Problem 1: Give a brief definition of each of the following.

• image processing
• rendering
• projection plane
• image window
• perspective projection
• polygonal mesh
• shading
• ray tracing

Problem 2: Suppose we have the rank of a card in the rank variable and the suit in the suit variable. Assume that both variables are ints and that there are constants such as RANK_KING and SUIT_CLUBS declared. Write a code fragment that determines the value the card would have in the game of Hearts and saves that value in the variable pointValue (also an int). pointValue should be 1 if the card is a heart, 13 if the card is the queen of spades, -10 if the card is the jack of diamonds, and 0 otherwise.

  int pointValue;

  if (suit == SUIT_HEARTS)
  {
    pointValue = 1;
  }
  else if (suit == SUIT_SPADES && rank == RANK_QUEEN)
  {
    pointValue = 13;
  }
  else if (suit == SUIT_DIAMONDS && rank == RANK_JACK)
  {
    pointValue = -10;
  }
  else
  {
    pointValue = 0;
  }

Problem 3: The following code fragment is intended to return the area, in hectares, of a rectangular plot of land given the coordinates (in meters) of either pair of opposite corners of the plot. As written, it does not accomplish its task. (Note: there are 10000 square meters in a hectare.)

  Scanner scan = new Scanner(System.in);
  System.out.println("Enter coordinates of opposite corners:");
  int x1 = scan.nextInt();
  int y1 = scan.nextInt();
  int x2 = scan.nextInt();
  int y2 = scan.nextInt();

  final int SQ_METERS_PER_HECTARE = 10000;

  int width = x2 - x1;
  int height = y2 - y1;

  if (width < 0)
  {
    width = -width;
  }
  else if (height < 0)
  {
    height = -height;
  }

  System.out.println(width / 10000 * height; (double)width * height / SQ_METERS_PER_HECTARE);
}

• When the user enters 0 0 100 50, the code fragment should display 0.5 What does it display? 0.0
• When the user enters 10000 10000 0 0, it should display 10000. What does it display? -10000.0
• Make corrections to the code so it works properly. Also, add the definiton of an appropriate constant and modify the code so it uses that constant.

Problem 4: The following code fragment was purposefully formatted poorly and has errors. Assume that x and y were properly declared and initialized.

  int z;
  z = 2;

  if (x == 1 || y > 0)

  if (y < x)
  {
  if (x > 0)
    z = 1;
  }
  else
    z = 0;

  System.out.println(z);

• What is displayed if x and y are both initialized to 1? 0
• Find a value of x and y for which the code displays a value different than your answer to (a). x = 2, y = 1
• Find a value of x and y for which the method does not reach an assignment to z. x = 2, y = 0
• Add one statement (any suitable statement will do) that will resolve the compiler error.

Problem 5: Rewrite the following code fragment using a switch statement. Assume that lines has been properly declared and initialized.

int score;
if (lines == 1)
{
  score = 100;
}
else if (lines == 2)
{
  score = 300;
}
else if (lines == 3)
{
  score = 700;
}
else if (lines == 4)
{
  score = 1500;
}
else
{
  score = 1500 + 1000 * (lines - 4);
}

int score;

switch (lines)
{
  case 1:
    score = 100;
    break;

  case 2:
    score = 300;
    break;

  case 3:
    score = 700;
    break;

  case 4:
    score = 1500;
    break;

  default:
    score = 1500 + 1000 * (lines - 4); // formula would work for case 4 too
    break;
}

Problem 6: Consider the following code fragment, which purportedly computes the sum of the first n odd integers. Assume that n has been properly declared and initialized.

  // compute the sum of the first n odd integers

  int sum = 0;
  int newSum = 0;
  for (int i = 0; i <= n; i++)
  {
    newSum = (i + 2) + sum;
    newSum += i;
  }
  System.out.println(newSum);

• 1 4
• 2 6
• 100 202

Problem 7: Consider the following code fragment, which purportedly computes the number of digits in the decimal representation of n. Assume that n was properly declared and initialized.

  // compute the number of digits in n

  int digits = 0;
  do
  {
    digits++;
  } while (n % 10 > 0);

  System.out.println(digits);

• For what values of n does the code display 0? none
• For what values of n does the code display 1? multiples of 10
• For what values of n does the code hang by going into an infinite loop? any other value

Problem 8: For each of the following sequences of statements, write loops that are equivalent. Each answer should have only one line that invokes the drawLine method.

g.drawLine(0, 0, 100, 100); g.drawLine(20, 0, 80, 100); g.drawLine(40, 0, 60, 100); g.drawLine(60, 0, 40, 100); g.drawLine(80, 0, 20, 100); g.drawLine(100, 0, 0, 100);

for (int l = 0; l < 6; l++)
{
  g.drawLine(l * 20, 0, 100 - l * 20, 100);
}

g.drawLine(0, 0, 20, 0); g.drawLine(0, 10, 20, 10); g.drawLine(0, 20, 20, 20); g.drawLine(40, 0, 60, 0); g.drawLine(40, 10, 60, 10); g.drawLine(40, 20, 60, 20); g.drawLine(0, 40, 20, 40); g.drawLine(0, 50, 20, 50); g.drawLine(0, 60, 20, 60); g.drawLine(40, 40, 60, 40); g.drawLine(40, 50, 60, 50); g.drawLine(40, 60, 60, 60);

for (int col = 0; col < 2; col++)
{
  for (int row = 0; row < 2; row++)
  {
    for (int line = 0; line < 3; line++)
    {
      g.drawLine(col * 40, row * 40 + line * 10, col * 40 + 20, row * 40 + line * 10);
    }
  }
}

Problem 9: Write code fragments to do each of the following.

• Compute and display the sum of the first n cubes. For example, if n is 4 your code should display 100 since 1³ + 2³ + 3³ + 4³ = 100. Do this using a loop, not a formula (there is one!).

    int sum = 0;
  	
    for (int i = 1; i <= n; i++)
      sum += Math.pow(i, 3);
  			
    System.out.println(sum);
  

• Compute and display the sum of the first n odd cubes. For example, if n is 3 your code should display 153 since 1³ + 3³ + 5³ = 153.

          int sum = 0;
          
          // here i is 1, 3, 5, ...
          // to get n terms, the bound must be i <= 2n-1
          
          for (int i = 1; i <= 2 * n - 1; i += 2)
                  sum += Math.pow(i, 3);
                  
          // alternate solution: have i = 1, 2, 3, ..., n
          // when i is 1 we want to add in 1^3
          // when i is 2 we want to add in 3^3
          // when i is 3 we want to add in 5^3
          // in general the term to add in is (2i-1)^3
                  
          /*      
          for (int i = 0; i < n; i++)
                  sum += Math.pow(2 * i - 1, 3);
          */
                  
          System.out.println(sum);
  

• Compute and display the integer i between 1 and 100 that minimizes tan i where i is in radians.

          // when doing a sum we start our accumulator at 0 (the
          // identity for +) and in the loop add the accumulator
          // and the next term
          
          // when doing a product we start our accumulator at 1
          // (the identity for *) and in the loop multiply the
          // accumulator and the next term
          
          // for a max we will start our accumulator at infinity
          // (the indentity for mub) and in the loop compute the
          // min of the accumulator and the next term
          
          double min = Double.POSITIVE_INFINITY;
          int iForMin = 0;
          
          for (int i = 1; i <= 100; i++)
          {
                  if (Math.tan(i) < min)
                  {
                          min = Math.min(min, Math.tan(i));
                          iForMin = i;
                  }
          }
          
          System.out.println(iForMin);
  

• Compute and display the index in the string s of the first lowercase letter.

          // start at the first character
          
          int loc = 0;
          
          // keep going until we're out of characters or we
          // find a lower case letter (or, in other words,
          // keep going while we're not out of characters and
          // we're not on a lower case letter
          
          while (loc < s.length() && !(s.charAt(loc) >= 'a' && s.charAt(loc) <= 'z'))
          {
                  loc++;
          }
                  
          // figure out why we stopped
          
          int result;
  
          if (loc == s.length())
          {
                  result = -1; // not found
          }
          else
          {
                  result = loc; // found at loc
          }
          
          System.out.println(result);
  

• Compute and display the index in the string s of the first pair of consecutive uppercase letters. For example, if s is "I used to kill for the CIA." then your code should display 23 since CI is the first pair of consecutive uppercase letters.

          // start at the first character
          
          int loc = 0;
          
          // keep going until we're out of characters or we
          // find the first of two straight upper case letters
          // (or, in other words, keep going while we're not out
          // of characters and we're not at the first of two upper
          // case letters
          
          // use s.length() - 1 instead of s.length() since the last
          // character can't be the first of two consecutive uppers
          
          while (loc < s.length() - 1 && !(s.charAt(loc) >= 'A' && s.charAt(loc) <= 'Z' && s.charAt(loc + 1) >= 'A' && s.charAt(loc + 1) <= 'Z'))
          {
                  loc++;
          }
  
          // figure out why we stopped
          
          int result;
          if (loc == s.length() - 1)
          {
                  result = -1; // not found
          }
          else
          {
                  result = loc; // found at loc
          }
          
          System.out.println(result);
  

• Given a string s that contains integers separated by spaces, compute and display the sum of all the integers. For example, if s is 5 10 100 -5 your code should display 110. This is harder if you do not use a Scanner object. Try it both ways.