Is that the correct way to plot the graph ?
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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
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采纳的回答
Star Strider
2021-12-21
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)
.
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