carmilumban_ ap186

Determination of grain size

2010-10-24T01:25:00.000-07:00

Image Processing

.. wala nang grades at stress this time

Plate 1. Unknown Sample

Jeffries Method

Table 1

Table 2

Mean Lineal Intercept Method (Heyn's Method)

Table 3

Table 4

Image processing using Scilab

Histogram

Table 5

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Spheroidized Steel

Jeffries Method

Table 1

Table 2

Mean Lineal Intercept Method (Heyn's Method)

Table 3

Table 4

Image processing using Scilab

Histogram

Table 5

A18 – Noise models and basic image restoration

2009-10-03T05:25:00.001-07:00

In this activity this is what I did:

Procedure:

1. Add noise to the original image:

Original Image

Noises:
I applied the following noises using genradn and imnoise in Scilab and obtained their corresponding PDF p(z):

2. Filter the noisy images

I made and then applied the following filters on each type of noise in (1).

Assuming a rectangular sub-image with of size n*m, n=m=3( for nearest neighbor perimeter)

1.ARITHEMETIC FILTER

2.GEOMETRIC FILTER

3.HARMONIC FILTER

4.CONTRAHARMNIC FILTER

Q=+1,- 1

3. Determined which type of filter suited for each type noise.

In the following images the arrangement is as follows:

from the top- left-right :
(1) NOISY image (2) Filtered Image with Arithmetic Filter, (3) Arithmetic Filter
bottom-left-right is :
(4) Harmonic Filter, (5) Contraharmonic Filter : Q=-1, (6) Contraharmonic Filter : Q=+1.

1. GAUSSIAN OR NORMAL NOISE, PDF AND FILTERING

All but the geometric filter was able to somehow remove the gaussian noise in the image.

2. RAYLEIGH NOISE , PDF AND FILTERING

From these results, all the filter worked but I think the contraharmonic filter with positive Q =+1 was most effective in removing the uniform noise since the filter at Q=+1 is beginning to restore the white circle at the center.

3.EXPONENTIAL NOISE, PDF AND FILTERING

All the filter worked except the arithmetic filter.

4.GAMMA NOISE AND FILTERING

The arithmetic and geometric filter cannot remove gamma noise.

5.UNIFORM NOISE , PDF AND FILTERING

The contra-harmonic filter was most effective since its beginning to restore the white circle at the center.

6.SALT AND PEPPER NOISE AND FILTERING

Again , the contra-harmonic filter wins.

7.SPECKLE NOISE AND FILTERING

Again , the contra-harmonic filter wins (again).

SOURCE: A18 – Noise models and basic image restoration manual

ASSESSMENT: 10/10 because I was able to code and implement all the image noise and image filters :) and use imnoise. The main problem was that for the results I only used the imwrite outputs. The imshow rendered different effects that seemed better than end results saved in imwrite :) ..odd

GRATITUDE: I would like to thank Mandee and Alva for discussing with me :) (salamat) and Gibert for the rayleigh.

A19 – Restoration of blurred and noisy image

2009-10-03T05:22:00.000-07:00

Objective: Be able to restore the original image from a degraded image using Wiener filter

Steps:

1.Motion blur the original image by taking

or the product of FT of f(image) and h (spatial degradation function) that is G(u,v):

where F(u,v) is the Fourier transform of the original image and H(u,v) is the trasfer function given by
where a and b is the total distance for which the image has been displaced in the x and y direction.
The blurred image g(x,y) is obtained by taking the ifft of G(u,v).This is shown below with a=b=0.01 and T=1.

2.Add noise to this blurred image

For the noise I used: n=grand(ix,iy,'nor',0.05,0.06);

The blurred and noisy image:

3.Use Wiener filter to restore the original image.

The Wiener filter aka minimum mean square error filter or least square error filter.

As stated earlier,the objective is to find an estimate of the original image f(x,y) from the of the corrupted image such that the mean square error between them is minimized. Based on these conditions, the minimum of the error function is given in the frequency domain by the expression :

where

From here the restored image f(x,y) can be retrieved as the ifft of F(u,v).

4. Effects of varying the values of a,b and at constant T=1(sec) and applied noise (normal / gaussian).

as expected , there will be no capture image as a=b=T=1.

Here we notice that as the distance a,b is reduced the image captured becomes less blurred and we can see that the text become more clear. Moreover, as the blurring is reduced the degradation due to the noise become more apparent and becomes a major factor in precluding the restoration of the image. So.

5.Effects of varying the noise added while at constant a=b=0.01, and T=1.

APPLYING RAYLEIGH NOISE:

APPLYING EXPONENTIAL NOISE:
APPLYING UNIFORM NOISE:
APPLYING NORMAL OR GAUSSIAN NOISE:
APPLYING SPECKLE NOISE:
The speckle noise was easily removed by the W.Filter as seen in the image above but the restored image still retained the degradation effects of the blur.

APPLYING SALT AND PEPPER NOISE:

Some errors and remarks:

I used the imwrite (Scilab) versions of the images for the blurred, noisy and restored and not the imshow.
I don't know but the resulting imshow and imwrite of the images are somewhat different:

(LEFT) imshow (RIGHT) imwrite

SOURCE: A19 Restoration of blurred and noisy image

ASSESSMENT: 10/10 since I was able to implement the motion blurr and Wiener filter in Scilab and get the results for various parameter :) I also enjoyed doing this activity.

A16 Neural Networks

2009-09-17T00:55:00.001-07:00

We are done with pattern recognition and classification by (1) mean distance method and (2) probabilistic classification by LDA. Let us now try another classification technique by way of (3) probabilistic neural networks. A neural network is a mathematical model of how neurons in the brain work. Among the many function and varied capabilities of the brain is classification.

Just like the brain, neural network can perform tasks based on experience. It can carry out mapping or classification only after learning the patterns of the classification scheme, something which can be taught by giving examples or training the program. Each example serves to improve the neural network program’s recognition processing.

The architecture of this neural network is illustrated as:

Input layer- is where the features of the set are fed. This first layer computes distances from the input vector to the training input vectors and produces a vector whose elements indicate how close this input is to a training input.

Hidden layer- acts contributions from the input layer by computing the summed probabilities for each class of inputs and passes the results to the output layer.

Output layer-outputs which class these set of feature vectors pertain/ belong to.

In Scilab, we use ANN or Artificial Neural Network to do the classification of the set of features extracted in activity 14. The program used was adapted from Cole’ Blog.

In Activity 14, the samples consisted of 2 class of chips, the Cheese- it labeled as (0) and Pillows as (1). Each class has 12 samples and each sample is described by 4 features namely: hor*ver or area 1, area 2(#pixels), perimeter, and the ratio of area2/perimeter.

Procedure:

1.Specify the no. of neurons in (4)input layer,(4) hidden layer and (1)output layer.

The ANN architecture

N = [4,4,1];

There are 4 features (normalized) extracted from each sample:
These features were specified as x:

x=read('data6train.txt',-1,4);

2. Training

Used 12 objects, 6 from each class in training process as the example classification

t=[0 0 0 0 0 0 1 1 1 1 1 1];

3. Learning

The learning rate and threshold are specified by:

lp = [0.1,0];

Here 0.1 and 0 is the threshold for the error tolerated by the network:

Training cycles or iterations:

T = 1000;

4. Run the classification using ann_FF_Std_online and ann_FF_run(x,N,W);

W = ann_FF_Std_online(x,t,N,W,lp,T);

c=ann_FF_run(x,N,W);

c=c'

5. Results and Discussion

The expected output values were supposed to be in 0’s and 1’s. However the actual results as seen in column 1 Table1 were in decimal. Still, values for each sample were very close to 0 and 1’s and from here we can say that the program has learned the proper grouping of each sample.

Next, I tested whether the program is now ready for the real classification.

This test worked well and the results show that the program worked with 100% correct classification.
Then to see whether or not the program is dependent on the arrangement of the samples, I shuffled the arrangement of the test samples.

Again the test worked well and the samples were classified correctly and showed that this time the program can work smoothly as it can adapt to the arrangement of the inputs.
Then I tested the effects of varying the parameter T which is the no. of cycles or iterations.

As expected lowering the T value reduced the accuracy of the classification while increasing it improves the accuracy, although doing that took a rather a longer runtime. Still the classifications for both cases were all correct.

Afterwards I shuffled the 24 samples to test whether this program can still sort them into their respected classes. Then I just rearranged the output in order.

These tests proved that the program has learned the classification pattern and is 100% accurate.

SOURCE: A16 Neural Networks Manual and Cole' Blog

ASSESSMENT: 10/10 because I was able to use NN in Scilab and make it classify my samples into their respective classes correctly.

A17 Photometric Stereo

2009-09-16T17:57:00.000-07:00

Fig1. Images I1,I2,I3 and I4

Based on the manual for this activity , from the point sources locations (1) and intensities obtained from the 4 images of the object we are able to compute for g via equation (2). Next we determine the normal vector(3), and then compute component surface normals(nx,ny,nz) from each row.

And relate these surface normals to the partial derivatives of f (4) and compute the elevation z=f(u,v) by taking the integral of in (5) via cumsum of (4) over x and y. This can be easily plotted in 3d using mesh.

Fig 2. 3D plot of the object shape using mesh

SOURCE: A17 Photometric Stereo Manual

ASSESSMENT: 10 because I was able to reconstruct the surface using the directions in the manual :) and I was able to use mesh :)

A15 Probabilistic Classification

2009-09-02T19:25:00.000-07:00

Linear discriminant analysis (LDA) is one of the techniques used pattern recognition to classify two or more classes of objects and in this case the separation of the 2 kinds of chips using the 4 features aextracte from previous activities.

The classification works by taking the combinations of features based on differences. Say the objects’ features are linearly separable, we can use linear discriminant model (LDA) formula given as:

SOURCE: http://people.revoledu.com/kardi/tutorial/LDA/LDA.html#LDA

This is a summary of the data used in activity 14:

Table 1.

Procedure:
1. LDA
After processing the data in Table 1 as input x in the implemented LDA formula in Scilab, each sample was easily classified as either belonging in class 1 or 2 depending on its value f.
2. Determine if f1>f2:
By comparison, a large value of f1 would mean that this object belonged to class 1 and and class 2 if f1 is small.
The LDA worked successfully in assigning the 24 samples to its respective class.

ASSESSMENT: 10/10 because I successfully implemented LDA in scilab and used EXCEL to summarize the data into tables :)

A14 Pattern Recognition

2009-08-26T10:11:00.000-07:00

PROCEDURE:
1. Trials
Pattern recognition by way of shape classification was implemented in this activity where I took an image full of various shapes(Fig.1a),convert it to black and white after thresholding, and took the edges of each shape, then labeled each shape and computed its perimeter by bwlabel and tabul.

a.original imageb. bw of the original im

c. thind. skeleton

Figure1.

Figure 2. Took the edge this time
Figure 3. Barplot of each shape tagged from 1-30 and its corresponding perimeter.

From Figure 3, the tally of the shapes are :DEW: 8,HEART: 5, petal :6, circle :10, leaf:1 and these results agree with the number of occurrence of those patterns in Figure 1,2. :)

2. Minimum Distance Classification
Applying this,we captured an image of 2 kinds(2 class) of chips and extracted the best features of the objects. After obtaining these features I separated them in half as training feature and test features.For each class, I computed mean feature vector mj and which I then used to classify the remaining test features. The one with the minimum distance was used determining which class an object belonged to. Afterward, the percent error of the classifier was then computed.

The original image (Fig4)has these vertical that I cannot remove using fft so I just used Gimp to remove it.Then I converted the image to grayscale then to bw after thresholding. I used dilation and erosion to finally clean the image. Here they are(Fig5).
I then extracted the following features:

1. AREA1 from the vertical and horizontal length of each chip,

hor=[];ver=[];this=1;

for i=2:n

[x,y]=find(L==i); hor(this)=max(y)-min(y); ver(this)=max(x)-min(x);

area1=ver.*hor; this=this+1;

end

area1

2. AREA from the length of the bwlabel(L,n) then the are itself by area=tabul(L,"i")

3. the perimeter using edge of the bwlabel(L,n) then the are itself by perimeter=tabul(L,"i")

4. ratio of area /perimeter by (area./perimeter)

5. rgb component of each chip by

r=im1(:,:,1);g=im1(:,:,2);b=im1(:,:,3);red=[];blue=[];green=[];

this=1;for i=1:n

[x,y]=find(L==i);

//at that chip lang

red(this)=mean(r(x));

green(this)=mean(g(x));

blue(this)=mean(b(x));

this=this+1;

end

//lhat ng chip

red

green

blue

After obtaining these features I separated them in half as training feature and test features.For each class, I computed mean feature vector mj and which I then used to classify the remaining test features. The one with the minimum distance was used determinining which class an object belonged to. Afterwards, the percent error of the classifier was then computed.

Fig 4 Image of the chips :), cheeseit and pillows

Fig 5 After removing the lines, graycaling, converting to bw, dilation and erosion .
From here got the area from the vertical*horizontal line running across each chip and area from the sum(pixels)

Fig 6 Got the perimeter from the edges.

Table1: Summary of the extracted features.

Table 2. Summary of the data after Minimum Distance Classification. In the classification I just used area1, area,perimeter, and area/perimeter in the classification(para madali).

SOURCE: A14 Pattern Recognition Manual

ASSESSMENT: 10/10 because I implemented my idea of using of using only bwlabel and tabul in getting the features from areas to RGB of each sample.I was also able to apply other techniques from the previous activity. Then the classifier was 100% error free in classifying the two chips. (yes:)

A13 – Correcting Geometric Distortions

2009-08-19T19:02:00.000-07:00

Two forms of distortion commonly encountered in image processing :

1.Barrel effect

This effect gives the image an inflated or blown up appearance .

2. Pincushion Effect

This effect gives the opposite of barrel distortion wherein the image suffers from deflated appearance.

PROCEDURE:
To correct for distortion we need to capture a reference image with a regularly occuring pattern so that we can visualize the distortion across the image frame.

The correction will be done on two image properties:
1. pixel coordinates and
2. graylevel values

1. From the most undistorted part of the image the ideal grid vertex points were located and the reference image for the corrected constructed.
2.For each rectangle compute c1 to c8 using pairs of equations (x's,y's) from

the four corner points. Per pixel in the ideal rectangle, the location of that point in the

distorted image was determined. If the computed distorted location is integer-valued, the grayscale value from the distorted image onto the blank pixel was used. If the location was not integer-valued, the interpolated grayscale value was computed instead using the equations in the manula provided in the source.

SOURCE: A13 Correcting Geometric Distortions Manual

Figure 1 . Distorted image

Figure 2. Reference image for the distorted image in figure1. pixel distances from a square in fig1.

Since it was really difficult to correct the distortion in Figure 1 at once, I just tried to correct the distortion in 1/2 the image.
So far

Figure 3. a. Half portion of image 1, (b) Corrected image distortion using linear approximation and (c) using bilinear interpolation. The crooked lines in b and c are partly due to the error in the location of the corner points in locate(and the sobrang grabeng distortion).

The quality of the corrected image using linear approximation was not as good as that of the bilinear interpolation since the image for the linear approximation using only the nearest neighbors resulted to more jagged corners for each square in the grid.

ASSESSMENT: 9/10, because I didn't get the complete distortion correction of the image though I was able to implement the distortion correction in Scilab. The main problem and the hardest part was the location of the distorted points xi and yi. Eventhough I used locate I still couldn't get the exact endpoint for each square. I should have limited myself in either barrel or pincushion distortion. :)

GRATITUDE:I would like to thank Mandee and Alva for teaching me how to do this activity :)

A10 Preprocessing Text

2009-08-06T22:46:00.000-07:00

Int this activity we tried to extract handwritten text from an imaged document with lines. This is often used in in handwriting recognition where individual examples of letters must be extracted.

Fig. 1 The problem

Fig. 2 Cropped image and mogrified to remove angle

Fig. 3 Histogram equalization 1

Fig. 4 Histogram equalization 2

Fig. 5. Used the enhanced image in Figure 4. So, from the left the image, its shifted Fourier tranform, its mask to remove the horizontal lines and the final image with only the text and the horizontal lines gone.

Figure 6. The histogram of the masked or cleaned image with handwriting. After thresholding , it was converted to a binary image to extract the text, but the text is still dirty.

In Figure 5 the desired text is already achieved and the image cleaned , but trying to extract the text requires more cleaning. But I think results inFigure 5 is already ok. I am still working on the threshoding and cleaning Figure 5 image... to get better results than what is shown in Figure 6.

A11 Color Image Processing

2009-08-05T19:29:00.000-07:00

Following the set procedure:
1.We take the picture of colorful objects and an ensemble of objects with different shades , in this case shades in green from leaves.
2. Under a fixed light source (sunlight, sunny) we took the their picture under proper and incorrect white balancing through the Auto, Cloudy, Daylight , Fluorescent and Incandescent settings.
3. We proscessed these images through the white and gray world algorithm.
4. For the WWA we took the the RGB values of pixels belonging to the white object of objects. Let these be Rw, Gw, Bw and implement the White Patch algorithm by dividing all the pixels in the red layer of the image by Rw, the green layer by Gw and the blue layer by Bw. After making sure that no pixel exceed 1.0, the resulting image is shown in the second column in Figure 1.
5. For the GWA we took the average red, green and bluevalue of the captured image and used them as the balancing constants and the resulting image is shown in the third column.

Figure 1.

6. Afterwards we did the same procedure for the ensemble of green leaves and put them agains a white background. And did both for the proper and improper white balancing for the outdoor illumination where AUTO nad DAYLIGHT provide the proper white balancing while CLOUDY, FLURESCENT and INCANDESCENT are the improper.

Haha I couldn't make the imwrite fcn work in this activity so I just resorted to PrtSc and the images are saturated since imshow has limited resolution and still saturated after multiplying the imshow by 0.5.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

A12 Color Image Segmentation

2009-08-05T18:54:00.000-07:00

Color segmentation can be performed by determining the probability that a pixel belongs to a color distribution of interest.This is how we can do just that.

1.The histogram of the region of interest (ROI) must first be extracted.

2.Tag a pixel as belonging to the ROI or not is to find its probability of belonging to the color of the ROI

3. We used the sample code in the manual to show just how histogram backprojection is implemented to tag the pixel and get its histogram value in chromaticity space.

Figure 1.Normalized chromaticity
space. X-axis is r and y-axis is g.[manual]

Figure 2. Image 1-Christmas balls

Figure 3. The top 3 RGB patches are taken from the RGB regions in figure 2. The 2D histograms at the bottom illustrate that these ROI coincide with the small region in their individual R G B space in the NCC shown on the left with the variation corresponding to the pixel chromaticity.

As expected, images of the 3D objects such as bright colored balls have shading variations . Given that its shading variation due to brightness is small, for the red green and blue ROI pathces its pixel chromaticities for RGB will occupy a small portion in its region in NCC as in Figure 3.

The next step was to see which method was more effective in the tagging of the pixels RGB value with the first using use of the red and green joint PDF by parametric segmentation with a Gaussian Distribution function and the other using non parametric segmenatation which takes the 2D histogram of the ROI that is then used segment the image using histogram backprojection. For both technique the result should yield white or high peaks (red in hotcolormap) for matches in the range of red and green of the ROI to that image pixel's RGB.

The results and code are shown below:

Figure 4. The ROI was the red patch in the left middle of the red ball.

Figure 5. The ROI was a blue patch in the left middle of the blue ball.

Figure 6. The ROI was a green patch in the left middle of the green ball.

Figure 7. Plots of Gaussian Probability Distribution of the red, blue and green ROI

It can be noted that both the parametric method and non parametric method were indeed effective in highlighting the image pixels with the same RGB probability distribution as seen in the rendered image.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Code:

1st part:
//////1. Capture a digital image of a 3D object with a single color (e.g. colored mugs,faces, hands or any exposed skin, brightly colored balls, etc.). You may also
//////use colored pictures you've captured in previous activities.
//
//

stacksize(4e7);
I=imread('red.jpg');

//
//R = I(:,:,1); G = I(:,:,2); B = I(:,:,3);
//Int= R + G + B;
//r=R./I;
//g=G./I;
//b=B./I;

I = double(I); //I is the image of the ROI
R = I(:,:,1); G = I(:,:,2); B = I(:,:,3);
Int= R + G + B;
Int(find(Int==0))=100000;
r = R./ Int; g = G./Int;
BINS = 32;
rint = round( r*(BINS-1) + 1);
gint = round (g*(BINS-1) + 1);
colors = gint(:) + (rint(:)-1)*BINS;
hist = zeros(BINS,BINS);
for row = 1:BINS
for col = 1:(BINS-row+1)
hist(row,col) = length( find(colors==( ((col + (row-1)*BINS)))));
end
end
scf(1);
subplot(221);
imshow(I)
subplot(222);
imshow(hist,[]);xset('colormap',jetcolormap(64));
//imwrite(hist,'redh.jpg');

/////////////////////////////////////////

I=imread('green.jpg');

//
//R = I(:,:,1); G = I(:,:,2); B = I(:,:,3);
//Int= R + G + B;
//r=R./I;
//g=G./I;
//b=B./I;

I = double(I); //I is the image of the ROI
R = I(:,:,1); G = I(:,:,2); B = I(:,:,3);
Int= R + G + B;
Int(find(Int==0))=100000;
r = R./ Int; g = G./Int;
BINS = 32;
rint = round( r*(BINS-1) + 1);
gint = round (g*(BINS-1) + 1);
colors = gint(:) + (rint(:)-1)*BINS;
hist = zeros(BINS,BINS);
for row = 1:BINS
for col = 1:(BINS-row+1)
hist(row,col) = length( find(colors==( ((col + (row-1)*BINS)))));
end
end
scf(2);
subplot(221);
imshow(I);
subplot(222);
imshow(hist,[]);xset('colormap',jetcolormap(64));

//imwrite(hist,'greenh.jpg');

/////////////////////////////////////////

I=imread('blue.jpg');

//
//R = I(:,:,1); G = I(:,:,2); B = I(:,:,3);
//Int= R + G + B;
//r=R./I;
//g=G./I;
//b=B./I;

I = double(I); //I is the image of the ROI
R = I(:,:,1); G = I(:,:,2); B = I(:,:,3);
Int= R + G + B;
Int(find(Int==0))=100000;
r = R./ Int; g = G./Int;
BINS = 32;
rint = round( r*(BINS-1) + 1);
gint = round (g*(BINS-1) + 1);
colors = gint(:) + (rint(:)-1)*BINS;
hist = zeros(BINS,BINS);
for row = 1:BINS
for col = 1:(BINS-row+1)
hist(row,col) = length( find(colors==( ((col + (row-1)*BINS)))));
end
end
//scf(6);
//subplot(221);
//imshow(hist,[]);
scf(3);
subplot(221);
imshow(I)
subplot(222);
imshow(hist,[]);xset('colormap',jetcolormap(64));
//imwrite(hist,'blueh.jpg');

2nd part:

stacksize(50000000);

//parametric segmentation

//ROI,image patch
ip=imread('red.jpg');//Imagepatches: red,green blue.jpg
Rp = ip(:,:,1); Gp= ip(:,:,2); Bp = ip(:,:,3);
Ip=Rp+Gp+Bp; Ip(find(Ip==0))=100000;

//normalize
Rp=Rp./Ip;Gp=Gp./Ip;Bp=Bp./Ip;

//for the probability destribution
//mean-nu
rm=mean(Rp);gm=mean(Gp);bm=mean(Bp);

//stdev-sigma
rs=stdev(Rp);gs=stdev(Gp);bs=stdev(Bp);

//whole image
wi=imread('balls.jpg');
R=wi(:,:,1);G=wi(:,:,2);B=wi(:,:,3);

I=R+G+B;I(find(I==0))=100000;

//normalize
R=R./I;G=G./I;B=B./I;//

//probability in r and g

pr=(1.0/(rs*sqrt(2*%pi))) *exp(-1.0*(R-rm).^2/(2*rs^2));

pg=(1.0/(gs*sqrt(2*%pi))) *exp(-1.0*(G-gm).^2/(2*gs^2));

p=pr.*pg;//joint prob.
//normalize
p=p/max(p);

scf(1);
imshow(p,[]);xset('colormap',hotcolormap(256));//parametric in Gaussian
imwrite(p,'prredpatch.jpg');xset('colormap',hotcolormap(256));

//hm=histogram of the patch

[row,col]=size(ip);

BINS=32;

r=0:1.0/32:1; g=0:1.0/32:1;

ht = zeros(BINS,BINS);//EMPTY

for i=1:row //of imagepatch
for j=1:col
//get
fr=find(r<=ip(i,j,1)); fg=find(g<=ip(i,j,2)); //find frr=size(fr,2); fgg=size(fg,2); ht( fr(frr), fg(fgg) )=ht( fr(frr) ,fg(fgg))+1 ; end end //the backprojection to the whole image [xi,yi]=size(wi);//whole image tag=zeros(xi,yi);// EMPTY for i=1:xi for j=1:yi //get fr=find(r<=wi(i,j,1)); fg=find(g<=wi(i,j,2)); //find frr=length(fr); fgg=length(fg); tag(i,j)=ht(fr(frr), fg(fgg)); end end scf(2); imshow(tag, []);xset('colormap',hotcolormap(256));// non paramertric imwrite(tag,'tagred.jpg');xset('colormap',hotcolormap(256));
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Acknowledgements: Mandee salamat sa pagturo ng parametric segmanetation :)

A9 – Binary Operations

2009-08-03T18:32:00.000-07:00

The activity goes like this:
From an image we must be able to obtain the a value of the cell area. Getting that means that we must be able to properly threshold the image before we can convert it to black and white and be able to separate as many cells as we can from the cluster.

First I crop a 256x256 patch from Figure 1 and called it co1. Converted it to graycale, then got the histogram to obtain a threshold so I can separate the cell from the background and turn co1 to a bw image. This is in Figure 2.

Afterwards, with no erosion nor dilation, I used paint to crop 5 distinct well separated cells and got its area using sum(im1).I used the mean area ai=540 as reference area before getting the areas from different regions of the image.

In order to removed the holes and other irregularity in cells and reduce some distinct cells from the clusters, I used a double erosion to remove holes and separate some cells and double dilation to reconstruct some separate cells.

However not all cells were completely separated and they remained in clusters. No matter, the areas of these cells or clusters are then computed as follows,

a.Identification and labelling by [L,n]=bwlabel(im)

b.Getting the area of the n labeled cells by getting the area via tabul(L,'i'). This way I get the frequency of each i in n which is equivalent to that cell or cluster area. I just set the limit the computed areas in the range of 500<600>

Finally, the average area obtained with this method is af= 531.38889 or 531 and the standard deviation of the values from this mean is 14.034155. Furthermore, the reference area mean is ai= 540, and the percent difference of the computed area from the reference is only 1.6%. This suggest that the values are very close and the calculation was indeed correct.

References:
[1]Activity 9 AP 186 Manual
[2]http://groups.google.com/group/face-rec/browse_thread/thread/0d29fb3a357a530a?pli=1
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

stacksize(50000000);

apat=4;

nito=1;//areas nito

for i=1:apat

cut=imread('C:\Users\USA\Desktop\co'+string(i)+'.png');

cut=im2gray(cut);

cut=im2bw(cut, 0.80);

//imwrite(cut,'C:\Users\USA\Desktop\cobw'+string(i)+'.bmp');

x= [-2.0:.5:2];

[X,Y] =ndgrid(x,x);

[sx,sy]= size(X);

r= sqrt(X.^2 + Y.^2);

c=zeros(sx,sy);

f=find(r<=2);

c(f)=1;

im=cut;

se=c;

//h=scf(i+1);

//1.erodeerode. to close holes

im=erode(im, se); im=erode(im, se);

im=dilate(im, se);im=dilate(im, se);

//h=scf(i+3);

//imshow(im);

//imwrite(im,'C:\Users\USA\Desktop\cobwdilate erode use tabul'+string(i+sho)+'.bmp');

[L,n]=bwlabel(im);

//find the area

mm=tabul(L,'i');// frequecny ng values na eun(0,1,2..n n cell)=area ng cell

xx=mm(:,2)//area

make it easy remove area na higher than 600 na

fi=find(xx<=600 & xx>=500);

xx=xx(fi)

end

-->Size: 258 rows X 258 columns

Truecolor Image

xx =

529.

554.

514.

523.

524.

533.

526.

Size: 258 rows X 258 columns

Truecolor Image

xx =

530.

521.

542.

522.

560.

559.

Size: 258 rows X 258 columns

Truecolor Image

xx =

533.

536.

523.

531.

526.

524.

521.

518.

529.

Size: 258 rows X 258 columns

Truecolor Image

xx =

528.

518.

515.

530.

areas=read('areas.txt',-1,1);

f=tabul(areas,'i');

x=f(:,1);

y=f(:,2);

bar(x,y);

areas=read('tabul areas.txt',-1,1);

f=tabul(areas,'i');

x=f(:,1);

fi=find(xx<=600)

fi=find(xx>=500);

x=x(fi);

y=f(:,2);

y=y(fi);

bar(x,y);

A8 – Morphological Operations

2009-08-03T17:58:00.000-07:00

Part I. Paper and pencil
First weee predicted the effects of erosion and dilation of different shapes ,namely, square (50×50) , a triangle (base = 50 , height= 30), a circle (radius 25), a hollow square (60×60, edges are 4 pixels thick), and a plus sign (8 pixels thick and 50 pixels long for each line). Each shape is individually dilated and eroded using the following structuring elements or SE :
1. 4×4 ones
2. 2×4 ones
3. 4×2 ones
4. cross, 5 pixels long, one pixel thick.

Figure 1. Triangle ( a.) top : original shape, (b.) middle: dilation and (c.) bottom: erosion.

Figure 2. Circle ( a.) top : original shape, (b.) middle: dilation and (c.) bottom: erosion.

Figure 3. Hollow square ( a.) top : original shape, (b.) middle: dilation and (c.) bottom: erosion.

Figure 4. Square ( a.) top : original shape, (b.) middle: dilation and (c.) bottom: erosion.

Figure 5. Plus ( a.) top : original shape, (b.) middle: dilation and (c.) bottom: erosion.

'under construction'

A7 – Enhancement in the Frequency Domain

2009-07-13T18:08:00.000-07:00

Filtering in Fourier Space
"Unwanted repetitive patterns in an image can be removed by masking their
frequencies in the Fourier domain. Alternatively, desired frequencies in the image
may also be enhanced."

A Convolution Theorem

Figure 1.Results for steps 1 to 3.

Figure 2. Variance

As you vary the variance the FT intensity changes in such a way that increasing the variance increase the radius of the FT.

Fig 3. Inversion

To invert the gaussian dots such that highest becomes 0, lowest becomes 255
or 1.0 , I just multiplied the original values by -1 *255 and then added 255.

B. Fingerprints : Ridge Enhancement

Fig 4. The problem

After converting the image to grayscale, I took the FFT and from here try to create a mask to remove the frequencies of the blotches which is ... unknown.
So what I did is start by removing the higher frequencies...obviously using a circular aperture didnt work as seen in Fig.5.

Fig 5. Removing the higher frequencies with a circular aperture... didnt work

Next I tried the other way around, I didnt mask the higher frquencies but just retained it, then I tried cleaning the frequencies close the center as shown in Fig 6 until I finally tried creating a mask from the FT and just clean the FT as shown in Fig 7. The final almost cleaned finger print is shown in Fig 6 (last).

Figure 5.
Figure 6.

before(original fingerprint)
after-the enhanced fingerprint with ridges more clear and less blotches. This final image is from Figure 6c.

C. Lunar Landing Scanned Pictures : Line removal
Remove the vertical lines in the image by filtering in the Fourier Domain.

Fig 7. Lunar image with vertical lines

a. b.

Fig 8. a.FT and horizontal mask to remove the vertical lines. b. the FFT of thw masked FT, the final image is inverted with no more vertical lines.

Fig. 9. The final cleaned image after rotation.

D Canvas Weave Modeling and Removal

Fig. 10 The canvas with weave pattern

Fig 11. The graycale of canvas image

Figure 12.FT of the image

Fig 12. Tried a combination of gaussian filtering and masking the regions shown in c... didint work , the resulting image in e was too blurred.

Figure 13. Made a mask that removes the symmetric peaks around the center.

The mask in figure 13 b, removed the frequencies of the weave pattern and a cleaned image is retrieved by taking the FT again. Also, the cleaned image has higher contrast such that the brushstrokes are enhanced.

Fig. 14. The unwanted canvas weave pattern is now removed(im from Fig 13 c).

Fig. 14. The inverted mask of the peaks

Fig. 15. The FT of inverted mask. It does look like the canvas weave pattern.

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Code:

a.Variance =2^2

b.Variance=3^2

nx = 100; ny = 100;

x = linspace(-100,1,nx);

y = linspace(-100,1,ny);

[X,Y] = ndgrid(x,y);

s1=4;//variance 1

s2=9;//variance 2

h=scf(1);

z1 = exp((-1*(sqrt((X+50).^2+(Y+50).^2)))/s1);

imshow(z1);

//xs2gif(1,'variance1.gif');

h=scf(2);

ft1=fftshift(abs(fft2(z1)));

imshow(ft1);

h=scf(3);

z2=exp((-1*(sqrt((X+50).^2+(Y+50).^2)))/s2);

imshow(z2);

//xs2gif(2,'variance2.gif');

h=scf(4);

ft2=fftshift(abs(fft2(z2)));

imshow(ft2);

A6 – Properties of the 2D Fourier Transform

2009-07-09T08:56:00.001-07:00

A .Familiarization with FT of different 2D patterns

Figure 1. a FFT of a circle, b. shifted FFT and c. FFT of FT resulting to the same circle.

Figure 2.

For Fig.2, the images above starting from the left are namely, an annulus, a square, square annulus, 2 slits and 2 dots all representing what we call apertures. The images on the bottom are their corresponding FT which is an Airy pattern for the annulus, a sinc function for the square apertures, double slit diffraction of maxima and minima in the horizontal axis for the vertical slits ans the sinusoid for the two dots.

a. b. c. d.

Figure 3. Comparison of a circular annulus and square annulus.

As expected the we have the a Airy pattern for a,b and a sinc function for c,d.
From these results we can say that the size of the aperture determines the intensity of the resulting diffraction pattern. It can be seen that the smaller the aperture is the brighter or more intense the FT or diffraction pattern is.

Figure 4. Variation in the dot size or pinhole radii

IT is shown in Fig.3 that the FT of 2 dots is a superposition of the FT of 2 small dots and FT of a circular aperture.

B Anamorphic property of the Fourier Transform

Anamorphic means something like producing different imaging effects along mutually perpendicular radii. We can see these in FT of perpendicular sinusoid signals.

Figure 5. Increase in the frquency

Increasing the frequency of the signal(sinusoid) resulted to the increase in the separation of the dots. We see here that the two dots corresponding to the frquency peaks shifted farther and father from the center, such that the lower the frequency the closer it is to the center while farther at higher frequency. The same with interferograms ,the spacing between the peaks of the FT of the signal gives corresponding frequency of the signal.

Fig.6 Effect of biasing

For a sinusoid of constant frequency and orientation, we expect that the peaks of the FT to remain the same since and it did and the position of the peaks did not vary. However the intensity of the dots was observe to change such that at low biasing the outer peaks were brighter and reduced at higher biasing which made (m=0) the central maximum peak brightest.

Fig.7

Since adding or subtracting the bias didn’t affect position of the peaks of the FT of the signal , we can just to choose to add the bias . For frequency=4, bias=0.1,1,10,100 was added to the sine wave and the resulting FT of the signal show that the only the intensity of the peaks varied and not the spacing.

Fig. 8

Rotating the signal(sinusoid) by an angle results to the rotation of the peaks of the FT.

Fig. 9

Top image shows the pattern and below is its Fourier transform .Now the combination of these sinusoids in X and Y with the formula

z = sin(2*%pi*4*X).*sin(2*%pi*4*Y);

has a FT that is again symmetric about the center and with peaks that correspond to the frequency of each sinusoid in their respective axis. The combined signal had the same frequency as illustrated in their perfect symmetry but once we changed one of the frequency of the signals, say in the X axis,

z4 = sin(2*%pi*8*X).*sin(2*%pi*4*Y);

the x separation in the vertical orientation increased. We recall that the higher the frequency the farther the peak from the center.

Fig 10

Addition of the combined sinusoids in the X and Y axis to 2 rotated sinusoids (theta=40,60deg) resulted to the image illustrated on top. Below is its FT which by inspection, is just the sum of the FT of each signal. The FT is thus useful for decomposing the frequency component in sampled signals.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Activity 5 Fourier transformers!

2009-07-06T20:22:00.000-07:00

-busy ang neurons...

Lens as Fourier Transformer

6.A – Familiarization with discrete FFT

Figure 1.

Figure 2.

Figure 3.

6.B Simulation of an imaging device

Figure 4.

Figure 5.

Figure 6.

6.C Template Matching using correlation

Figure 7.

Figure 8.

Figure 9.

Figure 10.

Figure 11.

6.D Edge detection using the convolution integral

Figure 12.

Activity 4 Enhancement by Histogram Manipulation

2009-06-29T18:34:00.000-07:00

The histogram of a grayscale image is equal to the graylevel probability
distribution function (PDF) if normalized by the total number of pixels. As such,
the grayscale PDF of an image can be modified in the same way one can modify
the PDF of random numbers.
Histogram manipulation is one way of improving the quality of an image,
enhancing certain image features, or mimicking the response of different imaging
systems, such as the human eye.
Given the cumulative distribution function (CDF) of a desired PDF, the grayscale
PDF of an image can be modified by backprojecting the grayscale values using
the CDF. Then
newly encountered functions in scilab:
tabul - frequency of values of a matrix or vector
tabul allows
you to compute the frequency of of occurence of a vector or matrix
[m]=tabul(X [,order])
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
•Activity 4:
•Two categories of image enhancement techniques include spatial and frequency domain techniques. Spatial techniques involve direct manipulation of image pixels while frequency domain techniques works by manipulation of fourier of the and wavelet transform of a signal.
•Here, the spatial technique will used by way of histogram enhancement. The histogram of an image illustrate the distribution of gray levels in the image in the range 0 to 255 where 0 is black and 255 is white and normalized when in the range [0.0, 1.0].
•The method of histogram enhancement works by equalization.This is one of the ways in which it can be done in Scilab:
•You get the histogram of the image by using the tabul(filename, ‘i’) command that gives the frequency of occurrence of the matrix or vectors, or in this case the no. of pixels having that graylevel and then plot it.
•The histogram:
•x axis for the graylevel
• y axis for the frequency value
•
•Next, normalize the greylevel and you distribute the frequency over the total no. of pixels and thus spread out the most frequent intensity values.

Results and discussions:

Using linear method of histrogram equalization, we see in Fig.1 that the resulting new image has a well distributed histogram as compared to the original image which has uneven probability distribution. The new image has a higher contrast than the original.

In Figure 2, using nonlinear histogram equalization using interp1(,,'linear') was used. However the resulting image has lower contrast than the original.

For this image the linear method was able to enhance the contrast in the image while the non linear method was not and only resulted to reduction in grayvalue of each graylevel.

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Code:
linear

im=imread('C:\Users\USA\Desktop\scilab and 186\grayscale parrot.png');

m=tabul(im,'i');hx=m(:,1);hy=m(:,2);

scf(0);

[px,py]=size(im);

np=px*py;

nx=hx/max(hx);

ny=hy;

ccy=cumsum(ny)/max(cumsum(ny));

subplot(321);

imshow(im,[]);

subplot(322);

plot(nx,ny);

title('PDF1- Histogram 1');xlabel('grayscale level');ylabel('frequency');

subplot(323);

plot(nx,ccy);

title('CDF1');

//2. backprojection of the image through the CDF original

newim= [];

for i = 1:px

for j = 1:py

newim(i,j) = ccy(im(i,j));

end

subplot(324);

imshow(newim,[]);

//scf(1);

//imshow(newim,[]);

//imwrite(good1, 'newim.jpg');

/////for this newimage get the PDF and CDF again

im=imread('C:\Users\USA\Desktop\scilab and 186\newim.jpg');

im=tabul(newim,'i');hx=m(:,1);hy=m(:,2);

scf(0);

[px,py]=size(im);

np=px*py;

nx=hx/max(hx);

ny=hy;

ccy=cumsum(ny)/max(cumsum(ny));

subplot(325);

plot(nx,ny);

title('PDF2- Histogram of Image 2');xlabel('grayscale level');ylabel('frequency');

subplot(326);

plot(nx,ccy)

title('CDF of Image 2');

xs2sgif(0,’figure1.gif’);

code: non linear:

im=imread('C:\Users\USA\Desktop\scilab and 186\grayscale parrot.png');

m=tabul(im,'i');hx=m(:,1);hy=m(:,2);

scf(0);

[px,py]=size(im);

np=px*py;

nx=hx/max(hx);

ny=hy;

ccy=cumsum(ny)/max(cumsum(ny));

subplot(321);

imshow(im,[]);

subplot(322);

plot(nx,ny);

title('PDF1- Histogram 1');xlabel('grayscale level');ylabel('frequency');

subplot(323);

plot(nx,ccy);

title('CDF1');

// for the equalization by backprojection of the image through a desired cdf

xx=[0:1:255];

yy=xx^2;

yy=yy/max(yy);

xxp=[];

yyp=[]

for i=1:px;

for j=1:py;

xxp(i, j)=ccy(im(i,j));

end

yyp=interp1(yy,xx, xxp,'linear');

//yyp=interp1(yy,xx, xxp,' spline');

//yyp=interp1(yy,xx, xxp,'nearest);

newim=yyp;

subplot(324);

imshow(newim, []);

newim=round(newim);

subplot(324);

imshow(newim,[]);

im=imread('C:\Users\USA\Desktop\scilab and 186\newim.jpg');

im=tabul(newim,'i');hx=m(:,1);hy=m(:,2);

scf(0);

[px,py]=size(im);

np=px*py;

nx=hx/max(hx);

ny=hy;

ccy=cumsum(ny)/max(cumsum(ny));

subplot(325);

plot(nx,ny);

title('PDF2- Histogram of Image 2');xlabel('grayscale level');ylabel('frequency');

subplot(326);

plot(nx,ccy)

title('CDF of Image 2');

xs2sgif(0,’figure1.gif’);

Activity 2:Area estimations of images with defined images

2009-06-29T18:14:00.000-07:00

Activity 2 Area estimations of images with defined images

For this activity the area of certain region in an image with a defined geometry is estimated using Green’s theorem. In summation (discrete) form, if there are Nb pixels in the boundary or contour of R then the area of R is

The parametrized x and y coordinates for this computation were obtained using th SCIlab’s SIP toolbox function follow() that has the command
[x,y] = follow(Img).The xy vectors stores the parametized contour and Img is a binary array of 0 and 1 where 0 is equivalent to background and 1 is the object. But before we can do this, the image data is first loaded in for processing.

Imread is used to read the image and obtain an array containing the image data and im2bw is used to convert these data values to binary by thresholding, say 0.5. Then the follow command is used to trace the x y pixel coordinates nearest the edge which we plug in to Green’s function to get the area.

Another way of obtaining the area is by taking the sum of the white region via sum(Img). Basically we sum the 1’s in the binary array. This value is equivalent to what we call the analytical sum and is used to check the experimental area value obtained using Green’s function. The closer these values coincide the lesser the error.

The code made is shown below:

I=imread('C:\Users\USA\Desktop\circle140d.bmp');
b=im2bw(I,0.5);
[x,y]=follow(b);
xsize=size(x);
ysize=size(y);
xlen=length(x);
ylen=length(y);
xi(1)=x(xlen);
yi(1)=y(ylen);
A_g=abs(0.5* sum( xi.*y - x.*yi));
A_sum=sum(b) ;
D= abs( (A_sum-A_g)/A_sum),"Error"

**************************************

I would like to acknowledge Rommel for helping me with scilab and code for loop code.. when it didn’t worked… thanks otherwise I wouldn’t have finished the work.
Grade: 9/10 because of the small error in the results.

A3 – Image types and basic image enhancement

2009-06-22T18:27:00.000-07:00

Images
"An image is represented by a two-dimensional table in which each cell is a pixel. To represent an image by means of computer, it is thus enough to create a pixel table in which each cell contains a value. The value stored in a cell is coded on a certain number of bits which determine the colour or the intensity of the pixel, This is called the coding depth (or is sometimes also called the colour depth. "[http://en.kioskea.net/contents/video/affich.php3]

Using imfinfo(filename, 'verbose') in Scilab , one can determine the type of an image and also determine other file information.

1.Grayscale:
Pixel color depth: 8-bit color

Grayscale images are monochromatic and carries only intensity information composed exclusively of shades of gray that vary from black (lowest intensity) to white (highest intensity) equivalent to 2^4 intensities( 256) or 8 bits per sampled pixel.

Image source:http://en.wikipedia.org/wiki/Grayscale

-->imfinfo('C:\Users\USA\Desktop\scilab and 186\grayscale parrot.png','verbose')
FileName: C:\Users\USA\Desktop\scilab and 186\grayscale parrot.png
FileSize: 29342
Format: PNG
Width: 150
Height: 200
Depth: 8
StorageType: indexed
NumberOfColors: 256
ResolutionUnit: centimeter
XResolution: 72.000000
YResolution: 72.000000
******
On part 6. Bonus – If you can make a program in Scilab to compute and plot the
histogram of a grayscale image, you get 2 points bonus.

im=imread('C:\Users\USA\Desktop\scilab and 186\grayscale parrot.png'); // read image
im=im*255;
scf(0);
m=tabul(im,'i');

hx=m(:,1);
hy=m(:,2);
h=scf(0);

[px,py]=size(im);
npixel=px*py;

nx= hx/max(hx);
ny= hy/npixel;

ccy=cumsum(ny).*(1/max(cumsum(ny)));

subplot(321);
plot(nx,ny);
title('norm hist');
title('nxy');

subplot(322);
plot(nx,ccy);
title('nxccy');

xs2gif(0,'C:\Users\USA\Desktop\histograms2.gif');
****************
2. True color
Pixel color depth: 24-bit color (True Color)

These consist of at least 256 shades of red, green, and blue resulting to around 16,777,216 color variations.

-->imfinfo('C:\Users\USA\Desktop\scilab and 186\color pizza.png','verbose')
FileName: C:\Users\USA\Desktop\scilab and 186\color pizza.png
FileSize: 246909
Format: PNG
Width: 302
Height: 302
Depth: 8
StorageType: truecolor
NumberOfColors: 0
ResolutionUnit: centimeter
XResolution: 23.230000
YResolution: 23.230000

3. Binary

These are two level images, wherein each pixel store only store two possible values, that is, either 0 or 1, black or white but any two color combination can be used.

Image source: http://en.wikipedia.org/wiki/File:Neighborhood_watch_bw.png

-->imfinfo('C:\Users\USA\Desktop\scilab and 186\binary.png','verbose')
FileName: C:\Users\USA\Desktop\scilab and 186\binary.png
FileSize: 2660
Format: PNG
Width: 200
Height: 140
Depth: 8
StorageType: indexed
NumberOfColors: 2
ResolutionUnit: centimeter
XResolution: 72.000000
YResolution: 72.000000

4. Indexed
Indexed images are made up of colors indexed by its position in the pallete or an array of color elements. The color information is not directly carried by the image pixel data.

Image source: http://en.wikipedia.org/wiki/File:IndexedColorSample_(Mosaic).png

-->imfinfo('C:\Users\USA\Desktop\indexed.png','verbose')
FileName: C:\Users\USA\Desktop\indexed.png
FileSize: 9789
Format: PNG
Width: 160
Height: 142
Depth: 8
StorageType: indexed
NumberOfColors: 256
ResolutionUnit: centimeter
XResolution: 72.000000
YResolution: 72.000000

********************************

Getting the area of scanned of images or objects
In this part, a rebisco cracker was photograph and the resulting image was scanned and the area of this biscuit was computed using the resulting plot in Scilab from the traced edges through the command follow.( Whahah I later realized that I should have scanned the rebisco directly in order to lessen the error :p .. oh well )

The procedure and results are as follows:
1. Take picture and scan it.
2. Approximate the physical area via an area of a circle with radius r=diameter/2.
3. Find the computed area from the ratio of the no. of pixels to the actual measurements in the photo or take the plot of the object via the follow() command.
4. The first one was difficult cause the x y coordinates in the photo matching the pixels in paint required alllot of data points and I soon give up this approach.
5. I then used the plot in the follow() command to trace the edges. and approximate the object as a circle and get the radius from the diameter.
6. Also I compared the results of the computed area from analytical sum and the Green's function but wasn't able to relate these values to the physical area. (whahaa)
7. The difference between the analytical area and are computed from green’s function is the error in this estimation. Taking that its area for a circle, the physical area is around 14.13 cm (d=4.5 cm) while the computed are is 10.6814 (w/ d=3.4 cm from the plot).
8. The area from the two physical and numerical methods do not coincide since area of the object was only approximated to that of a circle even thought the object is curved in the edges. The computed area of the image is more accurate since it has taken into account the curved edges but is affected by the distortion of the scanned image. The error is 24.42 image if we take the physical area to be more accurate while 32.30% for the computed area.
9. The sources of errors includes the distortion effect of the camera, lighting or background and the shades or shadow in the edges . These errors may have been avoided if the object was scanned directly (o_o) or photographed in a uniform background lighting to avoid unwanted shading in the edges.
Here are the pix:

The scanned picture:

The image after im2bw and thresholding, and converting the black to white and vv. to get the sum of the white area:

The plot:

.*.*.*.*.*.*.*.

Grade:

Activity 1 Re-plotting with Paint and Excel

2009-06-17T19:57:00.000-07:00

In this activity we scanned a handwritten plot and used Paint and Excel to replot this.
To do this:
1.A number of x and y coordinates in pixel positions of the scanned plot are obtained one by one starting from the left side and listed down in Excel. It must be noted that the origin in Paint is at the top left part.

2. Next, the origin of the scanned plot is noted down and the number of pixel per unit measure in the x and y axis of the original plot is computed to be able to convert the x-y pixel positions to x-y units--use this ratio as the conversion factor.

3. Third, the x-y coordinates for the replotting are computed using the relation of the listed x-y pixels coordinates to the origin of the scanned plot and multiplied by a conversion factor found in 2. This is how:
For the x axis, positive direction (to the right):

x plot=1* (x pixel-x origin) pixel *( measure/#pixel in x)

For the y axis, negative or downward direction:

y plot= -1* (y pixel-y origin) pixel *( measure/#pixel in y)

4. These xplot and y plot coordinates are then plotted in Excel and the picture of the scanned plot is used as the background in Plot area in Excel to superimpose the two to see if the resulting plot coincides with the original using the pixel and units relation.

The figure below illustrates the resulting plot and the table includes the datapoints for the pixels and the corresponding converted values using the formulas in 3.

This plot coincides with the original plot along with the grid and origin.

*************
Acknowledgement: I would like to give thanks to Kaye for teaching us and Gary in the CS lib on how to go about the pixel and ratio of the units while waiting for our library clearance. :) Also, thanks to Janno for rotation of the image in Nero.

Grade: 10/10. Because I was able to understand and enjoy the activity and superimpose the plots right away. :)

Book: Algebra and Calculus 1912, Palmer

A1 DIGITAL SCANNING

2009-06-15T20:30:00.000-07:00

Introduction (ahem...)

In this activity , we use ratio and proportion to fing the numerical values of a digitally scanned hand drawn plot. Using paint and XCel we reconstruct the graph ...kuno.

first day ng class at 186

2009-06-15T20:10:00.000-07:00

testing.....da- blog :)