CSci 6971: Image Registration Lecture 5: Feature-Base Regisration January 27, 2004 Prof. Chuck Stewart, RPI Dr.
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CSci 6971: Image Registration Lecture 5: Feature-Base Regisration January 27, 2004 Prof. Chuck Stewart, RPI Dr. Luis Ibanez, Kitware Overview What is feature-based (point-based) registration? Feature points The correspondence problem Solving for the transformation estimate Putting it all together: ICP Discussion and conclusion Image Registration Lecture 5 2 What is Feature-Based Registration? Images are described as discrete sets of point locations associated with a geometric measurement Locations may have additional properties such as intensities and orientations Registration problem involves two parts: Finding correspondences between features Estimating the transformation parameters based on these correspondences Image Registration Lecture 5 3 Feature Examples: Range Data Range image points: (x,y,z) values Triangulated mesh Surface normals are sometimes computed Notice: Some information (locations) is determined directly by the sensor (“raw data”) Some information is inferred from the data Image Registration Lecture 5 QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture. 4 Feature Examples: Vascular Landmarks Branching points pulmonary images: Lung vessels Airway branches Retinal image branches and cross-over points Typically augmented (at least) with orientations of vessels meeting to form landmarks Image Registration Lecture 5 5 Points Along Centers of Vessels and Airways Airways and vessels modeled as tubular structures Sample points spaced along center of tubes Note that the entire tube is rarely used as a unit Augmented descriptions: Orientation Radius Image Registration Lecture 5 6 “Interest” Points Locations of strong intensity variation in all directions Augmented with summary descriptions (moments) of surrounding intensity structures Recent work in making these invariant to viewpoint and illumination. We’ll discuss interest points during Lectures 16 and 17 Image Registration Lecture 5 Brown and Lowe, Int. Conf. On Computer Vision, 2003 7 Feature Points: Discussion Many different possible features Problem is reliably extracting features in all images This is why more sophisticated features are not used Feature extraction methods do not use all intensity values Use of features dominates range-image registration techniques where “features” are provided by the sensor Image Registration Lecture 5 8 Preamble to Feature-Based Registration: Notation Set of moving image features Set of fixed image features Each feature must include a point location in the coordinate system of its image. It may include more Set of correspondences Image Registration Lecture 5 9 Mathematical Formulation Error objective function depends on unknown transformation parameters and unknown feature correspondences Each may depend on the other! Transformation may include mapping of more than just locations Distance function, D, could be as simple as the Euclidean distance between location vectors. We are using the forward transformation model. Image Registration Lecture 5 10 Correspondence Problem Determine correspondences before estimating transformation parameters Based on rich description of features Error prone Determine correspondences at the same time as estimation of parameters “Chicken-and-egg” problem For the next few minutes we will assume a set of correspondences is given and proceed to the estimation of parameters Then we will return to the correspondence problem Image Registration Lecture 5 11 Example: Estimating Parameters 2d point locations: Similarity transformation: Euclidean distance: Image Registration Lecture 5 12 Putting This Together Image Registration Lecture 5 13 What Do We Have? Least-squares objective function Quadratic function of each parameter We can Take the derivative with respect to each parameter Set the resulting gradient to 0 (vector) Solve for the parameters through matrix inversion We’ll do this in two forms: component and matrix/vector Image Registration Lecture 5 14 Component Derivative (a) Image Registration Lecture 5 15 Component Derivative (b) At this point, we’ve dropped the leading factor of 2. It will be eliminated when this is set to 0. Image Registration Lecture 5 16 Component Derivatives tx and ty Image Registration Lecture 5 17 Gathering Setting each of these equal to 0 we obtain a set of 4 linear equations in 4 unknowns. Gathering into a matrix we have: Image Registration Lecture 5 18 Solving This is a simple equation of the form Provided the 4x4 matrix X is full-rank (evaluate SVD) we easily solve as Image Registration Lecture 5 19 Matrix Version We can do this in a less painful way by rewriting the following intermediate expression in terms of vectors and matrices: Image Registration Lecture 5 20 Matrix Version (continued) This becomes Manipulating: Image Registration Lecture 5 21 Matrix Version (continued) Taking the derivative of this wrt the transformation parameters (we didn’t cover vector derivatives, but this is fairly straightforward): Setting this equal to 0 and solving yields: Image Registration Lecture 5 22 Comparing the Two Versions Final equations are identical (if you expand the symbols) Matrix version is easier (once you have practice) and less error prone Sometimes efficiency requires handcalculation and coding of individual terms Image Registration Lecture 5 23 Resetting the Stage What we have done: Features Error function of transformation parameters and correspondences Least-squares estimate of transformation parameters for fixed set of correspondences Next: ICP: joint estimation of correspondences and parameters Image Registration Lecture 5 24 Iterative Closest Points (ICP) Algorithm Given an initial transformation estimate 0 t=0 Iterate until convergence: Establish correspondences: For fixed transformation parameter estimate, t, apply the transformation to each moving image feature and find the closest fixed image feature Estimate the new transformation parameters, For the resulting correspondences, estimate t+1 ICP algorithm was developed almost simultaneous by at least 5 research groups in the early 1990’s. Image Registration Lecture 5 25 Finding Correspondences Map feature into coordinate system of If Find closest point Image Registration Lecture 5 26 Finding Correspondences (continued) Enforce unique correspondences Avoid trivial minima of objective function due to having no correspondences Spatial data structures needed to make search for correspondences efficient K-d trees Digital distance maps More during lectures 11-15… Image Registration Lecture 5 27 Initialization and Convergence Initial estimate of transformation is again crucial because this is a minimization technique Determining correspondences and estimating the transformation parameters are two separate processes With Euclidean distance metrics you can show they are working toward the same minimum In general this is not true Convergence in practice is sometimes problematic and the correspondences oscillate between points. Image Registration Lecture 5 28 2d Retinal Example White = vessel centerline points from one image Black = vessel centerline points from second image Yellow line segments drawn between corresponding points Because of the complexity of the structure, initialization must be fairly accurate Image Registration Lecture 5 29 Comparison Intensity-Based Feature-Based For a given transformation estimate, we can only find a new, better estimate, not the best estimate, based on the gradient step. We then need to update the constraints and reestimate For given set of correspondences, we can directly (leastsquares) estimate the best transformation BUT, the transformation depends on the correspondences, so we generally need to reestablish the correspondences. Image Registration Lecture 5 30 Summary Feature-based registration Feature types and properties Correspondences Least-squares estimate of parameters based on correspondences ICP Comparison Image Registration Lecture 5 31 Looking Ahead to Lecture 6 Introduction to ITK and the ITK registration framework. Image Registration Lecture 5 32