Adaptive Huffman Coding And Decoding
Last Updated :
12 Jul, 2025
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 = 2e + r and 0 ≤ r ≤ 2e
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 condition (1)
So Fixed Code is Binary Representation of (k-1) = 0 as 5-bit representation
Fixed 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 condition (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 condition (1)
So Fixed Code is Binary Representation of (k-1 = 3) as 5-bit representation
Fixed 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 condition (2)
So Fixed Code is Binary Representation of (k-r-1 = 11) as 4-bit representation
Fixed 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 condition (1)
So Fixed Code is Binary Representation of (k-1 = 10) as 5-bit representation
Fixed 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
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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