Question

compare the order of growth n(n-1)/2 and n2 Co 2.5 (K5)​

Answer 1

Explanation:

Among exponentials, you can always convert them all to the same base and compare exponents; larger exponents beat smaller ones. Same for logarithms. so this is actually just a power. so that's the order of growth of the functions.

Answer 2

Answer:

$O(\frac{n(n-1)}{2} )$  growth rate is slower than $O(n^2)$.

Explanation:

The O notation, which is pronounced "big-oh," is used to express an asymptotic upper bound of a function, i.e., that a function f(n) cannot surpass another function g(n), and g(n) is therefore the upper bound of f(n).

As per the definition,

                   f(n) = O(g(n))                     , if f(n) $\leq$ c.g(n).

Analyzing the Time Complexity,  

$O(\frac{n(n-1)}{2} )$ ,    $\frac{n(n-1)}{2}=\frac{n^2-n}{2}$

The order of growth for varying input sizes of 'n' is as follows:-

$n^2$                                               $\frac{n(n-1)}{2}$

$1$                                                  $0$

$4$                                                  $1$  

$9$                                                  $3$                                    

$16$                                                $6$

We can clearly see here that the growth rate of  $O(\frac{n(n-1)}{2} )$  is slower than $O(n^2)$.

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