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Iteration and loops

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Iteration and loops

Multiple assignment

You can make more than one assignment to the same variable; the effect is to replace the old value with the new.

int liz = 5;
System.out.print(liz);
liz = 7;
System.out.println(liz);

The output of this program is 57, because the first time we print liz her value is 5, and the second time her value is 7.

This kind of multiple assignment is the reason I described variables as a container for values. When you assign a value to a variable, you change the contents of the container, as shown in the figure:

image

When there are multiple assignments to a variable, it is especially important to distinguish between an assignment statement and a statement of equality. Because Java uses the = symbol for assignment, it is tempting to interpret a statement like a = b as a statement of equality. It is not!

First of all, equality is commutative, and assignment is not. For example, in mathematics if \(a = 7\) then \(7 = a\). But in Java a = 7; is a legal assignment statement, and 7 = a; is not.

Furthermore, in mathematics, a statement of equality is true for all time. If \(a = b\) now, then \(a\) will always equal \(b\). In Java, an assignment statement can make two variables equal, but they don’t have to stay that way!

int a = 5;
int b = a;     // a and b are now equal
a = 3;         // a and b are no longer equal

The third line changes the value of a but it does not change the value of b, so they are no longer equal. In some programming languages a different symbol is used for assignment, such as <- or :=, to avoid this confusion.

Although multiple assignment is frequently useful, you should use it with caution. If the values of variables change often, it can make the code difficult to read and debug.

The while statement

Computers are often used to automate repetitive tasks. Repeating tasks without making errors is something that computers do well and people do poorly.

We have already seen methods like countdown and factorial that use recursion to perform repetition. This process is also called iteration. Java provides language features that make it easier to write these methods. In this chapter we are going to look at the while statement. Later on (in Section for loops) will check out the for statement.

Using a while statement, we can rewrite countdown:

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public static void countdown(int n) {
  while (n > 0) {
    System.out.println(n);
    n = n-1;
  }
  System.out.println("Blastoff!");
}

You can almost read a while statement like English. What this means is, “While n is greater than zero, print the value of n and then reduce the value of n by 1. When you get to zero, print the word ‘Blastoff!’”

More formally, the flow of execution for a while statement is as follows:

  1. Evaluate the condition in parentheses, yielding true or false.
  2. If the condition is false, exit the while statement and continue execution at the next statement.
  3. If the condition is true, execute the statements between the squiggly-brackets, and then go back to step 1.

This type of flow is called a loop because the third step loops back around to the top. The statements inside the loop are called the body of the loop. If the condition is false the first time through the loop, the statements inside the loop are never executed.

The body of the loop should change the value of one or more variables so that, eventually, the condition becomes false and the loop terminates. Otherwise the loop will repeat forever, which is called an infinite loop. An endless source of amusement for computer scientists is the observation that the directions on shampoo, “Lather, rinse, repeat,” are an infinite loop.

In the case of countdown, we can prove that the loop terminates if n is positive. In other cases it is not so easy to tell:

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public static void sequence(int n) {
  while (n != 1) {
    System.out.println(n);
    if (n%2 == 0) {           // n is even
      n = n / 2;
    } else {                  // n is odd
      n = n*3 + 1;
    }
  }
}

The condition for this loop is n != 1, so the loop will continue until n is 1, which will make the condition false.

At each iteration, the program prints the value of n and then checks whether it is even or odd. If it is even, the value of n is divided by two. If it is odd, the value is replaced by \(3n+1\). For example, if the starting value (the argument passed to sequence) is 3, the resulting sequence is 3, 10, 5, 16, 8, 4, 2, 1.

Since n sometimes increases and sometimes decreases, there is no obvious proof that n will ever reach 1, or that the program will terminate. For some particular values of n, we can prove termination. For example, if the starting value is a power of two, then the value of n will be even every time through the loop, until we get to 1. The previous example ends with such a sequence, starting with 16.

Particular values aside, the interesting question is whether we can prove that this program terminates for all values of n. So far, no one has been able to prove it or disprove it! For more information, see Wikipedia.

Tables

One of the things loops are good for is generating and printing tabular data. For example, before computers were readily available, people had to calculate logarithms, sines and cosines, and other common mathematical functions by hand.

To make that easier, there were books containing long tables where you could find the values of various functions. Creating these tables was slow and boring, and the results were full of errors.

When computers appeared on the scene, one of the initial reactions was, “This is great! We can use the computers to generate the tables, so there will be no errors.” That turned out to be true (mostly), but shortsighted. Soon thereafter computers were so pervasive that the tables became obsolete.

Well, almost. For some operations, computers use tables of values to get an approximate answer, and then perform computations to improve the approximation. In some cases, there have been errors in the underlying tables, most famously in the table the original Intel Pentium used to perform floating-point division [1].

Although a “log table” is not as useful as it once was, it still makes a good example of iteration. The following program prints a sequence of values in the left column and their logarithms in the right column:

double x = 1.0;
while (x < 10.0) {
  System.out.println(x + "   " + Math.log(x));
  x = x + 1.0;
}

The output of this program is

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1.0   0.0
2.0   0.6931471805599453
3.0   1.0986122886681098
4.0   1.3862943611198906
5.0   1.6094379124341003
6.0   1.791759469228055
7.0   1.9459101490553132
8.0   2.0794415416798357
9.0   2.1972245773362196

Looking at these values, can you tell what base the log method uses?

Since powers of two are important in computer science, we often want logarithms with respect to base 2. To compute them, we can use the formula:

\[\log_2 x = log_e x / log_e 2\]

Changing the print statement to

System.out.println(x + "   " + Math.log(x) / Math.log(2.0));

yields

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1.0   0.0
2.0   1.0
3.0   1.5849625007211563
4.0   2.0
5.0   2.321928094887362
6.0   2.584962500721156
7.0   2.807354922057604
8.0   3.0
9.0   3.1699250014423126

We can see that 1, 2, 4 and 8 are powers of two, because their logarithms base 2 are round numbers. If we wanted to find the logarithms of other powers of two, we could modify the program like this:

double x = 1.0;
while (x < 100.0) {
  System.out.println(x + "   " + Math.log(x) / Math.log(2.0));
  x = x * 2.0;
}

Now instead of adding something to x each time through the loop, which yields an arithmetic sequence, we multiply x by something, yielding a geometric sequence. The result is:

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1.0   0.0
2.0   1.0
4.0   2.0
8.0   3.0
16.0   4.0
32.0   5.0
64.0   6.0

Log tables may not be useful any more, but for computer scientists, knowing the powers of two is! When you have an idle moment, you should memorize the powers of two up to 65536 (that’s \(2^{16}\)).

Two-dimensional tables

A two-dimensional table consists of rows and columns that make it easy to find values at the intersections. A multiplication table is a good example. Let’s say you want to print a multiplication table for the values from 1 to 6.

A good way to start is to write a simple loop that prints the multiples of 2, all on one line.

int i = 1;
while (i <= 6) {
  System.out.print(2*i + "   ");
  i = i + 1;
}
System.out.println("");

The first line initializes a variable named i, which is going to act as a counter, or loop variable. As the loop executes, the value of i increases from 1 to 6; when i is 7, the loop terminates. Each time through the loop, we print the value 2*i and three spaces. Since we use System.out.print, the output appears on a single line.

In some environments the output from print gets stored without being displayed until println is invoked. If the program terminates, and you forget to invoke println, you may never see the stored output.

The output of this program is:

2   4   6   8   10   12

So far, so good. The next step is to encapsulate and generalize.

Encapsulation and generalization

Encapsulation means taking a piece of code and wrapping it up in a method, allowing you to take advantage of all the things methods are good for. We have seen two examples of encapsulation, when we wrote printParity in Section Alternative execution and isSingleDigit in Section Boolean methods.

Generalization means taking something specific, like printing multiples of 2, and making it more general, like printing the multiples of any integer.

Here’s a method that encapsulates the loop from the previous section and generalizes it to print multiples of n.

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public static void printMultiples(int n) {
  int i = 1;
  while (i <= 6) {
    System.out.print(n*i + "   ");
    i = i + 1;
  }
  System.out.println("");
}

To encapsulate, all I had to do was add the first line, which declares the name, parameter, and return type. To generalize, all I had to do was replace the value 2 with the parameter n.

If I invoke this method with the argument 2, I get the same output as before. With argument 3, the output is:

3   6   9   12   15   18

and with argument 4, the output is

4   8   12   16   20   24

By now you can probably guess how we are going to print a multiplication table: we’ll invoke printMultiples repeatedly with different arguments. In fact, we are going to use another loop to iterate through the rows.

int i = 1;
while (i <= 6) {
  printMultiples(i);
  i = i + 1;
}

First of all, notice how similar this loop is to the one inside printMultiples. All I did was replace the print statement with a method invocation.

The output of this program is

1   2   3   4   5   6
2   4   6   8   10   12
3   6   9   12   15   18
4   8   12   16   20   24
5   10   15   20   25   30
6   12   18   24   30   36

which is a (slightly sloppy) multiplication table. If the sloppiness bothers you, Java provides methods that give you more control over the format of the output, but I’m not going to get into that here.

Methods and encapsulation

In Section Adding new methods I listed some of the reasons methods are useful. Here are several more:

  • By giving a name to a sequence of statements, you make your program easier to read and debug.
  • Dividing a long program into methods allows you to separate parts of the program, debug them in isolation, and then compose them into a whole.
  • Methods facilitate both recursion and iteration.
  • Well-designed methods are often useful for many programs. Once you write and debug one, you can reuse it.

To demonstrate encapsulation again, I’ll take the code from the previous section and wrap it up in a method:

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public static void printMultTable() {
  int i = 1;
  while (i <= 6) {
    printMultiples(i);
    i = i + 1;
  }
}

The development process I am demonstrating is called encapsulation and generalization. You start by adding code to main or another method. When you get it working, you extract it and wrap it up in a method. Then you generalize the method by adding parameters.

Sometimes you don’t know when you start writing exactly how to divide the program into methods. This process lets you design as you go along.

Local variables

You might wonder how we can use the same variable i in both printMultiples and printMultTable. Didn’t I say that you can only declare a variable once? And doesn’t it cause problems when one of the methods changes the value of the variable?

The answer to both questions is “no,” because the i in printMultiples and the i in printMultTable are not the same variable. They have the same name, but they do not refer to the same storage location, and changing the value of one has no effect on the other.

Variables declared inside a method definition are called local variables because they only exist inside the method. You cannot access a local variable from outside its “home” method, and you are free to have multiple variables with the same name, as long as they are not in the same method.

Although it can be confusing, there are good reasons to reuse names. For example, it is common to use the names i, j and k as loop variables. If you avoid using them in one method just because you used them somewhere else, you make the program harder to read.

More generalization

As another example of generalization, imagine you wanted a program that would print a multiplication table of any size, not just the 6x6 table. You could add a parameter to printMultTable:

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public static void printMultTable(int high) {
  int i = 1;
  while (i <= high) {
    printMultiples(i);
    i = i + 1;
  }
}

I replaced the value 6 with the parameter high. If I invoke printMultTable with the argument 7, I get

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1   2   3   4   5   6
2   4   6   8   10   12
3   6   9   12   15   18
4   8   12   16   20   24
5   10   15   20   25   30
6   12   18   24   30   36
7   14   21   28   35   42

which is fine, except that I probably want the table to be square (same number of rows and columns), which means I have to add another parameter to printMultiples, to specify how many columns the table should have.

I also call this parameter high, demonstrating that different methods can have parameters with the same name (just like local variables):

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public static void printMultiples(int n, int high) {
  int i = 1;
  while (i <= high) {
    System.out.print(n*i + "   ");
    i = i + 1;
  }
  System.out.println("");
}

public static void printMultTable(int high) {
  int i = 1;
  while (i <= high) {
    printMultiples(i, high);
    i = i + 1;
  }
}

Notice that when I added a new parameter, I had to change the first line, and I also had to change the place where the method is invoked in printMultTable. As expected, this program generates a square 7x7 table:

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7
1   2   3   4   5   6   7
2   4   6   8   10   12   14
3   6   9   12   15   18   21
4   8   12   16   20   24   28
5   10   15   20   25   30   35
6   12   18   24   30   36   42
7   14   21   28   35   42   49

When you generalize a method appropriately, you often find that it has capabilities you did not plan. For example, you might notice that the multiplication table is symmetric, because \(ab = ba\), so all the entries in the table appear twice. You could save ink by printing only half the table. To do that, you only have to change one line of printMultTable. Change

printMultiples(i, high);

to

printMultiples(i, i);

and you get

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1
2   4
3   6   9
4   8   12   16
5   10   15   20   25
6   12   18   24   30   36
7   14   21   28   35   42   49

I’ll leave it up to you to figure out how it works.

Glossary

loop:
A statement that executes repeatedly while some condition is satisfied.
infinite loop:
A loop whose condition is always true.
body:
The statements inside the loop.
iteration:
One pass through (execution of) the body of the loop, including the evaluation of the condition.
encapsulate:
To divide a large complex program into components (like methods) and isolate the components from each other (for example, by using local variables).
local variable:
A variable that is declared inside a method and that exists only within that method. Local variables cannot be accessed from outside their home method, and do not interfere with any other methods.
generalize:
To replace something unnecessarily specific (like a constant value) with something appropriately general (like a variable or parameter). Generalization makes code more versatile, more likely to be reused, and sometimes even easier to write.
program development:
A process for writing programs. So far we have seen “incremental development” and “encapsulation and generalization”.

Exercises

Exercise

Consider the following code:

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public static void main(String[] args) {
    loop(10);
}

public static void loop(int n) {
    int i = n;
    while (i > 0) {
        System.out.println(i);
        if (i%2 == 0) {
            i = i/2;
        } else {
            i = i+1;
        }
    }
}
  1. Draw a table that shows the value of the variables i and n during the execution of loop. The table should contain one column for each variable and one line for each iteration.
  2. What is the output of this program?

Exercise

Let’s say you are given a number, \(a\), and you want to find its square root. One way to do that is to start with a very rough guess about the answer, \(x_0\), and then improve the guess using the following formula:

\[x_1 =(x_0 + a/x_0) / 2\]

For example, if we want to find the square root of 9, and we start with \(x_0 = 6\), then \(x_1 =(6 + 9/6) /2 = 15/4 = 3.75\), which is closer.

We can repeat the procedure, using \(x_1\) to calculate \(x_2\), and so on. In this case, \(x_2 = 3.075\) and \(x_3 = 3.00091\). So that is converging very quickly on the right answer(which is 3).

Write a method called squareRoot that takes a double as a parameter and that returns an approximation of the square root of the parameter, using this technique. You may not use Math.sqrt.

As your initial guess, you should use \(a/2\). Your method should iterate until it gets two consecutive estimates that differ by less than 0.0001; in other words, until the absolute value of \(x_n - x_{n-1}\) is less than 0.0001. You can use Math.abs to calculate the absolute value.

Exercise

In this Exercise we wrote a recursive version of power, which takes a double x and an integer n and returns \(x^n\). Now write an iterative method to perform the same calculation.

Exercise

Section More recursion presents a recursive method that computes the factorial function. Write an iterative version of factorial.

Exercise

One way to calculate \(e^x\) is to use the infinite series expansion

\[e^x = 1 + x + x^2 / 2! + x^3 / 3! + x^4 / 4! + ...\]

If the loop variable is named i, then the \(i\)th term is \(x^i / i!\).

  1. Write a method called myexp that adds up the first n terms of this series. You can use the factorial method from Section More recursion or your iterative version from the previous exercise.

  2. You can make this method much more efficient if you realize that in each iteration the numerator of the term is the same as its predecessor multiplied by x and the denominator is the same as its predecessor multiplied by i. Use this observation to eliminate the use of Math.pow and factorial, and check that you still get the same result.

  3. Write a method called check that takes a single parameter, x, and that prints the values of x, Math.exp(x) and myexp(x) for various values of x. The output should look something like:

    1.0     2.708333333333333       2.718281828459045
    

    HINT: you can use the String "\t" to print a tab character between columns of a table.

  4. Vary the number of terms in the series (the second argument that check sends to myexp) and see the effect on the accuracy of the result. Adjust this value until the estimated value agrees with the “correct” answer when x is 1.

  5. Write a loop in main that invokes check with the values 0.1, 1.0, 10.0, and 100.0. How does the accuracy of the result vary as x varies? Compare the number of digits of agreement rather than the difference between the actual and estimated values.

  6. Add a loop in main that checks myexp with the values -0.1, -1.0, -10.0, and -100.0. Comment on the accuracy.

Exercise

One way to evaluate \(\exp(-x^2)\) is to use the infinite series expansion

\[\exp(-x^2) = 1 - x^2 + x^4/2 - x^6/6 + \ldots\]

In other words, we need to add up a series of terms where the \(i\)th term is equal to \((-1)^i x^{2i} / i!\). Write a method named gauss that takes x and n as arguments and that returns the sum of the first n terms of the series. You should not use factorial or pow.

[1]See Wikipedia.

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