What is the order of each of the following functions?
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- GROWTH OF FUNCTIONS. Arrange the following mathematical terms from lowest to highest order. n3 n2 n! 2n 7n – 2 log n n log n 10n n5 + log n n2 + n log n Example: a < b < c < d < … < x < y < zarrow_forwardTrue or False? Justify your answer accordingly: 1. For any two functions f(n) and g(n), if f(n) = Θ(g(n)), then nf(n) = Θ(ng(n)). 2. O(f(n)+g(n))=O(max(f(n),g(n))).arrow_forwardWhat does the function return for any positive values of x and n ? The answer should be a general formula in terms of x and n. int what_do_I_do ( int x, int n ) { if ( n = = 1 ) return x; else return x + what_do_I_do ( x, n – 1 ); } C++ be quick pleasearrow_forward
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- If f = (log n)^2, g = n log n; what is the relationship between f and g? Choose all that applies. a. g= O(f) b. f = O(g) c. g= Ω(f) d. f = Ω(g)arrow_forwardPlace the following functions into their proper asymptotic order: f1(n) = n2log2n; f2(n) = n(log2n)2 ; f3(n) = 20 + 21 + .. + 2n ; f4(n) = log2 ( 20 + 21 + .. + 2n ).arrow_forwardSection A: Multiple Choice Questions Q8: Given a function f(n) = n2 + 20n on an input parameter n, which of the following is true a) f(n)=O(1) b) f(n)=O(n) c) f(n)=O(logn) d) f(n)=O(n3)arrow_forward
- Please show detailed explanation. Thanks. What is the order of n, O(f(n)) of the following function? n3+50n2+n What is the order of n, O(f(n)) of the following function? 2n+2n3+5n Suppose an algorithm takes exactly the given number of statements for each value below, in terms of the size of n, i.e., the order of n, O(f(n)). n2+logn+2narrow_forwardWhich function best represents the number of operations in the worst-case? start = 0;while (start < N) { ++start;} 1.f(N)=2N + 1 2.f(N)=2N + 2 3.f(N)=N + 2 4.f(N)=N + 3arrow_forwardWhich is a proper way to rewrite the following function using iteration? int f(int n) { if (n == 0) return 1; else return n * f(n - 1);} Group of answer choices for (int i=0, f=1; i<n; i++) f = f * i; for (int i=0, f=1; i<=n; i++) f = f * i; for (int i=1, f=1; i<n; i++) f = f * i; for (int i=1, f=1; i<=n; i++) f = f * i;arrow_forward
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