A mosaic is a matrix of nxn in which each cell is painted with some color. If we haven colors, we say that a mosaic is balanced if each color appears exactly once in each row andeach column. For example, the following is a balanced mosaic of 4x4:Figure 1: Balanced mosaic of 4x4Given a mosaic Mfrom n x n with some unpainted cells, we will say that Mit is colorable ifthere is a way to color the empty cells so that the resulting mosaic is balanced. Build a formulaPEL (P)such thatpit's satisfying + Mit is colorableand prove that your construction is correct.

As more solutions develop so do strategies for finding them: Some might notice the placement of…

Answered: Figure 1: Balanced mosaic of 4 x4 Given…

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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3. Given the following minterm expansion: F(a, b, c, d) = Em(0, 2, 3, 4, 8, 10, 11, 15) Draw the K-Map Find out all essential prime implicants, do it systematically. Do you need additional loops to cover all the minterms. If yes, find the additional prime implicants. Get the minimum SO.
Let U = {0, 1,2, ...,9}, A = {0,3,5, 8}, B = {0, 1, 2, 3, 6}, and C = {2, 3, 5, 8 9} %3D AUBUC= (0,1,2,3,5,6,8,9} AU (Bn C) = {0,2,3,5,8} (AU B) n C = | {2,3,5,8} C-(AUB) = {9} (C- A)n B = {2} B - (An C) = {0,1,2,6) B - (An C) = 2 %3D
Let m be a matrix with n rows and n columns whose entries are either 1 or 0. recall that the element of m on row i and column j is denoted by mij . the diagonal entries of m are {mii} for 1 ≤ i ≤ n. We call M permutable if it is possible to swap some of the rows and some of columns so that all diagonal entries are 1. Design a polynomial time algorithm that decides whether a binary matrix M is permutable or not. Note that we can swap any two rows or two columns. Also, the order in which these swaps are done is not important.
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) {//…
7. For n 2 1, in how many out of the n! permutations T = (T(1), 7(2),..., 7 (n)) of the numbers {1, 2, ..., n} the value of 7(i) is either i – 1, or i, or i +1 for all 1 < i < n? Example: The permutation (21354) follows the rules while the permutation (21534) does not because 7(3) = 5. Hint: Find the answer for small n by checking all the permutations and then find the recursive formula depending on the possible values for 1(n).
We have learned the algorithm that solves the Closest pair problem in 2D in Θ(n log n) time. (Closest pair problem in 2D: Given n points in the 2D plane, find a pair with smallest Euclidean distance between them.) Give an algorithm that solves the Closest pair problem in 3D in Θ(n log n) time. (Closest pair problem in 3D: Given n points in the 3D space, find a pair with smallest Euclidean distance between them.)
Consider an n by n matrix, where each of the n2 entries is a positive integer. If the entries in this matrix are unsorted, then determining whether a target number t appears in the matrix can only be done by searching through each of the n2 entries. Thus, any search algorithm has a running time of O(n²). However, suppose you know that this n by n matrix satisfies the following properties: • Integers in each row increase from left to right. • Integers in each column increase from top to bottom. An example of such a matrix is presented below, for n=5. 4 7 11 15 2 5 8 12 19 3 6 9 16 22 10 13 14 17 24 1 18 21 23 | 26 | 30 Here is a bold claim: if the n by n matrix satisfies these two properties, then there exists an O(n) algorithm to determine whether a target number t appears in this matrix. Determine whether this statement is TRUE or FALSE. If the statement is TRUE, describe your algorithm and explain why your algorithm runs in O(n) time. If the statement is FALSE, clearly explain why no…
Consider the problem of making change for n cents using the fewest number of coins. Assume that we live in a country where coins come in k dierent denominations c1, c2, . . . , ck, such that the coin values are positive integers, k ≥ 1, and c1 = 1, i.e., there are pennies, so there is a solution for every value of n. For example, in case of the US coins, k = 4, c1 = 1, c2 = 5, c3 = 10, c4 = 25, i.e., there are pennies, nickels, dimes, and quarters. To give optimal change in the US for n cents, it is sufficient to pick as many quarters as possible, then as many dimes as possible, then as many nickels as possible, and nally give the rest in pennies. Design a bottom-up (non-recursive) O(nk)-time algorithm that makes change for any set of k different coin denominations. Write down the pseudocode and analyze its running time. Argue why your choice of the array and the order in which you fill in the values is the correct one. Notice how it is a lot easier to analyze the running time of…

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Figure 1: Balanced mosaic of 4 x4 Given a mosaic M from n x nwith some unpainted cells, we will say that Mit is colorable if there is a way to color the empty cells so that the resulting mosaic is balanced. Build a formula $EL (P)such that