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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!');
Introduction
Comma-separated lists are really very simple. You use them all the time. Here is one:
a,b,c,d
That is a comma-separated list containing four variables, the variables a, b, c, and d. Every time you write a list separated by commas then you are writing a comma-separated list. Most commonly you would write a comma-separated list as inputs when calling a function:
fun(a,b,c,d)
or as arguments to the concatenation operator or cell construction operator:
[a,b,c,d]
{a,b,c,d}
or as function outputs:
[a,b,c,d] = fun();
It is very important to understand that in general a comma-separated list is NOT one variable (but it could be). However, sometimes it is useful to create a comma-separated list from one variable (or define one variable from a comma-separated list), and MATLAB has several ways of doing this from various container array types:
struct_array.field % all elements
struct_array(idx).field % selected elements
cell_array{:} % all elements
cell_array{idx} % selected elements
string_array{:} % all elements
string_array{idx} % selected elements
Note that in all cases, the comma-separated list consists of the content of the container array, not subsets (or "slices") of the container array itself (use parentheses to "slice" any array). In other words, they will be equivalent to writing this comma-separated list of the container array content:
content1, content2, content3, .. , contentN
and will return as many content arrays as the original container array has elements (or that you select using indexing, in the requested order). A comma-separated list of one element is just one array, but in general there can be any number of separate arrays in the comma-separated list (zero, one, two, three, four, or more). Here is an example showing that a comma-separated list generated from the content of a cell array is the same as a comma-separated list written explicitly:
>> C = {1,0,Inf};
>> C{:}
ans =
1
ans =
0
ans =
Inf
>> 1,0,Inf
ans =
1
ans =
0
ans =
Inf
How to Use Comma-Separated Lists
Function Inputs: Remember that every time you call a function with multiple input arguments you are using a comma-separated list:
fun(a,b,c,d)
and this is exactly why they are useful: because you can specify the arguments for a function or operator without knowing anything about the arguments (even how many there are). Using the example cell array from above:
>> vertcat(C{:})
ans =
1
0
Inf
which, as we should know by now, is exactly equivalent to writing the same comma-separated list directly into the function call:
>> vertcat(1,0,Inf)
ans =
1
0
Inf
How can we use this? Commonly these are used to generate vectors of values from a structure or cell array, e.g. to concatenate the filenames which are in the output structure of dir:
S = dir(..);
F = {S.name}
which is simply equivalent to
F = {S(1).name, S(2).name, S(3).name, .. , S(end).name}
Or, consider a function with multiple optional input arguments:
opt = {'HeaderLines',2, 'Delimiter',',', 'CollectOutputs',true);
fid = fopen(..);
C = textscan(fid,'%f%f',opt{:});
fclose(fid);
Note how we can pass the optional arguments as a comma-separated list. Remember how a comma-separated list is equivalent to writing var1,var2,var3,..., then the above example is really just this:
C = textscan(fid,'%f%f', 'HeaderLines',2, 'Delimiter',',', 'CollectOutputs',true)
with the added advantage that we can specify all of the optional arguments elsewhere and handle them as one cell array (e.g. as a function input, or at the top of the file). Or we could select which options we want simply by using indexing on that cell array. Note that varargin and varargout can also be useful here.
Function Outputs: In much the same way that the input arguments can be specified, so can an arbitrary number of output arguments. This is commonly used for functions which return a variable number of output arguments, specifically ind2sub and gradient and ndgrid. For example we can easily get all outputs of ndgrid, for any number of inputs (in this example three inputs and three outputs, determined by the number of elements in the cell array):
C = {1:3,4:7,8:9};
[C{:}] = ndgrid(C{:});
which is thus equivalent to:
[C{1},C{2},C{3}] = ndgrid(C{1},C{2},C{3});
Further Topics:
MATLAB documentation:
Click on these links to jump to relevant comments below:
Dynamic Indexing (indexing into arrays with arbitrary numbers of dimensions)
Summary
Just remember that in general a comma-separated list is not one variable (although they can be), and that they are exactly what they say: a list (of arrays) separated with commas. You use them all the time without even realizing it, every time you write this:
fun(a,b,c,d)
The line integral 
, where C is the boundary of the square 
oriented counterclockwise, can be evaluated in two ways:
, 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 Green’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:
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:- Use Symbolic Variables and Functions: Define your variables symbolically, including the parameters of your spherical coordinates θ and ϕ and the radius r . This allows MATLAB to handle the expressions symbolically, making it easier to manipulate and integrate them.
- 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 , θ and ϕ directly.
- 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.
Here’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
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:
- 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.
- 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.
- 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.
- 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.
- Advanced Features: MATLAB’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, MATLAB’s mathematical functions can handle a wide range of special cases and data types, providing more flexibility than custom implementations.
- 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.
The community is very helpful, yet I feel really powerless that I cannot find the appropriate way to code, nor find the problems with the codes I have written. I have read numerous books on MATLAB, mostly related with science and engineering applications. Any advice to improve would be greatly appreciated. Thanks.
Happy Pi Day!
3.14 π Day has arrived, and this post provides some very cool pi implementations and complete MATLAB code.
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:
% 获取pi前1500位小数
Pi=getPi(1500);
% 统计各个数字出现次数
numNum=find([diff(sort(Pi)),1]);
numNum=[numNum(1),diff(numNum)];
% 配色列表
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;
% 绘图并修饰
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
% 绘制图例并修饰
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:
% 获取pi前1500位小数
Pi=getPi(1500);
% 计算比例变化
Ratio=cumsum(Pi==(0:9)',2);
Ratio=Ratio./sum(Ratio);
D=1:length(Ratio);
% 配色列表
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
% 循环绘图
for i=1:10
plot(D(20:end),Ratio(i,20:end),'Color',[CM(i,:),.6],'LineWidth',1.8)
end
% 坐标区域修饰
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;
% 绘制图例并修饰
lgdHdl=legend(num2cell('0123456789'));
lgdHdl.NumColumns=5;
lgdHdl.FontWeight='bold';
lgdHdl.FontSize=11;
lgdHdl.TextColor=[.5,.5,.5];
3 stacked area diagram
% 获取pi前500位小数
Pi=getPi(500);
% 计算比例变化
Ratio=cumsum(Pi==(0:9)',2);
Ratio=Ratio./sum(Ratio);
% 配色列表
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;
% 绘制堆叠面积图
hold on
areaHdl=area(Ratio');
for i=1:10
areaHdl(i).FaceColor=CM(i,:);
areaHdl(i).FaceAlpha=.9;
end
% 图窗和坐标区域修饰
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 — 500 digits';
ax.Title.FontSize=14;
% 绘制图例并修饰
lgdHdl=legend(num2cell('0123456789'));
lgdHdl.NumColumns=5;
lgdHdl.FontSize=11;
lgdHdl.Location='southeast';
4 connected stacked bar chart
% 获取pi前100位小数
Pi=getPi(100);
% 计算比例变化
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
% 以下是生成连接的部分
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
% 图窗和坐标区域修饰
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 — 10-80 digits';
ax.Title.FontSize=14;
% 绘制图例并修饰
lgdHdl=legend(barHdl,num2cell('0123456789'));
lgdHdl.NumColumns=5;
lgdHdl.FontSize=11;
lgdHdl.Location='southeast';
5 bichord chart
Need to use this tool:
% 构建连接矩阵
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();
% 添加刻度
BCC.tickState('on')
% 修改字体，字号及颜色
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 
, 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;
% 基础配置
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;
% 初始数据
Pi=Pi(:);
N=Pi;
X=cos(T);Y=sin(T);
VX=T.*0;VY=T.*0;
PX=X;PY=Y;
% 未碰撞时初始质量
getM=@(x)(x+1).^K;
M=getM(N);
% 绘制初始圆圈
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
% 计算加速度
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;
% 计算速度及新位置
VX=VX+AX;X=X+VX;PX=[PX,X];
VY=VY+AY;Y=Y+VY;PY=[PY,Y];
% 检测是否有碰撞
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)
% 碰撞的点合为一体
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);
% 删除碰撞点并绘图
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,:)=[];
% 绘制圆形
for i=1:length(XC)
fill(cx+XC(i),cy+YC(i),CM(NC(i)+1,:),'EdgeColor','w','LineWidth',1)
end
% 补充合体点
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')
% 图窗及坐标区域修饰
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
The digits of π are shown as a forest. Each tree in the forest represents the digits of π 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 π 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 — 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 — 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 — 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 — the flower needs space!
function PiTree(X,pos,D)
lw=2;
theta=pi/2+(rand(1)-.5).*pi./12;
% 树叶及花朵颜色
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];
% 计算层级
Layer=0;
for i=1:length(X)
Layer=[Layer,ones(1,X(i)).*i];
end
% 计算树枝
if D
for i=1:length(X)-2
if X(i)==0 % 若数值为0则不长树枝
newSet=endSet(1,:);
elseif X(i)==1 % 若数值为1则一长一短两个树枝
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 % 其他情况数值为几长几个树枝
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
% 绘制树枝
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
% 计算叶子和花朵位置
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);
% 绘制树叶
for i=1:length(leafInd)
scatter(FLSet(leafInd(i),1),FLSet(leafInd(i),2),70,'filled','CData',CM(X(i)+1,:))
end
% 绘制花朵
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;
% 获取pi前n位小数
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')
% 图窗和坐标区域修饰
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];
% 绘制图例
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)];
% 配色数据
CM=[248,65,69;246,152,36;249,198,81;67,170,139;87,118,146]./255;
% 绘制圆圈
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 — 400 digits','Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','center','FontSize',25,'FontAngle','italic')
% 图窗和坐标区域修饰
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
% 配色数据
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;
% 绘制圆圈
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
% 绘制图例
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')
% 图窗和坐标区域修饰
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
% 循环绘制字母
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
% 图窗及坐标区域修饰
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);
% 配色列表
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;
% 绘制圆圈
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')
% 图窗和坐标区域修饰
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);
% 配色列表
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;
% 绘制圆圈
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';'—— 500 digits'},...
'Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','right','FontSize',25,'FontAngle','italic')
% 图窗和坐标区域修饰
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)];
% 配色数据
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);
% 绘制圆圈
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';'—— 1200 digits'},...
'Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','right','FontSize',23,'FontAngle','italic')
% 图窗和坐标区域修饰
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
% 构建连接矩阵
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
% 配色列表
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)';
% 循环绘图
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);
% 绘制连线
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
% 绘制圆点
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 — 400 digits','Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','center','FontSize',25,'FontAngle','italic')
% 图窗和坐标区域修饰
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:
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:
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));
code is here
You can also see the animated version of the competition here
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
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:
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)
%曲面数据计算
%==========================================================================
[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));
%颜色映射表
%==========================================================================
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
%曲面旋转及绘制
%==========================================================================
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
%axes属性调整
%==========================================================================
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];
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:
Also we will use the following parametric representation of the rose (rhodonea) curve:
This curve was named by the mathematician Guido Grandi.
Let us combine them in one curve and add possible shifts:
Now if we would like to create an ellipse, we will set 
and 
. If we would like to create a rose, we will set 
and 
. If we would like to shift our curve, we will set 
and 
to the required values. Of course, we could set all non-zero parameters to combine both chosen curves and use the shifts.
and . If we would like to create a rose, we will set and . If we would like to shift our curve, we will set and 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 
, with the step 0.1. The result of this code is the following ellipse curve with 
and 
.
, with the step 0.1. The result of this code is the following ellipse curve with and .
This ellipse is described by the following parametric equations:
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 
and 
.
and .
This rose is described by the following parametric equations:
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 
, 
, 
, 
, 
, and 
are written in the different submatrices of the matrix P. The code generates the following curves to illustrate the different parts of our sheep.
, , , , , and 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.
The following submatrix of the matrix P represents its parameters.
The parametric equations of the head are the following:
The following ellipses describe the eyes of the sheep.
The following submatrices of the matrix P represent their parameters.
The parametric equations of the left and right eyes correspondingly are the following:
The following ellipses describe the hoofs of the sheep.
The following submatrices of the matrix P represent their parameters.
The parametric equations of the right front, left front, right hind, and left hind hoofs correspondingly are the following:
The following ellipse combined with the rose describes the crown of the sheep.
The following submatrix of the matrix P represents its parameters.
The parametric equations of the crown are the following:
The following ellipse combined with the rose describes the body of the sheep.
The following submatrix of the matrix P represents its parameters.
The parametric equations of the body are the following:
The following ellipse combined with the rose describes the tail of the sheep.
The following submatrix of the matrix P represents its parameters.
The parametric equations of the tail are the following:
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!
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!
To enlarge an array with more rows and/or columns, you can set the lower right index to zero. This will pad the matrix with zeros.
m = rand(2, 3) % Initial matrix is 2 rows by 3 columns
mCopy = m;
% Now make it 2 rows by 5 columns
m(2, 5) = 0
m = mCopy; % Go back to original matrix.
% Now make it 3 rows by 3 columns
m(3, 3) = 0
m = mCopy; % Go back to original matrix.
% Now make it 3 rows by 7 columns
m(3, 7) = 0
MATLAB O/X Quiz
Answer BEFORE Googling!
- An infinite loop can be made using "for".
- "A == A" is always true.
- "round(2.5)" is 3.
- "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!
The MATLAB command window isn't just for commands and outputs—it can also host interactive hyperlinks. These can serve as powerful shortcuts, enhancing the feedback you provide during code execution. Here are some hyperlinks I frequently use in fprintf statements, warnings, or error messages.
1. Open a website.
msg = "Could not download data from website.";
url = "https://blogs.mathworks.com/graphics-and-apps/";
hypertext = "Go to website"
fprintf(1,'%s <a href="matlab: web(''%s'') ">%s</a>\n',msg,url,hypertext);
Could not download data from website. Go to website
2. Open a folder in file explorer (Windows)
msg = "File saved to current directory.";
directory = cd();
hypertext = "[Open directory]";
fprintf(1,'%s <a href="matlab: winopen(''%s'') ">%s</a>\n',msg,directory,hypertext)
File saved to current directory. [Open directory]
3. Open a document (Windows)
msg = "Created database.csv.";
filepath = fullfile(cd,'database.csv');
hypertext = "[Open file]";
fprintf(1,'%s <a href="matlab: winopen(''%s'') ">%s</a>\n',msg,filepath,hypertext)
Created database.csv. [Open file]
4. Open an m-file and go to a specific line
msg = 'Go to';
file = 'streamline.m';
line = 51;
fprintf(1,'%s <a href="matlab: matlab.desktop.editor.openAndGoToLine(which(''%s''), %d); ">%s line %d</a>', msg, file, line, file, line);
Go to streamline.m line 51
5. Display more text
msg = 'Incomplete data detected.';
extendedInfo = '\tFilename: m32c4r28\n\tDate: 12/20/2014\n\tElectrode: (3,7)\n\tDepth: ???\n';
hypertext = '[Click for more info]';
warning('%s <a href="matlab: fprintf(''%s'') ">%s</a>', msg,extendedInfo,hypertext);
<click>
- Filename: m32c4r28
- Date: 12/20/2014
- Electrode: (3,7)
- Depth: ???
6. Run a function
Similarly, you can also add hyperlinks in figures and apps
I would tell myself to understand vectorization. MATLAB is designed for operating on whole arrays and matrices at once. This is often more efficient than using loops.
and immeditaely everyone wanted the code! It turns out that this is the result of my remix of @Zhaoxu Liu / slandarer's entry on the MATLAB Flipbook Mini Hack.
I pointed people to the Flipbook entry but, of course, that just gave the code to render a single frame and people wanted the full code to render the animated gif. That way, they could make personalised versions
I just published a blog post that gives the code used by the team behind the Mini Hack to produce the animated .gifs https://blogs.mathworks.com/matlab/2024/02/16/producing-animated-gifs-from-matlab-flipbook-mini-hack-entries/
Thanks again to @Zhaoxu Liu / slandarer for a great entry that seems like it will live for a long time :)
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.
I found this list on Book Authority about the top MATLAB books: https://bookauthority.org/books/best-matlab-books
My favorite book is Accelerating MATLAB Performance - 1001 tips to speed up MATLAB programs. I always pick something up from the book that helps me out.
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