Activity 15: Color Image Processing

For this activity, we are to perform 2 different white balancing techniques, called the reference white algorithm and the gray world algorithm. In the reference white algorithm, a reference white object's pixel values are to be used to normalize the red, green and blue channels of the whole image. On the other hand, gray world algorithm uses the mean of each of the RGB channel values as the normalization factor.
We used two sets of objects for this activity. One set is a set of objects with different colors and another set is a set of objects with the same colors. For my second set I used a blue set of objects. The images were taken under a fluorescent lamp using the "tungsten", "daylight" and "cloudy" white balancing settings of a Canon Powershot A460 digital camera.

The following are the results for object with different colors:

TUNGSTEN

CLOUDY

DAYLIGHT

For all the images, Reference white algorithm seems to work better since the gray world algorithm requires a certain constant to be multiplied to reduce the brightness of the resulting image. For this part the constant used to limit the brightness is 0.75.

The following are the results for different blue objects under different white balancing conditions.

CLOUDY

DAYLIGHT


TUNGSTEN

For images with the same color, the reference white algorithm clearly performs better. The results of the gray world algorithm produced images with slightly redder image. A possible reason is because an average of an ensemble of colors may become white but an ensemble of objects having similar colors is not necessarily white. Hence, the white reference algorithm is expected to perform better.

Scilab Code:
I = imread("IMG_0921_cloudy.jpg");

//start of reference white algorithm
imshow(I);
pos = locate(1);
Rw = I(pos(1),pos(2),1);
Gw = I(pos(1),pos(2),2);
Bw = I(pos(1),pos(2),3);

//Start of gray world algorithm
//Rw = mean(I(:,:,1));
//Gw = mean(I(:,:,2));
//Bw = mean(I(:,:,3));

I(:,:,1) = I(:,:,1)/Rw;
I(:,:,2) = I(:,:,2)/Gw;
I(:,:,3) = I(:,:,3)/Bw;
//I = I*0.5;
I(I>1) = 1;

imwrite(I,"enhanced_cloudy.jpg");
----------------------------------------------

10/10 for this activity since I was able to finish the activity quickly and correctly.
Collaborator: Eduardo David

Activity 14: Stereometry

Activity 13: Photometric Stereo

In this activity we are to reconstruct an image using different images taken from the same point with different point source locations. The following are the images used with their corresponding point source positions



Using the following formula, we were able to obtain a 3d plot for the original image

loadmatfile("Photos.mat");
scf(0);imshow(I1,[]);
scf(1);imshow(I2,[]);
scf(2);imshow(I3,[]);
scf(3);imshow(I4,[]);
V1 = [0.085832, 0.17365, 0.98106];
V2 = [0.085832, -0.17365, 0.98106];
V3 = [0.17365, 0, 0.98481];
V4 = [0.16318, -0.34202, 0.92542];

I(1,:) = I1(:)';
I(2,:) = I2(:)';
I(3,:) = I3(:)';
I(4,:) = I4(:)';
//I = cat(1,I1(:)',I2(:)',I3(:)',I4(:)');
V = cat(1,V1,V2,V3,V4);

g = inv(V'*V)*V'*I;
N = size(g);

for i = 1:N(2)
gMag(i) = sqrt(g(1,i)**2+g(2,i)**2+g(3,i)**2+0.0000000000001);
end

nh(1,:) = (g(1,:)./gMag');
nh(2,:) = (g(2,:)./gMag');
nh(3,:) = (g(3,:)./gMag')+0.0000000000001;

fx = -nh(1,:)./nh(3,:);
fx = matrix(fx,[128,128])
fy = -nh(2,:)./nh(3,:);
fy = matrix(fy,[128,128]);
f = cumsum(fx,2)+cumsum(fy,1);

plot3d(1:128,1:128,f);


I will grade myself 10 for this activity since I was able to meet the objectives.
Collaborator: Eduardo David

Activity 12

Activity 11: Camera Calibration

In this activity, we are to obtain transformation of the camera coordinates with real world coordinates using a checkerboard pattern shown below as a calibration object
I assigned the bottom center of the image as my origin and the left part as the "x" axis and the right part as the "y" and the vertical direction as the "z". To calibrate the camera, first, we assigned 25 corner points chosen at random in the image (marked with yellow squares). The location of the points are obtained using the "locate" function of scilab.
To solve for the camera parameters, we use the following matrix:
Where xo, yo and zo correspond to real world coordinates and yi and zi correspond to image coordinates. aij correspond to the camera parameters.
The camera parameter matrix can be solved using:
Where Q is the matrix containing the real world coordinates and d is the matrix containing the image coordinates.
The camera parameters obtained are as follows:

-19.19404
9.7237133
-0.5773294
167.80985
-2.5034951
-3.831655
20.477795
32.311712
-0.0067168
-0.0120008
-0.0025509

To confirm if this camera parameters are correct, we use it to predict the locations of the selected points in the image from the real world coordinates. The following equation is applied:

Using the above equations I obtained the following results:
The 1st pair of data correspond to the actual image coordinates, 2nd pair correspond to the coordinates solved using the above equation and the 3rd pair corresponds to their difference.

Image Coordinates		Solved Coordinates		difference
yi	zi	yi	zi	yi	zi
29.096045	259.60452	29.300782	259.52771	0.204737	0.07681
89.548023	262.42938	89.604812	261.94118	0.056789	0.4882
216.66667	261.86441	216.84828	262.18338	0.18161	0.31897
166.38418	243.22034	166.27818	243.29594	0.106	0.0756
50	214.97175	50.653965	215.20451	0.653965	0.23276
147.74011	220.62147	147.806	220.65636	0.06589	0.03489
217.23164	216.66667	216.9003	216.65975	0.33134	0.00692
91.242938	194.63277	90.702993	195.35542	0.539945	0.72265
229.66102	192.37288	230.33101	192.45105	0.66999	0.07817
31.920904	169.20904	31.45059	169.10213	0.470314	0.10691
147.74011	176.55367	148.21666	177.51489	0.47655	0.96122
191.24294	175.9887	191.22007	175.33202	0.02287	0.65668
129.66102	155.08475	129.68482	154.61499	0.0238	0.46976
73.163842	127.9661	72.303439	128.12011	0.860403	0.15401
167.51412	137.00565	167.05391	136.44094	0.46021	0.56471
204.80226	130.22599	204.04286	129.52077	0.7594	0.70522
258.47458	119.49153	257.94833	119.43573	0.52625	0.0558
33.050847	81.073446	33.55399	80.628618	0.503143	0.444828
129.66102	90.677966	130.44029	90.649574	0.77927	0.028392
191.24294	88.983051	191.58995	88.895726	0.34701	0.087325
92.937853	65.254237	92.847265	65.342247	0.090588	0.08801
217.23164	60.169492	217.07798	61.190191	0.15366	1.020699
34.745763	37.00565	34.588753	37.10424	0.15701	0.09859
149.43503	50.564972	149.42329	50.756407	0.01174	0.191435
243.22034	32.485876	243.74758	32.200294	0.52724	0.285582

The average value of the difference is 0.359 with standard deviation of 0.318. This translates to an error of roughly 1%.

The scilab code I used is given below:
---------------------------------------------------------------------------------
I = imread("C:\Users\RAFAEL JACULBIA\Documents\subjects\186\activity11\marked.bmp");
I = im2gray(I);
//imshow(I,[]);
//d = [];
//d = locate(25,flag=1);

image = [7 4 0 0 6 1 0 4 0 7 1 0 2 5 0 0 0 7 2 0 4 0 7 1 0;
0 0 4 0 0 0 4 0 5 0 0 2 0 0 0 3 7 0 0 2 0 4 0 0 6;
11 11 11 10 9 9 9 8 8 7 7 7 6 5 5 5 5 3 3 3 2 2 1 1 1];


//image = image*72;

Q = [];
yi = [];
zi = [];
for i = 1:25
x = image(1,i);
y = image(2,i);
z = image(3,i);
yi = d(1,i);
zi = d(2,i);
Qi = [x y z 1 0 0 0 0 -(yi*x) -(yi*y) -(yi*z); 0 0 0 0 x y z 1 -(zi*x) -(zi*y) -(zi*z)];
Q = cat(1, Q, Qi);
end
d2 = [];
d2 = matrix(d, 50,1);
a = inv(Q'*Q)*Q'*d2;

for j = 1:25
yi2(j) = ((a(1)*image(1,j))+(a(2)*image(2,j))+(a(3)*image(3,j))+a(4))/((a(9)*image(1,j))+(a(10)*image(2,j))+(a(11)*image(3,j))+1);
zi2(j) = ((a(5)*image(1,j))+(a(6)*image(2,j))+(a(7)*image(3,j))+a(8))/((a(9)*image(1,j))+(a(10)*image(2,j))+(a(11)*image(3,j))+1);
end

for k = 1:25
yi(k) = d(1,k);
zi(k) = d(2,k);
end

tryx = 0;
tryy = 6;
tryz = 1;

yiTry = ((a(1)*tryx)+(a(2)*tryy)+(a(3)*tryz)+a(4))/((a(9)*tryx)+(a(10)*tryy)+(a(11)*tryz)+1)
ziTry = ((a(5)*tryx)+(a(6)*tryy)+(a(7)*tryz)+a(8))/((a(9)*tryx)+(a(10)*tryy)+(a(11)*tryz)+1)
-------------------------------------------------------------------------------------------

For this activity I will give myself a grade of 10 because I was able to obtain quite accurate results.
Thanks to Eduardo David for teaching me how to resize a matrix in scilab

Activity 10: Preprocessing Handwritten Text

For this activity, the goal is to extract text from an imaged documents with lines. The given document is as follows (image resized):

I cropped a small portion of the image for faster calculations. The cropped image and its FT is shown below:
To remove the vertical and horizontal lines I applied the following filter and the image beside it is the resulting image, also included is the binarized version of the image:


It can be seen that the lines are not actually removed totally, specially at the edges. In the image we can see that only the 1st line column can be read and last column is totally unreadable. To further enhance the image I performed opening and closing to isolate some of the letters using a 2x1 structuring element. And to make the letters 1 pixel thick, I performed erosion using the same structuring element. Afterwards, I labeled the text using bwlabel, the following image is the result:


As expected only some of the text are readable, almost all of the the numbers on the 1st column and the fractions on the second column are readable, but the third to fifth columns are pretty much unreadable.
The whole scilab code I used is shown below, it is a "mixture" of my older codes so it may look familiar. =)
------------------------------------------------------------------------------------
I = imread("C:\Users\RAFAEL JACULBIA\Documents\subjects\186\activity10\last\small.bmp");
I2 = im2gray(I);
C = imread("C:\Users\RAFAEL JACULBIA\Documents\subjects\186\activity10\last\cross.bmp");
C = im2gray(C);
Cshift = fftshift(C);

ftI = fft2(I2);
conv = Cshift.*ftI;
iconv = fft2(conv);
iconv = abs(fft2(fft2(iconv)));
scf(0);imshow(I2,[]);
scf(1);imshow(iconv,[]);
imwrite(iconv/max(iconv), "C:\Users\RAFAEL JACULBIA\Documents\subjects\186\activity10\last\noLines.bmp");
imwrite(abs(fft2(fft2(iconv)))/max(abs(fft2(fft2(iconv)))), "C:\Users\RAFAEL JACULBIA\Documents\subjects\186\activity10\last\res.bmp");
bw = im2bw(iconv,0.5);
imwrite(bw, "C:\Users\RAFAEL JACULBIA\Documents\subjects\186\activity10\last\bw.bmp");
scf(3);imshow(bw,[]);

bw = abs(bw-1);
se = [1,1];



OC = erode(dilate(dilate(erode(bw,se),se),se),se);

OC = erode(OC,se);


L = bwlabel(OC);
scf(4);imshow(L,[]);

imwrite(L/max(L), "C:\Users\RAFAEL JACULBIA\Documents\subjects\186\activity10\last\labeled.bmp");
------------------------------------------------------------------------------------------

For this activity, I will give myself a grade of 9/10 since I believe I was not able to accomplish the tasks perfectly. And for the first time I performed the activity without the help of others.

Activity 9: Binary Operations

In this activity, we are to measure the individual areas of a number of punched paper. It is assumed that the areas of the punched paper are uniform. The image I used is Circles001.jpg shown below:
For faster calculations, I subdivided the images into 10, 256x256pixel images. A typical subdivision is shown below:
Closing and Opening operators are necessary to be able to separate nearly touching circles and to clean the edges of broken circles. In general opening and closing operators are "mixtures" of dilation and erosion operator from the previous activity. Opening operations is achieved by dilating an eroded image. That is we are to perform erosion on an image then perform dilation on the resulting image. On the other hand, Closing is achieved in the opposite manner, in this case, performing erosion on a dilated image. Since erosion and dilation requires binary images, I binarized the subimages using 0.8 as a threshold. The code I used for the whole process is given below:

----------------------------------------------------------------------
I = imread("C:\Users\RAFAEL JACULBIA\Documents\subjects\186\activity9\C1_10.bmp");
se = imread("C:\Users\RAFAEL JACULBIA\Documents\subjects\186\activity9\disk.bmp");
bw = im2bw(I,0.8);

OC = erode(dilate(dilate(erode(bw,se),se),se),se);
L = bwlabel(OC);
imshow(L,[]);

imwrite(L/max(L), "C:\Users\RAFAEL JACULBIA\Documents\subjects\186\activity9\C1_10_labeled.bmp");
//imwrite(CO/max(CO), "C:\Users\RAFAEL JACULBIA\Documents\subjects\186\activity9\c2\C2_01_labeled2.bmp");
//imwrite(OC/max(OC), "C:\Users\RAFAEL JACULBIA\Documents\subjects\186\activity9\c2\C2_01_labeled3.bmp");
//imwrite(clos/max(clos), "C:\Users\RAFAEL JACULBIA\Documents\subjects\186\activity9\c2\C2_01_labeled4.bmp");
scircle = [];
for i=1:max(L)
circle = size(find(L==i));
scircle(i) = circle(2);
end

scircle

----------------------------------------------------------------------------
The structuring element I used was a 5x5 pixel "circle" which actually looks more like a cross than a circle drawn using GIMP. I performed opening and then closing operations on the images. In the code I displayed the resulting image with different circles "labeled" a typical image is shown below:
The areas of all subimages were tallied and a histogram of the values were obtained:

I selected only areas between 300 and 800, since I believe areas out of this range are either circles grouped together or small pixels which are the result of dividing the images or is an artifact of binarizing the images. The value with largest occurence was found at 547 pixels. To confirm if this value is correct, I measured the diameter of one circle and found it to be 26 pixels corresponding to an area of 530 pixels, hence the obtained areas was near the calculated values. Another method I used to confirm the area is isolating a single circle, binarizing it and counting the number of pixels. Using this method I was able to obtain an area equal to 536 pixels which is still near the calculated value.

I also tried entering the whole image into scilab. It is interesting to note that limiting the values for area is not neccessary if we performed the operation on the whole image. I was able to obtain an area of 557.8 which is near the value obtained if I subdivided the images. The histogram is given below:




For this activity I will give myself a grade of 10 since I believe I have performed the activity correctly.

Eduardo David Helped me in this activity.