We are only going to keep track of degrees and minutes, so we began with a class that has fields for each and a constructor:
public class Angle {
private int degrees;
private int minutes;
public Angle(int degrees, int minutes) {
this.degrees = degrees;
this.minutes = minutes;
}
}
We started with some client code that constructed an array of Angle objects and
that printed it out:
import java.util.*;
public class Mon {
public static void main(String[] args) {
Angle a1 = new Angle(23, 26);
Angle a2 = new Angle(15, 48);
List<Angle> list = new ArrayList<Angle>();
list.add(a1);
list.add(a2);
System.out.println(list);
}
}
When we ran this program, it produced output like the following:
[Angle@42719c, Angle@30c221]
The text being generated in each case includes the name of the class (Angle)
followed by an at sign followed by a hexadecimal (base 16) address. This is
what the built-in toString method produces. To get better output, we decided
to add a toString method to the class.Ideally we would show something like 23 degrees and 26 minutes using the standard symbols used for degrees and minutes:
23° 26′But those characters aren't easy to work with in Java, so instead we'll settle for "d" for degrees and "m" for minutes:
23d 26mWe wrote the following toString method to produce that output:
public String toString() {
return degrees + "d " + minutes + "m";
}
When we ran the client code again, it produced the following output:
[23d 26m, 15d 48m]
Then I discussed how to include a method that would add two of these angles
together. In a sense, what I'd like to be able to say in the client code is
something like this:
Angle a1 = new Angle(23, 26);
Angle a2 = new Angle(15, 48);
Angle a3 = a1 + a2;
We can't write it that way because the "+" operator is used only for numbers
and String concatenation. Some programming languages like C++ allow you to do
something called "operator overloading," but Java doesn't allow that. So we
have to settle for having a method called "add" that we can use to ask one of
these angles to add another one to it:
Angle a1 = new Angle(23, 26);
Angle a2 = new Angle(15, 48);
Angle a3 = a1.add(a2);
I pointed out that this is a common convention in the Java class libraries.
For example, there is a class called BigInteger for storing very large integers
that has a method called add that is similar.We wrote this as the first version of our add method:
public Angle add(Angle other) {
int d = degrees + other.degrees;
int m = minutes + other.minutes;
return new Angle(d, m);
}
I pointed out that we're referring to other.degrees and other.minutes, which
are private fields. This works because the understanding of the word private
is that it is "private to the class." This is not at all the way that we as
humans understand the meaning of private (if something is private to me, then
it shouldn't be available to other humans). But in Java, one Angle object can
access private elements of another Angle object because they are both of the
same class.I modified the client code to add this angle to the list as well:
Angle a1 = new Angle(23, 26);
Angle a2 = new Angle(15, 48);
Angle a3 = a1.add(a2);
List<Angle> list = new ArrayList<Angle>();
list.add(a1);
list.add(a2);
list.add(a3);
System.out.println(list);
When we ran it, we got the following output:
[23d 26m, 15d 48m, 38d 74m]
Clearly the program has added the two angles to get a third, but the third
angle is listed as having 74 minutes, which isn't right. The first thing we
did was to discuss the constructor. We added the following precondition:
// pre: minutes <= 59 and minutes >= 0 and degrees >= 0
public Angle(int degrees, int minutes) {
this.degrees = degrees;
this.minutes = minutes;
}
Then we added code to throw an exception when the precondition was not
satisfied:
// pre: minutes <= 59 and minutes >= 0 and degrees >= 0
// (throws IllegalArgumentException if not true)
public Angle(int degrees, int minutes) {
if (minutes < 0 || minutes > 59 || degrees < 0)
throw new IllegalArgumentException();
this.degrees = degrees;
this.minutes = minutes;
}
But we still needed to handle the case where the minutes become larger than 59.
We were able to fix this with a simple if statement:
public Angle add(Angle other) {
int d = degrees + other.degrees;
int m = minutes + other.minutes;
if (m >= 60) {
m -= 60;
d++;
}
return new Angle(d, m);
}
It also would have been possible to use integer division and the mod operator
to figure this out.When we ran the code again, we got this output, indicating that it had correctly added the two angles together:
[23d 26m, 15d 48m, 39d 14m]
Then I said that I wanted to explore how to modify the class so that we can put
a collection of Angle objects into sorted order. I added the following client
code to add a specific list of Angle values to our list and then called
Collections.sort to put them into sorted order:
int[][]data = {{30, 19}, {30, 12}, {30, 45}, {30, 8}, {30, 55}};
for (int[] coords : data) {
list.add(new Angle(coords[0], coords[1]));
}
System.out.println(list);
Collections.sort(list);
System.out.println(list);
Unfortunately, this code did not compile. That's because we haven't told Java
how to compare Angle objects to put them into order. For example, we know that
an angle of 45 degrees and 15 minutes is more than an angle of 30 degrees and
55 minutes, but how is Java supposed to figure that out?If you want to use utilities like Arrays.sort and Collections.sort, you have to indicate to Java how to compare values to figure out their ordering. There is an interface in Java known as the Comparable interface that captures this concept.
Not every class implements Comparable because not all data can be put into sorted order in an intuitive way. For example, the Point class does not implement the Comparable interface because it's not clear what would make one two-dimensional point "less than" another two-dimensional point. But many of the standard classes we have seen like String and Integer implement the Comparable interface. Classes that implement the Comparable interface can be used for many built-in operations like sorting and searching. Some people pronounce it as "come-pair-a-bull" and some people pronounce it as "comp-ra-bull". Either pronunciation is okay. The interface contains a single method called compareTo.
public interface Comparable<T> {
public int compareTo(T other);
}
So what does compareTo return? The rule in Java is that:
public class Angle implements Comparable<Angle> {
...
}
We completed the compareTo method fairly quickly. I pointed out that in
general the degrees field tells you which Angle is larger, so we can almost get
by with writing the method this way:
public int compareTo(Angle other) {
return degrees - other.degrees;
}
Consider the case where an Angle object has a value for degrees that is larger
than the other object's value of degrees. Then the expression above returns a
positive integer, which indicates that the object is greater. If an Angle
object has a lower value for degrees than the other one, then this expression
returns a negative value, which indicates that it is less than the other Angle
object.The only case where this fails is when the degree values are equal. In that case, we should use the minutes fields in a similar manner to break the tie. So the right way to code this is as follows:
public int compareTo(Angle other) {
if (degrees == other.degrees)
return minutes - other.minutes;
else
return degrees - other.degrees;
}
When we ran the client code with this new version of the code, we got output
like the following:
[23d 26m, 15d 48m, 39d 14m]
[23d 26m, 15d 48m, 39d 14m, 30d 19m, 30d 12m, 30d 45m, 30d 8m, 30d 55m]
[15d 48m, 23d 26m, 30d 19m, 30d 12m, 30d 45m, 30d 8m, 30d 55m, 39d 14m]
I said that we'd have several more examples in section.Then I began a new topic: recursion. We have seen how to write code using loops, which is a technique called iteration. Recursion is an alternative to iteration that is equally powerful. It might seem odd to study something equally powerful if it doesn't allow us to solve a new set of problems. There are several reasons for this. First, recursion versus iteration is almost a religious question for computer scientists with passionate advocates on both sides. Think of iteration as being like Christianity and recursion as being like Buddhism. It's not that one is right and one is wrong. They both are attempting to accomplish the same thing, they just do it in a slightly different way.
It's important to study recursion because some problems are naturally easier to express recursively or naturally easier to express iteratively. Also, you might find out that you're really a Buddhist and that you like programming recursively.
I started with a non-programming example. I picked out a student and asked him what row he was sitting in. He said, "row 3." I pointed out that he used a classic iterative solution. He started a counter at 0, moved his eyes to the front of the room and kept doing count++ while there were more rows left to count. So I asked people to imagine that the room is not full of people who have excellent vision but full of robots who can't see very far. How could we program the robots to answer this question?
The idea is to take advantage of the fact that there are many robots in the room. So I asked people to think about what the robot I talked to might ask the robot in front of him. I pointed out that it wouldn't be very helpful to ask, "Do you know what row I'm in?" Then you end up with a room full of robots all saying, "Do you know what row he's in?" No progress is being made.
Instead, the robot should ask the robot in front of him, "What row are you in?" Notice that it's the same question, but he's asking a robot that is closer to the front of the room. The fact that the question "recurs" (occurs again) is what leads us to call this recursion. The second robot asks the robot in front of him what row he is in. At that point, we have a robot who is in the first row. We assume that robots aren't so blind that they can't see when they're in the first row. So that robot can do something different. He can say, "I'm in row 1." That allows the robot just behind him to answer, "I'm in row 2." And that gets back to the robot I talked to, who can now say, "I'm in row 3."
That basic process is very much like recursion. We solve a large task by breaking it up into a series of smaller tasks that are somehow all the same. Another analogy that I said people should consider is the idea of descending a staircase. Suppose that someone is standing at the top of a staircase:
O
/|\
/ \
+----+
| |
+----+ |
| |
+----+ |
| |
+----+ |
| |
---------------+-------------------+
If the person wants to get down from the staircase, they need just two things:
they need to be able to get down from one step to the step below and they need
a way to step onto the ground when they get to the bottom of the staircase.All recursion problems involve different cases. The case that involves going from the current step to a lower step is referred to as the recursive case because it turns one stair-descending task into a simpler stair-descending task. The case that involves stepping onto the ground is called the base case. It's the case that stops the recursion from executing. Without a recursive case, you'd never get any closer to getting down the stairs. Without the base case, you're likely to keep trying to descend the staircase when you've run out of steps.
Then I turned our attention to a programming example. I decided to show something familiar first, so we looked at a method written with iteration that would be fairly familiar. The method produces a line of output with n stars on it (for some n greater than or equal to 0):
public void writeStars(int n) {
for (int i = 0; i < n; i++)
System.out.print("*");
System.out.println();
}
This is a classic iterative solution. There is some initialization before the
loop begins (int i = 0), a test for the while loop (i < n), some code that is
execute each time through the loop (print and i++) and some concluding code
that is executed after the loop finishes (the println).To write this recursively, we want to think in terms of cases. Recursion always involves case analysis. There is always at least one base case and at least one recursive case. Some people find it easier to think about the recursive case first, but I find it easier to think about the base case first.
So we have to think about the range of possible tasks we might be asked to perform. Of those, which is the simplest? Which writeStars task is the easiest of all? One person suggested that writing one star is easy and it is, but there's something even easier. Writing 0 stars is even easier. So we can begin by handling that case:
if (n == 0)
System.out.println();
else {
...
}
It is common to see if/else statements in recursive definitions because we are
so often distinguishing cases. So what do we do in the else part? This is
where your programming instincts are going to lead you astray. You're going to
want to solve the whole problem (a for loop starting at 0...). With recursive
solutions, we want to figure out how to solve just a little bit of the problem.
How can we get a little closer? Someone suggested writing out one star:
if (n == 0)
System.out.println();
else {
System.out.print("*");
...
}
This is where the second problem comes in. You have to believe that the method
you are writing actually works. I call this the "leap of faith" that you have
to make to be able to write recursive code (I said it was almost like a
religion). So after we've written one star, what we have left to do is to
write a line of (n-1) stars. How can we do that? If only we had a method that
would write out a line of stars, then we could call it. But we do have such a
method! It's the one we're writing:
public void writeStars2(int n) {
if (n == 0)
System.out.println();
else {
System.out.print("*");
writeStars2(n - 1);
}
}
This is the entire definition. It doesn't seem right to us because we're used
to solving the entire problem ourselves. And it doesn't seem like we should be
able to call writeStars2 inside of writeStars2. But that's what recursion is
all about.To understand this, I told people that you have to realize that a method like writeStars2 is not just a single method. Think of it as an entire army of robots all programmed with the writeStars2 code. We can have a series of robots solve one of these tasks. This is like the "what row are you in?" question where we had a group of robots solve the problem by working together.
Consider what happens when I call writeStars2(2). We bring out the first robot of the writeStars2 army and we tell it to write a line of 2 stars. It sees that 2 isn't 0, so it writes one star and calls up a second robot to write a line of 1 star. That robot sees that 1 isn't 0, so it writes out one star and calls up a third robot to write a line of 0 stars. That robot is the one that ends up with the easy task, the base case. It sees that it has been asked to write a line of 0 stars and 0 equals 0, so it does the println and finishes executing. So the first two robots each printed one star and the third did a println to produce our line of 2 stars. It is useful to draw diagrams of the series of calls, using indentation to keep track of the fact that one version called another:
writeStars2(2);
writeStars2(1);
writeStars2(0);
I spent a few more minutes reminding people to look at how these two methods
differ. The iterative version has a familiar loop structure. The recursive
version has an if/else that distinguishes between a base case and a recursive
case. The recursive case includes a call on the method we're writing, but with
a different value than the original (a smaller "n" than in the original
call).The writeStars example is a good one for seeing how recursion works, but it doesn't help you to see the power of recursion. So for a second example I said to consider the task of reversing the lines of text in a file. Suppose that we have a Scanner variable tied to a file that contains these four lines of text:
this
is
fun
no?
We want to write a method called reverse that takes the Scanner as a
parameter and that reads these lines and writes them in reverse order:
no?
fun
is
this
This isn't hard to do with a loop as long as you have some kind of data
structure like an ArrayList<String> to store the lines. But with recursion, we
can solve it without defining a data structure.I again began with the base case. What is the easiest file to reverse? A one-line file is pretty easy to reverse, but even easier would be a 0-line or empty file. Assume that the Scanner variable is called "input". With a Scanner, we know that there are no more lines to read if a call on hasNextLine() returns false. In other words, we could write code like the following for the base case:
if (!input.hasNextLine())
// handle base case
But in this case, there isn't anything to do. An empty file has no lines to
reverse. So in this case, it makes more sense to turn around the if/else so
that we test for the recursive case. That way we can make it a simple if that
has an implied "else there is nothing to do":
if (input.hasNextLine()) {
...
}
Again the challenge is to do a little bit. How do we get a bit closer to being
done? Someone suggested reading one line of text:
if (input.hasNextLine()) {
String line = input.nextLine();
...
}
So in our example, this would read the line "this" into the variable line and
would leave us with these three lines of text in the Scanner:
is
fun
no?
We're trying to produce this overall output:
no?
fun
is
this
You might be asking yourself questions like, "Is there another line of input to
process?" That's not how to think recursively. If you're thinking
recursively, you'd think about what a call on the method would get you. Since
the Scanner is positioned in front of the three lines: is/fun/no?, a call on
reverse should read in those lines and produce the first three lines of ouptut
that we're looking for. If that works, then we'd just have to write out the
line "this" afterwards to complete the output.This is where the leap of faith comes in. We have to believe that the reverse method actually works. Because if it does, then this code can be completed as follows:
public void reverse(Scanner input) {
if (input.hasNextLine()) {
String line = input.nextLine();
reverse(input);
System.out.println(line);
}
}
Then I went through a demonstration of how this would execute using overheads
that I would stack one on top of the other. Each overhead had the definition
for the reverse method and a place for storing the value of the variable
"line". In other words, they each looked like this:
+-----------------------------------------------+
| public void reverse(Scanner input) { |
| if (input.hasNextLine()) { |
| String line = input.nextLine(); |
| reverse(input); |
| System.out.println(line); |
| } |
| } |
| +------------------+ |
| line | | |
| +------------------+ |
+-----------------------------------------------+
By stacking them one on top of the other, I was giving a visual representation
of what is referred to as the "call stack". As methods are called, Java uses a
Stack to keep track of the next method invocation. The first time we call the
method, it reads the first line of text into its variable line and then it
reaches the recursive call on reverse:
+-----------------------------------------------+
| public void reverse(Scanner input) { |
| if (input.hasNextLine()) { |
| String line = input.nextLine(); |
| --> reverse(input); |
| System.out.println(line); |
| } |
| } |
| +------------------+ |
| line | "this" | |
| +------------------+ |
+-----------------------------------------------+
Then what happens? What happens whenever a method is called? Java sets aside
this method and turns its attention to the new method. So I took a second
overhead from my stack and put it on top of this one.
+-----------------------------------------------+
| public void reverse(Scanner input) { |
| +-----------------------------------------------+
| | public void reverse(Scanner input) { |
| | if (input.hasNextLine()) { |
| | String line = input.nextLine(); |
| | reverse(input); |
| | System.out.println(line); |
| | } |
| | } |
| | +------------------+ |
+--| line | | |
| +------------------+ |
+-----------------------------------------------+
This new version of the method has its own variable called "line" in which it
can store a line of text. And even though the previous version (the one
underneath this one) is in the middle of its execution, this new one is at the
beginning of its execution. Think back to the analogy of an entire army of
robots all programmed with the same code, but they're executing slightly
different tasks. This one finds that there is a line of text to be read and it
reads it in (the second line that contains "is"). Then it finds itself
executing a recursive call on reverse:
+-----------------------------------------------+
| public void reverse(Scanner input) { |
| +-----------------------------------------------+
| | public void reverse(Scanner input) { |
| | if (input.hasNextLine()) { |
| | String line = input.nextLine(); |
| | --> reverse(input); |
| | System.out.println(line); |
| | } |
| | } |
| | +------------------+ |
+--| line | "is" | |
| +------------------+ |
+-----------------------------------------------+
So Java sets aside this version of the method as well and brings up a third
version. I brought out a third overhead:
+-----------------------------------------------+
| public void reverse(Scanner input) { |
| +-----------------------------------------------+
| | public void reverse(Scanner input) { |
| | +-----------------------------------------------+
| | | public void reverse(Scanner input) { |
| | | if (input.hasNextLine()) { |
| | | String line = input.nextLine(); |
| | | reverse(input); |
| | | System.out.println(line); |
| | | } |
+--| | } |
| | +------------------+ |
+--| line | | |
| +------------------+ |
+-----------------------------------------------+
This is like the third robot of the "reverse" army that we've called up to work
on this problem. It finds that there is a line to reverse (the third line,
"fun"), so it reads it in and reaches a recursive call on reverse:
+-----------------------------------------------+
| public void reverse(Scanner input) { |
| +-----------------------------------------------+
| | public void reverse(Scanner input) { |
| | +-----------------------------------------------+
| | | public void reverse(Scanner input) { |
| | | if (input.hasNextLine()) { |
| | | String line = input.nextLine(); |
| | | --> reverse(input); |
| | | System.out.println(line); |
| | | } |
+--| | } |
| | +------------------+ |
+--| line | "fun" | |
| +------------------+ |
+-----------------------------------------------+
This brings up a fourth version of the method:
+-----------------------------------------------+
| public void reverse(Scanner input) { |
| +-----------------------------------------------+
| | public void reverse(Scanner input) { |
| | +-----------------------------------------------+
| | | public void reverse(Scanner input) { |
| | | +-----------------------------------------------+
| | | | public void reverse(Scanner input) { |
| | | | if (input.hasNextLine()) { |
| | | | String line = input.nextLine(); |
| | | | reverse(input); |
+--| | | System.out.println(line); |
| | | } |
+--| | } |
| | +------------------+ |
+--| line | | |
| +------------------+ |
+-----------------------------------------------+
This one finds a fourth line of input ("no?"), so it reads that in and reaches
the recursive call:
+-----------------------------------------------+
| public void reverse(Scanner input) { |
| +-----------------------------------------------+
| | public void reverse(Scanner input) { |
| | +-----------------------------------------------+
| | | public void reverse(Scanner input) { |
| | | +-----------------------------------------------+
| | | | public void reverse(Scanner input) { |
| | | | if (input.hasNextLine()) { |
| | | | String line = input.nextLine(); |
| | | | --> reverse(input); |
+--| | | System.out.println(line); |
| | | } |
+--| | } |
| | +------------------+ |
+--| line | "no?" | |
| +------------------+ |
+-----------------------------------------------+
This brings up a fifth version of the method:
+-----------------------------------------------+
| public void reverse(Scanner input) { |
| +-----------------------------------------------+
| | public void reverse(Scanner input) { |
| | +-----------------------------------------------+
| | | public void reverse(Scanner input) { |
| | | +-----------------------------------------------+
| | | | public void reverse(Scanner input) { |
| | | | +-----------------------------------------------+
| | | | | public void reverse(Scanner input) { |
| | | | | if (input.hasNextLine()) { |
+--| | | | String line = input.nextLine(); |
| | | | reverse(input); |
+--| | | System.out.println(line); |
| | | } |
+--| | } |
| | +------------------+ |
+--| line | | |
| +------------------+ |
+-----------------------------------------------+
This one turns out to have the easy task, like our robot sitting in the front
row or the call on writeStars2 asking for a line of 0 stars. This time around
the Scanner is empty (input.hasNextLine() returns false). This is our very
important base case that stops this process from going on indefinitely. This
version of the method recognizes that there are no lines to reverse, so it
simply terminates.Then what? What happens in general when a method terminates? We're done with it, so we throw it away and go back to where we were just before this call. In other words, we pop the call stack and return to the line of code right after the method call:
+-----------------------------------------------+
| public void reverse(Scanner input) { |
| +-----------------------------------------------+
| | public void reverse(Scanner input) { |
| | +-----------------------------------------------+
| | | public void reverse(Scanner input) { |
| | | +-----------------------------------------------+
| | | | public void reverse(Scanner input) { |
| | | | if (input.hasNextLine()) { |
| | | | String line = input.nextLine(); |
| | | | reverse(input); |
+--| | | --> System.out.println(line); |
| | | } |
+--| | } |
| | +------------------+ |
+--| line | "no?" | |
| +------------------+ |
+-----------------------------------------------+
We've finished the call on reverse, so now we're positioned at the println
right after it. So we print the text in our variable "line" (we print
"no?") and terminate. And where does that leave us? This method goes
away and we return to where we were just before (we pop the call stack and go
to the line of code right after the recursive call):
+-----------------------------------------------+
| public void reverse(Scanner input) { |
| +-----------------------------------------------+
| | public void reverse(Scanner input) { |
| | +-----------------------------------------------+
| | | public void reverse(Scanner input) { |
| | | if (input.hasNextLine()) { |
| | | String line = input.nextLine(); |
| | | reverse(input); |
| | | --> System.out.println(line); |
| | | } |
+--| | } |
| | +------------------+ |
+--| line | "fun" | |
| +------------------+ |
+-----------------------------------------------+
So we print our line of text, which is "fun" and this version goes away (we pop
the stack again):
+-----------------------------------------------+
| public void reverse(Scanner input) { |
| +-----------------------------------------------+
| | public void reverse(Scanner input) { |
| | if (input.hasNextLine()) { |
| | String line = input.nextLine(); |
| | reverse(input); |
| | --> System.out.println(line); |
| | } |
| | } |
| | +------------------+ |
+--| line | "is" | |
| +------------------+ |
+-----------------------------------------------+
So we execute this println for the text "is" and pop once more:
+-----------------------------------------------+
| public void reverse(Scanner input) { |
| if (input.hasNextLine()) { |
| String line = input.nextLine(); |
| reverse(input); |
| --> System.out.println(line); |
| } |
| } |
| +------------------+ |
| line | "this" | |
| +------------------+ |
+-----------------------------------------------+
Notice that we've written out three lines of text so far:
no?
fun
is
So our leap of faith was justified. The recursive call on reverse read in the
three lines of text that came after the first and printed them in reverse
order. We complete the task by printing our line of text, which leads to this
overall output:
no?
fun
is
this
Then this version of the method terminates and we're done.