use summation to analyze the running time (i.e. T(n)) of these functions and able to find some simple function f(n) such that T(n) = Θ(f(n)).
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use summation to analyze the running time (i.e. T(n)) of these functions and able to find some simple function f(n) such that T(n) = Θ(f(n)). show the steps, please
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- T(n) = 9T(n/3) + O(n2)int functionC (int n) { int i, j, sumC = 0; for (i=n; i > 0; i=i-5) for (j=1; j 0) { if (functionC(n) % 2 == 0) { for (i=m; i > 0; i=i/3) sumE++; } else (10) Asymptotic runtime of functionE { for (i=m; i > 0; i=i-3) sumE--; } n--; } return sumE;Construct a DFA A so that L(A) = L(N) where N is the following NFA:
- O(nlgn) means that there is function f(n) that is O(nlgn) which is an upper bound for the running time at large n Select one: True FalseFor each of the following program fragments: Give an analysis of the running time (Big-Oh). Justify your answer? A. public class GFG { // Linearly search x in att[]. if x is present then //return the index, otherwise return -1 static int search (int arr[], int n, intx) { int i; for (i = 0: iDesign the following LCG random number generators by choosing the following parameters ( a, c, m, and Zo ) carefully and writing the recursive formula for each of the following: a . LCG1: has Zo = 17. b. LCG2 : has 128 > m > 16 LCG3 : has c = 7 d . LCG4: has a = 7 e. LCG5: has c not equal to 3 C.F(A,B,C, D,E) =E ml0,1;3,6,8,9,14,ky1319,25,2931) DE 61 10 60 00 61 10sum= 0; for (int i = 0; i 1) { sum++; i= 1/2; } = 2*log2 (n) We denote by Ta(n), Tb (n), Te(n) the running time of the three fragments. 1. Give evaluations for Ta(n), Tb (n), Te(n). 2. Is T(n) = O(Ta(n)) ? Answer YES or NO and justify your answer. 3. Is Te(n) = (Ta(n)) ? Answer YES or NO and justify your answer.i = 1 while (i < n) do s = s + i i = i * 2 enddo Is the step count dependent on which term?Here &The maximal value of num = 16..Factorial of a number is defined as: n! = n(n-1)(n-2)(n-3)...(2)(1) For example, 4! = 4*3*2*1 The n! can be written in terms of (n-1)! as: n! = n* (n-1)! (n-1)! = (n-1)*(n-2) ! and so forth. Thus, in order to compute n!, we need (n-1)!, to have (n-1)!, we need (n-2)! and so forth. As you may immediately notice, the base case for factorial is 1 because 1! = 1. Write a program that uses a recursive function called factorial that takes an integer n as its argument and returns n! to the main. C++ PLEASEPlace 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 ).SEE MORE QUESTIONS