I want to plot a cube of side length 10, that would be symmetrical from -5 to 5 and not -6 to 4.
xc=1; yc=1; zc=1; % coordinated of the center
L=10; % cube size (length of an edge)
alpha=0.8; % transparency (max=1=opaque)
X = [0 0 0 0 0 1; 1 0 1 1 1 1; 1 0 1 1 1 1; 0 0 0 0 0 1];
Y = [0 0 0 0 1 0; 0 1 0 0 1 1; 0 1 1 1 1 1; 0 0 1 1 1 0];
Z = [0 0 1 0 0 0; 0 0 1 0 0 0; 1 1 1 0 1 1; 1 1 1 0 1 1];
C='blue'; % unicolor
X = L*(X-0.5) + xc;
Y = L*(Y-0.5) + yc;
Z = L*(Z-0.5) + zc;
fill3(X,Y,Z,C,'FaceAlpha',alpha); % draw cube
axis equal
I just set the center to (0,0,0) to get the desired result:
xc=0; yc=0; zc=0; % coordinates of the center
Related
I need to make an array of zeros and ones in this particular fractal pattern:
0 0 0 0 0 1 0 0 0 0 0 0
0 0 1 1 1 1 1 1 1 0 0 0
0 1 0 0 0 1 0 0 0 1 0 0
0 1 0 0 1 1 1 0 0 1 0 0
0 1 0 1 0 1 0 1 0 1 0 0
1 1 1 1 1 1 1 1 1 1 1 1
0 1 0 1 0 1 0 1 0 1 0 0
0 1 0 0 1 1 1 0 0 1 0 0
0 1 0 0 0 1 0 0 0 1 0 0
0 0 1 1 1 1 1 1 1 0 0 0
0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0 0
The actual array should be 100 x 100. The pattern should start from the middle (x,y) coordinate, and expand to look like the pattern (using loops).
So far, I have only managed to make a pattern that looks like a '+' sign. I am not sure how to continue it.
This is my code so far:
n = zeros(16); % using 16x16 array for practice
x = length(n)/2;
y = length(n)/2;
z = length(n) - 1;
xso = length(n)/2; % x axis south movement
xno = length(n)/2; % x axis north movement
yea = length(n)/2; % y axis east movement
ywe = length(n)/2; % y axis west movement
for i = 1:1:z
newyea = move('east', x, yea);
n(x, newyea) = 1;
yea = newyea;
newywe = move('west', x, ywe);
n(x, newywe) = 1;
ywe = newywe;
newxso = move('south', xso, y);
n(newxso, y) = 1;
xso = newxso;
newxno = move('north', xno, y);
n(newxno, y) = 1;
xno = newxno;
end
I also have a user defined function:
function newval = move(dir, x, y)
switch dir
case 'east'
newval = y + 1;
case 'west'
newval = y - 1;
case 'south'
newval = x + 1;
case 'north'
newval = x - 1;
end
Since there were no restrictions given on the appearance of the loop, I would propose the below solution. But, before, let's have a look at your given example:
Either you should restrict the desired dimension d to be odd, i.e. d = 11, 13, 15, ... or you should specify, how the pattern should be continued in case of an even dimension d, like here d = 12. For my solution, I decided to rely on the dimension d to be odd.
Here's the code:
d = 15; % Dimension
A = zeros(d); % Initialize output array A
c = (d + 1) / 2; % Calculate center index (row, column)
A(:, c) = 1; % Add: Cross
A(c, :) = 1;
J = 0; % Auxiliary index
for I = (c+2):2:d % For every second row (or column) from center to border
J = J + 1;
l = 4 * J - 1; % Calculate length of line to draw
s = c - (l-1)/2; % Calculate start point of line
e = c + (l-1)/2; % Calculate end point of line
A(I, s:e) = 1; % Add: "South" line
A(s:e, I) = 1; % Add: "East" line
A(c - 2*J, s:e) = 1; % Add: "North" line
A(s:e, c - 2*J) = 1; % Add: "West" line
end
figure(1); % Show image
imagesc(A);
Output for d = 15 (to compare to given example):
Output for d = 99:
Hope that helps!
If you have some more stricter limitations on the for loop, let me/us know. Then, I will try to modify my code accordingly.
I know the code below :
N = 5;
assert(N>1 && mod(N,2)==1);
A = zeros(N);
% diamond mask
N2 = fix(N/2);
[I,J] = meshgrid(-N2:N2);
mask = (abs(I) + abs(J)) == N2;
% fill with zeros
A(mask) = 1;
which transforms matrix A to this:
A=
0 0 1 0 0
0 1 0 1 0
1 0 0 0 1
0 1 0 1 0
0 0 1 0 0
But I want the diamond to be filled with 1.
What should I do?
Here's a vectorized approach using bsxfun -
Nh = (N+1)/2;
range_vec = [1:Nh Nh-1:-1:1];
out = bsxfun(#plus,range_vec(:),range_vec) > Nh
Sample runs -
1) N = 5 :
out =
0 0 1 0 0
0 1 1 1 0
1 1 1 1 1
0 1 1 1 0
0 0 1 0 0
2) N = 9 :
out =
0 0 0 0 1 0 0 0 0
0 0 0 1 1 1 0 0 0
0 0 1 1 1 1 1 0 0
0 1 1 1 1 1 1 1 0
1 1 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1 0
0 0 1 1 1 1 1 0 0
0 0 0 1 1 1 0 0 0
0 0 0 0 1 0 0 0 0
You can use tril and flip functions:
mat = tril(ones(N), round((N-1)/2)) - tril(ones(N), round((-N-1)/2));
out = mat & flip(mat)
Odd values of N:
% N = 5;
out =
0 0 1 0 0
0 1 1 1 0
1 1 1 1 1
0 1 1 1 0
0 0 1 0 0
Even values of N:
% N = 4;
out =
0 1 1 0
1 1 1 1
1 1 1 1
0 1 1 0
What you need is to return a 1 or a 0 based on the Manhattan distance from each array location to the center of your diamond
N = 5;
assert(N>1 && mod(N,2)==1);
A = false(N);
[m, n] = size(A); %dimensions of A
X = floor([m, n]/2); %floored division gives integer indices of center of array
x = X(1); y = X(2);
radius = m/2; %half the height gives the radius
for a = 1 : m
for b = 1 : n
A(a,b) = abs(a-x)+abs(b-y) <= radius; %test if manhatten distance <= radius
end
end
This naturally will need editing to suit your particular case... In particular, the center of your diamond can realistically be placed anywhere by modifying x, y, and the radius can be either smaller or larger than half the width of the array if you so choose.
Just add a for loop and fill all diagonals:
N = 5;
assert(N>1 && mod(N,2)==1);
A = zeros(N);
% diamond mask
N2 = fix(N/2);
[I,J] = meshgrid(-N2:N2);
for id = 0:N2
A((abs(I) + abs(J)) == id) = 1;
end
I have the following code for generating an adjacency matrix for a network.
How do I go about creating a distance matrix (Probably a two hop matrix)from this output.
function adj = AdjMatrixLattice4( N, M )
% Size of adjacency matrix
MN = M*N;
adj = zeros(MN,MN);
for i=1:N
for j=1:N
A = M*(i-1)+j; %Node # for (i,j) node
if(j<N)
B = M*(i-1)+j+1; %Node # for node to the right
C = M*(i-1)+j+2;
D = M*(i-1)+j+2;
adj(A,B) = 1;
adj(B,A) = 1;
adj(A,C) = 1;
adj(C,A) = 1;
adj(A,D) = 1;
adj(D,A) = 1;
end
if(i<M)
B = M*i+j;
C = M*i+j+1; %Node # for node below
D = M*i+j;
adj(A,B) = 1;
adj(B,A) = 1;
adj(A,C) = 1;
adj(C,A) = 1;
adj(A,D) = 1;
adj(D,A) = 1;
end
end
end
end
The output for the following program is
AdjMatrixLattice4(3,3)
ans =
0 1 1 1 1 0 0 0 0 0
1 0 1 1 1 1 0 0 0 0
1 1 0 0 0 1 1 0 0 0
1 1 0 0 1 1 1 1 0 0
1 1 0 1 0 1 1 1 1 0
0 1 1 1 1 0 0 0 1 1
0 0 1 1 1 0 0 1 1 0
0 0 0 1 1 0 1 0 1 1
0 0 0 0 1 1 1 1 0 0
0 0 0 0 0 1 0 1 0 0
I have a matrix m = zeros(1000, 1000). Within this matrix I want to draw an estimate of the line which passes through 2 points from my matrix. Let's say x = [122 455]; and y = [500 500];.
How can I do this in Matlab? Are there any predefined functions to do this? I am using Matlab 2012b.
I'll denote the two endpoints as p1 and p2 because I'm planning to use x and y for something else. I'm also assuming that the first coordinate of p1 and p2 is x and the second is y. So here's a rather simple way to do it:
Obtain the equation of the line y = ax + b. In MATLAB, this can be done by:
x = p1(1):p2(1)
dx = p2(1) - p1(1);
dy = p2(2) - p1(2);
y = round((x - p1(1)) * dy / dx + p1(2));
Convert the values of x and y to indices of elements in the matrix, and set those elements to 1.
idx = sub2ind(size(m), y, x);
m(idx) = 1;
Example
Here's an example for a small 10-by-10 matrix:
%// This is our initial conditon
m = zeros(10);
p1 = [1, 4];
p2 = [5, 7];
%// Ensure the new x-dimension has the largest displacement
[max_delta, ix] = max(abs(p2 - p1));
iy = length(p1) - ix + 1;
%// Draw a line from p1 to p2 on matrix m
x = p1(ix):p2(ix);
y = round((x - p1(ix)) * (p2(iy) - p1(iy)) / (p2(ix) - p1(ix)) + p1(iy));
m(sub2ind(size(m), y, x)) = 1;
m = shiftdim(m, ix > iy); %// Transpose result if necessary
The result is:
m =
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 1 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
Update: I have patched this algorithm to work when dy > dx by treating the dimension with the largest displacement as if it were the x-dimension, and then transposing the result if necessary.
Neither of the provided answers work for displacements in y greater than in x (dy > dx).
As pointed out, Bresenham's line algorithm is exactly meant for that.
The matlab file provided here works similarly than the examples provided in the other answers but covers all the use-cases.
To relate to the previously provided example, the script can be used like this:
% initial conditions
m = zeros(10);
p1 = [1, 4];
p2 = [5, 10];% note dy > dx
% use file provided on file exchange
[x y] = bresenham(p1(1),p1(2),p2(1),p2(2));
% replace entries in matrix m
m(sub2ind(size(m), y, x)) = 1;
result looks like this:
m =
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
For me (matlab R2013b) following line did not work, when p1(1)>p2(2) (":" can not count backwards):
x = p1(1):p2(1);
E.G.:
1:10
1 2 3 4 5 6 7 8 9 10
10:1
Empty matrix: 1-by-0
But it worked when I used linspac instead:
x = linspace(p1(1), p2(1), abs(p2(1)-p1(1))+1);
I am writing an algorithm in matlab to for bicubic interpolation of a surface Psi(x,y). I have a bug in the code and cannot seem to track it down. I am trying a test case with Psi=X^2-0.25 so that its easier to track down the bug. It seems as if my interpolation has an offset. My comments are included in my code. Any help would be appreciated.
Plot of Psi=X^2 in blue and interpolation in red
Countour lines of Psi are plotted and the red dot is the point I am computing the interpolation about. The thick red line is the interpolation which is offset quite a bit from the red dot.
function main()
epsilon=0.000001;
xMin=-1+epsilon;
xMax= 1+epsilon;
yMin=-1+epsilon;
yMax= 1+epsilon;
dx=0.1; Nx=ceil((xMax-xMin)/dx)+1;
dy=0.1; Ny=ceil((yMax-yMin)/dy)+1;
x=xMin:dx:xMax; x=x(1:Nx);
y=yMin:dy:yMax; y=y(1:Ny);
[XPolInX,XPolInY]=GetGhostMatricies(Nx,Ny); %Linear extrapolation matrix
[D0x,D0y]=GetDiffMatricies(Nx,Ny,dx,dy); %derivative matricies: D0x is central differencing in x
[X,Y]=meshgrid(x,y);
Psi=X.^2-0.25; %Note that my algorithm is being written for a Psi that may not have a analytic representation. This Psi is only a test case.
psi=zeros(Nx+2,Ny+2); %linearly extrapolate psi (for solving differential equation not shown here)
psi(2:(Nx+1),2:(Ny+1))=Psi';
psi=(XPolInY*(XPolInX*psi)')';
%compute derivatives of psi
psi_x =D0x*psi; psi_x =(XPolInY*(XPolInX*psi_x)')';
psi_y =(D0y*psi')'; psi_y =(XPolInY*(XPolInX*psi_y)')';
psi_xy=D0x*psi_y; psi_xy=(XPolInY*(XPolInX*psi_xy)')';
% i have verified that my derivatives are computed correctly
biCubInv=GetBiCubicInverse(dx,dy);
i=5; %lets compute the bicubic interpolation at this x(i), y(j)
j=1;
psiVoxel=[psi( i,j),psi( i+1,j),psi( i,j+1),psi( i+1,j+1),...
psi_x( i,j),psi_x( i+1,j),psi_x( i,j+1),psi_x( i+1,j+1),...
psi_y( i,j),psi_y( i+1,j),psi_y( i,j+1),psi_y( i+1,j+1),...
psi_xy(i,j),psi_xy(i+1,j),psi_xy(i,j+1),psi_xy(i+1,j+1)]';
a=biCubInv*psiVoxel; %a=[a00 a01 ... a33]; polynomial coefficients; 1st index is power of (x-xi), 2nd index is power of (y-yj)
xi=x(5); yj=y(1);
clear x y
x=(xi-.2):.01:(xi+.2); %this is a local region about the point we are interpolating
y=(yj-.2):.01:(yj+.2);
[dX,dY]=meshgrid(x,y);
Psi=dX.^2-0.25;
figure(2) %just plotting the 0 level contour of Psi here
plot(xi,yj,'.r','MarkerSize',20)
hold on
contour(x,y,Psi,[0 0],'r','LineWidth',2)
set(gca,'FontSize',14)
axis([x(1) x(end) y(1) y(end)])
grid on
set(gca,'xtick',(xi-.2):.1:(xi+.2));
set(gca,'ytick',(yj-.2):.1:(yj+.2));
xlabel('x')
ylabel('y')
[dX dY]=meshgrid(x-xi,y-yj);
%P is my interpolating polynomial
P = a(1) + a(5) *dY + a(9) *dY.^2 + a(13) *dY.^3 ...
+ a(2)*dX + a(6)*dX .*dY + a(10)*dX .*dY.^2 + a(14)*dX .*dY.^3 ...
+ a(3)*dX.^2 + a(7)*dX.^2.*dY + a(11)*dX.^2.*dY.^2 + a(15)*dX.^2.*dY.^3 ...
+ a(4)*dX.^3 + a(8)*dX.^3.*dY + a(12)*dX.^3.*dY.^2 + a(16)*dX.^3.*dY.^3 ;
[c h]=contour(x,y,P)
clabel(c,h)
figure(3)
plot(x,x.^2-.25) %this is the exact function
hold on
plot(x,P(1,:),'-r*')
%See there is some offset here
end
%-------------------------------------------------------------------------
function [XPolInX,XPolInY]=GetGhostMatricies(Nx,Ny)
XPolInX=diag(ones(1,Nx+2),0);
XPolInY=diag(ones(1,Ny+2),0);
XPolInX(1,1) =0; XPolInX(1,2) =2; XPolInX(1,3) =-1;
XPolInY(1,1) =0; XPolInY(1,2) =2; XPolInY(1,3) =-1;
XPolInX(Nx+2,Nx+2)=0; XPolInX(Nx+2,Nx+1)=2; XPolInX(Nx+2,Nx)=-1;
XPolInY(Ny+2,Ny+2)=0; XPolInY(Ny+2,Ny+1)=2; XPolInY(Ny+2,Ny)=-1;
fprintf('Done GetGhostMatricies\n')
end
%-------------------------------------------------------------------------
function [D0x,D0y]=GetDiffMatricies(Nx,Ny,dx,dy)
D0x=diag(ones(1,Nx-1),1)-diag(ones(1,Nx-1),-1);
D0y=diag(ones(1,Ny-1),1)-diag(ones(1,Ny-1),-1);
D0x(1,1)=-3; D0x(1,2)=4; D0x(1,3)=-1;
D0y(1,1)=-3; D0y(1,2)=4; D0y(1,3)=-1;
D0x(Nx,Nx)=3; D0x(Nx,Nx-1)=-4; D0x(Nx,Nx-2)=1;
D0y(Ny,Ny)=3; D0y(Ny,Ny-1)=-4; D0y(Ny,Ny-2)=1;
%pad with ghost cells which are simply zeros
tmp=D0x; D0x=zeros(Nx+2,Nx+2); D0x(2:(Nx+1),2:(Nx+1))=tmp; tmp=0;
tmp=D0y; D0y=zeros(Ny+2,Ny+2); D0y(2:(Ny+1),2:(Ny+1))=tmp; tmp=0;
%scale appropriatley by dx & dy
D0x=D0x/(2*dx);
D0y=D0y/(2*dy);
end
%-------------------------------------------------------------------------
function biCubInv=GetBiCubicInverse(dx,dy)
%p(x,y)=a00+a01(x-xi)+a02(x-xi)^2+...+a33(x-xi)^3(y-yj)^3
%biCubic*a=[psi(i,j) psi(i+1,j) psi(i,j+1) psi(i+1,j+1) psi_x(i,j) ... psi_y(i,j) ... psi_xy(i,j) ... psi_xy(i+1,j+1)]
%here, psi_x is the x derivative of psi
%I verified that this matrix is correct by setting dx=dy=1 and comparing to the inverse here http://en.wikipedia.org/wiki/Bicubic_interpolation
biCubic=[
%00 10 20 30 01 11 21 31 02 12 22 32 03 13 23 33
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
1 dx dx^2 dx^3 0 0 0 0 0 0 0 0 0 0 0 0;
1 0 0 0 dy 0 0 0 dy^2 0 0 0 dy^3 0 0 0;
1 dx dx^2 dx^3 dy dx*dy dx^2*dy dx^3*dy dy^2 dx*dy^2 dx^2*dy^2 dx^3*dy^2 dy^3 dx*dy^3 dx^2*dy^3 dx^3*dy^3;
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 1 2*dx 3*dx^2 0 0 0 0 0 0 0 0 0 0 0 0;
0 1 0 0 0 dy 0 0 0 dy^2 0 0 0 dy^3 0 0;
0 1 2*dx 3*dx^2 0 dy 2*dx*dy 3*dx^2*dy 0 dy^2 2*dx*dy^2 3*dx^2*dy^2 0 dy^3 2*dx*dy^3 3*dx^2*dy^3;
0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 1 dx dx^2 dx^3 0 0 0 0 0 0 0 0;
0 0 0 0 1 0 0 0 2*dy 0 0 0 3*dy^2 0 0 0;
0 0 0 0 1 dx dx^2 dx^3 2*dy 2*dx*dy 2*dx^2*dy 2*dx^3*dy 3*dy^2 3*dx*dy^2 3*dx^2*dy^2 3*dx^3*dy^2;
0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 1 2*dx 3*dx^2 0 0 0 0 0 0 0 0;
0 0 0 0 0 1 0 0 0 2*dy 0 0 0 3*dy^2 0 0;
0 0 0 0 0 1 2*dx 3*dx^2 0 2*dy 4*dx*dy 6*dx^2*dy 0 3*dy^2 6*dx*dy^2 9*dx^2*dy^2];
biCubInv=inv(biCubic);
end
%-------------------------------------------------------------------------
I found my error. I pad my matricies with ghost cells, however I forget that now the i in Psi without ghost cells is i+1 in psi with ghost cells. Hence, I should be evaluating my interpolating polynomial P at xi=x(6); yj=y(2), not xi=x(5); yj=y(1).