Question

Given a algorithm with f(n) 5n2 + 4n + 14 in the worst case, f(n) 3n2 + 17 log, n + 1in the average case, and f(n) in 17 the best case. Which of the following would be the tightest possible asymptotic descriptions of the algorithm? The following statement that would be tightest possible asymptotic description of the algorithm above A) O(n) B) o (n) C) (n?) D) On Log n) E) O(n) F) O (Log n) G) (n log n) H) 2 (1) 1) I cannot be defined J) (n) K) o (n? Log n) L) (n Log n) M) O (1) N) None of the other answers O) O cannot be defined P) 2 (nº) Q) (n) R) 2 (n? Log n) S) 2 (Log n) T) 0 (nº) U) 0 (nº) V) (Log n) W) (1) x) o (n? Log n) Y) O cannot be defined

Please provide solution/methods so I can understand how this work.

Answer 1

Lets consider theta notation since it provide asymptotic upper and lower bounds.
if f(n) = Theta(g(n)) then there exist c1, c2 and n0 such that

c1*g(n) <= f(n) <= c2*g(n) for all n>=n0

and c1, c2>0 , n0>=0

We know that logn and n terms are asymptotically smaller than n^2.

lets assume g(n) = n^2

So for all cases,

limit n-> infinity g(n)/f(n) = n^2/(con*n^2 + smaller terms than n^2)

limit n-> infinity g(n)/f(n) = 1/(con + (small terms) / n^2) = 1/con which is constant. That means they grow at similar rate

where con = 5, 3, 1/17 for worst, average and best case

Now let c1 = 1 and c2=10

c1*g(n) = n^2 / 18 <= f(n) in best, worst and average case
c2*g(n) = 10*n^2 >= f(n) in best, worst and average case
and n0 = 100 or 1000 maybe (works for both)

(10n^2 > 5n^2 + 4n + 14 and similarly for other cases)
(n^2/18 < 5n^2 + 4n + 14 and similarly for other cases)

so we have f(n) = Theta(n^2) which provide tight bounds for f(n)

Option B) is correct

to disprove other options, we can see that bounds can provide upper or lower limit correctly
like option K) Theta(n^2 logn) this can provide upper bound but not lower