Find the complexity of the below program: for (i=0; i<=n-1; i++){ for (j=i+1; j<=n-1; j++){ loop body }
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- java C++ Ackermann’s FunctionAckermann’s Function is a recursive mathematical algorithm that can be used to test how well a computer performs recursion. Write a function A(m, n) that solves Ackermann’s Function. Use the following logic in your function:If m = 0 then return n + 1If n = 0 then return A(m−1, 1) Otherwise, return A(m−1, A(m, n−1))Test your function in a driver program that displays the following values:A(0, 0) A(0, 1) A(1, 1) A(1, 2) A(1, 3) A(2, 2) A(3, 2) SAMPLE RUN #0: ./AckermannRF Hide Invisibles Highlight: Show Highlighted Only The·value·of·A(0,·0)=·1↵ The·value·of·A(0,·1)=·2↵ The·value·of·A(1,·1)=·3↵ The·value·of·A(1,·2)=·4↵ The·value·of·A(1,·3)=·5↵ The·value·of·A(2,·2)=·7↵ The·value·of·A(3,·2)=·29↵Explain the functionality of below recursive functions. static void fun1(int n) { int i = 0; if (n > 1) fun1(n - 1); for (i = 0; i<n; i++) System.out.print(" *"); } 2. int LIMIT = 1000; void fun2(int n) { if (n<=0)return; if (n>LIMIT)return; System.out.print( String.format("%d", n)); fun2(2*n); System.out.print( String.format("%d", n)) }Which is the base case of the following recursion function: def mult3(n): if n == 1: return 3 else: return mult3(n-1) + 3 else n == 1 mult3(n) return mult3(n-1) + 3
- Rewrite the following recursive function using a for loop. public class MyMain { public static int myFunct(int a) { if(a 0) return 0; == else return a + my Funct (a-1); } public static void main(String args []) { int k = 10; System.out.println(myFunct(k)); }sum= 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.For function addOdd(n) write the missing recursive call. This function should return the sum of all postive odd numbers less than or equal to n. Examples: addOdd(1) -> 1addOdd(2) -> 1addOdd(3) -> 4addOdd(7) -> 16 public int addOdd(int n) { if (n <= 0) { return 0; } if (n % 2 != 0) { // Odd value return <<Missing a Recursive call>> } else { // Even value return addOdd(n - 1); }}
- I Evaluate the time complexity of following code: int a-0; for (i-0; i< N; i++){ aa+i;) for (i 0; i< N; i+){ a-a+i;) 2void deleteRange( int from, int to) {int i, j = 0;for (i = 0; i < counter; i++) {if (i <= from - 1 || i >= to + 1) {A[j] = A[i];j++;}}for (int i = 0; i < j; i++)cout << A[i] << " ";} Above method deletes range of elements from an array. Consider array A[] globally declared and counter is its size. Please explain the logic of above code in simple english (algorithm and comments).For funX |C Solved xb Answer x+ CodeW X https://codeworko... 田) CodeWorkout X267: Recursion Programming Exercise: Cumulative Sum For function sumtok, write the missing recursive call. This function returns the sum of the values from1 to k. Examples: sumtok(5) -> 15 Your Answer: 1 public int sumtok(int k) { 2. } (0 => ) return 0; 3. } else { return > 6. { Check my answer! Reset Next exercise 1:09 AM
- void deleteRange( int from, int to) { int i, j = 0; for (i = 0; i < counter; i++) { if (i <= from - 1 || i >= to + 1) { A[j] = A[i]; j++; } } for (int i = 0; i < j; i++) cout << A[i] << " "; } Above method deletes range of elements from an array. Please explain the logic of above code in simple english (algorithm and comments).What type of recursion is used in the following function? int f(int n){ if (n==1) return 1; else return n+f(n-1); } Tail recursion Multiple recursion Indirect recursion Non-tail recursionQUESTION 2 Consider the below recursive function: public static void func (int n) { if (n == 0) return; else { System.out.println(n); func (n-2); } Draw activaiton frames with n=8 and determine the output produced by this method.