[SOLVED] Com s 228 project 4: archived message reconstruction
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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.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:
10110101011101101010100 is parsed as
1011|0|1010|111|0|110|1010|100
which is decoded as:
cadbard!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 payload3. 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 cadbard! message is encoded in the following file (“cadbard.arch”):
^a^^!^dc^rb
10110101011101101010100There are four test files in HW4S2021_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.4.1. 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;/*Can use a static char idx to the tree string for recursive
solution, but it is not strictly necessary*/
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. Note: 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() performs 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 classes in the edu.iastate.cs228.hw4 package. Turn in the zip file and not your
class files. Please follow the guideline in submission_guide.pdf.
Include the Javadoc tag @author in each class source file. Your zip file should be named
Firstname_Lastname_HW4.zip. No template files will be provided other than the skeleton in
Section 4.1.6. Extra credit (5% or 15%)
Print the following statistics after the rest of the program output:
8.0
1180
STATISTICS:
Avg bits/char:
Total characters:
Space savings: 50.0%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 extra effort compared to the
recursive solution.The bonus for early submission will stack with the 5/15% bonus.
Name your submission Firstname_Lastname_HW4_extra.zip if you completed
the iterative solution.