..on pattern recognition

Given an image, our goal is to correctly determine which class it belongs. Specifically, we consider 10 images of leaves of each kind of three different plant: Mango, an ornamental plant (I dont know the name), and Indian tree.






Mango Leaves














Ornamental Plant Leaves












Indian Tree Leaves









We first enhance the images as to make the classification process easier. Discussed in the previous activity, we used the Gray World Algorithm in white balancing our images.


















































For each class, we divided it into two groups: the training set and the test set. We use the training set to get the characteristic features of the leaves that would easily differentiate it from the other class. In this case, we used the RGB values and the eccentricity of the shape of the leaves as our distinguishing features. This would be our feature vector x. We take the mean m of our feature vector for the entire training set. This is given by:





where j is the class index and N is the number of samples in our training set for each class . Below is the table for the mean values for each class.





To classify which class images belongs to, we use the minimum distance classification. Taking the feature vector x of an unknown image, we determine which class mean m it is nearest to. This is given by:



where




The smallest D corresponds to the class where the unknown image belongs. We do this for the 15 remaining images in the test set. The table below is the summary of the results.














We have successfully classify all the 15 test images. The success rate is 100%.

I was able to successfully accomplish the activity. I want to give myself a 10.

Acknowledgment to Rica for uploading the images.

..on basic video processing

Our goal is to determine to moment of inertia (MOI) of a physical pendulum using only a video of its motion. We apply image processing techniques to extract necessary information needed to solve for the MOI. We used a flat metal ruler (mass=.0478kg, height=.635m, width=.03m) as the physical pendulum. Below is a sample video of its motion.














The MOI of a physical pendulum is given by:

I=mg(h/2)(T/2π)^2.
(source: http://cnx.org/content/m15585/latest/)

To solve for I, we have to determine the period of oscillation from the video. First, we convert the video into frames of images. In processing the images, we first apply the color image segmentation such that we pick out the moving physical pendulum away from the background. This technique has already been discussed from the previous activity. Below are the results.














Notice that the quality of the images are poor and distorted. We enhance the quality of the images using another technique previously discussed: morphological operations. We first binarized the images then apply dilation and erosion to enhance the shape of the physical pendulum.














Now that the images are enhanced, we determine the period T of the oscillation. Since the images are binarized, we could keep track of the motion of the pendulum by noting the x and y coordinates of the white pixels corresponding to the pendulum. We use the mean x and the mean y coordinate to approximate the motion of the center of mass. Below is the plot of the X axis coordinate as a function of time.














As the oscillation reaches a maximum, there would be a change in the directions of the coordinates of the center of mass. This periodic change in direction is the sinusoidal function we see in the plot above. Noting this and counting the number of frames between consecutive changes, we determine the number of frames corresponding to half the period. Since the frame rate of the video is 14.5 fps, the time interval per frame is 1/14.5=.069s. Thus the half period of the oscillation is just the number of frames times .069s.

From the processed images, the number of frames corresponding to half a period are: 10, 10, 9, 10, 9, 10, 9, 10, 10, 9. The mean value is: 9.6 frames corresponding to 0.6624s half period. Thus, the period of oscillation is 1.3248s. Using the equation above, we determine the MOI to be 6.612E-3.

To check, we determine to moment of inertia using another equation given as:

I=(m/2)(h)^2 + (m/12)(h)^2
(source: http://en.wikipedia.org/wiki/List_of_moments_of_inertia)

Noting the value of the mass=.0478kg, height=.635m, and width=.03m, the theoretical MOI is 6.428E-3. Comparing with the experimental MOI above, this corresponds to 2.86% error. The method is successful in determining the MOI of a physical pendulum using only a video of its motion.

I've successfully accomplished the activity. I want to give myself a 10.

Acknowledgment to my collaborators on the physical pendulum video gathering: Cole and Jeric.

..on color image segmentation

From a given image, our goal is to pick out the region of interest (ROI). To do this, we first have to get rid of the effect of brightness on the image (shadowing), and just consider the chromaticity. We represent the color space not by the RGB but by normalized chromaticity coordinates (NCC). Per pixel, we solve for the corresponding r,g,b values given by:




Since r+g+b=1, the chromaticity can be represented using only 2 values (r & g). Below is the normalized chromaticity space (x-axis is r and y-axis is g).











We now proceed with picking out the ROI. There are two methods: the Parametric and the Non Parametric.

Parametric

In this method, we first crop a subregion in the ROI. From this patch, we take the mean and standard deviation of r and g. We then compute for the distribution probability (we assume a Gaussian) using:





We do the same for p(g). The joint probability is just the product p(r)p(g). Below are sample the results:








Original Image


















Parametric
ROI: Blue Tiles

















Parametric
ROI: Yellow Tiles










Non Parametric (Histogram Backprojection)

In this method, we take the 2D r-g histogram of the crop subregion of the ROI. We then backproject the value to the pixel location of the original image. Below are sample the results:









Original Image
















Parametric
ROI: Blue Tiles



















Parametric
ROI: Yellow Tiles









Comparing between the two method, we could observe that the Parametric method has a better result.






Original Image














Parametric













Non Parametric









I've successfully accomplished the activity. I want to give myself a 10.

..on color image processing

Below are images taken at different white balancing.







Daylight setting under flourescent light











Cloudy setting under flourescent light















Flourescent setting under daylight







We notice that the images are wrongly balanced. The background which is supposed to be white appears otherwise. The white background for the daylight setting under flourescent light appears greenish. For the second image, the background appears grayish while for the last image, the background appears bluish. In order to correct these, we apply two popular algorithms for achieving automatic white balance: the reference white algorithm and the gray world algorithm.

Reference White Algorithm (RWA)
This method utilizes a known white image and used its RGB values as the divider. For our images, we used the background as the reference white. Below are the results of the enhancement. Notice that we have improved the color of the background as well as the colorful patches.





RWA (Daylight setting under flourescent light)














RWA (Cloudy setting under flourescent light)











RWA
(Flourescent setting under daylight)








Gray World Algorithm (GWA)
This method assumes that the world is essentially gray. Thus, we just take the average of the red, green, and blue value of the captured image and use them as the divider. Below are the resulting enhanced images. Again, the colors of the resulting images are greatly enhanced.







GWA (Daylight setting under flourescent light)













GWA (Cloudy setting under flourescent light)










GWA
(Flourescent setting under daylight)









Finally, we take an image of an ensemble of leaves having the same hue (green). This image is taken in flourescent setting under daylight. Notice that the white background appears bluish. Also, the color of the dark leaves appears black.






Flourescent setting under daylight







We improve this image using the reference white algorithm and gray world algorithm. Below are the resulting images. Both methods were able to enhance the image. The background now appears white and the color of the leaves are enhanced. Between the two methods, the resulting image from the gray world algorithm is better. The background appears really white and the color of the leaves are really distinct and clear.





RWA














GWA







I was able to successfully perform the activity. I want to give myself a 10.

Acknowledgment to Jeric for the rubik's cube and Rica for uploading the images.

..on stereometry

We try to reconstruct the image of a 3D object using images taken at different positions. Using a technique called stereometry, we will derive the depth z of the 3D object using 2D images (x,y). Consider the diagram below.





















Given that the 2 images have the same y coordinates, we could solve for z using:






where b is just the traverse distance between the 2 images and f is just the focal length of the camera. We could solve for f using the calibration technique discussed on the previous activity.

Below are the two different images of a rubik's cube taken with b=5cm.











































Using 25 different points (x,y), we calculated for the corresponding depth (z). Below are the 3D reconstruction using splin2d of scilab for not a knot, bilinear interpolation, natural, and monotone.
















Enlarging further, we see that the reconstructed 3D object depicts a cube. Though the resulting rendition is not a perfect cube, we were able to show the general shape of the 3D object.

















I want to give myself a 10.

..on photometric stereo

We use photometric stereo to extract the 3D shape of an object using only the information from the shadow. We estimate the shape of the object using the shading obtained from different images of different light source location.

Consider a point source of light at infinity.











The intensity I of the image is related to the vector position of the camera given by:








where N is the number of images used. We now solve for g using:




to get the normal vector n, we simply normalize it:





to derive the shape from the normals, we note that the surface elevation f(x,y) is related to the normals by:






Finally, we solve for the surface elevation using:







We apply this technique using four images of a sphere. The resulting 3D rendition is:















Indeed, the resulting shape is a sphere.

I've successfully accomplished the activity. I want to give myself a 10.

..on correcting geometric distortions

Below is an image of a "grid" capiz window. Notice that the image has a barrel distortion effect. The square grids located at the edges are much smaller than those found at the center. The center appears to be bloated while the sides are pinched. These are due to the "imperfect" lens of the camera that captured the image.

















Our goal is to correct this distortion. We use the center square grid as our reference since it is less distorted. We then determine the transformation matrix that caused the barrel effect. Let f(x,y) be the coordinates of the ideal image while g(x',y') are the coordinates of the the distorted image. To determine the transformation matrix C, we map the coordinates of the ideal image in the distorted image.










We then compute for the transformation matrix C using:






Now that we have determined the transformation matrix, we just simply copy the graylevel v(x',y') of the pixel located in g(x',y') into f(x,y). But since the calculated g(x',y') coordinates are real (pixel coordinates should be integral), we use bilinear interpolation. The graylevel of an arbitrary point is determined by the graylevel of the 4 nearest pixels encompassing that point.













We can no solve for the graylevel using:




For the remaining blank pixels in the ideal image, again we use interpolation of the four nearest corners to determine its graylevel. Below is a comparison of the original distorted image and the enhanced(ideal) image. Notice that at the lower left level, the size of the square grid for the enhanced image increased. The grid lines also become more parallel. The resulting image has lessen the effect of the distortion. The image is no more bloated.






Original distorted image











Enhanced ideal image









I think I've performed the activity successfully. The distorted image was enhanced. I want to give myself a 10.