Programming Exercises for Chapter 7
This page contains several exercises for Chapter 7 in Introduction to Programming Using Java. For each exercise, a link to a possible solution is provided. Each solution includes a discussion of how a programmer might approach the problem and interesting points raised by the problem or its solution, as well as complete source code of the solution.
Exercise 7.1:
An example in Subsection 7.2.4 tried to answer the question, How many random people do you have to select before you find a duplicate birthday? The source code for that program can be found in the file BirthdayProblemDemo.java. Here are some related questions:
- How many random people do you have to select before you find three people who share the same birthday? (That is, all three people were born on the same day in the same month, but not necessarily in the same year.)
- Suppose you choose 365 people at random. How many different birthdays will they have? (The number could theoretically be anywhere from 1 to 365).
- How many different people do you have to check before you've found at least one person with a birthday on each of the 365 days of the year?
Write three programs to answer these questions. Each of your programs should simulate choosing people at random and checking their birthdays. (In each case, ignore the possibility of leap years.)
Exercise 7.2:
Write a program that will read a sequence of positive real numbers entered by the user and will print the same numbers in sorted order from smallest to largest. The user will input a zero to mark the end of the input. Assume that at most 100 positive numbers will be entered.
Exercise 7.3:
A polygon is a geometric figure made up of a sequence of connected line segments. The points where the line segments meet are called the vertices of the polygon. The Graphics class includes commands for drawing and filling polygons. For these commands, the coordinates of the vertices of the polygon are stored in arrays. If g is a variable of type Graphics then
- g.drawPolygon(xCoords, yCoords, pointCt) will draw the outline of the polygon with vertices at the points (xCoords[0],yCoords[0]), (xCoords[1],yCoords[1]), ..., (xCoords[pointCt-1],yCoords[pointCt-1]). The third parameter, pointCt, is an int that specifies the number of vertices of the polygon. Its value should be 3 or greater. The first two parameters are arrays of type int[]. Note that the polygon automatically includes a line from the last point, (xCoords[pointCt-1],yCoords[pointCt-1]), back to the starting point (xCoords[0],yCoords[0]).
- g.fillPolygon(xCoords, yCoords, pointCt) fills the interior of the polygon with the current drawing color. The parameters have the same meaning as in the drawPolygon() method. Note that it is OK for the sides of the polygon to cross each other, but the interior of a polygon with self-intersections might not be exactly what you expect.
Write a panel class that lets the user draw polygons, and use your panel as the content pane in an applet (or standalone application). As the user clicks a sequence of points, count them and store their x- and y-coordinates in two arrays. These points will be the vertices of the polygon. Also, draw a line between each consecutive pair of points to give the user some visual feedback. When the user clicks near the starting point, draw the complete polygon. Draw it with a red interior and a black border. The user should then be able to start drawing a new polygon. When the user shift-clicks on the applet, clear it.
For this exercise, there is no need to store information about the contents of the applet. Do the drawing directly in the mousePressed() routine, and use the getGraphics() method to get a Graphics object that you can use to draw the line. (Remember, though, that this is considered to be bad style.) You will not need a paintComponent() method, since the default action of filling the panel with its background color is good enough.
You can try my solution here. Note that as the user is drawing the polygon, lines are drawn between the points that the user clicks. Click within two pixels of the starting point to see a filled polygon.
Exercise 7.4:
For this problem, you will need to use an array of objects. The objects belong to the class MovingBall, which I have already written. You can find the source code for this class in the file MovingBall.java. A MovingBall represents a circle that has an associated color, radius, direction, and speed. It is restricted to moving inside some rectangle in the (x,y) plane. It will "bounce back" when it hits one of the sides of this rectangle. A MovingBall does not actually move by itself. It's just a collection of data. You have to call instance methods to tell it to update its position and to draw itself. The constructor for the MovingBall class takes the form
new MovingBall(xmin, xmax, ymin, ymax)
where the parameters are integers that specify the limits on the x and y coordinates of the ball. (This sets the rectangle inside which the ball will stay.) In this exercise, you will want balls to bounce off the sides of the applet, so you will create them with the constructor call
new MovingBall(0, getWidth(), 0, getHeight())
The constructor creates a ball that initially is colored red, has a radius of 5 pixels, is located at the center of its range, has a random speed between 4 and 12, and is headed in a random direction. There is one problem here: You can't use this constructor until the width and height of the component are known. It would be OK to use it in the init() method of an applet, but not in the constructor of an applet or panel class. If you are using a panel class to display the ball, one slightly messy solution is to create the MovingBall objects in the panel's paintComponent() method the first time that method is called. You can be sure that the size of the panel has been determined before paintComponent() is called. This is what I did in my own solution to this exercise.
If ball is a variable of type MovingBall, then the following methods are available:
- ball.draw(g) -- draw the ball in a graphics context. The parameter, g, must be of type Graphics. (The drawing color in g will be changed to the color of the ball.)
- ball.travel() -- change the (x,y)-coordinates of the ball by an amount equal to its speed. The ball has a certain direction of motion, and the ball is moved in that direction. Ordinarily, you will call this once for each frame of an animation, so the speed is given in terms of "pixels per frame". Calling this routine does not move the ball on the screen. It just changes the values of some instance variables in the object. The next time the object's draw() method is called, the ball will be drawn in the new position.
- ball.headTowards(x,y) -- change the direction of motion of the ball so that it is headed towards the point (x,y). This does not affect the speed.
These are the methods that you will need for this exercise. There are also methods for setting various properties of the ball, such as ball.setColor(color) for changing the color and ball.setRadius(radius) for changing its size. See the source code for more information. A nice variation on the exercise would be to use random colors and sizes for the balls.
For this exercise, you should create an applet that shows an animation of balls bouncing around on a black background. Use a Timer to drive the animation. (See Subsection 6.5.1.) Use an array of type MovingBall[] to hold the data for the balls. In addition, your program should listen for mouse and mouse motion events. When the user presses the mouse or drags the mouse, call each of the ball's headTowards() methods to make the balls head towards the mouse's location. My solution uses 50 balls and a time delay of 50 milliseconds for the timer.
Here is my solution. Try clicking and dragging on the applet:
Exercise 7.5:
The sample program RandomArtPanel.java from Subsection 6.5.1 shows a different random "artwork" every four seconds. There are three types of "art", one made from lines, one from circles, and one from filled squares. However, the program does not save the data for the picture that is shown on the screen. As a result, the picture cannot be redrawn when necessary. In fact, every time paintComponent() is called, a new picture is drawn.
Write a new version of RandomArtPanel.java that saves the data needed to redraw its pictures. The paintComponent() method should simply use the data to draw the picture. New data should be recomputed only every four seconds, in response to an event from the timer that drives the program.
To make this interesting, write a separate class for each of the three different types of art. Also write an abstract class to serve as the common base class for the three classes. Since all three types of art use a random gray background, the background color can be defined in their superclass. The superclass also contains a draw() method that draws the picture; this is an abstract method because its implementation depends on the particular type of art that is being drawn. The abstract class can be defined as:
private abstract class ArtData {
Color backgroundColor; // The background color for the art.
ArtData() { // Constructor sets background color to be a random gray.
int x = (int)(256*Math.random());
backgroundColor = new Color( x, x, x );
}
abstract void draw(Graphics g); // Draws this artwork.
}
Each of the three subclasses of ArtData must define its own draw() method. It must also define instance variables to hold the data necessary to draw the picture. I suggest that you should create random data for the picture in the constructor of the class, so that constructing the object will automatically create the data for the random artwork. (One problem with this is that you can't create the data until you know the size of the panel, so you can't create an ArtData object in the constructor of the panel. One solution is to create an ArtData object at the beginning of the paintComponent() method, if the object has not already been created.) In all three subclasses, you will need to use several arrays to store the data.
The file RandomArtPanel.java only defines a panel class. A main program that uses this panel can be found in RandomArt.java, and an applet that uses it can be found in RandomArtApplet.java. You only need to modify RandomArtPanel.
Exercise 7.6:
Write a program that will read a text file selected by the user, and will make an alphabetical list of all the different words in that file. All words should be converted to lower case, and duplicates should be eliminated from the list. The list should be written to an output file selected by the user. As discussed in Subsection 2.4.5, you can use TextIO to read and write files. Use a variable of type ArrayList<String> to store the words. (See Subsection 7.3.4.) It is not easy to separate a file into words as you are reading it. You can use the following method:
/**
* Read the next word from TextIO, if there is one. First, skip past
* any non-letters in the input. If an end-of-file is encountered before
* a word is found, return null. Otherwise, read and return the word.
* A word is defined as a sequence of letters. Also, a word can include
* an apostrophe if the apostrophe is surrounded by letters on each side.
* @return the next word from TextIO, or null if an end-of-file is
* encountered
*/
private static String readNextWord() {
char ch = TextIO.peek(); // Look at next character in input.
while (ch != TextIO.EOF && ! Character.isLetter(ch)) {
// Skip past non-letters.
TextIO.getAnyChar(); // Read the character.
ch = TextIO.peek(); // Look at the next character.
}
if (ch == TextIO.EOF) // Encountered end-of-file
return null;
// At this point, we know the next character is a letter, so read a word.
String word = ""; // This will be the word that is read.
while (true) {
word += TextIO.getAnyChar(); // Append the letter onto word.
ch = TextIO.peek(); // Look at next character.
if ( ch == '\'' ) {
// The next character is an apostrophe. Read it, and
// if the following character is a letter, add both the
// apostrophe and the letter onto the word and continue
// reading the word. If the character after the apostrophe
// is not a letter, the word is done, so break out of the loop.
TextIO.getAnyChar(); // Read the apostrophe.
ch = TextIO.peek(); // Look at char that follows apostrophe.
if (Character.isLetter(ch)) {
word += "\'" + TextIO.getAnyChar();
ch = TextIO.peek(); // Look at next char.
}
else
break;
}
if ( ! Character.isLetter(ch) ) {
// If the next character is not a letter, the word is
// finished, so break out of the loop.
break;
}
// If we haven't broken out of the loop, next char is a letter.
}
return word; // Return the word that has been read.
}
Note that this method will return null when the file has been entirely read. You can use this as a signal to stop processing the input file.
Exercise 7.7:
The game of Go Moku (also known as Pente or Five Stones) is similar to Tic-Tac-Toe, except that it is played on a much larger board and the object is to get five squares in a row rather than three. Players take turns placing pieces on a board. A piece can be placed in any empty square. The first player to get five pieces in a row -- horizontally, vertically, or diagonally -- wins. If all squares are filled before either player wins, then the game is a draw. Write a program that lets two players play Go Moku against each other.
Your program will be simpler than the Checkers program from Subsection 7.5.3. Play alternates strictly between the two players, and there is no need to highlight the legal moves. You will only need two classes, a short panel class to set up the interface and a Board class to draw the board and do all the work of the game. Nevertheless, you will probably want to look at the source code for the checkers program, Checkers.java, for ideas about the general outline of the program.
The hardest part of the program is checking whether the move that a player makes is a winning move. To do this, you have to look in each of the four possible directions from the square where the user has placed a piece. You have to count how many pieces that player has in a row in that direction. If the number is five or more in any direction, then that player wins. As a hint, here is part of the code from my applet. This code counts the number of pieces that the user has in a row in a specified direction. The direction is specified by two integers, dirX and dirY. The values of these variables are 0, 1, or -1, and at least one of them is non-zero. For example, to look in the horizontal direction, dirX is 1 and dirY is 0.
int ct = 1; // Number of pieces in a row belonging to the player.
int r, c; // A row and column to be examined
r = row + dirX; // Look at square in specified direction.
c = col + dirY;
while ( r >= 0 && r < 13 && c >= 0 && c < 13
&& board[r][c] == player ) {
// Square is on the board, and it
// contains one of the players' pieces.
ct++;
r += dirX; // Go on to next square in this direction.
c += dirY;
}
r = row - dirX; // Now, look in the opposite direction.
c = col - dirY;
while ( r >= 0 && r < 13 && c >= 0 && c < 13
&& board[r][c] == player ) {
ct++;
r -= dirX; // Go on to next square in this direction.
c -= dirY;
}
Here is an applet version of my program It uses a 13-by-13 board. You can do the same or use a normal 8-by-8 checkerboard.