Consider the following function (assume n is non-negative): public static int factorial(int n) { if( n == 0) return 1; else return (n)* factorial(n+1); } a) recursion and Java (base case, recursive case, parameters, arguments). Explain why this program has infinite recursion using the terminology o b) How would you correct this function to correctly calculate factorials?
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- T/F 6. Consider the following recursive sum method:public int sum(int x){if (x == 0) return 0;else return sum(x - 1) + 1;}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))(Kindly note that language is in Java) The first examples of recursion are the mathematical functions factorial and fibonacci. These functions are defined for non-negative integers using the following recursive formulas:factorial(0) = 1factorial(N) = N*factorial(N-1) for N > 0fibonacci(0) = 1fibonacci(1) = 1fibonacci(N) = fibonacci(N-1) + fibonacci(N-2) for N > 1Write recursive functions to compute factorial(N) and fibonacci(N) for a given non-negative integer N, and write a main() routine to test your functions.(In fact, factorial and fibonacci are really not very good examples of recursion, since the most natural way to compute them is to use simple for loops. Furthermore, fibonacci is a particularly bad example, since the natural recursive approach to computing this function is extremely inefficient.)
- 1. Let product(n,m) be a recursive method that computes the product of two positive integers, using only addition and subtraction. To make this method a recursive one, you are to make a) a base case when m = 1, b) a general case when m ≠ 1. For a general case, the return value should be n plus the result of a recursive call to the method product() with parameters n and m - 1. Write a short Java code for this method, along with a test program.JAVA Question 2: For two integers m and n, their GCD (Greatest Common Divisor) can be computed by a recursive method. Write a recursive method gcd(m,n) to find their Greatest Common Divisor. Method body: If m is 0, the method returns n. If n is 0, the method returns m. If neither is 0, the method can recursively calculate the Greatest Common Divisor with two smaller parameters: One is n, the second one is m mod n (or m % n). The recursive method cannot have loops. Note: although there are other approaches to calculate Greatest Common Divisor, please follow the instructions in this question, otherwise you will not get the credit. main method: Prompt and read in two numbers to find the greatest common divisor. Call the gcd method with the two numbers as its argument. Print the result to the monitor. Example program run: Enter m: 12 Enter n: 28 GCD(12,28) = 4 And here is what I have so far, package CSCI1302;import java.util.*;public class RecursionDemo { public static void…Write the definition of a recursive function int simpleSqrt(int n) The function returns the integer square root of n, meaning the biggest integer whose square is less than or equal to n. You may assume that the function is always called with a nonnegative value for n. Use the following algorithm: If n is 0 then return 0. Otherwise, call the function recursively with n-1 as the argument to get a number t. Check whether or not t+1 squared is strictly greater than n. Based on that test, return the correct result. For example, a call to simpleSqrt(8) would recursively call simpleSqrt(7) and get back 2 as the answer. Then we would square (2+1) = 3 to get 9. Since 9 is bigger than 8, we know that 3 is too big, so return 2 in this case. On the other hand a call to simpleSqrt(9) would recursively call simpleSqrt(8) and get back 2 as the answer. Again we would square (2+1) = 3 to get back 9. So 3 is the correct return value in this case.
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- Hi can check my code below and see whether it has meet the following, if not kindly assist and correct it for me. a). Recursive function in Java that accepts an integer as input and returns 1 + 1/2 + 3 + 1/4 + ...(n or 1/n). b). The answer is the sum of the odd integers from 1 to n plus the sum of the reciprocals of the even integers. Thank you. public class Main {public static double sum(int n) {if (n <= 0) {return 0;} else if (n % 2 == 0) {return 1.0 / n + sum(n - 1);} else {return n + sum(n - 1);}}public static void main(String[] args) {System.out.println(sum(5));}}Identify the base case in this recursive function. Assume i> 0 when the function is invoked. (Line numbers are not part of the code.) 1. def add(i, j): 2. if i == 0: 3. return j 4. else: 5. return add(i - 1, j + 1) This function has no base case O line 5 line 1 line 4 O line 2 Question 6 The following recursive function is supposed to return a list that is the reverse of the list it was given. But it has a bug. What is the bug? (Line numbers are not part of the code.) MooP eok DrPlease can be handwritten. Question 2: Implementing a Recursive Function . Write recursive function, recursionprob(n), which takes a positive number as its argument and returns the output as shown below. The solution should clearly write the steps as shown in an example in slide number 59 and slide number 60 in lecture slides. After writing the steps, trace the function for “recursiveprob(5)” as shown in an example slide number 61. Function Output: >> recursionprob(1) 1 >> recursionprob(2) 1 4 >> recursionprob(3) 1 4 9 >>recrusionprob(4) 1 4 9 16