Order of Growth

Let f(n) and g(n) be the time taken by two algorithms where n >= 9 and f(n) and g(n) are also greater than equal to 0. A function f(n) is said to be growing faster than g(n) if g(n)/f(n) for n tends to infinity is 0 (or f(n)/g(n) for n tends to infinity is infinity).

Example 1: f(n) = 1000, g(n) = n + 1
For n > 999, g(n) would always be greater than f(n) because order of growth of g(n) is more than f(n).

Example 2: f(n) = 4n² , g(n) = 2n + 2000
f(n) has higher order of growth as it grows quadratically in terms of input size.

How do we Quickly find order of Growth?

When n >= 0, f(n) >= 0 and g(n) >= 0, we can use the below steps.

• Ignore the order terms.
• Ignore the constants

For example,

Example 1 : 4n² + 3n + 100
After ignoring lower order terms, we get
4n²
After ignoring constants, we get
n²
Hence order of growth is n²

Example 1 : 100 n Log n + 3n + 100 Log n + 2
After ignoring lower order terms, we get
100 n Log n
After ignoring constants, we get
n Log n
Hence order of growth is n Log n

How do we compare two order of growths?
The following are some standard terms that we must remember for comparison.

c < Log Log n < Log n < n^1/3 < n^1/2 < n < n Log n < n² < n² Log n < n³ < n⁴ < 2ⁿ < nⁿ

Here c is a constant

Next Read

• Asymptotic Analysis
• Worst, Average and Best Cases of Algorithms
• Asymptotic Notations

Order of Growth

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