**Prerequisite:** Huffman Coding, Huffman Decoding

Adaptive Huffman Coding is also known as Dynamic Huffman Coding. The implementation is done using Vitter Algorithm.

**Encoding**

Adaptive Huffman coding for a string containing alphabets:

Let m be the total number of alphabets. So m = 26.

For Vitter Algorithm, find a parameters e & r such that

m = 2^{e}+ r and 0 ≤ r ≤ 2^{e}Therefore, for m = 26 we get e = 4 & r = 10

There are two type of code NYT Code & Fixed Code.

NYT code = Traversing tree from the root node to that particular NYT node.

For Fixed Code, it can be calculated from the following two conditions:

- If 0 ≤ k ≤ 2r Then the letter Sk is encoded as the binary representation of (k-1) in (e+1) bits. (where k is position of alphabet in sorted order)
- Else the letter Sk is encoded as the binary representation of (k-r-1) in e bits.

**Tree Updation**

Tree Updation in Vitter Algorithm follows Implicit Numbering. In Implicit numbering,

- Nodes are numbered in increasing order i.e., by level and from left to right
- The Nodes that have the same weight and the type together form a block
- Blocks are related to each other as by increasing order of their weights
- Internal Node is represented by Oval shape. Weight of internal nodes = Sum of child node weights
- External Node is represented by a square shape. Weight of external nodes = Initially 1 and if repeated then increased the weight by 1

**Steps for Tree Updation:**

- Initialize the tree with the NYT Node
- For a symbol is recognized for the first time, the initial NYT node is further divided into an NYT Node and new Node initialize to that symbol and weight = 1.
- Assign the sum of the weight of child nodes to the parent node
- If a repeated symbol is encountered than weights are updated to that symbol.

**Note:** During Updation in Tree if the weight of the left subtree is greater than the right subtree, then nodes must be swapped.

**Example**

code= "aardvark" The final Code we get is: 00000 1 010001 0000011 0001011 0 10 110001010 a a r d v a r k

**Explanation:**

For string code = “aardvark”, e = 5, r = 10

As shown in the above image Tree is initialize with NYT Node with weight 0.

- For symbol ‘a’, k = 1.
NYT Code = "" (initially tree is empty)

For Fixed Code: As k < 2r i.e, 1 < 2*10, satisfy condtion (1)

So Fixed Code is Binary Representation of (k-1) = 0 as 5-bit representationFixed Code = "00000"

Huffman Code for symbol for 'a' is "00000"

- For symbol ‘a’ which already exists in the tree. Traversing Tree up to symbol ‘a’, we get code = “1”
Huffman Code for symbol for 'a' is "1"

- For symbol ‘r’, k = 18.
NYT Code = "0" (traversing up to NYT Node)

For Fixed Code: As k > 2r i.e, 18 > 2*10, satisfy condtion (2)

So Fixed Code is Binary Representation of (k-1 = 17) as 5-bit representation

Fixed Code = "10001"

Huffman Code for symbol for 'r' is "010001"

- For symbol ‘d’, k = 4.
NYT Code = "000" (traversing up to NYT Node)

For Fixed Code: As k < 2r i.e, 4 < 2*10, satisfy condtion (1)

So Fixed Code is Binary Representation of (k-1 = 3) as 5-bit representationFixed Code = "00011"

Huffman Code = "00000011"

- For symbol ‘v’, k = 22.
NYT Code = "000" (traversing up to NYT Node)

For Fixed Code: As k > 2r i.e, 22 > 2*10, satisfy condtion (2)

So Fixed Code is Binary Representation of (k-r-1 = 11) as 4-bit representationFixed Code = "1011"

Huffman Code = "0001011"

- Swap the node of left subtree and right as the tree is violating property
- For symbol ‘a’ which already exists in the tree. Traversing Tree up to symbol ‘a’, we get code = “0”
Huffman Code for symbol for 'a' is "0"

- For symbol ‘r’ which already exists in the tree. Traversing Tree up to symbol ‘a’, we get code = “10”
Huffman Code for symbol for 'r' is "10"

- For symbol ‘k’, k = 11.
NYT Code = "1100" (traversing up to NYT Node)

For Fixed Code: As k < 2r i.e, 11 < 2*10, satisfy condtion (1)

So Fixed Code is Binary Representation of (k-1 = 10) as 5-bit representationFixed Code = "01010"

Huffman Code for symbol for 'r' is "110001010"

**Decoding**

**Steps for Decoding:**

- Read Binary string
- If encountered leaf node is NYT
- Read next e bits
- If e bit value < r, Then to get required symbol convert (e+1) bits to decimal value of (e+1) bits + 1
- If e bit value > r, Then to get required symbol convert e bits to decimal value of e bits + r + 1

- Read next e bits

**Example:**

code= "00000101000100000110001011010110001010" We get final decoded code as 00000 1 0 10001 00 00011 000 1011 0 10 1100 01010 a a NYT r NYT d NYT v a r NYT k

**Explanation:**

- Begin decoding by reading first e bits. So the first 4 bits are 0000, converting into decimal = 0.

Now the value 0 < r , i.e, 0 < 10 satisfy condition (1).

Now according to the condition (1), convert first e+1 = 5 bit into decimal and add 1 to it.00000 = 0 0 + 1 = 1, which is value for alphabet a.

Update the tree and add a node for the symbol ‘a’ in the tree

- Read the next bit in the given code and traverse the tree. We reach the external leaf node ‘a’. So the next decoded symbol is ‘a’.
- Read the next set of bits given code and traverse the tree. We have 0 as NYT Node. After reaching the NYT Node, read e bits which are 1000. Convert 1000 to decimal is 8. As 8 < r satisfy condition (1).

Now Convert e+1 bits in decimal and add 1 to it.10001 = 17 17 + 1 = 18, which is value for alphabet r.

Update the tree and add a node for the symbol ‘r’ in the tree.

- Reading the next set of bits and traversing the Tree we reach NYT node at 00. Read e bits which are 0001. Convert 0001 to decimal is 1. As 1 < r satisfy condition (1).

Now Convert e+1 bits in decimal and add 1 to it.00011 = 3 3 + 1 = 4, which is value for alphabet d.

Update the tree and add a node for the symbol ‘d’ in the tree.

- Reading the next set of bits and traversing the Tree we reach NYT node at 000. Read e bits which are 1011. Convert 1011 to decimal is 11. As 11 > r satisfy condition (2).

Now Convert k+r+1 bits in decimal and decode the symbol.10110 = 22, which is value for alphabet v.

Update the tree and add a node for the symbol ‘v’ in the tree.

- Reading the next set of bits and traversing the Tree we get symbol ‘a’ at 0. Update the tree and add a node for the symbol ‘a’ in the tree.
- Reading the next set of bits and traversing the Tree we get symbol ‘r’ at 10. Update the tree and add a node for the symbol ‘a’ in the tree.
- Reading the next set of bits and traversing the Tree we reach NYT node at 1100. Read e bits which are 0101. Convert 0101 to decimal is 9. As 9 < r satisfy condition (1).

Now Convert e+1 bits in decimal and and add 1 to it.01000 = 8, 8 + 1 = 9. which is value for alphabet k.

Update the tree and add a node for the symbol ‘v’ in the tree.

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