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- Python Lab: Dr.D has invented yet another invention: the hateinator. He wants to test it on a group of N people (numbered 1 through N). The hateinator may be used any number of times; to use it once, Dr.D should divide these N people into two groups and press the fire button on the hateinator. We call each such grouping a Doofish set. Afterwards, there will be hatred between each two people who were in different groups. The hatred does not disappear ― any two people that hate each other before the hateinator is used still hate each other afterwards. The hateinator uses a lot of power. Let's denote the number of times it is used by K. Then, it consumes K⋅N units of power. Dr.D cannot afford to use the hateinator if this number exceeds 106. Dr.D has done the math and computed the most evil hatred system: a situation with some M pairs of people who hate each other. You are given these pairs. There must not be any other pair of people who hate each other. Initially, there is no hatred…* allSame returns true if all of the elements in list have the same value. * allSame returns false if any two elements in list have different values. * The array may be empty and it may contain duplicate values. * * Your solution should contain at most one loop. You may not use recursion. * Your solution must not call any other functions. * Here are some examples (using "==" informally): * * * * * * * * true == allSame (new double[] { }) true == allSame(new double[] {11}) true == allSame (new double[] { 11, 11, 11, 11 }) false == allSame(new double[] { 11, 11, 11, 22 }) false == allSame (new double[] { 11, 11, 22, 11 }) true == allSame (new double[] { 22, 22, 22, 22 }) * */ public static boolean allSame (double[] list) { return StdRandom.bernoulli(); //TODO: fix thisPython Programming Lab Dr.D has invented yet another invention: the hateinator. He wants to test it on a group of N people (numbered 1 through N). The hateinator may be used any number of times; to use it once, Dr.D should divide these N people into two groups and press the fire button on the hateinator. We call each such grouping a Doofish set. Afterwards, there will be hatred between each two people who were in different groups. The hatred does not disappear ― any two people that hate each other before the hateinator is used still hate each other afterwards. The hateinator uses a lot of power. Let's denote the number of times it is used by K. Then, it consumes K⋅N units of power. Dr.D cannot afford to use the hateinator if this number exceeds 106. Dr.D has done the math and computed the most evil hatred system: a situation with some M pairs of people who hate each other. You are given these pairs. There must not be any other pair of people who hate each other. Initially, there is no…
- def findOccurrences(s, ch): lst = [] for i in range(0, len(s)): if a==s[i]: lst.append(i) return lst Use the code above instead of enumerate in the code posted below. n=int(input("Number of rounds of Hangman to be played:")) for i in range(0,n): word = input("welcome!") guesses = '' turns = int(input("Enter the number of failed attempts allowed:")) def hangman(word): secrete_word = "-" * len(word) print(" the secrete word " + secrete_word) user_input = input("Guess a letter: ") if user_input in word: occurences = findOccurrences(word, user_input) for index in occurences: secrete_word = secrete_word[:index] + user_input + secrete_word[index + 1:] print(secrete_word) else: user_input = input("Sorry that letter was not found, please try again: ") def findOccurrences(s, ch): return [i for i, letter in enumerate(s) if letter == ch] *** enumerate not discussed in…x <- bubblesort(sample(1:6, size=5, replace=T)) if (x[1]==x[5]){ hand <-"Yahtzee" } else if (x[1]==x[4] ] x[2]==x[5]) { hand <- "4ofKind" } } else { hand <-"Chance" } The above code lays this out for Yahtzee and 4 of a kind. Find statements that work for the rest and put this into a function called get hand , which returns what type of hand was rolled from a sorted roll of five dice. Make sets of rolls and test them to make sure they work. Note that above we use an else if, which allows us to test multiple conditions in an if-else statement.Computer Science This is to be done in Java using Sets (HashSet). The task is to find out how many people you typically need to meet before you meet one person born each day of the year. You re assuming there are 365 days in a year (no leap years). This means a birthday can be represented by a random number from 1 - 365. Using a set, the task is to keep track of the birthdays you have already encountered. Run the whole trial 100 times to take an average. THE OUTPUT OF THE NUMBER OF PEOPLE will vary due to it being a random number but it should be in the 2000s.
- Let A={1,2,3). B={3,4,5,6}, and C={6,7,8}.Find (AUB)ᑎC.Write your answer using set notation, i.e. list all elements separated by comma inside braces{} or use the symbol of empty set, if needed Ø. Do not insert any spacesWrite a recursive function that finds the minimum value in an ArrayList. Your function signature should be public static int findMinimum(ArrayList<Integer>) One way to think of finding a minimum recursively is to think “the minimum number is either the last element in the ArrayList, or the minimum value in the rest of the ArrayList”. For example, if you have the ArrayList [1, 3, 2, 567, 23, 45, 9], the minimum value in this ArrayList is either 9 or the minimum value in [1, 3, 2, 567, 23, 45] Hint:The trick is to remove the last element each time to make the ArrayList a little shorter. import java.util.*; public class RecursiveMin{public static void main(String[] args){Scanner input = new Scanner(System.in);ArrayList<Integer> numbers = new ArrayList<Integer>();while (true){System.out.println("Please enter numbers. Enter -1 to quit: ");int number = input.nextInt();if (number == -1){break;}else {numbers.add(number);}} int minimum =…Write the following method that returns an ArrayList from a set:public static <E> ArrayList<E> setToList(Set<E> s)
- Let A = {0, b) and B = {a, d) and C = {c}. Compute the following sets: a. AX (BXC) b. AXCWrite proceedures for the following: 1) Given a 9x9 grid of numbers (lists of lists, so [[1,2,3,4,5,6,7,8,9],[3,4,5,1,2,6,7,8,9],....]), a column number, and a fixed number between 1 and 9, determine if the fixed number appears in the column. (Unimportant here, but missing numbers for an incomplete grid will be zeros). The program should return a true or false.Write a recursive function that finds the minimum value in an ArrayList. Your function signature should be public static int findMinimum(ArrayList<Integer>) One way to think of finding a minimum recursively is to think “the minimum number is either the last element in the ArrayList, or the minimum value in the rest of the ArrayList”. For example, if you have the ArrayList [1, 3, 2, 567, 23, 45, 9], the minimum value in this ArrayList is either 9 or the minimum value in [1, 3, 2, 567, 23, 45] ================================================ import java.util.*; public class RecursiveMin{public static void main(String[] args){Scanner input = new Scanner(System.in);ArrayList<Integer> numbers = new ArrayList<Integer>();while (true){System.out.println("Please enter numbers. Enter -1 to quit: ");int number = input.nextInt();if (number == -1){break;}else {numbers.add(number);}} int minimum = findMinimum(numbers);System.out.println("Minimum: " + minimum);}public static int…