Can someone explain how this recursive function output is 12? Recursion is hard for me
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Q: is confusing to me. def R(n): if n>=5: return 2 return R(n+1) + 2
A: Given : def R(n): if n>=5: return 2 return R(n+1) + 2 print(R(0))
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))
Step by step
Solved in 2 steps
- 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 recursionFor 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); }}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
- 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↵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 that takes as a parameter a nonnegative integer and generates the following pattern of stars. If the nonnegative integer is 4, then the pattern generated is: **** *** ** * ** *** ****
- •rewrite calculateSum function as a recursive function. m(i) = m(i-1) + i/(i+1), where i >=1Can 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))complete the identified statement such that the recursive function funx(n+1) = n*funx(n-1), any positive integer value of n, and funx(1)=1. 1 float funx (int n) 3 if (n <= 1) 4 return l; else 6 return // complete this statement 7
- 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 AMWhat does the following recursive function do? int f(int n){ if (n==1) return 1; else returm n*f(n-1); } O sum of numbers from 1 to n Factorial of numbern Square of numbers from 1 to n Print the numbers from 1 to n1. Given, nCr= n! (n-r)!r!" Write a function to compute nCr without recursion.