ECE 5578 Multimedia Communication
Lec 04 - Arithmetic Coding
Zhu Li Dept of CSEE, UMKC Office: FH560E, Email: [email protected], Ph: x 2346. http://l.web.umkc.edu/lizhu
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Z. Li,ECE 5578 Multimedia Communciation, 2020 p.1 Outline Lecture 04 ReCap Arithmetic Coding Tutorial for HW-1 (Anique)
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.2 JPEG Coding Block (8x8 pel) based coding DCT transform to find sparse * representation Quantization reflects human visual system Zig-Zag scan to convert 2D to 1D string Run-Level pairs to have even more = compact representation Hoffman Coding on Level Category Quant Table: Fixed on the Level with in the category
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.3 Coding of AC Coefficients
Zigzag scanning: Example
8 24 -2 0 0 0 0 0 -31 -4 6 -1 0 0 0 0 0 -12 -1 2 0 0 0 0 0 0 -2 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Example: zigzag scanning result 24 -31 0 -4 -2 0 6 -12 0 0 0 -1 -1 0 0 0 2 -2 0 0 0 0 0 -1 EOB (Run, level) representation: (0, 24), (0, -31), (1, -4), (0, -2), (1, 6), (0, -12), (3, -1), (0, -1), (3, 2), (0, -2), (5, -1), EOB
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.4 Coding of AC Coefficients
Run / Base Run / Base … Run / Base codeword Catg. codeword Catg. Codeword Cat. EOB 1010 - - … ZRL 1111 1111 001 0/1 00 1/1 1100 … 15/1 1111 1111 1111 0101 0/2 01 1/2 11011 … 15/2 1111 1111 1111 0110 0/3 100 1/3 1111001 … 15/3 1111 1111 1111 0111 0/4 1011 1/4 111110110 … 15/4 1111 1111 1111 1000 0/5 11010 1/5 11111110110 … 15/5 1111 1111 1111 1001 … … … … … … … ZRL: represent 16 zeros when number of zeros exceeds 15. Example: 20 zeros followed by -1: (ZRL), (4, -1). (Run, Level) sequence: (0, 24), (0, -31), (1, -4), …… Run/Cat. Sequence: 0/5, 0/5, 1/3, … 24 is the 24-th entry in Category 5 (0, 24): 11010 11000 -4 is the 3-th entry in Category 3 (1, -4): 1111001 011
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.5 Outline Lecture 04 ReCap Arithmetic Coding . Basic Encoding and Decoding . Uniqueness and Efficiency . Scaling and Incremental Coding . Integer Implementation About Homework-1 and Lab
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.6 Arithmetic Coding – The SciFi Story When I was in my 5th grade….
Aliens visit earth….
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.7 Huffman Coding
Replacing an input symbol with a codeword Need a probability distribution Hard to adapt to changing statistics Need to store the codeword table Minimum codeword length is 1 bit Arithmetic Coding Replace the entire input with a single floating-point number Does not need the probability distribution Adaptive coding is very easy No need to keep and send codeword table Fractional codeword length Z. Li,ECE 5578 Multimedia Communciation, 2020 p.8 Introduction
Recall table look-up decoding of Huffman code . N: alphabet size . L: Max code word length 1 . Divide [0, 2^L] into N intervals 00 . One interval for one symbol . Interval size is roughly 010 011 proportional to symbol prob. 000 010 011 100 Arithmetic coding applies this idea recursively Normalizes the range [0, 2L] to [0, 1]. Map an input sequence (multiple symbols) to a unique tag in [0, 1). abcd…..
dcba….. 0 1
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.9 Arithmetic Coding Symbol set and prob: a (0.8), b(0.02), c(0.18) Disjoint and complete partition of the range [0, 1) 0 1 [0, 0.8), [0.8, 0.82), [0.82, 1) Each interval corresponds to one symbol a b c Interval size is proportional to symbol probability
The first symbol restricts the tag 0 1 position to be in one of the intervals The reduced interval is partitioned 0 1 recursively as more symbols are processed. 0 1 Observation: once the tag falls into an interval, it never gets out of it
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.10 Some Questions to think about: Why compression is achieved this way? How to implement it efficiently? How to decode the sequence? Why is it better than Huffman code?
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.11 Example:
Symb Prob. ol 1 0.8 1 2 3 2 0.02
3 0.18 0 0.8 0.82 1.0
Map to real line range [0, 1) Order of symbols on the real line does not matter Decoder needs to use the same order • Disjoint but complete partition: • 1: [0, 0.8): 0, 0.799999…9 • 2: [0.8, 0.82): 0.8, 0.819999…9 • 3: [0.82, 1): 0.82, 0.999999…9
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.12 Concept of the encoding (not practical) Input sequence: “1321” 1 2 3 Range 1 0 0.8 0.82 1.0 1 2 3 Range 0.8 0 0.64 0.656 0.8
1 2 3 Range 0.144 0.656 0.7712 0.77408 0.8
1 2 3 Range 0.00288 0.7712 0.773504 0.7735616 0.77408 Termination: Encode the lower end or midpoint to signal the end. Difficulties: 1. Shrinking of interval requires very high precision for long sequence. 2. No output is generated until the entire sequence has been processed.
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.13 Encoder Pseudo Code
Cumulative Density Function (CDF) Probability Mass Function For continuous distribution: 0.4 x = £ = 0.2 0.2 0.2 FX (x) P(X x) ò p(x)dx -¥
For discrete distribution: 1 2 3 4 X
i F (i) = P(X £ i) = P(X = k) 1.0 X å 0.8 k =-¥ CDF P(X = i) = F (i) - F (i -1). 0.4 X X 0.2 Properties: Non-decreasing 1 2 3 4 X Piece-wise constant Each segment is closed at the lower end. Z. Li,ECE 5578 Multimedia Communciation, 2020 p.14 Encoder Pseudo Code
Keep track of LOW=0.0, HIGH=1.0; LOW, HIGH, RANGE while (not EOF) { Any two are sufficient, n = ReadSymbol(); e.g., LOW and RANGE. RANGE = HIGH - LOW; HIGH = LOW + RANGE * CDF(n); LOW = LOW + RANGE * CDF(n-1); } output LOW;
Input HIGH LOW RANGE Initial 1.0 0.0 1.0 1 0.0+1.0*0.8=0.8 0.0+1.0*0 = 0.0 0.8 3 0.0 + 0.8*1=0.8 0.0 + 0.8*0.82=0.656 0.144 2 0.656+0.144*0.82=0.77408 0.656+0.144*0.8=0.7712 0.00288 1 0.7712+0.00288*0=0.7712 0.7712+0.00288*0.8=0.773504 0.002304
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.15 Concept of the Decoding Arithmetic Decoding (conceptual) Suppose encoder encodes the lower end: 0.7712 Zoom 1 2 3 In by * CDF Decode 1 0 0.8 0.82 1.0 1 2 3 Decode 3 * CDF 0 0.64 0.656 0.8
1 2 3 Decode 2 * CDF 0.656 0.7712 0.77408 0.8
1 2 3 Decode 1 0.7712 0.773504 0.7735616 0.77408
Drawback: need to recalculate all thresholds each time. Z. Li,ECE 5578 Multimedia Communciation, 2020 p.16 Simplified Decoding (Floating Pt Ops) x - low Normalize RANGE to [0, 1) each time x ¬ range No need to recalculate the thresholds.
Receive 0.7712 1 2 3 Decode 1
x =(0.7712-0) / 0.8 0 0.8 0.82 1.0 = 0.964 1 2 3 Decode 3
0 0.8 0.82 1.0 x =(0.964-0.82) / 0.18 = 0.8 1 2 3 Decode 2 x =(0.8-0.8) / 0.02 0 0.8 0.82 1.0 = 0 Decode 1. 1 2 3 Stop.
0 0.8 0.82 1.0
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.17 Decoder Pseudo Code
Low = 0; high = 1; x = GetEncodedNumber(); While (x ≠ low) { n = DecodeOneSymbol(x); output symbol n; x = (x - CDF(n-1)) / (CDF(n) - CDF(n-1)); }; 1 00
010 011
000 010 011 100 But this method still needs high-precision floating point operations. It also needs to read all input at the beginning of the decoding.
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.18 Outline Lecture 04 ReCap Arithmetic Coding . Basic Encoding and Decoding . Uniqueness and Efficiency . Scaling and Incremental Coding About Homework-1 and Lab
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.19 Uniqueness
1 2 3 Range 0.00288 0.7712 0.773504 0.7735616 0.77408 Termination: Encode the midpoint (or lower end) to signal the end. How to represent the final tag uniquely and efficiently? Answer: Take the binary value of the tag T(X) and truncate to
(X is the sequence {x1, …, xm}, not individual symbol) é 1 ù 1 bit longer than l(X ) = êlog ú +1 bits. Shannon code ê p(X )ú 1 Proof: Assuming midpoint tag: T(X) = F(X -1) + p(X ), p(X) > 0. 2 First show the truncated code is unique, that is, code is within [F(X-1), F(X)).
1). ëT(X)û l( X ) £ T(X ) < F(X )
So ëT(X)û l( X ) is below the high end of the interval. Z. Li,ECE 5578 Multimedia Communciation, 2020 p.20 Uniqueness and Efficiency
æ é 1 ù ö æ 1 ö -ç êlog ú+1÷ -ç log +1÷ ç p( X ) ÷ ç ÷ 1 2). 2-l( X ) = 2 è ê ú ø £ 2 è p( X ) ø = p(X ) 2 1 By def, T(X) = F(X -1) + p(X ), p(X) > 0. 2 F(X) 1 T(X) - F(X -1) = p(X ) ³ 2-l(X) 2 -l( X ) Together with T(X)- êT(X)ú £ 2 ë û l( X ) T(X)
£ 2-l( X ) -l( X ) ëT(X)û l( X ) ³ F(X -1). ³ 2 ëT(X)û l( X )
F(X-1) Thus F(X -1) £ ëT(X)û l( X ) < F(X )
So the truncated code is still in the interval. This proves the uniqueness.
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.21 Uniqueness and Efficiency
F(X) Prove the code is uniquely decodable (prefix free): Any code with prefix ëT(X)û l( X ) ëT(X)û l( X ) is in 1 é 1 ö + ëT(X)û , ëT(X)û + ÷ 2l( X ) ëê l( X ) l( X ) 2l( X ) ø Need to show that this is in [F(X-1), F(X) ): ëT(X)û l( X ) We already show ëT(X)û l( X ) ³ F(X -1). F(X-1) 1 Only need to show F(X ) - ëT(X)û > l( X ) 2l( X ) p(X ) 1 F(X ) - ëT(X)û > F(X ) -T(X ) = > l( X ) 2 2l( X ) é 1 ù ëT(X)û is prefix free if l(X ) = êlog ú +1 bits. l( X ) ê p(X )ú
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.22 Uniqueness and Efficiency Efficiency of arithmetic code:
m é 1 ù X1 : {x1,..., xm } m = + l(X1 ) êlog m ú 1 bits. ê p(X1 )ú
æé 1 ù ö L = E p(X m )l(X m ) = P(X m )ç log +1 ÷ { 1 1 } å 1 çê m ú ÷ èê p(X1 )ú ø æ 1 ö £ m ç + + ÷ = m + å P(X1 )çlog m 1 1 ÷ H (X1 ) 2 è p(X1 ) ø
l(X) is the bits to code a sequence {x1,x2, …, xm}. m Assume iid sequence, H (X1 ) = mH (X ) L 2 H (X ) £ £ H (X ) + L/m H(X) for large m, stronger than prev m m results.
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.23 Uniqueness and Efficiency Comparison with Huffman code: . Expected length of Huffman code:
H (X ) £ L* £ H (X ) +1
Huffman code can reduce the overhead by jointly encoding more symbols, but needs much larger alphabet size. Arithmetic coding is more efficient for longer sequences.
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.24 Binary Arithmetic Coding
Arithmetic coding is slow in general: To decode a symbol, we need a series of decisions and multiplications: While (Tag > LOW + RANGE * Sum(n) / N - 1) { n++; }
The complexity is greatly reduced if we have only two symbols: 0 and 1. symbol 0 symbol 1
0 x 1 Only two intervals: [0, x), [x, 1)
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.25 Encoding of Binary Arithmetic Coding
HIGH ¬ LOW + RANGE ´CDF(n) LOW ¬ LOW + RANGE ´CDF(n -1) LOW = 0, HIGH = 1 Prob(0)=0.6. Sequence: 0110
LOW = 0, HIGH = 0.6 0 0.6 1
LOW = 0.36, HIGH = 0.6 0 0.36 0.6
LOW = 0.504, HIGH = 0.6 0.36 0.504 0.6
LOW = 0.504, HIGH = 0.5616 0.504 0.5616 0.6 Only need to update LOW or HIGH for each symbol.
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.26 Decoding of Binary Arithmetic Coding
Tag
0 0.6 1 General case (integer implementation): While (Tag > LOW + RANGE * Sum(n) / N - 1)) { n++; } Binary case: only one condition to check
if (Tag > LOW + RANGE * Sum(Symbol0) / N - 1) { n = 1; } else { n = 0; }
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.27 Applications of Binary Arithmetic Coding
Increasingly popular: . JBIG, JBIG2, JPEG2000, H.264 Convert non-binary signals into binary: . Golomb-Rice Code: used in H.264.
Bit-plane coding: used in JPEG2000. B = [B0, B1, ……, BK-1]: binary representation of B. Chain rule: H(B) = H(B0) +H(B1 | B0) + …… + H(BK-1 | B0, … BK-2). To code B0, needs P0(0): Prob(B0=0). To code B1, needs P1(0 | i): Prob(B1=0 | B0 = i), i = 0, 1. … … More details: . AVC Binary Adaptive Arithmetic Coding: CABAC . HEVC Arithmetic Coding:
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.28 Outline Lecture 04 ReCap Arithmetic Coding . Basic Encoding and Decoding . Uniqueness and Efficiency . Scaling and Incremental Coding About Homework-1 and Lab
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.29 Scaling and Incremental Coding
Problems of Previous examples: . Need high precision . No output is generated until the entire sequence is encoded. . Decoder needs to read all input before decoding. Key Observation: As the RANGE reduces, many MSB’s of LOW and HIGH become identical: Example: Binary form of 0.7712 and 0.773504: 0.1100010.., 0.1100011.., We can output identical MSB’s and re-scale the rest: Incremental encoding Can achieve infinite precision with finite-precision integers. Three scaling scenarios: E1, E2, E3 Important Rules: Apply as many scalings as possible before further encoding and decoding.
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.30 E1(lower half) and E2(higher half) Scaling
E1: [LOW HIGH) in [0, 0.5) 0 0.5 1.0 . LOW: 0.0xxxxxxx (binary), . HIGH: 0.0xxxxxxx.
0 0.5 1.0 Output 0, then shift left by 1 bit [0, 0.5) [0, 1): E1(x) = 2 x
E2: [LOW HIGH) in [0.5, 1) 0 0.5 1.0 LOW: 0.1xxxxxxx, HIGH: 0.1xxxxxxx.
Output 1, subtract 0.5, 0 0.5 1.0 shift left by 1 bit [0.5, 1) [0, 1): E2(x) = 2(x - 0.5) 0 0.5 1.0 The 3rd scaling, E3(mid), will be studied later: LOW < 0.5, HIGH > 0.5, but range < 0.5.
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.31 Encoding with E1 and E2 Symbol Prob. 1 0.8 Input 1 2 0.02 0 0.8 1.0 3 0.18 Input 3 0 0.656 0.8 E2: Output 1 Input 2 2(x – 0.5) 0.312 0.5424 0.54816 0.6 E2: Output 1
0.0848 0.09632 E1: 2x, Output 0
0.1696 0.19264 E1: Output 0
0.3392 0.38528 E1: Output 0
0.6784 0.77056 E2: Output 1 Input 1 Encode any value 0.3568 0.54112 in the tag, e.g., 0.5 Output 1 0.3568 0.504256 All: 1100011 (0.7734)
Z. Li,ECE 5578 Multimedia Communciation, 2020 32 To verify LOW = 0.5424 (0.10001010... in binary), HIGH = 0.54816 (0.10001100... in binary). So we can send out 10001 (0.53125) . Equivalent to E2E1E1E1E2 After left shift by 5 bits: . LOW = (0.5424 – 0.53125) x 32 = 0.3568 . HIGH = (0.54816 – 0.53125) x 32 = 0.54112 . Same as the result in the last page.
(In this example, suppose 7 bits are enough for the decoding)
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.33 Comparison with Huffman Rule: Complete all possible scaling before encoding the next symbol Symbol Prob. 1 0.8 2 0.02 Input Symbol 1 does not cause any output 3 0.18 Input Symbol 3 generates 1 bit Input Symbol 2 generates 5 bits Symbols with larger probabilities generates less number of bits. Sometimes no bit is generated at all Advantage over Huffman coding Large probabilities are desired in arithmetic coding Can use context-adaptive method to create larger probability and to improve compression ratio.
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.34 Incremental Decoding
If input is 1100011, the 1st symbol can be decoded without ambiguity after 5 bits are read. When 6 bits are read, the status is: Read 5 bits: Decode 1. After reading 6 bits: 0 0.8 1.0 Tag: 110001, 0.765625 No scaling.
0 0.656 0.8 Decode 3, E2 scaling: Shift out 1 bit, read 1 bit Tag: 100011 (0.546875) 0.312 0.5424 0.54816 0.6 Decode 2, E2 scaling Tag: 000110 (0.09375) 0.0848 0.09632 E1: Tag: 001100 (0.1875)
0.1696 0.19264 E1: Tag: 011000 (0.375)
0.3392 0.38528 E1: Tag: 110000 (0.75)
0.6784 0.77056 E2: Tag: 100000 (0.5)
0.3568 0.54112 Decode 1
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.35 Encoding Pseudo Code with E1, E2
Rule: Complete all possible scalings Symbol Prob. CDF before further decoding. Adjust LOW, 1 0.8 0.8 HIGH and Tag together. 2 0.02 0.82 (For floating-point implementation) 3 0.18 1 EncodeSymbol(n) { //Update variables RANGE = HIGH - LOW; HIGH = LOW + RANGE * CDF(n); LOW = LOW + RANGE * CDF(n-1);
//Apply all possible scalings before encoding the //next symbol while LOW, HIGH in [0, 0.5) or [0.5, 1) { Output 0 for E1 and 1 for E2 scale LOW, HIGH by E1 or E2 rule } }
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.36 Decoding Pseudo Code with E1, E2
(For floating-point implementation) DecodeSymbol(Tag) { RANGE = HIGH - LOW; n = 1; While ( (tag - LOW) / RANGE >= CDF(n) ) { n++; } Symbol Prob. CDF HIGH = LOW + RANGE * CDF(n); 1 0.8 0.8 LOW LOW RANGE CDF n = + * ( -1); 2 0.02 0.82
//keep scaling before decoding next symbol 3 0.18 1 while LOW, HIGH in [0, 0.5) or [0.5, 1) { scale LOW, HIGH by E1 or E2 rule read one more bit, update Tag } return n; }
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.37 E3 Scaling: [0.25, 0.75)[0, 1)
If RANGE straddles 1/2, E1 and E2 cannot be applied, but the range can be quite small Example: LOW=0.4999, HIGH=0.5001 Binary: LOW=0.01111…., HIGH=0.1000… We may not have enough bits to represent the interval.
0.75 0 0.25 0.5 1 E3 Scaling: [0.25, 0.75) [0, 1): 0.5 E3(x) = 2(x - 0.25) 0 1
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.38 Example Previous example without E3: Input 1 0 0.8 1.0 Input 3 0 0.656 0.8 (E2: Output 1) Input 2 0.312 0.5424 0.54816 0.6 E2: Output 1
0.0848 0.09632 E1: Output 0 0.1696 0.19264 With E3: 0.312 0.6
0.124 0 . 5 8 4 8 0 . 5 9 6 3 2 0.7 E3: (x-0.25)x2 Input 2 E2: Output 1 0.1696 0.19264 state after E2°E3 = state after E1°E2, but outputs are different… Z. Li,ECE 5578 Multimedia Communciation, 2020 39 Encoding Operation with E3
Without E3: Input 2 0.312 0.5424 0.54816 0.6 E2: Output 1
0.0848 0.09632 E1: Output 0
0.1696 0.19264 With E3: 0.312 0.6
0.124 0 . 5 8 4 8 0 . 5 9 6 3 2 0.7 E3 (no output) Input 2 E2: Output 1 0.1696 0.19264 Output 0 here! Don’t send anything when E3 is used, but send a 0 after E2: Same output, same final state Equivalent operations
Z. Li,ECE 5578 Multimedia Communciation, 2020 40 Decoding for E3 Input 1100011 Read 6 bits: Tag: 110001 (0.765625) 0 Without E3: 0.8 1.0 Decode 1
0 0.656 0.8 Decode 3, E2 scaling Tag: 100011 (0.546875)
0.312 0.5424 0.54816 0.6 Decode 2, E2 scaling Tag: 000110 (0.09375) 0.0848 0.09632 E1: Tag: 001100 (0.1875) 0.1696 0.19264 With E3: 0.312 0.6 Tag: 100011 (0.546875)
0.124 0 . 5 8 4 8 0 . 5 9 6 3 2 0.7 E3: Tag: 100110 (0.59375) Decode 2, E2 scaling Tag: 001100 (0.1875) 0.1696 0.19264 Apply E3 whenever it is possible, everything else is same.
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.41 Summary of Different Scalings
0 0.25 0.5 0.75 1.0 Need E1 scaling
Need E2 scaling
Need E3 scaling
No scaling is required. Continue to encode/decode the next symbol.
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.42 Tutorial for HW-1 Info Theory: Entropy, Conditional Entropy, Mutual Info, Relative Entropy (KL Divergence) Hoffman Coding Residual Image Error statistics and Golomb coding . Pixel value prediction filtering . Residual error distribution modeling . Optimal m selection in Golomb coding
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.43 Summary Arithmetic coding maps the sequence symbol prob into an interval on the real line the coding length l(X) is at most H(X) + 2 , for seq X=[x1,x2,...xn]
é 1 ù l(X ) = êlog ú +1 bits. ê p(X )ú
if have n symbols in the seq, we can achieve H(xk) + 2/n per symbol rate Implementation: Scaling to output bits on the go. Applications: after quantization and golomb coding, use AC.
Z. Li,ECE 5578 Multimedia Communciation, 2020 p.44