Lecture 4 Source Coding and Compression Dr.-Ing. Khaled Shawky Hassan Room: C3-222, ext: 1204, Email: khaled.shawky@guc.edu.eg 1 1

Static vs. Adaptive Coding Static (Two-Pass Model) Encoder Initialize the data model based on a first pass over the data (i.e., perform the probabilities analysis) Transmit the data model (encoder). Send data and while there is more data to send: -- Encode the next symbol using the existing data model and send it. Decoder Receive the data model (decoder). Receive the data and while there is more data to receive -- Decode the next symbol using the data model and output it. Summary about the Two-Pass procedure: 1. Collect statistics, generate codewords (1st pass round) 2. Perform actual encoding/compression (2nd pass round) 3. Not practical in many situations (e.g., compressing network transmissions) 2

Static vs. Adaptive Coding Adaptive (One-Pass Model) Encoder Initialize the data model as fixed probability and fixed code length. Send data first and while there is more data to send Encode the next symbol using the data model (if we have) and send it. Modify the existing data model based on the last symbol. Decoder Initialize the data model as per agreement. While there is more data to receive Decode the next symbol using the data model and output it. Modify the data model based on the decoded symbol. What Do We Find ? 3 No Encoder map to send!

Huffman Coding (e.g.: Lossless JPEG) Properties: Huffman codes are built from the bottom up, starting with the leaves of the tree and working progressively closer to the root Huffman coding will always at least work more efficient than Shannon-Fano coding, so it has become the predominate entropy coding method It was shown that Huffman coding cannot be improved or with any other integral bit-width coding stream Sibling Property: Defined by Gallager [Gallager 1978]: “A binary code tree has the sibling property if each node (except the root) has a sibling and if the nodes can be listed in order of nonincreasing (decreasing) weight with each node adjacent to its sibling.” Thus: 1- If A is the parent node of B (left) and C (right) is a child of B, then W(A) > W(B) > W(C) Thus if A is the parent node of B (left) and C (right), then W(B) < W(C) 4

W(#1) ≤ W(#3) ≤ W(#3) ≤ … ≤ W(#7) ≤ W(#8) ≤ W(#9) Huffman Coding Properties A binary tree is a Huffman tree if and only if it obeys the sibling property, i.e., W(#1) ≤ W(#3) ≤ W(#3) ≤ … ≤ W(#7) ≤ W(#8) ≤ W(#9) Non-Decreasing Order 5

Adaptive Huffman Coding Algorithm: Given: alphabet S = {s1, …, sn} (NO Probabilities !!!) Pick a fixed default binary codes for all symbols (block/quadratic code) Start with an empty “Huffman” tree (I said and I mean it – Empty ) Read symbol s from source If NYT(s) %% (//) Not Yet Transmitted Send NYT, default(s) (except for the first symbol) Update the tree (and keep it Huffman) Else Send codeword for s Update tree Repeat until done with all symbols in the source 6

Example (Adaptive Huffman) Assume we are encoding the message [a a r d v a r k] The total number of nodes in this tree will be (at most) 2*n – 1 + 2 = 2*26 -1 +2 = 53 where n is the number of usable alphabets and +2 is only for the “NYT” and its “node” The first letter to be transmitted is “a” As a does not yet exist in the tree, we send a binary code 00000 for a and then add a to the tree The NYT node gives birth to a new NYT node and a terminal node corresponding to “a” In this example, we will consider only 51 nodes and leaves (instead of 53!!). However, the correct is 53. The weight of the terminal node will be higher than the NYT node, so we assign the number 49 to the NYT node and 50 to the terminal node “a” The next symbol is a, and the transmitted code is 1 now (as a = 1 only now!) Lest see an example … (we first starts with a fixed code!) 7

Example: Adaptive Huffman Coding Input: aardvark To keep the rest of the slides as is, we started as the book with 51; however, the correct thing is to start with 53! Output: Symbol Code NYT a r d v k 8

Example: Adaptive Huffman Coding Input: aardvark Output: 00000 Symbol Code NYT a r d v k 9

Example: Adaptive Huffman Coding Input: aardvark 1 Output: 00000 Symbol Code NYT a 1 r d v k 1 10

Example: Adaptive Huffman Coding Input: aardvark Output: 000001 Symbol Code NYT a 1 r d v k 11

Example: Adaptive Huffman Coding Input: aardvark Output: 000001010001 Symbol Code NYT 00 a 1 r 01 d v k 12

Example: Adaptive Huffman Coding Input: aardvark Output:0000010100010000011 Symbol Code NYT 000 a 1 r 01 d 001 v k 13

Example: Adaptive Huffman Coding Input: aardvark Output:0000010100010000011000 Symbol Code NYT 0000 a 1 r 01 d 001 v 0001?? k 14

Example: Adaptive Huffman Coding Input: aardvark Output:0000010100010000011000 Symbol Code NYT 0000 a 1 r 01 d 001 v 0001?? k 15

Example: Adaptive Huffman Coding Input: aardvark Output:0000010100010000011000 Symbol Code NYT 0000 a 1 r 01 d 001 v 0101 ?? k 16

Example: Adaptive Huffman Coding Input: aardvark Output:0000010100010000011000 Symbol Code NYT 000 a 1 r 01 d 001 v ?? k 17

Example: Adaptive Huffman Coding Input: aardvark Output:000001010001000001100010101 Symbol Code NYT 1100 a r 10 d 111 v 1101 k 18

Example: Adaptive Huffman Coding Input: aardvark Output:0000010100010000011000101010 Symbol Code NYT 1100 a r 10 d 111 v 1101 k 19

Example: Adaptive Huffman Coding Input: aardvark Output:000001010001000001100010101010 Symbol Code NYT 1100 a r 10 d 111 v 1101 k 20

Example: Adaptive Huffman Coding Input: aardvark Output:000001010001000001100010101010110001010 Symbol Code NYT 11000? a r 10 d 111 v 1101 k 11001?? 21

Example: Adaptive Huffman Coding Input: aardvark Output:000001010001000001100010101010110001010 Symbol Code NYT 11000? a r 10 d 111 v 1101 k 11001 ?? 22

Example: Adaptive Huffman Coding Input: aardvark Output:000001010001000001100010101010110001010 Symbol Code NYT 11100 a r 10 d 110 v 1111 k 11101 23

Adaptive Huffman Decoding Input: a Output:000001010001000001100010101010110001010 Symbol Code NYT a r d v k 1 1 24

Adaptive Huffman Decoding Input: aa Output:000001010001000001100010101010110001010 Symbol Code NYT a 1 r d v k 25

Adaptive Huffman Decoding Input: aar Output:000001010001000001100010101010110001010 Symbol Code NYT 00 a 1 r 01 d v k 26

Adaptive Huffman Decoding Input: aard Output:000001010001000001100010101010110001010 Symbol Code NYT 000 a 1 r 01 d 001 v k 27

Adaptive Huffman Decoding Input: aardv Output:000001010001000001100010101010110001010 Symbol Code NYT 0000 ? a 1 r 01 d 001 v 0001?? k 28

Adaptive Huffman Decoding Input: aardv Output:000001010001000001100010101010110001010 Symbol Code NYT 0000 a 1 r 01 d 001 v 0001?? k 29

Adaptive Huffman Decoding Input: aardv Output:000001010001000001100010101010110001010 Symbol Code NYT 0000 a 1 r 01 d 001 v 0101 ?? k 30

Adaptive Huffman Decoding Input: aardv Output:000001010001000001100010101010110001010 Symbol Code NYT 000 a 1 r 01 d 001 v ?? k 31

Adaptive Huffman Decoding Input: aardv Output:000001010001000001100010101010110001010 Symbol Code NYT 1100 a r 10 d 111 v 1101 k 32

Adaptive Huffman Decoding Input: aardva Output:000001010001000001100010101010110001010 Symbol Code NYT 1100 a r 10 d 111 v 1101 k 33

Adaptive Huffman Decoding Input: aardvar Output:000001010001000001100010101010110001010 Symbol Code NYT 1100 a r 10 d 111 v 1101 k 34

Adaptive Huffman Decoding Input: aardvar Output:000001010001000001100010101010110001010 Symbol Code NYT 1100 a r 10 d 111 v 1101 k 35

Adaptive Huffman Decoding Input: aardvark Output:000001010001000001100010101010110001010 Symbol Code NYT 11000 a r 10 d 111 v 1101 k ?? 36

Adaptive Huffman Decoding Input: aardvark Output:000001010001000001100010101010110001010 ? Symbol Code NYT 11100 a r 10 d 110 v 1111 k 11101 37

Adaptive Huffman Exercise Try to solve the following! Find the adaptive Huffman encoder (compressor) for the following text: raaaabcbaacvkl Assuming 26 alphabet set! 38

Adaptive Huffman Notes To Follow the Text Book example: If the source has an alphabet {a1,a2, …, am} of size m , then pick e and r such that m = 2e+r and 0 ≤ r <2e . The letter ak is encoded as the ﴾e+1﴿-bit corresponds to k−1, iff 1≤ k ≤2r; else, ak is encoded as (only) the e-bit binary representation of k−r−1. Example: suppose m = 26, then e = 4, and r=10. Then symbol a1 is encoded as 00000, (“a” in English) the symbol a2 is encoded as 00001, (“b” in English) and the symbol a22 is encoded as 1011 (“b” in English) 39

Adaptive Huffman Applications Lossless Image Compression Steps to have lossless image compression: Generate a Huffman code for each uncompressed image (but already quantized and compressed with lossy methods) Encode the image using the Huffman code Save it in a file again !!! The original (uncompressed) image representation uses 8 bits/pixel. The image consists of 256 rows of 256 pixels, so the uncompressed representation uses 65,536 bytes. Compression ratio → number of bytes (uncompressed): number of bytes compressed 40 40

Adaptive Huffman Applications Lossless Image Compression 41

Adaptive Huffman Applications Lossless Image Compression Image Name Bits/Pixel Total Size (B) Compression Ratio Sena 7.01 57,504 1.14 Sensin 7.49 61,430 1.07 Earth 4.94 40,534 1.62 Omaha 7.12 58,374 1.12 Huffman (Lossless JPEG) Compression Based on Pixel value 42

Adaptive Huffman Applications Lossless Image Compression Image Name Bits/Pixel Total Size (B) Compression Ratio Sena 4.02 32,968 1.99 Sensin 4.70 38,541 1.70 Earth 4.13 33,880 1.93 Omaha 6.42 52,643 1.24 Huffman Compression Based on Pixel Difference value and Two-Pass Model 43

Adaptive Huffman Applications Lossless Image Compression Image Name Bits/Pixel Total Size (B) Compression Ratio Sena 3.93 32,261 2.03 Sensin 4.63 37,896 1.73 Earth 4.82 39,504 1.66 Omaha 6.39 52,321 1.25 Huffman Compression Based on Pixel Difference Value and One-Pass Adaptive Model 44

Adaptive Huffman Applications Lossless Image Compression Image Name Bits/Pixel Total Size (B) Compression Ratio Sena 3.93 32,261 2.03 Sensin 4.63 37,896 1.73 Earth 4.82 39,504 1.66 Omaha 6.39 52,321 1.25 Huffman Compression Based on Pixel Difference Value and One-Pass Adaptive Model 45

Optimality of Huffman Codes! The necessary conditions for an optimal variable-length binary code: Condition 1: Given any two letters aj and ak, if P(aj) ≥ P(ak), then lj ≤ lk, where lj is the number of bits in the codeword for aj. Condition 2: The two least probable letters have codewords with the same maximum length lm. Condition 3: In the tree corresponding to the optimum code, there must be two branches stemming from each intermediate node. Condition 4: Suppose we change an intermediate node into a leaf node by combining all the leaves descending from it into a composite word of a reduced alphabet. Then, if the original tree was optimal for the original alphabet, the reduced tree is optimal for the reduced alphabet. 46

Minimum Variance Huffman Codes By performing the sorting procedure in a slightly different manner, we could have found a different Huffman code. 47

Huffman Coding: Self Study! 3.2.1 Minimum Variance Huffman Codes (pp. 46 – 47 {redo the examples}) 3.2.3 Length of Huffman Codes (pp. 49 ~ 51 and the example 3.2.2) 3.2.3 Huffman Codes optimality condition!! 48