100_working code for conjugate gradient iterative algo for finding solution of Ax=b

%%100_working code for conjugate gradient iterative algo for finding

%%solution of Ax=b, where A is symmetric Positive definite

%%written by vikram bhayyA on 18-6-26 ~13:19-13:59

%%general

clc

clear all

A=input('enter a Symmetric Positive Definite matrix:');

%A=[4,1,1;1,3,-1;1,-1,3]; %has to be symmetric Positve definite

[m,n]=size(A);

n

b=input('enter the output vector of size n,1:');

%b=[6,3 3]';

%x0=zeros(n,1);

x0=[1 -1 0]';

%x0=[0 0 0]' ; % initial guess

%initialize

r0=b-A*x0; %residual

if norm(r0)<0.1

display('solution of Ax=b is:')

x0

else

p0=r0; %search direction

k=1;

while(k)

alpha0=(r0'*r0)/(p0'*A*p0); %step size

x1=x0+alpha0*p0; %update x1

r1=r0-alpha0*(A*p0); % update r1

if norm(r1)<0.01

k=0;

else

k=k+1;

end

beta0=(r1'*r1)/(r0'*r0); %improvement factor

p1=r1+beta0*p0;% new search direction

p0=p1;

r0=r1;

x0=x1;

end

display('solution of Ax=b is:')

x1

end

What is seen in Command window upon execution is:

enter a Symmetric Positive Definite matrix:[1 0 0;0 2 0; 0 0 4]

n =

     3

enter the output vector of size n,1:[1 4 12]'

Solution of Ax=b is: 

x1 =

     1
     2
     3