#exercise solutions

[maowebsolutions]

The Art and Science of Java, Chapter 4, Exercise 10, Solution

Can anyone help with the solution of the exercise 10. Here it is so far what I got. I'm not sure if it is right.

Modify the program in the preceding exercise so that instead of specifying the index of the final term, the program displays those terms in the Fibonacci sequence that are smaller than 10,000.

(I have included the preceding exercise below with code)

Exercise 10:

/* * FibonacciInverted.java * ---------------------- * This program displays the values in the Fibonacci sequence from F0 to F15. */

import acm.program.*;

public class FibonacciInverted extends ConsoleProgram{ public void run() { int n0 = 1; int n1 = 1; int n2; println("0" + "\n" + n0 + "\n" + n1); while(true) { // Loop nTimes terms n2 = n1 + n0; //The next term is the sum of the previous two terms if(n2 > N_TIMES)break; println(n2 + " "); n0 = n1; // The first previous number becomes the second previous number... n1 = n2; // ...and the current number becomes the previous number } } /*private constant*/ private final static int N_TIMES = 10000; }

Exercise 9:

Tn mathematics, there is a famous sequence of numbers called the Fibonacci sequence after the thirteenth-century Italian mathematician Leonardo Fibonacci. The first two terms in this sequence are 0 and I, and every subsequent term is the sum of the preceding two. Thus the first several terms in the Fibonacci sequence are as follows: F0=0

F1=1

F2 = I (0 + I) F3 = 2 (I + I) F4 = 3 (I + 2) F5 = 5 (2 + 3) F6 = 8 (3 + 5)

/* * FibonacciSequence.java * ---------------------- * This program displays the values in the Fibonacci sequence from F0 to F15. */

import acm.program.*;

public class FibonacciSequence extends ConsoleProgram{ public void run() { int n0 = 1; int n1 = 1; int n2; int nTimes = readInt("Enter the number of sequences: "); println("0" + "\n" + n0 + "\n" + n1); for (int i = 0; i < nTimes; i++) { // Loop nTimes terms n2 = n1 + n0; //The next term is the sum of the previous two terms println(n2 + " "); n0 = n1; // The first previous number becomes the second previous number... n1 = n2; // ...and the current number becomes the previous number } println(); } }

[maowebsolutions]

The Art and Science of Java Chapter 4 Exercise 9 Solution

Tn mathematics, there is a famous sequence of numbers called the Fibonacci sequence after the thirteenth-century Italian mathematician Leonardo Fibonacci. The first two terms in this sequence are 0 and I, and every subsequent term is the sum of the preceding two. Thus the first several terms in the Fibonacci sequence are as follows:

Fo = ° F, = F1 = I (0 + I) F3 = 2 (I + I) F4 = 3 (I + 2) F5 = 5 (2 + 3) F6 = 8 (3 + 5)

Write a program to display the values in this sequence from F0 through F15.

/* * FibonacciSequence.java * ---------------------- * This program displays the values in the Fibonacci sequnece from F0 to F15. */

import acm.program.*;

public class FibonacciSequence extends ConsoleProgram{ public void run() { int n0 = 1; int n1 = 1; int n2; println("0" + "\n" + n0 + "\n" + n1); for (int i = 0; i < 13; i++) { // Loop for the next 18 terms n2 = n1 + n0; //The next term is the sum of the previous two terms println(n2 + " "); n0 = n1; // The first previous number becomes the second previous number... n1 = n2; // ...and the current number becomes the previous number } println(); } }

[maowebsolutions]

The Art and Science of Java Chapter 4 Exercise 2 Solution

While we' re on the subject of silly songs, another old standby is "This Old Man." for which the fi rst verse is This old man, he played 1. He played knick-knack on my thumb. With a knick-knack, paddy-whack, Give your dog a bone. This old man came rolling home.

Each subsequent verse is the same, except for the number and the rhyming word at the end of the second line, which get replaced as follows:

1-thumb 2-shoe 3-knee 4-door 5-hive 6-sticks 7-heaven 8-pate 9 - spine lO -shin

/*

* OldManSong.java * --------------- * This program writes the lyrics of a song * and only replaces the number and the * rhyming word at the end of the second phrase. */

import acm.program.*;

public class OldManSong extends ConsoleProgram { public void run(){ String verse2 = "He played knick-knack on my "; for(int i = 1; i <= 10; i++){ println("This old man, he played " + i + "."); switch(i){ case 1: println(verse2 + "thumb.");break; case 2: println(verse2 + "shoe.");break; case 3: println(verse2 + "knee.");break; case 4: println(verse2 + "door.");break; case 5: println(verse2 + "hive.");break; case 6: println(verse2 + "sticks.");break; case 7: println(verse2 + "heaven.");break; case 8: println(verse2 + "pate.");break; case 9: println(verse2 + "spine.");break; case 10: println(verse2 + "shin.");break; } println("With a knick-knack, paddy-whack," + "\n" + "Give your dog a bone." + "\n" + "This old man came rolling home."); } } }

[maowebsolutions]

The Art and Science of Java - Programming Exercises - Chapter 4 - Question 1

Can someone advise if this is the best way to resolve this exercise?

I included the exercise and solution below.

1. As a way to pass the time on long bus trips, young people growing up in the United States have been known to sing the following rather repetitive song:

99 bottles of beer on the wall. 99 bottles of beer. You take one down, pass it around. 98 bottles of beer on the wall. 98 bottles of beer on the wall .... Anyway, you get the idea. Write a Java program to generate the lyrics to this song. In testing your program, it would make sense to use some constant other than 99 as the initial number of bottles.

Solution: 

/* * BottlesOfBeer.java * ------------------ * This program writes the lyrics for the * 99 bottles of beer on the wall song and * uses constants other then the initial 99 * number of bottles. */

import acm.program.*;

public class BottlesOfBeer extends ConsoleProgram{ public void run(){       String phrase1 = "bottles of beer";       String phrase2 = "on the wall.";       int bottles = INIT_BOTTLES;       println(bottles + " " + phrase1 + " " + phrase2);       println(bottles + " " + phrase1);       println("You take one down, pass it around.");       bottles -= 1;       println(bottles + " " + phrase1 + " " + phrase2);       println("\n" + bottles + " " + phrase1 + " " + phrase2 + ".."); }       /*Constants*/       private static final int INIT_BOTTLES = 99; }

[maowebsolutions]

The Art and Science of Java - Exercise Solutions - Chapter 3, Exercise 11

Can anyone help me with the Chapter - Exercise 11 Solution. This is what I have done so far. I'm wondering if there is a better way to resolve the exercise. 

If you don't have the book here it is the exercise:

11. Exercise 4 in Chapter 2 asks you to write a graphics program thal produces a diagram of a simple frame house. If you wanted to make the DrawHouse program easy to change, what named constants shou ld you define to provide the necessary flexibility? Rewrite the program so that it uses these named constants rather than explicit coordinate values.

Code from Exercise 4, Chapter 2:

import acm.graphics.*; import acm.program.*; import java.awt.*;

public class JHouse extends GraphicsProgram{ public void run(){ GRect rectWalls = new GRect(((getWidth() - 300)/2), ((getHeight()-120)/2), 300, 120); GRect rectDoor = new GRect(((getWidth() - 50)/2), ((getHeight()-80)/2+20), 50, 80); GRect windLeft = new GRect(((getWidth() - 40)/2-90), ((getHeight()-40)/2), 40, 40); GRect windRight = new GRect(((getWidth() - 40)/2+90), ((getHeight()-40)/2), 40, 40); GLine roofRight = new GLine(527,177,370,50); GLine roofLeft = new GLine(228,177,369,50); GLine windFrameR = new GLine(287,255,287,216); GLine windFrameL = new GLine(467,255,467,216); GOval doorHandle = new GOval(387,255,7,7); add(rectWalls); add(rectDoor); add(windLeft); add(windRight); add(roofRight); add(roofLeft); add(windFrameR); add(windFrameL); add(doorHandle); } }

Possible Solution for Exercise 11, Chapter 3

import acm.graphics.*;

import acm.program.*;

public class JHouseRevised extends GraphicsProgram{ public void run(){ add(new GRect(((getWidth()-HOUSE_SIZE)/2),((getHeight()-HOUSE_SIZE+ROOF_SIZE)/2),HOUSE_SIZE,HOUSE_SIZE/2)); add(new GRect(((getWidth()-(DOOR_SIZE/2))/2),((getHeight()-DOOR_SIZE)/2+(DOOR_SIZE/2)),DOOR_SIZE/2,DOOR_SIZE)); add(new GRect(((getWidth()-(DOOR_SIZE/2))/2-90),((getHeight()-(DOOR_SIZE/2))/2),(DOOR_SIZE/2),(DOOR_SIZE/2))); add(new GRect(((getWidth()-(DOOR_SIZE/2))/2+90),((getHeight()-(DOOR_SIZE/2))/2),(DOOR_SIZE/2),(DOOR_SIZE/2))); add(new GLine(((getWidth()+HOUSE_SIZE)/2),((getHeight()+(ROOF_SIZE-HOUSE_SIZE))/2),(getWidth()/2),(getHeight()-(HOUSE_SIZE*2)+ROOF_SIZE))); add(new GLine(((getWidth()-HOUSE_SIZE)/2),((getHeight()+(ROOF_SIZE-HOUSE_SIZE))/2),(getWidth()/2),(getHeight()-(HOUSE_SIZE*2)+ROOF_SIZE))); add(new GLine((getWidth()/2-(HOUSE_SIZE/4)-15),((getHeight()/2)+DOOR_SIZE/4),(getWidth()/2-(HOUSE_SIZE/4)-15),((getHeight()/2)-DOOR_SIZE/4))); add(new GLine((getWidth()/2+(HOUSE_SIZE/4)+15),((getHeight()/2)-DOOR_SIZE/4),(getWidth()/2+(HOUSE_SIZE/4)+15),((getHeight()/2)+DOOR_SIZE/4))); add(new GOval(((getWidth()-DOOR_HANDLE)/2+(DOOR_SIZE/6)),(getHeight()/2+(DOOR_SIZE/2)),DOOR_HANDLE,DOOR_HANDLE)); } /*Constant*/ private static final int HOUSE_SIZE = 300; private static final int ROOF_SIZE = 200; private static final int DOOR_SIZE = 100; private static final int DOOR_HANDLE = 7; }

[maowebsolutions]

The Art and Science of Java - Exercise Solutions - Chapter 3 Exercise 10

To position the tic tac toe board in the center you'll have to think there is a imaginary square which is the BOARD_SIZE.

With (getWidth( ) / 2) - (BOARD_SIZE / 2) 

* getWidth( ) / 2 - Gets the size of the window and divides by 2

* BOARD_SIZE / 2 - Board size divided by 2

You are getting the size of the window and dividing by 2 which centers the imaginary box. Now if we want to position the line inside the box we can divide the box by 2 and subtract by width of the window. 

You have to keep in mind:

new GLine( x1, y1, x2, y2 )

Creates a new GLine obiect connecting the points (x1, y1) & (x2,y2).

/* TicTacToeBoard.java

 * ---------------------

 * The TicTacToeBoard creates a board that is centered using

 * a private constant called BOARD_SIZE.

*/

import acm.program.*;

import acm.graphics.*;

public class TicTacToeBoard extends GraphicsProgram{

public void run(){

add(new GLine((getWidth()/2)-(BOARD_SIZE/2),(getHeight()/2)-(BOARD_SIZE/6),(getWidth()/2)+(BOARD_SIZE/2),(getHeight()/2)-(BOARD_SIZE/6)));

add(new GLine((getWidth()/2)-(BOARD_SIZE/2),(getHeight()/2)+(BOARD_SIZE/6),(getWidth()/2)+(BOARD_SIZE/2),(getHeight()/2)+(BOARD_SIZE/6)));

add(new GLine((getWidth()/2)-(BOARD_SIZE/6),(getHeight()/2)-(BOARD_SIZE/2),(getWidth()/2)-(BOARD_SIZE/6),(getHeight()/2)+(BOARD_SIZE/2)));

add(new GLine((getWidth()/2)+(BOARD_SIZE/6),(getHeight()/2)-(BOARD_SIZE/2),(getWidth()/2)+(BOARD_SIZE/6),(getHeight()/2)+(BOARD_SIZE/2)));

}

/*private constant*/

private static final int BOARD_SIZE = 100; 

}