清空筛选条件
清空筛选条件

I need help with a checker board pattern

2 次查看（过去 30 天）
Rafael Perales
Rafael Perales 2016-11-21
评论： Rafael Perales 2016-11-21
I have a code that lets the user input the size of the squares, the dimension of the pattern, and the two colors. The problem I am having is that I can not figure out how to properly have each square alternate between the colors and how to properly have all the squares in the dimension.
This is the code for the plot of the checkerboard pattern that I have so far:
N=4;
S=0;
theta=zeros(1,N);
for k=1:N
theta(k)=2*pi*(k-1)/N;
end
for m=1:D
x1=(R*cos(theta+pi/4)/sqrt(2))+S;
y1=(R*sin(theta+pi/4)/sqrt(2))+S;
plot(x1,y1);
fill(x1,y1,C1);
hold;
x2=(R*cos(theta+pi/4)/sqrt(2))+S;
y2=(R*sin(theta+pi/4)/sqrt(2));
plot(x2,y2);
fill(x2,y2,C2);
hold;
x3=(R*cos(theta+pi/4)/sqrt(2));
y3=(R*sin(theta+pi/4)/sqrt(2))+S;
plot(x3,y3);
fill(x3,y3,C2);
hold;
S=S+R;
end
R is the Size of the square
D is the Dimension Size
C1 and C2 are the colors
How would I go about fixing this code to make sure that I can satisfy all requirements?
  2 个评论
John D'Errico
John D'Errico 2016-11-21
Why are you working in polar coordinates to create a checkerboard?
Rafael Perales
Rafael Perales 2016-11-21
Because one of the requirements is to inscribe the square into a circle than to rotate by a phase to make a pattern out of those squares

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Image Analyst
Image Analyst 2016-11-21
Why not use the checkerboard() function?
  5 个评论
显示 3更早的评论
Image Analyst
Image Analyst 2016-11-21
checkerboard() creates an image. In my mind, and I think most people's minds, when we think of a checkerboard we think of an image, like if you were looking at a chessboard or checkerboard. Not sure what you mean by a plot. Can you show us what a plot might look like?
Rafael Perales
Rafael Perales 2016-11-21
Ok here is basically what I'm trying to do. Basically this is 2x2 version of what I'm trying to do.
The code I used for this is:
N=4;
R=2;
theta=zeros(1,N);
for k=1:N
theta(k)=2*pi*(k-1)/N;
end
x1=(R*cos(theta+pi/4)/sqrt(2))+R;
y1=(R*sin(theta+pi/4)/sqrt(2))+R;
plot(x1,y1);
fill(x1,y1,'y');
x2=(R*cos(theta+pi/4)/sqrt(2))+R;
y2=(R*sin(theta+pi/4)/sqrt(2));
plot(x2,y2);
fill(x2,y2,'b');
x3=(R*cos(theta+pi/4)/sqrt(2));
y3=(R*sin(theta+pi/4)/sqrt(2))+R;
plot(x3,y3);
fill(x3,y3,'b');
x4=(R*cos(theta+pi/4)/sqrt(2));
y4=(R*sin(theta+pi/4)/sqrt(2));
plot(x4,y4);
fill(x4,y4,'y');

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 General Applications 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by