Implement the logarithmBase22 method which, given a long integer number, returns the result of Log22(number), rounded down to the nearest floor integer. Design and implement this as a recursive method. Read the test cases in the JUnit tester file for more detailed specification.

Need python help. For problems 1 and 2, add your code to the file Lab2.java. Add your tests in the main function of Lab2.java. Do not use static variables in class Lab2 to implement recursive methods.  Problem 1: Implement a recursive method min that accepts an array and returns the minimum element in the array. The recursive step should divide the array into two halves and find the minimum in each half. The program should run in O(n) time and consume O(logn) memory. Demonstrate the output of min on the array int [] a = { 2, 3, 5, 7, 11, 13, 17, 19, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 23, 29, 31, 37, 41, 43 } Problem 2 You have been offered a job that pays as follows: On the first day, you are paid 1 cent, on the second day, 2 cents, on the third day, 4 cents and so on. In other words, your pay doubles every day. Write a recursive method computePay that for a given day number computes the pay in cents. Assume that you accumulate all the money that you are paid. Write a…

Write a recursive method that gets three parameters as input: an array of integers called nums, an integer called index,and an integer called The purpose of this method is to return true if value is stored in the array starting at nums[index]. That is, you have to check if value is equal to nums[index] or nums[index +1] or nums[index +2 ] …. nums[nums.length -1]. Do not use loops.(java code)

Write a RECURSIVE method called “sequence” that takes a single int parameter (n) and returns the int value of the nth element of the sequence S = 2, 4, 6, 12, 22, 40, 74, 136, 250, 460, … Where S is defined by the recursive formula:  For n >= 0S(0) = 2;   // Base case 1S(1) = 4;   // Base case 2S(2) = 6;   // Base case 3S(N) = 2 * (  S(N-1)/2 + S(N-2)/2 + S(N-3)/2)

Write a recursive method that returns the value of N! (N factorial) using the definition given in this chapter. Explain why you would not normally use recursion to solve this problem.

Write and test a Java/Python recursive method for finding the minimum element in an array, A, of n elements. What the running time? Hint: an array of size 1 would be the stop condition. The implementation is similar to linearSum method in lecture 5 examples. You can use the Math.min method for finding the minimum of two numbers.

Write a recursive method called add(int n). This method adds the integers from 0 to some value n. For example, the call to add(4) would return 10 (4+3+2+1=10).

• Write a recursive method to complete the class BaseRaiseToN. • The method: public static int raiseBToN(int base, int exp) computes and returns the value of Bn using recursion.• What value/values of n will stop the recursion (base case)?• A precondition of this method is that B be a positive number greater than zero Sample Output Enter the base4Enter the exponent 3 4 to 3 is 64

Trace binary search on the sorted dataset below. List first, last, and mid for each pass through the loop or each recursive method call. target = 33 index: 0 1 2 3 4 5 6 7 8 9 10 11 value: 12 15 19 23 25 27 36 39 47 58 62 67

The intName method in the "Problem Solving: Stepwise Refinement" section accepted arguments < 1,000. Using a recursive call, extend its range to 999,999. For example an input of 12,345 should return " twelve thousand three hundred forty five". Note that some number names may have extra spaces at the beginning or end. import java.util.Scanner; public class IntegerName { /** Turns a number into its English name. @param number a positive integer < 999,999 @return the name of the number (e.g. "two hundred seventy four") */ public static String intName(int number) { int part = number; // The part that still needs to be converted String name = ""; // The name of the number if (part >= 1000) { // Recursively call intName } if (part >= 100) { name = digitName(part / 100) + " hundred"; part = part % 100; } if (part >= 20) { name = name + " " + tensName(part); part = part % 10; } else if (part >= 10) { name = name + " " + teenName(part); part = 0; } if (part > 0) { name =…