Modern Medical Image Analysis 8DC00 Exam

Transcription

1 Parts of answers are inside square brackets [... ]. These parts are optional. Answers can be written in Dutch or in English, as you prefer. You can use drawings and diagrams to support your textual answers. You are not allowed to use the book at the exam. 1. Registration (a) (4 points) Explain the four essential aspects in general image registration. Solution: The feature space: the model- or image-based features based on which the correspondence of locations in two images are determined. (What information in the image are we using?) The similarity criterion with which the similarity of two images are compared. (How do we compute similarity?) The search space: the set of all possible transformations T mapping the moving image g to the reference/source image f = T (g). (What kind of movements are allowed?) The search strategy: the algorithm used to find the optimal transformation that tells how given an initial/current transformation the next transformation is found. (How do we move images so that they become more similar?) (b) (2 points) What is the difference between extrinsic markers and intrinsic markers in medical image registration? Solution: An extrinsic marker is a marker artificially attached to the body, e.g., on skin or skull. An intrinsic marker is a feature point that is either anatomically or geometrically distinguishable. (c) (2 points) Give an example of anatomical and an example of geometrical landmarks. Solution: An example of anatomical landmark: a line between the anterior and posterior commissura or feature points that are interactively identifiable by a medical expert. An example of geometrical landmark: corner/intersection/local curvature extrema or other geometrically salient point. (d) (2 points) Which of the following properties are characteristic to correlation and which to mutual information of two images f and g?(some may apply to both) 1. measures linear relationship; 2. not sensitive to absolute intensity; 3. measures the degree to which intensity values in image g at some location x may be predicted by intensity values in image f at the same location; 4. can detect nonlinear dependencies.

2 Solution: Correlation: 1, 2. Mutual information: 2,3,4. (e) (2 points) In intensity-based image registration, how can you use sign changes of intensity differences to match images? (Assuming that equal objects in both images have similar intensity.) Solution: Let the two images be f and g, and d their difference. If the images are in perfect alignment (with an assumption that the images have additive zero-mean noise) their difference d should be fluctuating around zero. This means that along an arbitrary row of image d the sign of the intensity values may look as follows: In case they are not well aligned, their will be areas that have either constant positive/negative value, e.g., One can then try to search for such transformation T that maximizes the number of sign changes in the new difference image d = f T (g). 2. Image enhancement (a) (1 point) What are edges in images? Solution: Edges correspond to object boundaries and are located at pixels where image intensity significantly changes. [Edges can be steps, ridges or roofs.] (b) (2 points) How can one enhance edges in images using image derivatives? Solution: By computing the magnitudes of the pixel-wise image gradients one gets high intensity values at the locations where large intensity difference takes place. Another way is to compute the zero-crossings of the (smoothed) Laplacian to obtain the central pixel of the edge. [Visual edge (contrast) enhancement on the final image level usually requires the additional steps of subtracting the laplacian from the image or methods like unsharp masking, but since these are not discussed in the book it is sufficient to answer in terms of first or second order derivatives (as pre-processing steps of edge-based methods).] (c) (2 points) What is the purpose of applying histogram equalization to an image? Solution: Histogram equalization is used for increase and even spread of contrast. For example an image with intensities concentrated in the mid-gray range is transformed to have all gray values between black and white. [ In case of an idealized continuous image, the histogram of a histogram equalized image is constant (each intensity value is equally probable). Continuous histogram equalization maximizes entropy.] (d) (2 points) What is a potential weakness of histogram equalization in medical image analysis? Solution: Histogram equalization tends to enhance all intensity values impartially, thus it also spreads the intensities of background. [Thereby actually decreasing contrast in the ROI (region of interest).]

3 (e) (1 point) Why is it useful to combine a Laplace filter with a smoothing filter? Solution: As a second order derivative operator, Laplace filter is very susceptible to noise. Therefore it is often necessary to combine it with a smoothing filter, e.g., a Gaussian filter (that evens out the tiny ripples on the image surface) to reduce noise. 3. Segmentation (a) (1 point) Why can foreground segmentation be easier in a medical image compared to a regular photograph? Solution: In medical imaging the acquisition technique is deliberately chosen to obtain information on the object of interest and the pixel intensities are more directly related to certain physical features in the voxel. (b) (1 point) How can assumptions about continuity in time be used to support thresholding a time sequence of body CT images to segment the spine? Solution: One may for example choose the reference time point from the time series of images and do initial segmentation on this particular data set. With the assumption of continuity in time the segmentations on the preceeding and subsequent images can be initiated from the boundaries of the current segment or the initial boundaries may be constrained to stay in the vicinity of the reference boundary. [One may also choose the best reference time point by comparing the delineation results according to the quality measure of choice. ] (c) (1 point) What is meant by Markov property of the intensity values of pixels/voxels? Solution: In this context Markov property states that the probability of pixel (p i,j ) having a certain intensity value (I i,j ) only depends on the intensity values of neighboring pixels (,e.g., p i±1,j±1 ). (d) (2 points) What is the difference between probability and likelihood? You can formulate your answer by using formulae or by giving an example. Solution: Probability can be seen as the relative frequency of an event (rain, PSV- AJAX 1-0,... ) to occur in long run of repeated trials [frequentist interpretation]. This requires the knowledge of the (hypothetical) distribution of the random variable of interest and refers to future events. Likelihood on the other hand is the likeliness of such a (hypothetical) distribution given that we have have some observations (3 sunny day, PSV-AJAX 1-1,0-2,2-1) of the random variable in question and refers to past events. [This is equivalent to the conditional probability of observing the particular events (e.g., 3 sunny days) given the (hypothetical) distribution (e.g., binomial distribution with p = 0.5 and n = 7).] (e) (3 points) Explain how the so-called graph cut can be used to segment images using the following key-words: sink, source, n-links, t-links, seed pixel, weights, Ford-Fulkerson algorithm.

4 Solution: One connects all pixels to their neighbors via n-links and to two artificial nodes: source and sink with t-links. Each link is assigned a flow capacity. The t-links are assigned a capacity according to the likeliness that a pixel belongs to foreground/background and the n-links are assigned a capacity according to how close the pixels and their intensity values are. Seed pixels are connected to the source (or sink) with probability one. Once all the edges (links) have a capacity assigned, the max-flow (equivalent to min-cut) can be computed using the Ford-Fulkerson algorithm. This min-cut separates the foreground from the background. 4. Feature detection (a) (1 point) How can principal component analysis (PCA) be useful in managing an image feature space? Solution: PCA can be used to reduce the feature space dimension. (b) (2 points) What is the basic idea of PCA? Solution: To describe the data in another orthogonal coordinate system where the axes correspond to the directions of maximal variation. One step in dimension reduction can be done as follows: compute the covariance matrix C of the feature vectors; compute the eigenvalues and eigenvectors of C; eliminate the dimension corresponding to the smallest eigenvalue. [The last step can be done by projecting all vectors to the (hyper-)plane {v v e = 0}, where e is the eigenvector corresponding to the smallest eigenvector. For a thorough intro to PCA see: (c) (2 points) Explain the two steps of Canny edge detector. Solution: Step 1: compute the potential edge locations with a gradient operator and discard pixels that are not local maxima (e.g., by computing the zero crossings of the Laplacian of Gaussian). Step 2: do edge tracking with hysteresis thresholding on the zero-crossings. That is, consider pixels with sufficient intensity t 1 as edge pixels. Add to this set of pixels also all adjacent pixels with intensity values above another (lower) threshold t Multiple choices Select the appropriate term/terms from the lists: (a) (1 point) An agglomerative clustering can be visualized with a 1. Feynman diagram

5 2. Venn diagram 3. heartagram 4. dendrogram 5. histogram Solution: 4. (b) (1 point) The so-called aperture problem is a property of 1. corners 2. homogeneous regions 3. edges Solution: 2 & 3. (c) (1 point) The graph in Figure 1 is a 1. tree graph 2. forest graph 3. fan graph 4. complete bipartite graph 5. complete tripartite graph Solution: 2. Figure 1: (d) (1 point) The so-called Dice coefficient is a common quality measure for 1. delineation task 2. detection task 3. registration task Solution: Affine registration

6 (a) (6 points) You have two images A and B = T (A), where T is an affine transformation. You have identified 4 representative feature points {(a 1x, a 1y ), (a 2x, a 2y ), (a 3x, a 3y ), (a 4x, a 4y )} in image A that correspond to points {(b 1x, b 1y ), (b 2x, b 2y ), (b 3x, b 3y ), (b 4x, b 4y )} in image B. Explain step by step using appropriate formulae, how to solve the T. Solution: Answer 1: Using homogeneous coordinates and simplifying the notation a 1x a 2x a 3x a 4x b 1x b 2x b 3x b 4x by putting a = a 1y a 2y a 3y a 4y, b = b 1y b 2y b 3y b 4y, t 11 t 12 t 13 and T = t 21 t 22 t 23 we obtain the equation Now the T can be solved as b = T a. T = b a T (a a T ) 1 Answer 2: Solve the first (t 1 ) and second (t 2 ) row of T separately using a least squares command in a mathematical software that gives the t i that minimizes the expression (a T t i b i ) 2, where i = 1, Classification (a) (2 points) How can you classify a sample using k-nearest neighborhood method? Solution: K-nearest neighborhood method assumes that there are reasonable amount of classified samples in the feature space. When an unclassified new sample is put in to this space, from this new sample we choose the k nearest neighbors and assign the new sample the most prevalent label within these neighbors. (b) (1 point) Why is it typically useful to have k > 1? Solution: If the labeling is done based on only one neighbor, there is the risk that it by chance happens to be an outlier sample of the wrong class. (c) (1 point) What is a decision boundary? Solution: A decision boundary in the context of classification is a (hyper)-plane or surface that divides the space into two mutually exclusive regions. [This can be computed from a given classified samples. The decision boundary is iteratively adjusted until it correctly classifies the training data.] (d) (1 point) Give an example of a typical usage of a decision boundary. Solution: It can be used to classify a new sample whose true class is unknown. [This can be used, e.g., in semi-automatic diagnosis.]

7 (e) (1 point) Are there other useful types of decision boundaries than lines, planes and hyperplanes? If so, give an example. Solution: Decision boundaries can be also quadratic or other non-linear surfaces. [On the other hand, usually these can be mapped into a higher dimensional linear space by introducing the mixed variables (such as x 2 y) as new coordinates. In the context of support vector machines also hyperplanes can be used through the so-called kernel tricks.] 8. Clustering (a) (1 point) What is the difference between clustering and classification? Solution: Clustering groups objects/feature vectors together based on their similarity. Classification assigns each object/feature vector a unique class that it belongs to. (b) (1 point) Select the correct description for each of the following three clustering methods: k-means clustering, mean-shift clustering and self-organizing maps (SOM): 1. For this clustering method the user has to initialize more clusters than the expected number of final clusters. 2. For this clustering method the user has to explicitly specify the number of clusters. 3. For this clustering method the user does not have to specify the number of clusters. Solution: k-means clustering 2, mean-shift clustering 3 and self-organizing maps 1. (c) (2 points) Describe in simple terms how self-organizing maps can be used for clustering in image analysis. Solution: The so-called neurons are either random feature vectors or samples of the feature vectors to be clustered. [The neurons can be connected to each other in a user-specified way.] Each sample is matched with the closest neuron and the neuron [and potentially its neighbors] is moved towards the sample. This is repeated for each sample[ and can be iterated for several rounds]. The resulting neurons represent the cluster centers to which finally all samples can be assigned. 9. Validation (a) (2 points) What is the difference between accuracy and precision? Solution: Accuracy is the degree of closeness of sample of a quantity to its actual (true) value. Precision is the degree to which repeated measurements under unchanged conditions show the same results. (b) (2 points) What is the difference between sensitivity and specificity?

8 Solution: Sensitivity is the proportion of correctly identified positives among the actual positives. Alternatively: sensitivity = T P T P + F N. Specificity is the proportion of correctly identified negatives among the actual negatives. Alternatively: T N specificity = T N + F P. (c) (2 points) Explain the so-called AUC of ROC in terms of sensitivity and specificity. Solution: ROC is the receiver operating curve which typically has on x-axis sensitivity and on y-axis 1-specificity (or 100 % -specificity). In case there are false-negatives or false-positives, as sensitivity increases the specificity decreases. The degree to which this happens is depicted by the ROC-curve. AUC refers to the area under the ROC curve and in case the algorithm is capable of perfect detection the AUC has value one. (d) (1 point) In the radiation therapy of cancerous tumors, what kind of risks are related to over-segmentation? Solution: In this case there is a risk that also healthy tissue is unnecessarily treated.