if(n==1) return 1;
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- Label the Recursion Requirements. Show the Activation Stack for n= 4. int fact (int n) int reaulti if (n-1) return li result - fact (n-1) return reult:Label the Recursion Requirements. Show the Activation Stack for n = 4. int fact (int n) { int result; if (n==1) return 1; result = fact (n-1) * return result; ni }Label the Recursion Requirements. Show the Activation Stack for n = 4. int fact (int n) int result; if (n==1) return 1; result = fact (n-1) * return result; n;
- Question:: Label the Recursion Requirements. Show the Activation Stack for n = 4. int fact (int n) int result:i if (n==1) return 1; result = fact (n-1). return result: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) + 3For 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); }}
- Modify the following operations into a recursive procedure. void ditui(int n) { int i; i=n; } while(i>1) print(i--);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.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))
- recursive solution oddEvenMatchRec:the method takes an integer array as a parameter and returns a boolean. The method returns trueif every odd index contains an odd integer AND every even index contains aneven integer(0 is even). Otherwise it returns falsIn what instances do you think it’s impractical to use recursion over loops?F+ TX X+ +EI# T+(E)l int Y T|# 3- Build Non-recursive Predictive Pursing table?