Description
1. Problem Description
The objective of this exercise is to decode/unzip a message archived with a binary-tree-based
algorithm. The program should ask for a single filename at the start: “Please enter filename to
decode: “, decode the message in the file and print it out to the console. The name of the
compressed message file will end in .arch, e.g. “monalisa.arch”. The file consists of two or three
lines: the first one or two lines contain the encoding scheme, and the second or third line
contains the archived message.
2. Encoding
The archival algorithm uses a binary tree. The edges of the tree represent bits, and the leaf
nodes contain one character each. Internal nodes are empty. An edge to a left child always
represents a 0, and an edge to a right child always represents a 1. Characters are encoded by
the sequence of bits along a path from the root to a particular leaf. The below tree serves as an
example.
The tree on the left encodes these characters:
Character Encoding
a 0
! 100
d 1010
c 1011
r 110
b 111
With the above encoding, the bit string:
10100101010110110111100 is parsed as
1010|0|1010|1011|0|110|111|100
which is decoded as:
dadcarb!
With this encoding, we can automatically infer where one character ends and another
begins.
That is because no character code can be the start of another character code. For example,
if
you have a character with the code 111, you cannot have the codes 1 and 11, as they
would be internal nodes.
The following steps decode one character from the bit string:
Start at root
Repeat until at leaf
Scan one bit
Go to left child if 0; else go to right child
Print leaf payload
3. Input Format
The archive file consists of two lines: the first line contains the encoding scheme, and the
second line contains the compressed string. For ease of development and to make the
archive file human-readable, each bit is represented as the character ‘0’ or ‘1’, rather
than as an actual bit from a binary file.
The encoding scheme can be represented as a string. For example, the tree from section 2
can be represented as:
^a^^!^dc^rb
where ^ indicates an internal node. The above code represents a preorder traversal of
the tree.
The dadcarb! message is encoded in the following file (“dadcarb.arch”):
^a^^!^dc^rb
10100101010110110111100
There are four test files in project3_Test_Files.zip. Note the encoding scheme representations
may include a space character and a newline character, thereby breaking the tree string into
two lines! The newline character needs to be parsed correctly if the encoding file has three
lines in total.
4. Task
4.1. Read in the first line (and possibly second line, if newline is part of the tree) of the
file and construct the character tree. Convert the line input into a MsgTree structure
using preorder traversal. The tree should be in a class MsgTree with the following members:
public class MsgTree{
public char payloadChar;
public MsgTree left;
public MsgTree right;
//Need static char idx to the tree string for recursive solution
private static int staticCharIdx = 0;
//Constructor building the tree from a string
public MsgTree(String encodingString){}
//Constructor for a single node with null children
public MsgTree(char payloadChar){}
//method to print characters and their binary codes
public static void printCodes(MsgTree root, String code){}
}
When building the tree, try a recursive solution where staticCharIdx tracks the location
within the tree string. You can pass the same tree string during recursive calls, and update
the staticCharIdx to point to the next character to be read.
If you decide to implement an iterative solution, you will receive a 15% bonus, as it is
considerably more difficult. In that case, you cannot get the 5% bonus for printing statistics.
printCodes() does a recursive preorder traversal of the MsgTree and prints all
the characters and their bit codes:
character code
————————-
c 1011
r 110
b 111
You are allowed to print the header of the table (character, code, —-) in main().
4.2. Write a method public void decode(MsgTree codes, String msg) to
decode the message. It would print the decoded message to the console:
MESSAGE:
The quick brown fox jumped over the lazy dog.
You are allowed to print “MESSAGE:” in main().
The overall output of the program should be the output of printCodes() followed by the
output of decode():
character code
————————-
c 1011
r 110
b 111
MESSAGE:
The quick brown fox jumped over the lazy dog.
5. Submission
Put your java class code in the edu.iastate.cs228.hw4 package. Do not submit your
class files. Please follow the guideline posted under Documents & Links on Canvas.
Include the Javadoc tag @author in each class source file. Your zip file should be
named Firstname_Lastname_HW4.zip.No template files are provided other than Section 4.1.
6. Extra credit (5% or 15%)
Print the following statistics after the rest of the program output:
6.8
179
STATISTICS:
Avg bits/char:
Total characters:
Space savings: 57.5%
The space savings calculation assumes that an uncompressed character is encoded
with 16 bits. It is defined as (1 – compressedBits/uncompressedBits)*100.
To earn a 15% non-cumulative bonus (either 5% for statistics or 15%), you can
create an non-recursive, iterative solution for building the tree, but be advised that it
will require hours of more effort than the recursive solution.
Name your submission Firstname_Lastname_HW4_extra.zip if you completed the
work for extra credit.
