Algorithms
October 12, 2023Algorithms
October 12, 2023Asymptotic-Complexity
| Question 1 |
Which one of the following options arranges the functions in the increasing order of asymptotic growth rate?
| f2, f3, f1 | |
| f3, f2, f1 | |
| f2, f1, f3 | |
| f1, f2, f3 |
The asymptotic notation order should be
Constant < logarithmic < linear < polynomial < exponential < factorial
F2 and F3 → Polynomial
F1 → Exponential
By the order of asymptotic notations F1 is definitely larger.
Method-1:
Consider n=100
F1 : 100 ^100 ⇒ 1.e+200
F2 : N^log(100) base 2 ⇒ 100 ^ log(100) base 2 ⇒ 100^6.6438561897747 = 1,93,96,00,91,15,564.181300016469223466
F3 : N^Sqrt(n) ====> 100^Sqrt(100) ⇒ 100^10 ⇒ 10,00,00,00,00,00,00,00,00,000
Method-2:
We can apply “log” on both sides.
log(F1)=nlog10 (base 2)
log(F2)=(logn)^2 = logn * logn (base 2)
log(F3)=sqrt(n)logn (base 2)
Answer: F2< F3< F1
The asymptotic notation order should be
Constant < logarithmic < linear < polynomial < exponential < factorial
F2 and F3 → Polynomial
F1 → Exponential
By the order of asymptotic notations F1 is definitely larger.
Method-1:
Consider n=100
F1 : 100 ^100 ⇒ 1.e+200
F2 : N^log(100) base 2 ⇒ 100 ^ log(100) base 2 ⇒ 100^6.6438561897747 = 1,93,96,00,91,15,564.181300016469223466
F3 : N^Sqrt(n) ====> 100^Sqrt(100) ⇒ 100^10 ⇒ 10,00,00,00,00,00,00,00,00,000
Method-2:
We can apply “log” on both sides.
log(F1)=nlog10 (base 2)
log(F2)=(logn)^2 = logn * logn (base 2)
log(F3)=sqrt(n)logn (base 2)
Answer: F2< F3< F1