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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↵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 AM8. Give a recursive definition of the sequence {an}, n =1, 2, 3, ... ifa) an = 4n − 2. b) an = 1 + (−1)n.c) an = n(n + 1). d) an = n2.
- 9. Ackermann's Function Ackermann's function is a recursive mathematical algorithm that can be used to test how well a computer performs recursion. Write a method ackermann (m, n), which solves Ackermann's function. Use the following logic in your method: If m = 0 then return n + 1 If n = 0 then return ackermann (m Otherwise, return ackermann(m 1, 1) 1, ackermann (m, n - 1))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)); }T/F 8. Without recursion, the Koch snowflake = 1 could be drawn.
- Alert dont submit AI generated answer. Consider the following implementation of function multiply.int multiply ( int x , int n ) {if ( n <= 0 ) {return 0 ;}else {return x+m u l t i p l y ( x , n −1 ) ;}}a) Briefly explain why this implementation is not tail-recursive.b) Give a tail-recursive implementation in the C language. You might need todefine a helper function.c) Give an equivalent loop-based implementation of the same function in the Clanguage. You are only allowed to use a while loop for this question.Convert the part to recursive way in Java: public boolean search(E data) { Node<E> curr = root; int cmp; while (curr != null) { if ((cmp = data.compareTo(curr.data)) == 0) { return true; } else if (cmp < 0) { curr = curr.left; } else { curr = curr.right; } } return false; }function recursion(B[0..n − 1], i) if n == 0 then return False if n == 1 then return (B[0] == i)x ← recursion(B[0........n/2 − 1], i) y ← recursion(B[n/2..........4 × n/2 − 1], i) z ← recursion(B[4 × n/2..........n − 1], i) return (x OR y OR z) input array is of length n = 2^p, p is a positive integer Write the recursive formula for above algorithm as of worst case inputs.
- 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) + 3Using the recursion three method find the upper and lower bounds for the following recur- rence (if they are the same, find the tight bound). T(n) = T(n/2) + 2T(n/3) + n.Many of the coding interview problems deal with permutations and combinations of a given set of components.1. Backtracking Method: When the subset is added to the resultset, each response is backtracked in this recursive process.An extra array subset's space and time complexity are O(n), respectively (2n). solve the aforementioned issue with Java