Convert the following Java method to a functionally equivalent iterative method without any recursion. public static double fun(int n) { 0) return 0; } if (n else return 1.0/(1.0+fun(n-1)); ==
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- In Java 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 - 1, 1) Otherwise, return ackermann(m - 1, ackermann(m, n - 1)) Test your method in a program that displays the return values of the following method calls: ackermann(0, 0) ackermann(0, 1) ackermann(1, 1) ackermann(1, 2)ackermann(1, 3) ackermann(2, 2) ackermann(3, 2)Consider the following recursive method: Public static int Fib(int a1, int a2, int n){ if(n == 1) return a1; else if (n == 2) return a2;else return Fib(a1, a2, n-1) + Fib(a1, a2, n-2);} Please draw the recursion trace for Fib(2,3,5)import java.util.Scanner; public class LabProgram { // Recursive method to draw the triangle public static void drawTriangle(int baseLength, int currentLength) { if (currentLength <= 0) { return; // Base case: stop when currentLength is 0 or negative } // Calculate the number of spaces needed for formatting int spaces = (baseLength - currentLength) / 2; if (currentLength == baseLength) { // If it's the first line, don't output spaces before the first '*' System.out.println("*".repeat(currentLength) + " "); } else { // Output spaces and asterisks System.out.println(" ".repeat(spaces) + "*".repeat(currentLength) + " "); } // Recursively call drawTriangle with the reduced currentLength drawTriangle(baseLength, currentLength - 2); } public static void drawTriangle(int baseLength) { drawTriangle(baseLength, baseLength); } public static…
- import java.util.Scanner; public class LabProgram { // Recursive method to draw the triangle public static void drawTriangle(int baseLength, int currentLength) { if (currentLength <= 0) { return; // Base case: stop when currentLength is 0 or negative } // Calculate the number of spaces needed for formatting int spaces = (baseLength - currentLength) / 2; if (currentLength == baseLength) { // If it's the first line, don't output spaces before the first '*' System.out.println(" ".repeat(spaces) + "*".repeat(currentLength)); } else { // Output spaces and asterisks System.out.println(" ".repeat(spaces) + "*".repeat(currentLength)); } // Recursively call drawTriangle with the reduced currentLength drawTriangle(baseLength, currentLength - 2); } public static void drawTriangle(int baseLength) { drawTriangle(baseLength, baseLength); } public…Convert the following Java method to a functionally equivalent iterative method without any recursion. public static double fun(int n) { if (n == 0) return 0; else return 1.0/(n+fun(n-1)); }Note: Java Consider the following recursive method: public static int Fun(int x) { if(x == 0) //line 1 return 0 //line 2 else if( x == 1) //line 3 return 1; //line 4 else //line 5 return (x*Fun(x-1)); //line 5 } a) Is Fun(4) a valid call? If so, what is the value? If not, explain why? Answer: b) Is Fun(-4) is a valid call? If so, what is the value? If not, explain why? Answer: (c) Consider the following method. public static int res(int[] list, int first, int last) { if (first == last) return list[first]; else return list[first] + res(list, first + 1, last); } Given the declaration int[] A = {2, 4, 6, 8, 10}; What is the output of the following statement? out.println(res(A, 0, 2)); out.println(res(A, 0,…
- Write a recursive method in java that sums all values between s and n inclusive /* assume s is less than or equal to n to start */ public static int sum(int s, int n) { }Java - Write an iterative method that calculates the SUM of all integers between 1 and a given integer N (input into the method). Write a corresponding recursive solution. (return answer, don’t print)I have two recursive methods in Java but I want to also implement a test class to see if both of them perform how can I do that? Method 1: Non Tail Recursive public class NonTailRecursiveFactorial { public static long factorial(int n) {if (n == 0) {return 1;}return n * factorial(n - 1);} public static void main(String[] args) {for (int i = 0; i <= 10; i++) {System.out.println(i + "! = " + factorial(i));}}} Method 2: Tail Recursive Method public class TailRecursiveFactorial { public static long factorial(int n) {return factorial(n, 1);} private static long factorial(int n, long accumulator) {if (n == 0) {return accumulator;}return factorial(n - 1, accumulator * n);} public static void main(String[] args) {for (int i = 0; i <= 10; i++) {System.out.println(i + "! = " + factorial(i));}}
- 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))1. Let product(n,m) be a recursive addition-subtraction method for multiplying two positive integers. Recursive cases for m = 1 and m < 1 make this method. The return value should be n plus a recursive product() call with n and m - 1. Test a Java method.in, java Assume a function g(x) is defined as follows where x is an int parameter:g(x) = g(x - 1) * g (x - 3) if x is even and x > 3 = g(x - 2) if x is odd and x > 3 = x otherwiseWrite a recursive method to compute g.