2. Efficient use of spectrum Less sensitive to noise & distortions Integration of digital services Data Encryption Digital Video

3. 576 lines 5.5MHz = 720 pixels Raw image 576 lines/frame 720 pixels/line 50 fields/second 8 bit per pixel Total: 576x720x25x3 x8 = 249Mbit/sec R,G andB R: 83Mbit/sG: 83Mbit/sB: 83Mbit/sTotal: 249Mbit/s Figure 2a

4. f VSB Chroma at 4.43 MHz -1.75 MHz Sound at 6 MHz Chrominance can be represented with a considerable narrower bandwidth (resolution) than luminance

5. 576 lines 5. 5MHz = 720 pixels PAL system 576 lines/frame 720 pixels/line 50 fields/second 8 bit per pixel Total: 576x720x25x8 = 83Mbit/sec Luminance Y Y: 83Mbit/sU:V:Total: Figure 2b

6. 288 lines 2.75MHz = 360 pixels PAL system 288 lines/frame 360 pixels/line 50 fields/second 8 bit per pixel Total: 288x360x25x8 = 21Mbit/sec Chrominance U Y: 83Mbit/sU: 21Mbit/sV:Total: Figure 2c

7. 288 lines 2.75MHz = 360 pixels PAL system 288 lines/frame 360 pixels/line 50 fields/second 8 bit per pixel Total: 288x360x25x8 = 21Mbit/sec Chrominance V Y: 83Mbit/sU: 21Mbit/sV: 21Mbit/sTotal: 125Mbit/s Figure 2d

8. Medium Quality : 1.2 Mbit/s Superior Quality : 6 Mbit/s Actual size - 249Mbit/s Result: Compression is necessary U,V downsampled - 125Mbit/s

9. Redundancy in image contents Adjacent pixels are similar Intensity variations can be predicted Sequential frames are similar Lossy compression: Removal of redundant information, resulting in distortion that is insensitive to Human Perception

10. Figure 3a Lenna Pixels within this region have similar but not totally identical intensity.

11. Figure 3b

12. Intensity position

13. 012345-2-3-4-5 Autocorrelation function 0.2 0.4 0.6 0.8 1.0 Figure 4

14. Interpolation 1.Pixel intensities usually varies in a smooth manner except at edge (dominant/salient) points 2.Record pixels at dominant points only. 3.Reconstruct the pixels between dominant points with “Interpolation”. 4.A straightforward method: Joining dominant points with straight lines. 5.High compression ratio for smooth varying intensity profile. 6.Difficulty: How to identify dominant points?

15. Intensity position Figure 5a Transmit only selected pixels predicted the rest

16. Prediction of current sample based on previous ones Quantizer (Q) Predictor (P) Input signal Predicted signal Error signal Quantized error signal Reconstructed signal Quantizer: representation of a continuous dynamic range with a finite number of discrete levels (will be discussed later) Error = Quantization error

17. Function of Predictive Coding: Data Compression Quantizer (Q) Predictor (P) 8 bits The better the predictor, the higher is the compression ratio 3 bits Prediction error

18. A simple example: 6 bitsLevelxy0 0 to 2 0 1 3 to 5 3 2 6 to 8 6 3 9 to 11 9 Quantizer (Q) Predictor (P) 2 bits Quantizer

19. Predictive Decoder Quantizer (Q) Predictor (P) Reconstructed signal Reconstruction error Quantization error Option: the quantized levels are transmitted instead of the actual errors Q -1

20. Predictive DecoderLevelxy0 0 to 2 0 1 3 to 5 3 2 6 to 8 6 3 9 to 11 9 2 bits Quantizer 6 bits Error = Quantization error Quantizer (Q) Predictor (P) Q -1

21. Predictive Decoder Quantizer (Q) Predictor (P) Levelxy0positive+S 1negative-S 1 bits Quantizer 6 bits Q -1 S = Fix step size

22. Where Prediction based on the linear combination of previously reconstructed samples Current sample = Optimal predictor design by minimizing the Mean Square Prediction Error

24. Intensity position Figure 5b Y A

25. e.g. Asin(  n/T)+Y Intensity position n Figure 5d A Y

26. 1. Select a basis - a set of fixed functions {f 0 (n), f 1 (n), f 2 (n), f 3 (n), ……………, f N (n)} 2. Assuming all types of signals can be approximated by a linear combination of these functions (i.e. A(n) = a 0 f 0 (n)+ a 1 f 1 (n)+ a 2 f 2 (n)+…+ a N f N (n) 3. Calculate the coefficients a 0, a 1, ….., a N 4. Represents the input signal with the coefficients instead of the actual data 5. Compression: Use less coefficients, e.g. a 0, a 1, ….., a K (K<N) 6. For example: the set of sine and cosine waves Major Steps

27. 1. Adopt the sine and cosine waves as a basis 2. Calculate the Fourier coefficients (Note: a sequence of N points will give N complex coefficients 3. Encoding (compression): Represents the signal with the first K coefficients, where K < N 4. Decoding (decompression): Reconstruct the signal with the K coefficients with inverse Fourier Transform. 5. Other Transforms (e.g. Walsh Transform) can be adopted Sinusoidal Waves

28. Set of basis functions

29. denotes Dot Product between A and B Transform from the “s” domain to the “S” domain

31. x(0) x(1) x(2) …….. x(N-2) x(N-1) X(k)X(k) W(0,k) W(1,k) W(2,k) W(N-2,k)W(N-1,k)

32. A. Orthogonal Property Delta function B. Orthonormal Property

33. s denotes Dot Product between A and B Inverse Transform from the “S” domain to the “s” domain are complex conjugates

35. X(0) X(1) X(2) …….. X(N-2) X(N-1) x(n)x(n) W * (n,0) W * (n,1) W * (n,2) W * (n,N-2)W * (n,N-1)

36. Note: X(k) is complex

37. Note: X(k) is real

38. x(n) = W’(n,k) = cos[(2n+1)k  Note: W k is real, therefore W’ k = W k k=0 N-1 X(k)W’(n,k) C(k)C(k) 2 C(k) = 2 -0.5 for k = 0 = 1 otherwise

39. Transform that are suitable for compression should exhibit the following properties: Optimal Transform : Karhunen-Loeve Transform(KLT) a. There exist an inverse transform b. Decorrelation c. Good Energy Compactness

40. x(0), x(1), x(2), x(3), x(4), x(5), x(6), x(7), ….., x(N-2), x(N-1) A sample can be predicted from its neighbor(s)

41. X(0)X(1)X(2)X(3)X(4)X(5)X(6)X(7) After DFT, a coefficient is less predictable from its neighbor(s) Magnitude of frequency components

42. x(0), x(1), x(2), x(3), x(4), x(5), x(6), x(7), ….., x(N-2), x(N-1) All samples are important

43. x(0), x(1), x(2), x(3), x(4), x(5), x(6), x(7), ….., x(N-2), x(N-1) All samples are important Any missing sample causes large distortion

44. X(0)X(1)X(2)X(3)X(4)X(5)X(6)X(7) x(0)x(1)x(2)x(3)x(4)x(5)x(6)x(7) DFT samples

45. X(0)X(1)X(2)X(3)X(4)X(5)X(6)X(7) x(0)x(1)x(2)x(3)x(4)x(5)x(6)x(7)

46. X(0)X(1)X(2)X(3)X(4)X(5)X(6)X(7) x(0)x(1)x(2)x(3)x(4)x(5)x(6)x(7) The signal can be constructed with the first 3 samples with good approximation

47. All information is concentrated in a small number of elements in the transformed domain DCT has very good Energy Compactness and Decorrelation Properties

48. X(j,k) = m=0 M-1 x(m,n)W(m,j) W(n,k) C(j)C(j) 2 n=0 C(k)C(k) 2 W(n,k) = cos[(2n+1)k  C(k), C(j) = 2 -0.5 for k = 0 and j = 0, respectively = 1 otherwise N-1 W(m,j) = cos[(2m+1)j 

49. x(0,0)x(0,1)x(0,2)x(0,N-1) x(1,0)x(1,1)x(1,2)x(1,N-1) x(M-1,0)x(M-1,1)x(M-2,2)x(M-1,N-1) X(0,0)X(0,1)X(0,2)X(0,N-1) X(1,0)X(1,1)X(1,2)X(1,N-1) X(M-1,0)X(M-1,1)X(M-2,2)X(M-1,N-1) 2-D DCT

50. x(m,n) = j=0 M-1 X(j,k)W(m,j) W(n,k) C(j)C(j) 2 C(k)C(k) 2 W(n,k) = cos[(2n+1)k  C(k), C(j) = 2 -0.5 for k = 0 and j = 0, respectively = 1 otherwise k=0 N-1 W(m,j) = cos[(2m+1)j 

51. x(0,0)x(0,1)x(0,2)x(0,N-1) x(1,0)x(1,1)x(1,2)x(1,N-1) x(M-1,0)x(M-1,1)x(M-2,2)x(M-1,N-1) X(0,0)X(0,1)X(0,2)X(0,N-1) X(1,0)X(1,1)X(1,2)X(1,N-1) X(M-1,0)X(M-1,1)X(M-2,2)X(M-1,N-1) 2-D IDCT

52. Importance

53. Given a signal and Assume f(n) is wide-sense stationary, i.e. its statistical properties are constant with changes in time Defineand (O1) (O2) f(n), define the mean and autocorrelation as

54. (O3) Equation O1 can be rewritten as The covariance of f is given by (O4) (O5)

55. The signal is transform to its spectral coefficients Comparing the two sequences: a. Adjacent terms are related b. Every term is important a. Adjacent terms are unrelated b. Only the first few terms are important

56. The signal is transform to its spectral coefficients similar to f, we can define the mean, autocorrelation and covariance matrix for 

57. a. Adjacent terms are relateda. Adjacent terms are unrelated Adjacent terms are uncorrelated if every term is only correlated to itself, i.e., all off-diagonal terms in the autocorrelation function is zero. Define a measurement on correlation between samples: (O6)

58. We assume that the mean of the signal is zero. This can be achieved simply by subtracting the mean from f if it is non- zero. The covariance and autocorrelation matrices are the same after the mean is removed.

59. b. Every term is important b. Only the first few terms are important Note: If only the first L-1 terms are used to reconstruct the signal, we have (O7)

60. If only the first L-1 terms are used to reconstruct the signal, the error is The energy lost is given by but, hence (O8) (O9) (O10)

61. Eqn. O10 is valid for describing the approximation error of a single sequence of signal data f. A more generic description for covering a collection of signal sequences is given by: (O11) An optimal transform mininize the error term in eqn. O11. However, the solution space is enormous and constraint is required. Noted that the basis functions are orthonormal, hence the following objective function is adopted.

62. (O12) The term r is known as the Lagrangian multiplier The optimal solution can be found by setting the gradient of J to 0 for each value of r, i.e., Eqn O13 is based on the orthonormal property of the basis functions. (O13)

63. The solution for each basis function is given by (O14)  r  is an eigenvector of R f and r is an eigenvalue Grouping the N basis functions gives an overall equation (O15) R  =  R f      =  (O16) which is a diagonal matrix. The decorrelation criteria is satisfied

64. The signal is transform to its spectral coefficients Given a signal The solution for each basis function is given by Determine the autocorrelation function R f

65. Redundancy in images Probability distribution of pixel values are uneven Assuming the pixel intensity (gray scale) ranges from 0 to 255 units Figure 6a 255 0

66. Pixel Intensity Probability of occurrence 01234100252253254255 Figure 6b 0.4 0.2 0.6 0.8 1.0 Use less bits to represent pixel intensity that occurs more often

67. A simple example: 720 pixels 576 pixels 8bit per pixels Total: 3.3Mbits Image size = 720x576x8 = 3.3Mbits 8bit per pixels

68. IntensityP r Pixel Intensity 01234100252253254255 0.004 0.002 0.500 0.098 1.000 PrPr 0 - 254 0.00196 255 0.500 720 pixels 576 pixels 8bit per pixels Total: 3.3Mbits

69. Pixel Intensity 01234100252253254255 0.004 0.002 0.500 0.098 1.000 Intensity P r # of bits Bit String PrPr 0 - 254 0.00196 91XXXXXXXX 255 0.500 10 Total = (720X576)X(0.500 + 0.002X255X9) = 2.1Mbits

70. Sequential frames are similar

71. Figure 7 P1 P2 P3 Only about 5-10% of the content had been changed between frames

72. Still picture - JPEG Joint Photographic Expert Group International Standard Organization (ISO) standards. Based on Discrete Cosine Transform (DCT). Motion picture - MPEG Motion Picture Expert Group

73. Image Image Vectors DCT Quantization Zig-Zag Coding Runlength Coding Entropy Coding Digitization JPEG Compressed Format

74. Image Image Vectors DCT Quantization Zig-Zag Coding Runlength Coding Entropy Coding Digitization JPEG Compressed Format

75. Digitization Figure 8

76. Figure 9 Image Digitization

77. Image Image Vectors DCT Quantization Zig-Zag Coding Runlength Coding Entropy Coding Digitization JPEG Compressed Format

78. Figure 10a

79. Figure 10b Image vectors

80. Figure 10c Image Vector - a magnified view

81. Figure 10d

82. Image Vector - a magnified view x(0,0)x(0,1)x(0,2)x(0,3)x(0,4)x(0,5)x(0,6)x(0,7) x(2,0)x(2,1)x(2,2)x(2,3)x(2,4)x(2,5)x(2,6)x(2,7) x(3,0)x(3,1)x(3,2)x(3,3)x(3,4)x(3,5)x(3,6)x(3,7) x(4,0)x(4,1)x(4,2)x(4,3)x(4,4)x(4,5)x(4,6)x(4,7) x(1,0)x(1,1)x(1,2)x(1,3)x(1,4)x(1,5)x(1,6)x(1,7) x(5,0)x(5,1)x(5,2)x(5,3)x(5,4)x(5,5)x(5,6)x(5,7) x(6,0)x(6,1)x(6,2)x(6,3)x(6,4)x(6,5)x(6,6)x(6,7) x(7,0)x(7,1)x(7,2)x(7,3)x(7,4)x(7,5)x(7,6)x(7,7) Figure 10e

83. Image Image Vectors DCT Quantization Zig-Zag Coding Runlength Coding Entropy Coding Digitization JPEG Compressed Format

84. Increasing horizontal frequency Increasing vertical frequency Figure 11a

85. Increasing horizontal frequency Increasing vertical frequency Figure 11b Because of the energy compactness of DCT, most of the information is concentrated in the low frequency corner

86. 200185170251313 1981801601711073 16515012551211109 30258135390 21019019512071558 29103621 55727115 492119230 Figure 11c

87. The DCT coefficients are normalised to 11 bits integer values Before the transform, the pixel intensity range is converted from [0,255] to [-128, 127] The process, known as ‘zero shift’, is performed by subtracting each pixel intensity by 128

88. Image Image Vectors DCT Quantization Zig-Zag Coding Runlength Coding Entropy Coding Digitization JPEG Compressed Format

89. Quantizer f 0 d1d1 d2d2 d3d3 r1r1 r2r2 -r 2 -r 1 d4d4 -d 4 -d 3 -d 2 -d 1 r3r3 -r 3 Uniform Symmetric Quantizers Input Output d i : Decision levels r i : Representation levels

90. Mean Square Quantization Error (MSQE) Mean Absolute Quantization Error (MAQE) Q1 Q2

91. Max-Lloyd Quantizer A method to determine the decision and representation levels Suppose Then Q3

92. Max-Lloyd Quantizer Consider two arbitrary adjacent reconstruction levels r k-1 and r k What will be the optimal value for d k so that error is minimized? d k-1 dkdk d k+1 r k-1 rkrk Q4

93. Max-Lloyd Quantizer Similarly Q5

94. Max-Lloyd Quantizer for uniform pdf Consider a uniform probability density function f 0 A/2 -A/2 1/A p(f)p(f)

95. Max-Lloyd Quantizer for uniform pdf From Q4, From Q5, Hence, Constant Step Size

96. Max-Lloyd Quantizer for uniform pdf Step size (SS) Q6 Q7 Variance =

97. Max-Lloyd Quantizer for uniform pdf Q8 For a b bits quantizer, Q9 SNR =

98. 200185170251313 1981801601711073 16515012551211109 30258135390 21019019512071558 29103621 55727115 492119230 Assign different quantization step size for each coefficients Figure 12

99. Consider a range of values from, lets say 0 to 255 0 - 7000000 0 8 - 15100001 8 16 - 23200010 16 24 - 31300011 24 32 - 39400100 32 248 - 255 3111111 248 If a step size = 8 is used, the range is divided into 256/8 = 32 levels 5 bits are required to represent each level in this range Value Level Bit string Quantized value

100. Consider a range of values from, lets say 0 to 255 0 - 15000000 0 16 - 31100001 16 32 - 47200010 32 48 - 63300011 48 64 - 79400100 64 240 - 255 1611111 240 If a step size = 16 is used, the range is divided into 256/16 = 16 levels 4 bits are required to represent each level in this range Value Level Bit string Quantized value

101. 16 levels The larger the step size, the smaller the number of quantized levels the smaller the number of bits, the larger the distortion in value and the other way round

102. Human Visual System is more sensitive to low frequency intensity (spatial) variation in an image Increasing horizontal frequency Increasing vertical frequency Figure 13

103. Human Visual System (HVS) is more sensitive to low frequency intensity (spatial) variation in an image Decreasing sensitivity to HVS Figure 14

104. Assign different quantization step size for each coefficients Figure 15 11148121620 148121622 25 48121620242530 812162022283032 114812202224 1214182425303540 1016202830354043 1220253032404548 DCT coefficientsQ Step Size

105. Assign different quantization step size for each coefficients Figure 16 200185170240000 1981801601680000 16414412000000 3224000000 2101901921200000 00000000 00000000 00000000 DCT coefficients Quantized DCT coefficients

106. After Quantization, a lot of high frequency DCT coefficients are truncated to ‘0’ Non-zero coefficients carry most of the image contents and those that are sensitive to the HVS Large number of ‘0’ value coefficients suggested runlength coding

107. For a continuous stream of numbers with identical values, it is only necessary to record 1. The value of the number 2. The number of duplication A sequence of 8 bytes of raw data s = [15, 15, 15, 15, 15, 15, 15, 15] Runlength representation: [ 15, 8 ] ValueRunlength Only 2 bytes are needed to represent ‘s’

108. The longer the string of duplicated numbers, the larger the Compression Ratio (CR) Runlength representation: [ 15, 4 ] ValueRunlengthCompression Ratio = 2 Runlength representation: [ 15, 16 ] ValueRunlengthCompression Ratio = 8 s = [15, 15, 15, 15] s = [15, 15, 15, 15, 15, 15, 15, 15,15,15,15,15,15,15,15,15]

109. 200185170240000 1981801601680000 16414412000000 3224000000 2101901921200000 00000000 00000000 00000000 Runlength of ‘0’ 4 4 4 5 6 8 8 8 CR 2 2 2 2.5 3 4 4 4 Figure 17

110. The compression ratio of horizontal scanning is always less than or equal to 4 A better approach is to adopted zig-zag scanning

111. Image Image Vectors DCT Quantization Zig-Zag Coding Runlength Coding Entropy Coding Digitization JPEG Compressed Format

112. 200185170240000 1981801601680000 16414412000000 3224000000 2101901921200000 00000000 00000000 00000000 Quantized DCT coefficients Runlength of ‘0’ = 47 CR = 23.5 Figure 18

113. Image Image Vectors DCT Quantization Zig-Zag Coding Runlength Coding Entropy Coding Digitization JPEG Compressed Format

114. Probability distribution of pixel values are uneven Use less bits to represent pixel intensity that occurs more often Remember this? This can be generalised to......

115. If probability distribution of data values are uneven Less bits can be used to represent values that occurs more often and vice versa

116. In JPEG, DC and other coefficients are encoded separately Figure 19 DCT coefficients DC All other coefficients are ‘AC’ terms

117. DC coefficients of adjacent image blocks are similar. DC coefficient represents the average intensity in an image block 8 pixels

118. Differential Pulse Code Modulation (DPCM) is applied to encode the ‘Quantized’ DC terms. Consider a row of image block 200190198202205200195220225 Image blocksQuantized DC coefficients -10+8+4 +3 -5 +25+5 DPCM

119. As adjacent DC terms are similar, the DPCM values are small in general, i.e., small values occur more often The DPCM values are divided in 16 classes according to their magnitude Each class had different probability of occurence

120. ClassDPCM difference values 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 [0] [-1][+1] [-3,-2][+2,+3] [-7, -6,...., -4][+4, +5,...., +7] [-15, -14,....,-9, -8][+8, +9,....,+14, +15] [-31, -30,....,-17, -16][+16, +17,....,+30, +31] [-63, -62,....,-33, -32][+32, +33,....,+62, +63] [-127, -126,......., -64][+64,......., +126, +127] [-255, -254,....., -128][+128, +129,....., +255] [-511, -511,....., -256][+256, +257,....., +511] [-1023,..., -513, -512][+512, +513,..., +1023] [-2047,........., -1024][+1024,..........., +2047] [-4095,........., -2048][+2048,..........., +4095] [-8191,........., -4096][+4096,..........., +8191] [-16383,........, -8192][+8192,.........., +16383] [-32767,......, -16384][+16384,........, +32767]

121. Small values, that occur more often, are grouped into classes that contain fewer members A class with fewer elements(s) require less bits to identify its members As a result, small values require less bits to represent

122. Any DPCM value is addressed by its class and a string of additional bits to identify its position in the class ClassDPCM difference values 6[-63, -62,....,-33, -32][+32, +33,....,+62, +63] For example, in class 6, there are 64 members, 6 additional bits is required 000000000001011111100000111111

123. Representation of DPCM data ClassAdditional bits 4 bitsAdaptive For most DC coefficients, the DPCM values are belonged to lower classes that require less additional bits

124. Nonzero AC terms are represented in the same way as DPCM coefficients ClassAdditional bits 4 bitsAdaptive Zero terms are encoded with zig-zag scanning followed by RLC How are these two items combined?

125. 2000-1400000 00000000 10000000 00000000 00000000 00000000 00000000 00000000 Quantized DCT coefficients V 015614152728 38121725304143 911182431404453 1019233239455254 2471316262942 2022333846515560 2134374750565961 3536484957586263 Zig-zag scanning index (I) 1234567891011126263 0000-14000100000 I V 4354RL

126. ClassAC coefficient values 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 [0] [-1][+1] [-3,-2][+2,+3] [-7, -6,...., -4][+4, +5,...., +7] [-15, -14,....,-9, -8][+8, +9,....,+14, +15] [-31, -30,....,-17, -16][+16, +17,....,+30, +31] [-63, -62,....,-33, -32][+32, +33,....,+62, +63] [-127, -126,......., -64][+64,......., +126, +127] [-255, -254,....., -128][+128, +129,....., +255] [-511, -511,....., -256][+256, +257,....., +511] [-1023,..., -513, -512][+512, +513,..., +1023] [-2047,........., -1024][+1024,..........., +2047] [-4095,........., -2048][+2048,..........., +4095] [-8191,........., -4096][+4096,..........., +8191] [-16383,........, -8192][+8192,.........., +16383] [-32767,......, -16384][+16384,........, +32767]

127. 1234567891011126263 0000-14000100000 I V 354RL 4Class1 AC coefficient values 4[-15, -14,....,-9, -8][+8, +9,....,+14, +15] 00000001011110001111 4 0001

128. ClassAC coefficient values 1[-1][+1] 01 1234567891011126263 0000-14000100000 I V 354RL 4Class1 4 00011

129. RL and Class are grouped into the RUN-SIZE Table 0001020304050F N/A11121314151F N/A21222324252F N/A31323334353F N/A41424344454F N/A51525354555F N/AF1F2F3F4F5FF 012345F 0 1 2 3 4 5 F RRRR SSSS 00 - End of Block

130. Each non-zero AC coefficient is represented by an 8- bit value ‘RRRRSSSS’ RRRR is the runlength of ‘zeros’ between current and previous AC coefficients If the runlength exceeds 15, a term ‘F0’ will be inserted to represent a runlength of 16 If all remaining coefficients are zero, a term ‘00’ (EOB) is inserted. A Few Points to Note

131. Additional bits RL Class 43 41 EOB EOB : End of Block RS443100Hexadecimal RS684900Decimal 1234567891011126263 0000-14000100000 I V 354RL 4Class1 4 00011

132. 1 Additional bits 68RS49 Encoded AC format 0001 1 6849 00 Number of bits: 8 + 4 + 8 + 1 + 8 = 29bits Number of bits for the 63 AC coefficients = 63 x 11 = 693 bits 1234567891011126263 0000-14000100000 I V 354RL 4Class1 4

133. The “Baboon” is one of the popular standard images that had been adopted for comparison purpose in image compression research. The difficult part is that the large amount of texture is pretty hard to compress with good fidelity. The easy part is the distortions are difficult to spot. Hi!, I am the famous Baboon, very nice to meet all of you.