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David
David
Last activity on 2 Apr 2024

Ghost page

I feel like no one at UC San Diego knows this page, let alone this server, is still live. For the younger generation, this is what the whole internet used to look like :)
http://www-pord.ucsd.edu/matlab/
In short: support varying color in at least the plot, plot3, fplot, and fplot3 functions.
This has been a thing that's come up quite a few times, and includes questions/requests by users, workarounds by the community, and workarounds presented by MathWorks -- examples of each below. It's a feature that exists in Python's Matplotlib library and Sympy. Anyways, given that there are myriads of workarounds, it appears to be one of the most common requests for Matlab plots (Matlab's plotting is, IMO, one of the best features of the product), the request precedes the 21st century, and competitive tools provide the functionality, it would seem to me that this might be the next great feature for Matlab plotting.
I'm curious to get the rest of the community's thoughts... what's everyone else think about this?
---
User questions/requests
User-provided workarounds
MathWorks-provided workarounds
David
David
Last activity on 1 Apr 2024

Prompts Matter

I was in a meeting the other day and a coworker shared a smiley face they created using the AI Chat Playground. The image looked something like this:
And I suspect the prompt they used was something like this:
"Create a smiley face"
I imagine this output wasn't what my coworker had expected so he was left thinking that this was as good as it gets without manually editing the code, and that the AI Chat Playground couldn't do any better.
I thought I could get a better result using the Playground so I tried a more detailed prompt using a multi-step technique like this:
"Follow these instructions:
- Create code that plots a circle
- Create two smaller circles as eyes within the first circle
- Create an arc that looks like a smile in the lower part of the first circle"
The output of this prompt was better in my opinion.
These queries/prompts are examples of 'zero-shot' prompts, the expectation being a good result with just one query. As opposed to a back-and-forth chat session working towards a desired outcome.
I wonder how many attempts everyone tries before they decide they can't anything more from the AI/LLM. There are times I'll send dozens of chat queries if I feel like I'm getting close to my goal, while other times I'll try just one or two. One thing I always find useful is seeing how others interact with AI models, which is what inspired me to share this.
Does anyone have examples of techniques that work well? I find multi-step instructions often produces good results.
The line integral , where C is the boundary of the square oriented counterclockwise, can be evaluated in two ways:
Using the definition of the line integral:
% Initialize the integral sum
integral_sum = 0;
% Segment C1: x = -1, y goes from -1 to 1
y = linspace(-1, 1);
x = -1 * ones(size(y));
dy = diff(y);
integral_sum = integral_sum + sum(-x(1:end-1) .* dy);
% Segment C2: y = 1, x goes from -1 to 1
x = linspace(-1, 1);
y = ones(size(x));
dx = diff(x);
integral_sum = integral_sum + sum(y(1:end-1).^2 .* dx);
% Segment C3: x = 1, y goes from 1 to -1
y = linspace(1, -1);
x = ones(size(y));
dy = diff(y);
integral_sum = integral_sum + sum(-x(1:end-1) .* dy);
% Segment C4: y = -1, x goes from 1 to -1
x = linspace(1, -1);
y = -1 * ones(size(x));
dx = diff(x);
integral_sum = integral_sum + sum(y(1:end-1).^2 .* dx);
disp(['Direct Method Integral: ', num2str(integral_sum)]);
Plotting the square path
% Define the square's vertices
vertices = [-1 -1; -1 1; 1 1; 1 -1; -1 -1];
% Plot the square
figure;
plot(vertices(:,1), vertices(:,2), '-o');
title('Square Path for Line Integral');
xlabel('x');
ylabel('y');
grid on;
axis equal;
% Add arrows to indicate the path direction (counterclockwise)
hold on;
for i = 1:size(vertices,1)-1
% Calculate direction
dx = vertices(i+1,1) - vertices(i,1);
dy = vertices(i+1,2) - vertices(i,2);
% Reduce the length of the arrow for better visibility
scale = 0.2;
dx = scale * dx;
dy = scale * dy;
% Calculate the start point of the arrow
startx = vertices(i,1) + (1 - scale) * dx;
starty = vertices(i,2) + (1 - scale) * dy;
% Plot the arrow
quiver(startx, starty, dx, dy, 'MaxHeadSize', 0.5, 'Color', 'r', 'AutoScale', 'off');
end
hold off;
Apply Green's Theorem for the line integral
% Define the partial derivatives of P and Q
f = @(x, y) -1 - 2*y; % derivative of -x with respect to x is -1, and derivative of y^2 with respect to y is 2y
% Compute the double integral over the square [-1,1]x[-1,1]
integral_value = integral2(f, -1, 1, 1, -1);
disp(['Green''s Theorem Integral: ', num2str(integral_value)]);
Plotting the vector field related to Greenโ€™s theorem
% Define the grid for the vector field
[x, y] = meshgrid(linspace(-2, 2, 20), linspace(-2 ,2, 20));
% Define the vector field components
P = y.^2; % y^2 component
Q = -x; % -x component
% Plot the vector field
figure;
quiver(x, y, P, Q, 'b');
hold on; % Hold on to plot the square on the same figure
% Define the square's vertices
vertices = [-1 -1; -1 1; 1 1; 1 -1; -1 -1];
% Plot the square path
plot(vertices(:,1), vertices(:,2), '-o', 'Color', 'k'); % 'k' for black color
title('Vector Field (P = y^2, Q = -x) with Square Path');
xlabel('x');
ylabel('y');
axis equal;
% Add arrows to indicate the path direction (counterclockwise)
for i = 1:size(vertices,1)-1
% Calculate direction
dx = vertices(i+1,1) - vertices(i,1);
dy = vertices(i+1,2) - vertices(i,2);
% Reduce the length of the arrow for better visibility
scale = 0.2;
dx = scale * dx;
dy = scale * dy;
% Calculate the start point of the arrow
startx = vertices(i,1) + (1 - scale) * dx;
starty = vertices(i,2) + (1 - scale) * dy;
% Plot the arrow
quiver(startx, starty, dx, dy, 'MaxHeadSize', 0.5, 'Color', 'r', 'AutoScale', 'off');
end
hold off;
To solve a surface integral for example the over the sphere easily in MATLAB, you can leverage the symbolic toolbox for a direct and clear solution. Here is a tip to simplify the process:
  1. Use Symbolic Variables and Functions: Define your variables symbolically, including the parameters of your spherical coordinates ฮธ and ฯ• and the radius r . This allows MATLAB to handle the expressions symbolically, making it easier to manipulate and integrate them.
  2. Express in Spherical Coordinates Directly: Since you already know the sphere's equation and the relationship in spherical coordinates, define x, y, and z in terms of r , ฮธ and ฯ• directly.
  3. Perform Symbolic Integration: Use MATLAB's `int` function to integrate symbolically. Since the sphere and the function are symmetric, you can exploit these symmetries to simplify the calculation.
Hereโ€™s how you can apply this tip in MATLAB code:
% Include the symbolic math toolbox
syms theta phi
% Define the limits for theta and phi
theta_limits = [0, pi];
phi_limits = [0, 2*pi];
% Define the integrand function symbolically
integrand = 16 * sin(theta)^3 * cos(phi)^2;
% Perform the symbolic integral for the surface integral
surface_integral = int(int(integrand, theta, theta_limits(1), theta_limits(2)), phi, phi_limits(1), phi_limits(2));
% Display the result of the surface integral symbolically
disp(['The surface integral of x^2 over the sphere is ', char(surface_integral)]);
% Number of points for plotting
num_points = 100;
% Define theta and phi for the sphere's surface
[theta_mesh, phi_mesh] = meshgrid(linspace(double(theta_limits(1)), double(theta_limits(2)), num_points), ...
linspace(double(phi_limits(1)), double(phi_limits(2)), num_points));
% Spherical to Cartesian conversion for plotting
r = 2; % radius of the sphere
x = r * sin(theta_mesh) .* cos(phi_mesh);
y = r * sin(theta_mesh) .* sin(phi_mesh);
z = r * cos(theta_mesh);
% Plot the sphere
figure;
surf(x, y, z, 'FaceColor', 'interp', 'EdgeColor', 'none');
colormap('jet'); % Color scheme
shading interp; % Smooth shading
camlight headlight; % Add headlight-type lighting
lighting gouraud; % Use Gouraud shading for smooth color transitions
title('Sphere: x^2 + y^2 + z^2 = 4');
xlabel('x-axis');
ylabel('y-axis');
zlabel('z-axis');
colorbar; % Add color bar to indicate height values
axis square; % Maintain aspect ratio to be square
view([-30, 20]); % Set a nice viewing angle
Hans Scharler
Hans Scharler
Last activity on 19 Apr 2024

Starter MATLAB Script

I am often talking to new MATLAB users. I have put together one script. If you know how this script works, why, and what each line means, you will be well on your way on your MATLAB learning journey.
% Clear existing variables and close figures
clear;
close all;
% Print to the Command Window
disp('Hello, welcome to MATLAB!');
% Create a simple vector and matrix
vector = [1, 2, 3, 4, 5];
matrix = [1, 2, 3; 4, 5, 6; 7, 8, 9];
% Display the created vector and matrix
disp('Created vector:');
disp(vector);
disp('Created matrix:');
disp(matrix);
% Perform element-wise multiplication
result = vector .* 2;
% Display the result of the operation
disp('Result of element-wise multiplication of the vector by 2:');
disp(result);
% Create plot
x = 0:0.1:2*pi; % Generate values from 0 to 2*pi
y = sin(x); % Calculate the sine of these values
% Plotting
figure; % Create a new figure window
plot(x, y); % Plot x vs. y
title('Simple Plot of sin(x)'); % Give the plot a title
xlabel('x'); % Label the x-axis
ylabel('sin(x)'); % Label the y-axis
grid on; % Turn on the grid
disp('This is the end of the script. Explore MATLAB further to learn more!');
I would like to propose the creation of MATLAB EduHub, a dedicated channel within the MathWorks community where educators, students, and professionals can share and access a wealth of educational material that utilizes MATLAB. This platform would act as a central repository for articles, teaching notes, and interactive learning modules that integrate MATLAB into the teaching and learning of various scientific fields.
Key Features:
1. Resource Sharing: Users will be able to upload and share their own educational materials, such as articles, tutorials, code snippets, and datasets.
2. Categorization and Search: Materials can be categorized for easy searching by subject area, difficulty level, and MATLAB version..
3. Community Engagement: Features for comments, ratings, and discussions to encourage community interaction.
4. Support for Educators: Special sections for educators to share teaching materials and track engagement.
Benefits:
- Enhanced Educational Experience: The platform will enrich the learning experience through access to quality materials.
- Collaboration and Networking: It will promote collaboration and networking within the MATLAB community.
- Accessibility of Resources: It will make educational materials available to a wider audience.
By establishing MATLAB EduHub, I propose a space where knowledge and experience can be freely shared, enhancing the educational process and the MATLAB community as a whole.
Zhaoxu Liu / slandarer
Zhaoxu Liu / slandarer
Last activity on 15 Mar 2024

Happy Pi Day !!

Happy Pi Day!
3.14 ฯ€ Day has arrived, and this post provides some very cool pi implementations and complete MATLAB code.
https://www.mathworks.com/matlabcentral/fileexchange/126210-happy-pi-day
Firstly, in order to obtain the first n decimal places of pi, we need to write the following code (to prevent inaccuracies, we need to take a few more tails and perform another operation of taking the first n decimal places when needed):
function Pi=getPi(n)
if nargin<1,n=3;end
Pi=char(vpa(sym(pi),n+10));
Pi=abs(Pi)-48;
Pi=Pi(3:n+2);
end
With this function to obtain the decimal places of pi, our visualization journey has begun~Step by step, from simple to complex~(Please try to use newer versions of MATLAB to run, at least R17b)
1 Pie chart
Just calculate the proportion of each digit to the first 1500 decimal places:
% ่Žทๅ–piๅ‰1500ไฝๅฐๆ•ฐ
Pi=getPi(1500);
% ็ปŸ่ฎกๅ„ไธชๆ•ฐๅญ—ๅ‡บ็Žฐๆฌกๆ•ฐ
numNum=find([diff(sort(Pi)),1]);
numNum=[numNum(1),diff(numNum)];
% ้…่‰ฒๅˆ—่กจ
CM=[20,164,199;43,187,170;53,165,81;189,190,28;248,167,22;
232,74,27;244,57,99;240,118,177;168,109,195;78,125,187]./255;
% ็ป˜ๅ›พๅนถไฟฎ้ฅฐ
pieHdl=pie(numNum);
set(gcf,'Color',[1,1,1],'Position',[200,100,620,620]);
for i=1:2:20
pieHdl(i).EdgeColor=[1,1,1];
pieHdl(i).LineWidth=1;
pieHdl(i).FaceColor=CM((i+1)/2,:);
end
for i=2:2:20
pieHdl(i).Color=CM(i/2,:);
pieHdl(i).FontWeight='bold';
pieHdl(i).FontSize=14;
end
% ็ป˜ๅˆถๅ›พไพ‹ๅนถไฟฎ้ฅฐ
lgdHdl=legend(num2cell('0123456789'));
lgdHdl.FontWeight='bold';
lgdHdl.FontSize=11;
lgdHdl.TextColor=[.5,.5,.5];
lgdHdl.Location='southoutside';
lgdHdl.Box='off';
lgdHdl.NumColumns=10;
lgdHdl.ItemTokenSize=[20,15];
title("VISUALIZING \pi 'Pi' Chart | 1500 digits",'FontSize',18,...
'FontName','Times New Roman','Color',[.5,.5,.5])
2 line chart
Calculate the change in the proportion of each number:
% ่Žทๅ–piๅ‰1500ไฝๅฐๆ•ฐ
Pi=getPi(1500);
% ่ฎก็ฎ—ๆฏ”ไพ‹ๅ˜ๅŒ–
Ratio=cumsum(Pi==(0:9)',2);
Ratio=Ratio./sum(Ratio);
D=1:length(Ratio);
% ้…่‰ฒๅˆ—่กจ
CM=[20,164,199;43,187,170;53,165,81;189,190,28;248,167,22;
232,74,27;244,57,99;240,118,177;168,109,195;78,125,187]./255;
hold on
% ๅพช็Žฏ็ป˜ๅ›พ
for i=1:10
plot(D(20:end),Ratio(i,20:end),'Color',[CM(i,:),.6],'LineWidth',1.8)
end
% ๅๆ ‡ๅŒบๅŸŸไฟฎ้ฅฐ
ax=gca;box on;grid on
ax.YLim=[0,.2];
ax.YTick=0:.05:.2;
ax.XTick=0:200:1400;
ax.YTickLabel={'0%','5%','10%','15%','20%'};
ax.XMinorTick='on';
ax.YMinorTick='on';
ax.LineWidth=.8;
ax.GridLineStyle='-.';
ax.FontName='Cambria';
ax.FontSize=11;
ax.XLabel.String='Decimals';
ax.YLabel.String='Proportion';
ax.XLabel.FontSize=13;
ax.YLabel.FontSize=13;
% ็ป˜ๅˆถๅ›พไพ‹ๅนถไฟฎ้ฅฐ
lgdHdl=legend(num2cell('0123456789'));
lgdHdl.NumColumns=5;
lgdHdl.FontWeight='bold';
lgdHdl.FontSize=11;
lgdHdl.TextColor=[.5,.5,.5];
3 stacked area diagram
% ่Žทๅ–piๅ‰500ไฝๅฐๆ•ฐ
Pi=getPi(500);
% ่ฎก็ฎ—ๆฏ”ไพ‹ๅ˜ๅŒ–
Ratio=cumsum(Pi==(0:9)',2);
Ratio=Ratio./sum(Ratio);
% ้…่‰ฒๅˆ—่กจ
CM=[231,98,84;239,138,71;247,170,88;255,208,111;255,230,183;
170,220,224;114,188,213;82,143,173;55,103,149;30,70,110]./255;
% ็ป˜ๅˆถๅ †ๅ ้ข็งฏๅ›พ
hold on
areaHdl=area(Ratio');
for i=1:10
areaHdl(i).FaceColor=CM(i,:);
areaHdl(i).FaceAlpha=.9;
end
% ๅ›พ็ช—ๅ’Œๅๆ ‡ๅŒบๅŸŸไฟฎ้ฅฐ
set(gcf,'Position',[200,100,720,420]);
ax=gca;
ax.YLim=[0,1];
ax.XMinorTick='on';
ax.YMinorTick='on';
ax.LineWidth=.8;
ax.FontName='Cambria';
ax.FontSize=11;
ax.TickDir='out';
ax.XLabel.String='Decimals';
ax.YLabel.String='Proportion';
ax.XLabel.FontSize=13;
ax.YLabel.FontSize=13;
ax.Title.String='Area Chart of Proportion โ€” 500 digits';
ax.Title.FontSize=14;
% ็ป˜ๅˆถๅ›พไพ‹ๅนถไฟฎ้ฅฐ
lgdHdl=legend(num2cell('0123456789'));
lgdHdl.NumColumns=5;
lgdHdl.FontSize=11;
lgdHdl.Location='southeast';
4 connected stacked bar chart
% ่Žทๅ–piๅ‰100ไฝๅฐๆ•ฐ
Pi=getPi(100);
% ่ฎก็ฎ—ๆฏ”ไพ‹ๅ˜ๅŒ–
Ratio=cumsum(Pi==(0:9)',2);
Ratio=Ratio./sum(Ratio);
X=Ratio(:,10:10:80)';
barHdl=bar(X,'stacked','BarWidth',.2);
CM=[231,98,84;239,138,71;247,170,88;255,208,111;255,230,183;
170,220,224;114,188,213;82,143,173;55,103,149;30,70,110]./255;
for i=1:10
barHdl(i).FaceColor=CM(i,:);
end
% ไปฅไธ‹ๆ˜ฏ็”Ÿๆˆ่ฟžๆŽฅ็š„้ƒจๅˆ†
hold on;axis tight
yEndPoints=reshape([barHdl.YEndPoints]',length(barHdl(1).YData),[])';
zeros(1,length(barHdl(1).YData));
yEndPoints=[zeros(1,length(barHdl(1).YData));yEndPoints];
barWidth=barHdl(1).BarWidth;
for i=1:length(barHdl)
for j=1:length(barHdl(1).YData)-1
y1=min(yEndPoints(i,j),yEndPoints(i+1,j));
y2=max(yEndPoints(i,j),yEndPoints(i+1,j));
if y1*y2<0
ty=yEndPoints(find(yEndPoints(i+1,j)*yEndPoints(1:i,j)>=0,1,'last'),j);
y1=min(ty,yEndPoints(i+1,j));
y2=max(ty,yEndPoints(i+1,j));
end
y3=min(yEndPoints(i,j+1),yEndPoints(i+1,j+1));
y4=max(yEndPoints(i,j+1),yEndPoints(i+1,j+1));
if y3*y4<0
ty=yEndPoints(find(yEndPoints(i+1,j+1)*yEndPoints(1:i,j+1)>=0,1,'last'),j+1);
y3=min(ty,yEndPoints(i+1,j+1));
y4=max(ty,yEndPoints(i+1,j+1));
end
fill([j+.5.*barWidth,j+1-.5.*barWidth,j+1-.5.*barWidth,j+.5.*barWidth],...
[y1,y3,y4,y2],barHdl(i).FaceColor,'FaceAlpha',.4,'EdgeColor','none');
end
end
% ๅ›พ็ช—ๅ’Œๅๆ ‡ๅŒบๅŸŸไฟฎ้ฅฐ
set(gcf,'Position',[200,100,720,420]);
ax=gca;box off
ax.YLim=[0,1];
ax.XMinorTick='on';
ax.YMinorTick='on';
ax.LineWidth=.8;
ax.FontName='Cambria';
ax.FontSize=11;
ax.TickDir='out';
ax.XTickLabel={'10','20','30','40','50','60','70','80'};
ax.XLabel.String='Decimals';
ax.YLabel.String='Proportion';
ax.XLabel.FontSize=13;
ax.YLabel.FontSize=13;
ax.Title.String='Area Chart of Proportion โ€” 10-80 digits';
ax.Title.FontSize=14;
% ็ป˜ๅˆถๅ›พไพ‹ๅนถไฟฎ้ฅฐ
lgdHdl=legend(barHdl,num2cell('0123456789'));
lgdHdl.NumColumns=5;
lgdHdl.FontSize=11;
lgdHdl.Location='southeast';
5 bichord chart
Need to use this tool:
https://www.mathworks.com/matlabcentral/fileexchange/121043-digraph-chord-chart
% ๆž„ๅปบ่ฟžๆŽฅ็Ÿฉ้˜ต
dataMat=zeros(10,10);
Pi=getPi(1001);
for i=1:1000
dataMat(Pi(i)+1,Pi(i+1)+1)=dataMat(Pi(i)+1,Pi(i+1)+1)+1;
end
BCC=biChordChart(dataMat,'Arrow','on','Label',num2cell('0123456789'));
BCC=BCC.draw();
% ๆทปๅŠ ๅˆปๅบฆ
BCC.tickState('on')
% ไฟฎๆ”นๅญ—ไฝ“๏ผŒๅญ—ๅทๅŠ้ขœ่‰ฒ
BCC.setFont('FontName','Cambria','FontSize',17)
set(gcf,'Position',[200,100,820,820]);
6 Gravity simulation diagram
Imagine each decimal as a small ball with a mass of
For example, if, the weight of ball 0 is 1, ball 9 is 1.2589, the initial velocity of the ball is 0, and it is attracted by other balls. Gravity follows the inverse square law, and if the balls are close enough, they will collide and their value will become
After adding, take the mod, add the velocity direction proportionally, and recalculate the weight.
Pi=[3,getPi(71)];K=.18;
% ๅŸบ็ก€้…็ฝฎ
CM=[239,32,120;239,60,52;247,98,32;255,182,60;247,235,44;
142,199,57;55,180,70;0,170,239;40,56,146;147,37,139]./255;
T=linspace(0,2*pi,length(Pi)+1)';
T=T(1:end-1);
ct=linspace(0,2*pi,100);
cx=cos(ct).*.027;
cy=sin(ct).*.027;
% ๅˆๅง‹ๆ•ฐๆฎ
Pi=Pi(:);
N=Pi;
X=cos(T);Y=sin(T);
VX=T.*0;VY=T.*0;
PX=X;PY=Y;
% ๆœช็ขฐๆ’žๆ—ถๅˆๅง‹่ดจ้‡
getM=@(x)(x+1).^K;
M=getM(N);
% ็ป˜ๅˆถๅˆๅง‹ๅœ†ๅœˆ
hold on
for i=1:length(N)
fill(cx+X(i),cy+Y(i),CM(N(i)+1,:),'EdgeColor','w','LineWidth',1)
end
for k=1:800
% ่ฎก็ฎ—ๅŠ ้€Ÿๅบฆ
Rn2=1./squareform(pdist([X,Y])).^2;
Rn2(eye(length(X))==1)=0;
MRn2=Rn2.*(M');
AX=X'-X;AY=Y'-Y;
normXY=sqrt(AX.^2+AY.^2);
AX=AX./normXY;AX(eye(length(X))==1)=0;
AY=AY./normXY;AY(eye(length(X))==1)=0;
AX=sum(AX.*MRn2,2)./150000;
AY=sum(AY.*MRn2,2)./150000;
% ่ฎก็ฎ—้€ŸๅบฆๅŠๆ–ฐไฝ็ฝฎ
VX=VX+AX;X=X+VX;PX=[PX,X];
VY=VY+AY;Y=Y+VY;PY=[PY,Y];
% ๆฃ€ๆต‹ๆ˜ฏๅฆๆœ‰็ขฐๆ’ž
R=squareform(pdist([X,Y]));
R(triu(ones(length(X)))==1)=inf;
[row,col]=find(R<=0.04);
if length(X)==1
break;
end
if ~isempty(row)
% ็ขฐๆ’ž็š„็‚นๅˆไธบไธ€ไฝ“
XC=(X(row)+X(col))./2;YC=(Y(row)+Y(col))./2;
VXC=(VX(row).*M(row)+VX(col).*M(col))./(M(row)+M(col));
VYC=(VY(row).*M(row)+VY(col).*M(col))./(M(row)+M(col));
PC=nan(length(row),size(PX,2));
NC=mod(N(row)+N(col),10);
% ๅˆ ้™ค็ขฐๆ’ž็‚นๅนถ็ป˜ๅ›พ
uniNum=unique([row;col]);
X(uniNum)=[];VX(uniNum)=[];
Y(uniNum)=[];VY(uniNum)=[];
for i=1:length(uniNum)
plot(PX(uniNum(i),:),PY(uniNum(i),:),'LineWidth',2,'Color',CM(N(uniNum(i))+1,:))
end
PX(uniNum,:)=[];PY(uniNum,:)=[];N(uniNum,:)=[];
% ็ป˜ๅˆถๅœ†ๅฝข
for i=1:length(XC)
fill(cx+XC(i),cy+YC(i),CM(NC(i)+1,:),'EdgeColor','w','LineWidth',1)
end
% ่กฅๅ……ๅˆไฝ“็‚น
X=[X;XC];Y=[Y;YC];VX=[VX;VXC];VY=[VY;VYC];
PX=[PX;PC];PY=[PY;PC];N=[N;NC];M=getM(N);
end
end
for i=1:size(PX,1)
plot(PX(i,:),PY(i,:),'LineWidth',2,'Color',CM(N(i)+1,:))
end
text(-1,1,{['Num=',num2str(length(Pi))];['K=',num2str(K)]},'FontSize',13,'FontName','Cambria')
% ๅ›พ็ช—ๅŠๅๆ ‡ๅŒบๅŸŸไฟฎ้ฅฐ
set(gcf,'Position',[200,100,820,820]);
ax=gca;
ax.Position=[0,0,1,1];
ax.DataAspectRatio=[1,1,1];
ax.XLim=[-1.1,1.1];
ax.YLim=[-1.1,1.1];
ax.XTick=[];
ax.YTick=[];
ax.XColor='none';
ax.YColor='none';
7 forest chart
The method comes from
http://mkweb.bcgsc.ca/pi/piday2021
The digits of ฯ€ are shown as a forest. Each tree in the forest represents the digits of ฯ€ up to the next 9. The first 10 trees are "grown" from the digit sets 314159, 2653589, 79, 3238462643383279, 50288419, 7169, 39, 9, 3751058209, and 749.
BRANCHES
The first digit of a tree controls how many branches grow from the trunk of the tree. For example, the first tree's first digit is 3, so you see 3 branches growing from the trunk.
The next digit's branches grow from the end of a branch of the previous digit in left-to-right order. This process continues until all the tree's digits have been used up.
Each tree grows from a set of consecutive digits sampled from the digits of ฯ€ up to the next 9. The first tree, shown here, grows from 314159. Each of the digits determine how many branches grow at each fork in the tree โ€” the branches here are colored by their corresponding digit to illustrate this. Leaves encode the digits in a left-to-right order. The digit 9 spawns a flower on one of the branches of the previous digit. The branching exception is 0, which terminates the current branch โ€” 0 branches grow!
LEAVES AND FLOWERS
The tree's digits themselves are drawn as circular leaves, color-coded by the digit.
The leaf exception is 9, which causes one of the branches of the previous digit to sprout a flower! The petals of the flower are colored by the digit before the 9 and the center is colored by the digit after the 9, which is on the next tree. This is how the forest propagates.
The colors of a flower are determined by the first digit of the next tree and the penultimate digit of the current tree. If the current tree only has one digit, then that digit is used. Leaves are placed at the tips of branches in a left-to-right order โ€” you can "easily" read them off. Additionally, the leaves are distributed within the tree (without disturbing their left-to-right order) to spread them out as much as possible and avoid overlap. This order is deterministic.
The leaf placement exception are the branch set that sprouted the flower. These are not used to grow leaves โ€” the flower needs space!
function PiTree(X,pos,D)
lw=2;
theta=pi/2+(rand(1)-.5).*pi./12;
% ๆ ‘ๅถๅŠ่Šฑๆœต้ขœ่‰ฒ
CM=[237,32,121;237,62,54;247,99,33;255,183,59;245,236,43;
141,196,63;57,178,74;0,171,238;40,56,145;146,39,139]./255;
hold on
if all(X(1:end-2)==0)
endSet=[pos,pos,theta];
else
kplot(pos(1)+[0,cos(theta)],pos(2)+[0,sin(theta)],lw./.6)
endSet=[pos,pos+[cos(theta),sin(theta)],theta];
% ่ฎก็ฎ—ๅฑ‚็บง
Layer=0;
for i=1:length(X)
Layer=[Layer,ones(1,X(i)).*i];
end
% ่ฎก็ฎ—ๆ ‘ๆž
if D
for i=1:length(X)-2
if X(i)==0 % ่‹ฅๆ•ฐๅ€ผไธบ0ๅˆ™ไธ้•ฟๆ ‘ๆž
newSet=endSet(1,:);
elseif X(i)==1 % ่‹ฅๆ•ฐๅ€ผไธบ1ๅˆ™ไธ€้•ฟไธ€็Ÿญไธคไธชๆ ‘ๆž
tTheta=endSet(1,5);
tTheta=linspace(tTheta+pi/8,tTheta-pi/8,2)'+(rand([2,1])-.5).*pi./8;
newSet=repmat(endSet(1,3:4),[X(i),1]);
newSet=[newSet.*[1;1],newSet+[cos(tTheta),sin(tTheta)].*.7^Layer(i).*[1;.1],tTheta];
else % ๅ…ถไป–ๆƒ…ๅ†ตๆ•ฐๅ€ผไธบๅ‡ ้•ฟๅ‡ ไธชๆ ‘ๆž
tTheta=endSet(1,5);
tTheta=linspace(tTheta+pi/5,tTheta-pi/5,X(i))'+(rand([X(i),1])-.5).*pi./8;
newSet=repmat(endSet(1,3:4),[X(i),1]);
newSet=[newSet,newSet+[cos(tTheta),sin(tTheta)].*.7^Layer(i),tTheta];
end
% ็ป˜ๅˆถๆ ‘ๆž
for j=1:size(newSet,1)
kplot(newSet(j,[1,3]),newSet(j,[2,4]),lw.*.6^Layer(i))
end
endSet=[endSet;newSet];
endSet(1,:)=[];
end
end
end
% ่ฎก็ฎ—ๅถๅญๅ’Œ่Šฑๆœตไฝ็ฝฎ
FLSet=endSet(:,3:4);
[~,FLInd]=sort(FLSet(:,1));
FLSet=FLSet(FLInd,:);
[~,tempInd]=sort(rand([1,size(FLSet,1)]));
tempInd=sort(tempInd(1:length(X)-2));
flowerInd=tempInd(randi([1,length(X)-2],[1,1]));
leafInd=tempInd(tempInd~=flowerInd);
% ็ป˜ๅˆถๆ ‘ๅถ
for i=1:length(leafInd)
scatter(FLSet(leafInd(i),1),FLSet(leafInd(i),2),70,'filled','CData',CM(X(i)+1,:))
end
% ็ป˜ๅˆถ่Šฑๆœต
for i=1:5
% if ~D
% tC=CM(X(end)+1,:);
% else
% tC=CM(X(end-2)+1,:);
% end
scatter(FLSet(flowerInd,1)+cos(pi*2*i/5).*.18,FLSet(flowerInd,2)+sin(pi*2*i/5).*.18,60,...
'filled','CData',CM(X(end-2)+1,:),'MarkerEdgeColor',[1,1,1])
end
scatter(FLSet(flowerInd,1),FLSet(flowerInd,2),60,'filled','CData',CM(X(end)+1,:),'MarkerEdgeColor',[1,1,1])
drawnow;%axis tight
% =========================================================================
function kplot(XX,YY,LW,varargin)
LW=linspace(LW,LW*.6,10);%+rand(1,20).*LW./10;
XX=linspace(XX(1),XX(2),11)';
XX=[XX(1:end-1),XX(2:end)];
YY=linspace(YY(1),YY(2),11)';
YY=[YY(1:end-1),YY(2:end)];
for ii=1:10
plot(XX(ii,:),YY(ii,:),'LineWidth',LW(ii),'Color',[.1,.1,.1])
end
end
end
main part:
Pi=[3,getPi(800)];
pos9=[0,find(Pi==9)];
set(gcf,'Position',[200,50,900,900],'Color',[1,1,1]);
ax=gca;hold on
ax.Position=[0,0,1,1];
ax.DataAspectRatio=[1,1,1];
ax.XLim=[.5,36];
ax.XTick=[];
ax.YTick=[];
ax.XColor='none';
ax.YColor='none';
for j=1:8
for i=1:11
n=i+(j-1)*11;
if n<=85
tPi=Pi((pos9(n)+1):pos9(n+1)+1);
if length(tPi)>2
PiTree(tPi,[0+i*3,0-j*4],true);
else
PiTree([Pi(pos9(n)),tPi],[0+i*3,0-j*4],false);
end
end
end
end
8 random walk
n=1200;
% ่Žทๅ–piๅ‰nไฝๅฐๆ•ฐ
Pi=getPi(n);
CM=[239,65,75;230,115,48;229,158,57;232,136,85;239,199,97;
144,180,116;78,166,136;81,140,136;90,118,142;43,121,159]./255;
hold on
endPoint=[0,0];
t=linspace(0,2*pi,100);
T=linspace(0,2*pi,11)+pi/2;
fill(endPoint(1)+cos(t).*.5,endPoint(2)+sin(t).*.5,CM(Pi(1)+1,:),'EdgeColor','none')
for i=1:n
theta=T(Pi(i)+1);
plot(endPoint(1)+[0,cos(theta)],endPoint(2)+[0,sin(theta)],'Color',[CM(Pi(i)+1,:),.8],'LineWidth',1.2);
endPoint=endPoint+[cos(theta),sin(theta)];
end
fill(endPoint(1)+cos(t).*.5,endPoint(2)+sin(t).*.5,CM(Pi(n)+1,:),'EdgeColor','none')
% ๅ›พ็ช—ๅ’Œๅๆ ‡ๅŒบๅŸŸไฟฎ้ฅฐ
set(gcf,'Position',[200,100,820,820]);
ax=gca;
ax.XTick=[];
ax.YTick=[];
ax.Color=[0,0,0];
ax.DataAspectRatio=[1,1,1];
ax.XLim=[-30,5];
ax.YLim=[-5,40];
% ็ป˜ๅˆถๅ›พไพ‹
endPoint=[1,35];
for i=1:10
theta=T(i);
plot(endPoint(1)+[0,cos(theta).*2],endPoint(2)+[0,sin(theta).*2],'Color',[CM(i,:),.8],'LineWidth',3);
text(endPoint(1)+cos(theta).*2.7,endPoint(2)+sin(theta).*2.7,num2str(i-1),'Color',[1,1,1].*.7,...
'FontSize',12,'FontWeight','bold','FontName','Cambria','HorizontalAlignment','center')
end
text(-15,35,'Random walk of \pi digits','Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','center','FontSize',25,'FontAngle','italic')
9 grid chart
Pi=[3,getPi(399)];
% ้…่‰ฒๆ•ฐๆฎ
CM=[248,65,69;246,152,36;249,198,81;67,170,139;87,118,146]./255;
% ็ป˜ๅˆถๅœ†ๅœˆ
hold on
t=linspace(0,2*pi,100);
x=cos(t).*.8.*.5;
y=sin(t).*.8.*.5;
for i=1:400
[col,row]=ind2sub([20,20],i);
if mod(Pi(i),2)==0
fill(x+col,y+row,CM(round((Pi(i)+1)/2),:),'LineWidth',1,'EdgeAlpha',.8)
else
fill(x+col,y+row,[0,0,0],'EdgeColor',CM(round((Pi(i)+1)/2),:),'LineWidth',1,'EdgeAlpha',.7)
end
end
text(10.5,-.4,'\pi on a grid โ€” 400 digits','Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','center','FontSize',25,'FontAngle','italic')
% ๅ›พ็ช—ๅ’Œๅๆ ‡ๅŒบๅŸŸไฟฎ้ฅฐ
set(gcf,'Position',[200,100,820,820]);
ax=gca;
ax.YDir='reverse';
ax.XLim=[.5,20.5];
ax.YLim=[-1,20.5];
ax.XTick=[];
ax.YTick=[];
ax.Color=[0,0,0];
ax.DataAspectRatio=[1,1,1];
10 scale grid diagram
Let's still put the numbers in the form of circles, but the difference is that six numbers are grouped together, and the pure purple circle at the end is the six 9s that we are familiar with decimal places 762-767
Pi=[3,getPi(767)];
% 762-767
% ้…่‰ฒๆ•ฐๆฎ
CM=[239,32,120;239,60,52;247,98,32;255,182,60;247,235,44;
142,199,57;55,180,70;0,170,239;40,56,146;147,37,139]./255;
% ็ป˜ๅˆถๅœ†ๅœˆ
hold on
t=linspace(0,2*pi,100);
x=cos(t).*.9.*.5;
y=sin(t).*.9.*.5;
for i=1:6:length(Pi)
n=round((i-1)/6+1);
[col,row]=ind2sub([10,13],n);
tNum=Pi(i:i+5);
numNum=find([diff(sort(tNum)),1]);
numNum=[numNum(1),diff(numNum)];
cumNum=cumsum(numNum);
uniNum=unique(tNum);
for j=length(cumNum):-1:1
fill(x./6.*cumNum(j)+col,y./6.*cumNum(j)+row,CM(uniNum(j)+1,:),'EdgeColor','none')
end
end
% ็ป˜ๅˆถๅ›พไพ‹
for i=1:10
fill(x./4+5.5+(i-5.5)*.32,y./4+14.5,CM(i,:),'EdgeColor','none')
text(5.5+(i-5.5)*.32,14.9,num2str(i-1),'Color',[1,1,1],'FontSize',...
9,'FontName','Cambria','HorizontalAlignment','center')
end
text(5.5,-.2,'FEYNMAN POINT of \pi','Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','center','FontSize',25,'FontAngle','italic')
% ๅ›พ็ช—ๅ’Œๅๆ ‡ๅŒบๅŸŸไฟฎ้ฅฐ
set(gcf,'Position',[200,100,600,820]);
ax=gca;
ax.YDir='reverse';
ax.Position=[0,0,1,1];
ax.XLim=[.3,10.7];
ax.YLim=[-1,15.5];
ax.XTick=[];
ax.YTick=[];
ax.Color=[0,0,0];
ax.DataAspectRatio=[1,1,1];
11 text chart
First, write a code to generate an image of each letter:
function getLogo
if ~exist('image','dir')
mkdir('image\')
end
logoSet=['.',char(65:90)];
for i=1:27
figure();
ax=gca;
ax.XLim=[-1,1];
ax.YLim=[-1,1];
ax.XColor='none';
ax.YColor='none';
ax.DataAspectRatio=[1,1,1];
logo=logoSet(i);
hold on
text(0,0,logo,'HorizontalAlignment','center','FontSize',320,'FontName','Segoe UI Black')
exportgraphics(ax,['image\',logo,'.png'])
close
end
dotPic=imread('image\..png');
newDotPic=uint8(ones([400,size(dotPic,2),3]).*255);
newDotPic(end-size(dotPic,1)+1:end,:,1)=dotPic(:,:,1);
newDotPic(end-size(dotPic,1)+1:end,:,2)=dotPic(:,:,2);
newDotPic(end-size(dotPic,1)+1:end,:,3)=dotPic(:,:,3);
imwrite(newDotPic,'image\..png')
S=20;
for i=1:27
logo=logoSet(i);
tPic=imread(['image\',logo,'.png']);
sz=size(tPic,[1,2]);
sz=round(sz./sz(1).*400);
tPic=imresize(tPic,sz);
tBox=uint8(255.*ones(size(tPic,[1,2])+S));
tBox(S+1:S+size(tPic,1),S+1:S+size(tPic,2))=tPic(:,:,1);
imwrite(cat(3,tBox,tBox,tBox),['image\',logo,'.png'])
end
end
Pi=[3,-1,getPi(150)];
CM=[109,110,113;224,25,33;244,126,26;253,207,2;154,203,57;111,150,124;
121,192,235;6,109,183;190,168,209;151,118,181;233,93,163]./255;
ST={'.','ZERO','ONE','TWO','THREE','FOUR','FIVE','SIX','SEVEN','EIGHT','NINE'};
n=1;
hold on
% ๅพช็Žฏ็ป˜ๅˆถๅญ—ๆฏ
for i=1:20%:10
STList='';
NMList=[];
PicListR=uint8(zeros(400,0));
PicListG=uint8(zeros(400,0));
PicListB=uint8(zeros(400,0));
% PicListA=uint8(zeros(400,0));
for j=1:6
STList=[STList,ST{Pi(n)+2}];
NMList=[NMList,ones(size(ST{Pi(n)+2})).*(Pi(n)+2)];
n=n+1;
if length(STList)>15&&length(STList)+length(ST{Pi(n)+2})>20
break;
end
end
for k=1:length(STList)
tPic=imread(['image\',STList(k),'.png']);
% PicListA=[PicListA,tPic(:,:,1)];
PicListR=[PicListR,(255-tPic(:,:,1)).*CM(NMList(k),1)];
PicListG=[PicListG,(255-tPic(:,:,2)).*CM(NMList(k),2)];
PicListB=[PicListB,(255-tPic(:,:,3)).*CM(NMList(k),3)];
end
PicList=cat(3,PicListR,PicListG,PicListB);
image([-1200,1200],[0,150]-(i-1)*150,flipud(PicList))
end
% ๅ›พ็ช—ๅŠๅๆ ‡ๅŒบๅŸŸไฟฎ้ฅฐ
set(gcf,'Position',[200,100,600,820]);
ax=gca;
ax.DataAspectRatio=[1,1,1];
ax.XLim=[-1300,1300];
ax.Position=[0,0,1,1];
ax.XTick=[];
ax.YTick=[];
ax.Color=[0,0,0];
ax.YLim=[-19*150-80,230];
12 spiral chart
Pi=getPi(600);
% ้…่‰ฒๅˆ—่กจ
CM=[78,121,167;242,142,43;225,87,89;118,183,178;89,161,79;
237,201,72;176,122,161;255,157,167;156,117,95;186,176,172]./255;
% ็ป˜ๅˆถๅœ†ๅœˆ
hold on
t=linspace(0,2*pi,100);
x=cos(t).*.8;
y=sin(t).*.8;
for i=1:600
X=i.*cos(i./10)./10;
Y=i.*sin(i./10)./10;
fill(X+x,Y+y,CM(Pi(i)+1,:),'EdgeColor','none','FaceAlpha',.9)
end
text(0,65,'The Circle of \pi','Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','center','FontSize',25,'FontAngle','italic')
% ๅ›พ็ช—ๅ’Œๅๆ ‡ๅŒบๅŸŸไฟฎ้ฅฐ
set(gcf,'Position',[200,100,820,820]);
ax=gca;
ax.XLim=[-60,60];
ax.YLim=[-60,70];
ax.XTick=[];
ax.YTick=[];
ax.Color=[0,0,0];
ax.DataAspectRatio=[1,1,1];
13 Archimedean spiral diagram
a=1;b=.227;
Pi=getPi(500);
% ้…่‰ฒๅˆ—่กจ
CM=[78,121,167;242,142,43;225,87,89;118,183,178;89,161,79;
237,201,72;176,122,161;255,157,167;156,117,95;186,176,172]./255;
% ็ป˜ๅˆถๅœ†ๅœˆ
hold on
T=0;R=1;
t=linspace(0,2*pi,100);
x=cos(t).*.7;
y=sin(t).*.7;
for i=1:500
X=R.*cos(T);Y=R.*sin(T);
fill(X+x,Y+y,CM(Pi(i)+1,:),'EdgeColor','none','FaceAlpha',.9)
T=T+1./R.*1.4;
R=a+b*T;
end
text(17.25,22,{'The Archimedes spiral of \pi';'โ€”โ€” 500 digits'},...
'Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','right','FontSize',25,'FontAngle','italic')
% ๅ›พ็ช—ๅ’Œๅๆ ‡ๅŒบๅŸŸไฟฎ้ฅฐ
set(gcf,'Position',[200,100,820,820]);
ax=gca;
ax.XLim=[-19,18.5];
ax.YLim=[-20,25];
ax.XTick=[];
ax.YTick=[];
ax.Color=[0,0,0];
ax.DataAspectRatio=[1,1,1];
14 proportional Archimedean spiral diagram
Pi=[3,getPi(1199)];
% ้…่‰ฒๆ•ฐๆฎ
CM=[239,32,120;239,60,52;247,98,32;255,182,60;247,235,44;
142,199,57;55,180,70;0,170,239;40,56,146;147,37,139]./255;
% CM=slanCM(184,10);
% ็ป˜ๅˆถๅœ†ๅœˆ
hold on
T=0;R=1;
t=linspace(0,2*pi,100);
x=cos(t).*.7;
y=sin(t).*.7;
for i=1:4:length(Pi)
X=R.*cos(T);Y=R.*sin(T);
tNum=Pi(i:i+3);
numNum=find([diff(sort(tNum)),1]);
numNum=[numNum(1),diff(numNum)];
cumNum=cumsum(numNum);
uniNum=unique(tNum);
for j=length(cumNum):-1:1
fill(x./4.*cumNum(j)+X,y./4.*cumNum(j)+Y,CM(uniNum(j)+1,:),'EdgeColor','none')
end
T=T+1./R.*1.4;
R=a+b*T;
end
text(14,16.5,{'The ratio of four numbers from \pi';'โ€”โ€” 1200 digits'},...
'Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','right','FontSize',23,'FontAngle','italic')
% ๅ›พ็ช—ๅ’Œๅๆ ‡ๅŒบๅŸŸไฟฎ้ฅฐ
set(gcf,'Position',[200,100,820,820]);
ax=gca;
ax.XLim=[-15,15.5];
ax.YLim=[-15,19];
ax.XTick=[];
ax.YTick=[];
ax.Color=[0,0,0];
ax.DataAspectRatio=[1,1,1];
15 graph
% ๆž„ๅปบ่ฟžๆŽฅ็Ÿฉ้˜ต
corrMat=zeros(10,10);
Pi=getPi(401);
for i=1:400
corrMat(Pi(i)+1,Pi(i+1)+1)=corrMat(Pi(i)+1,Pi(i+1)+1)+1;
end
% ้…่‰ฒๅˆ—่กจ
colorList=[0.3725 0.2745 0.5647
0.1137 0.4118 0.5882
0.2196 0.6510 0.6471
0.0588 0.5216 0.3294
0.4510 0.6863 0.2824
0.9294 0.6784 0.0314
0.8824 0.4863 0.0196
0.8000 0.3137 0.2431
0.5804 0.2039 0.4314
0.4353 0.2510 0.4392];
t=linspace(0,2*pi,11);t=t(1:10)';
posXY=[cos(t),sin(t)];
maxWidth=max(corrMat(corrMat>0));
minWidth=min(corrMat(corrMat>0));
ttList=linspace(0,1,3)';
% ๅพช็Žฏ็ป˜ๅ›พ
hold on
for i=1:size(corrMat,1)
for j=i+1:size(corrMat,2)
if corrMat(i,j)>0
tW=(corrMat(i,j)-minWidth)./(maxWidth-minWidth);
colorData=(1-ttList).*colorList(i,:)+ttList.*colorList(j,:);
CData(:,:,1)=colorData(:,1);
CData(:,:,2)=colorData(:,2);
CData(:,:,3)=colorData(:,3);
% ็ป˜ๅˆถ่ฟž็บฟ
fill(linspace(posXY(i,1),posXY(j,1),3),...
linspace(posXY(i,2),posXY(j,2),3),[0,0,0],'LineWidth',tW.*12+1,...
'CData',CData,'EdgeColor','interp','EdgeAlpha',.7,'FaceAlpha',.7)
end
end
% ็ป˜ๅˆถๅœ†็‚น
scatter(posXY(i,1),posXY(i,2),200,'filled','LineWidth',1.2,...
'MarkerFaceColor',colorList(i,:),'MarkerEdgeColor',[.7,.7,.7]);
text(posXY(i,1).*1.13,posXY(i,2).*1.13,num2str(i-1),'Color',[1,1,1].*.7,...
'FontSize',30,'FontWeight','bold','FontName','Cambria','HorizontalAlignment','center')
end
text(0,1.3,'Numerical adjacency of \pi โ€” 400 digits','Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','center','FontSize',25,'FontAngle','italic')
% ๅ›พ็ช—ๅ’Œๅๆ ‡ๅŒบๅŸŸไฟฎ้ฅฐ
set(gcf,'Position',[200,100,820,820]);
ax=gca;
ax.XLim=[-1.2,1.2];
ax.YLim=[-1.21,1.5];
ax.XTick=[];
ax.YTick=[];
ax.Color=[0,0,0];
ax.DataAspectRatio=[1,1,1];
16 circos chart
Need to use this tool:
https://www.mathworks.com/matlabcentral/fileexchange/118655-circos-plot
Class=getPi(1001)+1;
Data=diag(ones(1,1000),-1);
className={'0','1','2','3','4','5','6','7','8','9'};
colorOrder=[239,65,75;230,115,48;229,158,57;232,136,85;239,199,97;
144,180,116;78,166,136;81,140,136;90,118,142;43,121,159]./255;
CC=circosChart(Data,Class,'ClassName',className,'ColorOrder',colorOrder);
CC=CC.draw();
ax=gca;
ax.Color=[0,0,0];
CC.setClassLabel('Color',[1,1,1],'FontSize',25,'FontName','Cambria')
CC.setLine('LineWidth',.7)
YOU CAN GET ALL CODE HERE:
https://www.mathworks.com/matlabcentral/fileexchange/126210-happy-pi-day
Chord diagrams are very common in Python and R, but there are no related functions in MATLAB before. It is not easy to draw chord diagrams of the same quality as R language, But I created a MATLAB tool that could almost do it.
โ†“ โ†“ โ†“ โ†“ โ†“ โ†“ โ†“ โ†“ โ†“ โ†“
https://www.mathworks.com/matlabcentral/fileexchange/116550-chord-chart
โ†‘ โ†‘ โ†‘ โ†‘ โ†‘ โ†‘ โ†‘ โ†‘ โ†‘ โ†‘
Here is the help document:
1 Data Format
The data requirement is a numerical matrix with all values greater than or equal to 0, or a table array, or a numerical matrix and cell array for names. First, give an example of a numerical matrix:
1.1 Numerical Matrix
dataMat=randi([0,5],[5,4]);
% ็ป˜ๅ›พ(draw)
CC=chordChart(dataMat);
CC=CC.draw();
Since each object is not named, it will be automatically named Rn and Cn
1.2 Numerical Matrix and Cell Array for Names
dataMat=[2 0 1 2 5 1 2;
3 5 1 4 2 0 1;
4 0 5 5 2 4 3];
colName={'G1','G2','G3','G4','G5','G6','G7'};
rowName={'S1','S2','S3'};
CC=chordChart(dataMat,'rowName',rowName,'colName',colName);
CC=CC.draw();
RowName should be the same size as the rows of the matrix
ColName should be the same size as the columns of the matrix
For this example, if the value in the second row and third column is 1, it indicates that there is an energy flow from S2 to G3, and a chord with a width of 1 is needed between these two.
1.3 Table Array
A table array in the following format is required:
2 Decorate Chord
2.1 Batch modification of chords
Batch modification of chords can be done using the setChordProp function, and all properties of the Patch object can be modified. For example, modifying the color of the string, edge color, edge line sstyle, etc.:
CC.setChordProp('EdgeColor',[.3,.3,.3],'LineStyle','--',...
'LineWidth',.1,'FaceColor',[.3,.3,.3])
2.2 Individual Modification of Chord
The individual modification of chord can be done using the setChordMN function, where the values of m and n correspond exactly to the rows and columns of the original numerical matrix. For example, changing the color of the strings flowing from S2 to G4 to red:
CC.setChordMN(2,4,'FaceColor',[1,0,0])
2.3 Color Mapping of Chords
Just use function colormap to do so:
% version 1.7.0ๆ›ดๆ–ฐ
% ๅฏไฝฟ็”จcolormapๅ‡ฝๆ•ฐ็›ดๆŽฅไฟฎๆ”น้ขœ่‰ฒ
% Colors can be adjusted directly using the function colormap(demo4)
colormap(flipud(pink))
3 Arc Shaped Block Decoration
3.1 Batch Decoration of Arc-Shaped Blocks
use:
  • setSquareT_Prop
  • setSquareF_Prop
to modify the upper and lower blocks separately, and all attributes of the Patch object can be modified. For example, batch modify the upper blocks (change to black):
CC.setSquareT_Prop('FaceColor',[0,0,0])
3.2 Arc-Shaped Blocks Individually Decoration
use:
  • setSquareT_N
  • setSquareF_N
to modify the upper and lower blocks separately. For example, modify the second block above separately (changed to red):
CC.setSquareT_N(2,'FaceColor',[.8,0,0])
4 Font Adjustment
Use the setFont function to adjust the font, and all properties of the text object can be modified. For example, changing the font size, font, and color of the text:
CC.setFont('FontSize',25,'FontName','Cambria','Color',[0,0,.8])
5 Show and Hide Ticks
Usage:
CC.tickState('on')
% CC.tickState('off')
6 Attribute 'Sep' with Adjustable Square Spacing
If the matrix size is large, the drawing will be out of scale:
dataMat=randi([0,1],[20,10]);
CC=chordChart(dataMat);
CC=CC.draw();
% CC.tickState('on')
We can modify its Sep attribute:
dataMat=randi([0,1],[20,10]);
% use Sep to decrease space (separation)
% ไฝฟ็”จ sep ๅ‡ๅฐ็ฉบ้š™
CC=chordChart(dataMat,'Sep',1/120);
CC=CC.draw();
7 Modify Text Direction
dataMat=randi([0,1],[20,10]);
% use Sep to decrease space (separation)
% ไฝฟ็”จ sep ๅ‡ๅฐ็ฉบ้š™
CC=chordChart(dataMat,'Sep',1/120);
CC=CC.draw();
CC.tickState('on')
% version 1.7.0ๆ›ดๆ–ฐ
% ๅ‡ฝๆ•ฐlabelRatato็”จๆฅๆ—‹่ฝฌๆ ‡็ญพ
% The function labelRatato is used to rotate the label
CC.labelRotate('on')
8 Add Tick Labels
dataMat=[2 0 1 2 5 1 2;
3 5 1 4 2 0 1;
4 0 5 5 2 4 3];
colName={'G1','G2','G3','G4','G5','G6','G7'};
rowName={'S1','S2','S3'};
CC=chordChart(dataMat,'rowName',rowName,'colName',colName);
CC=CC.draw();
CC.setFont('FontSize',17,'FontName','Cambria')
% ๆ˜พ็คบๅˆปๅบฆๅ’Œๆ•ฐๅ€ผ
% Displays scales and numeric values
CC.tickState('on')
CC.tickLabelState('on')
% ่ฐƒ่Š‚ๆ ‡็ญพๅŠๅพ„
% Adjustable Label radius
CC.setLabelRadius(1.3);
% figure()
% dataMat=[2 0 1 2 5 1 2;
% 3 5 1 4 2 0 1;
% 4 0 5 5 2 4 3];
% dataMat=dataMat+rand(3,7);
% dataMat(dataMat<1)=0;
%
% CC=chordChart(dataMat,'rowName',rowName,'colName',colName);
% CC=CC.draw();
% CC.setFont('FontSize',17,'FontName','Cambria')
%
% % ๆ˜พ็คบๅˆปๅบฆๅ’Œๆ•ฐๅ€ผ
% % Displays scales and numeric values
% CC.tickState('on')
% CC.tickLabelState('on')
%
% % ่ฐƒ่Š‚ๆ ‡็ญพๅŠๅพ„
% % Adjustable Label radius
% CC.setLabelRadius(1.4);
9 Custom Tick Label Format
A function handle is required to input numeric output strings. The format can be set through the setTickLabelFormat function, such as Scientific notation:
dataMat=[2 0 1 2 5 1 2;
3 5 1 4 2 0 1;
4 0 5 5 2 4 3];
dataMat=dataMat+rand(3,7);
dataMat(dataMat<1)=0;
dataMat=dataMat.*1000;
CC=chordChart(dataMat);
CC=CC.draw();
CC.setFont('FontSize',17,'FontName','Cambria')
% ๆ˜พ็คบๅˆปๅบฆๅ’Œๆ•ฐๅ€ผ
% Displays scales and numeric values
CC.tickState('on')
CC.tickLabelState('on')
% ่ฐƒ่Š‚ๆ ‡็ญพๅŠๅพ„
% Adjustable Label radius
CC.setLabelRadius(1.4);
% ่ฐƒๆ•ดๆ•ฐๅ€ผๅญ—็ฌฆไธฒๆ ผๅผ
% Adjust numeric string format
CC.setTickLabelFormat(@(x)sprintf('%0.1e',x))
10 A Demo
rng(2)
dataMat=randi([1,7],[11,5]);
colName={'Fly','Beetle','Leaf','Soil','Waxberry'};
rowName={'Bartomella','Bradyrhizobium','Dysgomonas','Enterococcus',...
'Lactococcus','norank','others','Pseudomonas','uncultured',...
'Vibrionimonas','Wolbachia'};
CC=chordChart(dataMat,'rowName',rowName,'colName',colName,'Sep',1/80);
CC=CC.draw();
% ไฟฎๆ”นไธŠๆ–นๆ–นๅ—้ขœ่‰ฒ(Modify the color of the blocks above)
CListT=[0.7765 0.8118 0.5216;0.4431 0.4706 0.3843;0.5804 0.2275 0.4549;
0.4471 0.4039 0.6745;0.0157 0 0 ];
for i=1:5
CC.setSquareT_N(i,'FaceColor',CListT(i,:))
end
% ไฟฎๆ”นไธ‹ๆ–นๆ–นๅ—้ขœ่‰ฒ(Modify the color of the blocks below)
CListF=[0.5843 0.6863 0.7843;0.1098 0.1647 0.3255;0.0902 0.1608 0.5373;
0.6314 0.7961 0.2118;0.0392 0.2078 0.1059;0.0157 0 0 ;
0.8549 0.9294 0.8745;0.3882 0.3255 0.4078;0.5020 0.7216 0.3843;
0.0902 0.1843 0.1804;0.8196 0.2314 0.0706];
for i=1:11
CC.setSquareF_N(i,'FaceColor',CListF(i,:))
end
% ไฟฎๆ”นๅผฆ้ขœ่‰ฒ(Modify chord color)
for i=1:5
for j=1:11
CC.setChordMN(j,i,'FaceColor',CListT(i,:),'FaceAlpha',.5)
end
end
CC.tickState('on')
CC.labelRotate('on')
CC.setFont('FontSize',17,'FontName','Cambria')
Hope to have your Reviews and Stars!!!
โ†“ โ†“ โ†“ โ†“ โ†“ โ†“ โ†“ โ†“ โ†“ โ†“
https://www.mathworks.com/matlabcentral/fileexchange/116550-chord-chart
โ†‘ โ†‘ โ†‘ โ†‘ โ†‘ โ†‘ โ†‘ โ†‘ โ†‘ โ†‘
Most of my projects are the tension between what exists and what I must create to make the project work. The more I leverage what exists, the faster I can complete my project in a shorter amount of time. I often find out later in the process that a MATLAB function already exists and has existed for 10 releases. The faster that I discover built-in functions, the faster I can get to the new stuff that I have to invent.
I asked the AI Chat Playground why to use MATLAB's builtin functions and I got some interesting ideas. What are your thoughts?
AI Chat Playground:
  1. Efficiency and Performance: MATLAB's built-in functions are highly optimized for performance. They are written in optimized and compiled languages like C or C++, which means they run faster than equivalent functions written directly in MATLAB language. When you use these functions, your code can execute more efficiently, which is especially beneficial for handling large datasets or complex mathematical computations.
  2. Reliability and Stability: Built-in functions have been rigorously tested and debugged by the developers at MathWorks (the company behind MATLAB). They are less likely to contain errors compared to custom-written code, which means you can rely on them to produce accurate and consistent results.
  3. Ease of Use: Built-in functions are designed to be user-friendly, with clear documentation and support materials available. This can make it easier to perform complex operations without needing a deep understanding of the underlying algorithms. It also reduces the amount of code you need to write and debug, saving time and reducing the potential for errors.
  4. Compatibility: Built-in functions are updated alongside MATLAB itself, ensuring compatibility with new features and versions of the software. Using built-in functions can help ensure that your code continues to work correctly as MATLAB evolves.
  5. Advanced Features: MATLABโ€™s built-in functions often include advanced features and options that are not readily available or would be difficult to implement on your own. For example, MATLABโ€™s mathematical functions can handle a wide range of special cases and data types, providing more flexibility than custom implementations.
  6. Community and Support: Using built-in functions means you're working with standard tools recognized by the MATLAB community. This can make it easier to get help from other users or from MathWorks support, as more people will be familiar with the functionality and potential issues of these functions.
Given a vector v whose order we would like to randomly permute, many would perform the permutation by explicitly querying the length/size of v, e.g.,
I=randperm(numel(v));
v=v(I);
However, one can instead do as follows, avoiding the size query.
v=v(randperm(end))
Analogous things can be done with matrices, e.g.,
A=A(randperm(end), randperm(end));
The creativity comes from the copper sulfate crystal heart made in junior high school. Copper sulfate is a triclinic crystal, and the same structure was not used here for convenience in drawing.
Part 1. Coordinate transformation
To draw a crystal heart, one must first be able to draw crystal clusters. To draw a crystal cluster, one must first be able to draw a crystal. To draw a crystal, we need this kind of structure:
We first need a point with a certain distance from the straight line and a perpendicular point of cutPnt, which is very easy to find, for example, cutPnt=[x0, y0, z0]; The direction of the central axis is V=[x1, y1, z1]; If the distance to the straight line is L, the following points clearly meet the conditions:
v2=[z1,z1,-x1-y1];
v2=v2./norm(v2).*L;
pnt=cutPnt+v2;
But finding only one point is not enough. We need to find four points, and each point is obtained by rotating the previous point around a straight line by degrees. Therefore, we need to obtain our point rotation transformation matrix around a straight line
quite complex,right?
rotateMat=[u^2+(v^2+w^2)*cos(theta) , u*v*(1-cos(theta))-w*sin(theta), u*w*(1-cos(theta))+v*sin(theta), (a*(v^2+w^2)-u*(b*v+c*w))*(1-cos(theta))+(b*w-c*v)*sin(theta);
u*v*(1-cos(theta))+w*sin(theta), v^2+(u^2+w^2)*cos(theta) , v*w*(1-cos(theta))-u*sin(theta), (b*(u^2+w^2)-v*(a*u+c*w))*(1-cos(theta))+(c*u-a*w)*sin(theta);
u*w*(1-cos(theta))-v*sin(theta), v*w*(1-cos(theta))+u*sin(theta), w^2+(u^2+v^2)*cos(theta) , (c*(u^2+v^2)-w*(a*u+b*v))*(1-cos(theta))+(a*v-b*u)*sin(theta);
0 , 0 , 0 , 1];
Where [u, v, w] is the directional unit vector, and [a, b, c] is the initial coordinate of the axis:
Part 2. Crystal Cluster Drawing
function crystall
hold on
for i=1:50
len=rand(1)*8+5;
tempV=rand(1,3)-0.5;
tempV(3)=abs(tempV(3));
tempV=tempV./norm(tempV).*len;
tempEpnt=tempV;
drawCrystal([0 0 0],tempEpnt,pi/6,0.8,0.1,rand(1).*0.2+0.2)
disp(i)
end
ax=gca;
ax.XLim=[-15,15];
ax.YLim=[-15,15];
ax.ZLim=[-2,15];
grid on
ax.GridLineStyle='--';
ax.LineWidth=1.2;
ax.XColor=[1,1,1].*0.4;
ax.YColor=[1,1,1].*0.4;
ax.ZColor=[1,1,1].*0.4;
ax.DataAspectRatio=[1,1,1];
ax.DataAspectRatioMode='manual';
ax.CameraPosition=[-67.6287 -204.5276 82.7879];
function drawCrystal(Spnt,Epnt,theta,cl,w,alpha)
%plot3([Spnt(1),Epnt(1)],[Spnt(2),Epnt(2)],[Spnt(3),Epnt(3)])
mainV=Epnt-Spnt;
cutPnt=cl.*(mainV)+Spnt;
cutV=[mainV(3),mainV(3),-mainV(1)-mainV(2)];
cutV=cutV./norm(cutV).*w.*norm(mainV);
cornerPnt=cutPnt+cutV;
cornerPnt=rotateAxis(Spnt,Epnt,cornerPnt,theta);
cornerPntSet(1,:)=cornerPnt';
for ii=1:3
cornerPnt=rotateAxis(Spnt,Epnt,cornerPnt,pi/2);
cornerPntSet(ii+1,:)=cornerPnt';
end
F = [1,3,4;1,4,5;1,5,6;1,6,3;...
2,3,4;2,4,5;2,5,6;2,6,3];
V = [Spnt;Epnt;cornerPntSet];
patch('Faces',F,'Vertices',V,'FaceColor',[0 71 177]./255,...
'FaceAlpha',alpha,'EdgeColor',[0 71 177]./255.*0.8,...
'EdgeAlpha',0.6,'LineWidth',0.5,'EdgeLighting',...
'gouraud','SpecularStrength',0.3)
end
function newPnt=rotateAxis(Spnt,Epnt,cornerPnt,theta)
V=Epnt-Spnt;V=V./norm(V);
u=V(1);v=V(2);w=V(3);
a=Spnt(1);b=Spnt(2);c=Spnt(3);
cornerPnt=[cornerPnt(:);1];
rotateMat=[u^2+(v^2+w^2)*cos(theta) , u*v*(1-cos(theta))-w*sin(theta), u*w*(1-cos(theta))+v*sin(theta), (a*(v^2+w^2)-u*(b*v+c*w))*(1-cos(theta))+(b*w-c*v)*sin(theta);
u*v*(1-cos(theta))+w*sin(theta), v^2+(u^2+w^2)*cos(theta) , v*w*(1-cos(theta))-u*sin(theta), (b*(u^2+w^2)-v*(a*u+c*w))*(1-cos(theta))+(c*u-a*w)*sin(theta);
u*w*(1-cos(theta))-v*sin(theta), v*w*(1-cos(theta))+u*sin(theta), w^2+(u^2+v^2)*cos(theta) , (c*(u^2+v^2)-w*(a*u+b*v))*(1-cos(theta))+(a*v-b*u)*sin(theta);
0 , 0 , 0 , 1];
newPnt=rotateMat*cornerPnt;
newPnt(4)=[];
end
end
Part 3. Drawing of Crystal Heart
function crystalHeart
clc;clear;close all
hold on
% drawCrystal([1,1,1],[3,3,3],pi/6,0.8,0.14)
sep=pi/8;
t=[0:0.2:sep,sep:0.02:pi-sep,pi-sep:0.2:pi+sep,pi+sep:0.02:2*pi-sep,2*pi-sep:0.2:2*pi];
x=16*sin(t).^3;
y=13*cos(t)-5*cos(2*t)-2*cos(3*t)-cos(4*t);
z=zeros(size(t));
plot3(x,y,z,'Color',[186,110,64]./255,'LineWidth',1)
for i=1:length(t)
for j=1:6
len=rand(1)*2.5+1.5;
tempV=rand(1,3)-0.5;
tempV=tempV./norm(tempV).*len;
tempSpnt=[x(i),y(i),z(i)];
tempEpnt=tempV+tempSpnt;
drawCrystal(tempSpnt,tempEpnt,pi/6,0.8,0.14)
disp([i,j])
end
end
ax=gca;
ax.XLim=[-22,22];
ax.YLim=[-20,20];
ax.ZLim=[-10,10];
grid on
ax.GridLineStyle='--';
ax.LineWidth=1.2;
ax.XColor=[1,1,1].*0.4;
ax.YColor=[1,1,1].*0.4;
ax.ZColor=[1,1,1].*0.4;
ax.DataAspectRatio=[1,1,1];
ax.DataAspectRatioMode='manual';
function drawCrystal(Spnt,Epnt,theta,cl,w)
%plot3([Spnt(1),Epnt(1)],[Spnt(2),Epnt(2)],[Spnt(3),Epnt(3)])
mainV=Epnt-Spnt;
cutPnt=cl.*(mainV)+Spnt;
cutV=[mainV(3),mainV(3),-mainV(1)-mainV(2)];
cutV=cutV./norm(cutV).*w.*norm(mainV);
cornerPnt=cutPnt+cutV;
cornerPnt=rotateAxis(Spnt,Epnt,cornerPnt,theta);
cornerPntSet(1,:)=cornerPnt';
for ii=1:3
cornerPnt=rotateAxis(Spnt,Epnt,cornerPnt,pi/2);
cornerPntSet(ii+1,:)=cornerPnt';
end
F = [1,3,4;1,4,5;1,5,6;1,6,3;...
2,3,4;2,4,5;2,5,6;2,6,3];
V = [Spnt;Epnt;cornerPntSet];
patch('Faces',F,'Vertices',V,'FaceColor',[0 71 177]./255,...
'FaceAlpha',0.2,'EdgeColor',[0 71 177]./255.*0.9,...
'EdgeAlpha',0.25,'LineWidth',0.01,'EdgeLighting',...
'gouraud','SpecularStrength',0.3)
end
function newPnt=rotateAxis(Spnt,Epnt,cornerPnt,theta)
V=Epnt-Spnt;V=V./norm(V);
u=V(1);v=V(2);w=V(3);
a=Spnt(1);b=Spnt(2);c=Spnt(3);
cornerPnt=[cornerPnt(:);1];
rotateMat=[u^2+(v^2+w^2)*cos(theta) , u*v*(1-cos(theta))-w*sin(theta), u*w*(1-cos(theta))+v*sin(theta), (a*(v^2+w^2)-u*(b*v+c*w))*(1-cos(theta))+(b*w-c*v)*sin(theta);
u*v*(1-cos(theta))+w*sin(theta), v^2+(u^2+w^2)*cos(theta) , v*w*(1-cos(theta))-u*sin(theta), (b*(u^2+w^2)-v*(a*u+c*w))*(1-cos(theta))+(c*u-a*w)*sin(theta);
u*w*(1-cos(theta))-v*sin(theta), v*w*(1-cos(theta))+u*sin(theta), w^2+(u^2+v^2)*cos(theta) , (c*(u^2+v^2)-w*(a*u+b*v))*(1-cos(theta))+(a*v-b*u)*sin(theta);
0 , 0 , 0 , 1];
newPnt=rotateMat*cornerPnt;
newPnt(4)=[];
end
end
So, how to draw a roseball just like this ?
To begin with, we need to know how to draw a single rose in MATLAB:
function drawrose
set(gca,'CameraPosition',[2 2 2])
hold on
grid on
[x,t]=meshgrid((0:24)./24,(0:0.5:575)./575.*20.*pi+4*pi);
p=(pi/2)*exp(-t./(8*pi));
change=sin(15*t)/150;
u=1-(1-mod(3.6*t,2*pi)./pi).^4./2+change;
y=2*(x.^2-x).^2.*sin(p);
r=u.*(x.*sin(p)+y.*cos(p));
h=u.*(x.*cos(p)-y.*sin(p));
surface(r.*cos(t),r.*sin(t),h,'EdgeAlpha',0.1,...
'EdgeColor',[0 0 0],'FaceColor','interp')
end
Tts pretty easy, Now we are trying to dye it the desired color:
function drawrose
set(gca,'CameraPosition',[2 2 2])
hold on
grid on
[x,t]=meshgrid((0:24)./24,(0:0.5:575)./575.*20.*pi+4*pi);
p=(pi/2)*exp(-t./(8*pi));
change=sin(15*t)/150;
u=1-(1-mod(3.6*t,2*pi)./pi).^4./2+change;
y=2*(x.^2-x).^2.*sin(p);
r=u.*(x.*sin(p)+y.*cos(p));
h=u.*(x.*cos(p)-y.*sin(p));
map=[0.9176 0.9412 1.0000
0.8353 0.8706 0.9922
0.8196 0.8627 0.9804
0.7020 0.7569 0.9412
0.5176 0.5882 0.9255
0.3686 0.4824 0.9412
0.3059 0.4000 0.9333
0.2275 0.3176 0.8353
0.1216 0.2275 0.6471];
Xi=1:size(map,1);Xq=linspace(1,size(map,1),100);
map=[interp1(Xi,map(:,1),Xq,'linear')',...
interp1(Xi,map(:,2),Xq,'linear')',...
interp1(Xi,map(:,3),Xq,'linear')'];
surface(r.*cos(t),r.*sin(t),h,'EdgeAlpha',0.1,...
'EdgeColor',[0 0 0],'FaceColor','interp')
colormap(map)
end
I try to take colors from real roses and interpolate them to make them more realistic
Then, how can I put these colorful flowers on to a ball ?
We need to place the drawn flowers on each face of the polyhedron sphere through coordinate transformation. Here, we use a regular dodecahedron:
Move the flower using the following rotation formula:
We place a flower on each plane, which means that the angle between every two flowers is degrees. We can place each flower at the appropriate angle through multiple x-axis rotations and multiple z-axis rotations. The code is as follows:
function roseBall(colorList)
%ๆ›ฒ้ขๆ•ฐๆฎ่ฎก็ฎ—
%==========================================================================
[x,t]=meshgrid((0:24)./24,(0:0.5:575)./575.*20.*pi+4*pi);
p=(pi/2)*exp(-t./(8*pi));
change=sin(15*t)/150;
u=1-(1-mod(3.6*t,2*pi)./pi).^4./2+change;
y=2*(x.^2-x).^2.*sin(p);
r=u.*(x.*sin(p)+y.*cos(p));
h=u.*(x.*cos(p)-y.*sin(p));
%้ขœ่‰ฒๆ˜ ๅฐ„่กจ
%==========================================================================
hMap=(h-min(min(h)))./(max(max(h))-min(min(h)));
col=size(hMap,2);
if nargin<1
colorList=[0.0200 0.0400 0.3900
0 0.0900 0.5800
0 0.1300 0.6400
0.0200 0.0600 0.6900
0 0.0800 0.7900
0.0100 0.1800 0.8500
0 0.1300 0.9600
0.0100 0.2600 0.9900
0 0.3500 0.9900
0.0700 0.6200 1.0000
0.1700 0.6900 1.0000];
end
colorFunc=colorFuncFactory(colorList);
dataMap=colorFunc(hMap');
colorMap(:,:,1)=dataMap(:,1:col);
colorMap(:,:,2)=dataMap(:,col+1:2*col);
colorMap(:,:,3)=dataMap(:,2*col+1:3*col);
function colorFunc=colorFuncFactory(colorList)
xx=(0:size(colorList,1)-1)./(size(colorList,1)-1);
y1=colorList(:,1);y2=colorList(:,2);y3=colorList(:,3);
colorFunc=@(X)[interp1(xx,y1,X,'linear')',interp1(xx,y2,X,'linear')',interp1(xx,y3,X,'linear')'];
end
%ๆ›ฒ้ขๆ—‹่ฝฌๅŠ็ป˜ๅˆถ
%==========================================================================
surface(r.*cos(t),r.*sin(t),h+0.35,'EdgeAlpha',0.05,...
'EdgeColor',[0 0 0],'FaceColor','interp','CData',colorMap)
hold on
surface(r.*cos(t),r.*sin(t),-h-0.35,'EdgeAlpha',0.05,...
'EdgeColor',[0 0 0],'FaceColor','interp','CData',colorMap)
Xset=r.*cos(t);
Yset=r.*sin(t);
Zset=h+0.35;
yaw_z=72*pi/180;
roll_x=pi-acos(-1/sqrt(5));
R_z_2=[cos(yaw_z),-sin(yaw_z),0;
sin(yaw_z),cos(yaw_z),0;
0,0,1];
R_z_1=[cos(yaw_z/2),-sin(yaw_z/2),0;
sin(yaw_z/2),cos(yaw_z/2),0;
0,0,1];
R_x_2=[1,0,0;
0,cos(roll_x),-sin(roll_x);
0,sin(roll_x),cos(roll_x)];
[nX,nY,nZ]=rotateXYZ(Xset,Yset,Zset,R_x_2);
surface(nX,nY,nZ,'EdgeAlpha',0.05,...
'EdgeColor',[0 0 0],'FaceColor','interp','CData',colorMap)
for k=1:4
[nX,nY,nZ]=rotateXYZ(nX,nY,nZ,R_z_2);
surface(nX,nY,nZ,'EdgeAlpha',0.05,...
'EdgeColor',[0 0 0],'FaceColor','interp','CData',colorMap)
end
[nX,nY,nZ]=rotateXYZ(nX,nY,nZ,R_z_1);
for k=1:5
[nX,nY,nZ]=rotateXYZ(nX,nY,nZ,R_z_2);
surface(nX,nY,-nZ,'EdgeAlpha',0.05,...
'EdgeColor',[0 0 0],'FaceColor','interp','CData',colorMap)
end
%--------------------------------------------------------------------------
function [nX,nY,nZ]=rotateXYZ(X,Y,Z,R)
nX=zeros(size(X));
nY=zeros(size(Y));
nZ=zeros(size(Z));
for i=1:size(X,1)
for j=1:size(X,2)
v=[X(i,j);Y(i,j);Z(i,j)];
nv=R*v;
nX(i,j)=nv(1);
nY(i,j)=nv(2);
nZ(i,j)=nv(3);
end
end
end
%axesๅฑžๆ€ง่ฐƒๆ•ด
%==========================================================================
ax=gca;
grid on
ax.GridLineStyle='--';
ax.LineWidth=1.2;
ax.XColor=[1,1,1].*0.4;
ax.YColor=[1,1,1].*0.4;
ax.ZColor=[1,1,1].*0.4;
ax.DataAspectRatio=[1,1,1];
ax.DataAspectRatioMode='manual';
ax.CameraPosition=[-6.5914 -24.1625 -0.0384];
end
TRY DIFFERENT COLORS !!
colorList1=[0.2000 0.0800 0.4300
0.2000 0.1300 0.4600
0.2000 0.2100 0.5000
0.2000 0.2800 0.5300
0.2000 0.3700 0.5800
0.1900 0.4500 0.6200
0.2000 0.4800 0.6400
0.1900 0.5400 0.6700
0.1900 0.5700 0.6900
0.1900 0.7500 0.7800
0.1900 0.8000 0.8100
];
colorList2=[0.1300 0.1000 0.1600
0.2000 0.0900 0.2000
0.2800 0.0800 0.2300
0.4200 0.0800 0.3000
0.5100 0.0700 0.3400
0.6600 0.1200 0.3500
0.7900 0.2200 0.4000
0.8800 0.3500 0.4700
0.9000 0.4500 0.5400
0.8900 0.7800 0.7900
];
colorList3=[0.3200 0.3100 0.7600
0.3800 0.3400 0.7600
0.5300 0.4200 0.7500
0.6400 0.4900 0.7300
0.7200 0.5500 0.7200
0.7900 0.6100 0.7100
0.9100 0.7100 0.6800
0.9800 0.7600 0.6700
];
colorList4=[0.2100 0.0900 0.3800
0.2900 0.0700 0.4700
0.4000 0.1100 0.4900
0.5500 0.1600 0.5100
0.7500 0.2400 0.4700
0.8900 0.3200 0.4100
0.9700 0.4900 0.3700
1.0000 0.5600 0.4100
1.0000 0.6900 0.4900
1.0000 0.8200 0.5900
0.9900 0.9200 0.6700
0.9800 0.9500 0.7100];
Victoria
Victoria
Last activity on 1 Mar 2024

How to Draw a Happy Sheep in MATLAB!

Let us consider how to draw a Happy Sheep. A Happy Sheep was introduced in the MATLAB Mini Hack contest: Happy Sheep!
In this contest there was the strict limitation on the code length. So the code of the Happy Sheep is very compact and is only 280 characters long. We will analyze the process of drawing the Happy Sheep in MATLAB step by step. The explanations of the even more compact version of the code of the same sheep are given below.
So, how to draw a sheep? It is very easy. We could notice that usually a sheep is covered by crimped wool. Therefore, a sheep could be painted using several geometrical curves of similar types. Of course, then it will be an abstract model of the real sheep. Let us select two mathematical curves, which are the most appropriate for our goal. They are an ellipse for smooth parts of the sheep and an ellipse combined with a rose for woolen parts of the sheep.
Let us recall the mathematical formulas of these curves. A parametric representation of the standard ellipse is the following:
An ellipse
Also we will use the following parametric representation of the rose (rhodonea) curve:
A rose
This curve was named by the mathematician Guido Grandi.
Let us combine them in one curve and add possible shifts:
An ellipse combined with a rose
Now if we would like to create an ellipse, we will set a21 = 0 and a22 = 0. If we would like to create a rose, we will set a11 = 0 and a12 = 0. If we would like to shift our curve, we will set a31 and a32 to the required values. Of course, we could set all non-zero parameters to combine both chosen curves and use the shifts.
Let us describe how to create these curves using the MATLAB code. To make the code more compact, it is possible to program both formulas for the combined curve in one line using the anonymous function. We could make the code more compact using the function handles for sine and cosine functions. Then the MATLAB code for an example of the ellipse curve will be the following.
% Handles
s=@sin;
c=@cos;
% Ellipse + Polar Rose
F=@(t,a,f) a(1)*f(t)+s(a(2)*t).*f(t)+a(3);
% Angles
t=0:.1:7;
% Parameters
E = [5 7;0 0;0 0];
% Painting
figure;
plot(F(t,E(:,1),c),F(t,E(:,2),s),'LineWidth',10);
axis equal
The parameter t varies from 0 to 7, which is the nearest integer greater than 2pi, with the step 0.1. The result of this code is the following ellipse curve with a11 = 5 and a12 = 7.
Ellipse_5_7
This ellipse is described by the following parametric equations:
An ellipse_5_7
The MATLAB code for an example of the rose curve will be the following.
% Handles
s=@sin;
c=@cos;
% Ellipse + Polar Rose
F=@(t,a,f) a(1)*f(t)+s(a(2)*t).*f(t)+a(3);
% Angles
t=0:.1:7;
% Parameters
R = [0 0;4 4;0 0];
% Painting
figure;
plot(F(t,R(:,1),c),F(t,R(:,2),s),'LineWidth',10);
axis equal
The result of this code is the following rose curve with a21 = 4 and a22 = 4.
Rose_4_4
This rose is described by the following parametric equations:
A rose_4_4
Obviously, now we are ready to draw main parts of our sheep! As we reproduce an abstract model of the sheep, let us select the following main parts for the representation: head, eyes, hoofs, body, crown, and tail. We will use ellipses for the first three parts in this list and ellipses combined with roses for the last three ones.
First let us describe drawing of each part independently.
The following MATLAB code will be used to do this.
% Handles
s=@sin;
c=@cos;
% Ellipse + Polar Rose
F=@(t,a,f) a(1)*f(t)+s(a(2)*t).*f(t)+a(3);
% Angles
t=0:.1:7;
% Parameters
Head = 1;
Eyes = 2:3;
Hoofs = 4:7;
Body = 8;
Crown = 9;
Tail = 10;
G=-13;
P=[5 7 repmat([.1 .5],1,6) 6 4 14 9 3 3;zeros(1,14) 8 8 12 12 4 4;...
-15 2 G 3 -17 3 -3 G 0 G 9 G 12 G -15 12 4 3 20 7];
% Painting
figure;
hold;
for i=Head
plot(F(t,P(:,2*i-1),c),F(t,P(:,2*i),s),'LineWidth',10);
end
axis([-25 25 -15 20]);
figure;
hold;
for i=Eyes
plot(F(t,P(:,2*i-1),c),F(t,P(:,2*i),s),'LineWidth',10);
end
axis([-25 25 -15 20]);
figure;
hold;
for i=Hoofs
plot(F(t,P(:,2*i-1),c),F(t,P(:,2*i),s),'LineWidth',10);
end
axis([-25 25 -15 20]);
figure;
hold;
for i=Body
plot(F(t,P(:,2*i-1),c),F(t,P(:,2*i),s),'LineWidth',10);
end
axis([-25 25 -15 20]);
figure;
hold;
for i=Crown
plot(F(t,P(:,2*i-1),c),F(t,P(:,2*i),s),'LineWidth',10);
end
axis([-25 25 -15 20]);
figure;
hold;
for i=Tail
plot(F(t,P(:,2*i-1),c),F(t,P(:,2*i),s),'LineWidth',10);
end
axis([-25 25 -15 20]);
The parameters a11, a12, a21, a22, a31, and a32 are written in the different submatrices of the matrix P. The code generates the following curves to illustrate the different parts of our sheep.
The following ellipse describes the head of the sheep.
Head
The following submatrix of the matrix P represents its parameters.
P_Head
The parametric equations of the head are the following:
A head
The following ellipses describe the eyes of the sheep.
Eyes
The following submatrices of the matrix P represent their parameters.
P_Eyes
The parametric equations of the left and right eyes correspondingly are the following:
A left eye
A right eye
The following ellipses describe the hoofs of the sheep.
Hoofs
The following submatrices of the matrix P represent their parameters.
P_Hoofs
The parametric equations of the right front, left front, right hind, and left hind hoofs correspondingly are the following:
A right front hoof
A left front hoof
A right hind hoof
A left hind hoof
The following ellipse combined with the rose describes the crown of the sheep.
Crown
The following submatrix of the matrix P represents its parameters.
P_Crown
The parametric equations of the crown are the following:
A crown
The following ellipse combined with the rose describes the body of the sheep.
Body
The following submatrix of the matrix P represents its parameters.
P_Body
The parametric equations of the body are the following:
A body
The following ellipse combined with the rose describes the tail of the sheep.
Tail
The following submatrix of the matrix P represents its parameters.
P_Tail
The parametric equations of the tail are the following:
A tail
Now all the parts of our sheep should be put together! It is very easy because all the parts are described by the same equations with different parameters.
The following code helps us to accomplish this goal and ultimately draw a Happy Sheep in MATLAB!
% Happy Sheep!
% By Victoria A. Sablina
% Handles
s=@sin;
c=@cos;
% Ellipse + Rose
F=@(t,a,f) a(1)*f(t)+s(a(2)*t).*f(t)+a(3);
% Angles
t=0:.1:7;
% Parameters
% Head (1:2)
% Eyes (3:6)
% Hoofs (7:14)
% Crown (15:16)
% Body (17:18)
% Tail (19:20)
G=-13;
P=[5 7 repmat([.1 .5],1,6) 6 4 14 9 3 3;zeros(1,14) 8 8 12 12 4 4;...
-15 2 G 3 -17 3 -3 G 0 G 9 G 12 G -15 12 4 3 20 7];
% Painting
hold;
for i=1:10
plot(F(t,P(:,2*i-1),c),F(t,P(:,2*i),s),'LineWidth',10);
end
This code is even more compact than the original code from the contest. It is only 253 instead of 280 characters long and generates the same Happy Sheep!
Happy Sheep!
Our sheep is happy, because of becoming famous in the MATLAB community, a star!
Congratulations! Now you know how to draw a Happy Sheep in MATLAB!
Thank you for reading!
Several of the colormaps are great for a 256 color surface plot, but aren't well optimized for extracting m colors for plotting several independent lines. The issue is that many colormaps have start/end colors that are too similar or are suboptimal colors for lines. There are certainly many workarounds for this, but it would be a great quality of life to adjust that directly when calling this.
Example:
x = linspace(0,2*pi,101)';
y = [1:6].*cos(x);
figure; plot(x,y,'LineWidth',2); grid on; axis tight;
And now if I wanted to color these lines, I could use something like turbo(6) or gray(6) and then apply it using colororder.
colororder(turbo(6))
But my issue is that the ends of the colormap are too similar. For other colormaps, you may get lines that are too light to be visible against the white background. There are plenty of workarounds, with my preference being to create extra colors and truncate that before using colororder.
cmap = turbo(8); cmap = cmap(2:end-1,:); % Truncate the end colors
figure; plot(x,y,'LineWidth',2); grid on; axis tight;
colororder(cmap)
I think it would be really awesome to add some name-argument input pair to these colormaps that can specify the range you want so this could even be done inside the colororder calling if desired. An example of my proposed solution would look something like this:
cmap = turbo(6,'Range',[0.1 0.8]); % Proposed idea to add functionality
Where in this scenario, the resulting colormap would be 6 equally spaced colors that range from 10% to 80% of the total color range. This would be especially nice because you could more quickly modify the range of colors, or you could set the limits regardless of whether you need to plot 3, 6, or 20 lines.
lazymatlab
lazymatlab
Last activity on 28 Feb 2024

MATLAB O/X QUIZ

MATLAB O/X Quiz
Answer BEFORE Googling!
  1. An infinite loop can be made using "for".
  2. "A == A" is always true.
  3. "round(2.5)" is 3.
  4. "round(-0.5)" is 0.
MATLAB Support Package for Quantum Computing lets you build, simulate, and run quantum algorithms.
Check out the Cheat Sheet here!
2 x 2 ํ–‰๋ ฌ์˜ ํ–‰๋ ฌ์‹์€
  • ํ–‰๋ ฌ์˜ ๋‘ row ๋ฒกํ„ฐ๋กœ ์ •์˜๋˜๋Š” ํ‰ํ–‰์‚ฌ๋ณ€ํ˜•์˜ ๋ฉด์ ์ž…๋‹ˆ๋‹ค.
  • ๋ฌผ๋ก  ๋‘ column ๋ฒกํ„ฐ๋กœ ์ •์˜๋˜๋Š” ํ‰ํ–‰์‚ฌ๋ณ€ํ˜•์˜ ๋ฉด์ ์ด๊ธฐ๋„ ํ•ฉ๋‹ˆ๋‹ค.
  • ์ข€ ๋” ์ •ํ™•ํžˆ๋Š” signed area์ž…๋‹ˆ๋‹ค. ๋ฉด์ ์ด ์Œ์ˆ˜๊ฐ€ ๋  ์ˆ˜๋„ ์žˆ๋‹ค๋Š” ๋œป์ด์ฃ .
  • ํ–‰๋ ฌ์˜ ๋‘ ํ–‰(๋˜๋Š” ๋‘ ์—ด)์„ ๋งž๋ฐ”๊พธ๋ฉด ํ–‰๋ ฌ์‹์˜ ๋ถ€ํ˜ธ๋„ ๋ฐ”๋€Œ๊ณ  ๋ฉด์ ์˜ ๋ถ€ํ˜ธ๋„ ๋ฐ”๋€Œ์–ด์•ผํ•ฉ๋‹ˆ๋‹ค.
์ผ๋ฐ˜์ ์œผ๋กœ n x n ํ–‰๋ ฌ์˜ ํ–‰๋ ฌ์‹์€
  • ๊ฐ row ๋ฒกํ„ฐ(๋˜๋Š” ๊ฐ column ๋ฒกํ„ฐ)๋กœ ์ •์˜๋˜๋Š” N์ฐจ์› ๊ณต๊ฐ„์˜ ํ‰ํ–‰๋ฉด์ฒด(?)์˜ signed area์ž…๋‹ˆ๋‹ค.
  • ์ œ๋Œ€๋กœ ์ดํ•ดํ•˜๋ ค๋ฉด ๋Œ€์ˆ˜ํ•™์˜ ๊ฐœ๋…์„ ๋งŽ์ด ๊ฐ€์ง€๊ณ  ์™€์•ผ ํ•˜๋Š”๋ฐ ์ž์„ธํ•œ ์„ค๋ช…์€ ์ƒ๋žตํ•ฉ๋‹ˆ๋‹ค.(=์ €๋„ ๋ชจ๋ฅธ๋‹ค๋Š” ๋œป)
  • ๋” ์ž์„ธํžˆ ์•Œ๊ณ  ์‹ถ์œผ์‹œ๋ฉด ์ˆ˜ํ•™ํ•˜๋Š” ๋งŒํ™”์˜ '๋„“์ด ์ด์•ผ๊ธฐ' ํŽธ์„ ์ถ”์ฒœํ•ฉ๋‹ˆ๋‹ค.
  • ์ˆ˜ํ•™์ ์ธ ์ •์˜๋ฅผ ์•Œ๊ณ  ์‹ถ์œผ์‹œ๋ฉด ์œ„ํ‚คํ”ผ๋””์•„๋ฅผ ๋ณด์‹œ๋ฉด ๋ฉ๋‹ˆ๋‹ค.
  • ์ด๋ ‡๊ฒŒ ์ƒ๊ฒผ์Šต๋‹ˆ๋‹ค. ์ข€ ๋ฌด์„ญ์Šต๋‹ˆ๋‹ค.
์•„๋ž˜ ์ฝ”๋“œ๋Š”...
  • 2 x 2 ํ–‰๋ ฌ์— ๋Œ€ํ•ด์„œ ์ด๊ฒƒ์„ ์ˆ˜์‹ ์—†์ด ๊ทธ๋ฆผ๋งŒ์œผ๋กœ ์ฆ๋ช…ํ•˜๋Š” ๊ณผ์ •์ž…๋‹ˆ๋‹ค.
  • gif ์ƒ์„ฑ์—๋Š” ScreenToGif๋ฅผ ์‚ฌ์šฉํ–ˆ์Šต๋‹ˆ๋‹ค. (gif ๋งŒ๋“ค๊ธฐ์—” ์ด๊ฒŒ ํ‚น์™•์งฑ์ธ๋“ฏ)
Determinant of 2 x 2 matrix is...
  • An area of a parallelogram defined by two row vectors.
  • Of course, same one defined by two column vectors.
  • Precisely, a signed area, which means area can be negative.
  • If two rows (or columns) are swapped, both the sign of determinant and area change.
More generally, determinant of n x n matrix is...
  • Signed area of parallelepiped defined by rows (or columns) of the matrix in n-dim space.
  • For a full understanding, a lot of concepts from abstract algebra should be brought, which I will not write here. (Cuz I don't know them.)
  • For a mathematical definition of determinant, visit wikipedia.
  • A little scary, isn't it?
The code below is...
  • A process to prove the equality of the determinant of 2 x 2 matrix and the area of parallelogram.
  • ScreenToGif is used to generate gif animation (which is, to me, the easiest way to make gif).
% ๋‘ ์  (a, b), (c, d)์˜ ์ขŒํ‘œ
a = 4;
b = 1;
c = 1;
d = 3;
% patch ์ƒ‰ pre-define
lightgreen = [144, 238, 144]/255;
lightblue = [169, 190, 228]/255;
lightorange = [247, 195, 160]/255;
% animation params.
anim_Nsteps = 30;
% create window
figure('WindowStyle','docked')
ax = axes;
ax.XAxisLocation = 'origin';
ax.YAxisLocation = 'origin';
ax.XTick = [];
ax.YTick = [];
hold on
ax.XLim = [-.4, a+c+1];
ax.YLim = [-.4, b+d+1];
% create ad-bc patch
area = patch([0, a, a+c, c], [0, b, b+d, d], lightgreen);
p_ab = plot(a, b, 'ko', 'MarkerFaceColor', 'k');
p_cd = plot(c, d, 'ko', 'MarkerFaceColor', 'k');
p_ab.UserData = text(a+0.1, b, '(a, b)', 'FontSize',16);
p_cd.UserData = text(c+0.1, d-0.2, '(c, d)', 'FontSize',16);
area.UserData = text((a+c)/2-0.5, (b+d)/2, 'ad-bc', 'FontSize', 18);
pause
%% Is this really ad-bc?
area.UserData.String = 'ad-bc...?';
pause
%% fade out ad-bc
fadeinout(area, 0)
area.UserData.Visible = 'off';
pause
%% fade in ad block
rect_ad = patch([0, a, a, 0], [0, 0, d, d], lightblue, 'EdgeAlpha', 0, 'FaceAlpha', 0);
uistack(rect_ad, 'bottom');
fadeinout(rect_ad, 1, t_pause=0.003)
draw_gridline(rect_ad, ["23", "34"])
rect_ad.UserData = text(mean(rect_ad.XData), mean(rect_ad.YData), 'ad', 'FontSize', 20, 'HorizontalAlignment', 'center');
pause
%% fade-in bc block
rect_bc = patch([0, c, c, 0], [0, 0, b, b], lightorange, 'EdgeAlpha', 0, 'FaceAlpha', 0);
fadeinout(rect_bc, 1, t_pause=0.0035)
draw_gridline(rect_bc, ["23", "34"])
rect_bc.UserData = text(b/2, c/2, 'bc', 'FontSize', 20, 'HorizontalAlignment', 'center');
pause
%% slide ad block
patch_slide(rect_ad, ...
[0, 0, 0, 0], [0, b, b, 0], t_pause=0.004)
draw_gridline(rect_ad, ["12", "34"])
pause
%% slide ad block
patch_slide(rect_ad, ...
[0, 0, d/(d/c-b/a), d/(d/c-b/a)],...
[0, 0, b/a*d/(d/c-b/a), b/a*d/(d/c-b/a)], t_pause=0.004)
draw_gridline(rect_ad, ["14", "23"])
pause
%% slide bc block
uistack(p_cd, 'top')
patch_slide(rect_bc, ...
[0, 0, 0, 0], [d, d, d, d], t_pause=0.004)
draw_gridline(rect_bc, "34")
pause
%% slide bc block
patch_slide(rect_bc, ...
[0, 0, a, a], [0, 0, 0, 0], t_pause=0.004)
draw_gridline(rect_bc, "23")
pause
%% slide bc block
patch_slide(rect_bc, ...
[d/(d/c-b/a), 0, 0, d/(d/c-b/a)], ...
[b/a*d/(d/c-b/a), 0, 0, b/a*d/(d/c-b/a)], t_pause=0.004)
pause
%% finalize: fade out ad, bc, and fade in ad-bc
rect_ad.UserData.Visible = 'off';
rect_bc.UserData.Visible = 'off';
fadeinout([rect_ad, rect_bc, area], [0, 0, 1])
area.UserData.String = 'ad-bc';
area.UserData.Visible = 'on';
%% functions
function fadeinout(objs, inout, options)
arguments
objs
inout % 1์ด๋ฉด fade-in, 0์ด๋ฉด fade-out
options.anim_Nsteps = 30
options.t_pause = 0.003
end
for alpha = linspace(0, 1, options.anim_Nsteps)
for i = 1:length(objs)
switch objs(i).Type
case 'patch'
objs(i).FaceAlpha = (inout(i)==1)*alpha + (inout(i)==0)*(1-alpha);
objs(i).EdgeAlpha = (inout(i)==1)*alpha + (inout(i)==0)*(1-alpha);
case 'constantline'
objs(i).Alpha = (inout(i)==1)*alpha + (inout(i)==0)*(1-alpha);
end
pause(options.t_pause)
end
end
end
function patch_slide(obj, x_dist, y_dist, options)
arguments
obj
x_dist
y_dist
options.anim_Nsteps = 30
options.t_pause = 0.003
end
dx = x_dist/options.anim_Nsteps;
dy = y_dist/options.anim_Nsteps;
for i=1:options.anim_Nsteps
obj.XData = obj.XData + dx(:);
obj.YData = obj.YData + dy(:);
obj.UserData.Position(1) = mean(obj.XData);
obj.UserData.Position(2) = mean(obj.YData);
pause(options.t_pause)
end
end
function draw_gridline(patch, where)
ax = patch.Parent;
for i=1:length(where)
v1 = str2double(where{i}(1));
v2 = str2double(where{i}(2));
x1 = patch.XData(v1);
x2 = patch.XData(v2);
y1 = patch.YData(v1);
y2 = patch.YData(v2);
if x1==x2
xline(x1, 'k--')
else
fplot(@(x) (y2-y1)/(x2-x1)*(x-x1)+y1, [ax.XLim(1), ax.XLim(2)], 'k--')
end
end
end
Wolfgang Schwanghart
Wolfgang Schwanghart
Last activity on 16 Feb 2024

Render landscapes

I think that MATLAB's Flipbook Mini Hack had quite some inspiring entries. My work largely deals with digital elevation models (DEMs). Hence I really liked the random renderings of landscapes, in particular this one written by Tim which inspired me to adopt the code and apply to the example data that comes with my software TopoToolbox. The results and code are shown here.