Introduction to Computer Graphics using OpenGLCan someone tell me .pdf
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Introduction to Computer Graphics using OpenGL Can someone tell me what shall I do to connect between the LineClipping, drawing the line, and rectangle function? I wrote the code below for the above question: #include #define red makecol(255,0,0) #define green makecol(0,255,0) #include // use as needed for your system #include #include #include #include #include \"vector.h\" const int screenWidth = 800; const int screenHeight = 600; double xmax, xmin, ymax, ymin; typedef int OutCode; double xvmin, yvmin, xvmax, yvmax; int clicks = 0; Point positions[4]; bool flag = false; const int INSIDE = 0; // 0000 const int LEFT = 1; const int RIGHT = 2; const int BOTTOM = 4; // 0100 const int TOP = 8; OutCode ComputeOutCode(double x, double y) { OutCode code; code = INSIDE; if (x < xmin) code |= LEFT; else if (x > xmax) code |= RIGHT; if (y < ymin) code |= BOTTOM; else if (y > ymax) code |= TOP; return code; } void LineClipping(double x0, double y0, double x1, double y1) { OutCode outcode0 = ComputeOutCode(x0, y0); OutCode outcode1 = ComputeOutCode(x1, y1); bool accept = false; bool done = false; while (done) { if (!(outcode0 | outcode1)) // Trivially accept and get out of loop { accept = true; done = true; break; } else if (outcode0 & outcode1) // Trivially reject and get out of loop { done = true; break; } else { // failed both tests, so calculate the line segment to clip // from an outside point to an intersection with clip edge double x, y; // At least one endpoint is outside the clip rectangle; pick it. OutCode outcodeOut = outcode0 ? outcode0 : outcode1; // Now find the intersection point; if (outcodeOut & TOP) { // point is above the clip rectangle x = x0 + (x1 - x0) * (ymax - y0) / (y1 - y0); y = ymax; } else if (outcodeOut & BOTTOM) { // point is below the clip rectangle x = x0 + (x1 - x0) * (ymin - y0) / (y1 - y0); y = ymin; } else if (outcodeOut & RIGHT) { // point is to the right of clip rectangle y = y0 + (y1 - y0) * (xmax - x0) / (x1 - x0); x = xmax; } else if (outcodeOut & LEFT) { // point is to the left of clip rectangle y = y0 + (y1 - y0) * (xmin - x0) / (x1 - x0); x = xmin; } if (outcodeOut == outcode0) { x0 = x; y0 = y; outcode0 = ComputeOutCode(x0, y0); } else { x1 = x; y1 = y; outcode1 = ComputeOutCode(x1, y1); } } } if (accept) { // window to viewport mapping /* double sx = (xvmax - xvmin) / (xmax - xmin);// scale parameter in x direction double sy = (yvmax - yvmin) / (ymax - ymin);// scale parameter in y direction double vx0 = xvmin + (x0 - xmin)*sx; double vy0 = yvmin + (y0 - ymin)*sy; double vx1 = xvmin + (x1 - xmin)*sx; double vy1 = yvmin + (y1 - ymin)*sy;*/ //draw a red color viewport glColor3f(1.0, 0.0, 0.0); glBegin(GL_LINE_LOOP); glVertex2f(xmin, ymin); glVertex2f(xmax, ymin); glVertex2f(xmax, ymax); glVertex2f(xmin, ymax); glEnd(); glColor3f(0.0, 0.0, 1.0); glBegin(GL_LINES); glVertex2d(x0, y0); glVertex2d(x1, y1); glEnd(); } } //<<<<<<<<<<<<<<<<<<<<<<< myInit >>>>>>>>>>>>>>>>>>>> void myInit(void) { glClearColor(1.0, 1.0, .
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![Introduction to Computer Graphics using OpenGL
Can someone tell me what shall I do to connect between the LineClipping, drawing the line, and
rectangle function? I wrote the code below for the above question:
#include
#define red makecol(255,0,0)
#define green makecol(0,255,0)
#include // use as needed for your system
#include
#include
#include
#include
#include "vector.h"
const int screenWidth = 800;
const int screenHeight = 600;
double xmax, xmin, ymax, ymin;
typedef int OutCode;
double xvmin, yvmin, xvmax, yvmax;
int clicks = 0;
Point positions[4];
bool flag = false;
const int INSIDE = 0; // 0000
const int LEFT = 1;
const int RIGHT = 2;
const int BOTTOM = 4; // 0100
const int TOP = 8;
OutCode ComputeOutCode(double x, double y)
{
OutCode code;
code = INSIDE;
if (x < xmin)
code |= LEFT;
else if (x > xmax)
code |= RIGHT;](https://image.slidesharecdn.com/introductiontocomputergraphicsusingopenglcansomeonetellme-230707211325-d5ac2868/85/Introduction-to-Computer-Graphics-using-OpenGLCan-someone-tell-me-pdf-1-320.jpg)
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c++ code#include iostream#include string#include stdlib.h.pdf
c++ code
#include
#include
#include
using namespace std;
int bomb_count, checkx, checky;
const int sizex = 10, sizey = 10;
bool lose;
class Grid {
public:
bool has_bomb, marked;
Grid(void) {
has_bomb = false;
marked = false;
}
};
void drawBoad(Grid board[sizex][sizey])
{
// _0_1_2_3_4_5_6_7_8_9_
// 0|_|_|_|_|_|_|_|_|_|_|
// 1|_|_|_|_|_|_|_|_|_|_|
// 2|_|_|_|_|_|_|_|_|_|_|
// 3|_|_|_|_|_|_|_|_|_|_|
// 4|_|_|_|_|_|_|_|_|_|_|
// 5|_|_|_|_|_|_|_|_|_|_|
// 6|_|_|_|_|_|_|_|_|_|_|
// 7|_|_|_|_|_|_|_|_|_|_|
// 8|_|_|_|_|_|_|_|_|_|_|
// 9|_|_|_|_|_|_|_|_|_|_|
//This draws the top line
cout << \" _\";
for (int i = 0; i < sizex; i++)
{
cout << i << \"_\";
}
cout << endl;
//This draws the body
for (int y = 0; y < sizey; y++)
{
cout << y << \"|\";
for (int x = 0; x < sizex; x++)
{
if (board[x][y].has_bomb && board[x][y].marked)
{
cout << \"!|\";
}
if (board[x][y].has_bomb)
{
cout << \"b|\";
}
else if (board[x][y].marked)
{
cout << \"x|\";
}
else
{
cout << \"_|\";
}
}
cout << endl;
}
}
int main()
{
{
lose = false;
bomb_count = 15;
checkx = 0;
checky = 0;
Grid gameboard[sizex][sizey];
for (int i = 0; i < bomb_count; i++)
{
int xpos = rand() % 10;
int ypos = rand() % 10;
if (gameboard[xpos][ypos].has_bomb == false) {
gameboard[xpos][ypos].has_bomb = true;
}
else
{
i--;
}
}
drawBoad(gameboard);
cout << endl;
while (lose != true)
{
cout << \"Input x grid to check.\" << endl;
cin >> checkx;
cout << endl << \"Input y grid to check.\" << endl;
cin >> checky;
if (gameboard[checkx - 1][checky - 1].has_bomb == true) {
cout << \"Boom!\";
lose = true;
}
else {
cout << \"Try again.\" << endl;
}
gameboard[checkx - 1][checky - 1].marked = true;
cout << endl;
drawBoad(gameboard);
cout << endl;
}
return 0;
}
ithe lengthr size of standard array in java immutable it will have a fixed length.
java code
package {
import flash.display.Sprite;
import flash.events.MouseEvent;
public class Main extends Sprite {
public var zones:Array;
public var board:Sprite;
public var boardWidth:int;
public var boardHeight:int;
public var numberOfMines:int;
public function Main() { //constructor----
zones = new Array();
board = new Sprite();
boardWidth = 10;
boardHeight = 10;
numberOfMines = 20;
gameover_btn.visible = false;
again_btn.visible = false;
minesleft_txt.text = String(numberOfMines);
addChildAt(board, 0);
createBoard();
placeMines();
placeNumbers();
again_btn.addEventListener(MouseEvent.CLICK, makeAnotherGame);
} //end of constructor
private function createBoard():void{
for(var i:Number = 0; i < boardHeight; i++){
var hzones:Array = new Array();
zones[i] = hzones;
for(var j:Number = 0; j < boardWidth; j++){
var newzone:Zone = new Zone(j, i);
zones[i][j] = newzone;
board.addChild(zones[i][j]);
zones[i][j].x = j*40 + 20;
zones[i][j].y = i*40 + 20;
}
}
}
private function placeMines():void{
var bombsCounter:uint = numberOfMines;
while(bombsCounter > 0){
var randomNumber:Number = Math.round(Math.random()* ((boardHeight *
boardWidth) - 1));
var rowCount:Number = Math.floor(randomNumber / bo.
need help with code I wrote. This code is a maze gui, and i need hel.pdf
need help with code I wrote. This code is a maze gui, and i need help with the method
shorterstpath(); here is my code:
******mazeApp*******
package mazepackage;
import javafx.application.Application;
import javafx.application.Platform;
import javafx.scene.Group;
import javafx.scene.Scene;
import javafx.scene.canvas.Canvas;
import javafx.scene.canvas.GraphicsContext;
import javafx.scene.control.TextInputDialog;
import javafx.scene.paint.Color;
import javafx.stage.Stage;
public class MazeApp extends Application
{
// default canvas size is DEFAULT_SIZE-by-DEFAULT_SIZE
private static final int DEFAULT_SIZE = 768;
private int width = DEFAULT_SIZE;
private int height = DEFAULT_SIZE;
// The graphics context is needed to enable drawing on the canvas
private GraphicsContext gc;
// boundary of drawing canvas, 0% border
// private static final double BORDER = 0.05;
private static final double BORDER = 0.00;
private double xmin, ymin, xmax, ymax;
public static void main(String[] args)
{
launch(args);
}
@Override
public void start(Stage primaryStage)
{
Group root = new Group();
Canvas canvas = new Canvas(width, height);
gc = canvas.getGraphicsContext2D();
gc.setLineWidth(2);
gc.setFill(Color.WHITE);
gc.fillRect(0, 0, width, height);
root.getChildren().add(canvas);
TextInputDialog tid = new TextInputDialog();
tid.setTitle(\"Maze Size\");
tid.setHeaderText(\"Enter maze size between 10 and 50\");
tid.showAndWait();
int size = Integer.parseInt(tid.getResult());
if (size > 50)
size = 50;
if (size < 10)
size = 10;
primaryStage.setTitle(\"Maze Application\");
primaryStage.setScene(new Scene(root));
primaryStage.setResizable(false);
// Make sure that the application goes away when then window is closed
primaryStage.setOnCloseRequest(e -> System.exit(0));
primaryStage.show();
Maze maze = new Maze(this, size);
// Must solve the maze in a separate thread or else
// the GUI wont update until the end.....
Thread solver = new Thread(
new Runnable () {
public void run()
{
while(true)
{
maze.buildAndDrawMaze();
maze.findShortestPath();
try
{
Thread.sleep(5000);
}
catch(Exception e) { }
gc.setFill(Color.WHITE);
gc.fillRect(0, 0, width, height);
}
}
});
solver.start();
}
/**
* Sets the pen color to the specified color.
*
* @param color the color to make the pen
*/
public void setPenColor(Color color) {
gc.setStroke(color);
}
/**
* Sets the pen color to the specified color.
*
* @param color the color to make the pen
*/
public void setFillColor(Color color) {
gc.setFill(color);
}
/**
* Sets the x-scale to the specified range.
*
* @param min the minimum value of the x-scale
* @param max the maximum value of the x-scale
* @throws IllegalArgumentException if {@code (max == min)}
*/
public void setXscale(double min, double max) {
double size = max - min;
if (size == 0.0) {
throw new IllegalArgumentException(\"the min and max are the same\");
}
xmin = min - BORDER * size;
xmax = max + BORDER * size;
}
/**
* Sets the y-scale to the specified range.
*
* @param min the minimum valu.
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Computer graphics lab manual
Line drawing algorithms
circle generation algorithms
2D objects
3D objects
import java.awt.;import java.awt.event.MouseAdaptor;import java.pdf
import java.awt.*;
import java.awt.event.MouseAdaptor;
import java.awt.event.MouseEvent;
import javax.swing.*;
public class Prob1_4 extends JPanel
{
private static final long serialVersionUID = 1L;
private final int WIDTH = 1200;
private final int HEIGHT = 800;
int maxX ,maxY ;
static JFrame f;
float pixelWidth , pixelHeight;
rWidth = 10.0F;
rHeight = 7.5F;
xP = -1, yP;
FontMetrics metrics;
public Prob1_4() {
setPreferredSize(new Dimension(WIDTH, HEIGHT));
addMouseListener(new MouseAdapter()
{
public void mousePressed(MouseEvent evt)
{
xP = evt.getX(); yP = evt.getY();
f.repaint();
}
} );
}
void initgr()
{ Dimension d = getSize();
maxX = d.width - 1; maxY = d.height - 1;
pixelWidth = rWidth/maxX; pixelHeight = rHeight/maxY;
}
int iX(float x){return Math.round(x/pixelWidth);}
int iY(float y){return maxY - Math.round(y/pixelHeight);}
float fx(int x){return x * pixelWidth;}
float fy(int y){return (maxY - y) * pixelHeight;}
public void paint(Graphics g)
{
initgr();
Graphics2D g2d = (Graphics2D) g;
Point origin = new Point(WIDTH / 2, HEIGHT / 2);
metrics = g.getFontMetrics();
double value1 = Math.pow(xP-origin.x,2);
double value2 =Math.pow(yP- origin.y,2);
int radius = (int)Math.sqrt(value1+value2);
g.drawString(
\"Click anywhere on screen to generate hexagons of apprpriate radius! \"
, 40, 100);
double size = (radius/253.0)*7.0;
double rad = (radius/253.0)*30.0;
drawHexGridLoop(g2d, origin, (int)size-2, (int)rad+5, 0);
}
//Method to draw regular hexagon
private void drawHex(Graphics g, int posX, int posY, int x, int y, int r) {
Graphics2D g2d = (Graphics2D) g;
hexagon hex = new hexagon(x, y, r);
hex.draw(g2d, x, y, 0, 0x008844, false);
hex.draw(g2d, x, y, 4, 000000, false);
}
public static void main(String[] args) {
f = new JFrame();
Prob1_4 p = new Prob1_4();
f.setContentPane(p);
f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
f.pack();
f.setLocationRelativeTo(null);
f.setVisible(true);
}
}
int iX(float x){return Math.round(x/pixelWidth);}
int iY(float y){return maxY - Math.round(y/pixelHeight);}
float fx(int x){return x * pixelWidth;}
float fy(int y){return (maxY - y) * pixelHeight;}
public void paint(Graphics g)
{
initgr();
Graphics2D g2d = (Graphics2D) g;
Point origin = new Point(WIDTH / 2, HEIGHT / 2);
metrics = g.getFontMetrics();
double value1 = Math.pow(xP-origin.x,2);
double value2 =Math.pow(yP- origin.y,2);
int radius = (int)Math.sqrt(value1+value2);
g.drawString(
\"Click anywhere on screen to generate hexagons of apprpriate radius! \"
, 40, 100);
double size = (radius/253.0)*7.0;
double rad = (radius/253.0)*30.0;
drawHexGridLoop(g2d, origin, (int)size-2, (int)rad+5, 0);
}
//Method to draw regular hexagon
private void drawHex(Graphics g, int posX, int posY, int x, int y, int r) {
Graphics2D g2d = (Graphics2D) g;
hexagon hex = new hexagon(x, y, r);
hex.draw(g2d, x, y, 0, 0x008844, false);
hex.draw(g2d, x, y, 4, 000000, false);
}
public static void main(String[] args) {
f = new JFrame();
Prob1_4 p = new Prob1_4();
f.setContentPane(p);
f.
how can i build this as problem 2 of my code #include iostream.pdf
how can i build this as problem 2 of my code?
#include
#include
#include
#include
#include "./freeglut-3.4.0/include/GL/freeglut.h"
using namespace std;
bool leftDown = false, rightDown = false;
int lastPos[2];
float cameraPos[4] = { 0,1,4,1 };
int windowWidth = 640, windowHeight = 480;
double xRot = 0;
double yRot = 0;
int curProblem = 1; // TODO: change this number to try different examples
float specular[] = { 1.0, 1.0, 1.0, 1.0 };
float shininess[] = { 50.0 };
void problem1() {
// TODO
}
void problem2() {
//FIX HERE
}
void problem3() {
//TO DO
}
void problem4() {
//TO DO
}
void display()
{
glClearColor(0, 0, 0, 0);
glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
glDisable(GL_LIGHTING);
glEnable(GL_DEPTH_TEST);
glBegin(GL_LINES);
glColor3f(1, 0, 0); glVertex3f(0, 0, 0); glVertex3f(1, 0, 0); // x axis
glColor3f(0, 1, 0); glVertex3f(0, 0, 0); glVertex3f(0, 1, 0); // y axis
glColor3f(0, 0, 1); glVertex3f(0, 0, 0); glVertex3f(0, 0, 1); // z axis
glEnd(/*GL_LINES*/);
glEnable(GL_LIGHTING);
glShadeModel(GL_SMOOTH);
glMaterialfv(GL_FRONT, GL_SPECULAR, specular);
glMaterialfv(GL_FRONT, GL_SHININESS, shininess);
glEnable(GL_LIGHT0);
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
glViewport(0, 0, windowWidth, windowHeight);
float ratio = (float)windowWidth / (float)windowHeight;
gluPerspective(50, ratio, 1, 1000);
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();
gluLookAt(cameraPos[0], cameraPos[1], cameraPos[2], 0, 0, 0, 0, 1, 0);
glLightfv(GL_LIGHT0, GL_POSITION, cameraPos);
glRotatef(xRot, 1, 0, 0);
glRotatef(yRot, 0, 1, 0);
if (curProblem == 1) problem1();
if (curProblem == 2) problem2();
if (curProblem == 3) problem3();
if (curProblem == 4) problem4();
glutSwapBuffers();
}
void mouse(int button, int state, int x, int y)
{
if (button == GLUT_LEFT_BUTTON) leftDown = (state == GLUT_DOWN);
else if (button == GLUT_RIGHT_BUTTON) rightDown = (state == GLUT_DOWN);
lastPos[0] = x;
lastPos[1] = y;
}
void mouseMoved(int x, int y)
{
if (leftDown) {
xRot += (y - lastPos[1]) * .1;
yRot += (x - lastPos[0]) * .1;
}
if (rightDown) {
for (int i = 0; i < 3; i++)
cameraPos[i] *= pow(1.1, (y - lastPos[1]) * .1);
}
lastPos[0] = x;
lastPos[1] = y;
glutPostRedisplay();
}
void keyboard(unsigned char key, int x, int y)
{
curProblem = key - '0';
if (key == 'q' || key == 'Q' || key == 27) {
exit(0);
}
glutPostRedisplay();
}
void reshape(int width, int height)
{
windowWidth = width;
windowHeight = height;
glutPostRedisplay();
}
int main(int argc, char** argv)
{
glutInit(&argc, argv);
glutInitDisplayMode(GLUT_DOUBLE | GLUT_RGB | GLUT_DEPTH);
glutInitWindowSize(windowWidth, windowHeight);
glutCreateWindow("HW2");
glutDisplayFunc(display);
glutMotionFunc(mouseMoved);
glutMouseFunc(mouse);
glutReshapeFunc(reshape);
glutKeyboardFunc(keyboard);
glutMainLoop();
return 0;
}.
Computer Graphics and Multimedia lab report
According to NU of Bangladesh for BSC Hons in CSE 6th semester, here you can find the solution about this subject according to the board question respectively.
animation.#include windows.h#include GLglut.h#inclu.pdf
animation.
#include
#include
#include
#include
#include
#include
#include
animation.
#include
#include
#include
#include
#include
#include
#include
bool rotate1=false, rotate2=false;
double t; double dt=0.1;
double angle1=0; double dangle1 =20;
double angle2=0; double dangle2 =20;
double R=5;//radius spiral
double T=20;//time interval
double shiftx,shifty,shiftz;
bool origin=true;
void output_Text(float x, float y, char *text)
{ int len, i;
glColor3f(1, 0, 0); //red text
glRasterPos3f(x, y, 0);
len = (int)strlen(text);
for (i = 0; i < len; i++) { glutBitmapCharacter(GLUT_BITMAP_8_BY_13, text[i]); }
}
void Time_Delay(int dur) {
clock_t start_time;
start_time = clock();
while ((clock() - start_time) < dur) {
}
}
float ax[3]={25,0,0}; float ay[3]={0,25,0}; float az[3]={0,0,25};//axes void plot_Arrow(float
v1[3], float v2[3]) //plot arrow
{
glBegin(GL_LINES);
glVertex3fv(v1);
glVertex3fv(v2);
glEnd();
}
void plot_Vector(float v[3]) //plot vector
{
float zero[3]={0,0,0};
plot_Arrow(zero,v);
}
void plot_Axis()
{
glColor3f(0,0,0);
plot_Vector(ax);
plot_Vector(ay);
plot_Vector(az);
}
void Draw_Trajectory() {
//your code
}
void myInit3d(void)
{
glClearColor(1,1,1,0);
glColor3f(1.0, 0.0, 0.0); glClear(GL_COLOR_BUFFER_BIT); // clear the screen
glMatrixMode(GL_PROJECTION);
glLoadIdentity(); // reset the Current Modelview Matrix
bool rotate1=false, rotate2=false;
double t; double dt=0.1;
double angle1=0; double dangle1 =20;
double angle2=0; double dangle2 =20;
double R=5;//radius spiral
double T=20;//time interval
double shiftx,shifty,shiftz;
bool origin=true;
void output_Text(float x, float y, char *text)
{ int len, i;
glColor3f(1, 0, 0); //red text
glRasterPos3f(x, y, 0);
len = (int)strlen(text);
for (i = 0; i < len; i++) { glutBitmapCharacter(GLUT_BITMAP_8_BY_13, text[i]); }
}
void Time_Delay(int dur) {
clock_t start_time;
start_time = clock();
while ((clock() - start_time) < dur) {
}
}
float ax[3]={25,0,0}; float ay[3]={0,25,0}; float az[3]={0,0,25};//axes void plot_Arrow(float
v1[3], float v2[3]) //plot arrow
{
glBegin(GL_LINES);
glVertex3fv(v1);
glVertex3fv(v2);
glEnd();
}
void plot_Vector(float v[3]) //plot vector
{
float zero[3]={0,0,0};
plot_Arrow(zero,v);
}
void plot_Axis()
{
glColor3f(0,0,0);
plot_Vector(ax);
plot_Vector(ay);
plot_Vector(az);
}
void Draw_Trajectory() {
//your code
}
void myInit3d(void)
{
glClearColor(1,1,1,0);
glColor3f(1.0, 0.0, 0.0); glClear(GL_COLOR_BUFFER_BIT); // clear the screen
glMatrixMode(GL_PROJECTION);
glLoadIdentity(); // reset the Current Modelview Matrix
//the coordinate system, the window and glLookAt in your assigment //can be different
glMatrixMode(GL_MODELVIEW); // position and aim the camera glLoadIdentity();
gluPerspective(45, 1.33, .01, 1000);
gluLookAt(30.0, 30.0, 30.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0); }
void myDisplay(void) {
//your code
}
void Do_Nothing(void)
{}
void myMouse(int button, int state, int x, int y) //press left/right button
{ switch (button) {
case GLUT_LEFT_BUTTON:
if (state == GLUT_DOWN)
if.
import java.util.Scanner;public class Main { public static in.pdf
import java.util.Scanner;
public class Main {
public static int gridArray1[][] = null;
public static int gridArray2[][] = null;
public static boolean playerTurn = true;
public static int count = 0;
public static void setGridDimensions(int x_max, int y_max) {
Main.gridArray1 = new int[x_max][y_max];
Main.gridArray2 = new int[x_max][y_max];
for (int i = 0; i < x_max; i++) {
for (int j = 0; j < y_max; j++) {
gridArray1[i][j] = -1;
gridArray2[i][j] = -1;
}
}
}
static void placeShip(int starting_x, int starting_y, int length, int direction) {
count++;
int ships[] = new int[10];
int battleshipArrayPositions1[] = new int[4];
int cruiserArrayPositions1[] = new int[3];
int submarineArrayPositions1[] = new int[3];
int destroyerArrayPositions1[] = new int[2];
int carrierArrayPositions2[] = new int[5];
int battleshipArrayPositions2[] = new int[4];
int cruiserArrayPositions2[] = new int[3];
int submarineArrayPositions2[] = new int[3];
int destroyerArrayPositions2[] = new int[2];
if (playerTurn) {
if (direction == 0) {
for (int i = starting_x; i < starting_x + length; i++) {
Main.gridArray1[i][starting_y] = 0;
}
} else {
for (int i = starting_y; i < starting_y + length; i++) {
Main.gridArray1[starting_x][i] = 0;
}
}
} else {
if (direction == 0) {
for (int i = starting_x; i < starting_x + length; i++) {
Main.gridArray2[i][starting_y] = 0;
}
} else {
for (int i = starting_y; i < starting_y + length; i++) {
Main.gridArray2[starting_x][i] = 0;
}
}
}
}
boolean isConflictingShipPlacement(int starting_x, int starting_y, int length, int direction) {
return false;
}
static int shoot(int x, int y) {
if (playerTurn) {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (Main.gridArray2[i][j] == 0) {
return 0;
}
}
}
} else {
}
return 0;
}
boolean hasBeenAttempted(int x, int y) {
return false;
}
static void displayGrid(boolean showShips) {
int count = 0;
System.out.print(\" 0 1 2 3 4 5 6 7 8 9\");
if (showShips) {
if (playerTurn) {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (j % 10 == 0) {
System.out.print(\"\ \" + count + \" \");
count++;
}
if (Main.gridArray1[i][j] == -1) {
System.out.print(\"- \");
} else if (Main.gridArray1[i][j] == 0) {
System.out.print(\"@ \");
} else if (Main.gridArray1[i][j] == 1) {
System.out.print(\"+ \");
} else {
System.out.print(\"X \");
}
}
}
} else {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (j % 10 == 0) {
System.out.print(\"\ \" + count + \" \");
count++;
}
if (Main.gridArray2[i][j] == -1) {
System.out.print(\"- \");
} else if (Main.gridArray2[i][j] == 0) {
System.out.print(\"@ \");
} else if (Main.gridArray2[i][j] == 1) {
System.out.print(\"+ \");
} else {
System.out.print(\"X \");
}
}
}
}
} else {
if (playerTurn) {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (j % 10 == 0) {
System.out.print(\"\ \" + count + \" \");
count++;
}
if (Main.gridArray1[i][j] == -1) {
System.out.print(\"- \");
} else if (Main.gridArray1[i][j] == 0) {
System.out.print(\"- \");
} el.
Write a Python program that extracts 1000 unique links from Twitter .pdf
Write a program that draws a simple fence with vertical, equally spaced slats backed by two
horizontal support boards. Behind the fence show a simple house in the background. Make sure
the house is visible between the slats in the fence.
Solution
package app;
import java.awt.*;
import java.applet.*;
public class House extends Applet
{
public void paint (Graphics c)
{
house_wall (c);
window (c);
house_frame (c);
fence (c);
sup_boards (c);
}
public void fence (Graphics c)
{
c.setColor(Color.black);
c.fillRect (35,220,3,350);
c.setColor(Color.black);
c.fillRect (70,220,3,350);
c.setColor (Color.black); // slats of fence
c.fillRect (105,220,3,350);
c.setColor(Color.black);
c.fillRect (140,220,3,350);
c.setColor(Color.black);
c.fillRect (175,220,3,350);
c.setColor(Color.black);
c.fillRect (210,220,3,350);
c.setColor(Color.black);
c.fillRect (245,220,3,350);
c.setColor(Color.black);
c.fillRect (280,220,3,350);
c.setColor(Color.black);
c.fillRect (315,220,3,350);
c.setColor(Color.black);
c.fillRect (350,220,3,350);
c.setColor(Color.black);
c.fillRect (385,220,3,350);
c.setColor(Color.black);
c.fillRect (420,220,3,350);
c.setColor(Color.black);
c.fillRect (455,220,3,350);
c.setColor(Color.black);
c.fillRect (500,220,3,350);
c.setColor(Color.black);
c.fillRect (535,220,3,350);
}
public void sup_boards (Graphics c)
{
c.setColor (Color.black);
c.fillRect (20,350,550,8);
c.fillRect (20,460,550,8);
}
public void house_wall (Graphics c)
{
c.setColor (Color.RED);
c.fillRect (100,250,400,250);
c.setColor(Color.DARK_GRAY);
c.fillRect (245,380,110,70);
c.setColor (new Color(186,134,11));
c.fillOval (295,412,10,10);
}
public void window (Graphics c)
{
c.setColor (Color.blue);
c.fillRect (121,261,78,78);
c.fillRect (121,361,78,78);
c.fillRect (401,261,78,78);
c.fillRect (401,361,78,78);
c.setColor (Color.yellow);
c.fillRect (125,265,70,70);
c.fillRect (125,365,70,70);
c.fillRect (405,265,70,70);
c.fillRect (405,365,70,70);
}
public void house_frame (Graphics c)
{
c.setColor (Color.white);
c.fillRect (298,380,2,70);
c.setColor (Color.red);
c.fillRect (157,265,3,70);
c.fillRect (157,365,3,70);
c.fillRect (437,265,3,70);
c.fillRect (438,365,3,70);
c.fillRect (125,298,70,3);
c.fillRect (125,398,70,3);
c.fillRect (405,298,70,3);
c.fillRect (405,398,70,3);
c.fillRect (245,375,110,3);
}
}.
Which of the following statements about financial intermediaries is .pdf
Which of the following is true of a natural monopoly? Oa. If regulated, the firm will be allowed
to charge a price higher than its average cost. b. If regulated, the firm will haye a lower level of
output than if it was unregulated. Oc. If regulated, the firm will have a higher level of output than
if it was unregulated. Oc. If regulated, the firm will eam economic profit in the long run. De. If
regulated, the firm will experience diseconomies of scale.
Solution
Answer :- Option b). If regulated, the firm will have a lower level of output than if it was
unregulated.
Conclusion :- Option b)..
More Related Content
Similar to Introduction to Computer Graphics using OpenGLCan someone tell me .pdf
c++ code#include iostream#include string#include stdlib.h.pdf
c++ code
#include
#include
#include
using namespace std;
int bomb_count, checkx, checky;
const int sizex = 10, sizey = 10;
bool lose;
class Grid {
public:
bool has_bomb, marked;
Grid(void) {
has_bomb = false;
marked = false;
}
};
void drawBoad(Grid board[sizex][sizey])
{
// _0_1_2_3_4_5_6_7_8_9_
// 0|_|_|_|_|_|_|_|_|_|_|
// 1|_|_|_|_|_|_|_|_|_|_|
// 2|_|_|_|_|_|_|_|_|_|_|
// 3|_|_|_|_|_|_|_|_|_|_|
// 4|_|_|_|_|_|_|_|_|_|_|
// 5|_|_|_|_|_|_|_|_|_|_|
// 6|_|_|_|_|_|_|_|_|_|_|
// 7|_|_|_|_|_|_|_|_|_|_|
// 8|_|_|_|_|_|_|_|_|_|_|
// 9|_|_|_|_|_|_|_|_|_|_|
//This draws the top line
cout << \" _\";
for (int i = 0; i < sizex; i++)
{
cout << i << \"_\";
}
cout << endl;
//This draws the body
for (int y = 0; y < sizey; y++)
{
cout << y << \"|\";
for (int x = 0; x < sizex; x++)
{
if (board[x][y].has_bomb && board[x][y].marked)
{
cout << \"!|\";
}
if (board[x][y].has_bomb)
{
cout << \"b|\";
}
else if (board[x][y].marked)
{
cout << \"x|\";
}
else
{
cout << \"_|\";
}
}
cout << endl;
}
}
int main()
{
{
lose = false;
bomb_count = 15;
checkx = 0;
checky = 0;
Grid gameboard[sizex][sizey];
for (int i = 0; i < bomb_count; i++)
{
int xpos = rand() % 10;
int ypos = rand() % 10;
if (gameboard[xpos][ypos].has_bomb == false) {
gameboard[xpos][ypos].has_bomb = true;
}
else
{
i--;
}
}
drawBoad(gameboard);
cout << endl;
while (lose != true)
{
cout << \"Input x grid to check.\" << endl;
cin >> checkx;
cout << endl << \"Input y grid to check.\" << endl;
cin >> checky;
if (gameboard[checkx - 1][checky - 1].has_bomb == true) {
cout << \"Boom!\";
lose = true;
}
else {
cout << \"Try again.\" << endl;
}
gameboard[checkx - 1][checky - 1].marked = true;
cout << endl;
drawBoad(gameboard);
cout << endl;
}
return 0;
}
ithe lengthr size of standard array in java immutable it will have a fixed length.
java code
package {
import flash.display.Sprite;
import flash.events.MouseEvent;
public class Main extends Sprite {
public var zones:Array;
public var board:Sprite;
public var boardWidth:int;
public var boardHeight:int;
public var numberOfMines:int;
public function Main() { //constructor----
zones = new Array();
board = new Sprite();
boardWidth = 10;
boardHeight = 10;
numberOfMines = 20;
gameover_btn.visible = false;
again_btn.visible = false;
minesleft_txt.text = String(numberOfMines);
addChildAt(board, 0);
createBoard();
placeMines();
placeNumbers();
again_btn.addEventListener(MouseEvent.CLICK, makeAnotherGame);
} //end of constructor
private function createBoard():void{
for(var i:Number = 0; i < boardHeight; i++){
var hzones:Array = new Array();
zones[i] = hzones;
for(var j:Number = 0; j < boardWidth; j++){
var newzone:Zone = new Zone(j, i);
zones[i][j] = newzone;
board.addChild(zones[i][j]);
zones[i][j].x = j*40 + 20;
zones[i][j].y = i*40 + 20;
}
}
}
private function placeMines():void{
var bombsCounter:uint = numberOfMines;
while(bombsCounter > 0){
var randomNumber:Number = Math.round(Math.random()* ((boardHeight *
boardWidth) - 1));
var rowCount:Number = Math.floor(randomNumber / bo.
need help with code I wrote. This code is a maze gui, and i need hel.pdf
need help with code I wrote. This code is a maze gui, and i need help with the method
shorterstpath(); here is my code:
******mazeApp*******
package mazepackage;
import javafx.application.Application;
import javafx.application.Platform;
import javafx.scene.Group;
import javafx.scene.Scene;
import javafx.scene.canvas.Canvas;
import javafx.scene.canvas.GraphicsContext;
import javafx.scene.control.TextInputDialog;
import javafx.scene.paint.Color;
import javafx.stage.Stage;
public class MazeApp extends Application
{
// default canvas size is DEFAULT_SIZE-by-DEFAULT_SIZE
private static final int DEFAULT_SIZE = 768;
private int width = DEFAULT_SIZE;
private int height = DEFAULT_SIZE;
// The graphics context is needed to enable drawing on the canvas
private GraphicsContext gc;
// boundary of drawing canvas, 0% border
// private static final double BORDER = 0.05;
private static final double BORDER = 0.00;
private double xmin, ymin, xmax, ymax;
public static void main(String[] args)
{
launch(args);
}
@Override
public void start(Stage primaryStage)
{
Group root = new Group();
Canvas canvas = new Canvas(width, height);
gc = canvas.getGraphicsContext2D();
gc.setLineWidth(2);
gc.setFill(Color.WHITE);
gc.fillRect(0, 0, width, height);
root.getChildren().add(canvas);
TextInputDialog tid = new TextInputDialog();
tid.setTitle(\"Maze Size\");
tid.setHeaderText(\"Enter maze size between 10 and 50\");
tid.showAndWait();
int size = Integer.parseInt(tid.getResult());
if (size > 50)
size = 50;
if (size < 10)
size = 10;
primaryStage.setTitle(\"Maze Application\");
primaryStage.setScene(new Scene(root));
primaryStage.setResizable(false);
// Make sure that the application goes away when then window is closed
primaryStage.setOnCloseRequest(e -> System.exit(0));
primaryStage.show();
Maze maze = new Maze(this, size);
// Must solve the maze in a separate thread or else
// the GUI wont update until the end.....
Thread solver = new Thread(
new Runnable () {
public void run()
{
while(true)
{
maze.buildAndDrawMaze();
maze.findShortestPath();
try
{
Thread.sleep(5000);
}
catch(Exception e) { }
gc.setFill(Color.WHITE);
gc.fillRect(0, 0, width, height);
}
}
});
solver.start();
}
/**
* Sets the pen color to the specified color.
*
* @param color the color to make the pen
*/
public void setPenColor(Color color) {
gc.setStroke(color);
}
/**
* Sets the pen color to the specified color.
*
* @param color the color to make the pen
*/
public void setFillColor(Color color) {
gc.setFill(color);
}
/**
* Sets the x-scale to the specified range.
*
* @param min the minimum value of the x-scale
* @param max the maximum value of the x-scale
* @throws IllegalArgumentException if {@code (max == min)}
*/
public void setXscale(double min, double max) {
double size = max - min;
if (size == 0.0) {
throw new IllegalArgumentException(\"the min and max are the same\");
}
xmin = min - BORDER * size;
xmax = max + BORDER * size;
}
/**
* Sets the y-scale to the specified range.
*
* @param min the minimum valu.
10CSL67 CG LAB PROGRAM 5
Program to implement Cohen-Suderland Line Clipping Algorithm. Make provision to specify the input line, window for clipping and view port for displaying the clipped image.
C++ TUTORIAL 9
Numerical solution of ordinary differential equations by using Taylor method and Euler method
How to write the C++ codes?
Creating an Uber Clone - Part VIII - Transcript.pdf
This is a part of an online Codename One course published around 2017 see it all for free at https://debugagent.com/series/cn1
Computer graphics lab manual
Line drawing algorithms
circle generation algorithms
2D objects
3D objects
import java.awt.;import java.awt.event.MouseAdaptor;import java.pdf
import java.awt.*;
import java.awt.event.MouseAdaptor;
import java.awt.event.MouseEvent;
import javax.swing.*;
public class Prob1_4 extends JPanel
{
private static final long serialVersionUID = 1L;
private final int WIDTH = 1200;
private final int HEIGHT = 800;
int maxX ,maxY ;
static JFrame f;
float pixelWidth , pixelHeight;
rWidth = 10.0F;
rHeight = 7.5F;
xP = -1, yP;
FontMetrics metrics;
public Prob1_4() {
setPreferredSize(new Dimension(WIDTH, HEIGHT));
addMouseListener(new MouseAdapter()
{
public void mousePressed(MouseEvent evt)
{
xP = evt.getX(); yP = evt.getY();
f.repaint();
}
} );
}
void initgr()
{ Dimension d = getSize();
maxX = d.width - 1; maxY = d.height - 1;
pixelWidth = rWidth/maxX; pixelHeight = rHeight/maxY;
}
int iX(float x){return Math.round(x/pixelWidth);}
int iY(float y){return maxY - Math.round(y/pixelHeight);}
float fx(int x){return x * pixelWidth;}
float fy(int y){return (maxY - y) * pixelHeight;}
public void paint(Graphics g)
{
initgr();
Graphics2D g2d = (Graphics2D) g;
Point origin = new Point(WIDTH / 2, HEIGHT / 2);
metrics = g.getFontMetrics();
double value1 = Math.pow(xP-origin.x,2);
double value2 =Math.pow(yP- origin.y,2);
int radius = (int)Math.sqrt(value1+value2);
g.drawString(
\"Click anywhere on screen to generate hexagons of apprpriate radius! \"
, 40, 100);
double size = (radius/253.0)*7.0;
double rad = (radius/253.0)*30.0;
drawHexGridLoop(g2d, origin, (int)size-2, (int)rad+5, 0);
}
//Method to draw regular hexagon
private void drawHex(Graphics g, int posX, int posY, int x, int y, int r) {
Graphics2D g2d = (Graphics2D) g;
hexagon hex = new hexagon(x, y, r);
hex.draw(g2d, x, y, 0, 0x008844, false);
hex.draw(g2d, x, y, 4, 000000, false);
}
public static void main(String[] args) {
f = new JFrame();
Prob1_4 p = new Prob1_4();
f.setContentPane(p);
f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
f.pack();
f.setLocationRelativeTo(null);
f.setVisible(true);
}
}
int iX(float x){return Math.round(x/pixelWidth);}
int iY(float y){return maxY - Math.round(y/pixelHeight);}
float fx(int x){return x * pixelWidth;}
float fy(int y){return (maxY - y) * pixelHeight;}
public void paint(Graphics g)
{
initgr();
Graphics2D g2d = (Graphics2D) g;
Point origin = new Point(WIDTH / 2, HEIGHT / 2);
metrics = g.getFontMetrics();
double value1 = Math.pow(xP-origin.x,2);
double value2 =Math.pow(yP- origin.y,2);
int radius = (int)Math.sqrt(value1+value2);
g.drawString(
\"Click anywhere on screen to generate hexagons of apprpriate radius! \"
, 40, 100);
double size = (radius/253.0)*7.0;
double rad = (radius/253.0)*30.0;
drawHexGridLoop(g2d, origin, (int)size-2, (int)rad+5, 0);
}
//Method to draw regular hexagon
private void drawHex(Graphics g, int posX, int posY, int x, int y, int r) {
Graphics2D g2d = (Graphics2D) g;
hexagon hex = new hexagon(x, y, r);
hex.draw(g2d, x, y, 0, 0x008844, false);
hex.draw(g2d, x, y, 4, 000000, false);
}
public static void main(String[] args) {
f = new JFrame();
Prob1_4 p = new Prob1_4();
f.setContentPane(p);
f.
how can i build this as problem 2 of my code #include iostream.pdf
how can i build this as problem 2 of my code?
#include
#include
#include
#include
#include "./freeglut-3.4.0/include/GL/freeglut.h"
using namespace std;
bool leftDown = false, rightDown = false;
int lastPos[2];
float cameraPos[4] = { 0,1,4,1 };
int windowWidth = 640, windowHeight = 480;
double xRot = 0;
double yRot = 0;
int curProblem = 1; // TODO: change this number to try different examples
float specular[] = { 1.0, 1.0, 1.0, 1.0 };
float shininess[] = { 50.0 };
void problem1() {
// TODO
}
void problem2() {
//FIX HERE
}
void problem3() {
//TO DO
}
void problem4() {
//TO DO
}
void display()
{
glClearColor(0, 0, 0, 0);
glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
glDisable(GL_LIGHTING);
glEnable(GL_DEPTH_TEST);
glBegin(GL_LINES);
glColor3f(1, 0, 0); glVertex3f(0, 0, 0); glVertex3f(1, 0, 0); // x axis
glColor3f(0, 1, 0); glVertex3f(0, 0, 0); glVertex3f(0, 1, 0); // y axis
glColor3f(0, 0, 1); glVertex3f(0, 0, 0); glVertex3f(0, 0, 1); // z axis
glEnd(/*GL_LINES*/);
glEnable(GL_LIGHTING);
glShadeModel(GL_SMOOTH);
glMaterialfv(GL_FRONT, GL_SPECULAR, specular);
glMaterialfv(GL_FRONT, GL_SHININESS, shininess);
glEnable(GL_LIGHT0);
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
glViewport(0, 0, windowWidth, windowHeight);
float ratio = (float)windowWidth / (float)windowHeight;
gluPerspective(50, ratio, 1, 1000);
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();
gluLookAt(cameraPos[0], cameraPos[1], cameraPos[2], 0, 0, 0, 0, 1, 0);
glLightfv(GL_LIGHT0, GL_POSITION, cameraPos);
glRotatef(xRot, 1, 0, 0);
glRotatef(yRot, 0, 1, 0);
if (curProblem == 1) problem1();
if (curProblem == 2) problem2();
if (curProblem == 3) problem3();
if (curProblem == 4) problem4();
glutSwapBuffers();
}
void mouse(int button, int state, int x, int y)
{
if (button == GLUT_LEFT_BUTTON) leftDown = (state == GLUT_DOWN);
else if (button == GLUT_RIGHT_BUTTON) rightDown = (state == GLUT_DOWN);
lastPos[0] = x;
lastPos[1] = y;
}
void mouseMoved(int x, int y)
{
if (leftDown) {
xRot += (y - lastPos[1]) * .1;
yRot += (x - lastPos[0]) * .1;
}
if (rightDown) {
for (int i = 0; i < 3; i++)
cameraPos[i] *= pow(1.1, (y - lastPos[1]) * .1);
}
lastPos[0] = x;
lastPos[1] = y;
glutPostRedisplay();
}
void keyboard(unsigned char key, int x, int y)
{
curProblem = key - '0';
if (key == 'q' || key == 'Q' || key == 27) {
exit(0);
}
glutPostRedisplay();
}
void reshape(int width, int height)
{
windowWidth = width;
windowHeight = height;
glutPostRedisplay();
}
int main(int argc, char** argv)
{
glutInit(&argc, argv);
glutInitDisplayMode(GLUT_DOUBLE | GLUT_RGB | GLUT_DEPTH);
glutInitWindowSize(windowWidth, windowHeight);
glutCreateWindow("HW2");
glutDisplayFunc(display);
glutMotionFunc(mouseMoved);
glutMouseFunc(mouse);
glutReshapeFunc(reshape);
glutKeyboardFunc(keyboard);
glutMainLoop();
return 0;
}.
Computer Graphics and Multimedia lab report
According to NU of Bangladesh for BSC Hons in CSE 6th semester, here you can find the solution about this subject according to the board question respectively.
animation.#include windows.h#include GLglut.h#inclu.pdf
animation.
#include
#include
#include
#include
#include
#include
#include
animation.
#include
#include
#include
#include
#include
#include
#include
bool rotate1=false, rotate2=false;
double t; double dt=0.1;
double angle1=0; double dangle1 =20;
double angle2=0; double dangle2 =20;
double R=5;//radius spiral
double T=20;//time interval
double shiftx,shifty,shiftz;
bool origin=true;
void output_Text(float x, float y, char *text)
{ int len, i;
glColor3f(1, 0, 0); //red text
glRasterPos3f(x, y, 0);
len = (int)strlen(text);
for (i = 0; i < len; i++) { glutBitmapCharacter(GLUT_BITMAP_8_BY_13, text[i]); }
}
void Time_Delay(int dur) {
clock_t start_time;
start_time = clock();
while ((clock() - start_time) < dur) {
}
}
float ax[3]={25,0,0}; float ay[3]={0,25,0}; float az[3]={0,0,25};//axes void plot_Arrow(float
v1[3], float v2[3]) //plot arrow
{
glBegin(GL_LINES);
glVertex3fv(v1);
glVertex3fv(v2);
glEnd();
}
void plot_Vector(float v[3]) //plot vector
{
float zero[3]={0,0,0};
plot_Arrow(zero,v);
}
void plot_Axis()
{
glColor3f(0,0,0);
plot_Vector(ax);
plot_Vector(ay);
plot_Vector(az);
}
void Draw_Trajectory() {
//your code
}
void myInit3d(void)
{
glClearColor(1,1,1,0);
glColor3f(1.0, 0.0, 0.0); glClear(GL_COLOR_BUFFER_BIT); // clear the screen
glMatrixMode(GL_PROJECTION);
glLoadIdentity(); // reset the Current Modelview Matrix
bool rotate1=false, rotate2=false;
double t; double dt=0.1;
double angle1=0; double dangle1 =20;
double angle2=0; double dangle2 =20;
double R=5;//radius spiral
double T=20;//time interval
double shiftx,shifty,shiftz;
bool origin=true;
void output_Text(float x, float y, char *text)
{ int len, i;
glColor3f(1, 0, 0); //red text
glRasterPos3f(x, y, 0);
len = (int)strlen(text);
for (i = 0; i < len; i++) { glutBitmapCharacter(GLUT_BITMAP_8_BY_13, text[i]); }
}
void Time_Delay(int dur) {
clock_t start_time;
start_time = clock();
while ((clock() - start_time) < dur) {
}
}
float ax[3]={25,0,0}; float ay[3]={0,25,0}; float az[3]={0,0,25};//axes void plot_Arrow(float
v1[3], float v2[3]) //plot arrow
{
glBegin(GL_LINES);
glVertex3fv(v1);
glVertex3fv(v2);
glEnd();
}
void plot_Vector(float v[3]) //plot vector
{
float zero[3]={0,0,0};
plot_Arrow(zero,v);
}
void plot_Axis()
{
glColor3f(0,0,0);
plot_Vector(ax);
plot_Vector(ay);
plot_Vector(az);
}
void Draw_Trajectory() {
//your code
}
void myInit3d(void)
{
glClearColor(1,1,1,0);
glColor3f(1.0, 0.0, 0.0); glClear(GL_COLOR_BUFFER_BIT); // clear the screen
glMatrixMode(GL_PROJECTION);
glLoadIdentity(); // reset the Current Modelview Matrix
//the coordinate system, the window and glLookAt in your assigment //can be different
glMatrixMode(GL_MODELVIEW); // position and aim the camera glLoadIdentity();
gluPerspective(45, 1.33, .01, 1000);
gluLookAt(30.0, 30.0, 30.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0); }
void myDisplay(void) {
//your code
}
void Do_Nothing(void)
{}
void myMouse(int button, int state, int x, int y) //press left/right button
{ switch (button) {
case GLUT_LEFT_BUTTON:
if (state == GLUT_DOWN)
if.
import java.util.Scanner;public class Main { public static in.pdf
import java.util.Scanner;
public class Main {
public static int gridArray1[][] = null;
public static int gridArray2[][] = null;
public static boolean playerTurn = true;
public static int count = 0;
public static void setGridDimensions(int x_max, int y_max) {
Main.gridArray1 = new int[x_max][y_max];
Main.gridArray2 = new int[x_max][y_max];
for (int i = 0; i < x_max; i++) {
for (int j = 0; j < y_max; j++) {
gridArray1[i][j] = -1;
gridArray2[i][j] = -1;
}
}
}
static void placeShip(int starting_x, int starting_y, int length, int direction) {
count++;
int ships[] = new int[10];
int battleshipArrayPositions1[] = new int[4];
int cruiserArrayPositions1[] = new int[3];
int submarineArrayPositions1[] = new int[3];
int destroyerArrayPositions1[] = new int[2];
int carrierArrayPositions2[] = new int[5];
int battleshipArrayPositions2[] = new int[4];
int cruiserArrayPositions2[] = new int[3];
int submarineArrayPositions2[] = new int[3];
int destroyerArrayPositions2[] = new int[2];
if (playerTurn) {
if (direction == 0) {
for (int i = starting_x; i < starting_x + length; i++) {
Main.gridArray1[i][starting_y] = 0;
}
} else {
for (int i = starting_y; i < starting_y + length; i++) {
Main.gridArray1[starting_x][i] = 0;
}
}
} else {
if (direction == 0) {
for (int i = starting_x; i < starting_x + length; i++) {
Main.gridArray2[i][starting_y] = 0;
}
} else {
for (int i = starting_y; i < starting_y + length; i++) {
Main.gridArray2[starting_x][i] = 0;
}
}
}
}
boolean isConflictingShipPlacement(int starting_x, int starting_y, int length, int direction) {
return false;
}
static int shoot(int x, int y) {
if (playerTurn) {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (Main.gridArray2[i][j] == 0) {
return 0;
}
}
}
} else {
}
return 0;
}
boolean hasBeenAttempted(int x, int y) {
return false;
}
static void displayGrid(boolean showShips) {
int count = 0;
System.out.print(\" 0 1 2 3 4 5 6 7 8 9\");
if (showShips) {
if (playerTurn) {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (j % 10 == 0) {
System.out.print(\"\ \" + count + \" \");
count++;
}
if (Main.gridArray1[i][j] == -1) {
System.out.print(\"- \");
} else if (Main.gridArray1[i][j] == 0) {
System.out.print(\"@ \");
} else if (Main.gridArray1[i][j] == 1) {
System.out.print(\"+ \");
} else {
System.out.print(\"X \");
}
}
}
} else {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (j % 10 == 0) {
System.out.print(\"\ \" + count + \" \");
count++;
}
if (Main.gridArray2[i][j] == -1) {
System.out.print(\"- \");
} else if (Main.gridArray2[i][j] == 0) {
System.out.print(\"@ \");
} else if (Main.gridArray2[i][j] == 1) {
System.out.print(\"+ \");
} else {
System.out.print(\"X \");
}
}
}
}
} else {
if (playerTurn) {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (j % 10 == 0) {
System.out.print(\"\ \" + count + \" \");
count++;
}
if (Main.gridArray1[i][j] == -1) {
System.out.print(\"- \");
} else if (Main.gridArray1[i][j] == 0) {
System.out.print(\"- \");
} el.
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Creating an Uber Clone - Part VIII - Transcript.pdf
Creating an Uber Clone - Part VIII - Transcript.pdf
import java.awt.;import java.awt.event.MouseAdaptor;import java.pdf
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More from fathimafancyjeweller
Write a Python program that extracts 1000 unique links from Twitter .pdf
Write a program that draws a simple fence with vertical, equally spaced slats backed by two
horizontal support boards. Behind the fence show a simple house in the background. Make sure
the house is visible between the slats in the fence.
Solution
package app;
import java.awt.*;
import java.applet.*;
public class House extends Applet
{
public void paint (Graphics c)
{
house_wall (c);
window (c);
house_frame (c);
fence (c);
sup_boards (c);
}
public void fence (Graphics c)
{
c.setColor(Color.black);
c.fillRect (35,220,3,350);
c.setColor(Color.black);
c.fillRect (70,220,3,350);
c.setColor (Color.black); // slats of fence
c.fillRect (105,220,3,350);
c.setColor(Color.black);
c.fillRect (140,220,3,350);
c.setColor(Color.black);
c.fillRect (175,220,3,350);
c.setColor(Color.black);
c.fillRect (210,220,3,350);
c.setColor(Color.black);
c.fillRect (245,220,3,350);
c.setColor(Color.black);
c.fillRect (280,220,3,350);
c.setColor(Color.black);
c.fillRect (315,220,3,350);
c.setColor(Color.black);
c.fillRect (350,220,3,350);
c.setColor(Color.black);
c.fillRect (385,220,3,350);
c.setColor(Color.black);
c.fillRect (420,220,3,350);
c.setColor(Color.black);
c.fillRect (455,220,3,350);
c.setColor(Color.black);
c.fillRect (500,220,3,350);
c.setColor(Color.black);
c.fillRect (535,220,3,350);
}
public void sup_boards (Graphics c)
{
c.setColor (Color.black);
c.fillRect (20,350,550,8);
c.fillRect (20,460,550,8);
}
public void house_wall (Graphics c)
{
c.setColor (Color.RED);
c.fillRect (100,250,400,250);
c.setColor(Color.DARK_GRAY);
c.fillRect (245,380,110,70);
c.setColor (new Color(186,134,11));
c.fillOval (295,412,10,10);
}
public void window (Graphics c)
{
c.setColor (Color.blue);
c.fillRect (121,261,78,78);
c.fillRect (121,361,78,78);
c.fillRect (401,261,78,78);
c.fillRect (401,361,78,78);
c.setColor (Color.yellow);
c.fillRect (125,265,70,70);
c.fillRect (125,365,70,70);
c.fillRect (405,265,70,70);
c.fillRect (405,365,70,70);
}
public void house_frame (Graphics c)
{
c.setColor (Color.white);
c.fillRect (298,380,2,70);
c.setColor (Color.red);
c.fillRect (157,265,3,70);
c.fillRect (157,365,3,70);
c.fillRect (437,265,3,70);
c.fillRect (438,365,3,70);
c.fillRect (125,298,70,3);
c.fillRect (125,398,70,3);
c.fillRect (405,298,70,3);
c.fillRect (405,398,70,3);
c.fillRect (245,375,110,3);
}
}.
Which of the following statements about financial intermediaries is .pdf
Which of the following is true of a natural monopoly? Oa. If regulated, the firm will be allowed
to charge a price higher than its average cost. b. If regulated, the firm will haye a lower level of
output than if it was unregulated. Oc. If regulated, the firm will have a higher level of output than
if it was unregulated. Oc. If regulated, the firm will eam economic profit in the long run. De. If
regulated, the firm will experience diseconomies of scale.
Solution
Answer :- Option b). If regulated, the firm will have a lower level of output than if it was
unregulated.
Conclusion :- Option b)..
Why must each network router interface belong to a different network.pdf
Why is 89 and 93 (-infinity, infinity) and not [7/4,infinity) and [-7,infinity)?
Solution
89. k(x) = (4x -7)1/3 The function k(x) takes all the values on the real number line as x takes
various values on the real number line. Both the domain and the range of the function k(x) are R,
the set of all real numbers, or in interval notation, (- , ). When x = 7/4, k(x) = 0 and when x <
7/4, k(x) is negative. A negative real number can have a third root. For example, the 3rd root of -
1 is -1 as (-1)3 = -1. Also, if n is a positive number, then (-n)1/3 = (- 1*n)1/3 = -1(n)1/3 which
is real number. Hence the domain of k(x) is (- , ) and not [7/4, ).
93. F(x) = (x+7)1/3. The function F(x) takes all the values on the real number line as x takes
various values on the real number line. Both the domain and the range of the function k(x) are R,
the set of all real numbers, or in interval notation, (- , ). When x = -7, F(x) = 0 and when x < -7,
F(x) is negative. A negative real number can have a third root. For example, the 3rd root of -1 is
-1 as (-1)3 = -1. Also, if n is a positive number, then (-n)1/3 = (- 1*n)1/3 = -1(n)1/3 which is
real number. Hence the domain of F(x) is(- , ) and not [-7, )..
What are the similarities and differences between the seppuku of the.pdf
What are the consequences of stakeholder ethics with regards to organizational goals and
productivity
Solution
stakeholders are the people who contributes their services to the firm, they includes; customers,
employees, owners/promoters, media, society, government, suppliers and competitors.
if an organization is considering all the requirements of stakeholders as a whole, it means that it
is doing good. here it should treat all the stakeholders equally, and should not works for one
community or one kind of people to the organization. if the organization works to protect the
rights and requirements of all stakeholders, it means that they are doing good.
the ultimate goal of any organization is wealth maximization, in that process it has to satisfy all
its stakeholders. if not, they never reach the goal. so, in my words, if an organization is fulfilling
all its stakeholders requirements, there are no chances to violate ethics..
Which of the following are considered withdrawals from an economy.pdf
When using the enzyme coupled reaction for the glucose assay, it is important that the
absorbance value determined @415nm has a direct correlation to the amount of glucose in the
solution.
Glucose + O2 + H2O ----->Glucose Acid + H2O2
H2O2 + ABTS(reduced) [colourless] ------> ABTS+(oxidized) [blue]
After reviewing the glucose oxidase/peroxidase-coupled reaction above discuss the following.....
a, How would your unknown glucose result be affected if by some chance the enzyme, catalase,
was a cintaminant in your unknown glucose example? Explain ur reasoning.
b, How would your unknown glucose result be affected if by some chance Vitamin C, was a
cintaminant in your unknown glucose example? Explain ur reasoning.
Thank you!
Solution
a) Catalase is an antioxidant that makes the hydrogen peroxide less toxic by converting it into
water and oxygen gas. Absence of free hydrogen peroxide to oxidise the ABTS would not give
the blue color.
b) Vitamin C also serves as antioxidant by reducing the hydrogen peroxide. Therefore, no blue
color of the oxidised ABTS would be produced.
Hence, in both cases, presence of glucose would be masked by antioxidant activities of the
catalase and vitamin C..
What measures have the EU (or member nations) taken to mitigate the .pdf
What is the weighted average cost of capital for a corporation that finances an expansion project
using 35% retained earnings and the rest as debt capital? Assume the interest rates are 11% for
equity financing and 23% for debt financing. The weighted average cost of capital is 5785 0 %.
Solution
WACC = cost of equity x equity percentage + cost of debt x debt percentage
WACC = 23% x(1 - 35%) + 11% x 35% = 18.8%.
What types of powers are given to the states by the Constitution.pdf
What measures have the EU (or member nations) taken to mitigate the current economic
problems? What is your economic outlook for the EU nations (you may address the EU as a
whole or choose a specific member nation)?
Solution
In last few years,The EU financial and economic policies have been moulded under the
supervision and ratification of the member states to combat the series of economic challenges
that the region has confronted consistently.In this context,we can divide these recent challenges
according to the events that occured in recent past and discuss some of the economic remmedies
that had implemented by the member states either conjointly or separately.Evidently,the
challenges that the EU had faced recently majorly include Brexit issue,Financial Crisis that
culminated in 2008 and sustained,immigration issues and refugee crisis,capital/financial market
crisis and foreign and security policies and the recent policies implemented by the region against
these and potential crisis in the future will be discussed in further details.
Brexit Issue
Considering the apprehension of an impending economic collapse due to UK\'s exit from the
EU,there has been much emphasis on resistive policy mechanism from the part of the
administration.It had been apprehended that as Britain has traditionally been a strong economic
force of EU with considerably high contribution of overall GDP in EU,source of high export
revenue for EU as a whole,major source employment to the EU region,attracting significant
foreign investment due to the existence of so many top shot MNCs in Britain and its rapid GDP
growth also contributed notably to the growth and progress of the entire EU
region.Thus,expectedly,the apprehension became stronger that Brexit would lead to massive
economic collapse in EU region through decline in overall GDP nad GDP growth rate,lower
employment rate,restrictions on free flow of capital in the region,undersirable international trade
impacts,reduction in foreign capital and financial investment in the region and so forth.
However,the impact of Brexit has been clearly evident in UK with high inflation rate,decrease in
consumer spending and slow production and output in the country,some of the intial resistive
fiscal and monetary policies adopted by EU could resist the some of the expected
damages.Favorable monetary and labor market policies have restricted the extent of
macroeconomic crisis within the EU nations due to Brexit.In this regard,certain expansionary
monetary and fiscal relief meausers undertaken have been able to restore the production level
and consumer spending or at least ensure improvement amidst the economic threats of
brexit.Furthermore,low oil prices and consistent weaker Euro have safeguarded many EU nations
against the negative trade impacts of Brexit where other member nations are able to continue
high levels of trade volumes with the rest of the world.However,further challenges do lie in
strenghtening certain bilateral or reg.
Using the balanced equations given below, determine the overall mola.pdf
Using DOM manipulation to remove the Image link on a Bing.com search engine.
I am using firebug on Firefox. I did a search and now I need to DOM manipulate to remove the
Image link on the page. In the firebug code the Image is a link:
Images
If i use the document.
how can I remove the href link in the page?
Solution
To remove the image link using dom explorer we have to use the following code:
Images
\"retun flase is necessary as it will not allow the href to go anywhere except that page..
True or false Income statement profit and loss statement shows com.pdf
Today’s world seems to be moving beyond sovereignty and toward supranational leadership to
cooperate on issues of global importance. What are some of these issues? How might they be
solved through supranational cooperation? Does such cooperation impede the sovereignty of
independent nations? Please sure to include specific examples in supporting your points.
Solution
Answer:
The essential issues that are being experienced all inclusive incorporate battle against
psychological warfare, control of expansion, cultivating advancement, guaranteeing strength in
nations that don\'t have stable government. The sovereign obligation emergency and the euro
emergency have incited heads of state and government in Europe to uplift and heighten
supranational collaboration. Through supranational cooperation on matters to do with
intelligence the security agents among nations are equipped to able to share and providing
combined forces for peace keeping mission.
Control inflation by ensuring stability in the country currency and good trading relationships. For
development the nation is able to support investors from outside for its growth. Such
participation does not block national sway on the grounds that the administrations can help each
other in reinforcing their authority for example instance of United States and Russia on issues to
do with the utilization of atomic weapons in this way they have made assertion for peace
keeping..
Thomas Rinks and Joseph Shields developed Psycho Chihuahua, a carica.pdf
This Question: 2 pts 21 of 55 (0 complete) ? | > | This Test: 150 pts pos Which of the following
statements is true of process costing? O A. It assigns manufacturing overhead costs to products
only in the last production process. O B. It tracks and assigns both period costs and product costs
to units produced. O C. It accumulates product costs by production departments. O D. It uses one
Work-in-Process Inventory account.
Solution
Process costing accumulates product costs by production departments. Option C is correct.
The government uses part of payroll taxes to pay for Social Security..pdf
The future value at 4.75% interest, compounded continously for 6 Years, of the countinous
income steam with the rate of flow f(t)=1,350e^0.02t, is $8,857. Compute interest earnerd.
The future value at 4.75% interest, compounded continously for 6 Years, of the countinous
income steam with the rate of flow f(t)=1,350e^0.02t, is $8,857. Compute interest earnerd.
Solution
Answer : Given, Time (t) = 6 years;
Rate of interest (r) = 4.75%
Future value = $8857
Continuous income = 1350e^0.02t
= 1350 (2.7183)^0.02×6 ; where e is the mathematical constant and the value of e is 2.7183
(approximately).
= 1350 (2.7183)^0.12 = 1350 × 1.1274 = $1521.99
Therefore, earned interest amount is ( 8875 - 1521.99 ) = $7335.01.
The financial statements are prepared from the unadjusted trial balan.pdf
The expression Quadruple Play refers to the commercial bundle of voice, video, and data plus
mobility. Answer the four parts below about Quadruple Play; each worth 10 points.
Part 1: What are some examples of trends for commercial video communications as they relate
to Quadruple Play?
Part 2: What are some examples of how commercial VoIP can benefit a company?
Part 3: What are some examples of how Quadruple Play will affect commercial users of data?
Part 4: How will commercial customers benefit from the relationship between Quadruple Play
and unified messaging service?
Solution.
Should the US conform to international standardsSolutionYes, .pdf
Shell-and-tube heat exchangers with hundreds of tubes housed in a shell are commonly used in
practice for heat transfer between two fluids. Such a heat exchanger used in an active solar hot-
water system transfers heat from a water-antifreeze solution flowing through the shell and the
solar collector to fresh water flowing through the tubes at an average temperature of 60 degree C
at a rate of 15 L/s. The heat exchanger contains 80 brass tubes 1 cm in inner diameter and 1.5 m
in length. The roughness of brass tubing is epsilon 1.5 times 10^-6 m. Determine the pressure
drop across a single tube. Calculate the pumping power required by the tube-side fluid of the
heat exchanger. After operating for a long time. 1-mm-thick scale builds up on the inner
surfaces with an equivalent roughness of 0.4 mm. For the same pumping power input determine
the percent reduction in the flow rate of water through the tubes.
Solution
solution:
1)here for calculating pressure loss is given by
dP=f(L/D)*density*V^2/2
Q=15/80=.1875 l/s or 1.875*10^-4 m3/s
D=.01 m
L=1.5 m
e=1.5*106-6 m
velocity is given by
V=Q*4/pi*D^2=2.3873 m/s
Re=VD/kinematic viscosity=23873.241
here for terbulent flow friction factor is given by
f=1.325/{ln[e/3.7D+5.74/Re^.9]}^2
so we get
f=.02510
here pressure drop is given as
dP=f(L/D)*density*V^2/2
dP=10728.74 N/m2
where pumping power is
W=Q*dP/1000=2.011*10^-3 kw
3)when scale are deposited then flow velocity is given by
reduced diameter=d1=.01-.002=8*10^-3 m
velocity=1.875*10^-4*4/pi*d1=3.7301 m/s
reynolds number is
Re1=29840.8
friction factor is for e1=4*10^-4 m
f1=.07284
pressure drop=dP1=95016.55 N/m2
for same power flow rate is
W=Q1*dP1/1000
Q1=2.1164*10^-5 m3/s
hence percent reduction in flow rate is for single tube
%reduction=(Q-Q1/Q)*100=88.75%.
Research and describe a tool that can be used to test for web server.pdf
Requirement 1. Describe the difference between period costs and product costs. Period costs are
operating costs that are expensed in the accounting period in which they are incurred. Product
costs are all costs of a product that GAAS requires companies to treat as an asset for extemal
financial reporting. These costs are recorded as an asset (inventory) on the balance sheet until the
asset is sold. The cost is then transferred to an expense accountCost of Goods Manufactured on
the income statement. is subtracted from Sales Revenue to determine gross proft. The operating
costs are then On the income statement, Cost of Goods Sold subtracted to determine operating
income. Requirement 2. Classify Lawlor\'s costs as period costs or product costs. If the costs are
product costs, further classify as direct materials, direct labor, or Shaft and handle of weed
trimmer Motor of weed trimmer Factory labor for workers assembling weed trimmers Nylon
thread used by the weed trimmer (not traced to the product) Glue to hold housing together Plant
janitorial wages Depreciation on factory equipment Rent on plant Sales commissions
Administrative salaries Plant utilities Shipping costs to deliver finished weed trimmers to
customers Direct materials Direct materials Direct labor Manufacturing overhead Manufacturing
overhead Manufacturing overhead Manufacturing overhead Manufacturing overhead Period
costs Period costs Manufacturing overhead Period costs
Solution
Requirement 1
Period Costs are operating costs that are expensed in the accounting period in which they are
incurred.
Product Costs are all the cost of a product that GAAP requires companies to treat as an asset for
external financial reporting. These costs are recorded as an asset (inventory) on the balance sheet
until the asset is sold. The cost is then transferred to an expense account (Cost of Goods Sold) on
the income statement. Product costs include direct materials, direct labour, and manufacturing
overhead.
On the income statement, Cost of Goods Sold ( product Cost ) is subtracted from Sales Revenue
to determine gross profit. The period costs are then subtracted from gross profit to determine
operating income.
Requirement 2
Direct Materials
Direct Materials
Direct labour
Manufacturing Overhead
Manufacturing Overhead
Manufacturing Overhead
Manufacturing Overhead
Manufacturing Overhead
Period Cost
Period Cost
Manufacturing Overhead
Period Cost.
Question 11 of 24 In E. coli, three structural Deand enzymes A. D, an.pdf
Question 1. Dscribe the change you observed when you added 1 mL of 0.1 M KSCN to the 2 mL
portion of the diluted solution. Copy the equation from the procedure and explain your
observations in terms of LeChatelier’s Principle. Choose the best answer.
The additional thiocyantate ion shifted the equilibrium toward the products, producing
morehexathiocyanatoferrate(III) lightening the color.
Fe3+(aq) + 6SCN-(aq) Fe(SCN)63-(aq).
The additional thiocyantate ion shifted the equilibrium toward the reactants, producing
lesshexathiocyanatoferrate(III) and deepening the color.
Fe3+(aq) + 6SCN-(aq) Fe(SCN)63-(aq).
The additional thiocyantate ion shifted the equilibrium toward the reactants, producing
morehexathiocyanatoferrate(III) lightening the color.
Fe3+(aq) + 6SCN-(aq) Fe(SCN)63-(aq).
The additional thiocyantate ion shifted the equilibrium toward the products, producing
morehexathiocyanatoferrate(III) and deepening the color.
Fe3+(aq) + 6SCN-(aq) Fe(SCN)63-(aq).
Question 2. Show your calculations to determine the final volume of the 0.001 M FeCl3 solution.
Choose the best answer.
M1V1 = M2V2.
(0.001 M)V1=(0.1 M)(0.05 L)
V1=0.050 L = 50 mL
M1V1 = M2V2.
(0.001 M)V1=(0.1 M)(0.005 L)
V1=0.500 L = 500 mL
M1V1 = M2V2.
(0.1 M)(0.005 L)=(0.001 M)V2
V2=0.500 L = 500 mL
M1V1 = M2V2.
(0.1 M)(0.0005 L)=(0.001 M)V2
V2=0.050 L = 50 mL
Question 3. Describe the change you observed when you added 1 mL of 0.1 M FeCl3 to the
diluted mixture of 0.001 M FeCl3(aq) and 0.1 M KSCN(aq). Refer to the equation in step 1 and
explain your observations in terms of LeChatelier’s Principle. Choose the best answer.
Additional iron(III) ion shifted the equilibrium toward the products to form the red-colored
hexathiocyanatoferrate(III) complex and the intensity of color increased.color.
Fe3+(aq) + 6SCN-(aq) Fe(SCN)63-(aq).
Additional iron(III) ion shifted the equilibrium toward the products to form the red-colored
hexathiocyanatoferrate(III) chloride precipitate.
Fe3+(aq) + 6SCN-(aq) Fe(SCN)6Cl3(s).
Additional iron(III) ion caused the equilibrium to shift to the reactants to compensate for the
additional added iron(III), which increased the intensity of the red color.
Fe3+(aq) + 6SCN-(aq) Fe(SCN)63-(aq).
Additional iron(III) ion increased the concentration of the red-colored iron(III) ion.
Fe3+(aq) + 6SCN-(aq) Fe(SCN)63-(aq).
Question 4. Predict and then describe what happened when you added HCl(aq) to the copper
nitrate solution. Write down the equation for the reaction. Choose the best answer.
The added hydrogen ion shifted the equilibrium toward the products so that more CuCl42-(aq)
was createdand the solution changed from blue to green.
Cu(H2O)42+(aq) + 4Cl-(aq) CuCl42-(aq) + 4H2O(l).
The added chloride ion shifted the equilibrium toward the products so that more CuCl42-(aq)
was created and the solution changed from blue to green.
Cu(H2O)42+(aq) + 4Cl-(aq) CuCl42-(aq) + 4H2O(l).
The added chloride ion caused the precipitation of CuCl42-(s), which shifted the equilibrium
toward t.
Question 1. Dscribe the change you observed when you added 1 mL of 0.pdf
Question 1 5 pts Given the cash flow diagram below, evaluate the value of \"A\" if the \"F\"
value is $24.000 for an interest rate of 7%. ?? 4A
Solution
Future Value of investents: Year Investment Future Value of Investment of 1 Future
Value of Investment a b c=1.07^(5-a) d=b*c 1 A 1.3108
1.3108 A 2 2A 1.2250 2.4501 A 3 3A 1.1449
3.4347 A 4 4A 1.0700 4.2800 A
11.4756 A Now, As per question, 11.4756 A = $
24,000 or, A = $ 2,091.40 Thus, Value of A $ 2,091.40.
Prove that a group of order 255 is a cyclic.SolutionLet G be a .pdf
Project #8 requires you to find an escape path from a given starting point in a maze. The maze
(or swamp or mine field etc.) is a square 2D array of int. A value of zero in any cell represents a
wall (land mine or quicksand etc.) while a value of one represents a possible step to take. Each
input file will have a line at the top of the file that has the square dimension of the matrix
followed by the drop in point. The drop in point is the row/col pair indicating where you are
dropped into the swamp. You escape the swamp by taking one step at a time to any adjacent
square that is a 1. You are not allowed to step on a zero cell. You must somehow remember
which squares you have already stepped on in a path so that you don\'t get into an infinite loop.
The five input files you are given have cycles and dead ends and valid paths to the edge of the
swamp.
The starter file that you are given has two methods already written for you. The first given
method loads the grid values and generates the 2D int array to hold the 1s and 0s. The second
method echo\'s out the swamp to the screen before calling the method that you must fill in -
which will print all the escape paths.
Do not modify main any more than needed.
Do not modify the given methods that load or print the grid.
You are just to write and call the method(s) that print all escape paths.
the only output you are to write is one line per escape path and it must be in the exact format as
my sample output. If there are no escape paths then your code prints out no additional output.
Your escape path lines may be in a different order than mine but the row,col values must match
exactly in matching lines.
Your program must find and print ALL the escape paths. Lines may be in any order though.
Do not print any other output of any kind other than what my solution demo does.
DO NOT PRINT \"DEAD END\" OR *ANY* OUTPUT IF THERE ARE NO PATHS OUT.
JUST PRINT NOTHING!
input file:
output should look like:
Solution
input0.txt
10 2 3
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 1 0
0 0 0 1 0 1 0 1 0 0
0 0 0 1 0 1 1 0 0 0
0 0 1 0 1 0 0 1 0 0
0 1 0 0 1 1 1 1 0 0
0 0 1 0 1 0 1 0 1 0
0 0 0 1 0 1 0 1 0 0
0 0 1 0 1 0 1 0 1 0
0 0 0 0 0 0 0 0 0 0
Project2.java
import java.io.*;
import java.util.*;
public class Project2
{
public static void main(String[] args) throws Exception
{
int[] dropInPt = new int[2]; // row and col will be on the 2nd line of input file;
int[][] swamp = loadSwamp( args[0], dropInPt );
int row=dropInPt[0], col = dropInPt[1];
printSwamp( \"\ SWAMP: dropped in at: [\"+row+\",\"+col+\"]\ \",swamp );
System.out.println(\"\ ESCAPE PATHS:\ \");
// YOUR CODE HERE. DECLARE WHATEVER OBJECTS AND VARIABLES
NEEDED
// CALL YOUR METHOD(s) TO PRINT ALL ESCAPE PATHS
printPath(swamp, dropInPt);
} // END MAIN
// ###################################################
// DO NOT MODIFY THIS METHOD
// ----------------------------------------------------------------
private static void printSwamp(String label, int[][] swamp )
{
System.o.
Choose one field such as academia, health, consulting, business, man.pdf
Choose one field such as academia, health, consulting, business, management, and so on and
discuss the importance of followership in that field.
Solution
A follower can be a personality kind, as a foothold in a very hierarchy, as a role, or as a
collection of traits and behaviors. Studies of followership have created numerous theories
together with an attribute, activity attributes, role, and constructionist theories additionally to
exploring myths or misunderstandings concerning followership.
---Effective followership coaching within the school room is difficult due to media messages that
preference leadership, internal schemas command by students that ignore followership, and
cultural biases against it. undergrad and graduate students are proof against the thought of
followership and followership have been taken as leadership poorly enacted or as subsiding for a
lesser position. In recent years, attitudes have begun to change the students have noted that
following is an expected, healthy a part of a reciprocal relationship in social media which it
failed to carry negative connotations.
Although a student’s contribution within the schoolroom has such high significance, the school
admissions system has nevertheless to seek out the way to acknowledge and reward students who
have ceaselessly created these contributions. on condition that outstanding schoolroom
contributions are neglected, nevertheless play such an important role, it\'s the responsibility of
the school admissions system to seek out the way to spot them.
---In hospitals and tourism, being a good follower is vital for achieving the service-oriented goals
of the many operations.In hospital operations, it\'s typically necessary for followers to figure
severally of their leaders to holdout necessary tasks. it\'s been steered that incorporating
followership into coaching and education in intentional, purposeful ways in which may assist
operations in hospital and tourism.
---Followership, as outlined by Hurwitz (2008), is “accepting or enabling [italics original] the
goal accomplishment of one’s leader”. within the context of franchising, the franchise may well
be seen as an admirer as a result of he or she accepts the franchisor’s business plan and allows
the franchisor’s goal accomplishment through the individual franchise operations. Leaders will
begin by building structure price for followers and followership price could be a method of
incorporating the construct of followership into the organization’s culture, policies, and practices
as a result of leaders have followers it\'s their responsibility to line a vision, build trust, and
encourage the followers with passion and hope.
---The Followership Learning Community (FLC) could be a learning community inside the
International Leadership Organization (ILA) and is “dedicated to the event of information,
competencies, and programs regarding the leader-follower relationship. it\'s the primary such
educational or applies community dedi.
An intercalating agent 1. Inserts between base pairs of the two str.pdf
An intercalating agent: 1. Inserts between base pairs of the two strands of DNA 2. Makes double-
stranded lesions in the DNA 3. Inserts between adjacent base pairs on a single strand of DNA 4.
Commonly causes transversion mutations 5. Makes single-stranded lesions in the DNA
Solution
Chelating agent involves the formation or presence of two or more separate coordinate bonds
between a polydentate (multiple bonded) ligand and a single central atom. Usually these ligands
are organic compounds, and are called chelants, chelators, chelating agents, or sequestering
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a trait is considered discrete when 1-The traits are intermediat.pdf
a trait is considered: discrete when:
1-The traits are intermediate between the recessive and dominant cases
2- the traits are obviously either there or not there [what I guess the answer is]
3- the traits are variable in the population
4- the traits are a result of multiple genes.
when the allele of a trait is incompletely dominant:
1- the dominant trait masks the recessive trait
2- the recessive trait masks the dominant trait
3- the heterozygous individial is the same as the dominant.
4- the heterozygous individual is intermediate between the recessive and the dominant
5- the heterozygous individual expresses both traits.
Solution
A trait is considered discrete when the traits are either there or not there, eg Albinism, cystic
fibrosis, Huntingtons disease, -option 2.
Incomplete dominance is a form of intermediate inheritance where one allele for a specific trait
is not completely expressed over its paired allele. This results in a third phenotype in which the
expressed physical trait is a combination of the phenotypes of both the alleles. Option 4-the
heterozygous individual is intermediate between the recessive and the dominant.
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- 1. Introduction to Computer Graphics using OpenGL Can someone tell me what shall I do to connect between the LineClipping, drawing the line, and rectangle function? I wrote the code below for the above question: #include #define red makecol(255,0,0) #define green makecol(0,255,0) #include // use as needed for your system #include #include #include #include #include "vector.h" const int screenWidth = 800; const int screenHeight = 600; double xmax, xmin, ymax, ymin; typedef int OutCode; double xvmin, yvmin, xvmax, yvmax; int clicks = 0; Point positions[4]; bool flag = false; const int INSIDE = 0; // 0000 const int LEFT = 1; const int RIGHT = 2; const int BOTTOM = 4; // 0100 const int TOP = 8; OutCode ComputeOutCode(double x, double y) { OutCode code; code = INSIDE; if (x < xmin) code |= LEFT; else if (x > xmax) code |= RIGHT;
- 2. if (y < ymin) code |= BOTTOM; else if (y > ymax) code |= TOP; return code; } void LineClipping(double x0, double y0, double x1, double y1) { OutCode outcode0 = ComputeOutCode(x0, y0); OutCode outcode1 = ComputeOutCode(x1, y1); bool accept = false; bool done = false; while (done) { if (!(outcode0 | outcode1)) // Trivially accept and get out of loop { accept = true; done = true; break; } else if (outcode0 & outcode1) // Trivially reject and get out of loop { done = true; break; } else { // failed both tests, so calculate the line segment to clip // from an outside point to an intersection with clip edge double x, y; // At least one endpoint is outside the clip rectangle; pick it. OutCode outcodeOut = outcode0 ? outcode0 : outcode1; // Now find the intersection point; if (outcodeOut & TOP) { // point is above the clip rectangle x = x0 + (x1 - x0) * (ymax - y0) / (y1 - y0); y = ymax; } else if (outcodeOut & BOTTOM) { // point is below the clip rectangle
- 3. x = x0 + (x1 - x0) * (ymin - y0) / (y1 - y0); y = ymin; } else if (outcodeOut & RIGHT) { // point is to the right of clip rectangle y = y0 + (y1 - y0) * (xmax - x0) / (x1 - x0); x = xmax; } else if (outcodeOut & LEFT) { // point is to the left of clip rectangle y = y0 + (y1 - y0) * (xmin - x0) / (x1 - x0); x = xmin; } if (outcodeOut == outcode0) { x0 = x; y0 = y; outcode0 = ComputeOutCode(x0, y0); } else { x1 = x; y1 = y; outcode1 = ComputeOutCode(x1, y1); } } } if (accept) { // window to viewport mapping /* double sx = (xvmax - xvmin) / (xmax - xmin);// scale parameter in x direction double sy = (yvmax - yvmin) / (ymax - ymin);// scale parameter in y direction double vx0 = xvmin + (x0 - xmin)*sx; double vy0 = yvmin + (y0 - ymin)*sy; double vx1 = xvmin + (x1 - xmin)*sx; double vy1 = yvmin + (y1 - ymin)*sy;*/ //draw a red color viewport glColor3f(1.0, 0.0, 0.0);
- 4. glBegin(GL_LINE_LOOP); glVertex2f(xmin, ymin); glVertex2f(xmax, ymin); glVertex2f(xmax, ymax); glVertex2f(xmin, ymax); glEnd(); glColor3f(0.0, 0.0, 1.0); glBegin(GL_LINES); glVertex2d(x0, y0); glVertex2d(x1, y1); glEnd(); } } //<<<<<<<<<<<<<<<<<<<<<<< myInit >>>>>>>>>>>>>>>>>>>> void myInit(void) { glClearColor(1.0, 1.0, 1.0, 0.0); // set white background color glColor3f(0.4f, 0.7f, 0.2f); // set the drawing color glPointSize(8.0); // a ‘dot’ is 4 by 4 pixels } void setWindow(GLdouble left, GLdouble right, GLdouble bottom, GLdouble top) { glMatrixMode(GL_PROJECTION); glLoadIdentity(); gluOrtho2D(left, right, bottom, top); } // Set the screen viewport using thisfunction void setViewport(GLint left, GLint right, GLint bottom, GLint top) { glViewport(left, bottom, right - left, top - bottom); } //<<<<<<<<<<<<<<<<<<<<<<<< myDisplay >>>>>>>>>>>>>>>>> void myDisplay(void) { glClear(GL_COLOR_BUFFER_BIT); // clear the screen
- 5. setWindow(0, screenWidth, 0, screenHeight); setViewport(0, screenWidth, 0, screenHeight); xmax = double(positions[0].x); xmin = double(positions[1].x); ymax = double(positions[0].y); ymin = double(positions[1].y); if (clicks >1) { { glBegin(GL_LINE_LOOP); glVertex2i(positions[0].x, positions[0].y); glVertex2i(positions[1].x, positions[0].y); glVertex2i(positions[1].x, positions[1].y); glVertex2i(positions[0].x, positions[1].y); glEnd(); } } glColor3f(0.0, 1.0, 0.0); glBegin(GL_LINES); for (int i = 0; i < clicks; i++) { glVertex2i(positions[i].x, positions[i].y); } glEnd(); // draw a blue colored window for (int i = 0; i < clicks; i++) { glColor3f(0.0, 0.0, 1.0); glBegin(GL_POINTS); glVertex2d(positions[i].x, positions[i].y);
- 6. } glEnd(); glFlush(); // send all output to display glutSwapBuffers(); } void myKeyboard(unsigned char key, int x, int y) { if (key == 'i') flag = true; } void myMouse(int button, int state, int x, int y) { if (button == GLUT_RIGHT_BUTTON && state == GLUT_DOWN) { if (clicks == 4) clicks = 0; positions[clicks] = Point(x, screenHeight - y); clicks++; } if (button == GLUT_MIDDLE_BUTTON && state == GLUT_DOWN) clicks = 0; glutPostRedisplay(); } void main(int argc, char** argv) { glutInit(&argc, argv); // initialize the toolkit glutInitDisplayMode(GLUT_DOUBLE | GLUT_RGB | GLUT_MULTISAMPLE); // set display mode glutInitWindowSize(screenWidth, screenHeight); // set window size glutInitWindowPosition(100, 100); // set window position on screen glutCreateWindow("cohen sutherland clipping"); // open the screen window glutDisplayFunc(myDisplay); // register redraw function glutMouseFunc(myMouse); //glutMotionFunc(myMouseMove); //glutKeyboardFunc(myKeyboard); myInit();
- 7. glutMainLoop(); // go into a perpetual loop } Dynamic Line Clipping Input 2 points. Draw an axis aligned rectangle using the 2 points. Input 2 more points Draw the line segment between the two points Using the Cohen-Sutherland Clipping Algorithm find the clipped coordinates of the line segment with the rectangle. Draw the clipped line segment in a different color. Allow the user to move the rectangle, and calculate new clipping coordinate. Optional: Allow the user to also scale the rectangle using the mouse and calculate new clipping coordinates. Solution I will suggest you to go through this code,because the code you have written is very big and the thing with the programming is to write the small codes instead of big ones(which will create confusion). So here is the code:-