I setup and meshed a domain using Matlab's PDE toolbox. Along the boundary, is there someway to get the length of each element of the mesh? And the flux in the normal direction? (the bc are Dirichlet)
Edit:
See example code
RMax = 20;
RL = 1;
RU = 0.5;
HN = 5;
HL = 2;
HTT = 3;
HU = 1.5;
VL = -150;
p = [RL,0;RL,HN;0,HN+HL;0,HN+HL+HTT;RU,HN+HL+HTT+HU;RMax,HN+HL+HTT+HU;RMax,0];
t = [1;1;0;1;1;0;0];
v = [VL;VL;0;0;0;0;0];
dx = 0.5;
dy = 0.5;
bc = cell(size(t));
for i = 1:length(t)
if t(i) == 0
bc{i} = {'u', v(i)};
elseif t(i) == 1
bc{i} = {'g', v(i), 'q', 1};
else
error('Unrecognized boundary condition type.')
end
end
model = createpde;
gd = [2; size(p,1); p(:,1) ; p(:,2)];
ns = char('domain')';
sf = 'domain';
g = decsg(gd,sf,ns);
geometryFromEdges(model,g);
generateMesh(model, 'Hmax', min([dx,dy])/3, 'MesherVersion','R2013a');
for i = 1:size(bc,1)
applyBoundaryCondition(model, 'Edge', i, bc{i}{:});
end
u = assempde( model , 'x' , 0 , 0 );
pdemesh(model)
Edit: 2015-12-17 18:54 GMT
There are 2 points in the e shown at 1 and 2 in the figure below. I want to know the coordinate of 3 so I know which direction is into the domain.
You can answer your first question by using meshToPet to convert to [P,E,T] form:
[p,e,t] = meshToPet(model.Mesh);
x1 = p(1,e(1,:)); % x-coordinates of first point in each mesh edge
x2 = p(1,e(2,:)); % x-coordinates of second point in each mesh edge
y1 = p(2,e(1,:)); % y-coordinates of first point in each mesh edge
y2 = p(2,e(2,:)); % y-coordinates of second point in each mesh edge
% Plot first points of mesh edge
plot(x1,y1,'b.-',x1(1),y1(1),'go',x1(end),y1(end),'ro');
% Euclidean distance between first and second point in each edge
d = sqrt((x1-x2).^2+(y1-y2).^2);
I'm assuming you just want the lengths of the mesh edge/boundary. You can use similar methods to get the lengths of every single triangle using the t matrix.
As far as flux goes, there's pdecgrad. I think the following may work:
...
c = 'x';
u = assempde(model, c, 0, 0);
[p,e,t] = meshToPet(model.Mesh);
[cgxu,cgyu] = pdecgrad(p,t,c,u);
Related
This a program that I have, I already asked before for how to find the intersection on my image with a circle, and somebody has an answer it (thank you) and I have another problem...
a = imread('001_4.bmp');
I2 = imcrop(a,[90 5 93 180]);
[i,j]=size(I2);
x_hist=sum(I2,1);
y_hist=(sum(I2,2))';
x=1:j ; y=1:i;
centx=sum(x.*x_hist)/sum(x_hist);
centy=sum(y.*y_hist)/sum(y_hist);
BW = edge(I2,'Canny',0.5);
bw2 = imcomplement(BW);
circle = int32([centx,centy,40]);%<<----------
shapeInserter = vision.ShapeInserter('Fill',false);
release(shapeInserter);
set(shapeInserter,'Shape','Circles');
% construct binary image of circle only
bwCircle = step(shapeInserter,true(size(bw2)),circle);
% find the indexes of the intersection between the binary image and the circle
[i, j] = find ((bw2 | bwCircle) == 0);
figure
imshow(bw2 & bwCircle) % plot the combination of both images
hold on
plot(j, i, 'r*') % plot the intersection points
K = step(shapeInserter,bw2,circle);
[n,m]=size(i);
d=0;
k=1;
while (k < n)
d = d + sqrt((i(k+1)-i(k)).^2 + (j(k+1)-j(k)).^2);
k = k+1;
end
Q: How can I calculate all the existing intersection values(red *) in a clockwise direction?
Not sure, but I think that your teacher meant something different from the answer given in the previous question, because
CW direction can be a hint, not a additional problem
I've seen no clue that the circle should be drawn; for small radius circles are blocky and simple bw_circle & bw_edges may not work. For details please see "Steve on image processing" blog, "Intersecting curves that don't intersect"[1]
May be your teacher is rather old and wants Pascal-stile answer.
If my assumptions are correct the code below should be right
img = zeros(7,7); %just sample image, edges == 1
img(:,4) = 1; img(sub2ind(size(img),1:7,1:7)) = 1;
ccx = 4; % Please notice: x is for columns, y is for rows
ccy = 3;
rad = 2;
theta = [(pi/2):-0.01:(-3*pi/2)];
% while plotting it will appear CCW, for visual CW and de-facto CCW use
% 0:0.01:2*pi
cx = rad * cos(theta) + ccx; % gives slightly different data as for
cy = rad * sin(theta) + ccy; % (x-xc)^2 + (y-yc)^2 == rad^2 [2]
ccoord=[cy;cx]'; % Once again: x is for columns, y is for rows
[ccoord, rem_idx, ~] =unique(round(ccoord),'rows');
cx = ccoord(:,2);
cy = ccoord(:,1);
circ = zeros(size(img)); circ(sub2ind(size(img),cy,cx))=1;
cross_sum = 0;
figure, imshow(img | circ,'initialmagnification',5000)
hold on,
h = [];
for un_ang = 1:length(cx),
tmp_val= img(cy(un_ang),cx(un_ang));
if tmp_val == 1 %the point belongs to edge of interest
cross_sum = cross_sum + tmp_val;
h = plot(cx(un_ang),cy(un_ang),'ro');
pause,
set(h,'marker','x')
end
end
hold off
[1] https://blogs.mathworks.com/steve/2016/04/12/intersecting-curves-that-dont-intersect/
[2] http://matlab.wikia.com/wiki/FAQ#How_do_I_create_a_circle.3F
I want to find Orientation, MajorAxisLengthand MinorAxisLength of contour which is plotted with below code.
clear
[x1 , x2] = meshgrid(linspace(-10,10,100),linspace(-10,10,100));
mu = [1,3];
sigm = [2,0;0,2];
xx_size = length(mu);
tem_matrix = ones(size(x1));
x_mesh= cell(1,xx_size);
for i = 1 : xx_size
x_mesh{i} = tem_matrix * mu(i);
end
x_mesh= {x1,x2};
temp_mesh = [];
for i = 1 : xx_size
temp_mesh = [temp_mesh x_mesh{i}(:)];
end
Z = mvnpdf(temp_mesh,mu,sigm);
z_plat = reshape(Z,size(x1));
figure;contour(x1, x2, z_plat,3, 'LineWidth', 2,'color','m');
% regionprops(z_plat,'Centroid','Orientation','MajorAxisLength','MinorAxisLength');
In my opinion, I may have to use regionprops command but I don't know how to do this. I want to find direction of axis of contour and plot something like this
How can I do this task? Thanks very much for your help
Rather than trying to process the graphical output of contour, I would instead recommend using contourc to compute the ContourMatrix and then use the x/y points to estimate the major and minor axes lengths as well as the orientation (for this I used this file exchange submission)
That would look something like the following. Note that I have modified the inputs to contourc as the first two inputs should be the vector form and not the output of meshgrid.
% Compute the three contours for your data
contourmatrix = contourc(linspace(-10,10,100), linspace(-10,10,100), z_plat, 3);
% Create a "pointer" to keep track of where we are in the output
start = 1;
count = 1;
% Now loop through each contour
while start < size(contourmatrix, 2)
value = contourmatrix(1, start);
nPoints = contourmatrix(2, start);
contour_points = contourmatrix(:, start + (1:nPoints));
% Now fit an ellipse using the file exchange
ellipsedata(count) = fit_ellipse(contour_points(1,:), contour_points(2,:));
% Increment the start pointer
start = start + nPoints + 1;
count = count + 1;
end
orientations = [ellipsedata.phi];
% 0 0 0
major_length = [ellipsedata.long_axis];
% 4.7175 3.3380 2.1539
minor_length = [ellipsedata.short_axis];
% 4.7172 3.3378 2.1532
As you can see, the contours are actually basically circles and therefore the orientation is zero and the major and minor axis lengths are almost equal. The reason that they look like ellipses in your post is because your x and y axes are scaled differently. To fix this, you can call axis equal
figure;contour(x1, x2, z_plat,3, 'LineWidth', 2,'color','m');
axis equal
Thank you #Suever. It help me to do my idea.
I add some line to code:
clear
[X1 , X2] = meshgrid(linspace(-10,10,100),linspace(-10,10,100));
mu = [-1,0];
a = [3,2;1,4];
a = a * a';
sigm = a;
xx_size = length(mu);
tem_matrix = ones(size(X1));
x_mesh= cell(1,xx_size);
for i = 1 : xx_size
x_mesh{i} = tem_matrix * mu(i);
end
x_mesh= {X1,X2};
temp_mesh = [];
for i = 1 : xx_size
temp_mesh = [temp_mesh x_mesh{i}(:)];
end
Z = mvnpdf(temp_mesh,mu,sigm);
z_plat = reshape(Z,size(X1));
figure;contour(X1, X2, z_plat,3, 'LineWidth', 2,'color','m');
hold on;
% Compute the three contours for your data
contourmatrix = contourc(linspace(-10,10,100), linspace(-10,10,100), z_plat, 3);
% Create a "pointer" to keep track of where we are in the output
start = 1;
count = 1;
% Now loop through each contour
while start < size(contourmatrix, 2)
value = contourmatrix(1, start);
nPoints = contourmatrix(2, start);
contour_points = contourmatrix(:, start + (1:nPoints));
% Now fit an ellipse using the file exchange
ellipsedata(count) = fit_ellipse(contour_points(1,:), contour_points(2,:));
% Increment the start pointer
start = start + nPoints + 1;
count = count + 1;
end
orientations = [ellipsedata.phi];
major_length = [ellipsedata.long_axis];
minor_length = [ellipsedata.short_axis];
tet = orientations(1);
x1 = mu(1);
y1 = mu(2);
a = sin(tet) * sqrt(major_length(1));
b = cos(tet) * sqrt(major_length(1));
x2 = x1 + a;
y2 = y1 + b;
line([x1, x2], [y1, y2],'linewidth',2);
tet = ( pi/2 + orientations(1) );
a = sin(tet) * sqrt(minor_length(1));
b = cos(tet) * sqrt(minor_length(1));
x2 = x1 + a;
y2 = y1 + b;
line([x1, x2], [y1, y2],'linewidth',2);
I have three vectors x,y,t. For each combination x,y,t there is a (u,v) value associated with it. How to plot this in matlab? Actually I'm trying to plot the solution of 2d hyperbolic equation
vt = A1vx + A2vy where A1 and A2 are 2*2 matrices and v is a 2*1 vector. I was trying scatter3 and quiver3 but being new to matlab I'm not able to represent the solution correctly.
In the below code I have plot at only a particular time-level. How to show the complete solution in just one plot? Any help?
A1 = [5/3 2/3; 1/3 4/3];
A2 = [-1 -2; -1 0];
M = 10;
N = 40;
delta_x = 1/M;
delta_y = delta_x;
delta_t = 1/N;
x_points = 0:delta_x:1;
y_points = 0:delta_y:1;
t_points = 0:delta_t:1;
u = zeros(M+1,M+1,N+1,2);
for i=1:M+1,
for j=1:M+1,
u(i,j,1,1) = (sin(pi*x_points(i)))*sin(2*pi*y_points(j)) ;
u(i,j,1,2) = cos(2*pi*x_points(i));
end
end
for j=1:M+1,
for t=1:N+1,
u(M+1,j,t,1) = sin(2*t);
u(M+1,j,t,2) = cos(2*t);
end
end
for i=1:M+1
for t=1:N+1
u(i,1,t,1) = sin(2*t);
u(i,M+1,t,2) = sin(5*t) ;
end
end
Rx = delta_t/delta_x;
Ry = delta_t/delta_y;
for t=2:N+1
v = zeros(M+1,M+1,2);
for i=2:M,
for j=2:M,
A = [(u(i+1,j,t-1,1) - u(i-1,j,t-1,1)) ; (u(i+1,j,t-1,2) - u(i-1,j,t-1,2))];
B = [(u(i+1,j,t-1,1) -2*u(i,j,t-1,1) +u(i-1,j,t-1,1)) ; (u(i+1,j,t-1,2) -2*u(i,j,t-1,2) +u(i-1,j,t-1,2))];
C = [u(i,j,t-1,1) ; u(i,j,t-1,2)];
v(i,j,:) = C + Rx*A1*A/2 + Rx*Rx*A1*A1*B/2;
end
end
for i=2:M,
for j=2:M,
A = [(v(i,j+1,1) - v(i,j-1,1)) ; (v(i,j+1,2) - v(i,j-1,2)) ];
B = [(v(i,j+1,1) - 2*v(i,j,1) +v(i,j-1,1)) ; (v(i,j+1,2) - 2*v(i,j,2) +v(i,j-1,2))];
C = [v(i,j,1) ; v(i,j,2)];
u(i,j,t,:) = C + Ry*A2*A/2 + Ry*Ry*A2*A2*B/2;
end
end
if j==2
u(i,1,t,2) = u(i,2,t,2);
end
if j==M
u(i,M+1,t,1) = u(i,M,t,1);
end
if i==2
u(1,j,t,:) = u(2,j,t,:) ;
end
end
time_level = 2;
quiver(x_points, y_points, u(:,:,time_level,1), u(:,:,time_level,2))
You can plot it in 3D, but personally I think it would be hard to make sense of.
There's a quiver3 equivalent for your plotting function. z-axis in this case would be time (say, equally spaced), and z components of the vectors would be zero. Unlike 2D version of this function, it does not support passing in coordinate vectors, so you need to create the grid explicitly using meshgrid:
sz = size(u);
[X, Y, Z] = meshgrid(x_points, y_points, 1:sz(3));
quiver3(X, Y, Z, u(:,:,:,1), u(:,:,:,2), zeros(sz(1:3)));
You may also color each timescale differently by plotting them one at a time, but it's still hard to make sense of the results:
figure(); hold('all');
for z = 1:sz(3)
[X, Y, Z] = meshgrid(x_points, y_points, z);
quiver3(X, Y, Z, u(:,:,z,1), u(:,:,z,2), zeros([sz(1:2),1]));
end
I'm trying to solve the following question :
maximize x^2-5x+y^2-3y
x+y <= 8
x<=2
x,y>= 0
By using Frank Wolf algorithm ( according to http://web.mit.edu/15.053/www/AMP-Chapter-13.pdf ).
But after running of the following program:
syms x y t;
f = x^2-5*x+y^2-3*y;
fdx = diff(f,1,x); % f'x
fdy = diff(f,1,y); % y'x
x0 = [0 0]; %initial point
A = [1 1;1 0]; %constrains matrix
b = [8;2];
lb = zeros(1,2);
eps = 0.00001;
i = 1;
X = [inf inf];
Z = zeros(2,200); %result for end points (x1,x2)
rr = zeros(1,200);
options = optimset('Display','none');
while( all(abs(X-x0)>[eps,eps]) && i < 200)
%f'x(x0)
c1 = subs(fdx,x,x0(1));
c1 = subs(c1,y,x0(2));
%f'y(x0)
c2 = subs(fdy,x,x0(1));
c2 = subs(c2,y,x0(2));
%optimization point of linear taylor function
ys = linprog((-[c1;c2]),A,b,[],[],lb,[],[],options);
%parametric representation of line
xt = (1-t)*x0(1)+t*ys(1,1);
yt = (1-t)*x0(2)+t*ys(2,1);
%f(x=xt,y=yt)
ft = subs(f,x,xt);
ft = subs(ft,y,yt);
%f't(x=xt,y=yt)
ftd = diff(ft,t,1);
%f't(x=xt,y=yt)=0 -> for max point
[t1] = solve(ftd); % (t==theta)
X = double(x0);%%%%%%%%%%%%%%%%%
% [ xt(t=t1) yt(t=t1)]
xnext(1) = subs(xt,t,t1) ;
xnext(2) = subs(yt,t,t1) ;
x0 = double(xnext);
Z(1,i) = x0(1);
Z(2,i) = x0(2);
i = i + 1;
end
x_point = Z(1,:);
y_point = Z(2,:);
% Draw result
scatter(x_point,y_point);
hold on;
% Print results
fprintf('The answer is:\n');
fprintf('x = %.3f \n',x0(1));
fprintf('y = %.3f \n',x0(2));
res = x0(1)^2 - 5*x0(1) + x0(2)^2 - 3*x0(2);
fprintf('f(x0) = %.3f\n',res);
I get the following result:
x = 3.020
y = 0.571
f(x0) = -7.367
And this no matter how many iterations I running this program (1,50 or 200).
Even if I choose a different starting point (For example, x0=(1,6) ), I get a negative answer to most.
I know that is an approximation, but the result should be positive (for x0 final, in this case).
My question is : what's wrong with my implementation?
Thanks in advance.
i changed a few things, it still doesn't look right but hopefully this is getting you in the right direction. It looks like the intial x0 points make a difference to how the algorithm converges.
Also make sure to check what i is after running the program, to determine if it ran to completion or exceeded the maximum iterations
lb = zeros(1,2);
ub = [2,8]; %if x+y<=8 and x,y>0 than both x,y < 8
eps = 0.00001;
i_max = 100;
i = 1;
X = [inf inf];
Z = zeros(2,i_max); %result for end points (x1,x2)
rr = zeros(1,200);
options = optimset('Display','none');
while( all(abs(X-x0)>[eps,eps]) && i < i_max)
%f'x(x0)
c1 = subs(fdx,x,x0(1));
c1 = subs(c1,y,x0(2));
%f'y(x0)
c2 = subs(fdy,x,x0(1));
c2 = subs(c2,y,x0(2));
%optimization point of linear taylor function
[ys, ~ , exit_flag] = linprog((-[c1;c2]),A,b,[],[],lb,ub,x0,options);
so here is the explanation of the changes
ub, uses our upper bound. After i added a ub, the result immediately changed
x0, start this iteration from the previous point
exit_flag this allows you to check exit_flag after execution (it always seems to be 1 indicating it solved the problem correctly)
I am working on rotating image manually in Matlab. Each time I run my code with a different image the previous images which are rotated are shown in the Figure. I couldn't figure it out. Any help would be appreciable.
The code is here:
[screenshot]
im1 = imread('gradient.jpg');
[h, w, p] = size(im1);
theta = pi/12;
hh = round( h*cos(theta) + w*abs(sin(theta))); %Round to nearest integer
ww = round( w*cos(theta) + h*abs(sin(theta))); %Round to nearest integer
R = [cos(theta) -sin(theta); sin(theta) cos(theta)];
T = [w/2; h/2];
RT = [inv(R) T; 0 0 1];
for z = 1:p
for x = 1:ww
for y = 1:hh
% Using matrix multiplication
i = zeros(3,1);
i = RT*[x-ww/2; y-hh/2; 1];
%% Nearest Neighbour
i = round(i);
if i(1)>0 && i(2)>0 && i(1)<=w && i(2)<=h
im2(y,x,z) = im1(i(2),i(1),z);
end
end
end
end
x=1:ww;
y=1:hh;
[X, Y] = meshgrid(x,y); % Generate X and Y arrays for 3-D plots
orig_pos = [X(:)' ; Y(:)' ; ones(1,numel(X))]; % Number of elements in array or subscripted array expression
orig_pos_2 = [X(:)'-(ww/2) ; Y(:)'-(hh/2) ; ones(1,numel(X))];
new_pos = round(RT*orig_pos_2); % Round to nearest neighbour
% Check if new positions fall from map:
valid_pos = new_pos(1,:)>=1 & new_pos(1,:)<=w & new_pos(2,:)>=1 & new_pos(2,:)<=h;
orig_pos = orig_pos(:,valid_pos);
new_pos = new_pos(:,valid_pos);
siz = size(im1);
siz2 = size(im2);
% Expand the 2D indices to include the third dimension.
ind_orig_pos = sub2ind(siz2,orig_pos(2*ones(p,1),:),orig_pos(ones(p,1),:), (1:p)'*ones(1,length(orig_pos)));
ind_new_pos = sub2ind(siz, new_pos(2*ones(p,1),:), new_pos(ones(p,1),:), (1:p)'*ones(1,length(new_pos)));
im2(ind_orig_pos) = im1(ind_new_pos);
imshow(im2);
There is a problem with the initialization of im2, or rather, the lack of it. im2 is created in the section shown below:
if i(1)>0 && i(2)>0 && i(1)<=w && i(2)<=h
im2(y,x,z) = im1(i(2),i(1),z);
end
If im2 exists before this code is run and its width or height is larger than the image you are generating the new image will only overwrite the top left corner of your existing im2. Try initializing im2 by adding adding
im2 = zeros(hh, ww, p);
before
for z = 1:p
for x = 1:ww
for y = 1:hh
...
As a bonus it might make your code a little faster since Matlab won't have to resize im2 as it grows in the loop.