Please answer this only in python Alice and Bob play a game. At first, there is only one tree with roots. The players take turns, Alice begins. At one point, Alice has to pick up any empty underground tree roots (or shrubs) and remove them. Bob, on the other hand, has to pick out any part of the empty leaves and remove them from the tree (or trees). (Please note that Bob always takes some depth from the leaves of the first tree.) It is not allowed to skip the turn. The game ends when the player is unable to take a chance. Alice's goal is to reduce the number of turns made by both players, while Bob wants the number to be greater. Considering that they both played very well, how many turns did they make last... Input 1
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- You will be given a square chess board with one queen and a number of obstacles placed on it. Determine how many squares the queen can attack. A queen is standing on an chessboard. The chess board's rows are numbered from to , going from bottom to top. Its columns are numbered from to , going from left to right. Each square is referenced by a tuple, , describing the row, , and column, , where the square is located. The queen is standing at position . In a single move, she can attack any square in any of the eight directions (left, right, up, down, and the four diagonals). In the diagram below, the green circles denote all the cells the queen can attack from : There are obstacles on the chessboard, each preventing the queen from attacking any square beyond it on that path. For example, an obstacle at location in the diagram above prevents the queen from attacking cells , , and : Given the queen's position and the locations of all the obstacles, find and print the number of…In an astronomy board game, N planets in an imaginary universe do not follow the normal law of gravitation. All the planets are positioned in a row. The planetary system can be in a stable state only if the sum of the mass of all planets at even positions is equal to the sum of the mass of planets at the odd positions. Initially, the system is not stable, but a player can destroy one planet to make it stable. Find the planet that should be destroyed to make the system stable. If no such planet exists, then return -1. If there are multiple such planets, then destroy the planet with the smallest index and return the index of the destroyed planet. Example Let N-5 and planets = [2,4,6,3,4]. Destroying the fourth planet of mass 3 will result in planets = [2,4,6,4], and here, the sum of odd positioned planets is (2+6)=8, and the sum of even positioned planets is (4+4)=8, and both are equal now. Hence, we destroy the fourth planet. 11 MNBASK19922 13 14 15 16 17 18 20 * The function is…A hungry mouse wants to eat all four fruits in a maze such as the one below, in as few moves as possible.. At each turn the mouse can move any number of squares in one of the directions up, down, left or right, but it is not allowed to enter (or jump over) any walls (i.e., the black squares). Thus, the mouse moves just like a rook in chess. To eat a fruit, the mouse has to stop at that square. Assume that the maze has 4 fruits, and the size of b xh squares. 1. Give a suitable representatión of the states in this searching problem. 2. How many possible actions can the mouse perform at each move? (1.e., what is the branching factor?)
- Email me the answers to the following questions. If you are not familiar with Peg Solitaire, then look it up online. Peg Solitaire is a game consisting of a playing board with 33 holes together with 32 pegs. In the picture above, the hole in the center is empty and the remaining holes contain pegs. The goal is to remove all the pieces except one, which should be in the center. A piece can be removed by jumping an adjacent piece over it into an empty hole. Jumps are permitted horizontally or vertically, but not diagonally. Your assignment consists of one required part, plus one extra credit part: 1. Explain (in words) why Breadth First Search and Iterative Deepening are not good methods for this problem.Correct answer will be upvoted else downvoted. Computer science. You have w white dominoes (2×1 tiles, the two cells are hued in white) and b dark dominoes (2×1 tiles, the two cells are shaded in dark). You can put a white domino on the board in case both board's cells are white and not involved by some other domino. Similarly, you can put a dark domino if the two cells are dark and not involved by some other domino. Would you be able to put all w+b dominoes on the board if you can put dominoes both on a level plane and in an upward direction? Input The main line contains a solitary integer t (1≤t≤3000) — the number of experiments. The primary line of each experiment contains three integers n, k1 and k2 (1≤n≤1000; 0≤k1,k2≤n). The second line of each experiment contains two integers w and b (0≤w,b≤n). Output For each experiment, print YES in case it's feasible to put all w+b dominoes on the board and negative, in any case. You might print each letter…Kingdom of Trolls is celebrating their Kingdom Day and one of the activities that is taking place is a game where a player rolls a magic ball down the hill on a path with spikes. As the ball rolls down, it strikes a spike and bursts open to release a number of smaller balls (in our simulated game, the number of smaller balls is a randomly generated integer between 2 and 6, inclusive). As the smaller balls further roll down, when one strikes a spike, that ball and all its sibling balls burst and each generates another set of smaller balls (using the same random number already generated for the first roll). The balls keep rolling downhill and striking spikes and bursting into smaller balls until a golden ball is released by one of the bursts. At this time, the game is over and the player is told how many balls were generated during the last burst (including the golden ball). The game is played by two players at a time and the player who had the lowest number of balls generated on the…
- Othello is played as follows: Each Othello piece has a white side and a black side.When a component is encircled on both the left and right sides, or on both the top and bottom, it is said to be caught, and its colour is reversed. You must capture at least one of your opponent's pieces during your turn. When neither user has any more legitimate movements, the game is over. The winner is determined by who has the most pieces. Othello's object-oriented design should be implemented.Can you help me with this code because I am struggling. The Lights Out puzzle consists of an m x n grid of lights, each of which has two states: on and off. The goal of the puzzle is to turn all the lights off, with the caveat that whenever a light is toggled, its neighbors above, below, to the left, and to the right will be toggled as well. If a light along the edge of the board is toggled, then fewer than four other lights will be affected, as the missing neighbors will beignored. In this section, you will investigate the behavior of Lights Out puzzles of various sizes by implementing a LightsOutPuzzle class. Once you have completed the problems in this section, you can test your code in an interactive setting using the provided GUI. See the end of the section for more details. Task: A natural representation for this puzzle is a two-dimensional list of Boolean values, where True corresponds to the on state and False corresponds to the off state. In the LightsOutPuzzle class, write an…Consider a game that you want to develop to impress your friend. You design a game with rules: “A player rolls a die and a coin in a single throw. Each die has six faces. These faces contain 1, 2, 3, 4, 5, and 6 spots. A coin has two faces, e.g., Head and Tail (Head is considered as 1 and Tail as 0). After the die and coin have come to rest, the sum of the upward spots of die and coin is calculated. If the sum is 7 (makes 6 for the die and Head for coin) the player wins, if the sum is 1 (makes 1 for the die and Tail for coin) the player loses. The remaining point becomes the player's “point”. To win the game, the player must continue rolling the die and coin until the player makes his "point". The player loses by getting 7 or 1 before making his point.” Write down a Java program that simulates this game.
- There are four people who want to cross a rickety bridge; they all begin on the same side. You have 17 minutes to get them all across to the other side. It is night, and they have one flashlight. A maximum of two people can cross the bridge at one time. Any party that crosses, either one or two people, must have the flashlight with them. The flashlight must be walked back and forth; it cannot be thrown, for example. Person 1 takes 1 minute to cross the bridge, person 2 takes 2 minutes, person 3 takes 5 minutes, and person 4 takes 10 minutes. A pair must walk together at the rate of the slower person's pace. Write the specification of an algorithm that solves the problem.Make a MAZE game on Spyder, giving the user to choose their route. Each route will have either doors or obstacles that the user will solve or use logic to complete. Use a random function so the obstacles change with a new try. The user needs to get out of the Maze in three tries or he loses a life. The user will be asked a question whether they want to try it again or skip the level. Need help with the code!! Thank youBomberman lives in a rectangular grid. Each cell in the grid either contains a bomb or nothing at all. Each bomb can be planted in any cell of the grid but once planted, it will detonate after exactly 3 seconds. Once a bomb detonates, it's destroyed — along with anything in its four neighboring cells. This means that if a bomb detonates in cell , any valid cells and are cleared. If there is a bomb in a neighboring cell, the neighboring bomb is destroyed without detonating, so there's no chain reaction. Bomberman is immune to bombs, so he can move freely throughout the grid. Here's what he does: Initially, Bomberman arbitrarily plants bombs in some of the cells, the initial state. After one second, Bomberman does nothing. After one more second, Bomberman plants bombs in all cells without bombs, thus filling the whole grid with bombs. No bombs detonate at this point. After one more second, any bombs planted exactly three seconds ago will detonate. Here, Bomberman stands back and…