First Summer 2004


Lecture Notes Thirteen: More practice with loops. Loops, tokenizers, and Monte Carlo problems.
Do you like loops? I don't know: so far, so good.

Now we need to move on. First a few simple, basic exercises.

Can you explain this? I sure can.
frilled.cs.indiana.edu%cat Two.java
class Two {
    public static void main(String[] args) {
	for (int i = 0; i < 10; i++) {
	    System.out.print(i); 
	}
	i = 10; 
	System.out.println(i); 
    }
}
frilled.cs.indiana.edu%javac Two.java
Two.java:6: Undefined variable: i
	i = 10; 
	^
Two.java:7: Undefined variable: i
	System.out.println(i); 
	                   ^
2 errors
frilled.cs.indiana.edu%

Can you fix it? I sure can.
frilled.cs.indiana.edu%cat Two.java
class Two {
    public static void main(String[] args) {
	int i; 
	for (i = 0; i < 10; i++) {
	    System.out.print(i); 
	}
	i = 10; 
	System.out.println(i); 
    }
}
frilled.cs.indiana.edu%javac Two.java
frilled.cs.indiana.edu%java Two
012345678910
frilled.cs.indiana.edu%

How do you call that? I'd say: scope of a variable.
frilled.cs.indiana.edu%webster scope
scope \'sko^-p\ n
[It scopo purpose, goal, fr. Gk skopos; akin to Gk skeptesthai to
     watch, look at -- more at SPY]
(1555)
1: space or opportunity for unhampered motion, activity, or thought
2: INTENTION, OBJECT
3: extent of treatment, activity, or influence
4: range of operation
syn see RANGE 

frilled.cs.indiana.edu%
Sounds good. It's the range of operation for that variable.

Take a look at this: It doesn't compile!
class Wow {
    public static void main(String[] args) {
	{
	    int i; 
	    i = 3; 
	    System.out.println(i); 
	}
	System.out.println(i); 
    }
}

Why? The curly braces!

Indeed, they are defining the scope. And they do it in the same way for:
  • classes
  • if statements
  • for loops
  • while loops

As well as other kinds of loops. There's just one more, as we will see below.

In any event, having blocks of statements available as standalone entities that can be placed anywhere inside the program can sometimes be a source of confusion for the beggining Java programmer. Let me see some examples.

Here's one: That is not a source of any confusion!
frilled.cs.indiana.edu%cat Wow.java
class Wow {
    public static void main(String[] args) {
	int x = 1; 
	if (x > 2) {
	    System.out.println("Yes, " + x + " is greater than 2."); 
	}
    }
}
frilled.cs.indiana.edu%javac Wow.java
frilled.cs.indiana.edu%java Wow
frilled.cs.indiana.edu%

I know, but consider this: Ah, I see the semicolon!
frilled.cs.indiana.edu%cat Wow.java
class Wow {
    public static void main(String[] args) {
	int x = 1; 
	if (x > 2) ; {
	    System.out.println("Yes, " + x + " is greater than 2."); 
	}
    }
}
frilled.cs.indiana.edu%javac Wow.java
frilled.cs.indiana.edu%java Wow
Yes, 1 is greater than 2.
frilled.cs.indiana.edu%
I hope you do. That is an if with an empty body.

That's how we got an infinite loop last time. There's one thing to be learned from this.

Syntax rules. And people who use it properly, rock!

Let's look at other kind of loops. OK, here's a program that talks to the user.
frilled.cs.indiana.edu%cat Three.java
class Three {
    public static void main(String[] args) {
	ConsoleReader c = new ConsoleReader(System.in); 
	String line; 
	do {
	    System.out.print("Type something: "); 
	    line = c.readLine(); 
	    System.out.println("You typed: " + line); 
	} while (! line.equals("bye"));  
	System.out.println("Good bye!"); 
    } 
}
frilled.cs.indiana.edu%javac Three.java
frilled.cs.indiana.edu%java Three
Type something: I am here
You typed: I am here
Type something: You are there
You typed: You are there
Type something: Your name is Echo
You typed: Your name is Echo
Type something: Bye
You typed: Bye
Type something: bye
You typed: bye
Good bye!
frilled.cs.indiana.edu%

Using while is just a bit longer. Yes, since the test comes first.
frilled.cs.indiana.edu%cat Three.java
class Three {
    public static void main(String[] args) {
	ConsoleReader c = new ConsoleReader(System.in); 
	String line; 
	System.out.print("Type something: ");
	line = c.readLine();
	System.out.println("You typed: " + line);
        while (! line.equals("bye")) { 
	    System.out.print("Type something: "); 
	    line = c.readLine(); 
	    System.out.println("You typed: " + line); 
	}	
	System.out.println("Good bye!"); 
    } 
}
frilled.cs.indiana.edu%javac Three.java
frilled.cs.indiana.edu%java Three
Type something: Works the same, doesn't it?
You typed: Works the same, doesn't it?
Type something: It does seem so. 
You typed: It does seem so. 
Type something: I am happy. 
You typed: I am happy. 
Type something: BYe
You typed: BYe
Type something: byE
You typed: byE
Type something: bye
You typed: bye
Good bye!
frilled.cs.indiana.edu%

In both cases we work with whole loops. Yes, we know we're done either at the beginning or at the end of the loop's body of statements.

Sometimes we'd like to allow for being able to realize we're done halfway through the loop. And end your processing in mid-loop? Doesn't this coding situation have a specific name?

Yes, it's called the loop and a half problem. Very good. Here's an example: A program that reads lines from a file, then reverses them.
frilled.cs.indiana.edu%cat Four.java
class Four {
    public static void main(String[] args) {
	ConsoleReader c = new ConsoleReader(System.in); 
	boolean done = false; 
	String line; 
	while (! done) {
	    System.out.print("Echo> "); 
	    line = c.readLine(); 
	    if (line == null) { // EOF or ctrl-D for that
		done = true; 
	    } else {
		System.out.println(line); 
	    } 
	} 
    } 
} 
frilled.cs.indiana.edu%javac Four.java
frilled.cs.indiana.edu%java Four
Echo> Hello!
Hello!
Echo> I am here.
I am here.
Echo> How are you?
How are you?
Echo> I am fine, how about you?
I am fine, how about you?
Echo> You don't say...
You don't say...
Echo> I am going to type control-D now
I am going to type control-D now
Echo> Bye!
Bye!
Echo> frilled.cs.indiana.edu%
Where's the file? You can pipe one into your program! (See your friendly TA for more details).

Alright. So we read lines, one by one. When the line we read is empty there's nothing to be reversed. So the program simply quits the loop. Oh, I see: and we skip the lower half of the loop.

Indeed, at that point we simply quit it. How do you do that?

One can do that with break (where's that in the book?), or... ... using a boolean variable, and an if statement.

We choose the second approach, because one can argue it's a bit more structured. But the first one is also often used, and it greatly simplifies code on occasion.

So now we have a basic echo program. Yes, but as of right now nothing is being reversed. Wasn't that what we set out to do?

OK, let's make this more exciting, as if through a looking glass. Yes, s'tel esrever sretcarahc ni sdrow!

Well, yletanutrofnu ew t'nac od taht. ?neht ,elbissop si siht spahrep tuB

That, we can do.
frilled.cs.indiana.edu%cat Four.java
class Four {
    public static void main(String[] args) {
	ConsoleReader c = new ConsoleReader(System.in); 
	boolean done = false; 
	String line; 
	while (! done) {
	    System.out.print("Echo> "); 
	    line = c.readLine(); 
	    if (line == null) { // EOF or ctrl-D for that
		done = true; 
	    } else {
		String rev = ""; 
		int i; 
		for (i = line.length() - 1; i >= 0; i--) {
		    rev += line.charAt(i);
		}
		System.out.println(rev); 
	    } 
	} 
    } 
} 
frilled.cs.indiana.edu%javac Four.java
frilled.cs.indiana.edu%java Four
Echo> Hello!
!olleH
Echo> I say, that's Spanish.
.hsinapS s'taht ,yas I
Echo> Arabic?
?cibarA
Echo> What's going on, can you please tell me. 
 .em llet esaelp uoy nac ,no gniog s'tahW
Echo> Hmmm... 
 ...mmmH
Echo> :-)
)-:
Echo> ...ees I
I see...
Echo> .ahctoG
Gotcha.
Echo> frilled.cs.indiana.edu%

I like this guy, Gotcha. Once upon a time there was a Polish carpenter, by the name of Zbigniew Gotcha, who lived in Kracow. You know the story?

No. I don't either, but I like the way it starts.

Time to wrap up. Let's do random numbers.

Yes, we kept for dessert. Do you like sherbet?

Let's program the following simulation:
Darts are thrown at random points onto the square with corners (1 ,1) and (-1, -1). If the dart lands inside the unit circle, that is, the circle with center (0, 0) and radius 1 it is a hit. Otherwise it is a miss. Run this simulation to determine an experimental value for the fraction of hits in the total number of attempts, multiplied by 4.
Oh, this is so easy! Easy as !

Fine, if it's easy, why don't you do it first? Relax, here it is:
frilled.cs.indiana.edu%cat Pi.java
import java.util.Random;

public class Pi {
    public static void main(String[] args) {
	Random r = new Random();
	double x, y, d;
	int i, count = 0;
	for (i=0; i < 100000 ; i++) {
	    x = r.nextDouble() * 2 - 1;
	    y = r.nextDouble() * 2 - 1;
	    d = Math.sqrt(x * x + y * y);
	    if (d < 1) count++;
	}
	System.out.println("Pi is approximately " + 4.0 * count / i);
    }
}

frilled.cs.indiana.edu%javac Pi.java
frilled.cs.indiana.edu%java Pi
Pi is approximately 3.13648
frilled.cs.indiana.edu%java Pi
Pi is approximately 3.15064
frilled.cs.indiana.edu%java Pi
Pi is approximately 3.14424
frilled.cs.indiana.edu%java Pi
Pi is approximately 3.1422
frilled.cs.indiana.edu%java Pi
Pi is approximately 3.1438
frilled.cs.indiana.edu%

Can you explain it? The probability of a hit is the fraction that the circle represents of the total area.

It is also close to the ratio of the measured frequencies: hits divided by attempts. So we write the formulas, and the radius simplifies, as it appears on both sides.

But there is a factor of half (1/2) which... ... is squared, so it participates as a fourth (1/4) in the end, and there you have it.

Pretty good. I want to do more problems.

I was hoping you would. I'm in great shape today.

Here's a program that helps with Lab Seven
Exactly what does it do?
import java.util.*;
import java.io.*; 

class Six {
    public static void main(String[] args) {
	ConsoleReader console = new ConsoleReader(System.in); 
	System.out.print("Hello> "); 
        String line = console.readLine(); 
        while (line != null) { // ^D would do it 
	    StringTokenizer tokenizer = new StringTokenizer(line); 
            while (tokenizer.hasMoreTokens()) {
		String token = tokenizer.nextToken(); 
		System.out.println(token.toUpperCase()); 
	    } 
            System.out.print("Hello> "); 
            line = console.readLine(); // what if we take this out? 
        }
	System.out.println("End of program.");
    } 
} 

Reads lines, one by one. When does it end?

When you type Control-D. Which is EOF (end of file).

Yes, in that case the line is null. What does it do, line by line?

Looks at the tokens, and prints them back, but converted into uppercase. Not much different from the one about Zbigniew, except this one uses this StringTokenizer.

Exactly, that's the main difference. How does that work?

Take a closer look:
StringTokenizer tokenizer = new StringTokenizer(line); 
while (tokenizer.hasMoreTokens()) {
  String token = tokenizer.nextToken(); 
  System.out.println(token.toUpperCase()); 
}
Better look this class up.

It's like a machine gun loaded with words. Or like a stapler, if you don't mind.

A stapler would also be a good analogy. With staples of variable length.

Glued together by blank spaces. Staples are tokens.
frilled.cs.indiana.edu%webster token
to-ken \'to^--ken\ n
[ME, fr. OE ta^-cen, ta^-cn sign, token; akin to OHG zeihhan sign, Gk
     deiknynai to show -- more at DICTION]
(bef. 12c)
1: an outward sign or expression 
2a: SYMBOL, EMBLEM <a white flag is a token of surrender>
2b: an instance of a linguistic expression 
3: a distinguishing feature: CHARACTERISTIC
[...]

frilled.cs.indiana.edu%

Here's a problem: I have another one:
Write a program that asks the user to enter today's exchange rate between U.S. dollars and the Euro. Then the program reads U.S. dollar values and converts each to Euro values. Write a program that asks for the rate, then reads and converts amounts in dollars. Make it such it can detect when you end the dollar inputs. Then the program reads a sequence of Euro amounts which it converts, one by one.

For both of these problems the trick is to interpret (read and promptly evaluate) input. I think I can handle that.

You're going to have to read
  • a rate,
  • then numbers,
  • finished by zero...
... then numbers again, ending with EOF.
Which is Control-D (in Unix).

Or some such thing. I could even end it with a keyword, such as "quit" or "bye", or some other meaningful word.

Really, how?
Take a look.
import java.util.*;
import java.io.*; 

class TwoAndThree {
    public static void main(String[] args) {

	ConsoleReader console = new ConsoleReader(System.in); 

	System.out.print("Rate>"); 
	String line = console.readLine(); 

	StringTokenizer tokenizer = new StringTokenizer(line); 
	double rate = Double.parseDouble(tokenizer.nextToken()); 

	double amount; 

	do {

	    System.out.print("Dollars>"); 
	    line = console.readLine(); 

	    tokenizer = new StringTokenizer(line); 

	    amount = Double.parseDouble(tokenizer.nextToken()); 

	} while (amount > 0); 

	do {

	    System.out.print("Euros>"); 
	    line = console.readLine(); 

	    if (line == null || line.equalsIgnoreCase("quit")) 
		break; 

	    tokenizer = new StringTokenizer(line); 

	    amount = Double.parseDouble(tokenizer.nextToken()); 

	} while (true); 
	System.out.println("Thank you for using this program."); 
    } 
} 
This should get you started.

Nice, but I see that you're assuming the user will never type more than one number on a line. So I could get by without a tokenizer. I agree, but I used one for the sake of practice.

I have a larger set of problems for next time. Can't wait. Let's press on with the ones for today. Next one up: the Drunkard problem.

There are many ways of solving the Drunkard problem. Yes, but empathy would not be one of them.

Of course. What I meant was that you need to generate random directions. And you have more than one way to produce random numbers.

Let's use whatever the book uses. OK, let's see the text of the problem.

Simulate the wandering of an intoxicated person in a square street grid. For 100 times, have the simulated drunkard randomly pick a direction (east, west, north, south) and move one block in the chosen direction. After the iterations, display the distance that the drunkard has covered. One might expect that on average the person might not get anywhere because the moves to different directions cancel another out in the long run, but in fact it can be shown that with probability 1 (certainty) the person eventually moves outside any finite region.

OK, I think I know what I'll do: I will use the Random class from the java.util package. Sounds like a good idea.

In addition to that, I would like to use an object oriented approach. And bring a drunkard into the picture.

Exactly. A drunkard is like a BankAccount.

I was going to say.
Well, here's how I'd get started.
import java.util.*;

class Drunkard {
    int x, y; 
    Drunkard(int x, int y) {
	this.x = x; 
	this.y = y;
    } 
    void moveNorth() {
	this.y -= 1; 
    }
    void moveEast() {
	this.x += 1; 
    }
    void report() {
	System.out.println("Hiccup: " + x + ", " + y); 
    } 
} 

class Four {
    public static void main(String[] args) {
	Random generator = new Random(); 
	Drunkard drunkard = new Drunkard(100, 100); 
	int direction; 
	for (int i = 0; i < 100; i++) {
	    direction = Math.abs(generator.nextInt()) % 4; 

	    if        (direction == 0) { // N
		drunkard.moveNorth();

	    } else if (direction == 1) { // E
		drunkard.moveEast(); 

	    } else if (direction == 2) { // S
		System.out.println("Should move South."); 

	    } else if (direction == 3) { // W
		System.out.println("Should move West."); 

	    } else {
		System.out.println("Impossible!"); 

	    } 
	    drunkard.report(); 
	} 
    }
} 
Not bad at all. Of course, one needs to finish it first.

Reminds me of homework assignment two. Somewhat. Now let's look at two more problems.

Here's the first problem:
Suppose a cannonball is propelled vertically into the air with a starting velocity v0.

Any calculus book will tell us that the position of the ball after t seconds is

s(t) = -0.5 * g * t2 + v0 * t
where g = 9.81 m/s2 is the gravitational force of the earth.

No calculus book ever mentions why someone would want to carry out such an obviously dangerous experiment, so we will do it in the safety of the computer. In fact, we will confirm the theorem from calculus by a simulation. In our simulation, we will consider how the ball moves in very short time intervals deltaT. In a short time interval the velocity v is nearly constant, and we can compute the distance the ball moves as deltaS = v * deltaT.

In our program, we will simply set

double deltaT = 0.01;
and update the position by
s = s + v * deltaT;
The velocity changes constantly---in fact it is reduced by the gravitational force of the earth.

In a short time interval,

v = -g * deltaT
and we must keep the velocity updated as
v = v - g * deltaT
In the next iteration the new velocity is used to update the distance. Now run the simulation until the cannonball falls back onto earth. Get the initial velocity as an input (100 m/s is a good value). Update the position and velocity 100 times per second, but print out the position only every full second. Also print out the the values from the exact formula (for comparison):
s(t) = -0.5 * g * t2 + v0 * t

What is the benefit of this kind of simulation when an exact formula is not available? Well, the formula from the calculus book is not exact. Actually the gravitational force diminishes the further the cannonball is away from the surface of the earth. This complicates the algebra sufficiently that it is not possible to give an exact formula for the actual motion, but the computer simulation can simply be extended to apply a variable gravitational force. For cannonballs, the calculus book formula is actually good enough, but computers are necessary to compute accurate trajectories for higher-flying objects such as ballistic missiles.

Problem Number Two:

Now to complete the picture we need to say that most cannonballs are not shot upright but at an angle.

If the starting velocity has magnitude v and the starting angle is alpha, then the velocity is actually a vector with components

vx = v * Math.cos(alpha)
and
vy = v * Math.sin(alpha)
In the x-direction the velocity does not change.

In the y-direction the gravitational force takes its toll.

Repeat the simulation from the previous exercise,

This kind of problem is of historical interest. The first computers were designed to carry out just such ballistic calculations, taking into account the diminishing gravity for high-flying projectiles and wind speeds.

Isn't a cannonball like a drunkard? Yes, they're both tiggers.

They all deal directly with gravitational fields.
Cannonballs require more physics, though.
class Cannonball {
    double x; 
    double y; 
    double vx; 
    double vy;
    final double g = 9.81; 

    Cannonball(double speed, double angle) {
	x = 0; 
	y = 0; 
	vx = Math.cos(angle) * speed; 
	vy = Math.sin(angle) * speed; 
    } 

    void move() {
	double deltaT = 0.01; 
	x += vx * deltaT;
	y += vy * deltaT; 
	vy -= g * deltaT;
    }

    void report() {
	System.out.println("Located at: (" + x + ", " + y + ")"); 
    }

    double height() {
	return y; 
    } 

}

class FiveAndSix {
    public static void main(String[] args) {

	Cannonball c = new Cannonball(10, Math.PI / 4);

	for (int i = 0; i < 10 * 100; i++) { // flying 10 seconds 

	    c.move(); 
	    c.report(); 

	    if (c.height() < 0) {
		System.out.println("SPLOOOF! The cannonball landed!"); 
		break; 
	    } 

	}

    }
} 
Homework assignment one! (Spring 2003). Almost, only in finer steps.

Could you also compute the highest point that the cannonball gets to? That would touch on the Report problem, wouldn't it?

It would. We'd have to update our notion of a current maximum every time we are looking at a height.

Come to think of it---the cannonball could do it. Intelligent cannonball.

Intelligent, but misguided. Well, here it is anyway:
class Cannonball {
    double x; 
    double y; 
    double vx; 
    double vy;
    final double g = 9.81; 

    double max; 

    Cannonball(double speed, double angle) {
	x = 0; 
	y = 0; 
	vx = Math.cos(angle) * speed; 
	vy = Math.sin(angle) * speed; 

	max = 0; 

    } 

    void move() {
	double deltaT = 0.01; 
	x += vx * deltaT;
	y += vy * deltaT; 
	vy -= g * deltaT;

	if (y > max) { max = y; } 
    }

    void report() {
	System.out.println("Located at: (" + x + ", " + y + ") "); 

        System.out.println("     max altitude so far: " + max); 

    }

    double height() {
	return y; 
    } 

}

class Max {
    public static void main(String[] args) {

	Cannonball c = new Cannonball(10, Math.PI / 4);

	for (int i = 0; i < 10 * 100; i++) { // flying 10 seconds 

	    c.move(); 
	    c.report(); 

	    if (c.height() < 0) {
		System.out.println("The cannonball landed!"); 
		break; 
	    } 

	}

    }
}

You just added three lines? Four, and yes, that was all.

But how often do those get executed? They're part of the cannonball's movement.

Let's move on to the Fibonacci problem.

The Fibonacci sequence is defined by the following rule(s):

Write a program that

This problem looks real easy.

Yes, from two values we compute a third. And we keep doing this over and over again. Now only this new value and the most recent of the two it was computed from should be kept.

And these two values are then added to compute a new value. And the whole process is repeated, as follows:
for (int i = 3; i <= n; i++) {    
  fNew = fOld + fOlder; 
  fOlder = fOld; 
  fOld = fNew; 
} 

Yes, that's it. This reminds me of another problem:

What problem is that? See if you can figure it for yourself.
import java.io.*; 

class Mistery {
    public static void main(String[] args) {

	ConsoleReader console = new ConsoleReader(System.in);
	System.out.print("Pass the salt please: "); 
	double a = console.readDouble(); 

	System.out.print("And the butter: "); 
	int n = console.readInt(); 

	double xold, xnew; 
	xold = ... ; 
	do {

	  xnew = xold - (Math.pow(xold, n) - a) / (n * Math.pow(xold, n - 1));
	  xold = xnew; 
	  System.out.println("     " + (Math.pow(xnew, n) - a)); 

	} while (Math.abs(Math.pow(xnew, n) - a) > 0.001); 

	System.out.println(xnew); 
	System.out.println(Math.pow(xnew, n) + " " + a); 
        System.out.println("Thank you!"); 

    } 
} 

Netwon-Raphson perhaps? But what's the ...? It doesn't matter, if it has a value.

For this problem, at least. Problems 609, 610, and 615 are easy, although for 610 and 615 we need to wait until we look at applets to do any graphics.

Let's do number 610. (Crazy numbers!) OK, but I won't draw circles, I'll just create them.

Drunkards, circles, cannonbals: they're all the same.
Especially drunkards. Here's the code:
import java.util.*; 

class Ten {
    public static void main(String[] args) { 

	System.out.print("Enter number of circles: "); 

        ConsoleReader console = new ConsoleReader(System.in); 
	int n = console.readInt(); 

	System.out.println("Generating " + n + " circles.");

	Random generator = new Random(); 

	for (int i = 0; i < n; i++) {

	    int x = 100 + Math.abs(generator.nextInt()) % 200; 
	    int y = 100 + Math.abs(generator.nextInt()) % 200; 

	    int r = 10 + Math.abs(generator.nextInt()) % 40;

	    Circle e = new Circle(x, y, r); 

	    System.out.println(e); // looks better when you draw it

	} 

	System.exit(0); 

    }
}

class Circle {
  double xCenter, yCenter, radius;
  Circle (double x, double y, double r) {
    xCenter = x; 
    yCenter = y;
    radius = r; 
  } 
  public String toString() {
    return "I am a circle at: ("     + 
           xCenter + ", " + yCenter  + 
           ") with a radius of "     +  
           radius;
  } 
} 

Why is 615 easy? Because it's like 610, and in addition you have complete information about the position of the squares. You'll see, later.

Why is 615 hard? Because you should come up with a formula for the position of a square given its row and column (or line and column).

This way you can use a for loop for the lines, ... and another one for the columns.

One, inside each other. Like for the patterns we developed last week.

And why is 609 easy? Because you just have to compute two sums, and take a square root at the end.

Why is 609 hard? 609 is not hard, but you have to read the problem carefully and use the right formula.

Which is the last one (second of two). The last one for today anyway.

We'll do a lot more next time. As always, I can hardly wait.


Here's a review of our approach of producing random numbers.

Say we need random numbers

Let y be such a random number.

Then we can calculate how far into the segment y is, as a percentage.

double p = (y - a) / (b - a); // [1]
Since y is random that means p can be anywhere between 0 and 1.

Well, such numbers can be generated with Math.random().

So, let's turn [1] on its head, to obtain y as a function of p.

Then, we have:

y = p * (b - a) + a; 
Now replace p by Math.random():
y = (b - a) * Math.random() + a;
Same as stretching followed by a translation. End of story.


Appendix: the (mentioned, numbered) problems.

Problem 614

Write a program that asks the user for an integer and then prints out all its factors.

For example, when the user enters 150, the program should print:

2 3 5 5
Problem 609
Write a program that reads a set of floating-point data values from the input.

When the end of file is reached, print out the count of the values, the average, and the standard deviation.

The average of a data set

is
The standard deviation is
However that formula is not suitable for our task.

By the time you have computed the mean, the individual xi's are long gone.

Until you know how to save these values, use the numerically less stable formula

You can compute this quantity by keeping track of the count, the sum, and the sum of squares as you process the input values.

Problem 610
Write a graphical applet (?!---see if you get around that)
Problem 615
Write a program that For example, when the user enters 20, the program should print:
2 3 5 7 11 13 17 19
Recall that a number is a prime number
Problem Netwon-Raphson

The best known iterative methods for computing the roots of a function f
(that is the x values for which f(x) is 0)
is Newton-Raphson approximation.

To find the zero of a function whose derivative is also known, compute

xnew = xold - f(xold) / f'(xold)
For this exercise, write a program to compute nth roots of floating point numbers.

Prompt the user for a and n, then obtain

by computing a zero of the function f(x) = xn - a.


Last updated: May 13, 2004 by Adrian German for A201