Write a function that diagonally orders numbers in a n x n matrix, depending onwhich of the four corners you originate from: upper-left ( u1 ), upper-right (ur),lower-left (11), lower-right (1r).Examplesdiagonalize(3, "ul")[0, 1, 2],[1, 2, 3],[2, 3, 4](Ctrl) -[

Algorithm - Take input from user. Now use the below logic - ul: n => Array.from({ length: n },…

Answered: Write a function that diagonally orders…

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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Create a 3-by-3 matrix: A = [1 2 3;4 5 6;7 8 10] (a)Extract a submatrix B consisting of rows 1 and 3 and columns 1 and 2 of the matrix A (b)Interchange rows 1 and 3 of A and put it in a variable C. Hint: use the vector of row indices together with the colon operator (c)Create a vector version of the original matrix A. (d)Delete the 2nd column of the original matrix A and put it in a variable called A. (e)Insert a 2nd column to the new matrix A to restore it back to its original form A. (f) Extract those entries of the original matrix A that are greater than 5
Find a sorted matrix: The input is a real number x and a matrix A[1..n, 1..m] of nm real numbers, with each row A[i, 1..m] and column A[1..n, j] sorted. The objective is to locate the largest array entry A[i, j] that is less than or equal to x, or to report that all components of A are greater than x. Create and test an iterative method that analyses as few matrix elements as feasible. Be cautious if you assume a simple binary search would fix the problem. Find a sorted matrix: The input is a real number x and a matrix A[1..n, 1..m] of nm real numbers, with each row A[i, 1..m] and column A[1..n, j] sorted. The objective is to locate the largest array entry A[i, j] that is less than or equal to x, or to report that all components of A are greater than x. Create and test an iterative method that analyses as few matrix elements as feasible. Be cautious if you assume a simple binary search would fix the problem.
The pseudo-code to create the matrix:Procedure AdjacencyMatrix(N): //N represents the number of nodesMatrix[N][N]for i from 1 to N for j from 1 to N Take input -> Matrix[i][j] endforendfor Make Java code implemented using above pseudo-code:

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Write a function that diagonally orders numbers in a n x n matrix, depending on which of the four corners you originate from: upper-left (ul), upper-right (ur), lower-left (11), lower-right ( 1r ). Examples diagonalize (3, "ul") → [ [0, 1, 2], [1, 2, 3], [2, 3, 4] (Ctrl)