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- Python: Shopping PlanExample: assert best_shopping_plan([], 1) == 0 # Alice cannot buy anything as there is no item in the store assert best_shopping_plan([1], 1) == 1 # Alice can only buy that one item of price 1 given her budget amount assert best_shopping_plan([1,2], 1) == 1 # Alice can only buy that one item of price 1 given her budget amount assert best_shopping_plan([1,1,2], 2) == 2 # Alice can buy one item of price 2, or she can buy the two items of price 1 assert best_shopping_plan([1,2,3], 3) == 3 # Alice can buy one item of price 3, or she can buy the two items of price 1 and 2 assert best_shopping_plan([1,2,3,4], 10) == 7 # The best Alice can buy is for items of prices 3 and 4 assert best_shopping_plan([1,2,2,3,3,3], 10) == 6 # The best Alice can buy is for items of prices 3 and 3 assert best_shopping_plan([1,1,1,3,5,5], 7) == 6 # The best Alice can buy is for items of prices 1 and 5Description: There are 5 philosophers sitting around a table and try to eat from the center of the table. 5 chopsticks also lay on the table. Let us call the 5 philosophers in clockwise P1, P2, ...P5. There also 5 chopsticks clockwise S1, S2, ..., S5 on the table. There are only one chopstick between every two philosophers. Every right hand chopstick will have the same index as the philosopher. For example, on the right hand side of P1, the chopstick is called S1. And every left hand side chopstick number is 1+number of the philosopher. A philosopher spend random time to think, then he feel hungry and try to eat. The middle dish can provide enough food for everyone at the same time. But a philosopher only can start to eat when he picked up two chopsticks from left hand side and right hand side to form a pair of chopsticks. If a philosopher take one chopsticks, he will try to fight with neighbours to get another one, and never back off to put down the one in his hand. Once the…public LLNode secondHalf(LLNode head) { }
- dict1 = {(1,1,1):"red", (2,1,-1):"green", (0,-1,1):"red", (0,0,-2):"blue"}In this case, there are 2 red points, and their centroid is at ( (1+0)/2, (1-1)/2, (1+1)/2 ) = (0.5, 0, 1)Considering that in this example there is only one green and one blue point, they are representing the respective color's centroid. Hence, your function should return the following dictionary:{"red":(0.5,0.0,1.0), "green":(2.0,1.0,-1.0), "blue":(0.0,0.0,-2.0)}please create a class diagram for the 2048 game can you please draw the diagram2048 is a single-player puzzle game created by Gabriele Cirulli1. It is played on a 4×4 grid that contains integers ≥2 that are powers of 2. The player can use a keyboard arrow key (left/up/right/down) to move all the tiles simultaneously. Tiles slide as far as possible in the chosen direction until they are stopped by either another tile or the edge of the grid. If two tiles of the same number collide while moving, they will merge into a tile with the total value of the two tiles that collided. The resulting tile cannot merge with another tile again in the same move. Please observe this merging behavior carefully in all Sample Inputs and Outputs. Input The input is always a valid game state of a 2048 puzzle. The first four lines of input, that each contains four integers, describe the 16 integers in the 4×4 grid of 2048 puzzle. The j-th integer in the i-th line denotes the content of the cell located at the i-th row and the j -th cell. For this problem, all integers in the input will…
- #include "TerminalPlayer.h" Card TerminalPlayer::playCard(const Card& opponentCard) { // if the opponentCard is a Joker we are going first // if the opponentCard is not a Joker we are going second and opponentCard is what our opponent played // Display the player's hand // prompt them to choose a card // remove that card from the hand and return that card@return index of the point that is closest to the origin, which is (0, 0) * In case of a tie, return the lowest index */ public int closestToOriginIndex() { return 0; }Blue-Eyed Island: A bunch of people are living on an island, when a visitor comes with a strangeorder: all blue-eyed people must leave the island as soon as possible. There will be a flight out at8:00pm every evening. Each person can see everyone else's eye color, but they do not know theirown (nor is anyone allowed to tell them). Additionally, they do not know how many people haveblue eyes, although they do know that at least one person does. How many days will it take theblue-eyed people to leave?
- Hi i got this in my APS class You are playing a 2 person matchstick game. Players take turns to remove 1, 5 or 6 matches from the pile of matches. We want you to analyse this game, clearly explaining everything you are doing and develop a strategy to ensure that you would always win the game If you are the person answering with the handwritten answer, could you type it out please as your handwriting is a little difficult to read, thank you !Chiralitydef is_left_handed(pips): Even though this has no effect on fairness, pips from one to six are not painted on dice any which way, but so that the pips on the opposite faces always add up to seven. (This convention makes it easier to tell when someone tries to use gaffed dice that leave out some spot values.) In each corner of the cube, one value from each pair of opposites 1-6, 2-5 and 3-4 meets two values from the other two pairs of opposites. The math works out correctly in for the 23 = 8 corners of the cube. This still leaves two possibilities how to paint these pips. Look at the corner shared by the faces 1, 2, and 3, and read these numbers clockwise around that corner starting from 1. If these three numbers read out as 1-2-3, that die is left-handed, and if they read out as 1-3-2, that die is right-handed. Analogous to a pair of shoes made separately for the left and right foot, left- and right-handed dice are in one sense identical, yet no matter how you twist and turn,…Crack the crag def crag_score(dice): Crag (see the Wikipedia page for the scoring table needed in this problem) is a dice game similar to the more popular games of Yahtzee and Poker dice in style and spirit, but with much simpler combinatorics of roll value calculation due to this game using only three dice. Players repeatedly roll three dice and assign the resulting patterns to scoring categories so that once some roll has been assigned to a category, that category is considered to have been spent and cannot be used again for any future roll. These tactical choices between safety and risk-taking give this game a little bit more tactical flair on top of merely relying on the favours of Lady Luck for rolling the bones. Given the list of pips of the three dice of the first roll, this function should compute and return the highest possible score available when all categories of the scoring table are still available for you to choose from, so that all that matters is maximizing this…