Activity18: Pattern recognition

Pattern recognition under the domain of machine learning. It has many applications in medicine, gait analysis, speech recognition and text classification. Pattern recognition technique is done through feature extraction and then classification of the gathered features. The test set will then be subjected for comparison based from the features obtained from the training set.

In this activity, we use different objects belonging to different categories which are to be extracted from the data. Below are samples of object images used for pattern recognition technique.

Pillow


Piatos

Fishball

Kwek kwek

For each image, I used R,G,B values, shape and area as classifiers. Each feature is normalized such that the bias to large values or very small values will be removed. 4 samples of each set is used as training set and the other 4 is used as test set.

The table below shows the percent of success of recognition based from the chosen features.

	Pillow	Piatos	Fishball	Kwek kwek
Pillow	4/4	1/4	0/4	0/4
Piatos	1/4	3/4	0/4	0/4
Fishball	0/4	1/4	3/4	2/4
Kwek kwek	0/4	1/4	1/4	4/4

Based from the classifiers,the accuracy of recognition is 87.5%. To increase the accuracy of recognition, we should add more features for classification.


Rating: I will give myself 9.0 points. I was not able to achieve the 100%recognition accuracy due to the limited number of features that I can extract based from the image processing techniques that I know.

Activity 17: Basic Video Processing

For this activity, we applied basic video and image processing techniques to a diffusive system. We used a drop of a very concentrated black ink and allow it to diffuse in warm water (warm for faster diffusion rate). We used a transparent beaker and a white background for easier thresholding of the resulting images which we will convert to a binary image. Water level is kept low to assume a 2D diffusion. The threshold used is the threshold for the image where the system has uniformly distributed concentration. Pixel values above this threshold is included in the pixel counting which corresponds to the total concentration at that time. The area of diffusing ink is expected to increase (Ref. S.Lee et.al.,"Ink diffusion in water",EJP).

In our experiment,we obtained the same behavior of area as a function of time which can fit with the right fitting parameters.



The ROI is the diffusing ink in the middle. Some boundaries reached or surpassed the threshold which adds to the error in area calculation. This was minimized by labeling the blobs, considering only the middle blob to be counted.

Rating: Based from literature existing, I was able to reproduce the resulting dynamics of an ink diffusing in water. Also, I was able to familiarize myself with some basic video processing techniques and integrate the previous area calculation method. Therefore,I'll give myself 10.0 points.

Code:
t=110/255;
c=[];
for i=1:9
im=imread('C:\Documents and Settings\gpedemonte\Desktop\ap186_a17\vid000'+string(i)+'.png');
imb=im2bw(im, t);
imwrite(imb,'C:\Documents and Settings\gpedemonte\Desktop\recon\vid000'+string(i)+'.png');
c(i)=sum(abs(1-imb));
end
for i=10:99
im=imread('C:\Documents and Settings\gpedemonte\Desktop\ap186_a17\vid00'+string(i)+'.jpg');
imb=im2bw(im, t);
imwrite(imb,'C:\Documents and Settings\gpedemonte\Desktop\recon\vid00'+string(i)+'.png');
c(i)=sum(abs(1-imb));
end
plot(c);

Activity 16:Image Segmentation

Similar to previous activities, this time we have to reconstruct a colored image using histogram of RGB.

Figure 1 Sample image used

Figure 2 Cropped image

Figure 3 Reconstructed from the cropped image histogram

Figure 4 RG Histogram

Figure 5 Reconstructed from back projection

Rating: I'll give myself 9.8 points. I used jeric's code,but the algorithm is the same if i use my code. I find it better to use his code.

Code:
ROI=imread('C:\Documents and Settings\gpedemonte\Desktop\ap187-2\activity 3\sample1.jpg');
I=ROI(:,:,1)+ROI(:,:,2)+ROI(:,:,3);
I(find(I==0))=100;
ROI(:,:,1)=ROI(:,:,1)./I;
ROI(:,:,2)=ROI(:,:,2)./I;
ROI(:,:,3)=ROI(:,:,3)./I;

ROI_sub=imread('C:\Documents and Settings\gpedemonte\Desktop\ap187-2\activity 3\sample2.jpg');
I=ROI_sub(:,:,1)+ROI_sub(:,:,2)+ROI_sub(:,:,3);
ROI_sub(:,:,1)=ROI_sub(:,:,1)./I;
ROI_sub(:,:,2)=ROI_sub(:,:,2)./I;
ROI_sub(:,:,3)=ROI_sub(:,:,3)./I;

//probability estimatation
mu_r=mean(ROI_sub(:,:,1)); st_r=stdev(ROI_sub(:,:,1));
mu_g=mean(ROI_sub(:,:,2)); st_g=stdev(ROI_sub(:,:,2));

Pr=1.0*exp(-((ROI(:,:,1)-mu_r).^2)/(2*st_r^2))/(st_r*sqrt(2*%pi));
Pg=1.0*exp(-((ROI(:,:,2)-mu_g).^2)/(2*st_g^2))/(st_r*sqrt(2*%pi));
P=Pr.*Pg;
P=P/max(P);
scf(1);
x=[-1:0.01:1];
Pr=1.0*exp(-((x-mu_r).^2)/(2*st_r))/(st_r*sqrt(2*%pi));
Pg=1.0*exp(-((x-mu_g).^2)/(2*st_g))/(st_g*sqrt(2*%pi));
plot(x,Pr, 'r-', x, Pg, 'g-');
scf(2);
roi=im2gray(ROI);
subplot(211);
imshow(roi, []);
subplot(212);
imshow(P,[]);

//<————-histogram backprohection————————————>
//histogram
r=linspace(0,1, 32);
g=linspace(0,1, 32);
prob=zeros(32, 32);
[x,y]=size(ROI_sub);
for i=1:x
for j=1:y
xr=find(r<=ROI_sub(i,j,1));
xg=find(g<=ROI_sub(i,j,2));
prob(xr(length(xr)), xg(length(xg)))=prob(xr(length(xr)), xg(length(xg)))+1;
end
end
prob=prob/sum(prob);
imshow(prob,[]); xset('colormap', hotcolormap(256));
scf(3)
surf(prob);

//backprojection
[x,y]=size(ROI);
rec=zeros(x,y);
for i=1:x
for j=1:y
xr=find(r<=ROI(i,j,1));
xg=find(g<=ROI(i,j,2));
rec(i,j)=prob(xr(length(xr)), xg(length(xg)));
end
end
scf(4);
imshow(rec, []);

Activity 15: White Balancing

Correcting images captured under different light conditions such that white appears white is a technique called white balancing. To do this, we have to subject an object to different lighting conditions and compare the whiteness of both object and image.

tunsten

automatic

cloudy

daylight

fluorescent

There is variation in the whiteness in the image for different lighting conditions. There are two methods that can be used in performing white balancing, namely, the Reference White and the Gray World algorithm. Here, we determined first the RGB values of white which is used as a reference. The gray world algorithm on the other hand works by assuming the image is gray. On the other hand, the gray world algorithm is implemented by averaging the R-G-B of each channel in the unbalanced image. The red channel of the unbalanced image is then divided by the R above. The same is done for the blue and green channels.


Rating: I'll give myself 9.0 points. Since I was able to do the activity with a good degree of acccuracy. Thanks to Mark and Jeric for the sample images.

Activity 14: Stereometry

In this activity, we wish to recover the depth information and reconstruct the 3D shape of an object via stereometry. This method requires multiple images of an object (at least 2) where the camera position is varied and the object location is fixed or we can change the position of the object and the camera position is fixed. It should be noted that the image plane is the same as well as the object plane.

Here is the images of a rubik's cube which i used as a sample. I exhausted points in between the grids of the cube for the upper face because it is the only part of the image with an expected variation in depth.

The value of f=14mm and b=10mm. For my case, the resulting image is just the upper face inverted due to arrangement of points in the matrix used and data plotting order. The deeper part is the upper portion of the face.

Rating: I'm convinced that i was able to retrieve the depth information based from the results so I'll give myself a 9.50 points. Manual point location perhaps gave rise to errors which resulted to non-uniformities of points in the reconstruction.

The Code:
x1=[183 214 248 283 320 352 385; 182 214 248 283 320 352 386; 181 213 248 283 321 354 387; 180 212 247 283 321 355 389; 179 211 247 283 322 356 391; 178 210 247 283 323 358 393; 176 209 247 283 323 358 395];
x2=[144 174 209 243 280 313 346; 143 173 209 243 280 313 347; 141 172 209 244 280 314 349; 140 171 208 244 281 315 350; 137 169 208 244 281 316 352; 135 167 207 244 282 317 353; 133 165 206 243 281 318 353];
f=14.;
b=10,;
z=[];
z=(b*f)./(x1-x2);
n = 7 // a regular grid with n x n interpolation points
// will be used
x = linspace(0,1,n); y = x;
//z = cos(x')*cos(y);z = cos(x')*cos(y);
C = splin2d(x, y, z, "periodic");
m = 10; // discretisation parameter of the evaluation grid
xx = linspace(0,1,m); yy = xx;
[XX,YY] = ndgrid(xx,yy);
zz = interp2d(XX,YY, x, y, C);
emax = max(abs(zz - cos(xx')*cos(yy)));
xbasc()
plot3d(xx, yy, zz);//, flag=[2 4 4])
[X,Y] = ndgrid(x,y);
param3d1(X,Y,list(z,-9*ones(1,n)), flag=[0 0])
str = msprintf(" with %d x %d interpolation points. ermax = %g",n,n,emax);
//xtitle("spline interpolation of cos(x)cos(y)"+str)n

Activity 13: Photometric Stereo

Using Photometric stereo technique, we reconstruct the shape of a 3D object by capturing the image of the object at different positions of the light source.

We have to compute for the matrix of reflectance g using the equation


where V is the matrix of the light source positions and I is the image matrix.
Plotting the value of the elevations given by equations



the result is a hemisphere.


Rating: I'll give myself 10. since the reconstructed shape is similar to the original object.


The Code:
loadmatfile(’Photos.mat’);

V(1,: ) = [0.085832 0.17365 0.98106];
V(2,: ) = [0.085832 -0.17365 0.98106];
V(3,: ) = [0.17365 0 0.98481];
V(4,: ) = [0.16318 -0.34202 0.92542];

I(1, : ) = I1(: )’;
I(2, : ) = I2(: )’;
I(3, : ) = I3(: )’;
I(4, : ) = I4(: )’;

g = inv(V’*V)*V’*I;
g = g;
ag = sqrt((g(1,:).*g(1,: ))+(g(2,:).*g(2,: ))+(g(3,: ).*g(3,: )))+1e-6;

for i = 1:3
n(i,: ) = g(i,: )./(ag);
end

//get derivatives

dfx = -n(1,: )./(nz+1e-6);
dfy = -n(2,: )./(nz+1e-6);

//get estimate of line integrals

int1 = cumsum(matrix(dfx,128,128),2);
int2 = cumsum(matrix(dfy,128,128),1);
z = int1+int2;
plot3d(1:128, 1:128, z);

A12: Correcting Geometric Distortion