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- Create an ABM function that takes the following parameters: n := number of paths to be simulated m := number of discretization points per path S0 := initial starting point dS=μdt+σdW Program the function by using two nested "for loops" def ABM(n,m,S0,mu,sigma,dt): np.random.seed(999) arr = # create 2D zeros array with the correct dimensions arr[,] = #initialize column 0 # fill in array entries for i in : for j in : arr[i,j] = return arrGiven a matrix of m x n elements (m rows, n columns),return all elements of the matrix in spiral order.For example,Given the following matrix:[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ]]You should return [1,2,3,6,9,8,7,4,5].""" def spiral_traversal(matrix): res = [] if len(matrix) == 0: return res row_begin = 0 row_end = len(matrix) - 1 col_begin = 0 col_end = len(matrix[0]) - 1 while row_begin <= row_end and col_begin <= col_end: for i in range(col_begin, col_end+1): res.append(matrix[row_begin][i]) row_begin += 1 for i in range(row_begin, row_end+1): res.append(matrix[i][col_end]) col_end -= 1 if row_begin <= row_end: for i in range(col_end, col_begin-1, -1): res.append(matrix[row_end][i]) row_end -= 1 if col_begin <= col_end: for i in range(row_end, row_begin-1, -1): res.append(matrix[i][col_begin]) col_begin += 1Search a sorted matrix: The input consists of a real number x and a matrix A[1..n, 1..m] of nm real values, where A[i, 1..m] and A[1..n, j] are the rows and columns, respectively, of the matrix. The objective is to either state that all of the members of A are greater than x or to locate the largest array entry A[i, j] that is less than or equal to x. Create and evaluate an iterative approach for this issue that looks at the least amount of matrix entries feasible. If you think the issue can be resolved with a straightforward binary search, think twice.
- Write a code in "C++"that can swap all the values of its first row with its last row in a [n×m] matrix After swapping, add all the values of the last column.(using C++ only)In elementwise multiplication A.*B each value in one matrix is paired up with a buddy value in the other and they are multiplied together.In matrix multiplication A*B each row in matrix A is dot-producted with each column in matrix B.The value in the upper left corner of the matrix C is the same as which of the following?C = A*B; options: A(1,1) * B(1,1) dot(A(1,:),B(:,1)) A = [3 1; 5 2];B = [1 -1; 4 0];C = A*Bthe value in the upper left corner of the matrix C is which of the following? % Starting with this code:a = [1 2 3]b = [-1 0 1]% All of the following are identical, except which one? Question 4 options: dot(a,b) a*transpose(b) b*transpose(a) sum(a.*b) They are all identical Fill in the blank to calculate the dot product of amounts and costs.amounts = [2 1 2 5 1]costs = [3.5 1.25 4.25 1.55 3.15]____________ Question 5 options:…Write a code in "C"that can swap all the values of its first row with its last row in a [n×m] matrix After swapping, add all the values of the last column.
- Write a program which will read the non zero elements of a sparse matrix from the keyboard in the triplet form and the program needs to print the corresponding sparse matrix.Consider a list of data components L[0:5] = 23, 14, 98, 45, 67, and 53. Let's look for the K = 53 key. Naturally, the search moves down the list, comparing key K with each element until it discovers it as the final element in the list. When looking for the key K = 110, the search moves forward but eventually drops off the list, making it a failed search. Write the algorithmic steps for both sorted and unordered linear searches, along with the times involved.## Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)# def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…
- #plea# Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)# def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…#com# Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)# def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…The Burrows-Wheeler transform (BWT) is a transformation that is used in data compression algorithms, including bzip2, and in high-throughput sequencing in genomics. Write a SuffixArray client that computes the BWT in linear time, as follows: Given a string of length N (terminated by a special end-of-file character $ that is smaller than any other character), consider the N-by-N matrix in which each row contains a different cyclic rotation of the original text string. Sort the rows lexicographically. The Burrows-Wheeler transform is the rightmost column in the sorted matrix. For example, the BWT of Mississippi is ipssm$pissii. The Burrows-Wheeler inverse transform (BWI) inverts the BWT. For example, the BWI of ipssm$pissii is mississippi$. Also write a client that, given the BWT of a text string, computes the BWI in linear time.