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Analysis of algorithms | little o and little omega notations

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The main idea of asymptotic analysis is to have a measure of the efficiency of algorithms that don’t depend on machine-specific constants, mainly because this analysis doesn’t require algorithms to be implemented and time taken by programs to be compared. We have already discussed Three main asymptotic notations. The following 2 more asymptotic notations are used to represent the time complexity of algorithms.

Little ? asymptotic notation

Big-? is used as a tight upper bound on the growth of an algorithm’s effort (this effort is described by the function f(n)), even though, as written, it can also be a loose upper bound. “Little-?” (?()) notation is used to describe an upper bound that cannot be tight. 

Definition: Let f(n) and g(n) be functions that map positive integers to positive real numbers. We say that f(n) is ?(g(n)) (or f(n) ? ?(g(n))) if for any real constant c > 0, there exists an integer constant n0 ? 1 such that 0 ? f(n) < c*g(n).  little o and little omega notations

 Thus, little o() means loose upper-bound of f(n). Little o is a rough estimate of the maximum order of growth whereas Big-? may be the actual order of growth. 
In mathematical relation, f(n) = o(g(n)) means lim  f(n)/g(n) = 0 n?? 

Examples:

Is 7n + 8 ? o(n2)? 

In order for that to be true, for any c, we have to be able to find an n0 that makes 

f(n) < c * g(n) asymptotically true. 

lets took some example, 

If c = 100,we check the inequality is clearly true. If c = 1/100 , we’ll have to use 

a little more imagination, but we’ll be able to find an n0. (Try n0 = 1000.) From 

these examples, the conjecture appears to be correct. 

then check limits, 

lim  f(n)/g(n) = lim  (7n + 8)/(n2) = lim  7/2n = 0 (l’hospital) 

n?? n?? n?? 

hence 7n + 8 ? o(n2)

Little ? asymptotic notation

Definition : Let f(n) and g(n) be functions that map positive integers to positive real numbers. We say that f(n) is ?(g(n)) (or f(n) ? ?(g(n))) if for any real constant c > 0, there exists an integer constant n0 ? 1 such that f(n) > c * g(n) ? 0 for every integer n ? n0. 

f(n) has a higher growth rate than g(n) so main difference between Big Omega (?) and little omega (?) lies in their definitions.In the case of Big Omega f(n)=?(g(n)) and the bound is 0<=cg(n)<=f(n), but in case of little omega, it is true for 0<=c*g(n)<f(n). 

The relationship between Big Omega (?) and Little Omega (?) is similar to that of Big-? and Little o except that now we are looking at the lower bounds. Little Omega (?) is a rough estimate of the order of the growth whereas Big Omega (?) may represent exact order of growth. We use ? notation to denote a lower bound that is not asymptotically tight. And, f(n) ? ?(g(n)) if and only if g(n) ? ?((f(n)). 

In mathematical relation, 

if f(n) ? ?(g(n)) then, 

lim  f(n)/g(n) = ? 

n?? 

Example: 

Prove that 4n + 6 ? ?(1); 

the little omega(?) running time can be proven by applying limit formula given below. 

if lim  f(n)/g(n) = ? then functions f(n) is ?(g(n)) 

n?? 

here,we have functions f(n)=4n+6 and g(n)=1 

lim   (4n+6)/(1) = ? 

n?? 

and,also for any c we can get n0 for this inequality 0 <= c*g(n) < f(n), 0 <= c*1 < 4n+6 

Hence proved. 

 


Last Updated : 18 Sep, 2023
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