Is that the correct way to plot the graph ?

I have two functions and i would like to plot a graph.

My concern is that when i call Ja function it wont take the correct t because, it might think that when i call Ja it outputs only one output x and not t. the second function works the same so i will post only one.

n1 = 10:1:5000;           %Range vector of n for Jacobi function.
n2 = 10:100:5010;         %Range vector of n for GaussPartialPivoting2 function.   
ts1 = zeros(size(n1));    %Vector of the time for Jacobi function.
ts2 = zeros(size(n2));    %Vector of the time for GaussPartialPivoting2 function.  
for i = 1:numel(n1)
    [x,ts1(i)]=Ja(n1(i));
    
end
    for j=1:numel(n2)
        [x,ts2(j)]=GaussPartialPivoting2(n2(j));
    end
%plot graph  
subplot(2,1,1)
plot(n1,ts1,'LineWidth',1);
title('Jacobi')
xlabel('number of equations')
ylabel('time for solutions [sec]')
subplot(2,1,2)
plot(n2,ts2,'r','LineWidth',1)
title('Gauss Partial Pivoting')
xlabel('number of equations')
ylabel('time for solutions [sec]')

function [x,t]=Ja(n)
tic
v4=repelem(4,n);                        %Creats a vector with value of 4 n(input) times.
v1=repelem(-1,n-1);                     %Creats a vector with value of -1, n-1 times(input-1) times.
A=diag(v4)+diag(v1,-1)+diag(v1,1);      %Creats a matrix with v4 vector as diagonal and v1 as as diagonal under and on top the v4 diagonal.
 A(1,n)=-1;                             %assigns value -1 to the (1:n) position in the matrix.
 A(n,1)=-1;                             %assigns value -1 to the (n:1) position in the matrix.
 b=1:n;                                 % creats b a row vector of solutions.
 B=b'/n;                                % Changing b from row vector to a column vector.

x = zeros(n,1);                         %Initial guess.
epsilon = 1e-6;                         %required accurecy/Maximum error.
flag = 0;
counter = 0;                            %Set initial iteration counter.
%Disassemble marix A
L = tril(A,-1);                         %Builds lower triangular of matrix A without the diagonal.
D = diag(A);                            %Builds only the diagonal of matrix A.
U = triu(A,1);                          %Builds upper triangular of matrix A without the diagonal.
BB = -1./D.*(L+U);                      %Build new Matrix BB  from the decomposition of matrix A. 
c = (1./D).*B;
while flag == 0
    counter = counter + 1; 
    if counter > 10000
        error('Too many iterations');
    end
    
    x_new = BB*x+c;                       %Updating solution  until the stop condition is met.
    
    if max(abs(x_new-x)./abs(x)) < epsilon %stop condition
        flag = 1;
    end
    
    x = x_new;                             %Final Solution 
end
t=toc;
sprintf('Time elapsed : %g',t);
 end

Answer 1

[Star Strider]

It plots ‘t’ correctly and plots it —

plot(n1,ts1,'LineWidth',1);

however the sprintf is not going to produce anything because no output is assigned to it and the terminal semicloln (;) will suppress the output —

sprintf('Time elapsed : %g',t);

The semicolon will suppress output to the Command Window, so removing it will allow sprintf to pprint to the Command Window —

sprintf('Time elapsed : %g',t)

.