ω(N^0+ε) for &epsilon=1 and af(N/b) = (1)2(N/4) = N/2 = (1/4)f(N)

Therefore T(N) = Θ(N)

Example-4:  Find the Time complexity for T(N) = 9T(N/3) + N^2.5

a=9, b=3, f(N)=N^2.5

log_ba = 2, so N^log_b = N²

Using case 3 since f(N) is Ω(N^0+ε), epsilon=.5 and af(N/b)=9f(N/3)= 9(N/3)^2.5=(1/3)^0.5f(n).

Therefore T(n) is O(N^2.5)