Numpy has its own black-box linear system solver numpy.linalg.solve(): https://numpy.org/doc/stable/reference/generated/numpy.linalg. solve.html. Generate a random, square matrix and random vector using numpy and then solve using numpy.linalg.solve(). Confirm it is a solution using 'A.dot (x)' where x is the solution from solve(). Numpy.linalg import numpy as np n=100 A-np.random.rand (n,n) b=np.random.rand (n,1)

One can manually count path lengths in a graph using adjacency matrices. Using the simple example below, produces the following adjacency matrix: A B A 1 1 B 1 0 This matrix means that given two vertices A and B in the graph above, there is a connection from A back to itself, and a two-way connection from A to B. To count the number of paths of length one, or direct connections in the graph, all one must do is count the number of 1s in the graph, three in this case, represented in letter notation as AA, AB, and BA. AA means that the connection starts and ends at A, AB means it starts at A and ends at B, and so on. However, counting the number of two-hop paths is a little more involved. The possibilities are AAA, ABA, and BAB, AAB, and BAA, making a total of five 2-hop paths. The 3-hop paths starting from A would be AAAA, AAAB, AABA, ABAA, and ABAB. Starting from B, the 3-hop paths are BAAA, BAAB, and BABA. Altogether, that would be eight 3-hop paths within this graph. Write a program…

Generate all permutations of {3, 8, 2} using the Johnson-Trotter algorithm. Show the steps.

PYTHON/JUPYTER NOTEBOOKS  Program a backward solver (i.e. write your code) for an upper triangular system Ux=b. The algorithm is attached for i and j with offset 1. Recall that NumPy and Python have offset 0 for their sequence data types.