Based on your experience with the Ratio class, write a new class, ComplexNumber (and ComplexNumberDemo), where a complex number is a pair of doubles, "a" and "b", representing the number (a + bi), where i equals the square root of -1. Your ComplexNumber class should include a reasonable constructor (that takes two doubles, "a" and "b") and public methods to add, subtract, multiply, divide, and invert (which are explained succinctly in the Operations section of the Wikipedia page on Complex Numbers). Also include test code for each public method.
Indicate what each of the following will print or (if it is graphical) draw. Do not run these programs. Figure this out by hand. Remember to show your work.
Hint: No loop runs for more than 6 iterations (so if your answer is doing so, stop and check your work!).
| a) |
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| b) |
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| c) |
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State what the following program does in general, and in just a few words of plain English. (No credit will be awarded for stating what it does line-by-line, expression-by-expression -- instead, find the general pattern and state is in just a few word.)
Note: some of these methods may seem a bit confusing, and in fact some are, in that there may be some much easier, clearer, and more sensible ways to do the same thing.
-
public static boolean f(int x) { boolean b = false; while (x != 0) { if (x % 10 == 3) b = !b; x = x/10; } return b; } -
public static int f(int
x, int y) {
int a = x - y;
int b = 1 - (3*a)/(3*a-1);
return b*x + (1-b)*y;
}
-
public static int f(int x) { int a = x%10; int b = (x/10)%10; int c = x/100; int d = 100*a + 10*b + c; return d; } -
public static int f(int x, int y) { return (x % 2) * (y % 2); } - public static int
f(int n) {
int result = 0;
for (int i=1; i<=n; i++) {
int sum = 0;
for (int j=1; j<=n; j++)
sum += j;
result += 2*sum - n;
}
return result;
}
- public static
boolean f(int n) {
int s = (int)Math.round(Math.sqrt(n));
if (s*s != n) return false;
int i = s/n;
while (i*i < s)
i++;
return (i*i == s);
}
- public static int
f(String s) {
int result = 0;
while (s.length() > 0) {
String t = "";
for (int i=s.length()-1; i>=0; i--) {
char c = s.charAt(i);
if ((c >= '0') && (c < '9'))
t += ((char)(c+1));
else if ((c > '0') && (c <= '9'))
result++;
}
s = t;
}
return result;
}
-
Note: This method uses a "switch" statement. While
you are not responsible for writing code that uses "switch" statements,
you should be able to briefly read about switch statements and
understand enough to solve this problem under lab conditions.
public static String f(int n) {
char b = (char)('a'+3);
for (int i=n; i/2*2==i; i--) b++;
String c = "", d = "mccain";
switch (b/'e') {
case 3: c = "obama";
case 2: d = "huckabee";
case 0:
break;
case 1: char g = 'z';
for (int i='a'; i<b; i++) g--;
c += g;
default:
d = "clinton";
}
return ((n % 2 == 1) ? "o" : "") + b + c + b + d.substring(6);
}
-
public String
f(String s, int i) {
if ((s == null) || (i < 0) || (i >= s.length()))
return s;
String result = "";
for (int j=0; j<s.length(); j++)
if (i != j)
result += s.charAt(j);
return result;
}
-
public boolean f(String s1, String s2) {
if ((s1 == null) || (s2 == null))
return false;
for (int i=0; i<s1.length(); i++)
if (s2.equals(f(s1,i)))
return true;
return false;
}
Write and test this graphics method:
public void fillAstroid(Graphics page, Color color, int cx, int cy, int r)
This method takes a page, a color, a center point (cx,cy), and a radius, and draws a single filled astroid (the type of hypocycloid in the Steelers logo) in the given color that is just contained by the circle with the given center and radius. Do this as follows:
1. Draw "x" and "y" axes centered inside the circle.
2. Place n evenly-spaced points along each of the x and y axes. (You do not have to actually draw these points, but you will use them in the next step...)
3. In the first quadrant:
* Draw a line from (0, n) to (1, 0)
(That is, from the highest point on the y-axis to the first point to the right on the x-axis)
* Draw a line from (0, n-1) to (2, 0)
(That is, from the second highest point on the y-axis to the second point to the right on the x-axis)
* And so on (as you move down the y axis, you move right on the x axis). Of course, you must use a loop to draw all these lines!
4. Do the same for the other three quadrants.
Here is a picture of an astroid drawn in this manner, with about 10 points on each positive or negative coordinate axis:
