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Adaptive Huffman Code Lecture7: Huffman Code

0 1

1

0

0 1

Ex:

49 50

Ex:

43 44

45 46

47 48

49 50

Then increment node

number 48 to weight 2

instead of 1

Then increment node

number 50 to weight 3

instead of 2

Then increment node

number 51 to weight 5

instead of 4

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Adaptive Huffman Code Lecture7: Huffman Code

Encoding

Initially, the tree at both the encoder and

decoder consists of a single node; the NYT node

The codeword for the very first symbol that

appears is a previously agreed upon fixed code

Send the code for the NYT node, followed by the

previously agreed upon fixed code for the symbol

The code for the NYT node is obtained by

traversing the Huffman tree from the root to the

NYT node

If a symbol to be encoded has a corresponding

node in the tree, then the code for the symbol is

generated by traversing the tree from the root to

the external node corresponding to the symbol

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Ex: Encoding the message [a a r d v a r k], where our alphabet consists of the 26 lowercase

letters of the English alphabet

The total number of nodes in this tree will be =2m-1=2×26−1

= 51, so we start numbering backwards from 51 with the

number of the root node being 51

We begin with only the NYT node with weight 0, nothing

transmitted so far

Input: aardvark

Symbol Code

NYT 0

a

r

d

v

K

Output:

Adaptive Huffman Code Lecture7: Huffman Code

LECTURES 11/4/2017 3

Starting the tree

code

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The first symbol to be encoded is the letter ‘a’

As ‘a’ is the first letter of the alphabets order, k =1. As 1 is less than 20, ‘a’ is encoded as the 5 bit

binary representation of k−1=0, which is 00000

We send the code for the NYT node, by traversing the Huffman tree from the root to the NYT node

(0) followed by the previously agreed upon fixed code for the symbol “a” (00000)

r d v a r k a aInput: Output: 00000

Adaptive Huffman Code Lecture7: Huffman Code

11/4/2017

Encoding the first symbol

Transmission

The NYT node gives birth to an external node corresponding to

the element ‘a’ and new NYT node

The weight of the terminal node should be higher than the

NYT node and since, ‘a’ has occurred once, its external node

has a weight of one, hence, the weight of the NYT node is 1

The parent node also has a weight of 1, as its weight is the sum

of the weights of its offspring

Also we assign the number 49 to the NYT node and 50 to the

terminal node corresponding to the letter ‘a’

The node corresponding to ‘a’ has the highest number (if we

do not consider its parent), so we do not need to swap nodes

Updating

Symbol Code

NYT 0

a 1

r

d

v

k

0 1

1

0

0 1

SOURCE CODING PROF. A.M.ALLAM

After a symbol has been encoded, the external node

corresponding to the symbol is examined to see if it has

the largest node number in its block. If the external

node does not have the largest node number, it is

exchanged with the node that has the largest node

number in the block, as long as the node with the

higher number is not the parent of the of the node

being updated, so no exchange is done, then increment

the weight of the external node and its parent

The second letter to be transmitted is also ‘a’

It is not a new symbol. Its code is obtained by traversing the tree from the root node to the external node

corresponding to it. This traversal consists of a single right branch, therefore, the Huffman code for the symbol

‘a’ is 1

Symbol Code

NYT 0

a 1

r

d

v

k

Output: 000001

Adaptive Huffman Code Lecture7: Huffman Code

11/4/2017 5

Encoding the next symbol

Transmission

Updating

r d v a r k aa Input:

We send the code for the NYT node, by traversing the Huffman tree from the root to the NYT node (0) followed by

the code for the symbol “a” (000001)

0 1

0 1

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Adaptive Huffman Code Lecture7: Huffman Code

11/4/2017 6

Transmission

Encoding the next symbol

We send the code for the NYT node , which is 0 (we obtain it by traversing the tree from the root

to the NYT node), followed by the previously arranged binary representation for ‘r’ . The letter ‘r’

is the 18th letter of the alphabet; therefore, the binary representation of ‘r’ is 10001

The third symbol to be encoded is the letter ‘r’

As ‘r’ is the 18th letter of the alphabets order, , the binary representation of ‘r’ is 10001

d v a r k r aa Input: Output: 000001010001

0 1

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11/4/2017 7

Adaptive Huffman Code Lecture7: Huffman Code

Updating

The NYT node gives birth to an external node corresponding to the element ‘r’ and new NYT

node. The weight of the terminal node should be higher than the NYT node and since, ‘r’ has

occurred once, its external node has a weight of one, hence, the weight of the NYT node is 0. The

parent node also has a weight of 1, as its weight is the sum of the weights of its offspring. Also we

assign the number 47 to the NYT node and 48 to the terminal node corresponding to the letter ‘r’

The node corresponding to ‘r’ has the highest number (if we do not consider its parent), so we do

not need to swap nodes . The parent node is not the root node, hence we go to the grandparent,

if that node has the highest number in its block , then increment its weight , else swap within the

block and increment its weight . Continue till the root node

Symbol Code

NYT 00

a 1

r 01

d

v

k

0 1

1

1

0 1

3

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11/4/2017 8

Adaptive Huffman Code Lecture7: Huffman Code

We send the code for the NYT node , which is 00 (we obtain it by traversing the tree from the

root to the NYT node), followed by the previously arranged binary representation for ‘d’

Transmission

Encoding the next symbol

It is the 4th symbol in the alphabet. As this is less than 20, the 5 bit binary representation of

4−1 = 3 is 00011

The next symbol to be transmitted is ‘d’

v a r k r d aa Input: Output: 0000010100010000011

SOURCE CODING PROF. A.M.ALLAM

The NYT node gives birth to an external node

corresponding to the element ‘d’ and new

NYT node. The weight of the external node

should be higher than the NYT node and

since, ‘d’ has occurred once, its external node

has a weight of one, hence, the weight of the

NYT node is zero. The parent node also has a

weight of one, as its weight is the sum of the

weights of its offspring. Also we assign the

number 45 to the NYT node and 46 to the

terminal node corresponding to the letter ‘d’

Adaptive Huffman Code Lecture7: Huffman Code

9

Updating

2

4

The node corresponding to ‘d’ has the highest

number (if we do not consider its parent), so we

do not need to swap nodes . The parent node is

not the root node, hence we go to the

grandparent, if that node has the highest number

, then increment its weight , else swap within the

block and increment its weight . Continue till the

root node

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Symbol Code

NYT 000

a 1

r 01

d 001

v

k

Adaptive Huffman Code Lecture7: Huffman Code

11/4/2017 10

SOURCE CODING PROF. A.M.ALLAM

The next symbol to be transmitted is ‘v’

As this is the first appearance of this symbol,. It is the 22nd symbol in the alphabet. As this is

greater than 20, the 4 bit binary representation of 22−10−1 = 11 is 1011

Adaptive Huffman Code Lecture7: Huffman Code

11

Transmission

Encoding the next symbol

a r k r d v aa Input:

We send the code for the NYT node , which is

000 (we obtain it by traversing the tree from

the root to the NYT node), followed by the

previously arranged binary representation for

‘v’ which is 1011

Output: 00000101000100000110001011

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Adaptive Huffman Code Lecture7: Huffman Code

12

Updating

The NYT node gives birth to an external

node corresponding to the element ‘v’ and

new NYT node. The weight of the terminal

node should be higher than the NYT node

and since, ‘v’ has occurred once, its

external node has a weight of one, hence,

the weight of the NYT node is 0. The

internal node also has a weight of 1, as its

weight is the sum of the weights of its

offspring. Also we assign the number 43 to

the NYT node and 44 to the terminal node

corresponding to the letter ‘v’. The node

corresponding to ‘v’ has the highest

number (if we do not consider its parent),

so we do not need to swap nodes

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13

-We examine the grandparent node of v (node 47) to see if it has the largest number in its block. As

it does not, we swap it with node 48, which has the largest number in its block. We then increment

node 48 and move to its parent, which is node 49

Adaptive Huffman Code Lecture7: Huffman Code

1 2

In the block containing node 49, the largest number belongs to node 50. Therefore, we swap

nodes 49 and 50 and then increment node 50. We then move to the parent node of node 50, which

is node 51. As this is the root node, all we do is increment node 51

SOURCE CODING PROF. A.M.ALLAM

11/4/2017 14 LECTURES

Adaptive Huffman Code Lecture7: Huffman Code

Symbol Code

NYT 1100

a 0

r 10

d 111

v 1101

k

50 49

47 48

46 45

44 43

4

2 3

5

SOURCE CODING PROF. A.M.ALLAM

11/4/2017 15

Adaptive Huffman Code

-The next letter to be encoded is ‘a’ . It has an external node , we simply traverse the tree from the

root node to the external node corresponding to ‘a’ which consists of a single left branch.

Therefore, the Huffman code for the symbol ‘a’ is 0, send it . No swap

Symbol Code

NYT 1100

a 0

r 10

d 111

v 1101

k

48 47

50 49

Lecture7: Huffman Code

r k r d v a aa Input:

Output: 0000010100010000011000101111000

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11/4/2017 16

Adaptive Huffman Code

Symbol Code

NYT 1100

a 0

r 10

d 001

v 1101

k

48 47

50 49

Lecture7: Huffman Code

-The next letter to be encoded is ‘r’ . It has an external node , we simply traverse the tree from the

root node to the external node corresponding to ‘r’ which consists of a signle right branch followed

by a single left branch. Therefore, the Huffman code for the symbol ‘r’ is 10, send it . No swap

k r d v a r aa Input:

Output: 0000010100010000011000101111000110010

SOURCE CODING PROF. A.M.ALLAM

17

-The next symbol to be transmitted is ‘k’. As this is the first appearance of this symbol, we send

the code for the NYT node , which is 1100 (we obtain it by traversing the tree from the root to

the NYT node), followed by the previously arranged binary representation for ‘k’ . It is the 11th

symbol in the alphabet. the 5 bit binary representation of 11−1 = 10 is 01010

Adaptive Huffman Code

Symbol Code

NYT 11000

a 0

r 10

d 111

v 1101

k

Lecture7: Huffman Code

r d v a r k aa Input:

1

48 47

50 49

Output: 0000010100010000011000101111000110010110001010

SOURCE CODING PROF. A.M.ALLAM

11/4/2017 18

Adaptive Huffman Code

50 49

Lecture7: Huffman Code

r d v a r k aa Input:

Symbol Code

NYT 11100

a 0

r 10

d 110

v 1111

k 11101

Output: 0000010100010000011000101111000110010110001010