Arabesque

Activity 19: Restoration of Blurred Image

2009-10-01T11:00:00.000-07:00

"... Twilight on the frozen lake
North wind about to break
On footprints in the snow
Silence down below."

This activity initially distorts an image through blurring and adding of gaussian noise. It aims to restore the distorted image based on previous knowledge on the nature of distortions introduced.

Say we initially have an input image f(x,y) (refer to the image below) introduced by a degradation function h(x,y) and Gaussian noise g(x,y), restoration filters then act on the distorted image to produce an image close to the original.

The restored image can be represented by the following equation in spatial domain

and by the following equation in frequency domain.

where H in fourier space is given by the following formula. I used a=0.0005 (vertical blur) and b=0.05 (horizontal blur). The more I increase a and b values, the more blurry the image becomes. Hence, the choice of a and b values correspond to a somewhat blurred image with poor image visibility.

The original image as shown below represents the function f(x,y) and F(u,v) in real space and fourier space respectively.

Original Image

(http://www.musicroom.fr/images/catalogue/fullsize/OP52811.jpg)

Blurring (left) and adding gaussian noise (right) is mathematically represented by the equations presented above.

Left: Blurred Image, Right: Blur+Gaussian Noise

Details on the applied distortions are known beforehand. Restoration is then possible though utilizing functions to obtain the target restored image. Restoration comes in two forms, one is represented by the equation below (6) and the other one is given by equation 7.

After several attempts of trial and error, the best suited K value was 0.03. Surprisingly, the ratio between Sn and Sf is 25.194144. By virtue of comparison, K is equal to this ratio.

And now we obtain the following restored images. It can be oberved that using the ration between Sn and Sf for each pixel in the image turned out to be a better restoration option than K estimates.

Left: Restored (eqn 6), Right: Restored (eqn 7)

Equation 6 is better than equation 7 since the latter only estimates a fixed value for K.

I give myself a grade of 10 for this activity. I think I have understood all concepts here.
Thanks to Mandee and Carmen for the very interesting discussions.

And since this is my last blog for this course, Thanks Ma'am Jing and Ma'am Gay for the patience, and brilliant mentoring. :D

Lyrics: Portishead, Bob Dylan

Activity 18: Noise Models and Basic Image Restorations

2009-10-01T09:36:00.000-07:00

"You're gonna cry for pills

And your head's gonna be in a knot,
But the pills are gonna cost more
Than what you've got."

This activity aims to create distorted images by adding different types of noises to them namely, Gaussian, Exponential, Rayleigh, Salt and Pepper, Erling and Normal noise. After which, the distorted images are restored through also different types of filters namely: Arithmetic, Contraharmonic positive, Contaharmonic negative, Exponential, Geometric, Harmonic.

(http://medisinmusicforthemasses.files.wordpress.com/2008/10/terry_bozzio_drumset2.jpg)

Original Image (left), Histogram (right)
sample image(down)

We note that the visual differences are more often than not unnoticeable and thus negligible. Hence, there is a need to plot histograms for each of the images and interpret from them.

SALT AND PEPPER
(Salt and Pepper, Arithmetic, Contraharmonic positive, Contraharmonic negative, Geometric, Harmonic)

All types of filters except geometric filter seem to exhibit the same results.

SALT
(No filter, Arithmetic, Contraharmonic positive, Contaharmonic negative, Geometric, Harmonic)

All filters seem to have similar effect on this type of noise. Based on peaks presented in the histogram, Geometric filtering is the best among other types.

RAYLEIGH
(No filter, Arithmetic, Contraharmonic positive, Contaharmonic negative, Geometric, Harmonic)

Arithmetic filtering works best for this type of noise as seen on the histogram. Peaks are best portrayed in this type of filtering method.

PEPPER
(No filter, Arithmetic, Contraharmonic positive, Contaharmonic negative, Geometric, Harmonic)

Nothing seems to work for the Pepper noise.

NORMAL
(No filter, Arithmetic, Contraharmonic positive, Contaharmonic negative,Geometric, Harmonic)

Positive Contraharmonic exhibited the best type of filtering method for this type of noise based on peaks of histogram. Geometric on the other hand shows more traces of leftover noise than the rest of the filters used.

GAUSS
(No filter, Arithmetic, Contraharmonic positive, Contaharmonic negative,Geometric, Harmonic)

Positive Contraharmonic is the best type of filtering for this type of noise as shown by the peaks of the histogram. Geometric filter on the other hand has a problem on leftover noise.

EXPONENTIAL
(No filter, Arithmetic, Contraharmonic positive, Contaharmonic negative, Geometric, Harmonic)

Geometric filtering is the best method here based on peaks of histogram.

ERLANG
(No filter, Arithmetic, Contraharmonic positive, Contaharmonic negative, Geometric, Harmonic)

Positive Contraharmonic works best for this type of noise.

Different types of filtering methods work for different types of noise. Hence, it is best to identify the type of noise beforehand and pick the appropriate filtering technique afterward.

Im not quite sure if the results i have obtained are correct. However, I still give myself a grade of 10 for this activity because I think i have understood all concepts.

Thanks to Mandee and Carmen for pruductive discussions. And to Gilbert (of course) for the shortcut on Rayleigh Noise.

Activity 17: Photometric Stereo

2009-09-09T19:11:00.000-07:00

"... You're gonna need
You're gonna need my help someday
Well, if you can't quit your shinnin'
Please quit your low down ways."

Photometric Stereo is a technique used by computers to view object surfaces through exposing them to different types of lighting conditions [1]. This activity aims to extract shapes of surfaces illuminated under four types of lighting conditions.

We can estimate shapes of surfaces by taking several pictures of the surface of the image exposed to light source at different locations. Shadows on the images contain the necessaty infromation for surface rescontrcution. In our case, four images of a spherical object exposed under four types of lighting conditions brought upon by four different locations the light source is located which are imbedded in a matlab file were initially extracted. Images of I1, I2, I3 and I4 are respectively shown below.

I1, I2

I3, I4

Which leads to the construction of a matrix of locations of the following form:

V ( already provided) is equal to the following matrix in this activity

0.085832 0.17365 0.98106 0.085832 - 0.17365 0.98106 0.17365 0. 0.98481 0.16318 - 0.34202 0.92542

Each row corresponds to the N number light sources while columns indicate correspodning x,y and z coordinates. I and V are already provided. We can now compute for g:

and solve for the normal vector n:

To get the shape of the sphere from the normal, the elevation z whcih is also a function of x and y is computed through the following formula

where

The final 3d constrcution obtained is shown below.

Reference:
[1] http://en.wikipedia.org/wiki/Photometric_Stereo

Activity 16: Neural Networks

2009-09-06T09:16:00.000-07:00

".. The sweet pretty things are in bed now of course The city fathers they're trying to endorse"

Numerous processes are known to classify objects among classes they belong to. Linear Discriminant Analysis was already implemented in the previous experiment. This acivity introduces yet another effective classifier known as a neural network. It is a computational model of how neurons in the brain work. The latter method is known more effective than LDA since it is independent of recognition rules to perform classification. The neural network is initially fed with information through which it learns from these set of examples. Once a new set of information is tested on the neural network, it then readily classifies these elements based on previously acquired knowledge.

Again, we use the same sets of data values used in the previous activities. Through the code provided by Cole, a student of AP 186 a year ago, computed class was approximated and was found to perfectly correspond to the actual classes the elements belong to. Thus, we can now say that the network has indeed learned from the information fed to it. It is now ready to classify the second half of the data set (test set).

Following the same procedures as the training set, we get the following results as summarized in the table below.

Once again, predictions were found to be 100% accurate for all elements in the set.

Rating: 10
I have understood all concepts in this activity.

Activity15: Probabilistic Classification

2009-09-03T11:36:00.000-07:00

"...If the wine don't come to the top of yer cup
If the wind's got you sideways with with one hand holdin' on
And the other starts slipping and the feeling is gone"

This activity used similar samples as in the previous activity (sample group 1: Pillows which are wafer covered chocolate and sample group 2: Cheezit which are cheese flavored chips). Classification among the two major classes was already done in the previous exercise. However, a problem arises if a classifier wrongly groups certain samples into their appropriate classes. The aim here is to minimize the risk or loss in classification through a statistical method known as Linear Discrimant Analysis (LDA).

LDA is a method used in Statistics to find linear combination of features or characteristics which best separate two or more classes of objects or events [1]. In other words, LDA is an effective classifier of objects based on features they exhibit.

Now we start with a training set composed of 4 Pillows assigned class 1 and 4 Cheezit assigned class 2. Following all steps provided by the LDA tutorial in [2] and implementing them in Scilab, we come up with values together with discriminant functions shown below.

Discriminant functions (f1 and f2) take full responsibility for classifying between classes Pillows (1) and Cheezit (2). For each element in the set, a comparison is made between the two functions wherein the larger valued function takes over (e.g. if f1>f2 means object belongs to class 1). Results show that LDA exhibited 0% classification error. Indeed, the first 4 elements highlighted in brown were precisely classified belonging to the class Pillows and correspondingly the last four elements highlighted in yellow as Cheezit.

Now we use the same method for the test samples. Results are shown below.

All of the samples in the test set were classified correctly. LDA is thus an effective method for classifying objects to their respective classes.

Reference:
[1] http://en.wikipedia.org/wiki/Linear_discriminant_analysis
[2] http://people.revoledu.com/kardi/tutorial/LDA/Numerical%20Example.html

Rating: 10
I think I have fully understood all concepts discussed here. All elements were classified precisely.

Thanks to mandee and carmen for useful discussions, and the company as well.

Activity 14: Pattern Recognition

2009-08-30T11:38:00.000-07:00

"... But I can hear another drum beating for the dead that rise
Whom nature's beast fear as they come and all I see are dark eyes."

A pattern refers to a set of features or characteristics of objects. Humans readily classify objects around them based on common sense and intuition. On the other hand, Computers do not possess this complex ability despite the drastic advancement of technology these days.

This activity aims to classify certain objects through pattern recognition. 12 pieces of pillows and cheezit were classified based on two features: RGB values and Area. Six elements of each class were used as training elements while the rest served as test elements.

Pictures of 12 pieces for each class (Pillows and Cheezit) placed side by side were initially taken.

Pillows vs Cheezit

AREA ESTIMATION

For area computation, images of test sets (6 elements for each class)were thresholded in Scilab resulting to images shown below (left: Pillows; right: Cheezit).

Similar methods used in Activity 9 were implemented on these images. The function bwlabel in scilab was used to label pixels which correspond to a specific element (piece of Pillow or Cheezit) in the image. After which, pixels with the same labels were added. This sum corresponds to the estimated area of a single element. The same methods were implemented on the other half (test set).

The table below summarizes area of samples 1-6 (training set) and 7-12 (test set).

ESTIMATED AREA

Pillows
(Training Set)
Mean: 7424.1667
Std dev: 501.19115

Cheezit
(Training Set)
Mean: 16737.167
Std dev: 1413.6567

Red-Green-Blue

RGB values of each of the elements for both classes (training and test set) were sampled through image patches and summarized in the table shown below.

RGB VALUES

Pillows
(MEAN)
R: 0.514855633
G: 0.3820315
B: 0.243684817

Cheezit
(MEAN)
R: 0.514855633
G: 0.3820315
B: 0.243684817

CLASSIFICATION PROPER

RGB values and Area estimates were then plotted for all test elements. including mean values of training elements. The yellow data point corresponds to the mean value for Cheezit while the brown one corresponds to Pillows. Based on the figures shown, we infer that an imaginary line drawn along the diagonal would separate the two classes into Pillows and Cheezit. Clearly, the blue data points here correspond to Cheezit while the red ones to Pillows.

Rating: 10
I have understood all concepts and did what was asked in this activity.

Thanks to mandee and carmen for the pictures (and snack).

Activity 13: Correcting Geometric Distortions

2009-08-30T03:04:00.000-07:00

"... Blue moon, you saw me standing alone
Without a dream in my heart, without a love of my own."

This activity aims to correct distortions on images. Among the most common distortion is barrel distortion where the image is often viewed somewhat distorted in mid areas and pinched on its sides.

There are many ways to restore distorted images, among others is the interpolation technique categorized into three groups namely: Nearest neighbor, Cubic convolution and Bilinear. The last two techniques were compared in this activity and we neglect the first technique since it renders low quality images.

Original Image

Starting with four corner points of the best rectangle in the distorted image, an ideal image can then be generated.

Arrows connect corresponding vertex points of ideal (right image) and distorted (left image). These sets of corner points for both ideal and distorted images were collected with the use of the function locate in Scilab.

Point Locations in x and y coordinates

The above matrices were used to numerically relate ideal and distorted coordinates. This relationship is best described by constants c defined by equations below.

Equations shown below represent locations for the distorted image in terms of x and y coordinates in the ideal image.

Grayscale values corresponding to the computed x and y coordinates are immediately copied into a blank pixel in the ideal grid if locations are integer-values. Interpolation techniques are performed otherwise.

Nearest Neighbor Interpoation technique oly refers to rounding off of non-integer values to nearest values (hence nearest neighbor) and assume that these locations correspond to the correct grayscale pixel values ready to be copied. Bilinear interpolation technique on the other hand requires a more rigorous approach than the former technique since it computes the pixel value based on four nearest pixel values, hence the equation shown.

Results shown below have proven that Bilinear interpolation technique renders a clearer thus better image than Nearest Neighbor Technique.

Nearest Neighbor Interpolation Technique

Bilinear Interpolation Technique

I give myself a grade of 10 for completing the task and understanding the concepts quite well in this activity.

Thanks to Mandee and Carmen for very useful discussions regarding misconceptions on the problem and the code itself.

Activity 12: Color Image Segmentation

2009-08-06T14:38:00.000-07:00

"...After time the bitter taste
Of innocence decent or race
Scattered seeds, buried lives
Mysteries of our disguise revolve
Circumstance will decide ...."

Image Segmentation requires one to pick out any region of interest(ROI) from an image to make way for further image processing applications. We have have already tried this in previous exercises which mostly involves grayscale images binarized based on threshold values. This acitivity deals with segmentation of colored 3D objects in two ways: Parametric segmentation and Non-parametric segmenation. Parametric Segmentation involves the derivation of Gaussian PDF based on normalized chromaticity coordinates. After having known these sets of values, the joint probability of pixels to belong to r or to g are computed together with respective mean and standard deviations. Non-parametric segmentation on the other hand makes use of the 2d histogram of the ROI. The histogram itself is then used to segment the image through the process of backprojection (as in Activity4).

Patches are are cropped from the image below. One patch was taken from the upper half of the gutar whle the other patch was taken from its lower half.

Original Image

(http://www.bassguitars.us/four-string-guitar-636.jpg)

Patch1
(from guitar's upper half)

Patch2
(from guitar's upper half)

PARAMETRIC DISTRIBUTION APPROXIMATION
(patch1, patch2,average patch1 and patch2,total)

Application of Parametric segmentation for patch 1 and patch 2 lead to the detection of respective locations in the image as to where the patches were taken from. It is als interesting to know that the application of the average of the two patches lead to the detection of the middle part of the guitar. These results are logical since light conditions under which the picture was taken made a darker to lighter transition from top to bottom of the guitar. Summing obtained segmentaions and normalizing them detects the total area of color red.

NON-PARAMETRIC DISTRIBUTION APPROXIMATION
(patch1, patch2, average patch1 and patch2, total)

Nonparametric segmentation follows the same result as the parametric segmentation. The difference lies on the quality of segmentation. Parametric Segmentation appeared to produce better results than nonparametric segmenation since the former covered finer details in segmentation as compared to the latter. This however is not much of a difference.

(histogram: patch 1, patch 2,average patch 1 and 2, Normalized Chromaticity Space)

The peak of the 2d histogram perfectly matches the red hue on the normalized chromaticity space, thus proving its validity. Note that the triangle here is inverted since images in scilab start counting at the upperleft part of the canvas.

Rating:10
I have thoroughly understood all concepts in this activity.

Thanks to Irene for sharing the idea behind the inverted triangle.

Refernce:
[1] M.Soriano, A12 – Color Image Segmentation, 2009.

Activity 11: Color Image Processing

2009-08-06T14:14:00.000-07:00

"...All the muse in myself
My desire I can't hide
No reason and I thought..."

White balance on cameras allows one to select white balancing constants appropriate for different capturing conditions. Typical icons found on cameras are sun, cloud, light bulb and a fluorescent lamp. Two popular algorithms for achieving automatic white balance are known: White Patch Algorithm and Gray World Algorithm. White Patch algorithm makes use of the RGB values of a white patch in an image and uses it as a divider. The Gray World Algorithm on the other hand utilizes the average of Red, Green and Blue values of the image and use them as constant dividers.

The following object is a combination of different colors: red, green, yellow hues of red etc.

Auto
(original, White Patch, Grayworld)

The original image is bluish. Application of the white Patch algorithm (where the white patch here is represented by the wall) resulted to a brownish image. Application of Gray World algorithm resulted to a blue-gray image also but it appeared brighter than the original image.

Cloudy
(original, White Patch, Grayworld)

Cloudy image is more of brownish yellow. Application of White Patch algorithm brightened the image and whitened the wall. Gray World algorithm on the other hand made the image appear purple-blue.

Daylight
(original, White Patch, Grayworld)

Image in daylight setting is somewhat similar with the cloudy option only that it is brighter. Application of White Patch algorithm brightened the image and made the wall appear white. Colors in the image appear clearer and more visible than the original image. Application of Gray World Algorithm resulted to a purple-blue image.

Fluorescent
(original, White Patch, Grayworld)

Image under fluorescent setting appeared grayish blue. Application of White Patch Algorithm brightened up the image and made the wall appear close to white. Application of Gray World Algorithm made the image appear blue-gray.

Incandescent
(original, White Patch, Grayworld)

Under incandescent light, orginal object appeared blue. Application of White Patch Algorithm made the image appear whiter. Blue here vanished. The same obervation applies to Gray World Algorithm.

Red (Cloudy)
(original, White Patch, Grayworld)

Objects of similar hues (red plastic bag and red marker) are placed together and photographed under cloudy conditions. It is observed that white patch algorithm did a better job than gray world algorithm in terms of white balancing.

White patch algorithm has the advantage of neutralizing whatever bias is found on the image. Since a white patch found on the object is also exposed under the same light conditions, thus, it serves as a better dividing constant than the average of RGB values.

Rating: 9
I'm unsure of the Grayworld Algorithm results. Nevertheless, i have understood all other concepts.

Reference:
[1] M. Soriano, A11 – Color Image Processing, 2009.

Activity 10: Preprocessing Text

2009-08-06T13:57:00.000-07:00

"...Divine upper reaches
Still holding on
This ocean will not be grasped
All for nothing..."

This activity aims to extract handwritten text from an imaged document with lines. Examples of each letter is necessary for handwriting recognition.

Starting with the original document shown below, a sample subimage was cropped and processed.

Original Document

A sample subimage containing handwritten or printed text on top of lines (vertical or horizontal) was cropped from the main document. The cropped image and it FT is shown below.

Cropped Image, FT

It is observed that the image is slightly tilted. The mogrify command in scilab allows the rotation of an image to some arbitrary angle. In this case, angle is 0.9. This however is only an approximation. The rotated image and its fourier transform is shown below.

Rotated Image, FT

Previous exercises have shown that the construction of vertical and horizontal filters tend to eliminate horizontal and vertical components in the image respectively. The filter used and the final outcome after filtration is shown below.

Filter, Eliminated vertical and horizontal lines

At this point, it is necessary to binarize the image in order to clean it for further image processing operations. Binarization relies heavily on the threshold values interpreted from its histogram. This procedure is however very inaccurate since we only estimate this value based on how we examine the histogram.

Histogram, Binarized Image

As we examine the binarized image shown above, we observe that letters stand very close to each other. It is thus necessary to do erosion operation using a 2x1 structuring element. This size is necessary since we aim to come up with letter 1 pixel thick. This initial operation increases the distance between letters. Closing operation by a 2x2 structuring element is then implemented. This operation reconstructs a vivid shape of each of the letter in the image. The result is shown below.

Final Image
(Erosion --> Dilation-->Erosion)

The final image is an unsatisfactory reconstruction. Application of bwlabel on the image outputs and estimated number of letters equal to 31. This however is far from the 24 letter/numbers found in the image (im not sure of this number since the image itself is blurry and handwritten text stands very close to one another). The following image represents labeled letter/numbers in the image.

Labels
(31)

This part of the activity aims to locate "Description" in the document. We have done this before (recognition of A in a phrase).

Finding "Description"
(template image)

We start with an inverted template, since objects in scilab should always be in white. The word "Description" is isolated from the document as shown below. Once again, it is detrimental that the font style and font size of the test word ("Description") perfectly matches that of the template. Also, there is no need to rotate the image since the test word used is also rotated in the same angle.

"Description"

After undergoing convolution, the final image shown below locates the exact locations of "Description" in the document as indicated by the three white spots encircled yellow.

Locations of "Description" on Template

Rating: 9
I have understood all the concepts, however, i wasn't able to perfectly clean the handwritten image.

Reference:
[1] M. Soriano, A10 – Preprocessing Text, 2009.

Activity 9: Binary Operations

2009-08-03T10:54:00.000-07:00

"...Your softly spoken words
Release my whole desire
Undenied
Totally..."

More often than not, segmenting the desired region of interest (ROI) is necessary in order to make it more visible against poor background. Binarizing the image (eg. im2bw command in scilab) offers an easier step in order to separate the background from the ROI. However, choosing the appropriate threshold values is detrimental. Binarizing through histogram examination is thus necessary.

This activity aims to estimate the best area of simulated "cells" through closing (dilation followed by erosion) and opening (erosion followed by dilation). The closing and opening operations make it possible to eliminate holes and separate overlapping blobs.

Starting of with the original image shown below, 7 equally sized subimages (256x256) were cropped from it. Image processing operations such as binarizing, opening and closing were applied unto the image. All subimages used a circular structuring element of varying radi made suitable for every case shown below.

original image

subimage1

(open)

Examining the plotted histogram of the subimage makes it convenient to estimate threshold values which would attempt to "clean" the image. The binarized image shown above contains numerous overlapping blobs. The opening operation (erosion followed by dilation) separated most of the blobs from each other. Even though a certain number of blobs disappeared after the opening operation, the result is still considered valid since we only aim for the estimation of a single blob's area, and not the total area of all the blobs in this particular image. The same concept holds for the rest of the subimages.

subimage2

(open --> open)

The opening operation in this particular subimage is used twice. The first output (left) contains a blob which lies very close to the right boundary of the subimage. A second application of the opening operation on the previous image eliminated this discrepancy.

subimage3

(open)

subimage4

(open)

subimage5

(close --> open)

Unlike the rest of the subimages which require application of the opening operation, solely ,this particular subimage finds it necessary to close first before opening. It is observed that no overlapping of blobs occur in this subimage, Thus, opening as an initial step is no longer necessary. The closing operation eliminated the occurrences of holes inside the blobs. The first output after the closing operation however resulted to imperfect circles. Opening the image with a circular structuring element restored the original shape of the blobs.

subimage6

(open --> open)

The opening operation is applied twice since initial application creates neighboring with connected boundaries. A second application of the opening operation cuts this connection.

subimage7

(open)

List of Areas and corresponding Frequencies
(with tabul command in scilab)

522.-- 3.
521.-- 1.
515.-- 1.
514.-- 4.
513. --1.
510. --1.
508. --1.
506. --4.
505. --2.
500. --1.
498. --1.
497. --1.
492. --3.
489.-- 7.
485. --1.
472. --1.
466. --1.
456. --1.
444. --5.
441. --1.
437. --4.
421. --3.

Mean: 482.66667
Standard Deviation: 31.930998

Majority (7 occurences) of the measured circles have area equal to 489.

The best area estimate lies on the mean of the set. Thus , blob area area is approximately 482.66667 +/- 31.930998 .

Rating: 10
I have understood all the concepts tackled in this activity.

Reference:
[1] M. Soriano, A9 – Binary Operations,2009.

Activity 8: Morphological Operations

2009-07-27T06:17:00.000-07:00

"...The taste of life I can't describe It's chocking on my mind Reaching out I can't believe Faith it can't decide..."

Morphology in image processing requires reshaping and reforming images which may be beneficial for adding or extracting information from it. All morphological processes are said to affect an object's shape. Holes in images may be filled in, or contours on images may be enhanced. This activity initially aims to explore two types of morphological operations dilation and erosion. Dilating an image leads to the elongation of an image based on the shape of the structuring element operated in it. The opposite happens when an image is subjected to erotion. The image in this case is reduced/"eroded" in reference to the shape of the structuring element used.

Shapes (in binary format) tested include square (50x50), triangle (base=50, height=30), circle (radius 25), hollow square (60x60, edges are 4 pixels thick) and a plus sign (8 pixels thick and 50 pixels long for each line). These images were dilated and eroded with the following structuring elelements: 4x4 ones, 2x4 ones, 4x2 ones and a cross 5pixels long and one pixel thick. Results are as follows.

Majority of my predictions came out right. The hollow square confused me most.

1. Trangle

2. Square

3. Hollow Square

4. Cross

5. Circle

Explorations of the same images in the previous set ere applied with built-in functions thin and skel in scilab.

Thin and Skel

[Code]

triangle=im2bw(gray_imread('C:\Users\Public\alva\186\A8\images\triangle.bmp'),0.5); circle=im2bw(gray_imread('C:\Users\Public\alva\186\A8\images\circle.bmp'),0.5); cross=im2bw(gray_imread('C:\Users\Public\alva\186\A8\images\cross.bmp'),0.5); Hsquare=im2bw(gray_imread('C:\Users\Public\alva\186\A8\images\Hsquare.bmp'),0.5); Square=im2bw(gray_imread('C:\Users\Public\alva\186\A8\images\square.bmp'),0.5); s24=ones(2,4); s42=ones(4,2); s44=ones(4,4); tiny=zeros(5,5); tiny(3,:)=1; tiny(:,3)=1;

Rating: 10
I think i have done exactly what was being asked for in this activity.

Activity 7: Enhancement in the Frequency Domain

2009-07-27T03:00:00.000-07:00

"...Wild, white horses
They will take me away
And the tenderness I feel
Will send the dark underneath
Will I follow?..."

Masking is a technique done in the Fourier domain and is used to remove unwanted patterns in images. Filter masks block unwanted frequencies. [1]

Convolution Theorem
a. dots

b. circles
r=0

r=2

r=5

r=10

r=20

c. squares
s=1

s=2

s=5

s=10

d. gaussian
variance=50, distance=5

variance=50, distance=50

variance=50, distance=100

variance=100, distance=50

e. inverse gaussian
variance=300, distance=50

variance=500, distance=50

Fingerprints: Ridge Enhancement

Original Fingerprint [2]

Method 1: Through the use of a High Pass Filter (Inverse Gaussian)

Filter used: Inverse Gaussian with Variance=25 (right).

High-pass filters allow transmission of high frequency signals such as edges in images. Results should show obvious enhancement of edges. The downfall however for this method is that this filter also allows transmission of noise since noise is usually of higher frequency A possible solution to this is through subtracting the high-passed image with the original image resulting to a noiseless image. The image can again be passed through this same filter to enhance the edges some more.

Result:
The image is noiseless to begin with, so there is no need to subtract the filtered image with the original image.

Method 2: Use of a filter which matches the fourier tranform of the image

The function im2bw binarized the original image (middle). The latter served as the template for the filter used (right).

This type of filter specially made for the image allows the transmission of information which lies only within the central ranges. All signals located at the sides are attenuated.

Result:

Fingerprint images compared: original(left) inverse gaussian(middle) filter(right)

Lunar Landing Scanned Pictures : Line removal

The regularly spaced vertical lines on the below are the result of combining individually digitized framelets to create a composite photograph[1]. In order to eliminate these lines, a filter patterned on the FT of the original image is drawn using paint. Sometimes, it is more efficient to draw it by hand rather than to utilize threshold through the im2bw command since information may be lost as threshold values are adjusted.

Previous exercises have shown that a vertical line in real space corresponds to a horizontal counterpart in fourier space. It is thus necessary to block only the horizontal component in fourier space since this is what we are after. Vertical lines have been successfully deleted in the final image.

Original Image [3]:

FT of image(left) Filter(right)

Final Image(left) FT of final image(right)

Images compared: Original (left) Without Vertical lines (right)

Canvas Weave Modeling and Removal

The canvas weave pattern below obstructs a proper view of the painting. Once again, the FT of the original image was patterned for creating the filter. We have seen this pattern before. As a matter of fact, it represents sinusoids of different frequencies. It is necessary to block the dot signals shown in the FT of the image to succesfully eliminate the pattern.

The final image is an enhanced version of the original one. This time, brush strokes are rendered clearly.

Original Image [1]:

Image FFT(left) Filter(right)

Final Image (without the canvas pattern)

Weaving the Canvas

Retrieving the canvas requires the initial step of inverting the filter. Once the image is convolved with the inverse filter, only the canvas passes through leaving behind the rstructure of the painting.

Inverse Filter(Left) Weaved Canvas(Right)

Rating: 10
I have fully understood the concepts tackled in this activity.

References:
[1] M. Soriano, "A7 – Enhancement in the Frequency Domain", 2009.
[2] http://www.cs.tau.ac.il/~turkel/fingerprint_2.jpg
[3] http://www.lpi.usra.edu/lunar/missions/apollo/apollo_11/images

Activity 6: Properties of the 2D Fourier Transform

2009-07-09T01:57:00.000-07:00

"... in the days of golden days
when everybody knew what they wanted
it aint here today"

Revised Version: The "[]" went missing in imshow. This command allows scilab to normalize the matrices to be displayed. It is noted that images should be of values between 0 and 1. Going beyond or below this range results to oversaturation or absence of visual information. Thanks Ma'am Jing for pointing this out.

Familiarization with FT of different 2d patterns

Through different shapes (square, annulus/donut, square annulus, two slits and two dots) generated in scilab, the following FT moduli were obtained.

legend: (a) aperture (b) fft

A. Square
side=5
a. b

side=20
a. b.

side=80
a. b.

Observation: An opposite phenomenon happens in fourier space. Increasing the size of an aperture in real space results to a corresponding decrease in fourier space.

mat=ones(256,256);
mat_s=size(mat);
cen=round(mat_s(1)/2);
big=5;
for i=1:1:mat_s(1),
for j=1:1:mat_s(1),
if abs(cen-i)>=big then
mat(i,j)=0;
elseif abs(cen-j)>=big then
mat(i,j)=0;
end;
end;
end
xset('window',0);
imshow(mat);
mat=abs(fft2(mat));
xset('window',1);
imshow(fftshift(mat),[])

B. Donut
outer radius= 20; inner radius=19
a. b.

outer radius= 20; inner radius=10
a. b.

outer radius= 80; inner radius=79
a. b.

outer radius= 80; inner radius=50
a. b.

mat=ones(256,256);
mat_s=size(mat);
cen=round(mat_s(1)/2);
big_r=80;
small_r=50;
for i=1:1:mat_s(1),
for j=1:1:mat_s(1),
rsq=(i-cen)**2+(j-cen)**2;
r=round(rsq**(1/2));
if r>=big_r then
mat(i,j)=0;

C. Square Annulus

outer square=20, inner square=19 a.
a. b.

outer square=20, inner square=10
a. b.

outer square=80, inner square=79
a. b.

outer square=80, inner square=50
a. b.

Obsevation: Interestingly, a square annulus with thin edges behaves like two sets of parallel slits placed perpendicular with respect to the other. This is evident by examining its fourier transform.

mat=ones(256,256);
mat_s=size(mat);
cen=round(mat_s(1)/2);
big=80;
small=50;
for i=1:1:mat_s(1),
for j=1:1:mat_s(1),
if abs(cen-i)>=small then
mat(i,j)=0;
elseif abs(cen-j)>=small then
mat(i,j)=0;
end;
end;
end

dummy=ones(256,256);
for i=1:1:mat_s(1),
for j=1:1:mat_s(1),
if abs(cen-i)>=big then
dummy(i,j)=0;
elseif abs(cen-j)>=big then
dummy(i,j)=0;
end;
end;
end

MAT=dummy-mat;
//imshow(MAT)
xset('window',0);
imshow(MAT);
xset('window',1);
imshow((abs(fftshift(fft2(MAT)))),[]);

D. Two Slits

distance between slits: 8
a. b.

distance between slits: 20

a. b.

Observation: A slit spanning the vertical length (horizontal) corresponds to an orientation perpendicular to it in fourier space. As shown earlier, when distance between two slits is increased, spaces found in its fourier transform (sinusoid) is decreased.

mat=zeros(256,256);
s=size(mat);
cen=round(s(1)/2);
j1=cen-20;
j2=cen+20;
for i=1:1:s(1),
mat(i,j1)=1;
mat(i,j2)=1;end
//imshow(mat)
xset('window',0);
imshow(mat,[]);
mat=abs(fft2(mat));
xset('window',1);
imshow(fftshift(mat),[]);

E. Two Dots
distance between dots: 8
a. b .

distance between dots: 40
a. b.

Obsevation: Increase in distance in real space corresponds to a decrease in fourier space.

mat=zeros(256,256);
s=size(mat);
cen=round(s(1)/2);
j1=cen-20;
j2=cen+20;
mat(cen,j1)=1;
mat(cen,j2)=1;
xset('window',0);
imshow(mat);
mat=abs(fft2(mat));
xset('window',1);
imshow(mat);

Anamorphic property of the Fourier Transform

1. 2D sinusoid using Scilab in the X direction

sinusoid at f=4

2. Fourier Transform of 2D sinusoid shown in (1)

3. Frequency change:

(a) sinusoid at f=1 (b) fft
a.b.

(a) sinusoid at f=40 (b) fft

a.b.

Obsevation: Increasing the frequency in fourier space decreases the gaps in real space. It is however expected that the spaces between the dots in fourier space should also increase since this is in the frequency domain.

4. Addition of constant bias
Digital images have no negative values. Intuition tells us that the sinusoid generated in (1) has values ranging from -1 to 1. It is therefore necessary to introduce bias in order to achieve positive image values.

I. constant bias (by subtracting a constant value which corresponds to the minimum data point of the original sinusoid)

(a) sinusoid with constant bias (b) fft
a. b.

Observation:
Constant bias added(or subtracted) in real space did not change the appearance of the sinusoid. The imshow command together with the "[]" symbol automatically normalizes the image making it appear the same way as the unbiased sonusoid. Thus, scilab displays the images already biased (all values are positive). In fourier space however, three dots appeared instead of the original two shown in number 2. The appearance of the brighter center dot emphasizes this constant bias.

II. non-constant bias (subtracting another sinusoid with freq 1 from the original sinusoid)
(a) sinusoid with nonconstant bias (b) fft
a. b.

An interferogram is a CCD camera recording of an optical interferenc pattern caused from laser light beam shearing between incoming and outgoing beam of the interferometer[1]. For Young's double slit experiment, wavelength can be readily calculated based on the equation lamba=(fringe distance)(separation distance between slits)/(slit-screen distance) [2]. From computed wavelength values, actual frequencies can be obtained.

Conclusion:
5. rotated sinusoid
(a) sinusoid rotated at 30 degrees (b) fft
a. b.

Observation: A rotated image in real space also produced a rotated fourier transform.

6. combination of sinusoids in X and Y

(a) combination sinusoid (b) fft
a.b.

Obervation: Combination of sinusoids in which each span a specific axis (x and y) outputs a checkered image when plotted. Corresponding orientation of dots were formed as shown above. Recall that two dots lying in the x axis (E, two dots, previously presented) outputs vertical sinusoids in fourier space. The opposite happens when a vertical sinusoid in real space is projected unto fourier space. A horizontal sinusoid traversing in real space however projects dots which lie along the y axis in fourier space. The above image is thus a combination of the said properties.

7.Addition of Several Rotated sinusoids to 6

Right below are the FT images of (6),(5), sinusoid rotated at 120 degrees and the sum of all FTs consecutively.
+ + =

Images below are the corresponding images of sinusoids with FTs given above.

+ + =

Observation:
It was easily predicted that the sum of all three types of sinusoids is just the superposition of all waves in real and fourier space.

nx = 100; ny = 100;
x = linspace(-1,1,nx);
y = linspace(-1,1,ny);
[X,Y] = ndgrid(x,y);
f = 4
z = sin(2*%pi*f*X);

//xset('window',0);
//imshow(z,[]);
zfft=abs(fft2(z));
//xset('window',1);
//imshow(fftshift(zfft),[]);

zb = z-min(z);
//xset('window',2);
//imshow(zb,[]);
zfftb=fft2(zb);
//xset('window',3);
//imshow(abs(fftshift(zfftb)),[]);

fnew = 1
znew = z-sin(2*%pi*fnew*X);
//xset('window',4);
//imshow(znew,[]);
zfftnew=fft2(znew);
//xset('window',5);
//imshow(abs(fftshift(zfftnew)),[]);

theta = 30;
ftheta=4;
z = sin(2*%pi*ftheta*(Y*sin(theta) + X*cos(theta)));
//xset('window',6);
//imshow(abs(fftshift(fft2(z))),[]);

z = sin(2*%pi*4*X).*sin(2*%pi*4*Y);
//xset('window',7);
//imshow(abs(fftshift(fft2(z))),[]);

Rating: 9
I have understood the concepts behind two dimensional fourier transforms fairly well. However, Im not sure if I have answered the Double Slit problem correctly.

Reference:
[1] http://www.opticalguidancesystems.com/interferogram.htm
[2] http://en.wikipedia.org/wiki/Double-slit_experiment

Activity 5: Fourier Transform Model of Image Formation

2009-07-06T20:28:00.000-07:00

"...From this time unchained
We're all looking at a different picture
Thru this new frame of mind
A thousand flowers ould bloom
Move over, and give us some room"

Revised version:
Special thanks to Ma'am Jing for pointing out what seemed to be the problem in the code. Apparently, it was the absence of "[ ]" in the imshow command.

For this activity, we investigate the effect of fourier transformation, convolution and correlation among images.

Familiarization with discrete FFT
a. CIRCLE

a. b.
c. d.

(a) original image (b) fft (c) shifted fft (d) twice fft

Igray = gray_imread('F:\Documents and Settings\avabs\My Documents\papel\physics\physics 186\A5\Images\sssscircle.bmp');
xset('window',0);
imshow(Igray);

xset('window',1);
imshow(abs(fft2(Igray)),[]);

xset('window',2);
imshow(abs(fftshift(fft2((Igray)))),[]);

xset('window',3);
imshow(abs(fft2(fft2(Igray))),[]);

b. CIRCLE drawn in scilab
a. b .
c. d.

(a) original image (b) fft (c) shifted fft (d) twice fft

x = [-1:0.1:1];
[X,Y] = meshgrid(x);
r = sqrt(X.^2 + Y.^2);
circle = zeros(size(X,1), size(X,2));
circle(find (r <=0.5)) = 1.0; xset('window',0); imshow(circle,[]); xset('window',1); imshow(abs(fft2(circle)),[]); xset('window',2); imshow(abs(fftshift(fft2(circle))),[]); xset('window',3); imshow(fft2(fft2(circle)));

A
a. b.
c. d.

(a) original image (b) fft (c) shifted fft (d) twice fft

Igray = gray_imread('F:\Documents and Settings\avabs\My Documents\papel\physics\physics 186\A5\Images\A.bmp');
Igray=1-Igray;
xset('window',0);
imshow(Igray);

xset('window',1);
imshow(abs(fft2(Igray)),[]);

xset('window',2);
imshow(abs(fftshift(fft2(Igray))),[]);

xset('window',3);
imshow(fft2(fft2(Igray)));

Observations:
1. Applying fftshift displaces the image in the area of focus. Periodic boundary conditions also apply.
2. Applying fft unto the image results to an indecipherable image.
3. Applying the fft twice changes the image back to its original form. Only this time, the image is inverted.

Simulation of an imaging device.
VIP and Circle

a.
b.
c.
d.
(a) original image (b) big aperture (c) medium aperture (d) small aperture

r=gray_imread('F:\Documents and Settings\avabs\My Documents\papel\physics\physics 186\A5\Images\bigcircle.bmp');
a=gray_imread('F:\Documents and Settings\avabs\My Documents\papel\physics\physics 186\A5\Images\VIP.bmp');
a=1-a;
xset('window',0);
imshow(r);
Fr = fftshift(r);
Fa = fft2(a);
FRA = Fr.*(Fa);
IRA = fft2(FRA);
FImage = abs(IRA);
xset('window',1);
imshow(FImage, [ ]);

r=gray_imread('F:\Documents and Settings\avabs\My Documents\papel\physics\physics 186\A5\Images\smallc.bmp');
a=gray_imread('F:\Documents and Settings\avabs\My Documents\papel\physics\physics 186\A5\Images\VIP.bmp');
a=1-a;
xset('window',2);
imshow(r);
Fr = fftshift(r);
Fa = fft2(a);
FRA = Fr.*(Fa);
IRA = fft2(FRA);
FImage = abs(IRA);
xset('window',3);
imshow(FImage, [ ]);

r=gray_imread('F:\Documents and Settings\avabs\My Documents\papel\physics\physics 186\A5\Images\sssscircle.bmp');
a=gray_imread('F:\Documents and Settings\avabs\My Documents\papel\physics\physics 186\A5\Images\VIP.bmp');
a=1-a;
xset('window',4);
imshow(r);
Fr = fftshift(r);
Fa = fft2(a);
FRA = Fr.*(Fa);
IRA = fft2(FRA);
FImage = abs(IRA);
xset('window',5);
imshow(FImage, [ ]);

Observation:
Correlation allows one to look for a duplicate of an image in an even larger image. The white spots indicate the positions of the letter A in the phrase "the rain in spain stays mainly in the plain". It is however crucial that both font and fontsize of both images perfectly match.

Edge detection using the convolution integral
VIP and Pattern

a.
b.
c.

(a) [-1 -1 -1; 2 2 2; -1 -1 -1] (b) [-1 2 -1; -1 2 -1; -1 2 -1] (c) [-1 -1 -1; -1 8 -1; -1 -1 -1]

pattern = [-1 -1 -1; 2 2 2; -1 -1 -1];
xset('window',3);
imshow(pattern);
a=gray_imread('F:\Documents and Settings\avabs\My Documents\papel\physics\physics 186\A5\Images\VIP.bmp');

C=imcorrcoef(a,pattern);
xset('window',0);
imshow(C);

pattern = [-1 2 -1; -1 2 -1; -1 2 -1];
xset('window',4);
imshow(pattern);
C=imcorrcoef(a,pattern);
xset('window',1);
imshow(C);

pattern = [-1 -1 -1; -1 8 -1; -1 -1 -1];
xset('window',5);
imshow(pattern);
C=imcorrcoef(a,pattern);
xset('window',2);
imshow(C);

Observation:
Similar numbers used to form horizontal or vertical pattern result to images with enhanced horizontal and vertical edges respectively. Numbers of the same value located at the contour of the matrix used as basis for patterns also result to images with enhanced overall edges.

Thanks to martin for suggesting fftshift and interpreting the final image in edge detection.

Rating: 9, code error.

Activity 4: Enhancement by Histogram Manipulation

2009-06-29T18:31:00.000-07:00

"...I wonder why
I don't know what you see
Of course I care
I won't pretend"

Image enhancement, one of the most basic methods in image processing makes details more obvious[1]. Planetary scientists use the technology of image processing in order to enhance images of planets such as Mars, Venus and other planets. It is also important to doctors since they use it to manipulate CAT scans and MRI images [2]. More often than not, photographs need more enhancing in order for them to be appreciated and thus need to be enhanced. One way to enhance an image is through the manipulation of its histogram. Given the cumulative distribution function (CDF) of a desired PDF, the grayscale PDF of an image can be modified by backprojecting the grayscale calues using the CDF.

The following process was implemented on the image.

Once the image's CDF is plotted and an equation for the ideal CDF is known, the easiest way to do backprojection is through the input of original image's CDF Y values into the ideal equation (linnear, exponential, logarithmic) to compute their corresponding X values. These x values are then exchanged with the pixel values known to represent the original image depending on the position of the original image's pixel values (x coordinates). The new image is then plotted. This gives the enhanced version of the origianal image.

Results:
(please click on the images below to see its clearer version)

I. Linear Ideal CDF

Ideal Linear CDF

http://photogiftsworld.com/imagen es/c atalogo/resta urar/antes_faded-1-final.jpg.jpg

(http://www.joyfulimages.com/im ages/emma%20an d%20charles2.jpg)

(http://www.neuroguidance.com/Li_M RI_faded.jpg)

(http://www.neurosurgicalassociates.com.au/images/MRI-2.jpg)

II. Exponential Ideal CDF

Ideal Exponential CDF

III. Logarithmic Ideal CDF (human eye)

Logarithmic Ideal CDF

IV. Code

//automatic plotting
stacksize(3e7);
image=gray_imread('C:\Documents and Settings\avabs\My Documents\papel\physics\physics 186\A4\images\MRI-2.jpg');
subplot(241);
imshow(image);
title('Original Image');

image=image*255;
subplot(242);
histplot(255,image);
title('Histogram of Original Image');

//PDF original
tick=1;
temp=0;
x_pdf=[];
y_pdf=[];
for i=0:1:255,
[x,y]=find(image<=i);
x_pdf(tick)=i;
y_pdf(tick)=length(x)-temp;
temp=length(x);
tick=tick+1;end
subplot(243);
y_pdfnorm=y_pdf/max(y_pdf);
plot(x_pdf, y_pdf);
title('PDF of Original Image');

//CDF original
x_cdf=0:1:255;
y_cdf=cumsum(y_pdf);
subplot(244);
plot(x_cdf,y_cdf);
title('CDF of Original Image');

//CDF ideal Linear
x_cdfi=0:1:255;
y_cdfi=(max(y_cdf)/max(x_cdf))*x_cdfi;

//Backtrack Linear
image_copy=image;
image_r=int(image);
x_int=min(x_pdf):(max(x_pdf)-min(x_pdf))/255:max(x_pdf);
for i = 1:1:length(x_cdf),
x_seek=int((max(x_cdf)/max(y_cdf))*y_cdf(i));
loc=find(image_r==i);
image_copy(loc)=x_seek;end
subplot(245);
imshow(image_copy/255);
title('Enhanced Image');

//CDF ideal Exp
//x_cdfi=0:2/255:2;
//y_cdfi=1-exp(-x_cdfi**3);
//x_cdfi=x_cdfi*255/2;
//y_copy=y_cdfi;
//y_cdfi=y_cdfi*max(y_cdf)/max(y_cdfi);

//Backtrack Exp
//image_copy=image;
//image_r=int(image);
//y_prep=y_cdf*max(y_copy)/max(y_cdf);
//x_int=min(x_pdf):(max(x_pdf)-min(x_pdf))/255:max(x_pdf);
//for i = 1:1:length(x_cdf),
// x_seek=(-log(1-y_prep(i)))**(1/3);
// loc=find(image_r==i);
// image_copy(loc)=int(x_seek*255/2);end
//subplot(245);
//imshow(image_copy/255);
//title('Enhanced Image');

//CDF ideal Log
//x_cdfi=1:1:255;
//y_cdfi=log(x_cdfi);
//y_copy=y_cdfi;
//y_cdfi=y_cdfi*max(y_cdf)/max(y_cdfi);

//Backtrack Log
//image_copy=image;
//image_r=int(image);
//y_prep=y_cdf*max(y_copy)/max(y_cdf);
//x_int=min(x_pdf):(max(x_pdf)-min(x_pdf))/255:max(x_pdf);
//for i = 1:1:length(x_cdf),
// x_seek=exp(y_prep(i));
// loc=find(image_r==i);
// image_copy(loc)=int(x_seek);end
//subplot(245);
//imshow(image_copy/255);
//title('Enhanced Image');

subplot(246);
histplot(255,image_copy);
title('Histogram of Enhanced Image');

//PDF of New Image
tick=1;
temp=0;
x_pdf=[];
y_pdf=[];
for i=0:1:255,
[x,y]=find(image_copy<=i);
x_pdf(tick)=i;
y_pdf(tick)=length(x)-temp;
temp=length(x);
tick=tick+1;end
subplot(247);
y_pdfnorm=y_pdf/max(y_pdf);
plot(x_pdf, y_pdf);
title('PDF of Enhanced Image');

//CDF of New Image
x_cdf=0:1:255;
y_cdf=cumsum(y_pdf);
subplot(248);
plot(x_cdf,y_cdf);
title('CDF of Enhanced Image');

xset('window',1);
plot(x_cdfi,y_cdfi);
title('Ideal CDF (linear)');

Results show that most images are enhanced better when a linear ideal CDF was used as reference rather than exponential. Original images used were of low contrast. PDF's plotted above do not span the range of values for all pixels. Most images have pixel values (in terms of brightness) skewed either to the left or to the right. This means that the image might be too bright or too dark thus making it hard to look through the image. Plotted CDF's also vary in shape depending on their PDFs. In order to enhance the images, CDFs should follow an ideal form: (1) linear and (2) exponential. Through backprojecting, enhanced images achieved PDFs which are scattered/stretched over all ranges of pixel values (0-1). Their CDF's imitates that of the ideal CDFs.

The human eye mimics the CDF of a logarithmic plot. Backprojecting the sample images above resulted to even darker images. In summary, exponential less than linear less than garithmic where exponential exhibited more of the smaller pixel values (light colors).

Rating: 10
I have enhanced all images properly. This activity is a fulfilling one.

Thanks to mandee for useful discussions and insights on problem analysis.

reference:
[1] http://radiographics.rsnajnls.org/cgi/reprint/8/4/771.pdf
[2] http://www.the-aps.org/education/k12curric/activities/pdfs/rapp.pdf

Activity 3: Image Types and Basic Image Enhancement

2009-06-22T19:41:00.000-07:00

"...Wandering stars, for whom it is reserved
The blackness of darkness forever"

I. Image Types

Binary Image
Have pixel values of either 1 or 0. We usually see them as black or white images [1].

(source: http://www.mattmahoney.net/ocr/plaid_bw200.bmp)

PROPERTIES:
Bit Depth: 1
Pixel dimensions: 853 x 1253 pixels
Print Size: 300.92 x 442.03 millimeters
Resolution: 72 x 72 ppi
Color Space: Indexed color (2 colors)
File Size: 132.2 KB
File Type: Windows BMP image

(source: http://www.20x200.com/blog/blogimages/miyazaki_kevin_blog.jpg)

PROPERTIES:
Pixel dimensions: 610 x 375 pixels
Print Size: 84.72 x 52.08 millimeters
Resolution: 7.2 x 7.2 pixels/mm
Color Space: Grayscale
File Size: 101.5 KB
File Type: Windows JPEG image
Horizontal Resolution: 182 dpi
Vertical Resolution: 182 dpi
Bit Depth: 8
Frame Coount: 1
Equipment Make: Canon
Camera Model: Canon EOS 5D
Copyright: Kevin J. Miyazaki

True Color

(source: http://www.20x200.com/blog/blogimages/padberg_verlag3_500px_artworkimage.jpg)

PROPERTIES:
Bit Depth: 24
pixel dimensions: 500 x 501 pixels
print size: 1.667 x 1.670 inches
resolution: 300 x 300 ppi
color space: RGB color
File Size: 49.9 KB
File Type: JPEG image

Index Image

(source: http://www-zeus.desy.de/img/home.05.i.gif)

PROPERTIES:
Bit Depth: 8
pixel dimensions: 506 x 433 pixels
print size: 7.028 x 6.014 inches
resolution: 72 x 72 ppi
color space: indexed color (3 colors)
File Size: 3.65 KB
File Type: GIF image

Scanned Image:

a.
b.
(a) original image (b) image with contour

I used the following code for area calculation. This is a recycled code from actvity 2, only with the addition of imfinfo.

AREA CALCULATION
//contour
stacksize(3e7);
Img=gray_imread('C:\Documents and Settings\avabs\My Documents\papel\physics\physics 186\A3\images\lock.bmp');
info=imfinfo('C:\Documents and Settings\avabs\My Documents\papel\physics\physics 186\A3\images\lock.bmp');
imshow(Img)
bw=im2bw(Img,0.7);
imshow(bw);
[x,y]=follow(bw);
plot(x,y);

//area computation
xsize=size(x);
xs=xsize(1);
ysize=size(y);
ys=ysize(1);
total=0;
for i = 1:xs-1, total = total+0.5*(x(i)*y(i+1)-y(i)*x(i+1));end;
actual=total+0.5*(x(xsize)*y(1)-y(ysize)*x(1))
theory=((max(y)-min(y))/2)^2*%pi;
total=0;
for i = 1:xs-1, total = total+0.5*(x(i)*y(i+1)-y(i)*x(i+1));end;
actual=total+0.5*(x(xsize)*y(1)-y(ysize)*x(1))

The code shows how to do automatic and manual plotting of the image histogram in scilab. To start with, the image is binary. So we expect that the histogram would have a wide dent within its middle x range. Minimum and maximum values corresponding to pixel values (0-1 when normalized, but in my case, since i prefer not to normalize it to allow easy viewing, range is at 0-255) are the only ones occupied as seen below.

a. b.
(a) automatic plotting (b) manual plotting

HISTOGRAM (bonus)
//automatic histogram image=gray_imread('C:\Documents and Settings\avabs\My Documents\papel\physics\physics 186\A3\images\lock.bmp');
xset('window',0);
imshow(image);
xset('window',1);
image=image*255; histplot(255,image); //manual histogram
tick=1;
x_pdf=[];
y_pdf=[];
temp=0;
for i=0:1:255, [x,y]=find(image<=i);
x_pdf(tick)=i;
y_pdf(tick)=length(x)-temp;
temp=length(x);
tick=tick+1;
end
xset('window',2);
y_pdfnorm=y_pdf/max(y_pdf);
plot(x_pdf, y_pdfnorm);

Image information (through iminfo):
FileName FileSize Format Width Height Depth StorageType NumberOfColors

Filename: C:\Documents and Settings\avabs\My Documents\papel\physics\physics 186\A3\images\lock.bmp
Format: BMP
Filesize: 384438.
Width: 352.
Height: 364.
Depth: 8.
StorageType: truecolor

Area using Green's Theorem: 19235.5
Area through Pixel Count: 19590
Theoretical Area (rectangle+circle): 19294.138

plus equals

We can also apply this to other images like the one below.
a. b.
c. d.
e.

(a) original image (b) grayscale (c) contour (d) automatic histogram (e) manual plot

Area using Green's Theorem: 138381.5
Area through Pixel Count: 145489.

Rating: 9
I think i did what was being asked and have understood all concepts fully. This blog however lacks more discussions.

Reference:

[1] Activity 3 Handout, M. Soriano, 2009

Activity 2: Area Estimation of Images with Defined Edges

2009-06-22T19:38:00.000-07:00

"...Tempted in our minds
Tormented inside lie
Wounded and afraid
Inside my head
Falling through changes"

Here are some examples of black and white regular geometric images I have drawn in paint. All images are of 256x256 pixels.

circle

rectangle

square

The aim of this activity is to be able to estimate the area of a geometrical image with defined edges using Green's theorem. Knowledge on the image's edge would suffice and allow one to estimate it's area. After which, results can be verified analytically since the geometrical images listed above have known area. Green's theorem is summarized by the following formula:

Green's Theorem

xi: x coordinates
yi: y coordinates
Nb: total number of pixels around a contour

There is a function in scilab called "follow" which gives the contour values in x and y coordinates. Executing this command into the images listed above only resulted to a contour around the square template of the image. Two options remain, it's either to invert the colors of the image manually (using Paint or GIMP) or to invert them digitally using scilab SIP toolbox. I chose to do it digitally. So here's a snippet of the code:

Program Code for Area Estimation of a Circle
(click to zoom in)

The first half of the program imports the image, inverts the black color to white and vice versa, converts the image to binary, outputs the original image and plots its edge contour over it. Here's how it looked like after program execution:

The following image shows the contour (blue circle) along the edge of the circle image. However, it can be observed that the contour is not a perfect superposition of the original edge of the image. The rest of the shapes also followed this trend.

A. contour for circle

B. contour for rectangle

C. contour for square

The bottom half of the program aims to compute the area bounded by the contour by implementing Green's Theorem. After which, results were compared with the theoretical areas of each of the images. The table below summarizes these results.

Summary Table

r: radius of circle
s: side of square
l: length
w:width

Theoretical values were computed by calculating the "lengths" of both the x and y axis and substituting them into corresponding area equations shown above.

Area estimation using Green's Theorem exhibited the greatest error among the three shapes. Figure A shows that the contour (blue line along the circle's edge) does not perfectly match the circle's edge. As we zoom into the image further, the contour actually does not follow a perfect circular path. Instead, it is composed of a series of short straight lines which appear jagged when zoomed in but appear circular when zoomed out. This observation might have contributed to this minimal error. Furthermore, contours drawn in figures B and C also do not match the edges made by the images. However, it appears (based on the table above) that area computation using Green's theorem precisely matches the theoretical calculation. Thus, offsets observed in figures B and C might have allowed transitions of contours into another location while the original geometry of the images were retained.

For this activity, i give myself a grade of 9 because i wasn't able to minimize the offset exhibited by the contour. However, this could serve as a future project.

Thanks to Mandee and Irene for sharing their knowledge on the syntax of Scilab.

Activity 1: Digital Scanning

2009-06-15T20:01:00.000-07:00

"... I stand on the edge of a broken sky

And I will come down; don't know why"

Introduction ( as the worksheet indicates)

In this activity, we use ratio and proportion to find the numerical value of a digitally scanned hand-drawn plot (of course we didn't follow this since hand-drawn graphs are EXTREMELY difficult to find). For now, you will need the following software, Paint (missed paint after photoshop monopolized the editing community haaahhh) for finding the pixel locations of points on the graph, and Excel or OpenOffice Spreadsheet to tabulate the values of the graph.

On the Graph

I found the graph (a and b) in a 1912 book (c) :). It is entitled "Advanced Calculus" by Edwin Bidwell Wilson (c).

a. scanned page

b. a closer look on the graph

c. the book

Moving the cursor on the graph (c/o Paint) gives corresponding specific locations in X and Y coordinates as to where a certain pixel is located.

Now, look at picture b, the picture looks a bit slanted. So, i rotated the image (using Microsoft Word since my photoshop is currently "out of order") to give this: (d)

d. a better oriented image (rather assume it is)

Origin is located at (43,359). But the thing is, the graph doesn't start at this point. Instead, it starts at (91, 298).

Then after which, in my case, with the use of both Paint and another program (of which I have forgotten the name), locations specifying the x and y coordinates of each pixel in the graph were tabulated in Microsoft Excel and in Open Office. I REALLY had a hard time deciphering the icons of the latter program.

You see, the thing about Paint is it displays its y coordinates in an inverse manner. Meaning, y=0 is located at the topmost part of the image while y=100 is located somewhere at the bottom. X coordinates are displayed normal.

After manually encoding list of pairs of x and y coordinates in Microsoft Excel and Open Office, these points were finally plotted. An upside down image of the plot is not a shocking nor surprising revelation. Paint only displays the y coordinates quite uniquely. The trick here is to multiply the y axis with -1.

Now, the BIGGEST part of the problem lies in the BONUS. This part took almost ALL of my time.

So what was the given problem? Simple. How does one insert an image and make it the background of a chart/plot.

Steps (based on how i remembered it):

1. Click on the 'area' icon located at the toolbar.
2. Click Bitmap.
3. Import image. Now it becomes part of your choices for Area Background.
4. Right Click on the plot. Go to Properties.
5. Change Area Background by clicking on bitmap.
6. Choose your saved image.
7. There you have it, after 100 errors of doing so.

Now, here's the fun part! You have to fit the manually plotted points into the image. This part of the activity confused me the most. Because every time I adjust the x and y coordinates by multipliers and addends, the graph automatically shifts itself into some weird position which doesn't make sense at all. After some time I realized that both the X and Y axis were set to automatic adjustments. That is, the program makes its own decisions and automatically expands or narrows the range of x and y axis points. In order keep track and control these changes, I fixed the origin and the end points of both coordinates to some fixed values. In my case, I fixed the x axis at 0 to 1000 and the y axis at 0 to -350.

Now, this is where grade school math comes in handy: the concept of ratio and proportion.

To begin with, the scanned image of the graph does not start at the origin. So I need to ASSUME the horizontal length of the graph. Physical observation via the human eye suggests that the scanned graph starts somewhere at 90 and ends at 990. SO that's a total horizontal length of 900. Vertical length on the other hand is approximately 245 (from -40 to -285).

The plot of x and y coordinates (red plot in figure e) has a horizontal length of 507 (pixel locations start at 91 and end at 698) and vertical length of 225 (pixel location 298 to 72).

Ratio and Proportion Proper:

So here's the general equation for both horizontal and vertical lengths:

Factor = Image Length/Plot Length
Factor provides numerical relationship between pixel and physical distance.

Based on approximations previously done,

Horizontal Length Factor: approximately 1.7751
Vertical Length Factor: approximately 1.0841

How to we fit the plot into the image?

Answer: trial and error + Equations

Basic step:
Coincide the FIRST point of the plot with the leftmost point of the scanned graph by adding or subtracting values to it.

But before anything else, we realize that points in the plot are at RANDOM positions relative to one another. Thus, if we intend to "stretch" the plot uniformly to fit snugly into the image, distances between consecutive sample points should be taken into account.

Let

Factor=Image Length/Plot Length
Addend=match initial point of scanned graph

a=sample point 1
b=sample point 2
c=sample point 3

A=adjusted sample point 1
B=adjusted sample point 2
C=adjusted sample point 3

A=a +/- Addend
B=A+((b-a)*Factor)
C=B+((c-b)*Factor)

So here's the set of equations I used in order to fit the graph (e):

X Coordinate: C+((b-a)*1.781)
Y Coordinate: C-((b-a)*1.085)

e. plot

Factor values assumed earlier are only based on physical assumptions based on approximations made subjectively. The factor given above best fit the plot into the image. So a rate of 10 would suffice. :)