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If there are n bits, the AND computational problem determines if at least one of them is a one. This is the same game where you have to figure out which of the n doors hides a prize. The difference between this and the parity problem is that the
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- Computer science. Correct answer will be upvoted else downvoted. Think about a n by n chessboard. Its columns are numbered from 1 to n from the top to the base. Its sections are numbered from 1 to n from the passed on to one side. A cell on a convergence of x-th line and y-th section is indicated (x,y). The fundamental corner to corner of the chessboard is cells (x,x) for all 1≤x≤n. A stage of {1,2,3,… ,n} is composed on the fundamental slanting of the chessboard. There is actually one number composed on every one of the cells. The issue is to segment the cells under and on the principle askew (there are by and large 1+2+… +n such cells) into n associated areas fulfilling the accompanying imperatives: Each district ought to be associated. That implies that we can move from any cell of a locale to some other cell of a similar area visiting just cells of a similar district and moving from a cell to a neighboring cell. The x-th area ought to contain cell on the fundamental…For a given problem with inputs of size n, algorithms running time,one of the algorithm is o(n), one o(nlogn) and one o(n2). Some measured running times of these algorithm are given below: Identify which algorithm is which and explain the observed running Times which algorithm would you select for different value of n?Al and Bob are arguing about their algorithms. Al claims his O(nlogn)-time method is always faster than Bob's O(n²)-time method. To settle the issue, they perform a set of experiments. To Al's dismay, they find that if n 100 is the O(nlogn)-time one better. Explain how this is possible.
- Consider two algorithms for the same problem: • Algorithm A, which runs in O(n) and produces a correct answer with probability 0.7, and a wrong answer with probability 0.3. • Algorithm B, which runs in O(n log n) and produces a correct answer with probability 0.99, and a wrong answer with probability 0.01. Which of the two algorithms should you use to build a more asymptotically (in n) efficient algorithm with probability 0.99 of producing a correct answer? Justify. Assume that you can always check if an answer is correct in constant time.Correct answer will be upvoted else downvoted. Computer science. Polycarp recalled the 2020-th year, and he is content with the appearance of the new 2021-th year. To recall such a great second, Polycarp needs to address the number n as the amount of a specific number of 2020 and a specific number of 2021. For instance, if: n=4041, then, at that point, the number n can be addressed as the total 2020+2021; n=4042, then, at that point, the number n can be addressed as the total 2021+2021; n=8081, then, at that point, the number n can be addressed as the total 2020+2020+2020+2021; n=8079, then, at that point, the number n can't be addressed as the amount of the numbers 2020 and 2021. Assist Polycarp with seeing if the number n can be addressed as the amount of a specific number of numbers 2020 and a specific number of numbers 2021. Input The primary line contains one integer t (1≤t≤104) — the number of experiments. Then, at that point, t experiments follow.…2. For a problem we have come up with three algorithms: A, B, and C. Running time of Algorithm A is O(n1000), Algorithm B runs in 0(2¹) and Algorithm C runs in O(n!). How do these algorithms compare in terms of speed, for large input? Explain why.
- Given the following Truth Table: Inputs ABC Outputs Y 000 001 010 011 100 101 110 111 Y= 1. Construct a K-Map on Scrap Paper 2. Perform Grouping on Scrap Paper 3. Give the SIMPLEST Boolean Expression for x in the answer box below. 1 X 1 0 X 1 1 1 4. Give sum of minterms and sum of don't cares. Em( )+ d( Should you have an empty set please answer with N/A )According to the loop invariant theorem, a valid loop invariant I(n) must have the property that Select one: O a. After the loop, P I(N) where P is the postcondition. O b. I(n) =0 after a finite number of steps. %3D O c. None of the other answers are correct O d. I(0) → Q, where Q is the precondition.Given is a strictly increasing function, f(x). Strictly increasing meaning: f(x)< f(x+1). (Refer to the example graph of functions for a visualization.) Now, define an algorithm that finds the smallest positive integer, n, at which the function, f(n), becomes positive. The things left to do is to: Describe the algorithm you came up with and make it O(log n).
- You are given a N*N maze with a rat placed at maze[0][0]. Find whether any path exist that rat can follow to reach its destination i.e. maze[N-1][N-1]. Rat can move in any direction ( left, right, up and down).Value of every cell in the maze can either be 0 or 1. Cells with value 0 are blocked means rat cannot enter into those cells and those with value 1 are open.Input FormatLine 1: Integer NNext N Lines: Each line will contain ith row elements (separated by space)Output Format :The output line contains true if any path exists for the rat to reach its destination otherwise print false.Sample Input 1 :31 0 11 0 11 1 1Sample Output 1 :trueSample Input 2 :31 0 11 0 10 1 1Sample Output 2 : false Solution: //// public class Solution { public static boolean ratInAMaze(int maze[][]){ int n = maze.length; int path[][] = new int[n][n]; return solveMaze(maze, 0, 0, path); } public static boolean solveMaze(int[][] maze, int i, int j, int[][] path) {//…Please, answer the whole question. Suppose you toss n biased coins independently. Given positive integers n and k, along with a set of non-negative real numbers p1,..., pn in [0, 1], where pi is the probability that the ith coin comes up head, your goal is to compute the probability of obtaining exactly k heads when tossing these n biased coins. Design an O(nk)-time algorithm for this task. Explain the algorithm, write down the pseudo code and do run time analysis.Correct answer will be upvoted else Multiple Downvoted. Computer science. Gildong has a square board comprising of n lines and n sections of square cells, each comprising of a solitary digit (from 0 to 9). The cell at the j-th section of the I-th line can be addressed as (i,j), and the length of the side of every cell is 1. Gildong prefers enormous things, so for every digit d, he needs to find a triangle with the end goal that: Every vertex of the triangle is in the focal point of a cell. The digit of each vertex of the triangle is d. Somewhere around one side of the triangle is corresponding to one of the sides of the board. You might expect that a side of length 0 is corresponding to the two sides of the board. The space of the triangle is boosted. Obviously, he can't simply be content with tracking down these triangles with no guarantees. Along these lines, for every digit d, he will change the digit of precisely one cell of the board to d, then, at that point, track…
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