Description
COSC 117 Homework 6
2 Exercises
1. The first program is to construct a Fahrenheit to Celsius conversion chart that will print out a chart
of Fahrenheit to Celsius conversions. The output should look like the following.
Fahrenheit Celsius
-50 -45.56
-45 -42.78
-40 -40.00
-35 -37.22
-30 -34.44
-25 -31.67
-20 -28.89
-15 -26.11
-10 -23.33
-5 -20.56
0 -17.78
5 -15.00
10 -12.22
15 -9.44
20 -6.67
25 -3.89
30 -1.11
35 1.67
40 4.44
45 7.22
50 10.00
55 12.78
60 15.56
65 18.33
70 21.11
75 23.89
80 26.67
85 29.44
90 32.22
95 35.00
100 37.78
105 40.56
110 43.33
115 46.11
120 48.89
125 51.67
130 54.44
135 57.22
140 60.00
145 62.78
1
COSC 117 Homework #6: Methods
150 65.56
The main program that you must use is,
public static void main(String[] args) {
System.out.println(“Fahrenheit Celsius”);
for (int i = -50; i <= 150; i += 5) {
System.out.printf(“%6d %10.2f\n”, i, Celsius(i));
}
}
You need to write the method Celsius that takes in a single parameter, a double data type, of the
Fahrenheit temperature and return the corresponding Celsius temperature, also a double data type.
The conversion formula for Fahrenheit to Celsius is
C =
5
9
(F − 32)
where C is the Celsius temperature and F is the Fahrenheit temperature.
2. Most computer games use physics engines to take care of many aspects of game play. The most work
that a physics engine usually does in a game is collision detection. Determining when the user (or
object) hits a wall, when the laser blast or bullet in a FPS hits a target, etc. Collision detection is a
fairly difficult problem to solve and is beyond the scope of this class, but another, more accessible, use
of a physics engine in a game is to apply gravity to an object.
You have probably seen this in high school physics or possibly as an example in a mathematics class
like Calculus. If not, we will discuss the equation fully. Say you take an object, like a tennis ball or
small rock, and through it up into the air. The object will have a particular height off the ground at
any time after you toss it until it eventually hits the ground. The object will increase in height, until
gravity overcomes its initial velocity, and then it will start to fall to the ground. To find the height of
the object at any time t in seconds after you toss the object is given by the equation,
d = −
1
2
gt2 + v0t + d0
where d is the current height of the object, g is the acceleration of gravity, v0 is the initial velocity and
d0 is the initial height. The acceleration of gravity on Earth in English units is 32.17405 feet/sec2
, so
velocity is in feet/sec, distance is in feet and time is in seconds. Also note that we usually consider the
positive direction as up. So a positive d0 would indicate that the initial position of the object is above
the ground and a positive v0 would indicate that the initial velocity is in the upward direction. Along
these lines, gravity would be pulling the object down and hence the negative sign.
A sample run of the program is below.
Input the Initial Height (in feet): 10
Input the Initial Velocity (in feet/sec.): 25
Time Height
0.0 10.000
0.1 12.339
0.2 14.357
0.3 16.052
0.4 17.426
0.5 18.478
0.6 19.209
0.7 19.617
0.8 19.704
0.9 19.470
1.0 18.913
1.1 18.035
1.2 16.835
2
1.3 15.313
1.4 13.469
1.5 11.304
1.6 8.817
1.7 6.008
1.8 2.878
1.9 Hit the ground.
The main program that you must use is,
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
System.out.print(“Input the Initial Height (in feet): “);
Double d0 = keyboard.nextDouble();
System.out.print(“Input the Initial Velocity (in feet/sec.): “);
Double v0 = keyboard.nextDouble();
double d = d0;
double t = 0;
System.out.println();
System.out.println(” Time Height”);
while (d > 0) {
d = height(d0, v0, t);
if (d > 0)
System.out.printf(“%5.1f %15.3f \n”, t, d);
else
System.out.printf(“%5.1f %20s \n”, t, “Hit the ground.”);
t += 0.1;
}
}
You need to write the method height that takes in three parameters all double values. The first is
the initial height, the second the initial velocity, and the third is the time. The return value is the
current height of the object, also a double data type.
3. Nice output for a blackjack program. The output of the program is to look like the following.
Player 1
Card 1: King of Spades
Card 2: 8 of Spades
Player 2
Card 1: King of Hearts
Card 2: Queen of Clubs
For this program the main must be the following.
public static void main(String[] args) {
int p1f1 = (int) (Math.random() * 13) + 1;
int p1s1 = (int) (Math.random() * 4) + 1;
int p1f2 = (int) (Math.random() * 13) + 1;
int p1s2 = (int) (Math.random() * 4) + 1;
int p2f1 = (int) (Math.random() * 13) + 1;
int p2s1 = (int) (Math.random() * 4) + 1;
int p2f2 = (int) (Math.random() * 13) + 1;
int p2s2 = (int) (Math.random() * 4) + 1;
System.out.println(“Player 1”);
System.out.println(“Card 1: ” + CardString(p1f1, p1s1));
System.out.println(“Card 2: ” + CardString(p1f2, p1s2));
System.out.println();
System.out.println(“Player 2”);
System.out.println(“Card 1: ” + CardString(p2f1, p2s1));
System.out.println(“Card 2: ” + CardString(p2f2, p2s2));
}
You need to create the method CardString that takes in two integer parameters. The first parameter
is the face value of the card (1–13 with 1 being the Ace and 11, 12, and 13 being Jack, Queen, and
3
King respectively). The second parameter is the suit value of the card (1–4 which represent Hearts,
Diamonds, Clubs, and Spades respectively). The method is to return the string of the card’s value and
suit so that the output matches the examples.
4. High-low game: The output of the program is to look like the following. These are three separate runs.
Player 1: 5 of Spades
Player 2: Ace of Clubs
Player 2 Wins
Player 1: Queen of Hearts
Player 2: Queen of Spades
Player 2 Wins
Player 1: 8 of Clubs
Player 2: 8 of Clubs
It is a draw.
For this program the main must be the following.
public static void main(String[] args) {
int player1CardFace = getCardFace();
int player1CardSuit = getCardSuit();
int player2CardFace = getCardFace();
int player2CardSuit = getCardSuit();
System.out.println(“Player 1: ” + CardString(player1CardFace, player1CardSuit));
System.out.println(“Player 2: ” + CardString(player2CardFace, player2CardSuit));
int win = Winner(player1CardFace, player1CardSuit, player2CardFace, player2CardSuit);
if (win == 1)
System.out.println(“Player 1 Wins”);
else if (win == 2)
System.out.println(“Player 2 Wins”);
else
System.out.println(“It is a draw.”);
}
Copy over your CardString method from the previous exercise, then update it so that the face
values go from 2–14 with 11, 12, 13, and 14 being Jack, Queen, King, and Ace respectively. Create the
following three methods,
(a) getCardFace() that returns a random number between 2 and 14.
(b) getCardSuit() that returns a random number between 1 and 4.
(c) Winner(int p1f, int p1s, int p2f, int p2s) that takes in four parameters, the first
two are face value and suit of the card for player 1 and the last two are the face value and suit
of the card for player 2. The method should return 1 if player 1 wins, 2 if player 2 wins and 0 if
it is a draw. Remember that the rules are that the higher face value wins (with Ace high), if the
face values are the same then the suits are compared with Spades winning over Clubs which wins
over Diamonds which wins over Hearts.
5. Factorial: The output of the program is to look like the following. These are five separate runs.
n = 5
5! = 120
n = 13
13! = 6227020800
n = 20
20! = 2432902008176640000
n = 0
n! = 1
n = -3
Invalid input!
For this program the main must be the following.
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
System.out.print(“n = “);
int n = keyboard.nextInt();
if (n < 0)
System.out.print(“Invalid input!”);
else if (n == 0)
System.out.print(“n! = 1”);
else
System.out.print(n + “! = ” + factorial(n));
}
4
You need to create the method factorial that takes in a long integer parameter n and returns a
long with the value of n!.
6. Double Factorial: The output of the program is to look like the following. These are five separate runs.
n = 5
5!! = 15
n = 6
6!! = 48
n = 20
20!! = 3715891200
n = 0
n!! = 1
n = -3
Invalid input!
For this program the main must be the following.
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
System.out.print(“n = “);
int n = keyboard.nextInt();
if (n < 0)
System.out.print(“Invalid input!”);
else if (n == 0)
System.out.print(“0!! = 1”);
else
System.out.print(n + “!! = ” + doubleFactorial(n));
}
You need to create the method doubleFactorial that takes in a long integer parameter n and
returns a long with the value of n!!.
7. Character Counts: The output of the program is to look like the following.
Input a phrase: This is a test.
The A count is 1
The E count is 1
The H count is 1
The I count is 2
The S count is 3
The T count is 3
The most frequent letters are S and T with a count of 3.
For this program the main must be the following.
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
System.out.print(“Input a phrase: “);
String str = keyboard.nextLine();
str = str.toUpperCase();
String MaxCharacters = “”;
int maxCount = 0;
int numMax = 0;
int maxCharsFound = 0;
System.out.println();
for (char ch = ’A’; ch <= ’Z’; ch++) {
int count = countLetters(str, ch);
if (count > 0)
System.out.println(“The ” + ch + ” count is ” + count);
if (count > maxCount) {
maxCount = count;
}
}
System.out.println();
for (char ch = ’A’; ch <= ’Z’; ch++) {
int count = countLetters(str, ch);
5
COSC 117 Homework #6: Methods
if (count == maxCount)
numMax++;
}
for (char ch = ’A’; ch <= ’Z’; ch++) {
int count = countLetters(str, ch);
if (count == maxCount) {
MaxCharacters += ch;
maxCharsFound++;
if (numMax > 1) {
if (maxCharsFound < numMax – 1)
MaxCharacters += “, “;
else if (maxCharsFound == numMax – 1)
MaxCharacters += ” and “;
}
}
}
if (maxCount > 0) {
if (numMax == 1)
System.out.println(“The most frequent letter is ” + MaxCharacters + ” with a count of ” +
maxCount + “.”);
else
System.out.println(“The most frequent letters are ” + MaxCharacters + ” with a count of ” +
maxCount + “.”);
}
}
You need to create the method countLetters that takes in a string and a character and returns the
number of times that character is in the string.
6
