This is a makeup of the first midterm exam. You have two hours (120 minutes) to complete the test. The test is open-book.

• Please read the questions carefully and provide an answer for each question.

• Please justify ALL your answers.

Write your name or username here:______________________

1. (10 points) Use De Morgan's law to simplify the following Boolean expression.

!(x <= 10 || x <= 5)

A better form for this would have been, for example

!(x <= 10 || ! (x > 5))

In any event the law we're using is this:

!(a || b) == (!a && !b)

So that gives

!(x <= 10) && !(x <= 5))

which in turn gives

x > 10 && x > 5

which gives

x > 10

which is the answer we were looking for.

if (x > y) z = z + 1; else if (x <= y) z = z + 2;

This question has been discussed in lecture.

The second if statement is redundant.

That is, it's else (not shown) will never be run.

Also, changing z does not affect x and y.

Therefore the answer is NO.

if (! (x > y)) z = z + 2; else z = z + 1;

The answer this time is: YES.

If I remove the else the second assignment becomes unconditional.

That means the if statement becomes:

if (! (x > y)) { 
  z = z + 2; 
  z = z + 1; 
} else {
  z = z + 1; 
} 

So we see that the then branch changes.

if (x > 3) if (x <= 5) y = 1;
           else if (x != 6) y = 2; 
                else y = 3;
else y = 4;

The answer is

x >= 7

There are four paths through the program and and only one of them results in y getting a value of 2. To follow that path we need x > 3 and !(x <= 5) and x != 6 and we know that x is an integer. So x must be strictly bigger than 3 and x must be strictly bigger than 5 and x is not 6.

For the next two questions assume that x and y are integer variables.

5. (10 points) Consider the following code fragment

if (x > 3) { if (x <= 5) y = 1; 
             else if (x != 6) y = 2;
           } 
else y = 3;

This question is extremely easy.

Because of the curly braces we immediately reach the y = 3; statement.

But watch out for

• what values of x we reach the same statement
• without curly braces (in the next question).

As we shall see, x < 3 no longer takes us there.

if (x > 3) if (x <= 5) y = 1;
           else if (x != 6) y = 2; 
                else y = 3;

The answer is that x must be 6.

If x is less than or equal to 3 then we don't know what y ends up as.

All the other paths set y to a different value.

For y to be 3 we need

• an x strictly greater than 3,
• such that !(x <= 5),
• and such that !(x != 6) as well.

That means (x >3 ) and (x > 5) and x == 6.

Which makes the final answer x == 6.

Quick question: what's the value of (x > 5 && x < 3)?

class Rectangle{ double x, y, w, h; 
  Rectangle (double x, double y, double w, double h) { 
    this.x = x; this.y = y; this.w = w; this.h = h; 
  } 
} 

class Rectangle{ 
  double x, y, w, h; 
  Rectangle (double x, double y, double w, double h) { 
    this.x = x; this.y = y; this.w = w; this.h = h; 
  } 
  double area() { 
    return this.w * this.h; 
  }
} 

class Circle { double x, y, radius;
  boolean inside(Point target) {
    Point center = new Point(x, y); 
    if (target.distanceTo(center) >= radius) return false; 
    else return true; 
  }
} 

class Point {
  double x, y;
  Point(double a, double b) { this.x = a; this.y = b; }
  double distanceTo(Point other) {
    double dX = this.x - other.x,
           dY = this.y - other.y;
    return Math.sqrt( dX * dX + dY * dY);
  }
}

So we could have:

class Rectangle{ 
  double x, y, w, h; 
  Rectangle (double x, double y, double w, double h) { 
    this.x = x; 
    this.y = y; 
    this.w = w; 
    this.h = h; 
  } 
  double area() { 
    return this.w * this.h; 
  }
  boolean inside(Point a) {
    if (a.x <= this.x && a.x <= (this.x + this.w) &&
        a.y <= this.y && a.y <= (this.y + this.h)) {
      return true; 
    } else { 
      return false; 
    } 
  }
} 

1. a Point (for the center), and
2. a (double) radius

class Circle { 
  Point center;
  double radius;
  boolean inside(Point target) {
    if (target.distanceTo(center) >= radius) 
      return false; 
    else 
      return true; 
  }
  Circle(Point center, double radius) {
    this.center = center; 
    this.radius = radius; 
  }
  void move(Point newLocation) {
    this.center = newLocation; 
  }
  double area() {
    return Math.PI * radius * radius; 
  }
} 

class Experiment {
  public static void main(String[] args) {
    Circle a = new Circle(new Point(2, 3), 5); 
    System.out.println(a.area()); 
    Point b = new Point(10, 10); 
    System.out.println(a.inside(b)); 
    a.move(9, 9); 
    System.out.println(a.inside(b)); 
  }
}