Description
Problem 1: Texture Analysis (50%)
In this problem, you will implement texture analysis and segmentation algorithms based on the
5×5 Laws filters constructed by the tensor product of the five 1D kernels in Table 1.1:
Tabel 1.1. 1D Kernel for 5×5 Laws Filters
Name Kernel
L5 (Level) [ 1 4 6 4 1]
E5 (Edge) [-1 -2 0 2 1]
S5 (Spot) [-1 0 2 0 -1]
W5(Wave) [-1 2 0 -2 1]
R5 (Ripple) [ 1 -4 6 -4 1]
1(a) Texture Classification (Basic: 15%)
Twelve texture images[1], from texture1.raw to texture12.raw (size 128×128) are provided for the
texture classification task in this part. Samples of these images are shown in Figure 1.1.
Figure 1.1: Four types of textures: bark, straw, brick, and bubbles
Please cluster them into four texture types with the following steps below to complete this problem.
1. Feature Extraction: Use the twenty-five 5×5 Laws Filters to extract feature vectors from
each pixel in the image (use appropriate boundary extensions).
2. Feature Averaging: Average the feature vectors of all image pixels, leading to a 25-D
feature vector for each image. Which feature dimension has the strongest discriminant
power? Which has the weakest? Please justify your answer.
EE 569 Digital Image Processing: Homework #4
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3. Reduce the feature dimension from 25 to 3 using the principal component analysis (PCA).
Plot the reduced 3-D feature vector in the feature space. (You may use built-in C++/Matlab
functions of PCA.)
4. Clustering: Use the K-means algorithm for image clustering based on the 25-D and 3-D
obtained in Steps 2 and 3, respectively. Discuss the effectiveness of feature dimension
reduction over K-means. Report your results and compare them with the reality (by eyes).
1(b) Texture Segmentation (Basic: 20%)
In this part, apply the twenty-five 5×5 Laws Filters to texture segmentation for the image shown
in Figure 1.2 with the following steps.
1. Laws feature extraction: Apply all 25 Laws filters to the input image and get 25 gray–scale
images.
2. Energy feature computation: Use a window approach to computer the energy measure for
each pixel based on the results from Step 1. You may try a couple of different window
sizes. After this step, you will obtain 25-D energy feature vector for each pixel.
3. Energy feature normalization: All kernels have a zero-mean except for 𝐿5𝑇 𝐿5. Actually,
the feature extracted by the filter 𝐿5𝑇 𝐿5 is not a useful feature for texture classification and
segmentation. Use its energy to normal all other features at each pixel.
4. Segmentation: Use the k–means algorithm to perform segmentation on the composite
texture images given in Figure 1.2 based on the 25-D energy feature vectors.
Figure 1.2: Composite texture image (comb.raw)
In order to denote segmented regions, if there are K textures in the image, your output image will
be of K gray levels, with each level represents one type of texture. For example, there are 7 types
of texture in Figure 1.2 you can use 7 gray levels (0, 42, 84, 126, 168, 210, 255) to do denotation.
EE 569 Digital Image Processing: Homework #4
Professor C.-C. Jay Kuo 3
1(c) Advanced Texture Segmentation Techniques (Advanced 15%)
You may not get good texture segmentation results for the complicated image in Figure 1.2. Please
develop various techniques to enhance your segmentation result. Several ideas are sketched below.
1. Adopt the PCA for feature reduction and, thus, cleaning.
2. Develop a post-processing technique to merge small holes.
3. Enhance the boundary of two adjacent regions by focusing on the texture properties in these
two regions only.
Problem 2: Image Feature Extractor (50%)
Image feature extractors are useful for representing the image information in a low dimensional
form.
2(a) SIFT (Basic: 20%)
In this problem, you are asked to read the original SIFT paper in [2] and answer the following
questions.
I. From the paper abstract, the SIFT is robust to what geometric modifications?
II. How does SIFT achieves its robustness to each of them?
III. How does SIFT enhances its robustness to illumination change?
IV. What are the advantages that SIFT uses difference of Gaussians (DoG) instead of
Laplacian of Gaussians (LoG)?
V. What is the SIFT’s output vector size in its original paper?
2(b) Image Matching (Basic: 20%)
One implementation of SIFT is image retrieval. You can do nearest neighbor search in the
searching database for the query image which is represented as a SIFT extracted feature vector.
Figure 2.1 River images
Use open source SIFT tool (OpenCV, VLFeat, etc.) to find the key-points in the two river images
shown above. Pick the key-point with the largest scale in river image 1 and find its closest
neighboring key-point in river image 2. Discuss your results, esp. the orientation of each keypoints.
EE 569 Digital Image Processing: Homework #4
Professor C.-C. Jay Kuo 4
2(c) Bag of Words (Advanced: 10%)
You are given samples of zeros and ones from the MNIST [3] dataset as shown below
Figure 2.2 Images of zeros and ones
Use these images as your training dataset to form a codebook with bin size of 2 (two clusters for
k-means clustering). Then extract the SIFT feature vector for the image of eight and show your
histogram of the Bag of Words for this image. Discuss your observation.
Figure 2.3 Image of eight
Appendix:
Problem 1: Texture Analysis
texture1~12.raw 128×128 8-bit gray
comb. raw 510×510 8-bit gray
Problem 2: Image Matching
river1.raw 1024x768x3 24-bit color
river2.raw 1024x768x3 24-bit color
zero_1~5.raw 28×28 8-bit gray
one_1~5.raw 28×28 8-bit gray
eight. raw 28×28 8-bit gray
Reference:
[1] http://sipi.usc.edu/database/database.php?volume=textures&image=19#top
[2] Lowe, David G. “Object recognition from local scale-invariant features.” iccv. Ieee, 1999.
[3] http://yann.lecun.com/exdb/mnist/
