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Davies: Computer Vision, 5th edition: Solutions to selected problems 18 © E. R. Davies 2017 10.1 (a) The first part is simple coordinate geometry: gy/gx = yf/xf = a, and (x – xf)xf + (y – yf)yf = 0 Eliminating […]

Davies: Computer Vision, 5th edition: Solutions to selected problems 23 11.1 (a) The Hough transform operates by taking a parameter space (often congruent to image space), clearing it, accumulating votes to build up evidence that particular sets of parameter values […]

Davies: Computer Vision, 5th edition: Solutions to selected problems 31 13.1 Equation (13.15) gives L(Ci|Cj) = 1 – ij, where ij is the Kronecker delta. Substituting in equation (13.13) simplifies the summation hugely, as all but one of the terms […]

Davies: Computer Vision, 5th edition: Solutions to selected problems 32 © E. R. Davies 2017 14.5 The point nearest to the origin on the ROC curve is the position where the function Pn = (PFP2 + PFN2)1/2 is a minimum. […]

Davies: Computer Vision, 5th edition: Solutions to selected problems 33 © E. R. Davies 2017 16.1 Consider a point P in image 1 whose centre of projection is C1. The epipolar line L for image 2 occurs where plane PC1C2 […]

Davies: Computer Vision, 5th edition: Solutions to selected problems 37 17.1 The complete table of pose ambiguities that arise for weak perspective projection, for various numbers of object points identified in the image, is: arrangement of the points N no. […]

Davies: Computer Vision, 5th edition: Solutions to selected problems 39 © E. R. Davies 2017 18.2 With no loss of generality we can take the conic to be an ellipse. If the two points are inside the ellipse, draw a […]

Davies: Computer Vision, 5th edition: Solutions to selected problems 42 19.1 From the main text in Chapter 16, the standard formulae for depth from disparity are: f/Z = x1/(X + b/2) = x2/(X – b/2) D = x1 – x2 […]

Davies: Computer Vision, 5th edition: Solutions to selected problems 2 2.1 The required edge location algorithm is: shrink objects in image; subtract shrunken objects from original image; // the result is to leave the edges To prove this, note that […]

Davies: Computer Vision, 5th edition: Solutions to selected problems 44 20.1 If more sources of error are applied to the same data, the variances have to be added. Furthermore, if each source contributes equal amounts of error, the variance has […]

Davies: Computer Vision, 5th edition: Solutions to selected problems 3 3.3 This type of filter emulates the mode filter, and as such will enhance images by taking a strong decision at each pixel as to which side of an edge […]

Davies: Computer Vision, 5th edition: Solutions to selected problems 7 © E. R. Davies 2017 4.3 Two solutions exist because one solution represents a threshold in the area of overlap between the two Gaussians; the other solution is necessary mathematically […]

Davies: Computer Vision, 5th edition: Solutions to selected problems 8 © E. R. Davies 2017 8.2 The definition of a North point is that it is a 1 with a 0 to the North of it, and a 1 to […]

Davies: Computer Vision, 5th edition: Solutions to selected problems 15 © E. R. Davies 2017 9.1 If a boundary has already been found using the EDGE routine, it will be no more than 2 pixels wide, but will be 4-connected. […]

Davies: Computer Vision, 5th edition: Solutions to selected problems 45 A.1 (a) The breakdown point is the proportion of a distribution on one side or the other which can be disregarded by a data analysis method and hence the proportion […]