Suppose the game is programmed so that the computer uses a binary search strat- egy for making its guesses. What is the maximum number of guesses the com- puter could make before guessing the user's number? (A) 50 (B) 25 (C) 10 (D) 7 (E) 6
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- Mastermind is a code-breaking game for two players. In the original real-world game, one player A selects 4 pegs out of 6 colors and puts them in a certain fixed order; multiples of colors are possible (for example, red-green red-green). His opponent B does not know the colors or order but has to find out the secret code. To do so, B makes a series of guesses, each evaluated by the first player. A guess consists of an ordered set of colors which B believes is the code. The first player A evaluates the guess and feeds back to B how many positions and colors are correct. A position is correct ("black") if the guess and the secret code have the same color. Additional colors are correct ("white"), if they are in the guess and the code, but not at the same location. For example1 2 3 4secret: red-green red greenguess: red blue green purpleresults in one correct position ("black = 1") for the red peg at position one and one additional correct color ("white=1") for the green peg in the guess.…Mastermind is a code-breaking game for two players. In the original real-world game, one player A selects 4 pegs out of 6 colors and puts them in a certain fixed order; multiples of colors are possible (for example, red-green red-green). His opponent B does not know the colors or order but has to find out the secret code. To do so, B makes a series of guesses, each evaluated by the first player. A guess consists of an ordered set of colors which B believes is the code. The first player A evaluates the guess and feeds back to B how many positions and colors are correct. A position is correct ("black") if the guess and the secret code have the same color. Additional colors are correct ("white"), if they are in the guess and the code, but not at the same location. For example1 2 3 4secret: red-green red greenguess: red blue green purpleresults in one correct position ("black = 1") for the red peg at position one and one additional correct color ("white=1") for the green peg in the guess.…Bus timetables specify to the second the exact arrival and departure time of each bus on each stop. You need to pay for the full fare of every bus you ride and different bus lines charge different fees , but they are flat fees (independent of distance travelled on the line) A travel plan is a sequence of stop-time pairs where stop is a location of a bus stop and time is when we arrive at that stop. The plan is feasible if for any two consecutive pairs (a, t) and (b, t′) in the plan there exists a bus that departs after t and arrives at b at exactly t′. That is, a travel plan does not allow us to walk between stops. Assuming that no two buses arrive at the same time at the same stop, a feasible plan uniquely identifies the bus lines that we need to take to realize the plan. The cost of the plan is the sum of the fares we need to pay. Your task is to design an efficient algorithm that given a departure time t, an arrival time t′, an origin stop a and a destination stop b, finds the…
- Correct answer will be upvoted else downvoted. Computer science. All arbitrary choices that are made during the title are made similarly plausible and autonomously. For instance, assuming n=4, a=[1,2,4,3], one of the choices for the game (there could be different choices) is: during the main game, the first and fourth players were chosen. The fourth player has more tokens, so he takes the main player's tokens. Presently a=[0,2,4,4]; during the subsequent game, the fourth and third players were chosen. They have similar number of tokens, yet arbitrarily, the third player is the champ. Presently a=[0,2,8,0]; during the third game, the second and third players were chosen. The third player has more tokens, so he requires the second player's tokens. Presently a=[0,0,10,0]; the third player is proclaimed the victor of the title. Title champs will get customized prizes. Consequently, the adjudicators need to know ahead of time which players get an opportunity of winning,…Correct answer will be upvoted else Multiple Downvoted. Don't submit random answer. Computer science. Two cells are adjoining in the event that they share a side. Thusly, every cell (x, y) has precisely three neighbors: (x+1, y) (x−1, y) (x+1, y−1) in case x is even and (x−1, y+1) in any case. At first a few cells are contaminated, all the others are sound. The course of recuperation starts. Each second, for precisely one cell (despite the fact that there may be different cells that could change its state) one of the accompanying occurs: A sound cell with something like 2 contaminated neighbors likewise becomes tainted. A contaminated cell with something like 2 solid neighbors likewise becomes sound. In the event that no such cell exists, the course of recuperation stops. Patient is considered recuperated if the course of recuperation has halted and every one of the cells are solid. We're keen on a most dire outcome imaginable: is it conceivable that the patient…Correct answer will be upvoted else Multiple Downvoted. Don't submit random answer. Computer science. Today the kindergarten has another gathering of n kids who should be situated during supper. The seats at the table are numbered from 1 to 4n. Two children can't sit on a similar seat. It is realized that two children who sit on seats with numbers an and b (a≠b) will enjoy if: gcd(a,b)=1 or, a partitions b or b separates a. gcd(a,b) — the greatest number x with the end goal that an is distinct by x and b is detachable by x. For instance, if n=3 and the children sit on seats with numbers 2, 3, 4, then, at that point, they will enjoy since 4 is isolated by 2 and gcd(2,3)=1. On the off chance that children sit on seats with numbers 4, 6, 10, they won't enjoy. The educator truly doesn't need the wreck at the table, so she needs to situate the children so there are no 2 of the child that can enjoy. All the more officially, she needs no pair of seats an and b that the children…
- Bum 0; for (int i = 1; i6. The Jaccard coefficient between two sets {a, b, c, d, e, f, g} and {a, b, g, h, i, j} is: Group of answer choices 0.2 0.3 0.4 0.5If A={8,12,16,20} and B={12,16,20,24,28}, what is the value of P (A and B)?Dingyu is playing a game defined on an n X n board. Each cell (i, j) of the board (1 2, he may only go to (2, n).) The reward he earns for a move from cell C to cell D is |value of cell C – value of cell D|. The game ends when he reaches (n, n). The total reward - is the sum of the rewards for each move he makes. For example, if n = 1 2 and A = 3 the answer is 4 since he can visit (1, 1) → (1, 2) → (2, 2), and no other solution will get a higher reward. A. Write a recurrence relation to express the maximum possible reward Dingyu can achieve in traveling from cell (1, 1) to cell (n, n). Be sure to include any necessary base cases. B. State the asymptotic (big-O) running time, as a function of n, of a bottom-up dynamic programming algorithm based on your answer from the previous part. Briefly justify your answer. (You do not need to write down the algorithm itself.)In computer science and mathematics, the Josephus Problem (or Josephus permutation) is a theoretical problem. Following is the problem statement: There are n people standing in a circle waiting to be executed. The counting out begins at some point (rear) in the circle and proceeds around the circle in a fixed direction. In each step, a certain number (k) of people are skipped and the next person is executed. The elimination proceeds around the circle (which is becoming smaller and smaller as the executed people are removed), until only the last person remains, who is given freedom. Given the total number of persons n and a number k which indicates that k-1 persons are skipped and kth person is killed in circle. The task is to choose the place in the initial circle so that you are the last one remaining and so survive. For example, if n = 5 and k = 2, then the safe position is 3. Firstly, the person at position 2 is killed, then person at position 4 is killed, then person at position 1…A wrestling tournament has 256 players. Each match includes 2 players. The winner each match will play another winner in the next round. The tournament is single elimination, so no one will wrestle after they lose. The 2 players that are undefeated play in the final game, and the winner of this match wins the entire tournament. How would you determine the winner? Here is one algorithm to answer this question. Compute 256/2 = 128 to get the number of pairs (matches) in the first round, which results in 128 winners to go on to the second round. Compute 128/2 = 64, which results in 64 matches in the second round and 64 winners, to go on to the third round. For the third round compute 64/2 = 32, so the third round has 64 matches, and so on. The total number of matches is 128 + 64 + 32+ .... Finish this process to find the total number of matches.SEE MORE QUESTIONS