Convert the following recursive functions into iterative ones: 1. def sum_odd (n, total): if n = 2. 3. == 1 return total elif n % 2 == 0: return sum_odd (n - 1, total) else return sum_odd (n - 2, total + n ) def max (I, n): if I []: return n elif [0] > n return max (1[1:], [0] ) else return max (1[1:], n) : == def mylen (I, n): if | == : return n else return mylen (1[1:], n +1)

Convert the following recursive functions into iterative ones: 1. def sum_odd (n, total): if n = 2. 3. == 1 return total elif n % 2 == 0: return sum_odd (n - 1, total) else return sum_odd (n - 2, total + n ) def max (I, n): if I []: return n elif [0] > n return max (1[1:], [0] ) else return max (1[1:], n) : == def mylen (I, n): if | == : return n else return mylen (1[1:], n +1)

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter15: Recursion

Section: Chapter Questions

Problem 6PE

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2. Sum: a recursive function that computes the sum of integers 1, 2, 3, …., n for a given number n. So Sum(6) should return 1 + 2 + 3 + 4 + 5 + 6 , i.e. 21.sum(n) = n + sum(n-1)
4. CodeW. X b For fun X Solved x b Answer x+ Ohttps://codeworko... CodeWorkout X264: Recursion Programming Exercise: Multiply For function multiply,write the missing base case condition and action. This function will multiply two numbers x and y.You can assume that both x and y are positive. Examples: multiply(2, 3) -> 6 Your Answer: 1 public int multiply(int x, int y) { 2. if > { > } else { return multiply(x 1, y) + y; 3. 5. { 7. 1:08 AM 50°F Clear 日

Python Test: import recursive_functionsimport mathdef main():# Test factorialprint('Testing factorial.')assert recursive_functions.factorial(0) == 1assert recursive_functions.factorial(1) == math.factorial(1)== 1assert recursive_functions.factorial(2) == math.factorial(2)== 2assert recursive_functions.factorial(5) == math.factorial(5)== 120assert recursive_functions.factorial(7) == math.factorial(7)== 5040print('All tests pass for `factorial` ()\n')# Test sum_recursivelyprint('Testing sum_recursively.')assert recursive_functions.sum_recursively(0) == 0assert recursive_functions.sum_recursively(1) ==sum(range(1+1)) == 1assert recursive_functions.sum_recursively(2) ==sum(range(2+1)) == 3assert recursive_functions.sum_recursively(10) ==sum(range(10+1)) == 55print('All tests pass for `sum_recursively` () ')# Test sumlist_recursively(l)print('Testing sumlist_recursively.')assert recursive_functions.sumlist_recursively([1,2,3])…
Consider the following function: void fun_with_recursion(int x) { printf("%i\n", x); fun_with_recursion(x + 1); } What will happen when this function is called by passing it the value 0?
PYTHON RECURSIVE FUNCTION Write a python program that lists all ways people can line up for a photo (all permutations of a list of strings). The program will read a list of one word names, then use a recursive method to create and output all possible orderings of those names, one ordering per line. When the input is: Julia Lucas Mia then the output is (must match the below ordering): Julia Lucas Mia Julia Mia Lucas Lucas Julia Mia Lucas Mia Julia Mia Julia Lucas Mia Lucas Julia
Question 5: 2.1 Sum Problem View Past Answers Given a positive number n, the sum of all digits is obtained by adding the digit one-by-one. For example, the sum of 52634 is 5+ 2 +6 +3 + 4 = 20. Write a recursive and iterative function sum (n) to compute the sum of all the digits in n. You may assume that n > 0. template.py 1 def sum R(n): if len(str(n)) == 1: 2- return n else: return 20 7. def sum_I(n): add = 0 8 while n: 10 add += n % 10 n //= 10 return add 11 12
How can I apply this python code? def createList(n): #Base Case/s #TODO: Add conditions here for your base case/s #if <condition> : #return <value> #Recursive Case/s #TODO: Add conditions here for your recursive case/s #else: #return <operation and recursive call> #remove the line after this once you've completed all the TODO for this function return [] def removeMultiples(x, arr): #Base Case/s #TODO: Add conditions here for your base case/s #if <condition> : #return <value> #Recursive Case/s #TODO: Add conditions here for your recursive case/s #else: #return <operation and recursive call> #remove the line after this once you've completed all the TODO for this function return [] def Sieve_of_Eratosthenes(list): #Base Case/s if len(list) < 1 : return list #Recursive Case/s else: return [list[0]] + Sieve_of_Eratosthenes(removeMultiples(list[0],…
CodeWorkout Gym Course Search exercises... Q Search kola shreya@colum X459: Review- Fibonacci In mathematics, the Fibonacci numbers are the numbers in the following integer sequence, characterized by the fact that every number after the first two is the sum of the two preceding ones: e, 1, 1, 2, 3, 5, 8, 13, Write a recursive function that the returns the nth fibonacci number. Examples: fibonacci(0) -> 0 fibonacci(1) -> 1 fibonacci(7) -> 13 Your Answer: 1 public int fibonacci(int n) { 2 3} 4 CodeWorkout © Virginia Tech About License Privacy Contact
CodeW For fun X C Solved https://codeworkou... 臺亂 CodeWorkout X272: Recursion Programming Exercise: Is Reverse For function isReverse, write the two missing base case conditions. Given two strings, this function returns true if the two strings are identical, but are in reverse order. Otherwise it returns false. For example, if the inputs are "tac" and "cat", then the function should return true. Examples: isReverse("tac", "cat") -> true Your Answer: 1 public boolean isReverse(String s1, String s2) { 2. if > 3. 4. else if > return true; return false; 5. 6. else { String s1first = String s2last return s1first.equals (s2last) && 51. substring(0, 1); s2, substring(s2.length() 1); 7. 8. 6. isReverse(s1.substring(1), s2.substring(0, s2.length() 1)); { 12} 1:11AM 50°F Clear 12/4/2021
how to convert this code to fulfill this condition: recursive function to display the sequence: 1 2 3 ... n-2 n-1 n n-1 n-2 ... 3 2 1 Code: void rec1(int n) {if(n==1) { cout<<1;return; }cout<<n<<" ";rec1(n-1);cout<<n<<" ";}
Python Test Program: import recursive_functionsimport mathdef main():# Test factorialprint('Testing factorial.')assert recursive_functions.factorial(0) == 1assert recursive_functions.factorial(1) == math.factorial(1)== 1assert recursive_functions.factorial(2) == math.factorial(2)== 2assert recursive_functions.factorial(5) == math.factorial(5)== 120assert recursive_functions.factorial(7) == math.factorial(7)== 5040print('All tests pass for `factorial` ()\n')# Test sum_recursivelyprint('Testing sum_recursively.')assert recursive_functions.sum_recursively(0) == 0assert recursive_functions.sum_recursively(1) ==sum(range(1+1)) == 1assert recursive_functions.sum_recursively(2) ==sum(range(2+1)) == 3assert recursive_functions.sum_recursively(10) ==sum(range(10+1)) == 55print('All tests pass for `sum_recursively` () ')# Test sumlist_recursively(l)print('Testing sumlist_recursively.')assert recursive_functions.sumlist_recursively([1,2,3])…
Submission: Submit python code Problem: If n is a positive integer, then n factorial (written n!) is the product of the numbers from 1 through n. Write a recursive function to calculate n factorial. Hints: n! = (n – 1) · (n– 2) 2: 1 As written, n! can be calculated iteratively with a for loop-however, when rewritten as n! = n · ((n – 1) · (n – 2) 3 · 2 · 1) = n · (n – 1)! n! is expressed in terms of (n – 1)! and can be calculated recursively with n = 1 as the base case. | Example input and output: >>> factorial(5) 120 >>>