what I know about IMAGE Processing

Contents by Date

2009-04-13T18:46:00.002+08:00

Do find the following links (on the right of the subjects) for compatible SCILAB code (last update - 13 April 2009). Examples coverted are highlighted in GREEN.

Video Processing: From Offline to Online

Download Examples by Author

Maple for Image Processing (II)

Measuring Object Length (with a Reference Object)

Character Recognition Example: An Explanation On the Simple Concept Used in This Demo

Hough Transform For Hyperbola Detection - with known eccentricity

Hough Transform For Circle Detection - with unknown radius (II) (compatible code in SCILAB)

Fill The Area Between 2 Lines

Converting Pixels to Line

Detecting Objects' Motion in 2 Subsequence Images

Hough Transform For Circle Detection - with unknown radius (I) (compatible code in SCILAB)

Hough Transform For Circle Detection - with known radius (II) (compatible code in SCILAB)

Hough Transform For Circle Detection - with known radius (I) (compatible code in SCILAB)

Simple Code for Hough Transform (Coming Next)

Signature of Binary Objects

Object Detection using Hough Transform (II)

Object Detection using Hough Transform (I)

Maple for Image Processing?

Which Software to Use?

Drawing Circles Around Round Object

Drawing Transparent Object

Detecting Objects' Shape(III) : Triangle Object

Detecting Objects' Shape(II) : Square Object

Detecting Objects' Shape(I) : Round Object

Playing with “Edge Detection”

Panoramic Picture Creation

Character Recognition Example (III):Training a Simple NN for classification (Coming Soon)

Character Recognition Example (II):Automating Image Pre-processing (Coming Soon)

Character Recognition Example (I): Image Pre-processing (Coming Soon)

Detecting Object in an Image

Drawing Shapes by Overwriting Pixel Value

What is Image Compression?

Useful Resources:-

Useful software could be found from the link at the sidebar

Useful keyword to seacrh for this blogs: image processing, image analysis, edge detection, matlab, maple, mathematica, mathcad

Migrating to SCILAB

2009-04-03T13:38:00.002+08:00

Hi all,

I'm currently migrating to SCILAB for image processing, stay tuned at http://scilab-imageprocessing.blogspot.com/ for the SCILAB version of this page.

CL

Video Processing: From Offline to Online

2009-02-24T16:06:00.003+08:00

I’ve chance to use LabVIEW for image and video processing, while having fear in converting from the software I used to, I found a lot of fun and excitement playing around with the LabVIEW Vision Development Module and Vision Assistant.
Here how it goes:
1. Using LabVIEW Vision for offline Image Processing

2. Having a little bit modification, it could be turned to real-time using webcam.

Required Components: LabVIEW, Vision Module, USB WebCam

Contents by Categories

2007-05-23T23:38:00.003+08:00

Most of the examples in this blog using Matlab as the programming tool for this moment, however, other software examples will be uploaded in the future.

The following list is the full content by categories, and it will be updated everytime a new post is posted. The full content sorted by date could be found by clicking on the link at the side panel of this page or at the bottom of every page.

p/s : Matlab is the registered trademark of The Mathworks Inc. The link to the page could be found at the side panel as well.

LabVIEW Examples:-

Video Processing: From Offline to Online

Downloads:-

Download Examples by Author

General:-

Maple for Image Processing (II)

Maple for Image Processing?

Which Software to Use?

What is Image Compression?

Adding Object on Images:-

Fill The Area Between 2 Lines

Converting Pixels to Line

Drawing Circles Around Round Object

Drawing Transparent Object

Drawing Shapes by Overwriting Pixel Value

Object Detection:-

Measuring Object Length (with a Reference Object)

Hough Transform For Hyperbola Detection - with known eccentricity

Hough Transform For Circle Detection - with unknown radius (II)

Detecting Objects' Motion in 2 Subsequence Images

Hough Transform For Circle Detection - with unknown radius (I)

Hough Transform For Circle Detection - with known radius (II)

Hough Transform For Circle Detection - with known radius (I)

Simple Code for Hough Transform

Signature of Binary Objects

Object Detection using Hough Transform (II)

Object Detection using Hough Transform (I)

Detecting Objects' Shape(III) : Triangle Object

Detecting Objects' Shape(II) : Square Object

Detecting Objects' Shape(I) : Round Object

Detecting Object in an Image

Character Recognition:-

Character Recognition Example: An Explanation On the Simple Concept Used in This Demo

Character Recognition Example (III):Training a Simple NN for classification

Character Recognition Example (II):Automating Image Pre-processing

Character Recognition Example (I): Image Pre-processing

Other Examples:-

Playing with “Edge Detection”

Panoramic Picture Creation

Useful Resources:-

Useful software could be found from the link at the sidebar

Useful keyword to seacrh for this blogs: image processing, image analysis, edge detection, matlab, maple, mathematica, mathcad

Code Udpates/Error Corrections:-

Do notes that for all the Circle Det

Download Examples by Author

2007-05-23T21:16:00.001+08:00

For those who are not aware, most of the examples descript here had been post in MATLAB Central:

1. Speech Analysis using LPC

2. Character Recognition Example (III):Training a Simple NN for classification

3. Simple DIP demos

4. Video Surveillance Using MATLAB and Image Acquisition Toolbox

5. Icon Menu

6. Character Recognition Example (I&II): Pre-processing of Image

7. Simple Shape Detection using Hough Transform

8. DTMF generator using TI C6711 DSK

9. Using RTW and Compiler

10. Hough Transform for Circle Detection (GUI)

11. Drawing Transparent Object on Image

Maple for Image Processing (II)

2007-05-11T21:25:00.000+08:00

More Functions for Image Processing in Symbolic Tools

Found more image processing functions in Maple 11, however, doesn't have much time to explore much.

http://www.maplesoft.com/applications/app_center_view.aspx?AID=2059&CID=10&SCID=149

Well, since this is the first article after one year, at least it is a new start. :)

Measuring Object Length (with a Reference Object)

2006-05-25T20:26:00.000+08:00

It has been a long long time I've not update this blog due to my new notebook modem spoiled just after I bought it! The support staff came 3 times and now I am waiting for a replacement model...

Further more, life become more and more busy, and sometimes I really out of idea what to post here. :P

Until I receive some e-mail from some reader...

This is rather a simple example to measure a length of an object in an image. We could easily get the reading in pixels, and with a reference, we could just use a simple mathematic to compute the lenght in some standard unit.

Let's see this simple example:

1. Reading image and show the true color image.

clear all;clc;
I = imread('pic29.tif');
imshow (I);

2. Assuming that we know the length of the red object (as reference) which is 5 cm, locate the red object with simple command and find the length in pixels:

Ired_labeled = bwlabel(Ired);
Ired_props = regionprops(Ired_labeled);
Ired_length_in_pixel = Ired_props.BoundingBox(3);
disp(Ired_length_in_pixel);

212

3. To measure the length of the green object, extract the object:

Igreen_labeled = bwlabel(Igreen);
Igreen_props = regionprops(Igreen_labeled);
Igreen_length_in_pixel = Igreen_props.BoundingBox(3);
disp(Igreen_length_in_pixel);

146

4. Calculate the length of green object in cm

Igreen_length_in_cm = Igreen_length_in_pixel/Ired_length_in_pixel*5;
disp(Igreen_length_in_cm);

3.4434

Character Recognition Example: An Explanation On the Simple Concept Used in This Demo

2006-04-06T20:11:00.000+08:00

After I posted these series of examples, I’ve received a few comments from readers. Some of them using different version of software and can’t run the program successfully, some were happy with the simple code in which they had applied it to some other application, while a few readers were asking for more explanations on how these code work.

Well, I believe I’ve answered the doubt on the different version issue, but not the doubt on more details explanations. Here, I tried to draft a few paragraph to explain the concept hidden behind the code, and hope it helps to answer the readers who ask me about it though email.

1. Image Preprocessing

The image is first being converted to grayscale image follow by the threshing technique, which make the image become binary image. The binary image is then go through connectivity test in order to check for the maximum connected component, which is, the box of the form. After locating the box, the individual characters are then cropped into different sub images that are the raw data for the following feature extraction routine.

The size of the sub-images are not fixed since they are expose to noises which will affect the cropping process to be vary from one to another. This will causing the input of the network become not standard and hence, prohibit the data from feeding through the network. To solve this problem, the sub-images have been resize to 50 by 70 and then by finding the average value in each 10 by 10 blocks, the image can be down to 5 by 7 matrices, with fuzzy value, and become 35 inputs for the network. However, before resize the sub-images, another process must be gone through to eliminate the white space in the boxes.

2. Feature Extraction

The sub-images have to be cropped sharp to the border of the character in order to standardize the sub-images. The image standardization is done by finding the maximum row and column with 1s and with the peak point, increase and decrease the counter until meeting the white space, or the line with all 0s. This technique is shown in figure below where a character “S” is being cropped and resize.

The image pre-processing is then followed by the image resize again to meet the network input requirement, 5 by 7 matrices, where the value of 1 will be assign to all pixel where all 10 by 10 box are filled with 1s, as shown below:

Finally, the 5 by 7 matrices is concatenated into a stream so that it can be feed into network 35 input neurons. The input of the network is actually the negative image of the figure, where the input range is 0 to 1, with 0 equal to black and 1 indicate white, while the value in between show the intensity of the relevant pixel.

3. Neural Network Training

Well, I have just met a lecturer in my country and we had a short discussion on the NN for classification purpose. Both of us agree that while using simple FF-BP-NN for classification, the more important thing is the pre-processing of the data. “Rubbish-in, rubbish-out”, always true… So if you were to use NN after this process, it should be quite straight forward after you get the features, which is 5 by 7 = 35 values.

Hough Transform For Hyperbola Detection - with known eccentricity

2006-02-21T20:59:00.000+08:00

More Hough Transform examples!

In last few examples, the Hough Transform are used to detect different shape, such as line and circle. For the circle detection, examples of circles with known radius detection and circles with unknown radius detection have been illustrated.

Besides, different techniques of implemeting the code are shown, as in "for" loops and vectorized code.

In this example, example of how the Hough Transform could be used for hyperbola detection would be demonstrated.

Firstly, let's see the equation of the hyperbola:

(x-xo)^2/a^2 - (y-yo)^2/b^2 = 1

Let's start with a simple example to find the asymptotes and the crossing point of a hyperbola with known eccentricity, e^2 = a^2 + b^2, and to simplified the problem, let's make the a = b = 29.

1. Reading image and the convert to binary image. After that, the location of the value '1' is found using 'find' function. (the image is in white background and black object, so the '~' is used to get the negative image.

clear all;clc;
I = imread('pic27.bmp');
I =im2bw(I);
imshow(I);
I = ~I;
[y,x]=find(I);
[sy,sx]=size(I);
imshow(I);

2. Find all the require information for the transformatin. the 'totalpix' is the numbers of '1' in the image

totalpix = length(x);

3. Preallocate memory for the Hough Matrix. In this case, the a and b are known as 29 (in pixels).

HM = zeros(sy,sx);
a = 29;
b = 29;

4. Performing Hough Transform to transform the x-y domain to xo-yo domain.

yo = 1:sy;

for cnt = 1:totalpix

% a. Four lines of equation here to detect the hyperbola shapes in 4
% different direction
xo(:,1) = round(x(cnt) + sqrt(a.^2.*(((y(cnt)-yo).^2/(b.^2))+1)));
xo(:,2) = round(x(cnt) - sqrt(a.^2.*(((y(cnt)-yo).^2/(b.^2))+1)));
xo(:,3) = round(x(cnt) + sqrt(a.^2.*(((y(cnt)-yo).^2/(b.^2))-1)));
xo(:,4) = round(x(cnt) - sqrt(a.^2.*(((y(cnt)-yo).^2/(b.^2))-1)));

% b. Since there are 4 equations, Hough Matrix must consists of the
% combination for 4 equation.
for cnt3 = 1:4
for cnt2 = 1:sy

HM(cnt2,xo(cnt2,cnt3)) = HM(cnt2,xo(cnt2,cnt3)) + 1;

end
end
end

end

5. Showing the Hough Matrix

imshow(HM,[]);

6. Finding the asymptotes and the crossing point of a hyperbola

[maxval, maxind] = max(max(HM));
[Yo,Xo] = find(HM==maxval);
imshow(I); hold on;
plot(mean(Xo),mean(Yo),'xr')
c1 = mean(Yo) - mean(Xo);hold on;
plot(1:100, [1:100] + c1);
c2 = mean(Yo) + mean(Xo); hold on;
plot(1:100, -[1:100] + c2);

Y = 1:sy;
X = -sqrt(((Y-mean(Yo)).^2/a.^2 + 1).*b.^2) + mean(Xo);
hold on;
plot(X,real(Y),'r');

Hough Transform For Circle Detection - with unknown radius (II)

2006-02-06T21:55:00.000+08:00

The detection of circle with unknown radius has been shown in previous post - Hough Transform For Circle Detection - with unknown radius (I) , by using a lot of "for" loops.

This example shows the same function with "vectorized" code, which run 25% faster than the previous code, in a Pentium M 1.5 GHz computer with 512 M RAM.

Let's see how to perform this:

1. Reading image and the convert to binary image. After that, the location of the value '1' is found using 'find' function. (Assume the image is already preprocess, if not, the 'edge' function might help)

clear all;clc;
I = imread('pic26.bmp');
I =im2bw(I);
[y,x]=find(I);
[sy,sx]=size(I);
imshow(I);

2. Find all the require information for the transformation. The 'totalpix' is the numbers of '1' in the image.

totalpix = length(x);

3. Pre-allocate memory for the Hough Matrix and other variables. The ‘R’ is known in the range from 1 to 50, so we reserve 50 "layers" for the HM matrix.

HM = zeros(sy*sx*50,1);
R = 1:50;
R2(1,1,:) = R.^2;
R2 = repmat(R2,[sy,totalpix,1]);
y = repmat(y',[sy,1,50]);
x = repmat(x',[sy,1,50]);
b = 1:sy;
b = repmat(b',[1,totalpix,50]);
a = zeros(sy,totalpix);

4. Performing Hough Transform. Do notice that no "for" loops are used here.

a. Find the a parameter

a = (round(x - sqrt(R2 - (y - b).^2)));

b. Find the location of ‘a’ and ‘b’, as well as the valid value both.

[xx,yy,zz]=find((imag(a)==0 & a>0));
b = b(imag(a)==0 & a>0);
a = a(imag(a)==0 & a>0);

c. Find the indices and remap it to different layers

ind = sub2ind([sy,sx],b,a);
ind = ind + (ceil(yy/totalpix)-1)*sx*sy;

d. Construct the Hough Matrix of 50 layers

val = ones(length(ind),1);
data=accumarray(ind,val);
HM(1:length(data)) = data;
HM2 = reshape(HM,[sy,sx,50]);

5. Find for the maximum value for each layer, or in other words, the layer with maximum value will indicate the corresponding ‘R’ for the circle.

for cnt = 1:50
H(cnt) = max(max(HM2(:,:,cnt)));
end
plot(H,'*-');

6. Extract the information from the layer with maximum value, and overlap with the original image.

[maxval, maxind] = max(H);
[B,A] = find(HM2(:,:,maxind)==maxval);
imshow(I); hold on;
plot(mean(A),mean(B),'xr')
text(mean(A),mean(B),num2str(maxind),'color','green')

7. This example use the vectorized code, and the speed of the program is 25 % faster than the previous example using "for" loops.

Fill The Area Between 2 Lines

2006-02-02T01:07:00.000+08:00

This example shows how to fill the area between 2 lines with color.

1. Assuming we have two lines on a plot

x1 = 0:10;
y1 = 2*x+3;
y2 = 2*x+7;
plot(x,y1);
hold on;
plot(x,y2);

2. One of the simple way to fill the are between 2 lines is:

a. Find 4 points for the 2 lines - start point and end point, assuming 2 lines are within certain boundary, so the x vector will be (do note that the points must be defined in order, either clock-wise or anti clock-wise):

x = [0 10 10 0];

b. Replacing x in to y1 and y2 to get y vector:

y = [3 23 27 7];

3. Use the "fill" or "patch" function to fill the area between 2 lines.

patch(x,y,'r')

Converting Pixels to Line

2006-01-30T17:37:00.000+08:00

Converting pixels to line could be a complicated job. A simple example is shown to serve as the basis of a complicated problem.

1. Reading image

clear all;clc;
I = imread('pic24.jpg');
I = im2bw(I);
imshow(I);

2. Finding all the pixels with the value of '1'

[y,x] = find(I==1);
imshow(I);
hold on;
plot(x,y);

3. Use the "polyfit" function to fit polynomial to data, in our case, since it is a straight line, we use 1st order polynomial or a linear equation Y = M*X + C

p = polyfit(x,y,1);
disp(p);

-0.7873 158.3164

4. Find the image size and create the X and Y based on the coefficients of "p".

sz = size(I);
X = 1:sz(2);
Y = p(1) * X + p(2);

5. Display the result

imshow(I);
hold on;
plot(X,Y);

Detecting Objects' Motion in 2 Subsequence Images

2006-01-29T20:35:00.000+08:00

I've been posting examples of Hough Transform since last month, and now is the time to switch to other examples.

1. Reading image and comparing 2 images side by side

clear all;clc;
Ia = imread('pic23a.jpg');
Ib = imread('pic23b.jpg');
subplot(121);imshow(Ia);
subplot(122);imshow(Ib);

2. Finding the location of green object in 2 images

a. Set the threshold value for green color

p/s: I can't make the "&" appear in text, so the text aboved is in image format.

b. Find the location of the green object

[y1,x1] = find(Ia_green==1);
[y2,x2] = find(Ib_green==1);

c. Find the centroid of the green object

x1 = round(mean(x1));
y1 = round(mean(y1));
x2 = round(mean(x2));
y2 = round(mean(y2));

3. Putting 2 images together and show the movement of the objects

figure;
imshow(Ia); hold on;
imshow(Ib);
alpha(.5);
plot(x1,y1,'r*');
plot(x2,y2,'ro');
plot([x1,x2],[y1,y2]);

The '*' indicates the start point of the object while the 'o' is the stop point of the object. Other objects' movement could be found by modifying the step number 2.

Hough Transform For Circle Detection - with unknown radius (I)

2006-01-23T19:47:00.000+08:00

In previous 2 examples, the generalized Hough Transform has been applied to find for the circle with known radius. We were using the circle equation --> (x-a)^2 + (y-b)^2 = R^2 to map the pixel to a-b domain.

For detection circle with unknown radius, another parameter need to be make as variable, which is, the radius R.

Let's see how to perform this:

1. Reading image and the convert to binary image. After that, the location of the value '1' is found using 'find' function. (assume the image is already preprocess, if not, the 'edge' function might help)

clear all;clc;
I = imread('pic22.bmp');
I =im2bw(I);
[y,x]=find(I);
[sy,sx]=size(I);
imshow(I);

2. Find all the require information for the transformatin. the 'totalpix' is the numbers of '1' in the image.

totalpix = length(x);

3. Preallocate memory for the Hough Matrix. The R is known in the range from 1 to 50, so we reserve 50 "layers" for the HM matrix.

HM = zeros(sy,sx,50);
R = 1:50;
R2 = R.^2;
sz = sy*sx;

4. Performing Hough Transform. Notice the accumulator is located in the inner for loop.

for cnt = 1:totalpix
for cntR = 1:50
b = 1:sy;
a = (round(x(cnt) - sqrt(R2(cntR) - (y(cnt) - [1:sy]).^2)));
b = b(imag(a)==0 & a>0);
a = a(imag(a)==0 & a>0);
ind = sub2ind([sy,sx],b,a);
HM(sz*(cntR-1)+ind) = HM(sz*(cntR-1)+ind) + 1;
end
end

5. Find for the maximum value for each layer, or in other words, the layer with maximum value will indicate the correspond R for the circle.

for cnt = 1:50
H(cnt) = max(max(HM(:,:,cnt)));
end
plot(H,'*-');

6. Extract the information from the layer with maximum value, and overlap with the original image.

[maxval, maxind] = max(H);
[B,A] = find(HM(:,:,maxind)==maxval);
imshow(I); hold on;
plot(mean(A),mean(B),'xr')
text(mean(A),mean(B),num2str(maxind),'color','green')

7. This example use the "for loop" extensively, and the speed of the program is rather slow. Vectorized the code might speed up the execution time, we wee see it in the next example. :)

Hough Transform For Circle Detection - with known radius (II)

2006-01-18T20:13:00.000+08:00

This example doing the same as the previous example, with the difference that this example does not use any "for loops" in the code. The vectorized code make the algorithm runs faster but with the trade-off of high memory consumtion. Try to compare this with the previous example:

1. Reading image and the convert to binary image. After that, the location of the value '1' is found using 'find' function. (assume the image is already preprocess, if not, the 'edge' function might help)

clear all;clc;
I = imread('pic21.bmp');
I =im2bw(I);
[y,x]=find(I);
[sy,sx]=size(I);
imshow(I);

2. Find all the require information for the transformatin. the 'totalpix' is the numbers of '1' in the image.

totalpix = length(x);

3. Preallocate memory for the Hough Matrix. Try to play around with the R, or the radius to see the different results.

HM = zeros(sy*sx,1);
R = 36;
R2 = R^2;

4. Performing Hough Transform. Notice that no "for-loop" in this portion of code.

a. Preparing all the matrices for the computation without "for-loop"

b = 1:sy;
a = zeros(sy,totalpix);

y = repmat(y',[sy,1]);
x = repmat(x',[sy,1]);

b = repmat(b',[1,totalpix]);

b. The equation for the circle

a = (round(x - sqrt(R2 - (y - b).^2)));

c. Removing all the invalid value in matrices a and b

b = b(imag(a)==0 & a>0);
a = a(imag(a)==0 & a>0);
ind = sub2ind([sy,sx],b,a);

d. Reconstruct the Hough Matrix

val = ones(length(ind),1);
data=accumarray(ind,val);
HM(1:length(data)) = data;
HM2 = reshape(HM,[sy,sx]);

5. Showing the Hough Matrix
cla
imshow(HM2,[]);

6. Finding the location of the circle with radius of R

[maxval, maxind] = max(max(HM2));
[B,A] = find(HM2==maxval);
imshow(I); hold on;
plot(mean(A),mean(B),'xr')

Hough Transform For Circle Detection - with known radius (I)

2006-01-13T18:54:00.000+08:00

Some examples using Hough Transform are shown in previous few examples. Firstly, we have look into how to use standard Hough Transform function from new version of image processing toolbox to classified different objects.(Refer to Object Detection using Hough Transform (I) and (II) ) Secondly, we have look into how to implement our own code for Hough Transform (Refer to Simple Code for Hough Transform).

The second approach would be more flexible especially when we want to modified the Hought Transform to detect other shape other than line. In this example, a modified version of Hough Transform is used to detect the circle instead of line.

In this case, instead of transforming the pixel to rho and theta domain, we use the circle equation --> (x-a)^2 + (y-b)^2 = R^2 to map the pixel to a-b domain. Let's see how to perform this:

1. Reading image and the convert to binary image. After that, the location of the value '1' is found using 'find' function. (assume the image is already preprocess, if not, the 'edge' function might help)

clear all;clc;
I = imread('pic20.bmp');
I =im2bw(I);
[y,x]=find(I);
[sy,sx]=size(I);
imshow(I);

2. Find all the require information for the transformatin. the 'totalpix' is the numbers of '1' in the image.

totalpix = length(x);

3. Preallocate memory for the Hough Matrix. Try to play around with the R, or the radius to see the different results.

HM = zeros(sy,sx);
R = 34;
R2 = R^2;

4. Performing Hough Transform. Notice the accumulator is located in the inner for loop. This portion of codes will map the original image to the a-b domain.

b = 1:sy;

for cnt = 1:totalpix
a = (round(x(cnt) - sqrt(R2 - (y(cnt) - [1:sy]).^2)));
for cnt2 =1:sy
if isreal(a(cnt2)) & a(cnt2)>0
HM(cnt2,a(cnt2)) = HM(cnt2,a(cnt2)) + 1;
end
end
end

5. Showing the Hough Matrix

imshow(HM,[]);

6. Finding the location of the circle with radius of R

[maxval, maxind] = max(max(HM));
[B,A] = find(HM==maxval);
imshow(I); hold on;
plot(mean(A),mean(B),'xr')

Simple Code for Hough Transform

2006-01-05T19:39:00.000+08:00

Previous few examples using the Hough Transform function to perform the operation. However, Hough Transform function only available in new version of the tools. For those who are still using older version, this example shows a simple way of performing Hough Transform with a few lines of codes.

1. Reading image and the convert to binary image. After that, the location of the value '1' is found using 'find' function. (assume the image is already preprocess, if not, the 'edge' function might help)

I = imread('pic20.jpg');
I =im2bw(I);
[y,x]=find(I);
[sy,sx]=size(I);
imshow(I);

2. Find all the require information for the transformatin. the 'totalpix' is the numbers of '1' in the image, while the 'maxrho' is used to find the size of the Hough Matrix

totalpix = length(x);
maxrho = round(sqrt(sx^2 + sy^2));

3. Preallocate memory for the Hough Matrix. The resolution for both rho and theta are set to one.

HM = zeros(2*maxrho,180);

4. Performing Hough Transform. Notice the accumulator is located in the inner for loop.

for cnt = 1:totalpix
cnt2 = 1;
for theta = -pi/2:pi/180:pi/2-pi/180
rho = round(x(cnt).*cos(theta) + y(cnt).*sin(theta));
HM(rho+maxrho,cnt2) = HM(rho+maxrho,cnt2) + 1;
cnt2 = cnt2 + 1;
end
end

theta = rad2deg(-pi/2:pi/180:pi/2-pi/180);
rho = -maxrho:maxrho-1;
imshow(uint8(HM),[],'xdata',theta,'ydata',rho);
xlabel('\theta'),ylabel('\rho')
axis on, axis normal;
title('Hough Matrix');

5. Done! However, the code is quite slow, anyway, it works!

Signature of Binary Objects

2006-01-02T01:31:00.000+08:00

This example illustrates a simple way of converting the boundary of binary object into a 1-dimentional representation, or signature. One of the purpose of this conversion is to reduce the complexity of the representation for classification purpose. Let's have a look on how to convert the object to signature and compare the signature for different object shapes.

1. Original image (value 0 represent background while 1 represents object).

S = imread('pic19.jpg');
S = im2bw(S);
imshow(S);

2. Get the boundary and plot them on top of the original image.

[B,L,N,A] = bwboundaries(S);
imshow(S);
for cnt = 1:N
hold on;
boundary = B{cnt};
plot(boundary(:,2), boundary(:,1),'r');
hold on;
text(mean(boundary(:,2)), mean(boundary(:,1)),num2str(cnt));
end

figure;
subplotrow = ceil(sqrt(N));

3. Plot the signature of each object.

for cnt = 1:N
boundary = B{cnt};
[th, r]=cart2pol(boundary(:,2)-mean(boundary(:,2)), ...
boundary(:,1)-mean(boundary(:,1)));
subplot(subplotrow,N/subplotrow,cnt);
plot(th,r,'.');
axis([-pi pi 0 50]);
xlabel('radian');ylabel('r');
title(['Object ', num2str(cnt)]);
end

More details explanation could be found from the image processing reference book which could be found at the site bar of this page

Object Detection using Hough Transform (II)

2005-12-06T13:51:00.000+08:00

In previous post we've compare the Hough Matrix of a single dot, 2 dots and a single line. In this post, we'll look into more shapes:

a. Square Object

The right hand plot shows the frequencies of an intensity occurs in the Hough Matrix. It can also interpreted as: Number of lines that cross the number of points shown in the x-axis. For e.g.: for a dot, number of lines that cross one point is 180, for 2 dots image, number of lines that cross one point is 358, while number of lines that cross 2 points are 1. (refer to previous post)

For a square object, numbers of lines that cross 1 point, 2 points.....until n points in which n is the lenght of the square (approximately) would be almost the same. This could be noticed in the right figure.

b. Round Object

For a round object, numbers of lines that cross 1 point, 2 points.....until n points in which n is the diameter of the round object (approximately) would be increase in exponently. This could be noticed in the right figure.

Well, there are much more information we could get from the Hough Matrix, and what I written here might be wrong in someway, so try it out by using different images with 1 dot, 2 dots, 3 dots... using the code provided in the previous post, you would find more and more useful information from the plots!

Object Detection using Hough Transform (I)

2005-12-05T19:03:00.000+08:00

Hough Transform convert the binary image from x-y plane to rho-theta plane. This series of examples are going to show some properties that we can observe from the hough matrix and classify some simple object shapes based on it. For more information on the Hough Transform, you can use any searh engine such as google or yahoo to searh for the keyword: "Hough Transform".

1. Original image and convert to binary image
S = imread('img1.bmp');
S = im2bw(S);
imshow(S);

2. Perform Hough Transform and show the result
[H, theta,rho]=hough(S);
figure;
imshow(H,[],'xdata',theta,'ydata',rho);
xlabel('\theta'),ylabel('\rho')
axis on, axis normal;
title('Hough Matrix');

3. Count the "frequency" of the intensity of the Hough Matrix.
clear data;
for cnt = 1:max(max(H))
data(cnt) = sum(sum(H == cnt));
end
figure;
plot(data,'.');

The codes are applied to different images to compare the differences of
the Hough Matrix on different objects' shapes.

Some simple objects' shape and their Hough Matrix are as follow:

a. Single Point

To obtain the Hough Transform, draw lines from -90 to 90 degree that cross the dot, follow by drawing a Perpendicular line from the upper left of the image to the line, and measure the lenght(rho) and the angle (theta) of the Perpendicular line.

The "hough" function will perform this for every 1 degree, and hence, for a single dot, the there will be 180 of '1' in the "Hough Matrix".

b. Two Points

When there are two points on an image, the Hough Matrix consist of 2 lines, and interect at a point which is a line that link the 2 points together in x-y plane (left figure).

Since there is only one intersection (only one line that link 2 points together), the frequencies of '1' in Hough Matrix is 358, while the frequencies of '2' is 1 in the Hough Matrix. (the '1' and '2' indicates how many time a rho-theta appear in the Hough Matrix)

c. Line

A line consists of numbers of points, ideally (the line is 45 degree) the intersection in the Hough Matrix will have the intensity same as the line lenght.

Hough Transform always used to detect lines in a real image, in which the edges always distorted. A higher value in the Hough Matrix indicates the higher possibility of the existence of a straight line.

More objects shapes will be shown in the next post...

Maple for Image Processing?

2005-11-18T19:01:00.000+08:00

Can Maple used for Image Processing?

Maple is the tool for solving mathematical problems and creating interactive technical applications. My experience with Maple is that --> a symbolic software.

However, when I did a search on "maple image processing", I found the following link:
Image Processing with Maple

Seems like the new version of Maple (or even older version) support basic image processing function, and it might be supporting more and more in the future...

Some other software for image processing could be found at the sidebar, or you can simply do a search using google or yahoo..

Keyword to search for : Matlab, Maple, Mathematica, Math Software, Image Processing Software, Image Analysis...

Which Software to Use?

2005-11-17T20:35:00.000+08:00

What Software Should be Used for Image Processing?

There are numbers of different software for engineering, each has its' capability and weakness. Most of the examples here use Matlab because I deal alot with Matlab in University.

However, I won't say there are no other software better than Matlab, but i still believe:-

"Choose the Software you're familiar with, know your direction and how far the software can accompany you along the way..."

A good article comparing 3 famous software:

3Ms for Instruction: Reviews of Maple, Mathematica, and Matlab

Happy programming.

Drawing Circles Around Round Object

2005-11-05T01:26:00.000+08:00

Drawing shapes to highlight certain objects could be done in several ways. Previous examples illustrate few ways of drawing object by overwriting pixel value as well as by adding a "patch" on top of an image. This example shows how to draw circles around a round object just simply using "plot" command

1. Original image

S = imread('pic13.jpg');
S = im2bw(S);
S2 = ~S;
imshow(S2);

2. Extracting required objects' properties. Required properties are centroid and the area of the round objects. The area property is used to compute the radius of the object.

S3 = bwlabel(S2);
S4 = regionprops(S3,'Centroid','area');
R = sqrt(S4(1).Area/pi);
h = S4(1).Centroid(1);
k = S4(1).Centroid(2);

3. Drawing circles around 1st object with different radius by refering to the radius of the object.

ratio = [2 1.5 1];
nR = round(R*ratio);
t = 0:pi/20:2*pi;
x1 = (h+nR'*cos(t))';
y1 = (k+nR'*sin(t))';
hold on;
plot(x1,y1);

Drawing Transparent Object

2005-11-05T01:13:00.000+08:00

In one of the previous post (Drawing Shapes by Overwriting Pixel Value), a method of drawing transparent object had been illustrated using "Overwriting Pixel Value" technique. In this example, another method will be introduced without overwriting pivel value, but drawing a "Patch" on top of the image.

1. Original image

S = imread('pic12.jpg');
S = im2bw(S);
S2 = S;
imshow(S2);

2. Define transparent object properties

transparency = 0.5;
facecolor = [0 1 0];
edgecolor = [0 1 0];

3. Creating and drawing transparent object

x = [0 20 20 0];
y = [0 0 20 20];
z = [1 1 1 1];
tcolor(1,1,1:3) = facecolor;

h = patch(x,y,z,tcolor,'FaceAlpha',transparency,'edgecolor',edgecolor);

4. To move the object, simply set the "Xdata" and "Ydata" of the object without deleting current object and redrawing a new onject

set(h,'Xdata',x+20);
set(h,'Ydata',y+20);

Detecting Objects' Shape(III) : Triangle Object

2005-11-04T14:27:00.000+08:00

This example illustrates a simple method in detecting triangle object. This method is slightly different from the methods used for round object and square object detection in which the area and the perimeter of the objects are used to calculate the "score" of similarity. Instead, this method uses the "extrema" points of an object to determine whether the object is triangle.

1. Original image and the negative image (value 0 represent background while 1 represents object).

S = imread('pic11.jpg');
S = im2bw(S);
S2 = ~S;
imshow(S2);

2. Labeling the objects in the image. The advance programming language will support this function and label the objects.

S3 = bwlabel(S2);
imshow(S3,[]);

3. In this example of triangle object detection, only one object property is extracted --> "Extrema".

S4 = regionprops(S3,'Extrema','Centroid');

The “Centriod” is just used for labeling purpose. Following figure shows what “Extrema” of an object is:
4. Ideally, a triangle will only have 3 extrema points. However, due to the imperfection of a binary object, a few of close-by points might exist instead of one.
For example eight extrema points of a triangle are: (135.5,11.5),(136.5,11.5),(157.5,45.5),(157.5,47.5),(157.5,47.5),(112.5,4 7.5),(112.5,47.5),(112.5,45.5) first 2 points are referring to the upper corner of the triangle but the values are different. This applied to point 3 to 5 and points 6-8 too.

In order to solve this problem, we need to cluster the points to their group based on the location of the points. Different ways could be used for clustering but since our topic is about image processing, image processing method will be used to group the similar points together.

In this method, a morphology technique --> dilation will be used. Firstly, the extrema points will be used to construct an image. The coordinates of the extrema points will be converted to points in an image. Let's look into one of the object for illustration:

a. Create morphological structuring element.

se = strel('disk',5);

b. Round the extrema points for 5th object (pixel value must be integer)

temp = round(S4(5).Extrema);

c. Remap the coordinates by referring to the minimum values of x and y

temp(:,1) = temp(:,1) - min(temp(:,1)) + 1;
temp(:,2) = temp(:,2) - min(temp(:,2)) + 1;

d. Create a zero matrix (black image)

mask = zeros(max(temp(:,1)),max(temp(:,2)));

e. Convert the coordinates to "objects" in the black image

for cnt2 = 1:8
mask(temp(cnt2,1),temp(cnt2,2))=1;
end
imagesc(mask);

f. Do note that some of the points are close together; use the image dilation technique to group close-by points.

mask2 = imdilate(mask,se);
imagesc(mask2);

g. Count number of real extrema points in the image

[labeled,numObjects] = bwlabel(mask2,8);
disp(numObjects);

3

5. The following codes perform calculation for all objects

se = strel('disk',5);
for cnt = 1:length(S4)
temp = round(S4(cnt).Extrema);
temp(:,1) = temp(:,1) - min(temp(:,1)) + 1;
temp(:,2) = temp(:,2) - min(temp(:,2)) + 1;
mask = zeros(max(temp(:,1)),max(temp(:,2)));
for cnt2 = 1:8
mask(temp(cnt2,1),temp(cnt2,2))=1;
end
mask2 = imdilate(mask,se);
[labeled,numObjects(cnt)] = bwlabel(mask2,8);
end

disp(numObjects);
5 3 4 6 3 4

score = (numObjects == 3);

6. Display the result visually

imshow(S2)
for cnt = 1:length(S4)
text(S4(cnt).Centroid(1),S4(cnt).Centroid(2),num2str(score(cnt)),'color','red');
end

Detecting Objects' Shape(II) : Square Object

2005-10-30T11:35:00.000+08:00

Detecting square object could be done in similar way we used to detect the round object. What we need to know is just the the formula for the area and the perimeter of a square object which is A = w^2 and P = 4*w.

1. Original image and the negative image (value 0 represent background while 1 represent oject).
S = imread('pic10.jpg');
S = im2bw(S);
S2 = ~S;
imshow(S2);

2. Labeling the objects in the image. The advance programming language will support this function and label the objects.
S3 = bwlabel(S2);
imshow(S3,[]);

3. Some properties of the objects are extracted using the command. In this example, we just use the area and perimeter for our calculation.
S4 = regionprops(S3,'Area','Perimeter','centroid');

4. Now, evaluate the "squareness" of the objects. this is evaluated by only 1 criteria in this example:
a. Calculate the length or width of the object by using the formulas of area and perimeter, comparing the result and the square objects would have higher score than others. Do note that instead of for loop, the code below is a vectorized code.
score = (min(sqrt([S4.Area]),[S4.Perimeter]/4)./(max(sqrt([S4.Area]), ...[S4.Perimeter]/4))).^2;

5. Display the result visually
imshow(S2)
for cnt = 1:length(S4)
text(S4(cnt).Centroid(1),S4(cnt).Centroid(2),num2str(score(cnt)),'color','red');
end

Do note that this example assumes perfect images. For more accurate results, more properties can be compared by using more criteria. The example above is done using MATLAB®.

MATLAB® is registered trademarks of The MathWorks.

Detecting Objects' Shape(I) : Round Object

2005-10-27T21:42:00.000+08:00

With the powerful technical computing language, the job of detection an object's shape no longer require lengthy codes. What important is the fundamental of mathematic that define the properties of the shape of objects. Let's look into the detection of round object in this section.

1. Original image and the negative image (value 0 represent background while 1 represents object).

S = imread('pic9.jpg');
S = im2bw(S);
S2 = ~S;
imshow(S2);

2. Labeling the objects in the image. The advance programming language will support this function and label the objects automatically.
S3 = bwlabel(S2);
imagesc(S3);

3. Getting the properties of the objects that require for calculation, which are: the longest and shortest diameter for the object, the area and the perimeter of the object, and the centroid of the object.
S4 = regionprops(S3,'MinorAxisLength','MajorAxisLength','Area','Perimeter','centroid');

4. Now, evaluate the "roundness" of the objects. this is evaluated by 3 criteria:
a. Comparing the longest and shortest diameter for the object,
b. Comparing the Area with the formula pi*r^2.
c. Comparing the Perimeter with the formula 2*pi*r
The "scores" are normalized so that 1 represent "best fit"

for cnt = 1:length(S4)
score1(cnt) = abs(1-(S4(cnt).MajorAxisLength-S4(cnt).MinorAxisLength)...
/max([S4(cnt).MajorAxisLength,S4(cnt).MinorAxisLength]));
score2(cnt) = abs(1 - abs(pi*((mean([S4(cnt).MajorAxisLength,...
S4(cnt).MinorAxisLength]))/2)^2-S4(cnt).Area)/S4(cnt).Area);
score3(cnt) = abs(1 - abs(pi*(max([S4(cnt).MajorAxisLength,...
S4(cnt).MinorAxisLength]))-S4(cnt).Perimeter)/S4(cnt).Perimeter);
end
score = mean([score1;score2;score3]);

5. Display the result visually

imshow(S2)
for cnt = 1:length(S4)
text(S4(cnt).Centroid(1),S4(cnt).Centroid(2),num2str(score(cnt)),'color','red');
end

Playing with “Edge Detection”

2005-10-26T19:33:00.000+08:00

A lot of articles explain what is edge detection, how to apply it, and the usage of it. Different edge detection techniques such as sobel, prewitt, and canny are easily found from any image processing software.

However, when we really want to perform edge detection for certain purpose, the problems arise. Look at the following example:

1. Original image
S = imread('pic8.jpg');
imshow(S);

2. Edge detection using different methods
S2 = rgb2gray(S);
figure,edge(S2,'sobel');
figure,edge(S2,'prewitt');
figure, edge(S2,'canny');

It seems like great but the usage of these output are useless until we really know what we want to do. For exmaple, we want to find the white rabbit, we need to perform some pre-processing before the edge detection so that we can fully utilize the function to suite to our needs.

3. Edge detection for the white rabbit
Since we know the RGB value for white color is [255,255,255], we can pre-process the image to extract the white color region from the image.

Sw = S(:,:,1)>200 & S(:,:,2)>200 & S(:,:,3)>200;
edge(double(Sw));

Well, it seems like make more sense. Let's have a look on how to perform edge detection on the black rabbit:

4. Edge detection for the black rabbit

Sb = S(:,:,1)<40 & S(:,:,2)<40 & S(:,:,3)<40;

The post processing can be used to remove unwanted noise.

Panoramic Picture Creation

2005-10-16T00:28:00.000+08:00

This example illustrates simple method of panoramic picture creation using sum of absolute differences. It assumes no zooming, rotational variation in 2 images.

1. Reading Images for processing
S1 = imread('p1.jpg');
S2 = imread('p2.jpg');

2. Showing Images
1st Image

imshow(S1);

2nd Image

imshow(S2);

3. Extract red layer of the image for processing using Sum of Absolute Differences (SAD)
S1r = S1(:,:,1);
S2r = S2(:,:,1);

4. Sum of Absolute Differences (SAD) of a selected block is from each image are computed.
sz = 50;
roi2 = S2r(1:sz,1:sz);
[x,y]=size(S2r);
for cnt = y:-1:x/2
roi1 = S1r(1:sz,cnt-sz+1:cnt);
data(y-cnt+1)=sum(sum(abs(double(roi2)-double(roi1))));
end

5. Finding minimum error among all comparison
plot(data)
[minval,ind] = min(data);
hold on;
plot(ind,minval,'r*');

6. Combine the images at the minimum error region.
Sc = S1(1:end,1:y-sz-ind+1,:);
Sfinal = [Sc S2];

7. Final Result
figure;
imshow(Sfinal);

Character Recognition Example (III):Training a Simple NN for classification

2005-10-06T23:59:00.000+08:00

After preprocess the image, the training dataset is used to train "classification engine" for recognition purpose. In this example, a multi-layer feed-forward back-propagation Neural Network will be used.

1. Read the image
I = imread('sample.bmp');

2. Image Preprocessing
This cell of codes will extract all the components (objects) as descript in section II.

img = edu_imgpreprocess(I);
for cnt = 1:50
bw2 = edu_imgcrop(img{cnt});
charvec = edu_imgresize(bw2);
out(:,cnt) = charvec;
end

3. Creating Vectors data for the Neural Network (objects)
These few line of codes creates training vector and testing vector for the neural network. This is to match the input accepted by the neural network function. The front 4 rows will be used to train the network, while the last row will be used to evaluate the performance of the network. (Please refer to part I and II for the dataset)

P = out(:,1:40);
T = [eye(10) eye(10) eye(10) eye(10)];
Ptest = out(:,41:50);

4. Creating and training of the Neural Network
Create and Train NN. The "edu_createnn" is a function file to create and train the network by accepting the training-target datasets

net = edu_createnn(P,T);

TRAINGDX, Epoch 0/5000, SSE 100.603/0.1, Gradient 46.9458/1e-006
TRAINGDX, Epoch 20/5000, SSE 38.3704/0.1, Gradient 1.1137/1e-006
TRAINGDX, Epoch 40/5000, SSE 39.4562/0.1, Gradient 0.485735/1e-006
TRAINGDX, Epoch 60/5000, SSE 39.5979/0.1, Gradient 0.387782/1e-006
TRAINGDX, Epoch 80/5000, SSE 39.6081/0.1, Gradient 0.387929/1e-006
TRAINGDX, Epoch 100/5000, SSE 39.5559/0.1, Gradient 0.445316/1e-006
TRAINGDX, Epoch 120/5000, SSE 39.2765/0.1, Gradient 0.758668/1e-006
TRAINGDX, Epoch 140/5000, SSE 35.941/0.1, Gradient 1.34484/1e-006
TRAINGDX, Epoch 160/5000, SSE 23.9786/0.1, Gradient 2.73872/1e-006
TRAINGDX, Epoch 180/5000, SSE 1.57322/0.1, Gradient 0.898383/1e-006
TRAINGDX, Epoch 196/5000, SSE 0.0909575/0.1, Gradient 0.0769913/1e-006
TRAINGDX, Performance goal met.

5. Testing the Neural Network
Simulate the testing dataset

[a,b]=max(sim(net,Ptest));
disp(b);

1 2 3 4 5 6 7 8 9 10

Do note that there might be some misclassifications on the simulated result with different training properties. The accuracy could be improved by better features extraction techniques as well as the selection of NN's properties.

* MATLAB® is the registered trademarks for The MathWorks, Inc

The relevant files can be found at:
(link to the files will be provided soon…)

Character Recognition Example (II):Automating Image Pre-processing

2005-09-30T22:56:00.000+08:00

Most of the time we would like to automate our image pre-processing stage especially during the data extraction for database. In other words, the program should be able to extract the characters one by obe and map the target output for training purpose. The following code shows the example:

1. Read Image
This cell of codes read the image to MATLAB workspace

I = imread('sample.bmp');
imshow(I)

2. Convert to grayscale image
This cell of codes convert the RGB to gray

Igray = rgb2gray(I);
imshow(Igray)

3. Convert to binary image
This cell of codes convert the gray to binary image

Ibw = im2bw(Igray,graythresh(Igray));
imshow(Ibw)

4. Edge detection
This cell of codes detect the edge of the image

Iedge = edge(uint8(Ibw));
imshow(Iedge)

5. Morphology
This cell of codes perform the image dilation and image filling on the image

Image Dilation

se = strel('square',2);
Iedge2 = imdilate(Iedge, se);
imshow(Iedge2);

Image Filling

Ifill= imfill(Iedge2,'holes');
imshow(Ifill)

6. Blobs analysis
This cell of codes find all the objects on the image, and find the properties of each object.

[Ilabel num] = bwlabel(Ifill);
disp(num);
Iprops = regionprops(Ilabel);
Ibox = [Iprops.BoundingBox];
Ibox = reshape(Ibox,[4 50]);
imshow(I)

50

7. Plot the Object Location
This cell of codes plot the object locations

hold on;
for cnt = 1:50
rectangle('position',Ibox(:,cnt),'edgecolor','r');
end

By this, we are able to extract the character and pass to another stage for "classification" or "training" purpose. (To be continue...)

* MATLAB® is the registered trademarks for The MathWorks, Inc

The required files can be found at:
(link for the required files will be provided soon...)

Character Recognition Example (I): Image Pre-processing

2005-09-30T22:45:00.000+08:00

This example illustrates simple way of character recognition. It just serves as a kick start for beginers by introducing them simple coding in MATLAB for character recognition.

Some useful example for image pre-processing before the recognition stage is shown as follow:

1. Manual Cropping
This cell of codes allow the user to crop the image manually, which is important proccess especially the programmer would like to manipulate the data more in details.

img = imread('sample.bmp');
imshow(img)
imgGray = rgb2gray(img);
imgCrop = imcrop(imgGray);
imshow(imgCrop)

2. Resizing
This cell of codes magnify the image for 5 times. Resize function is important when the size of the character are not standard, especially in the handwritting recognition program.

imgLGE = imresize(imgCrop, 5, 'bicubic');
imshow(imgLGE)

3. Rotation
This cell of codes rotate the image for 35 degree. Some of the printed documents might not in the right degree as expected. This operation coulf make the character to our desired angle.

imgRTE = imrotate(imgLGE, 35);
imshow(imgRTE)

4. Binary Image
This cell of codes convert the image to binary image. Definitely, the speed and also the accuracy will increse if this operation is perform properly.

imgBW = im2bw(imgLGE, 0.90455);
imshow(imgBW)

Detecting Object in an Image

2005-09-27T19:31:00.000+08:00

1. How to detect an object in an image?
Determining the features from the object that you want to detect is the key. The feature in this case is something that differentiates the object from others, such as color, shape, size, etc…

2. What are the techniques for object detection?
The image processing techniques such as morphology or color processing usually did this job. A simple example in MATLAB® of detecting ‘white rabbit’ is shown as follow, in this case, ‘color’ is the feature used to distinguish the white rabbit from other:

% Original Image
S = imread('pic3b.jpg');
imshow(S);

% Gray scale image
S2 = rgb2gray(S);
imshow(S2);

% Find the white color
S3 = S2>180;
imshow(S3);

% Morphology technique, image erosion to erase the unwanted components
se = strel('line',10,90);
S4 = imerode(S3,se);
se = strel('line',10,0);
S4 = imerode(S4,se);
imshow(S4)

% Labeled the component(s), and plot the centroid on the original image
S5 = bwlabel(S5);
[x,y]=find(S6==1);
imshow(S)
hold on;plot(mean(y),mean(x),'r*')

More information of image processing techniques can be found by searching "morphology", "morphological" from the search engine or the google search bar at the right hand site of this page.

* MATLAB® is the registered trademarks for The MathWorks, Inc

Drawing Shapes by Overwriting Pixel Value

2005-09-23T23:30:00.000+08:00

1. How to highlight certain portion on an image?
The basic concept of highlighting part of an image is “Overwriting Pixel Value” on the image. We start from the basic idea on how to mask portion of image with blank sub-image (black color sub-image). Fig 1 shows the image with the size of 128x128x3 in which 3 represent s RGB. (In grayscale image 3 layers are the same). The image at the right hand site shows a blank sub-image is placed at the upper left of the original image. The MATLAB® code to perform the operation are as follow:

S = imread('t3.jpg');
imshow(S);
Sblack = uint8(zeros(20,20,3));
S2 = S;
S2(1:20,1:20,:) = Sblack;
imshow(S2);

Fig 1

2. How to create a color mask rather than black color mask?
Since the 3 layers of image matrix represent RGB layers of the image, we can create the red color mask using following command, and the results are shown in fig 2.

Sred = Sblack;
Sred(:,:,1)=255;
S2 = S;
S2(1:20,1:20,:) = Sred;
imshow(S2);

Fig 2

3. How to create a transparent mask?
Simple enough, just play with the values R value, and perform the image addition rather than overwriting the value as follow:

Sred = Sblack;
Sred(:,:,1)=200;
S2 = S;
S2(1:20,1:20,:) = S2(1:20,1:20,:) + Sred;
imshow(S2);

Fig 3

4. Finally, how to create an outline for the image?

linelength = 10;
Sblack = uint8(zeros(128+linelength*2,128+linelength*2,3));
S2 = Sblack;
S2((1+linelength):(end-linelength),(1+linelength):(end-linelength),:)=S;
imshow(S2);

Fig 4

The “linelength” variable is the outline length in pixel.

* MATLAB® is the registered trademarks for The MathWorks, Inc

What is Image Compression?

2005-09-21T16:58:00.000+08:00

This blog is not for the expert, but for those who like to ask, “what is..?” “how to..?” and “…so what?”. If you want the prove in the form of mathematic equations, this is not the right place you should come.

No mathematic equation will be shown, but prove of the methods and the basic concepts of an algorithm and how to apply it in research, work, or “just for fun”.

The very first topic that I would like to share is the concept of simple image compression. So the first question you might ask is:

1. What is image compression and how to perform image compression?
Image compression is the technique to compress the image to save storage space as well as before transmitting it through some transmission medium. The most common method of image compression is discrete cosine transform (DCT). By taking the DCT of an original image, the low frequencies contents will be transform into the upper left of the DCT matrix… why…? Don’t ask me why, again, this blob is just for those what want to know “how to…?” and “…so what?” but not for “why…?”. If you would like to learn more on the image compression, you can use any search engine, or you might want to use the Google search bar on the right panel of this page to search for the “image compression”. :)

2. So what after I know the DCT concept? How to apply this into image compression?
The low frequencies components, which also carry most of the information of the image, are our focus. The fig 1 shows the original image while fig 2 shows the DCT matrix for the image. Notice that the “white color” is concentrated at the upper left of the DCT matrix? That’s the “information (energy) for the image which also represents the low frequency components of the image. If we perform a inverse DCT for the fig 2, we will get back the fig 1.

3. How to compress the image?
Fig 2 is the DCT matrix of the image. If we through away the high frequencies components, which contain not much information (dark color portion), we can achieve certain level of compression. For example, if I just take quarter of the DCT matrix, I save 75% of the storage!

4. So what is the tradeoff for this storage saving?
We will get some distortion of the image after the reconstruction of the image using the Inversed DCT as shown in fig 4. The level of the distortion is depends on the level of compression you choose. Some visual comparisons are shown in the following figures.

5. What software can I use to work on the above example?
You can use any search engine or the Google search bar on the right panel of this page and search for the “math software”, you would find numbers of useful softwares to perform this.