Answered Essay: Arrange the functions below in non-decreasing order such that if f_i appears before f_j then f_i(n) elementof O(f_j(n)).

1. Arrange the functions below in non-decreasing order such that if f, appears before f then fi(n) O(J,(n)). [15 points] f1(n) = 1020 ½(n) = (lg n)4 fs(n) = 4n (n) = n lg n fs(n) = n. 100n2 fe(n) = n + lg n fr(n) = lg lg n fs(n) = no.1 Jo(n) = lgn5

Arrange the functions below in non-decreasing order such that if f_i appears before f_j then f_i(n) elementof O(f_j(n)). f1(n) = 10^20 f_2(n) = (lg n)^4 f_3(n) = 4^n f_4(n) = n lg n f_5(n) = n^3 – 100n^2 f_6(n) = n + lg n f_7(n) = lg lg n f_8(n) = n^0.1 f_9(n) = lg n^5

Expert Answer

 

Solution:

We know that, f(n) epsilon O(g(n)), if f(n) <= c*g(n), where c is a positive constant.

f1(n) is smallest of them all because 10^20 is almost equivalent to constant (Irrespective of value of n)

then comes f7(n)= lg lg n= lg lg (2^512)= lg 512= 9

then f9(n)= lg n^5, 5*(log 2^512)= 512*5

then f2(n)= (log n)^4, for n=2^512, 512^4 will be the outcome

then comes f8(n)= n^0.1, we can check the value the function is producing by putting a very large value of n (2^512)

(2^512)^0.1= 2.5866387e+15

then f6(n)= n + lg n

then f4(n)= n lg n

then comes f5(n)= n^3-100*n^2,

and at last f3(n)= 4^n,

So the order will go like this,

f1(n), f7(n), f9(n), f2(n), f8(n), f6(n), f4(n), f5(n), and f3(n).

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