# IMAGE PROCESSING

" Two roads diverged in a wood, and I,
I took the one less traveled by,
And that has made all the difference "-Robert Frost

## Recent Posts

### Optical Illusion – Circle

Let us define a circle and assign different color to each one of the four quadrants.

MATLAB CODE:

color_cir= uint8([size(nh,1) size(nh,2) 3]);
nh =uint8(nh);

%Color Blue

%color pink

%color Green

%Color Red

imshow(color_cir);

Concentric Circles:
Colored:
MATLAB CODE:
clc
clear all

color_cir =  uint8([size(nh,1) size(nh,2) 3]);
nh  = uint8(nh);

color_cir(1:size(nh,1),1:size(nh,2),1) = nh*10;
color_cir(1:size(nh,1),1:size(nh,2),2) = nh*1;
color_cir(1:size(nh,1),1:size(nh,2),3) = nh*2;

nh  = uint8(nh);

%Color Red

%Color Green

%Color Blue

%Color yellow

end

figure,imshow(color_cir);

EXPLANATION:
1.      Draw a circle of radius 600.
2.      Fill the circle with black color.
3.      Draw another circle of radius 550 and fill the first quadrant will green, second quadrant with red, third quadrant with blue and the fourth quadrant with yellow color.
4.      Place the smaller circle (radius 550) on the bigger circle (radius 600).
5.      Draw another circle of radius 500 and repeat the same process till the radius of the smaller circle is 50.

Black and White:
MATLAB CODE:
clc
clear all

color_cir =  double([size(nh,1) size(nh,2) 3]);
nh  = double(nh);

color_cir(1:size(nh,1),1:size(nh,2),1) = nh*1;
color_cir(1:size(nh,1),1:size(nh,2),2) = nh*1;
color_cir(1:size(nh,1),1:size(nh,2),3) = nh*1;

c1 = 0;
c2 = 255;

nh  = double(nh);

if(c2 > 1)
c1 = 255;
c2 = -255;
else
c1 = -255;
c2 = 255;
end

end

color_cir = uint8(color_cir);
figure,imshow(color_cir);

EXPLANATION:
1.      Draw a circle of radius 1250.
2.      Fill it with black color.
3.      Draw another circle with radius 1200.
4.      Fill second and fourth quadrant of the smaller circle with white color.
5.      Fill First and Third quadrant of the smaller circle with black color.
6.      Place the smaller circle on the bigger circle.
7.      Define another circle of radius 1150.
8.      Fill second and fourth quadrant of the smaller circle with black color.
9.      Fill First and Third quadrant of the smaller circle with white color.
10.  Now place this circle on the bigger one. Repeat this process with smaller circles and by interchanging black and white color on the quadrants for each circle.

Concentric circle - 2:

MATLAB CODE:

%Define the matrix
[x,y] = meshgrid(-600:600);

%Preallocate
ccircle = zeros(size(x));
for i = 500:-10:1

Tmp = x.^2 + y.^2 <= i*i;

%Find the gradient of the image

ccircle = ccircle + Gmag;
end

figure,imshow(ccircle);

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### Some Random effects

• Negative Image
Subtract 255 from the RGB image, to obtain negative image.
MATLAB CODE:
%Here the  I am subtracting 255 from R,G and B.
imshow(255-A);

 Original Image
 Negative Image

• Sine Wave

First find the mid point of the RGB image.
Convert the points from Cartesian to polar co_ordinates
Find the sine value for the angle 'theta'
Again convert the points to Cartesian co_ordinates

SAMPLE CODE:
x2=zeros([size(A,1) size(A,2)]);
y2=zeros([size(A,1) size(A,2)]);
for i=1:size(A,1)
x=i-midx;
for j=1:size(A,2)
%Cartesian to Polar co-ordinates
[theta1,rho1]=cart2pol(i,j-midy);

phi=sin(theta1);

%Polar to Cartesian co-ordinates
[l,m]=pol2cart(phi,rho1);
x2(i,j)=ceil(l)+midx;
y2(i,j)=ceil(m)+midy;

end
end

 Sine wave

 Original Image

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### Tiling effect in MATLAB

Tiling effect

 original Image
Let’s make some random rectangles from the image.

MATLAB CODE:

C=uint8(zeros(size(A)));
i=1; j=1;
col=0;
row=0;
m=1;n=1;
%change the rsz and csz value to obtain pieces in different sizes
rsz=15;
csz=36;

%change the tsz_row and tsz_col values to make some gaps between the pieces.

tsz_row=3;
tsz_col=5;

while ( i < size(A,1))
%check whether i is less than size of (A,1)

while(m+i+row < size(A,1)&& n+j+col < size(A,2))

%size of the tiles varies depending upon the random number
C(m+i:m+i+row,n+j:n+j+col,:)=A(i:i+row,j:j+col,:);

m=ceil(rand(1)*tsz_row);
n=ceil(rand(1)*tsz_col);
col=ceil(rand(1)*csz);
row=ceil(rand(1)*rsz);
j=j+col;

end
i=i+row;
j=1;
end
figure,imshow(C);
 With small gaps

 With Large gaps
Subscribe to 'Image processing' to get the complete code.

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### Cone effect in MATLAB

The image shrinks towards the center, forming a cone effect. Here, I first found the mid point of the image, since the image expands from the center. Then the angle and the radius of the points are found by converting into polar co-ordinates from Cartesian.  The square root (radius * K) is obtained, where K  can be changed to make the image size differ.
Again convert from polar to Cartesian and display the final image.
For more fun, try this effect with your own photography image .

MATLAB CODE:

B=uint8(zeros(size(A)));
figure,imshow(A);

%FIND THE MID VALUES
midx=ceil((size(A,1)+1)/2);
midy=ceil((size(A,2)+1)/2);

%CHANGE THE VALUE OF 'K'
K=180;
x2=zeros([size(A,1) size(A,2)]);
y2=zeros([size(A,1) size(A,2)]);

%COMPUTATION TO GET CONE EFFECT
for i=1:size(A,1)
x=i-midx;
for j=1:size(A,2)
[theta,rho]=cart2pol(x,j-midy);
sqtrho=sqrt(rho*K);
[l,m]=pol2cart(theta,sqtrho);
x2(i,j)=ceil(l)+midx;
y2(i,j)=ceil(m)+midy;
end
end

% IF THE ARRAY CONTAINS VALUES LESS THAN 1 REPLACE IT WITH 1 AND
% IF GREATER THAN 1 REPLACE WITH SIZE OF THE ARRAY
x2(x2<1)=1;
x2(x2>size(A,1))=size(A,1);

y2(y2<1)=1;
y2(y2>size(A,2))=size(A,2);

for i=1:size(A,1)
for j=1:size(A,2)
B(i,j,:)=A(x2(i,j),y2(i,j),:);
end
end

figure,   imshow(B);

 K=100
 K=180

SEE ALSO : Oil Painting  Swirl effect   Glassy effect   Tiling effect   Paint application
Reference:

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### Glassy effect in MATLAB

To obtain a glassy effect, define a window of size m by n. Obtain a random pixel value from the window matrix and add it with the current position value.

MATLAB CODE:
%WINDOW SIZE
m=6;
n=7;
Image=uint8(zeros([size(A,1)-m,size(A,2)-n,3]));

for i=1:size(A,1)-m
for j=1:size(A,2)-n

%Select a pixel value from the neighborhood.
x2=ceil(rand(1)*m);
y2=ceil(rand(1)*n);

end
end

figure,imshow(Image);
We can change the matrix size and see the difference in the result.

SEE ALSO : Oil Painting  Swirl effect   Cone effect     Tiling effect   Paint application

Reference:
FOR FUN:
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### Oil Painting in MATLAB

To obtain a painting-like effect , define a small window matrix of size m by n. Copy the original image pixel values into the matrix and find the histogram of each value. Find the maximum occurring pixel value and replace the current position with the maximum occurrence value.

 Original Image

 Oil Paint Effect

MATLAB CODE:

%Define the matrix size of  your convience.
m=5;
n=6;
Image=uint8(zeros([size(A,1)-m,size(A,2)-n,3]));
%Calculate the histogram for each RGB value.
for v=1:3

for i=1:size(A,1)-m
for j=1:size(A,2)-n
h=zeros(1,256);
for x=1:(m*n)
end
%Maximum occurring value and the position is obtained
[maxvalue,pos]=max(h);
Image(i,j,v)=pos-1;
end
end
end
figure,imshow(Image);

SEE ALSO : Swirl effect  Cone effect    Glassy effect   Tiling effect   Paint application(coloring)
Reference:
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### Swirl effect in MATLAB

After reading a couple of image processing books, I tried to implement some Photoshop effects in MATLAB.  First, I tried to implement the twirl effect in MATLAB.

The mid point of the image is stored in the variables ‘midx’ and ‘midy’.
The angle and the radius for each point in the image are found by converting the co-ordinates from Cartesian to polar co-ordinates.

The effect can be obtained by using the following formula:
new[rho , theta] = old[rho , theta + rho /K] .

By changing the value of K, the swirl can be increased or decreased.

For more fun, try this effect with your own photography image.

Additional Information: Learn to make your own Photoshop effects so that you can code your own effects.

For beginners in photoshop: Read photoshop for dummies .

MATLAB CODE:

B=uint8(zeros(size(A)));
figure,imshow(A);

%Mid point of the image
midx=ceil((size(A,1)+1)/2);
midy=ceil((size(A,2)+1)/2);
 K=150
K=100;
x2=zeros([size(A,1) size(A,2)]);
y2=zeros([size(A,1) size(A,2)]);
for i=1:size(A,1)
x=i-midx-K;
for j=1:size(A,2)
%Cartesian to Polar co-ordinates
[theta1,rho1]=cart2pol(x,j-midy+K);

phi=theta1+(rho1/K);

%Polar to Cartesian co-ordinates
[l,m]=pol2cart(phi,rho1);
x2(i,j)=ceil(l)+midx;
y2(i,j)=ceil(m)+midy;

end
end

%The result may produce value lesser than 1 or greater than the image size.

x2=max(x2,1);
x2=min(x2,size(A,1));

y2=max(y2,1);
y2=min(y2,size(A,2));

for i=1:size(A,1)
for j=1:size(A,2)
B(i,j,:)=A(x2(i,j),y2(i,j),:);
end
end

figure,   imshow(B);
 K=100

Reference:

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