Using recursion, write a Python function def countOdds(A) which which takes an array of integers A as input and returns the number of elements in the array that are odd. For example, if A is [1,2,5,3,6,5,3,5,5,4] the the function should return 7. If A is empty then it should return 3. Hint: To check if n is odd it suffices to check if dividing it by 2 gives remainder 1 (in Python, this remainder is calculated by n%2).

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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.

Write a recursive function that displays the number of even and odd digits in an integer using the following header: void evenAndOddCount(int value) Write a test program that prompts the user to enter an integer and displays the number of even and odd digits in it.

Write code to complete raise_to_power(). Note: This example is for practicing recursion; a non-recursive function, or using the built-in function math.pow(), would be more common.Sample output with inputs: 4 24^2 = 16   def raise_to_power(base_val, exponent_val):   if exponent_val == 0:      result_val = 1   else:      result_val = base_val * ''' Your solution goes here '''    return result_val user_base = int(input())user_exponent = int(input()) print('{}^{} = {}'.format(user_base, user_exponent,      raise_to_power(user_base, user_exponent)))

Prompt: In python langauge, write a function that adds a random noise to all elements of a NumPy array Code:  import numpy as np import math import matplotlib.pyplot as plt import pandas as pd   def addNoise(inputArray):   modifiedArray = np.zeros(len(inputArray))   #YOUR CODE HERE:   return(modifiedArray) def test(): inputs = np.arange(-5, 5,0.2) outputs = addNoise(inputs) plt.figure(1) plt.plot(inputs) plt.title('Input') plt.xlabel('Index') plt.ylabel('Value') plt.show() plt.figure(2) plt.plot(outputs, 'black') plt.title('Output') plt.xlabel('Index') plt.ylabel('Value') plt.show() test()

Java, I am not displaying my results correct. It should add up the digits in the string. input string |result for the sumIt Recursion functions and a findMax function that finds the largest number in a string "1d2d3d" | 6 total "55" |10 total "xx" | 0 total "12x8" |12 Max number "012x88" |88 Max Number "012x88ttttt9xe33ppp100" |100 Max Number public class Finder { //Write two recursive functions, both of which will parse any length string that consists of digits and numbers. Both functions //should be…

Use Python Write a function, print_integers_less_than(n), which takes an integer parameter n and prints each integer k which is at least 0 and is less than n, in ascending order.  Hint: use a simple for loop. For example: Test Result print_integers_less_than(2) 0 1 print_integers_less_than(5) 0 1 2 3 4 print_integers_less_than(-3)

python3 Write a recursive function called is_palindrome(string) that takes a string parameter and checks if it is a palindrome ignoring the spaces, if any, and returns True/False. Sample output:>>> print(is_palindrome("never odd or even"))True>>> print(is_palindrome("step on no pets"))True

Using recursion, write a function sum that takes a single argument n and computes the sum of all integers between 0 and n inclusive. Do not write this function using a while or for loop. Assume n is non-negative. def sum(n): """Using recursion, computes the sum of all integers between 1 and n, inclusive. Assume n is positive. >>> sum(1) 1 >>> sum(5) # 1 + 2 + 3+ 4+ 5 15 "*** YOUR CODE HERE ***"