Solution for
Programming Exercise 9.2
THIS PAGE DISCUSSES ONE POSSIBLE SOLUTION to
the following exercise from this online
Java textbook.
Exercise 9.2:
As discussed in Section 1, values of type
int are limited to 32 bits. Integers that are too large
to be represented in 32 bits cannot be stored in an int
variable. Java has a standard class, java.math.BigInteger,
that addresses this problem. An object of type BigInteger
is an integer that can be arbitrarily large. (The maximum size is
limited only by the amount of memory on your computer.) Since
BigIntegers are objects, they must be manipulated using
instance methods from the BigInteger class. For example,
you can't add two BigIntegers with the + operator.
Instead, if N and M are variables that refer
to BigIntegers, you can compute the sum of N
and M with the function call N.add(M). The value
returned by this function is a new BigInteger object that
is equal to the sum of N and M.
The BigInteger class has a constructor
new BigInteger(str),
where str is a string. The string must represent an
integer, such as "3" or "39849823783783283733". If the string
does not represent a legal integer, then the constructor
throws a NumberFormatException.
There are many instance methods in the BigInteger class.
Here are a few that you will find useful for this exercise. Assume
that N and M are variables of type BigInteger.
N.add(M)  a function that
returns a BigInteger representing the sum of N and M.
N.multiply(M)  a function that
returns a BigInteger representing the result of multiplying
N times M.
N.divide(M)  a function that
returns a BigInteger representing the result of dividing N
by M.
N.signum()  a function that
returns an ordinary int. The returned value represents the
sign of the integer N. The returned value is 1 if N
is greater than zero. It is 1 if N is less than zero.
And it is 0 if N is zero.
N.equals(M)  a function that
returns a boolean value that is true if N
and M have the same integer value.
N.toString()  a function that
returns a String representing the value of N.
N.testBit(k)  a function
that returns a boolean value. The parameter k is an
integer. The return value is true if the kth
bit in N is 1, and it is false if the kth
bit is 0. Bits are numbered from right to left, starting with 0.
Testing "if (N.testBit(0))" is an easy way to
check whether N is even or odd. N.testBit(0)
is true if and only if N is an odd number.
For this exercise, you should write a program that prints
3N+1 sequences with starting values specified by the user.
In this version of the program, you should use BigIntegers
to represent the terms in the sequence. You can read the user's
input into a String with the TextIO.getln() function.
Use the input value to create the BigInteger object that
represents the starting point of the 3N+1 sequence.
Don't forget to catch and handle the NumberFormatException
that will occur if the user's input is not a legal integer! You should
also check that the input number is greater than zero.
If the user's input is legal, print out the 3N+1 sequence.
Count the number of terms in the sequence, and print the count at the
end of the sequence. Exit the program when the user inputs
an empty line.
Discussion
My solution uses a subroutine, printThreeNSequence(N), to
print out the 3N+1 sequence starting from the BigInteger,
N.
The subroutine assumes that N is not null and that it
represents a value that is greater than one. (These conditions are
ensured by the main program which calls the subroutine.) Given these
assumptions, the subroutine cannot generate any errors. The only
interesting aspect of the subroutine is that all operations on
N must be performed using instance methods from the
BigInteger class. For example, to multiply N by
2, I use a statement "N = N.multiply(TWO);",
where TWO is a BigInteger that represents the
integer 2. My program defines TWO as a constant,
along with several other BigIntegers that represent
values that I need:
static final BigInteger THREE = new BigInteger("3");
static final BigInteger ONE = new BigInteger("1");
static final BigInteger TWO = new BigInteger("2");
With these constants, the code for computing the next term
in a 3N+1 sequence becomes:
if (N.testBit(0) == false) {
// N is even. Divide N by 2.
N = N.divide(TWO);
}
else {
// N is odd. Multiply N by 3, then add 1.
N = N.multiply(THREE);
N = N.add(ONE);
}
You can find the complete subroutine in the solution that is given
below.
In the main() routine, the user's input is read into a
variable, line, of type String. The input is used
to construct a BigInteger with the statement
"N = new BigInteger(line);". Since this statement can
produce a NumberFormatException, it is placed in a try
statement that can catch and handle the error. The test
"if (N.signum() == 1)" is used to make sure that N >= 1.
The value of N.signum() is 1 if and only if N is a
positive integer. Here is the loop form the main() routine
that processes the users input:
while (true) {
TextIO.putln();
TextIO.putln("Enter the starting value, or press return to end.");
TextIO.put("? ");
line = TextIO.getln().trim();
if (line.length() == 0)
break;
try {
N = new BigInteger(line);
if (N.signum() == 1)
printThreeNSequence(N);
else
TextIO.putln("Error: The starting value cannot be less than or equal to zero.");
}
catch (NumberFormatException e) {
TextIO.putln("Error: \"" + line + "\" is not a legal integer.");
}
}
The Solution
/*
This program prints out 3N+1 sequences for starting values of N that
are entered by the user. Since integers are represented as objects of
type BigInteger, it will work for arbitrarily large integers. The
starting value specified by the user must be greater than zero. The
program continues to read input from the user and print 3N+1 sequences
until the user inputs an empty line. If the user's input is illegal,
the program will print an error message and continue.
*/
import java.math.BigInteger;
public class BigThreeN {
static final BigInteger THREE = new BigInteger("3");
static final BigInteger ONE = new BigInteger("1");
static final BigInteger TWO = new BigInteger("2");
public static void main(String[] args) {
String line; // A line of input from the user.
BigInteger N; // The starting point for the 3N+1 sequence,
// as specified by the user.
TextIO.putln("This program will print 3N+1 sequences for positive starting values");
TextIO.putln("that you enter. There is no preset limit on the number of");
TextIO.putln("digits in the numbers that you enter. The program will end when");
TextIO.putln("you enter an empty line.");
while (true) {
TextIO.putln();
TextIO.putln("Enter the starting value, or press return to end.");
TextIO.put("? ");
line = TextIO.getln().trim();
if (line.length() == 0)
break;
try {
N = new BigInteger(line);
if (N.signum() == 1)
printThreeNSequence(N);
else
TextIO.putln("Error: The starting value cannot be less than or equal to zero.");
}
catch (NumberFormatException e) {
TextIO.putln("Error: \"" + line + "\" is not a legal integer.");
}
}
TextIO.putln();
TextIO.putln("OK. Bye for now!");
} // end main()
static void printThreeNSequence(BigInteger N) {
// Print the 3N+1 sequence starting from N, and count the number
// of terms in the sequence. It is assumed that N is not null and
// that it is greater than zero.
int count; // The number of terms in the sequence.
TextIO.putln();
TextIO.putln("The 3N+1 sequence starting with " + N + " is:");
TextIO.putln();
TextIO.putln(N.toString()); // Print N as the first term of the sequence
count = 1;
while ( ! N.equals(ONE) ){ // Compute and print the next term
if (N.testBit(0) == false) {
// N is even. Divide N by 2.
N = N.divide(TWO);
}
else {
// N is odd. Multiply N by 3, then add 1.
N = N.multiply(THREE);
N = N.add(ONE);
}
TextIO.putln(N.toString());
count++;
}
TextIO.putln();
TextIO.putln("There were " + count + " terms in the sequence.");
} // end printThreeNSequence
} // end BigThreeN
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