I need a solution of this complicated problem to finalize my final project

Introduction
This project revolves around an important text processing task, text compression. In particular, you will be required to encode a sequence of words read from a source file into binary strings (using only the characters 0 and 1). It is important to note that text compression makes it possible to minimize the time needed to transmit text over a low-bandwidth channel, such as infrared connection. Moreover, text compression is helpful in storing large documents more efficiently. The coding scheme explored in this project is the Huffman Coding Scheme. While standard encoding schemes, such as Unicode and ASCII, use fixed-length binary strings to encode characters, Huffman coding assigns variable-length codes to characters. The length of a Huffman code depends on the relative frequency of its associated character. Specifically, Huffman coding capitalizes on the fact that some characters are used more frequently than others to use short codewords when encoding high-frequency characters and long codewords to encode low-frequency characters. Huffman coding saves space over state of the art fixed-length encoding and is therefore at the heart of file compression techniques in common use today. Figure 1 shows the relative frequencies of the letters of the alphabet as they appear in a representative sample of English documents.
Letter     Frequency     Letter     Frequency
A     77     N     67
B     17     O     67
C     32     P     20
D     42     Q     5
E     120     R     59
F     24     S     67
G     17     T     85
H     50     U     37
I     76     V     12
J     4     W     22
K     7     X     4
L     42     Y     22
M     24     Z     2
Figure 1. Relative frequencies for the 26 letters of the alphabet.
Huffman coding and decoding
Huffman’s algorithm for producing optimal variable-length codes is based on the construction of a binary tree T that represents the code. In other words, the Huffman code for each character is derived from a full binary tree known as the Huffman coding tree, or simply the Huffman tree. Each edge in the Huffman tree represents a bit in a codeword, with each edge connecting a node with its left child representing a “0” and each edge connecting a node with its right child representing a “1”. Each external node in the tree is associated with a specific character, and the Huffman code for a character is defined by the sequence of bits in the path from the root to the leaf corresponding to that character. Given codes for the characters, it is a simple matter to use these codes to encode a text message. You will have simply to replace each letter in the string with its binary code (a lookup table can be used for this purpose).
In this project, you are not going to use the table given in Figure 1 to determine the frequency of occurrence per character. Instead, you will derive the frequency corresponding to a character by counting the number of times that character appears in an input file. For example, if the input file contains the following line of text “a fast runner need never be afraid of the dark”, then the frequencies listed in the table given in Figure 2 should be used per character:
Character          a     b     d     e     f     H     i     k     n     O     r     s     t     u     v
Frequency     9     5     1     3     7     3     1     1     1     4     1     5     1     2     1     1
Figure 2. The frequency of each character of the String X.
Based on the frequencies shown in Figure 2, the Huffman tree depicted in Figure 3 can be constructed:
Figure 3. Huffman tree for String X.
The code for a character is thus obtained by tracing the path from the root of the Huffman tree to the external node where that character is stored, and associating a left edge with 0 and a right edge with 1. In the context of the considered example for instance, the code for “a” is 010, and the code for “f” is 1100.
Once the Huffman tree is constructed and the message obtained from the input file has been encoded, decoding the message is done by looking at the bits in the coded string from left to right until all characters are decoded. This can be done by using the Huffman tree in a reverse process from that used to generate the codes. Decoding the bit string begins at the root of the tree. Branches are taken depending on the bit value – left for ‘0’ and right for ‘1’ – until reaching a leaf node. This leaf contains the first character in the message. The next bit in the code is then processed from the root again to start the next character. The process is repeated until all the remaining characters are decoded. For example, to decode the string “0101100” in the case of the example under study, you begin at the root of the tree and take a left branch for the first bit which is ‘0’. Since the next bit is a ‘1’, you take a right branch. Then, you take a left branch (for the third bit ‘1’), arriving at the leaf node corresponding to the letter a. Thus, the first letter of the coded word is a. You then begin again at the root of the tree to process the fourth bit, which is a ‘1’. Taking 2 right branches then two left branches, you reach the leaf node corresponding to the letter f.
Problem statement
You are required to implement the Huffman coding/decoding algorithms. After you complete the implementation of the coding/decoding processes, you are asked to use the resulting Java code to:
1.     Read through a source file called “in1.dat” that contains the following paragraph:
“the Huffman coding algorithm views each of the d distinct characters of the string X as being in separate Huffman trees initially with each tree composed of a single leaf node these separate trees will eventually be joined into a single Huffman tree in each round the algorithm takes the two binary trees with the smallest frequencies and merges them into a single binary tree it repeats this process until only one tree is left.”
2.     Determine the actual frequencies for all the letters in the file.
3.     Use the frequencies from the previous step to create a Huffman coding tree before you assign codes to individual letters. Use the LinkedBinaryTree class that we developed in class to realize your Huffman coding trees.
4.     Produce an encoded version of the input file “in1.dat” and then store it in an output file called “out.dat”.
5.     Finally, the decoding algorithm will come into play to decipher the codes contained in “out.dat”. The resulting decoded message should be written to an output file called “in2.dat”. If nothing goes wrong, the text stored in “in1.dat” and the one in “in2.dat” must match up correctly.

jschell wrote:
I need a solution of this complicated problem to finalize my final project The solution:
1. Write code
2. Test code3. If test fails, debug code, then go to step 2.

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    192.168.0.2         e6-1f-6d-6c-db-da      dynamic ]<-NOT
    192.168.0.3         e6-1f-6d-6c-db-da      dynamic ]<-NOT
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    224.0.0.252         01-00-5e-00-00-fc      static
    224.1.1.1             01-00-5e-01-01-01     static
    255.255.255.255   ff-ff-ff-ff-ff-ff                static
    The XP TCP/IP stack does not do this and you might think that my windows 7 has a problem it does not because the TCP/IP stack in windows 7 is a rebuild of what the TCP/IP stack was like in XP but whats done is done and that how the windows 7 TCP/IP stack is and that should be a eye opener as to why nothing has been done about this but thats just me saying that but its not really a TCP/IP stack problem. So is that my ISP problem? The answer is no because even if my ISP did not reply you still can send requests from a valid IP like 77.96.238.3 to your ISP a ARP with a target IP that does not and should not exist out to your ISP gateway like 192.168.0.1 yet you can.
    The simple fact is this there is no, none and nothing to make a ACL for ARP to drop the target/sender IP's for the 0806 Ethertype.
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    Maybe just maybe if the DHCP Snooping looked at the Option 1 & 3 for the Subnet Mask & Router to know the range that the ARP Inspection can drop ARP for target IP's outside that range because like I said to send from your IP to an IP outside the subnet your in you send to the ISP gateway MAC where it will be routed only then would that Dynamic ARP Inspection work as I was hoping for but sadly no.
    So please tell me why I can't simply drop ARP for given target/sender IP's is their another way (and NAT is not a solution) that ideally makes ARP for from 77.96.238.3 as the sender for a request for a target IP of 10. for all of 172.16. to 172.31. for all of 192.168 and 169.254. NOT reach my ISP BUT allows ARP from 77.96.238.3 as the sender for a request for a target IP within the given subnet to my ISP for a reply.
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    I really do not get why providing a config would help because if you have a config that does what I need should it not be you to provide a config or tell me how? Why not tell me how if you know how? because I need to understand how if you know what I need to do.
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    VLAN Management
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    Verify MAC Address: Enable - ticked
    Interface Settings
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    DHCP Snooping Trusted Interfaces
    GE9 Trusted Interface Yes
    GE1-GE8 and GE10 Trusted Interface No
    ARP Inspection
    ARP Inspection Status: Enable - ticked
    ARP Packet Validation: Enable- ticked
    Interface Settings
    GE9 Trusted Interface Yes
    GE1-GE8 and GE10 Trusted Interface No
    VLAN Settings
    Enabled VLANs VLAN 10
    Access Control
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    Dropsetin
    Dropsetintoout
    IPv4-Based ACE for Dropsetin
    Priority---------Action--Protocol------Source IP Address----------Destination IP Address
    100--------------Deny---Any (IP)------Any-----------Any --------------192.168.0.0--0.0.255.255
    101--------------Deny---Any (IP)------192.168.0.0--0.0.255.255-------Any----------Any
    102--------------Deny---Any (IP) -----Any------------Any--------------10.0.0.0-------0.255.255.255
    104--------------Deny---Any (IP)------Any------------Any--------------172.16.0.0----0.15.255.255
    105--------------Deny---Any (IP)------172.16.0.0----0.15.255.255----Any------------Any
    106--------------Deny---Any (IP)------Any------------Any--------------169.254.0.0---0.0.255.255
    107--------------Deny---Any (IP)------169.254.0.0---0.0.255.255-----Any------------Any
    2147483647---Permit--Any (IP) -----Any------------Any---------------Any-----------Any
    IPv4-Based ACE for Dropsetintoout
    Priority---------Action--Protocol------Source IP Address----------Destination IP Address
    100--------------Deny---Any (IP)------Any-----------Any --------------192.168.0.0--0.0.255.255
    101--------------Deny---Any (IP)------192.168.0.0--0.0.255.255-------Any----------Any
    102--------------Deny---Any (IP) -----Any------------Any--------------10.0.0.0-------0.255.255.255
    103--------------Deny---Any (IP) -----10.0.0.0-------0.255.255.255---Any------------Any
    104--------------Deny---Any (IP)------Any------------Any--------------172.16.0.0----0.15.255.255
    105--------------Deny---Any (IP)------172.16.0.0----0.15.255.255----Any------------Any
    106--------------Deny---Any (IP)------Any------------Any--------------169.254.0.0---0.0.255.255
    107--------------Deny---Any (IP)------169.254.0.0---0.0.255.255-----Any------------Any
    2147483647---Permit--Any (IP) -----Any------------Any---------------Any-----------Any
    ACL Binding for Dropsetin
    GE9
    ACL Binding for Dropsetintoout
    GE8
    And added a IPv6-Based ACL not that thiers any piont yet.
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    Message was edited by: Peter __

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