Introduction to Programming with C++, 3rd edition
Introduction to Programming with C++, 3rd edition
3rd Edition
ISBN: 9780133377477
Author: Y. Daniel Liang
Publisher: Pearson Academic Computing
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1.[30 pts] Computers generate color pictures on a video screen or liquid crystal display by mixing three different colors of light: red, green, and blue. Imagine a simple scheme, with three different lights, each of which can be turned on or off, projecting onto a glass screen: We can create eight different colors based on the absence (0) or presence (1) of light sources R,G and B: R G B Color 0 0 0 Black 0 0 1 Blue 0 1 0 Green 0 1 1 Cyan 1 0 0 Red 1 0 1 Magenta 1 1 1 0 Yellow 1 White 1 Each of these colors can be represented as a bit vector of length 3, and we can apply Boolean operations to them. a. The complement of a color is formed by turning off the lights that are on and turning on the lights that are off. What would be the complement of each of the eight colors listed above? b. Describe the effect of applying Boolean operations on the following colors: Λ 1. Red(100) ^ Magenta(101)= Blue(001) 2. Bue(001) | Green(010)= 3. Yellow(100) & Cyan(011)= 2.[30 pts] Perform the following…
D. S. Malik, Data Structures Using C++, 2nd Edition, 2010
Methods (Ch6) - Review 1. (The MyRoot method) Below is a manual implementation of the Math.sqrt() method in Java. There are two methods, method #1 which calculates the square root for positive integers, and method #2, which calculates the square root of positive doubles (also works for integers). public class SquareRoot { public static void main(String[] args) { } // implement a loop of your choice here // Method that calculates the square root of integer variables public static double myRoot(int number) { double root; root=number/2; double root old; do { root old root; root (root_old+number/root_old)/2; } while (Math.abs(root_old-root)>1.8E-6); return root; } // Method that calculates the square root of double variables public static double myRoot(double number) { double root; root number/2; double root_old; do { root old root; root (root_old+number/root_old)/2; while (Math.abs (root_old-root)>1.0E-6); return root; } } Program-it-Yourself: In the main method, create a program that…
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