Archives: Solution Manual

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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 […]

1 Engineering Software Products: Notes for Instructors I have designed this book for an introductory, single-semester, course in software engineering. As I said the book’s preface, my motivation was to write about the type of software that students can relate […]

Chapter 16 Answers to Concept Questions 4. The COMBINE multiple event policy allows the event listener to fire again, even though the listener may have already fired and the procedure that was called to respond to that previous event has […]

Chapter 15 Answers to Concept Questions 1. Alice uses the term “item” and Java uses the term “element” for an object or value stored in an indexed position in an array. In common usage, the terms are interchangeable. 2. The […]

Chapter 14 Answers to Concept Questions 2. An example shown here declares the array to be of type Quadruped. Note: The array could have been declared with an ancestor class, for example SModel. 3. Not at runtime – the array, […]

Chapter 13 Answers to Concept Questions 2. (a) SJoint (b) SJoint [ ] 3. (a) A joint configuration is a specific arrangement of the joints — in essence, a pattern. A joint configuration can be but is not necessarily an […]

Chapter 12 Answers to Concept Questions 1. The center of an object in Alice 3 is typically located at the object’s base, not the center of mass. So, if the object is to be rotated around the center of mass, […]

Chapter 11 Answers to Concept Questions 1. From the perspective of the user: A traditional sequential program runs in a pre-programmed 2. At least three threads. 3. End conditions are generally stated as rules that determine when the game ends. […]

Chapter 10 Answers to Concept Questions (e) String 2. (a) boolean (b) char (c) double (d) int (e) char (f) double (g) int 3. String is not a primitive data type. 4. (a) value stores an Integer, but true is […]

Chapter 9 Answers to Concept Questions 3. a. DecimalNumber (Double, in Java) b. WholeNumber (Integer, in Java) c. WholeNumber d. DecimalNumber e. Boolean f. TextString (String, in Java) g. TextString 4. a. 22 b. 5 c. 5.5 d. 1 e. […]

Chapter 8 Answers to Concept Questions 4. In Alice, a property declaration statement is typically for the purpose of adding an object to the scene or creating a variable to store a value. In each case, a name and a […]

Chapter 7 Answers to Concept Questions 1. In NetBeans, each class has one tab that contains a text-based definition of all components in the class, including declaration statements for properties, constructors, and methods (both procedural and functional). 2. The main […]

Chapter 6 Answers to Concept Questions 8. Table: 1. Stepwise refinement is the repeated use of decomposition to break down a complex task into simpler and simpler steps. 2. C. Long procedures are difficult to debug and difficult to revise […]

Chapter 5 Answers to Concept Questions 1. 4. a) together b) in order c) in order d) together e) in order f) together g) in order 5. Group 1: do in order, sequential, series (all perform one action at a […]

Chapter 4 Answers to Concept Questions 1. To implement is to write statements that translate algorithm steps into program code that, when 2. Design is the process of planning a solution for a given problem. Implementation is the process of […]

Chapter 3 Answers to Concept Questions 1. A design algorithm provides a plan for implementing the code. 2. Two kinds of tabs in the Alice Code editor: 3. A procedure is an action that can be performed by an object […]

Chapter 2 Answers for Concept Questions 1. The purpose of set design is to plan the visual aspects of the scene, including relative positioning 2. Assuming the fish is currently positioned with its “belly” downward, use a one-step turn procedure […]

Chapter 1 Answers to Concept Questions 1. (a) Analyze the task (understand the problem task, identify objects and actions) 2. Background information describes the setting, or environment in which objects will perform actions. Object-actions identifies which objects will perform specific […]