Question

2.1 Summary In this part, you will create a figure, and use linear transformations (matrices) to move the figure around the screen. In the end, your figure should move up 8 steps, then turn and face left. Reference material for this part can be found in Linear Algebra, and its applications, David Lay, Section 2.7 starting at the beginning of the section up to, but not incluing,3D Graphics. Also, this poster presentation does a pretty good job explaining the same concepts http://web.csulb.edu/~jchang9/m247/m247sp12 Daniel Kris James.Walter.pdf 2.2 Create Figure in Plane Recreate the image below by creating a matrix whose first row is the x-coordinates of the vertices in clockwise order, and whose second row are the corresponding y-coordinates. Place the following function at the bottom of your script to plot the points in the matrix function plot-points(points) figure(1) fill (points(1,), points(2, . ' set (gca, 'color', [o, o, o): axis ([-10, 10, -10, 10]); end You can then use this function in your script by using the command plot-points(P) Where P is the matrix created above. (It will look a bit different than shown). 1.8 1.6 1.2 08 0.4 02 1 08 06 04 02 02 04 08 8 2.3 Homogeneous Coordinates Convert the points stored in your matrix to homogeneous coordinates by adding a row of all ones to the bottom of your matrix of coordinates. 2.4 Translation Create a matrix that represents the following transformation i.e. Translation in the y direction by one unit. You can test that you have the correct matrix by applying the transformation to each column in the matrix of homogeneous coordinates and then plotting the new points using the function from above. Use a for-loop to apply this transformation 8 times and plot the result each iteration using the function from part 1 2.5 Find Center Write a function (at the bottom of your script) that will compute the center of your figure The x-coordinate of the center is just the average of each of the x-coordinates (of the vertices) similarly the y-coordinate of the center is the average of each of the y-coordinates. You should be able to use the function in the following way [centerx, center-y] -FindCenter (P) Where P is the matrix of homogeneous coordinates. 2.6 Translate to Origin Create a matrix that represents the following transformation Tenter((r, y,, y- cy,) Where ccy are the coordinates of the center found in the previous part. You can test that you have the correct matrix by applying the transformation to each column in the matrix of homogeneous coordinates and then plotting the new points using the function from part 1 you should see the figure move to be centered at the origin. Comment out any tests, before moving on to the next section 2.7 Rotation Create a matrix (Linear Transformation) that has the effect of rotating the figure by 90 Apply the following transformation to each of the columns in the matrix of homogeneous coordinates and then plot the new points: center Too Toenter Aside: Check what happens if only Too is used. The above transformation is used so that the center is kept in the same place 2.8 Bonus Make the figure travel in a complete square

Answer 1

MATLAB CODE:

For conjugate gradient method
clear all
close all

%Running the program with given parameters
%All A, x0, b and epsilon values

A=[4 3 0;3 4 -1;0 -1 4];
b=[2;1;0];
epsi=10^-8;
x0=[0;0;0];

fprintf('Example code for Conjugate gradient \n')
fprintf('The A matrix is \n')
disp(A)

fprintf('The b matrix is \n')
disp(b)

x=x0;
r_old=b-A*x;
p=r_old;
epsi=epsi*norm(r_old);
cnt=0;

%loop for conjugate gradient method
while norm(r_old)>epsi
  
    cnt=cnt+1;
    alpha=(r_old'*r_old)/(p'*A*p);
    x=x+alpha*p;
    r_new=r_old-alpha*A*p;
    beta=(r_new'*(r_new-r_old))/(r_old'*r_old);
    p=r_new+beta*p;
    r_old=r_new;
  
end

fprintf('The solution matrix using manual code is \n')
disp(x)

%Running the program with conjugate gradient function
[x,iter]=conjugategradient(A,x0,b);

fprintf('The solution matrix using conjugategradient function is \n')
disp(x)

fprintf('Error in conjugate gradient method is %e.\n',norm(x-A\b))

%Running program for iteration count
r=4;
fprintf('\nSolution using conjugate gradient method for different n.\n')

for i=1:6
  
    n=2^(r+i);
    %creating tridiagonal matrix A
    A=full(gallery('tridiag',n,-1,2,-1));
    %considering exact solution value as 1
    x=ones(n,1);
    %creating b vector of length n
    b=A*x;
    %initial guess
    x0=zeros(n,1);
  
    %finding solution using conjugategradient solution
    [x1,iter]=conjugategradient(A,x0,b);
    fprintf('\tFor n=2^%d the iteration count is %d.\n',(r+i),iter)
    fprintf('\tError in conjugate gradient method is %e.\n\n',norm(x1-A\b))
  
end
  
  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%Function for Conjugate Gradient method
function [x,iter]=conjugategradient(A,x0,b)
    epsi=10^-8;
    x=x0;
    r_old=b-A*x;
    p=r_old;
    epsi=epsi*norm(r_old);
    cnt=0;
    while norm(r_old)>epsi

        cnt=cnt+1;
        alpha=(r_old'*r_old)/(p'*A*p);
        x=x+alpha*p;
        r_new=r_old-alpha*A*p;
        beta=(r_new'*(r_new-r_old))/(r_old'*r_old);
        p=r_new+beta*p;
        r_old=r_new;

    end
    iter=cnt;
end