Results for
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));
s = ['M','A','T','L','A','B']
9%
char([77,65,84,76,65,66])
7%
"MAT" + "LAB"
21%
upper(char('matlab' - '0' + 48))
17%
fliplr("BALTAM")
17%
rot90(rot90('BALTAM'))
30%
2929 votes
I've read thru the API refernce doc and the information in the right-hand side panel on "API Keys" tab and so I 'm pretty sure the answer is "no" but throught it was worth double-checking.
My application for this is that I have several devices, each with their own channel. For security, I'd like them to be able to deduce or calculate the write API key to their channel based on their device serial number (which each device knows) so that I don't have to store the API key in their firmware.
thanks...
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];
Hello everyone,
I am currently working on a project to simulate an autonomous energy production plant integrating renewable sources and a hydroelectric generator using MATLAB.
If anyone has experience in this area or can provide advice I would be extremely grateful. Thank you in advance for your precious help !
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!
The File Exchange team is thrilled to introduce a more streamlined approach to working with GitHub and File Exchange - the MATLAB and Simulink Integration for GitHub!
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Starting April 16, 2024, your File Exchange submissions will no longer update automatically unless you take the following steps:
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4. Return to the My File Exchange page and verify the installation.
If you prefer your File Exchange submission not to update automatically from GitHub, no action is required. Users will still be able to find and download your submissions. However, to release a new version of your code, you must either install the GitHub App or disconnect from GitHub and manually upload new versions of your code.
Should you have any questions or encounter issues with the App, please feel free to comment on this post!
Hello, I am using an Arduino UNO R3 with an ESP8266 WiFi Shield to connect 2 DHT-11 sensors to upload their data to ThingSpeak. Sometimes ThingSpeak would only post data on a certain sensor for a bit and then upload to the other randomly. I want ThingSpeak to pick up the sensors within 30-40 seconds of each other consistently. I believe there might be something wrong with my code. I was hoping to get some help on it. Anything is appreciated.
Below is provided code
SoftwareSerial esp8266(RX,TX);
void setup() {
Serial.begin(9600);
esp8266.begin(38400);
sendCommand("AT",5,"OK");
sendCommand("AT+CWMODE=1",5,"OK");
sendCommand("AT+CWJAP=\""+ AP +"\",\""+ PASS +"\"",20,"OK");
}
void loop() {
String getData = "GET /update?api_key="+ API +"&field1="+getTemperatureValue()+"&field2="+getHumidityValue();
sendCommand("AT+CIPMUX=1",5,"OK");
sendCommand("AT+CIPSTART=0,\"TCP\",\""+ HOST +"\","+ PORT,15,"OK");
sendCommand("AT+CIPSEND=0," +String(getData.length()+4),4,">");
esp8266.println(getData);delay(6000);countTrueCommand++;
String getData2 = "GET /update?api_key="+ API +"&field3="+getTemperatureValue2()+"&field4="+getHumidityValue2();
sendCommand("AT+CIPMUX=1",5,"OK");
sendCommand("AT+CIPSTART=0,\"TCP\",\""+ HOST +"\","+ PORT,15,"OK");
sendCommand("AT+CIPSEND=0," +String(getData2.length()+4),4,">");
esp8266.println(getData2);delay(6000);countTrueCommand++;
sendCommand("AT+CIPCLOSE=0",5,"OK");
}
String getTemperatureValue(){
dhtObject.read(dht_apin);
Serial.print(" Temperature(C)= ");
int temp = dhtObject.temperature;
Serial.println(temp);
delay(50);
return String(temp);
}
String getHumidityValue(){
dhtObject.read(dht_apin);
Serial.print(" Humidity in %= ");
int humidity = dhtObject.humidity;
Serial.println(humidity);
delay(50);
return String(humidity);
}
String getTemperatureValue2(){
dhtObject2.read(dht_bpin);
Serial.print(" Temperature(C)= ");
int temp2 = dhtObject2.temperature;
Serial.println(temp2);
delay(50);
return String(temp2);
}
String getHumidityValue2(){
dhtObject2.read(dht_bpin);
Serial.print(" Humidity in %= ");
int humidity2 = dhtObject2.humidity;
Serial.println(humidity2);
delay(50);
return String(humidity2);
}
void sendCommand(String command, int maxTime, char readReplay[]) {
Serial.print(countTrueCommand);
Serial.print(". at command => ");
Serial.print(command);
Serial.print(" ");
while(countTimeCommand < (maxTime*1))
{
esp8266.println(command);//at+cipsend
if(esp8266.find(readReplay))//ok
{
found = true;
break;
}
countTimeCommand++;
}
if(found == true)
{
Serial.println("Pass");
countTrueCommand++;
countTimeCommand = 0;
}
if(found == false)
{
Serial.println("Fail");
countTrueCommand = 0;
countTimeCommand = 0;
}
found = false;
}
Due to temporary problem MathWorks account was unable to login. Kindly resolve this issue.
how to generate CA signed certificate for mqtt
I asked my question in the general forum and a few minutes later it was deleted. Perhaps this is a better place?
Rather than using my German regional forum (as I do not speak German), I want to ask questions in an international English-speaking forum. Presumably there should be an international English forum for everyone around the world, as English is the first or second language of everyone who has gone to school. Where is it?
Big congratulations to @VBBV for achieving the remarkable milestone of 3,000 reputation points, earning the prestigious title of Editor within our community.
This achievement is a testament to @VBBV's exceptional contributions and steadfast commitment to the community. These efforts have also been endorsed by fellow top contributors, underscoring the value and impact of @VBBV's expertise.
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eye(3) - diag(ones(1,3))
11%
0 ./ ones(3)
9%
cos(repmat(pi/2, [3,3]))
16%
zeros(3)
20%
A(3, 3) = 0
32%
mtimes([1;1;0], [0,0,0])
12%
3009 votes
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.
When I embed Matlab windows into C#, I use C# to call the mouse functions such as drawline() and getpts() which are encapsulated in Matlab's dll, and then the program crashes, how can I solve this problem?
ps : The functions that I call in C# can be executed in Matlab.
ps : If I don't embed the window into C#, but just let C# call the Matlab function, it can execute the mouse control normally in the window, but once embedded into the C# window, it will crash when execute the mouse command.
MATLAB Support Package for Quantum Computing lets you build, simulate, and run quantum algorithms.
Check out the Cheat Sheet here!
I have a channel updating every 10 minutes. The last 8000 entries therefore go back to late December (around 55 days). I want to extend the range of data downloaded by setting a timescale between results, so say if I request a timescale of 20 minutes should I not be able to get data from twice as long ago? e.g.
https://api.thingspeak.com/channels/xxxxxx/fields/1.json?api_key=XXXXXXXXXXX×cale=20&start=2022-12-30%2022:00:00
No matter what timescale or averaging I use, the last 8000 entries are the only ones processed for the downloaded data, so I just get fewer results. Unless I use a timescale of 1440; then I can download results from the very first entry in 2022. A timescale of 720 or lower however will only go back 55 days.
So I see the desired result when I use a timescale of 1440, but no other timescale will work. Is this a bug?
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.
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 :)
