Can someone explain how the output of this recursive function is 18? Recursion is confusing to me. def R(n): if n>=5: return 10 return R(n+1) + 2 print(R(1))
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Can someone explain how the output of this recursive function is 18? Recursion is confusing to me.
def R(n):
if n>=5:
return 10
return R(n+1) + 2
print(R(1))
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Solved in 2 steps
- Can someone explain how the output of this recursive function is 12? Recursion is confusing to me. def R(n): if n>=5: return 2 return R(n+1) + 2 print(R(0))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 recursionjava 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 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); }}Can someone explain how this recursive function output is 12? Recursion is hard for me def R(n): if n>=5: return 2 return R(n+1) + 2 print(R(0))What does the following recursive function do? int f(int n)X if (n==1) return 1; else return n+f(n-1); } Sum of numbers from 1 to n Factorial of number n Square of numbers from 1 to n Print the numbers from 1 to 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) + 3Write a recursive function called that takes a string of single names separated by spaces and prints out all possible combinations (permutations), each combination on a new line. When the input is: Alice Bob Charlie then the output is: Alice Bob Charlie Alice Charlie Bob Bob Alice Charlie Bob Charlie Alice Charlie Alice Bob Charlie Bob Alice Here is the original code in Python: def all_permutations(permList, nameList): # TODO: Implement method to create and output all permutations of the list of names. pass if __name__ == "__main__": nameList = input().split(' ') permList = [] all_permutations(permList, nameList)Compute f(6) for the recursive function below. def f(n): if n == 0: return 1 if n == 1: return 2 else: return f(n-1)+n*f(n-2)-n
- Ackermann’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 + 1 If 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)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 calculateSum function as a recursive function. m(i) = m(i-1) + i/(i+1), where i >=1