Circle

      Did you come here by searching for ‘How to draw a circle in MATLAB’ then definitely at the end of the article you will draw a circle. 

 Let me first discuss how to draw a circle and then we will see about concentric circles.

The equation of the circle: (x-x1).^2 +(y-y1).^2=r^2.

where x and y are the centre of the circle and r is the radius.

Here I am not using any plot function, just the equation of the circle.

Initialize the image with 255 and find the centre of the image. The variable ‘sz’ changes the size of the image. The image is a square matrix. Then ‘rad’ contains the radius of the circle.

sz=300;

rad=100;

clear RGB

RGB(1:sz,1:sz,1:3)=255;

 I am storing the x co-ordinates in x and y co-ordinates in y.

[x y]= find(RGB==255);

‘xc’ and ‘yc’ contains the midpoint of the circle.

xc=ceil((sz+1)/2);

yc=ceil((sz+1)/2);

r=rad.^2;

 Here I am finding the points which satisfy the equation. To avoid for loop I am using the ‘find’ function.

d=find(((x-xc).^2+(y-yc).^2) <= r);

The points that are stored in d contains the index value for x and y that satisfies the equation.

        for i=1:size(d,1)

      

         RGB(x(d(i)),y(d(i)),:)=0;

      

        end

Before edge detection

           

       

   The rest of the procedure is same. I found the edge and I strengthen the edge by dilating.

B=rgb2gray(RGB);

ED=edge(B);

SE=strel('disk',1);

cir=~imdilate(ED,SE);

figure,imshow(cir);

Concentric circles:

The procedure is same but I use loop to produce ‘n’ no.of circles.

for i=1:8

    radius=(rad-10*i).^2;

    r=find(((x-xc).^2+(y-yc).^2)<=radius);

   

for j=1:size(r,1)

    if(mod(i,2)==0)

                RGB(x(r(j)),y(r(j)),:)=255;

    end

            if(mod(i,3)==1)

               RGB(x(r(j)),y(r(j)),:)=0;

            end

        

end

end

Before edge detection

final image

Check some optical Illusion concept on circle using MATLAB: https://www.imageeprocessing.com/2014/06/optical-illusion-circle.html

If you find any difficulty in the code, Mail me, I will send the code.

Rainbow

Last night I was watching dragon tales, a cartoon series. I saw a scene with rainbow and my mind started to think to implement that rainbow in MATLAB. To my surprise, what I learned in the tenth grade mathematics came in handy, the equation of a circle.  Many times our lower class basic mathematics will come throughout our life. As a result I have started to read tenth grade mathematics once again ;-)

I followed the same procedure, 'How to draw a circle'. Instead of finding the midpoint for x co-ordinate I took the image size, so that I may get a half a circle.

sz=400;

rad=200;

clear RGB

RGB(1:sz,1:sz,1:3)=255;

RGB=uint8(RGB);

[x y z]= find(RGB==255);

xc=sz;

yc=ceil((sz+1)/2);

for i=1:12

    radius=(rad-10*i).^2;

    r=find(((x-xc).^2+(y-yc).^2)<=radius);

   

for j=1:size(r,1)

    if(mod(i,2)==0)

                RGB(x(r(j)),y(r(j)),:)=255;

    end

            if(mod(i,3)==1)

               RGB(x(r(j)),y(r(j)),:)=0;

            end

        

end

end

B=rgb2gray(RGB);

D=edge(B);

SE=strel('disk',2);

D1=~imdilate(D,SE);

I colored the image based on label

[BW ,L]=bwlabel(D1,8);

    mycolor(:,1)=[128;128;255];

    mycolor(:,2)=[215;0;0];

    mycolor(:,3)=[255;128;0];

    mycolor(:,4)=[255;255;0];

    mycolor(:,5)=[0;128;64];

    mycolor(:,6)=[10;160;220];

    mycolor(:,7)=[0;0;70];

    mycolor(:,8)=[65;0;128];

    mycolor(:,9)=[128;128;255];

for n=1:9

[r c]=find(BW==n);

for i=1:size(r,1)

            RGB(r(i,1),c(i,1),:)=mycolor(:,n);

end

end

imshow(RGB);

Checkerboard

                 I came across the matlab built in function checkerboard. And I tried to implement my own code.

First I declared the size of each square and then the number of rows and columns. Then I declared two matrixes initialized with zeros and ones. Using mod 2 I changed the colors or ones and zeros alternatively.

MATLAB CODE:

%Size of the square

sz=45;

%Number of rows

xvalue=8;

%Number of columns

yvalue=8;

%Intialize matrix A with zeros and matrix B with ones

A=zeros([sz sz]);

B=ones([sz sz]);

clear C

 m=sz;

 n=1;

 num=2;

for i=1:xvalue

    n1=1;

    m1=sz;

   

    for j=1:yvalue

        if(mod(num,2)==0)

        C(n:m,n1:m1)=A;

        num=num+1;

        else

        C(n:m,n1:m1)=B;

        num=num+1;

        end

       

        m1=m1+sz;

        n1=n1+sz;

    end

    if(mod(yvalue,2)==0)

    num=num+1;

    end

    n=n+sz;

    m=m+sz;

end

imshow(C)

8X8 size:45

8X20 size:15