Question

Matlab

==text only not images==

==correct solution only plz==

a) Write a nested for loops to test if an input Matrix A is a sparse matrix or not. A sparse matrix is a matrix in which most of the elements are zero. By contrast, if most of the elements are nonzero, then the matrix is considered dense. The number of zero-valued elements divided by the total number of elements is called the sparsity of the matrix, so matrix A will be sparse when its sparsity is greater than 0.5.

b) Search the Matlab documentation for a built-in function which check the sparsity of the matrix and rewrite the code using it. (Do NOT use loops)

c) Compute and show the time required to execute the code in (a) and the code in (b). Which code is faster? What conclusion can be drawn from this result? Hint: Use tic…toc commands to compute the time elapsed.

text only not images

correct solution only plz

Answer 1

(a) MATLAB CODE:

clc
clear all
close all

tic

A = [1 2 3;0 1 0;1 0 0]; % Matrix A
S = size(A); % Size of matrix A
R = S(1); % Number of rows of matrix A
C = S(2); % Number of coloumns of matrix A

Z = 0; % Initializing zeros count
N = 0; % Initializing non-zeros count

for i = 1:R
for j = 1:C
if A(i,j) == 0
Z = Z+1;
else
N = N+1;
end
end
end
fprintf('Number of zeros present in Matrix-A : %d\n',Z)
fprintf('Number of non-zeros present in Matrix-A : %d\n',N)

T = R*C; % Total number of elements in Matrix-A
Sparsity = Z/T; % Sparsity means number of zero valued elements divide by totla number of values in matrix-A
fprintf('Sparsity of Matrix-A:%d\n',Sparsity)

if Sparsity > 0.5
fprintf('Matrix-A is Sparse Matrix\n')
else
fprintf('Matrix-A is Dense Matrix\n')
end

toc

OUTPUT:

(b) MATLAB CODE:

clc
clear all
close all

tic

A = [1 2 3;0 1 0;1 0 0]; % Matrix-A
S = issparse(A); % Checking Matrix-A is sparse or not
if S == 1
fprintf('Matrix-A is Sparse Matrix\n')
else
fprintf('Matrix-A is Dense Matrix\n')
end

toc

OUTPUT:

(c) If we see the elapsed time of part(a) and part(b), part(b) Elapsed time is very very less than that of part(a) elapsed time.That means code in part(b) is faster