12.18 LAB: Binary search Binary search can be implemented as a recursive algorithm. Each call makes a recursive call on one-half of the list the call received as an argument. Complete the recursive method binarySearch() with the following specifications: 1. Parameters: 。 a target integer 。 an ArrayList of integers 。 lower and upper bounds within which the recursive call will search 2. Return value: 。 the index within the ArrayList where the target is located 。 -1 if target is not found The template provides main() and a helper function that reads an ArrayList from input. The algorithm begins by choosing an index midway between the lower and upper bounds. 1. If target == integers.get(index) return index 2. If lower == upper, return -1 to indicate not found 3. Otherwise call the function recursively on half the ArrayList parameter: 。 If integers.get(index) < target, search the ArrayList from index + 1 to upper 。 If integers.get(index) > target, search the ArrayList from lower to index - 1 The ArrayList must be ordered, but duplicates are allowed. Once the search algorithm works correctly, add the following to binarySearch(): 4. Count the number of calls to binarySearch(). 5. Count the number of times when the target is compared to an element of the ArrayList. Note: lower Hint: Use a static variable to count calls and comparisons. The input of the program consists of: 1. the number of integers in the ArrayList 2. the integers in the ArrayList 3. the target to be located Ex: If the input is: == upper should not be counted. 9 1 2 3 4 56789 2 the output is: index: 1, recursions: 2, comparisons: 3

Write up Java source codes for compiling and execution of program below. Draw UML diagram for codes. There is a very simple solution to keeping Tic-tac-toe fresh and interesting, though, and it has been thought up by a group of mathematicians. It’s being called Ultimate Tic-tac-toe and simply embeds a Tic-tac-toe board in each of the nine squares of the original game. Now instead of just winning the main board, you also need to win each of the smaller boards first until you have a line of three. New Rule: You can only place a mark on the board determined by the position of your opponent’s last placed mark. So, if they put an X or O in the top right corner of a square, your next move must occur in the top right board. By adding that rule the game is no longer about winning a single board, it’s about tactically managing up to 9 games at once and plotting ahead. You still win by marking three squares in a row, but that now involves winning three games.   The effects of the New Rule: 1:…

In Java, hiding implementation details is done by marking data and methods as : a   private b   protected c   public d   default

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