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4. Quantization and Data Compression
ECE 302 Spring 2012 Purdue University, School of ECE
Prof. Ilya Pollak
What is data compression? • Reducing the file size without compromising the
quality of the data stored in the file too much (lossy compression) or at all (lossless compression).
• With compression, you can fit higher-quality data (e.g., higher-resolution pictures or video) into a file of the same size as required for lower-quality uncompressed data.
Ilya Pollak
Why data compression?
• Our appetite for data (high-resolution pictures, HD video, audio, documents, etc) seems to always significantly outpace hardware capabilities for storage and transmission.
Ilya Pollak
Data compression: Step 0 • If the data is continuous-time (e.g., audio) or
continuous-space (e.g., picture), it first needs to be discretized.
Ilya Pollak
Data compression: Step 0 • If the data is continuous-time (e.g., audio) or
continuous-space (e.g., picture), it first needs to be discretized.
• Sampling is typically done nowadays during signal acquisition (e.g., digital camera for pictures or audio recording equipment for music and speech).
Ilya Pollak
Data compression: Step 0 • If the data is continuous-time (e.g., audio) or
continuous-space (e.g., picture), it first needs to be discretized.
• Sampling is typically done nowadays during signal acquisition (e.g., digital camera for pictures or audio recording equipment for music and speech).
• We will not study sampling. It is studied in ECE 301, ECE 438, and ECE 440.
• We will consider compressing discrete-time or discrete-space data.
Ilya Pollak
Example: compression of grayscale images
• An eight-bit grayscale image is a rectangular array of integers between 0 (black) and 255 (white).
• Each site in the array is called a pixel.
Ilya Pollak
Example: compression of grayscale images
• An eight-bit grayscale image is a rectangular array of integers between 0 (black) and 255 (white).
• Each site in the array is called a pixel. • It takes one byte (eight bits) to store one pixel value,
since it can be any number between 0 and 255.
Ilya Pollak
Example: compression of grayscale images
• An eight-bit grayscale image is a rectangular array of integers between 0 (black) and 255 (white).
• Each site in the array is called a pixel. • It takes one byte (eight bits) to store one pixel value,
since it can be any number between 0 and 255. • It would take 25 bytes to store a 5x5 image.
Ilya Pollak
Example: compression of grayscale images
• An eight-bit grayscale image is a rectangular array of integers between 0 (black) and 255 (white).
• Each site in the array is called a pixel. • It takes one byte (eight bits) to store one pixel value,
since it can be any number between 0 and 255. • It would take 25 bytes to store a 5x5 image. • Can we do better?
Ilya Pollak
Example: compression of grayscale images
255 255 255 255 255
255 255 255 255 255
200 200 200 200 200
200 200 200 200 200
200 200 200 200 100
Can we do better than 25 bytes?
Ilya Pollak
Two key ideas • Idea #1:
– Transform the data to create lots of zeros.
Ilya Pollak
Two key ideas • Idea #1:
– Transform the data to create lots of zeros. For example, we could rasterize the image, compute the differences, and store the top left value along with the 24 differences [in reality, other transforms are used, but they work in a similar fashion]
Ilya Pollak
Two key ideas • Idea #1:
– Transform the data to create lots of zeros. For example, we could rasterize the image, compute the differences, and store the top left value along with the 24 differences [in reality, other transforms are used, but they work in a similar fashion]:
– 255,0,0,0,0,0,0,0,0,0,−55,0,0,0,0,0,0,0,0,0,0,0,0,0,−100
Ilya Pollak
Two key ideas • Idea #1:
– Transform the data to create lots of zeros. For example, we could rasterize the image, compute the differences, and store the top left value along with the 24 differences [in reality, other transforms are used, but they work in a similar fashion]:
– 255,0,0,0,0,0,0,0,0,0,−55,0,0,0,0,0,0,0,0,0,0,0,0,0,−100 – This seems to make things worse: now the numbers can
range from −255 to 255, and therefore we need two bytes per pixel!
Ilya Pollak
Two key ideas • Idea #1:
– Transform the data to create lots of zeros. For example, we could rasterize the image, compute the differences, and store the top left value along with the 24 differences [in reality, other transforms are used, but they work in a similar fashion]:
– 255,0,0,0,0,0,0,0,0,0,−55,0,0,0,0,0,0,0,0,0,0,0,0,0,−100 – This seems to make things worse: now the numbers can
range from −255 to 255, and therefore we need two bytes per pixel!
• Idea #2: – when encoding the data, spend fewer bits on frequently
occurring numbers and more bits on rare numbers.
Ilya Pollak
Entropy coding
value of X 0 255 −55 −100 probability 22/25 1/25 1/25 1/25
Suppose we are encoding realizations of a discrete random variable X such that
Ilya Pollak
Entropy coding
value of X 0 255 −55 −100 probability 22/25 1/25 1/25 1/25
value of X 0 255 −55 −100 codeword 00 01 10 11
Suppose we are encoding realizations of a discrete random variable X such that
Consider the following fixed-length encoder:
Ilya Pollak
Entropy coding
value of X 0 255 −55 −100 probability 22/25 1/25 1/25 1/25
value of X 0 255 −55 −100 codeword 00 01 10 11
Suppose we are encoding realizations of a discrete random variable X such that
Consider the following fixed-length encoder:
For a file with 25 numbers, E[file size] = 25*2*(22/25+1/25+1/25+1/25) = 50 bits
Ilya Pollak
Entropy coding
value of X 0 255 −55 −100 probability 22/25 1/25 1/25 1/25
value of X 0 255 −55 −100 codeword 00 01 10 11
Suppose we are encoding realizations of a discrete random variable X such that
Consider the following fixed-length encoder:
For a file with 25 numbers, E[file size] = 25*2*(22/25+1/25+1/25+1/25) = 50 bits
value of X 0 255 −55 −100 codeword 1 01 000 001
Now consider the following encoder:
Ilya Pollak
Entropy coding
value of X 0 255 −55 −100 probability 22/25 1/25 1/25 1/25
value of X 0 255 −55 −100 codeword 00 01 10 11
Suppose we are encoding realizations of a discrete random variable X such that
Consider the following fixed-length encoder:
For a file with 25 numbers, E[file size] = 25*2*(22/25+1/25+1/25+1/25) = 50 bits
value of X 0 255 −55 −100 codeword 1 01 000 001
Now consider the following encoder:
For a file with 25 numbers, E[file size] = 25(22/25 + 2/25 + 3/25 + 3/25) = 30 bits!
Ilya Pollak
Entropy coding • A similar encoding scheme can be devised for a
random variable of pixel differences which takes values between −255 and 255, to result in a smaller average file size than two bytes per pixel.
Ilya Pollak
Entropy coding • A similar encoding scheme can be devised for a
random variable of pixel differences which takes values between −255 and 255, to result in a smaller average file size than two bytes per pixel.
• Another commonly used idea: run-length coding. I.e., instead of encoding each 0 individually, encode the length of each string of zeros.
Ilya Pollak
Back to the four-symbol example
value of X 0 255 −55 −100 probability 22/25 1/25 1/25 1/25 codeword 1 01 000 001
Can we do even better than 30 bits?
Ilya Pollak
Back to the four-symbol example
value of X 0 255 −55 −100 probability 22/25 1/25 1/25 1/25 codeword 1 01 000 001
Can we do even better than 30 bits? What about this alternative encoder?
value of X 0 255 −55 −100 probability 22/25 1/25 1/25 1/25 codeword 0 01 1 10
Ilya Pollak
Back to the four-symbol example
value of X 0 255 −55 −100 probability 22/25 1/25 1/25 1/25 codeword 1 01 000 001
Can we do even better than 30 bits? What about this alternative encoder?
value of X 0 255 −55 −100 probability 22/25 1/25 1/25 1/25 codeword 0 01 1 10
E[file size] = 25(22/25 + 2/25 + 1/25+2/25) = 27 bits
Ilya Pollak
Back to the four-symbol example
value of X 0 255 −55 −100 probability 22/25 1/25 1/25 1/25 codeword 1 01 000 001
Can we do even better than 30 bits? What about this alternative encoder?
value of X 0 255 −55 −100 probability 22/25 1/25 1/25 1/25 codeword 0 01 1 10
E[file size] = 25(22/25 + 2/25 + 1/25+2/25) = 27 bits Is there anything wrong with this encoder?
Ilya Pollak
The second encoding is not uniquely decodable!
value of X 0 255 −55 −100 probability 22/25 1/25 1/25 1/25 codeword 0 01 1 10
Encoded string ‘01’ could either be 255 or 0 followed by −55
Ilya Pollak
The second encoding is not uniquely decodable!
value of X 0 255 −55 −100 probability 22/25 1/25 1/25 1/25 codeword 0 01 1 10
Encoded string ‘01’ could either be 255 or 0 followed by −55
Therefore, this code is unusable! It turns out that the first code is uniquely decodable.
Ilya Pollak
What kinds of distributions are amenable to entropy coding?
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
a b c d0
0.1
0.2
0.3
a b c d
Cannot do better than two bits per symbol
Can do a lot better than two bits per symbol
Ilya Pollak
What kinds of distributions are amenable to entropy coding?
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
a b c d0
0.1
0.2
0.3
a b c d
Cannot do better than two bits per symbol
Can do a lot better than two bits per symbol
Conclusion: the transform procedure should be such that the numbers fed into the entropy coder have a highly concentrated histogram (a few very likely values, most values unlikely).
Ilya Pollak
What kinds of distributions are amenable to entropy coding?
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
a b c d0
0.1
0.2
0.3
a b c d
Cannot do better than two bits per symbol
Can do a lot better than two bits per symbol
Conclusion: the transform procedure should be such that the numbers fed into the entropy coder have a highly concentrated histogram (a few very likely values, most values unlikely). Also, if we are encoding each number individually, they should be independent or approximately independent.
Ilya Pollak
What if we are willing to lose some information?
253 253 255 254 255
254 254 254 255 254
252 255 255 254 252
253 253 254 254 254
252 255 253 252 253
Ilya Pollak
What if we are willing to lose some information?
253 253 255 254 255
254 254 254 255 254
252 255 255 254 252
253 253 254 254 254
252 255 253 252 253
Quantization
Ilya Pollak
253.5 253.5 253.5 253.5 253.5
253.5 253.5 253.5 253.5 253.5
253.5 253.5 253.5 253.5 253.5
253.5 253.5 253.5 253.5 253.5
253.5 253.5 253.5 253.5 253.5
Some eight-bit images
The five stripes contain random values from (left to right): {252,253,254,255}, {188,189,190,191}, {125,126,127,128}, {61,62,63,64}, {0,1,2,3}.
The five stripes contain random integers from (left to right): {240,…,255}, {176,…,191}, {113,…,128}, {49,…,64 }, {0,…,15}.
Ilya Pollak
Converting continuous-valued to discrete-valued signals
• Many real-world signals are continuous-valued. – audio signal a(t): both the time argument t and the intensity value
a(t) are continuous; – image u(x,y): both the spatial location (x,y) and the image
intensity value u(x,y) are continuous; – video v(x,y,t): x,y,t, and v(x,y,t) are all continuous.
Ilya Pollak
Converting continuous-valued to discrete-valued signals
• Many real-world signals are continuous-valued. – audio signal a(t): both the time argument t and the intensity value
a(t) are continuous; – image u(x,y): both the spatial location (x,y) and the image
intensity value u(x,y) are continuous; – video v(x,y,t): x,y,t, and v(x,y,t) are all continuous.
• Discretizing the argument values t, x, and y (or sampling), is studied in ECE 301, 438, and 440.
Ilya Pollak
Converting continuous-valued to discrete-valued signals
• Many real-world signals are continuous-valued. – audio signal a(t): both the time argument t and the intensity value
a(t) are continuous; – image u(x,y): both the spatial location (x,y) and the image
intensity value u(x,y) are continuous; – video v(x,y,t): x,y,t, and v(x,y,t) are all continuous.
• Discretizing the argument values t, x, and y (or sampling), is studied in ECE 301, 438, and 440.
• However, in addition to descretizing the argument values, the signal values must be discretized as well in order to be digitally stored.
Ilya Pollak
Quantization • Digitizing a continuous-valued signal into a discrete and
finite set of values. • Converting a discrete-valued signal into another discrete
-valued signal, with fewer possible discrete values.
Ilya Pollak
How to compare two quantizers?
Ilya Pollak
• Suppose data X(1),…,X(N) is quantized using two quantizers, to result in Y1(1),…,Y1(N) and Y2(1),…,Y2(N).
• Suppose both Y1(1),…,Y1(N) and Y2(1),…,Y2(N) can be encoded with the same number of bits.
• Which quantization is better? • The one that results in less distortion. But how to measure distortion?
– In general, measuring and modeling perceptual image similarity and similarity of audio are open research problems.
– Some useful things are known about human audio and visual systems that inform the design of quantizers.
Sensitivity of the Human Visual System to Contrast Changes, as a
Function of Frequency
Ilya Pollak
Sensitivity of the Human Visual System to Contrast Changes, as a
Function of Frequency
[From Mannos-Sakrison IEEE-IT 1974]
Ilya Pollak
Sensitivity of the Human Visual System to Contrast Changes, as a
Function of Frequency
High and low frequencies may be quantized more coarsely
[From Mannos-Sakrison IEEE-IT 1974]
Ilya Pollak
But there are many other intricacies in the way human
visual system computes similarity…
Ilya Pollak
Are these two images similar?
Ilya Pollak
What about these two?
Ilya Pollak
What about these two?
• Performance assessment of compression algorithms and quantizers is complicated, because measuring image fidelity is complicated. • Often, very simple distortion measures are used such as mean-square error.
Ilya Pollak
255
Scalar vs Vector Quantization
r 0 127
• quantize each value separately • simple thresholding
• quantize several values jointly • more complex
255 s
r 0 95 255
95 127
s 255
Ilya Pollak
255
What kinds of joint distributions are amenable to scalar quantization?
r 0 127
If (r,s) are jointly uniform over green square (or, more generally, independent), knowing r does not tell us anything about s. Best thing to do: make quantization decisions independently.
127
s 255
Ilya Pollak
255
What kinds of joint distributions are amenable to scalar quantization?
r 0 127
If (r,s) are jointly uniform over yellow region, knowing r tells us a lot about s.
Best thing to do: make quantization decisions jointly.
255 s
r 0 95 255
95 127
s 255
Ilya Pollak
If (r,s) are jointly uniform over green square (or, more generally, independent), knowing r does not tell us anything about s. Best thing to do: make quantization decisions independently.
255
What kinds of joint distributions are amenable to scalar quantization?
r 0 127
If (r,s) are jointly uniform over yellow region, knowing r tells us a lot about s.
Best thing to do: make quantization decisions jointly.
255 s
r 0 95 255
95 127
s 255
Conclusion: if the data is transformed before quantization, the transform procedure should be such that the coefficients fed into the quantizer are independent (or at least uncorrelated, or almost uncorrelated), in order to enable the simpler scalar quantization.
Ilya Pollak
If (r,s) are jointly uniform over green square (or, more generally, independent), knowing r does not tell us anything about s. Best thing to do: make quantization decisions independently.
• Does it make sense to do scalar quantization with different quantization bins for different variables?
More on Scalar Quantization
255 r 0 127
127
s 255
Ilya Pollak
• Does it make sense to do scalar quantization with different quantization bins for different variables? – No reason to do this if we are
quantizing grayscale pixel values.
More on Scalar Quantization
255 r 0 127
127
s 255
Ilya Pollak
• Does it make sense to do scalar quantization with different quantization bins for different variables? – No reason to do this if we are
quantizing grayscale pixel values. – However, if we can decompose the
image into components that are less perceptually important and more perceptually important, we should use larger quantization bins for the less important components.
More on Scalar Quantization
255 r 0 127
127
s 255
Ilya Pollak
Structure of a Typical Lossy Compression Algorithm for Audio,
Images, or Video
transform quantization entropy coding
compressed bitstream data
Ilya Pollak
Structure of a Typical Lossy Compression Algorithm for Audio,
Images, or Video
transform quantization entropy coding
compressed bitstream data
Let’s more closely consider quantization and entropy coding. (Various transforms are considered in ECE 301 and ECE 438.)
Ilya Pollak
Quantization: problem statement
Ilya Pollak
Source (e.g., image, video, speech signal)
Sequence of discrete or continuous random variables X(1),…,X(N) (e.g., transformed image pixel values).
Quantization: problem statement
Ilya Pollak
Source (e.g., image, video, speech signal) Quantizer
Sequence of discrete or continuous random variables X(1),…,X(N) (e.g., transformed image pixel values).
Sequence of discrete random variables Y(1),…,Y(N), each distributed over a finite set of values (quantization levels)
Quantization: problem statement
Errors: D(1),…,D(N) where D(n) = X(n) − Y(n)
Ilya Pollak
Source (e.g., image, video, speech signal) Quantizer
Sequence of discrete or continuous random variables X(1),…,X(N) (e.g., transformed image pixel values).
Sequence of discrete random variables Y(1),…,Y(N), each distributed over a finite set of values (quantization levels)
MSE is a widely used measure of distortion of quantizers
Ilya Pollak
• Suppose data X(1),…,X(N) are quantized, to result in Y(1),…,Y(N).
E X(n) −Y (n)( )2
n=1
N
∑⎡⎣⎢
⎤⎦⎥= E D(n)( )2
n=1
N
∑⎡⎣⎢
⎤⎦⎥
If D(1),...,D(N ) are identically distributed, this is the same as NE D(n)( )2⎡⎣ ⎤⎦, for any n.
Scalar uniform quantization
• Use quantization intervals (bins) of equal size [x1,x2), [x2,x3),…[xL,xL+1].
• Quantization levels q1, q2,…, qL. • Each quantization level is in the middle of
the corresponding quantization bin: qk=(xk+xk+1)/2.
Ilya Pollak
Scalar uniform quantization
• Use quantization intervals (bins) of equal size [x1,x2), [x2,x3),…[xL,xL+1].
• Quantization levels q1, q2,…, qL. • Each quantization level is in the middle of
the corresponding quantization bin: qk=(xk+xk+1)/2.
• If quantizer input X is in [xk,xk+1), the corresponding quantized value is Y = qk.
Ilya Pollak
Uniform vs non-uniform quantization
• Uniform quantization is not a good strategy for distributions which significantly differ from uniform.
Ilya Pollak
Uniform vs non-uniform quantization
• Uniform quantization is not a good strategy for distributions which significantly differ from uniform.
• If the distribution is non-uniform, it is better to spend more quantization levels on more probable parts of the distribution and fewer quantization levels on less probable parts.
Ilya Pollak
Scalar Lloyd-Max quantizer • X = source random variable with a known distribution. We assume it to be a
continuous r.v. with PDF fX(x)>0.
Ilya Pollak
Scalar Lloyd-Max quantizer • X = source random variable with a known distribution. We assume it to be a
continuous r.v. with PDF fX(x)>0. – The results can be extended to discrete or mixed random variables, and to
continuous random variables whose density can be zero for some x.
Ilya Pollak
Scalar Lloyd-Max quantizer • X = source random variable with a known distribution. We assume it to be a
continuous r.v. with PDF fX(x)>0. – The results can be extended to discrete or mixed random variables, and to
continuous random variables whose density can be zero for some x. • Quantization intervals (x1,x2), [x2,x3),…[xL,xL+1) and levels q1, …, qL such that
– x1 = −∞ – xL+1 = ∞ –
Ilya Pollak
−∞ < q1 < x2 ≤ q2 < x3 ≤ q3 <… ≤ qL−1 < xL ≤ qL < +∞I.e., qk ∈k-th quantization interval
Scalar Lloyd-Max quantizer • X = source random variable with a known distribution. We assume it to be a
continuous r.v. with PDF fX(x)>0. – The results can be extended to discrete or mixed random variables, and to
continuous random variables whose density can be zero for some x. • Quantization intervals (x1,x2), [x2,x3),…[xL,xL+1) and levels q1, …, qL such that
– x1 = −∞ – xL+1 = ∞ –
• Y = the result of quantizing X, a discrete random variable with L possible outcomes, q1, q2,…, qL, defined by
Ilya Pollak
Y = Y (X) =
q1 if X < x2
q2 if x2 ≤ X < x3
qL−1 if xL−1 ≤ X < xLqL X ≥ xL
⎧
⎨
⎪⎪⎪
⎩
⎪⎪⎪
−∞ < q1 < x2 ≤ q2 < x3 ≤ q3 <… ≤ qL−1 < xL ≤ qL < +∞I.e., qk ∈k-th quantization interval
Scalar Lloyd-Max quantizer: goal
• Given the pdf fX(x) of the source r.v. X and the desired number L of quantization levels, find the quantization interval endpoints x2,…,xL and quantization levels q1,…, qL to minimize the mean-square error, E[(Y−X)2].
Ilya Pollak
Scalar Lloyd-Max quantizer: goal
• Given the pdf fX(x) of the source r.v. X and the desired number L of quantization levels, find the quantization interval endpoints x2,…,xL and quantization levels q1,…, qL to minimize the mean-square error, E[(Y−X)2].
• To do this, express the mean-square error in terms of the quantization interval endpoints and quantization levels, and find the minimum (or minima) through differentiation.
Ilya Pollak
Scalar Lloyd-Max quantizer: derivation
Ilya Pollak
E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞
∞
∫
Ilya Pollak
E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞
∞
∫ = y(x) − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑
Scalar Lloyd-Max quantizer: derivation
Ilya Pollak
E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞
∞
∫ = y(x) − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑ = qk − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑
Scalar Lloyd-Max quantizer: derivation
Ilya Pollak
E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞
∞
∫ = y(x) − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑ = qk − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑
Minimize w.r.t. qk : ∂∂qk
E Y − X( )2⎡⎣ ⎤⎦ = 2 qk − x( ) fX (x)dxxk
xk+1
∫ = 0
Scalar Lloyd-Max quantizer: derivation
Ilya Pollak
E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞
∞
∫ = y(x) − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑ = qk − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑
Minimize w.r.t. qk : ∂∂qk
E Y − X( )2⎡⎣ ⎤⎦ = 2 qk − x( ) fX (x)dxxk
xk+1
∫ = 0
qk fX (x)dxxk
xk+1
∫ = xfX (x)dxxk
xk+1
∫
Scalar Lloyd-Max quantizer: derivation
Ilya Pollak
E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞
∞
∫ = y(x) − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑ = qk − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑
Minimize w.r.t. qk : ∂∂qk
E Y − X( )2⎡⎣ ⎤⎦ = 2 qk − x( ) fX (x)dxxk
xk+1
∫ = 0
qk fX (x)dxxk
xk+1
∫ = xfX (x)dxxk
xk+1
∫ , therefore qk =xfX (x)dx
xk
xk+1
∫
fX (x)dxxk
xk+1
∫
Scalar Lloyd-Max quantizer: derivation
Ilya Pollak
E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞
∞
∫ = y(x) − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑ = qk − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑
Minimize w.r.t. qk : ∂∂qk
E Y − X( )2⎡⎣ ⎤⎦ = 2 qk − x( ) fX (x)dxxk
xk+1
∫ = 0
qk fX (x)dxxk
xk+1
∫ = xfX (x)dxxk
xk+1
∫ , therefore qk =xfX (x)dx
xk
xk+1
∫
fX (x)dxxk
xk+1
∫= E X | X ∈k-th quantization interval[ ]
Scalar Lloyd-Max quantizer: derivation
Ilya Pollak
E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞
∞
∫ = y(x) − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑ = qk − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑
Minimize w.r.t. qk : ∂∂qk
E Y − X( )2⎡⎣ ⎤⎦ = 2 qk − x( ) fX (x)dxxk
xk+1
∫ = 0
qk fX (x)dxxk
xk+1
∫ = xfX (x)dxxk
xk+1
∫ , therefore qk =xfX (x)dx
xk
xk+1
∫
fX (x)dxxk
xk+1
∫= E X | X ∈k-th quantization interval[ ]
This is a minimum, since ∂2
∂qk2 E Y − X( )2⎡⎣ ⎤⎦ = 2 fX (x)dx
xk
xk+1
∫ > 0.
Scalar Lloyd-Max quantizer: derivation
Ilya Pollak
E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞
∞
∫ = y(x) − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑ = qk − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑
Minimize w.r.t. xk , for k = 2,…,L
Scalar Lloyd-Max quantizer: derivation
Ilya Pollak
E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞
∞
∫ = y(x) − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑ = qk − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑
Minimize w.r.t. xk , for k = 2,…,L:
∂∂xk
E Y − X( )2⎡⎣ ⎤⎦ =∂∂xk
qk−1 − x( )2 fX (x)dxxk−1
xk
∫ + qk − x( )2 fX (x)dxxk
xk+1
∫⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
Scalar Lloyd-Max quantizer: derivation
Ilya Pollak
E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞
∞
∫ = y(x) − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑ = qk − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑
Minimize w.r.t. xk , for k = 2,…,L:
∂∂xk
E Y − X( )2⎡⎣ ⎤⎦ =∂∂xk
qk−1 − x( )2 fX (x)dxxk−1
xk
∫ + qk − x( )2 fX (x)dxxk
xk+1
∫⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
= qk−1 − xk( )2 fX (xk ) − qk − xk( )2 fX (xk )
Scalar Lloyd-Max quantizer: derivation
Ilya Pollak
E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞
∞
∫ = y(x) − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑ = qk − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑
Minimize w.r.t. xk , for k = 2,…,L:
∂∂xk
E Y − X( )2⎡⎣ ⎤⎦ =∂∂xk
qk−1 − x( )2 fX (x)dxxk−1
xk
∫ + qk − x( )2 fX (x)dxxk
xk+1
∫⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
= qk−1 − xk( )2 fX (xk ) − qk − xk( )2 fX (xk ) = qk−1 − qk( ) qk−1 + qk − 2xk( ) fX (xk ) = 0.By assumption, fX (x) ≠ 0 and qk−1 ≠ qk .
Scalar Lloyd-Max quantizer: derivation
Ilya Pollak
E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞
∞
∫ = y(x) − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑ = qk − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑
Minimize w.r.t. xk , for k = 2,…,L:
∂∂xk
E Y − X( )2⎡⎣ ⎤⎦ =∂∂xk
qk−1 − x( )2 fX (x)dxxk−1
xk
∫ + qk − x( )2 fX (x)dxxk
xk+1
∫⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
= qk−1 − xk( )2 fX (xk ) − qk − xk( )2 fX (xk ) = qk−1 − qk( ) qk−1 + qk − 2xk( ) fX (xk ) = 0.By assumption, fX (x) ≠ 0 and qk−1 ≠ qk . Therefore,
xk =qk−1 + qk
2, for k = 2,…,L.
Scalar Lloyd-Max quantizer: derivation
Ilya Pollak
E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞
∞
∫ = y(x) − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑ = qk − x( )2 fX (x)dxxk
xk+1
∫k=1
L
∑
Minimize w.r.t. xk , for k = 2,…,L:
∂∂xk
E Y − X( )2⎡⎣ ⎤⎦ =∂∂xk
qk−1 − x( )2 fX (x)dxxk−1
xk
∫ + qk − x( )2 fX (x)dxxk
xk+1
∫⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
= qk−1 − xk( )2 fX (xk ) − qk − xk( )2 fX (xk ) = qk−1 − qk( ) qk−1 + qk − 2xk( ) fX (xk ) = 0.By assumption, fX (x) ≠ 0 and qk−1 ≠ qk . Therefore,
xk =qk−1 + qk
2, for k = 2,…,L.
This is a minimum, since ∂2
∂xk2 E Y − X( )2⎡⎣ ⎤⎦ = 2 qk − qk−1( ) fX (xk ) > 0.
Scalar Lloyd-Max quantizer: derivation
Nonlinear system to be solved
Ilya Pollak
qk =xfX (x)dx
xk
xk+1
∫
fX (x)dxxk
xk+1
∫= E X | X ∈k-th quantization interval[ ], for k = 1,…,L
xk =qk−1 + qk
2, for k = 2,…,L
⎧
⎨
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
Nonlinear system to be solved
Ilya Pollak
qk =xfX (x)dx
xk
xk+1
∫
fX (x)dxxk
xk+1
∫= E X | X ∈k-th quantization interval[ ], for k = 1,…,L
xk =qk−1 + qk
2, for k = 2,…,L
⎧
⎨
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
• Closed-form solution can be found only for very simple PDFs. – E.g., if X is uniform, then Lloyd-Max quantizer = uniform quantizer.
Nonlinear system to be solved
Ilya Pollak
qk =xfX (x)dx
xk
xk+1
∫
fX (x)dxxk
xk+1
∫= E X | X ∈k-th quantization interval[ ], for k = 1,…,L
xk =qk−1 + qk
2, for k = 2,…,L
⎧
⎨
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
• Closed-form solution can be found only for very simple PDFs. – E.g., if X is uniform, then Lloyd-Max quantizer = uniform quantizer.
• In general, an approximate solution can be found numerically, via an iterative algorithm (e.g., lloyds command in Matlab).
Nonlinear system to be solved
Ilya Pollak
qk =xfX (x)dx
xk
xk+1
∫
fX (x)dxxk
xk+1
∫= E X | X ∈k-th quantization interval[ ], for k = 1,…,L
xk =qk−1 + qk
2, for k = 2,…,L
⎧
⎨
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
• Closed-form solution can be found only for very simple PDFs. – E.g., if X is uniform, then Lloyd-Max quantizer = uniform quantizer.
• In general, an approximate solution can be found numerically, via an iterative algorithm (e.g., lloyds command in Matlab).
• For real data, typically the PDF is not given and therefore needs to be estimated using, for example, histograms constructed from the observed data.
Vector Lloyd-Max quantizer?
Ilya Pollak
X = X(1),…,X(N )( ) = source random vector with a given joint distribution.L = a desired number of quantization points.
Vector Lloyd-Max quantizer?
Ilya Pollak
X = X(1),…,X(N )( ) = source random vector with a given joint distribution.L = a desired number of quantization points.We would like to find:(1) L events A1,…,AL that partition the joint sample space of X(1),…,X(N ), and(2) L quantization points q1 ∈A1,…,qL ∈AL
Vector Lloyd-Max quantizer?
Ilya Pollak
X = X(1),…,X(N )( ) = source random vector with a given joint distribution.L = a desired number of quantization points.We would like to find:(1) L events A1,…,AL that partition the joint sample space of X(1),…,X(N ), and(2) L quantization points q1 ∈A1,…,qL ∈AL ,
such that the quantized random vector, defined byY = qk if X ∈Ak , for k = 1,…,L,minimizes the mean-square error,
E Y − X 2⎡⎣ ⎤⎦ = E Y (n) − X(n)( )2
n=1
N
∑⎡⎣⎢
⎤⎦⎥
Vector Lloyd-Max quantizer?
Ilya Pollak
X = X(1),…,X(N )( ) = source random vector with a given joint distribution.L = a desired number of quantization points.We would like to find:(1) L events A1,…,AL that partition the joint sample space of X(1),…,X(N ), and(2) L quantization points q1 ∈A1,…,qL ∈AL ,
such that the quantized random vector, defined byY = qk if X ∈Ak , for k = 1,…,L,minimizes the mean-square error,
E Y − X 2⎡⎣ ⎤⎦ = E Y (n) − X(n)( )2
n=1
N
∑⎡⎣⎢
⎤⎦⎥
Difficulty: cannot differentiate with respect to a set Ak , and so unless the set of all allowedpartitions is somehow restricted, this cannot be solved.
Hopefully, prior discussion gives you some idea about various
issues involved in quantization. And now, on to entropy coding…
Ilya Pollak
transform quantization entropy coding
compressed bitstream data
Problem statement
Ilya Pollak
Source (e.g., image, video, speech signal, or quantizer output)
Sequence of discrete random variables X(1),…,X(N) (e.g., transformed image pixel values), assumed to be independent and identically distributed over a finite alphabet {a1,…,aM}.
Problem statement
Requirements: • minimize the expected length of the binary string; • the binary string needs to be uniquely decodable, i.e., we need to be able
to infer X(1),…,X(N) from it!
Ilya Pollak
Source (e.g., image, video, speech signal, or quantizer output)
Encoder: mapping between source
symbols and binary strings (codewords)
Sequence of discrete random variables X(1),…,X(N) (e.g., transformed image pixel values), assumed to be independent and identically distributed over a finite alphabet {a1,…,aM}. Binary string
Problem statement
• Since X(1),…,X(N) are assumed independent in this model, we will encode each of them separately.
• Each can assume any value among {a1,…,aM}. • Therefore, our code will consist of M codewords, one for each symbol
a1,…,aM.
Ilya Pollak
Source (e.g., image, video, speech signal, or quantizer output)
Encoder: mapping between source
symbols and binary strings (codewords)
Sequence of discrete random variables X(1),…,X(N) (e.g., transformed image pixel values), assumed to be independent and identically distributed over a finite alphabet {a1,…,aM}. Binary string
symbol codeword
a1 w1
… …
aM wM
Unique Decodability
• How to decode the following string: 0001? • It could be aaab or aad or acb or cab or cd. • Not uniquely decodable!
Ilya Pollak
symbol codeword
a 0
b 1
c 00
d 01
A condition that ensures unique decodability
Ilya Pollak
• Prefix condition: no codeword in the code is a prefix for any other codeword.
A condition that ensures unique decodability
Ilya Pollak
• Prefix condition: no codeword in the code is a prefix for any other codeword.
• If the prefix condition is satisfied, then the code is uniquely decodable. – Proof. Take a bit string W that corresponds to two different
strings of symbols, A and B. If the first symbols in A and B are the same, discard them and the corresponding portion of W. Repeat until either there are no bits left in W (in this case A=B) or the first symbols in A and B are different. Then one of the codewords corresponding to these two symbols is a prefix for the other.
A condition that ensures unique decodability
Ilya Pollak
• Prefix condition: no codeword in the code is a prefix for any other codeword.
• Visualizing binary strings. Form a binary tree where each branch is labeled 0 or 1. Each codeword w can be associated with the unique node of the tree such that string of 0’s and 1’s on the path from the root to the node forms w.
A condition that ensures unique decodability
Ilya Pollak
• Prefix condition: no codeword in the code is a prefix for any other codeword.
• Visualizing binary strings. Form a binary tree where each branch is labeled 0 or 1. Each codeword w can be associated with the unique node of the tree such that string of 0’s and 1’s on the path from the root to the node forms w.
• Prefix condition holds if an only if all the codewords are leaves of the binary tree.
A condition that ensures unique decodability
Ilya Pollak
• Prefix condition: no codeword in the code is a prefix for any other codeword.
• Visualizing binary strings. Form a binary tree where each branch is labeled 0 or 1. Each codeword w can be associated with the unique node of the tree such that string of 0’s and 1’s on the path from the root to the node forms w.
• Prefix condition holds if an only if all the codewords are leaves of the binary tree---i.e., if no codeword is a descendant of another codeword.
Example: no prefix condition, no unique decodability, one word is not a leaf
Ilya Pollak
symbol codeword
a 0
b 1
c 00
d 01
• Codeword 0 is a prefix for both codeword 00 and codeword 01
Example: no prefix condition, no unique decodability, one word is not a leaf
Ilya Pollak
symbol codeword
a 0
b 1
c
d
• Codeword 0 is a prefix for both codeword 00 and codeword 01
1
0 wa=0
wb=1
Example: no prefix condition, no unique decodability, one word is not a leaf
Ilya Pollak
symbol codeword
a 0
b 1
c 00
d
• Codeword 0 is a prefix for both codeword 00 and codeword 01
0
1
0 wa=0
wb=1
wc=00
Example: no prefix condition, no unique decodability, one word is not a leaf
Ilya Pollak
symbol codeword
a 0
b 1
c 00
d 01
• Codeword 0 is a prefix for both codeword 00 and codeword 01
0
1
1 0
wa=0
wb=1
wd=01
wc=00
Example: prefix condition, all words are leaves
Ilya Pollak
symbol codeword
a 1
b
c
d
1
0
wa=1
Example: prefix condition, all words are leaves
Ilya Pollak
symbol codeword
a 1
b 01
c
d
0
1
1 0
wa=1
wb=01
Example: prefix condition, all words are leaves
Ilya Pollak
symbol codeword
a 1
b 01
c 000
d 001
0
1
1 0
wa=1
wb=01
wd=001
0 wc=000
1
Example: prefix condition, all words are leaves
Ilya Pollak
symbol codeword
a 1
b 01
c 000
d 001
0
1
1 0
wa=1
wb=01
wd=001
0 wc=000
1
• No path from the root to a codeword contains another codeword. This is equivalent to saying that the prefix condition holds.
Example: prefix condition, all words are leaves => unique decodability
Ilya Pollak
symbol codeword
a 1
b 01
c 000
d 001
0
1
1 0
wa=1
wb=01
wd=001
0 wc=000
1
Decoding: traverse the string left to right, tracing the corresponding path from the root of the binary tree. Each time a leaf is reached, output the codeword and go back to the root.
Example: prefix condition, all words are leaves => unique decodability
Ilya Pollak
0
1
1 0
wa=1 wb=01
wd=001
0 wc=000
1
000001101 How to decode the following string?
Example: prefix condition, all words are leaves => unique decodability
Ilya Pollak
0
1
1 0
wa=1 wb=01
wd=001
0 wc=000
1
000001101
Example: prefix condition, all words are leaves => unique decodability
Ilya Pollak
0
1
1 0
wa=1 wb=01
wd=001
0 wc=000
1
000001101
Example: prefix condition, all words are leaves => unique decodability
Ilya Pollak
0
1
1 0
wa=1 wb=01
wd=001
0 wc=000
1
000001101
Example: prefix condition, all words are leaves => unique decodability
Ilya Pollak
0
1
1 0
wa=1 wb=01
wd=001
0 wc=000
1
000001101
output: c
Example: prefix condition, all words are leaves => unique decodability
Ilya Pollak
0
1
1 0
wa=1 wb=01
wd=001
0 wc=000
1
000001101
output: c
Example: prefix condition, all words are leaves => unique decodability
Ilya Pollak
0
1
1 0
wa=1 wb=01
wd=001
0 wc=000
1
000001101
output: c
Example: prefix condition, all words are leaves => unique decodability
Ilya Pollak
0
1
1 0
wa=1 wb=01
wd=001
0 wc=000
1
000001101
output: c
Example: prefix condition, all words are leaves => unique decodability
Ilya Pollak
0
1
1 0
wa=1 wb=01
wd=001
0 wc=000
1
000001101
output: cd
Example: prefix condition, all words are leaves => unique decodability
Ilya Pollak
0
1
1 0
wa=1 wb=01
wd=001
0 wc=000
1
000001101
output: cd
Example: prefix condition, all words are leaves => unique decodability
Ilya Pollak
0
1
1 0
wa=1
wb=01
wd=001
0 wc=000
1
000001101
output: cda
Example: prefix condition, all words are leaves => unique decodability
Ilya Pollak
0
1
1 0
wa=1
wb=01
wd=001
0 wc=000
1
000001101
output: cda
Example: prefix condition, all words are leaves => unique decodability
Ilya Pollak
0
1
1 0
wa=1
wb=01
wd=001
0 wc=000
1
000001101
output: cda
Example: prefix condition, all words are leaves => unique decodability
Ilya Pollak
0
1
1 0
wa=1
wb=01
wd=001
0 wc=000
1
000001101
output: cdab
Example: prefix condition, all words are leaves => unique decodability
Ilya Pollak
0
1
1 0
wa=1
wb=01
wd=001
0 wc=000
1
000001101
final output: cdab
Prefix condition and unique decodability
• There are uniquely decodable codes which do not satisfy the prefix condition (e.g., {0, 01}).
Ilya Pollak
Prefix condition and unique decodability
• There are uniquely decodable codes which do not satisfy the prefix condition (e.g., {0, 01}). For any such code, a prefix condition code can be constructed with an identical set of codeword lengths. (E.g., {0, 10} for {0, 01}.)
Ilya Pollak
Prefix condition and unique decodability
• There are uniquely decodable codes which do not satisfy the prefix condition (e.g., {0, 01}). For any such code, a prefix condition code can be constructed with an identical set of codeword lengths. (E.g., {0, 10} for {0, 01}.)
• For this reason, we can consider just prefix condition codes.
Ilya Pollak
Entropy coding • Given a discrete random variable X with M possible outcomes
(“symbols” or “letters”) a1,…,aM and with PMF pX, what is the lowest achievable expected codeword length among all the uniquely decodable codes? – Answer depends on pX; Shannon’s source coding theorem provides
bounds.
• How to construct a prefix condition code which achieves this expected codeword length? – Answer: Huffman code.
Ilya Pollak
Huffman code • Consider a discrete r.v. X with M possible outcomes a1,…,aM and with PMF
pX. Assume that pX(a1) ≤ … ≤ pX(aM). (If this condition is not satisfied, reorder the outcomes so that it is satisfied.)
Ilya Pollak
Huffman code • Consider a discrete r.v. X with M possible outcomes a1,…,aM and with PMF
pX. Assume that pX(a1) ≤ … ≤ pX(aM). (If this condition is not satisfied, reorder the outcomes so that it is satisfied.)
• Consider “aggregate outcome” a12 = {a1,a2} and a discrete r.v. X’ such that
Ilya Pollak
X ' = a12 if X = a1 or X = a2
X otherwise
⎧⎨⎪
⎩⎪
Huffman code • Consider a discrete r.v. X with M possible outcomes a1,…,aM and with PMF
pX. Assume that pX(a1) ≤ … ≤ pX(aM). (If this condition is not satisfied, reorder the outcomes so that it is satisfied.)
• Consider “aggregate outcome” a12 = {a1,a2} and a discrete r.v. X’ such that
Ilya Pollak
X ' = a12 if X = a1 or X = a2
X otherwise
⎧⎨⎪
⎩⎪
pX ' a( ) =pX a1( ) + pX a2( ) if a = a12
pX a( ) if a = a3,…,aM
⎧⎨⎪
⎩⎪
Huffman code • Consider a discrete r.v. X with M possible outcomes a1,…,aM and with PMF
pX. Assume that pX(a1) ≤ … ≤ pX(aM). (If this condition is not satisfied, reorder the outcomes so that it is satisfied.)
• Consider “aggregate outcome” a12 = {a1,a2} and a discrete r.v. X’ such that
Ilya Pollak
X ' = a12 if X = a1 or X = a2
X otherwise
⎧⎨⎪
⎩⎪
• Suppose we have a tree, T’, for an optimal prefix condition code for X’. A tree T for an optimal prefix condition code for X can be obtained from T’ by splitting the leaf a12 into two leaves corresponding to a1 and a2.
pX ' a( ) =pX a1( ) + pX a2( ) if a = a12
pX a( ) if a = a3,…,aM
⎧⎨⎪
⎩⎪
Huffman code • Consider a discrete r.v. X with M possible outcomes a1,…,aM and with PMF
pX. Assume that pX(a1) ≤ … ≤ pX(aM). (If this condition is not satisfied, reorder the outcomes so that it is satisfied.)
• Consider “aggregate outcome” a12 = {a1,a2} and a discrete r.v. X’ such that
Ilya Pollak
X ' = a12 if X = a1 or X = a2
X otherwise
⎧⎨⎪
⎩⎪
• Suppose we have a tree, T’, for an optimal prefix condition code for X’. A tree T for an optimal prefix condition code for X can be obtained from T’ by splitting the leaf a12 into two leaves corresponding to a1 and a2.
• We won’t prove this.
pX ' a( ) =pX a1( ) + pX a2( ) if a = a12
pX a( ) if a = a3,…,aM
⎧⎨⎪
⎩⎪
Example
Ilya Pollak
letter pX(letter)
a1 0.10
a2 0.10
a3 0.25
a4 0.25
a5 0.30
Example
Ilya Pollak
letter pX(letter)
a1 0.10
a2 0.10
a3 0.25
a4 0.25
a5 0.30
Step 1: combine the two least likely letters.
letter pX’(letter)
a12 0.20
a3 0.25
a4 0.25
a5 0.30
Example
Ilya Pollak
letter pX(letter)
a1 0.10
a2 0.10
a3 0.25
a4 0.25
a5 0.30
Step 1: combine the two least likely letters.
letter pX’(letter)
a12 0.20
a3 0.25
a4 0.25
a5 0.30
0
1 a1
a2
a12
Example
Ilya Pollak
letter pX(letter)
a1 0.10
a2 0.10
a3 0.25
a4 0.25
a5 0.30
Step 1: combine the two least likely letters.
letter pX’(letter)
a12 0.20
a3 0.25
a4 0.25
a5 0.30
0
1 a1
a2
a12 Tree for X:
Tree for X’ (still to be constructed)
Example
Ilya Pollak
Step 2: combine the two least likely letters from the new alphabet.
letter pX’’(letter)
a123 0.45
a4 0.25
a5 0.30
letter pX’(letter)
a12 0.20
a3 0.25
a4 0.25
a5 0.30
Example
Ilya Pollak
Step 2: combine the two least likely letters from the new alphabet.
letter pX’’(letter)
a123 0.45
a4 0.25
a5 0.30
letter pX’(letter)
a12 0.20
a3 0.25
a4 0.25
a5 0.30
0
1 a1
a2
a12
a123
a3
1
0
Example
Ilya Pollak
Step 2: combine the two least likely letters from the new alphabet.
letter pX’’(letter)
a123 0.45
a4 0.25
a5 0.30
letter pX’(letter)
a12 0.20
a3 0.25
a4 0.25
a5 0.30
0
1 a1
a2
a12 Tree for X:
Tree for X’’
a123
a3
1
0
Example
Ilya Pollak
Step 2: combine the two least likely letters from the new alphabet.
letter pX’’(letter)
a123 0.45
a4 0.25
a5 0.30
letter pX’(letter)
a12 0.20
a3 0.25
a4 0.25
a5 0.30
0
1 a1
a2
a12 Tree for X:
Tree for X’’
a123
a3
Tree for X’
1
0
Example
Ilya Pollak
Step 3: again combine the two least likely letters
letter pX’’’(letter)
a123 0.45
a45 0.55
0
1 a1
a2
a12
a123
a3
letter pX’’(letter)
a123 0.45
a4 0.25
a5 0.30
1
0
a4 a45
a5
1
0
Example
Ilya Pollak
Step 3: again combine the two least likely letters
letter pX’’’(letter)
a123 0.45
a45 0.55
0
1 a1
a2
a12 Tree for X:
a123
a3
letter pX’’(letter)
a123 0.45
a4 0.25
a5 0.30
1
0
a4 a45
a5
1
0
Tree for X’’’
Example
Ilya Pollak
Step 3: again combine the two least likely letters
letter pX’’’(letter)
a123 0.45
a45 0.55
0
1 a1
a2
a12 Tree for X:
a123
a3
letter pX’’(letter)
a123 0.45
a4 0.25
a5 0.30
1
0
a4 a45
a5
1
0
Tree for X’’
Tree for X’’’
Example
Ilya Pollak
Step 3: again combine the two least likely letters
letter pX’’’(letter)
a123 0.45
a45 0.55
0
1 a1
a2
a12 Tree for X:
a123
a3
Tree for X’
letter pX’’(letter)
a123 0.45
a4 0.25
a5 0.30
1
0
a4 a45
a5
1
0
Tree for X’’
Tree for X’’’
Example
Ilya Pollak
Step 4: combine the last two remaining letters letter pX’’’(letter)
a123 0.45
a45 0.55 Done!
0
1 a1
a2
a12 Tree for X:
a123
a3
1
0
a4 a45
a5
1
0
a12345 1
0
Example
Ilya Pollak
Step 4: combine the last two remaining letters letter pX’’’(letter)
a123 0.45
a45 0.55
Done! The codeword for each leaf is the sequence of 0’1 and 1’s along the path from the root to that leaf.
0
1 a1
a2
Tree for X:
a3
1
0
a4
a5
1
0
1
0
Example
Ilya Pollak
0
1 a1
a2
Tree for X:
a3
1
0
a4
a5
1
0
1
0
letter pX(letter) codeword
a1 0.10 111
a2 0.10
a3 0.25
a4 0.25
a5 0.30
Example
Ilya Pollak
0
1 a1
a2
Tree for X:
a3
1
0
a4
a5
1
0
1
0
letter pX(letter) codeword
a1 0.10 111
a2 0.10 110
a3 0.25
a4 0.25
a5 0.30
Example
Ilya Pollak
0
1 a1
a2
Tree for X:
a3
1
0
a4
a5
1
0
1
0
letter pX(letter) codeword
a1 0.10 111
a2 0.10 110
a3 0.25 10
a4 0.25
a5 0.30
Example
Ilya Pollak
0
1 a1
a2
Tree for X:
a3
1
0
a4
a5
1
0
1
0
letter pX(letter) codeword
a1 0.10 111
a2 0.10 110
a3 0.25 10
a4 0.25 01
a5 0.30
Example
Ilya Pollak
0
1 a1
a2
Tree for X:
a3
1
0
a4
a5
1
0
1
0
letter pX(letter) codeword
a1 0.10 111
a2 0.10 110
a3 0.25 10
a4 0.25 01
a5 0.30 00
Example
Ilya Pollak
0
1 a1
a2
Tree for X:
a3
1
0
a4
a5
1
0
1
0
letter pX(letter) codeword
a1 0.10 111
a2 0.10 110
a3 0.25 10
a4 0.25 01
a5 0.30 00
Expected codeword length: 3(0.1) + 3(0.1) + 2(0.25) + 2(0.25) + 2(0.3) = 2.2 bits
Self-information • Consider again a discrete random variable X with M possible
outcomes a1,…,aM and with PMF pX.
Ilya Pollak
Self-information • Consider again a discrete random variable X with M possible
outcomes a1,…,aM and with PMF pX. • Self-information of outcome am is I(am) = −log2 pX(am) bits.
Ilya Pollak
Self-information • Consider again a discrete random variable X with M possible
outcomes a1,…,aM and with PMF pX. • Self-information of outcome am is I(am) = −log2 pX(am) bits. • E.g., pX(am) = 1 then I(am) = 0. The occurrence of am is not at
all informative, since it had to occur. The smaller the probability of an outcome, the larger its self-information.
Ilya Pollak
Self-information • Consider again a discrete random variable X with M possible
outcomes a1,…,aM and with PMF pX. • Self-information of outcome am is I(am) = −log2 pX(am) bits. • E.g., pX(am) = 1 then I(am) = 0. The occurrence of am is not at
all informative, since it had to occur. The smaller the probability of an outcome, the larger its self-information.
• Self-information of X is I(X) = −log2 pX(X) and is a random variable.
Ilya Pollak
Self-information • Consider again a discrete random variable X with M possible
outcomes a1,…,aM and with PMF pX. • Self-information of outcome am is I(am) = −log2 pX(am) bits. • E.g., pX(am) = 1 then I(am) = 0. The occurrence of am is not at
all informative, since it had to occur. The smaller the probability of an outcome, the larger its self-information.
• Self-information of X is I(X) = −log2 pX(X) and is a random variable.
• Entropy of X is the expected value of its self-information:
Ilya Pollak
H (X) = E I(X)[ ] = − pX (am )log2m=1
M
∑ pX (am )
Source coding theorem (Shannon)
For any uniquely decodable code, the expected codeword length is ≥ H (X).Moreover, there exists a prefix condition code for which the expected codewordlength is < H (X) +1.
Ilya Pollak
Example
Ilya Pollak
• Suppose that X has M=2K possible outcomes a1,…,aM.
Example
Ilya Pollak
• Suppose that X has M=2K possible outcomes a1,…,aM. • Suppose that X is uniform, i.e., pX (a1) = … = pX (aM) = 2−K.
Example
Ilya Pollak
• Suppose that X has M=2K possible outcomes a1,…,aM. • Suppose that X is uniform, i.e., pX (a1) = … = pX (aM) = 2−K. Then
H (X) = E I(X)[ ] = − 2−K log2k=1
2K
∑ 2−K( ) = 2K −2−K( ) −K( ) = K
Example
Ilya Pollak
• Suppose that X has M=2K possible outcomes a1,…,aM. • Suppose that X is uniform, i.e., pX (a1) = … = pX (aM) = 2−K. Then
H (X) = E I(X)[ ] = − 2−K log2k=1
2K
∑ 2−K( ) = 2K −2−K( ) −K( ) = K
• On the other hand, observe that there exist 2K different K-bit sequences. Thus, a fixed-length code for X that uses all these 2K K-bit sequences as codewords for all the 2K outcomes of X, will have expected codeword length of K.
Example
Ilya Pollak
• Suppose that X has M=2K possible outcomes a1,…,aM. • Suppose that X is uniform, i.e., pX (a1) = … = pX (aM) = 2−K. Then
H (X) = E I(X)[ ] = − 2−K log2k=1
2K
∑ 2−K( ) = 2K −2−K( ) −K( ) = K
• On the other hand, observe that there exist 2K different K-bit sequences. Thus, a fixed-length code for X that uses all these 2K K-bit sequences as codewords for all the 2K outcomes of X, will have expected codeword length of K.
• I.e., for this particular random variable, this fixed-length code achieves the entropy of X, which is the lower bound given by the source coding theorem.
Example
Ilya Pollak
• Suppose that X has M=2K possible outcomes a1,…,aM. • Suppose that X is uniform, i.e., pX (a1) = … = pX (aM) = 2−K. Then
H (X) = E I(X)[ ] = − 2−K log2k=1
2K
∑ 2−K( ) = 2K −2−K( ) −K( ) = K
• On the other hand, observe that there exist 2K different K-bit sequences. Thus, a fixed-length code for X that uses all these 2K K-bit sequences as codewords for all the 2K outcomes of X, will have expected codeword length of K.
• I.e., for this particular random variable, this fixed-length code achieves the entropy of X, which is the lower bound given by the source coding theorem.
• Therefore, the K-bit fixed-length code is optimal for this X.
Lemma 1: An auxiliary result helpful for proving the source coding theorem
Ilya Pollak
• log2α ≤ (α−1) log2e for log2 α > 0. • Proof: differentiate g(α) = (α−1) log2e − log2α and show that
g(1) = 0 is its minimum.
Another auxiliary result: Kraft inequality
Ilya Pollak
If integers d1,…,dM satisfy the inequality
2−dm
m=1
M
∑ ≤ 1, (1)
then there exists a prefix condition code whose codeword lengths are these integers.Conversely, the codeword lengths of any prefix condition code satisfy this inequality.
Some useful facts about full binary trees
Ilya Pollak
A full binary tree of depth D has 2D leaves.
Some useful facts about full binary trees
Ilya Pollak
A full binary tree of depth D has 2D leaves. (Here, depth is D=4 and the number of leaves is 24=16.)
Tree depth D = 4
Some useful facts about full binary trees
Ilya Pollak
A full binary tree of depth D has 2D leaves. (Here, depth is D=4 and the number of leaves is 24=16.)
In a full binary tree of depth D, each node at depth d has 2D−d leaf descendants. (Here, D=4, the red node is at depth d=2, and so it has 24−2 = 4 leaf descendants.)
Tree depth D = 4
Depth of red node = 2
Kraft inequality: proof of
Ilya Pollak
Suppose d1 ≤… ≤ dM satisfy (1). Consider the full binary tree of depth dM , and consider all itsnodes at depth d1. Assign one of these nodes to symbol a1.
⇒
Kraft inequality: proof of
Ilya Pollak
Suppose d1 ≤… ≤ dM satisfy (1). Consider the full binary tree of depth dM , and consider all itsnodes at depth d1. Assign one of these nodes to symbol a1. Consider all the nodes at depth d2 whichare not a1 and not descendants of a1. Assign one of them to symbol a2 .
⇒
Kraft inequality: proof of
Ilya Pollak
Suppose d1 ≤… ≤ dM satisfy (1). Consider the full binary tree of depth dM , and consider all itsnodes at depth d1. Assign one of these nodes to symbol a1. Consider all the nodes at depth d2 whichare not a1 and not descendants of a1. Assign one of them to symbol a2 . Iterate like this M times.
⇒
Kraft inequality: proof of
Ilya Pollak
Suppose d1 ≤… ≤ dM satisfy (1). Consider the full binary tree of depth dM , and consider all itsnodes at depth d1. Assign one of these nodes to symbol a1. Consider all the nodes at depth d2 whichare not a1 and not descendants of a1. Assign one of them to symbol a2 . Iterate like this M times.If we have run out of tree nodes to assign after r < M iterations, it means that every leaf in the fullbinary tree of depth dM is a descendant of one of the first m symbols, a1,…,ar .
⇒
Kraft inequality: proof of
Ilya Pollak
Suppose d1 ≤… ≤ dM satisfy (1). Consider the full binary tree of depth dM , and consider all itsnodes at depth d1. Assign one of these nodes to symbol a1. Consider all the nodes at depth d2 whichare not a1 and not descendants of a1. Assign one of them to symbol a2 . Iterate like this M times.If we have run out of tree nodes to assign after r < M iterations, it means that every leaf in the fullbinary tree of depth dM is a descendant of one of the first m symbols, a1,…,ar . But note that everynode at depth dm has 2dM −dm descendants. Note also that the full tree has 2dM leaves. Therefore, ifevery leaf in the tree is a descendant of a1,…,ar , then
2dM −dm
m=1
r
∑ = 2dM
⇒
Kraft inequality: proof of
Ilya Pollak
Suppose d1 ≤… ≤ dM satisfy (1). Consider the full binary tree of depth dM , and consider all itsnodes at depth d1. Assign one of these nodes to symbol a1. Consider all the nodes at depth d2 whichare not a1 and not descendants of a1. Assign one of them to symbol a2 . Iterate like this M times.If we have run out of tree nodes to assign after r < M iterations, it means that every leaf in the fullbinary tree of depth dM is a descendant of one of the first m symbols, a1,…,ar . But note that everynode at depth dm has 2dM −dm descendants. Note also that the full tree has 2dM leaves. Therefore, ifevery leaf in the tree is a descendant of a1,…,ar , then
2dM −dm
m=1
r
∑ = 2dM ⇔ 2−dm
m=1
r
∑ = 1
⇒
Kraft inequality: proof of
Ilya Pollak
Suppose d1 ≤… ≤ dM satisfy (1). Consider the full binary tree of depth dM , and consider all itsnodes at depth d1. Assign one of these nodes to symbol a1. Consider all the nodes at depth d2 whichare not a1 and not descendants of a1. Assign one of them to symbol a2 . Iterate like this M times.If we have run out of tree nodes to assign after r < M iterations, it means that every leaf in the fullbinary tree of depth dM is a descendant of one of the first m symbols, a1,…,ar . But note that everynode at depth dm has 2dM −dm descendants. Note also that the full tree has 2dM leaves. Therefore, ifevery leaf in the tree is a descendant of a1,…,ar , then
2dM −dm
m=1
r
∑ = 2dM ⇔ 2−dm
m=1
r
∑ = 1
Therefore, 2−dm
m=1
M
∑ = 2−dm
m=1
r
∑ + 2−dm
m= r+1
M
∑ > 1. This violates (1).
⇒
Kraft inequality: proof of
Ilya Pollak
Suppose d1 ≤… ≤ dM satisfy (1). Consider the full binary tree of depth dM , and consider all itsnodes at depth d1. Assign one of these nodes to symbol a1. Consider all the nodes at depth d2 whichare not a1 and not descendants of a1. Assign one of them to symbol a2 . Iterate like this M times.If we have run out of tree nodes to assign after r < M iterations, it means that every leaf in the fullbinary tree of depth dM is a descendant of one of the first m symbols, a1,…,ar . But note that everynode at depth dm has 2dM −dm descendants. Note also that the full tree has 2dM leaves. Therefore, ifevery leaf in the tree is a descendant of a1,…,ar , then
2dM −dm
m=1
r
∑ = 2dM ⇔ 2−dm
m=1
r
∑ = 1
Therefore, 2−dm
m=1
M
∑ = 2−dm
m=1
r
∑ + 2−dm
m= r+1
M
∑ > 1. This violates (1).
Thus, our procedure can in fact go on for M iterations. After the M -th iteration, we will haveconstructed a prefix condition code with codeword lengths d1,…,dM .
⇒
Kraft inequality: proof of
Ilya Pollak
Suppose d1 ≤… ≤ dM , and suppose we have a prefix condition code with there codeword lengths.Consider the binary tree corresponding to this code.
⇐
Kraft inequality: proof of
Ilya Pollak
Suppose d1 ≤… ≤ dM , and suppose we have a prefix condition code with there codeword lengths.Consider the binary tree corresponding to this code. Complete this tree to obtain a full tree ofdepth dM .
⇐
Kraft inequality: proof of
Ilya Pollak
Suppose d1 ≤… ≤ dM , and suppose we have a prefix condition code with there codeword lengths.Consider the binary tree corresponding to this code. Complete this tree to obtain a full tree ofdepth dM . Again use the following facts:the full tree has 2dM leaves;the number of leaf descendants of the codeword of length dm is 2dM −dm .
⇐
Kraft inequality: proof of
Ilya Pollak
Suppose d1 ≤… ≤ dM , and suppose we have a prefix condition code with there codeword lengths.Consider the binary tree corresponding to this code. Complete this tree to obtain a full tree ofdepth dM . Again use the following facts:the full tree has 2dM leaves;the number of leaf descendants of the codeword of length dm is 2dM −dm .The combined number of all leaf descendants of all codewords must be less than or equal tothe total number of leaves in the full tree:
2dM −dm
m=1
M
∑ ≤ 2dM
⇐
Kraft inequality: proof of
Ilya Pollak
Suppose d1 ≤… ≤ dM , and suppose we have a prefix condition code with there codeword lengths.Consider the binary tree corresponding to this code. Complete this tree to obtain a full tree ofdepth dM . Again use the following facts:the full tree has 2dM leaves;the number of leaf descendants of the codeword of length dm is 2dM −dm .The combined number of all leaf descendants of all codewords must be less than or equal tothe total number of leaves in the full tree:
2dM −dm
m=1
M
∑ ≤ 2dM ⇔ 2−dm
m=1
M
∑ ≤ 1.
⇐
Source coding theorem: proof of H(X)≤E[C]
Ilya Pollak
Let dm be the codeword length for am , and let random variable C be the codeword length for X.
Source coding theorem: proof of H(X)≤E[C]
Ilya Pollak
Let dm be the codeword length for am , and let random variable C be the codeword length for X.
H (X) − E[C] = − pX (am )log2m=1
M
∑ pX (am ) − pX (am )dmm=1
M
∑
Source coding theorem: proof of H(X)≤E[C]
Ilya Pollak
Let dm be the codeword length for am , and let random variable C be the codeword length for X.
H (X) − E[C] = − pX (am )log2m=1
M
∑ pX (am ) − pX (am )dmm=1
M
∑ = pX (am ) log21
pX (am )⎛⎝⎜
⎞⎠⎟− log2 2dm
⎡
⎣⎢
⎤
⎦⎥
m=1
M
∑
Source coding theorem: proof of H(X)≤E[C]
Ilya Pollak
Let dm be the codeword length for am , and let random variable C be the codeword length for X.
H (X) − E[C] = − pX (am )log2m=1
M
∑ pX (am ) − pX (am )dmm=1
M
∑ = pX (am ) log21
pX (am )⎛⎝⎜
⎞⎠⎟− log2 2dm
⎡
⎣⎢
⎤
⎦⎥
m=1
M
∑
= pX (am ) log21
pX (am )2dm
⎛⎝⎜
⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥
m=1
M
∑
Source coding theorem: proof of H(X)≤E[C]
Ilya Pollak
Let dm be the codeword length for am , and let random variable C be the codeword length for X.
H (X) − E[C] = − pX (am )log2m=1
M
∑ pX (am ) − pX (am )dmm=1
M
∑ = pX (am ) log21
pX (am )⎛⎝⎜
⎞⎠⎟− log2 2dm
⎡
⎣⎢
⎤
⎦⎥
m=1
M
∑
= pX (am ) log21
pX (am )2dm
⎛⎝⎜
⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥
m=1
M
∑
≤ pX (am ) 1pX (am )2dm
−1⎛⎝⎜
⎞⎠⎟m=1
M
∑ log2 e (by Lemma 1)
Source coding theorem: proof of H(X)≤E[C]
Ilya Pollak
Let dm be the codeword length for am , and let random variable C be the codeword length for X.
H (X) − E[C] = − pX (am )log2m=1
M
∑ pX (am ) − pX (am )dmm=1
M
∑ = pX (am ) log21
pX (am )⎛⎝⎜
⎞⎠⎟− log2 2dm
⎡
⎣⎢
⎤
⎦⎥
m=1
M
∑
= pX (am ) log21
pX (am )2dm
⎛⎝⎜
⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥
m=1
M
∑
≤ pX (am ) 1pX (am )2dm
−1⎛⎝⎜
⎞⎠⎟m=1
M
∑ log2 e (by Lemma 1)
= 12dm
m=1
M
∑ − pX (am )m=1
M
∑⎛⎝⎜
⎞⎠⎟
log2 e
Source coding theorem: proof of H(X)≤E[C]
Ilya Pollak
Let dm be the codeword length for am , and let random variable C be the codeword length for X.
H (X) − E[C] = − pX (am )log2m=1
M
∑ pX (am ) − pX (am )dmm=1
M
∑ = pX (am ) log21
pX (am )⎛⎝⎜
⎞⎠⎟− log2 2dm
⎡
⎣⎢
⎤
⎦⎥
m=1
M
∑
= pX (am ) log21
pX (am )2dm
⎛⎝⎜
⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥
m=1
M
∑
≤ pX (am ) 1pX (am )2dm
−1⎛⎝⎜
⎞⎠⎟m=1
M
∑ log2 e (by Lemma 1)
= 12dm
m=1
M
∑ − pX (am )m=1
M
∑⎛⎝⎜
⎞⎠⎟
log2 e
= 2−dm
m=1
M
∑ −1⎛⎝⎜
⎞⎠⎟
log2 e ≤ 0
Source coding theorem: proof of H(X)≤E[C]
Ilya Pollak
Let dm be the codeword length for am , and let random variable C be the codeword length for X.
H (X) − E[C] = − pX (am )log2m=1
M
∑ pX (am ) − pX (am )dmm=1
M
∑ = pX (am ) log21
pX (am )⎛⎝⎜
⎞⎠⎟− log2 2dm
⎡
⎣⎢
⎤
⎦⎥
m=1
M
∑
= pX (am ) log21
pX (am )2dm
⎛⎝⎜
⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥
m=1
M
∑
≤ pX (am ) 1pX (am )2dm
−1⎛⎝⎜
⎞⎠⎟m=1
M
∑ log2 e (by Lemma 1)
= 12dm
m=1
M
∑ − pX (am )m=1
M
∑⎛⎝⎜
⎞⎠⎟
log2 e
= 2−dm
m=1
M
∑ −1⎛⎝⎜
⎞⎠⎟
log2 e ≤ 0
By Kraft inequality, this holds for any prefix condition code. But it is also true for any uniquelydecodable code.
Source coding theorem: how to satisfy E[C] < H(X)+1?
Ilya Pollak
Choose dm = − log2 pX (am )⎡⎢ ⎤⎥ (where x⎡⎢ ⎤⎥ stands for the smallest integer which is ≥ x). Thendm ≥ − log2 pX (am )
Source coding theorem: how to satisfy E[C] < H(X)+1?
Ilya Pollak
Choose dm = − log2 pX (am )⎡⎢ ⎤⎥ (where x⎡⎢ ⎤⎥ stands for the smallest integer which is ≥ x). Then
dm ≥ − log2 pX (am ) ⇒ − dm ≤ log2 pX (am ) ⇒ 2−dm ≤ pX (am )
Source coding theorem: how to satisfy E[C] < H(X)+1?
Ilya Pollak
Choose dm = − log2 pX (am )⎡⎢ ⎤⎥ (where x⎡⎢ ⎤⎥ stands for the smallest integer which is ≥ x). Then
dm ≥ − log2 pX (am ) ⇒ − dm ≤ log2 pX (am ) ⇒ 2−dm ≤ pX (am ) ⇒ 2−dm
m=1
M
∑ ≤ pX (am )m=1
M
∑ = 1.
Source coding theorem: how to satisfy E[C] < H(X)+1?
Ilya Pollak
Choose dm = − log2 pX (am )⎡⎢ ⎤⎥ (where x⎡⎢ ⎤⎥ stands for the smallest integer which is ≥ x). Then
dm ≥ − log2 pX (am ) ⇒ − dm ≤ log2 pX (am ) ⇒ 2−dm ≤ pX (am ) ⇒ 2−dm
m=1
M
∑ ≤ pX (am )m=1
M
∑ = 1.
Therefore, Kraft inequality is satisfied, and we can construct a prefix condition code with codewordlengths d1,…,dM .
Source coding theorem: how to satisfy E[C] < H(X)+1?
Ilya Pollak
Choose dm = − log2 pX (am )⎡⎢ ⎤⎥ (where x⎡⎢ ⎤⎥ stands for the smallest integer which is ≥ x). Then
dm ≥ − log2 pX (am ) ⇒ − dm ≤ log2 pX (am ) ⇒ 2−dm ≤ pX (am ) ⇒ 2−dm
m=1
M
∑ ≤ pX (am )m=1
M
∑ = 1.
Therefore, Kraft inequality is satisfied, and we can construct a prefix condition code with codewordlengths d1,…,dM . Also, by construction,dm −1 < − log2 pX (am ) ⇒ dm < − log2 pX (am ) +1
Source coding theorem: how to satisfy E[C] < H(X)+1?
Ilya Pollak
Choose dm = − log2 pX (am )⎡⎢ ⎤⎥ (where x⎡⎢ ⎤⎥ stands for the smallest integer which is ≥ x). Then
dm ≥ − log2 pX (am ) ⇒ − dm ≤ log2 pX (am ) ⇒ 2−dm ≤ pX (am ) ⇒ 2−dm
m=1
M
∑ ≤ pX (am )m=1
M
∑ = 1.
Therefore, Kraft inequality is satisfied, and we can construct a prefix condition code with codewordlengths d1,…,dM . Also, by construction,dm −1 < − log2 pX (am ) ⇒ dm < − log2 pX (am ) +1 ⇒ pX (am )dm < − pX (am )log2 pX (am ) + pX (am )
Source coding theorem: how to satisfy E[C] < H(X)+1?
Ilya Pollak
Choose dm = − log2 pX (am )⎡⎢ ⎤⎥ (where x⎡⎢ ⎤⎥ stands for the smallest integer which is ≥ x). Then
dm ≥ − log2 pX (am ) ⇒ − dm ≤ log2 pX (am ) ⇒ 2−dm ≤ pX (am ) ⇒ 2−dm
m=1
M
∑ ≤ pX (am )m=1
M
∑ = 1.
Therefore, Kraft inequality is satisfied, and we can construct a prefix condition code with codewordlengths d1,…,dM . Also, by construction,dm −1 < − log2 pX (am ) ⇒ dm < − log2 pX (am ) +1 ⇒ pX (am )dm < − pX (am )log2 pX (am ) + pX (am )
⇒ pX (am )dmm=1
M
∑ < − pX (am )log2 pX (am ) + pX (am )( )m=1
M
∑
Source coding theorem: how to satisfy E[C] < H(X)+1?
Ilya Pollak
Choose dm = − log2 pX (am )⎡⎢ ⎤⎥ (where x⎡⎢ ⎤⎥ stands for the smallest integer which is ≥ x). Then
dm ≥ − log2 pX (am ) ⇒ − dm ≤ log2 pX (am ) ⇒ 2−dm ≤ pX (am ) ⇒ 2−dm
m=1
M
∑ ≤ pX (am )m=1
M
∑ = 1.
Therefore, Kraft inequality is satisfied, and we can construct a prefix condition code with codewordlengths d1,…,dM . Also, by construction,dm −1 < − log2 pX (am ) ⇒ dm < − log2 pX (am ) +1 ⇒ pX (am )dm < − pX (am )log2 pX (am ) + pX (am )
⇒ pX (am )dmm=1
M
∑ < − pX (am )log2 pX (am ) + pX (am )( )m=1
M
∑
⇒ E[C] < − pX (am )log2 pX (am )( )m=1
M
∑ + pX (am )m=1
M
∑
Source coding theorem: how to satisfy E[C] < H(X)+1?
Ilya Pollak
Choose dm = − log2 pX (am )⎡⎢ ⎤⎥ (where x⎡⎢ ⎤⎥ stands for the smallest integer which is ≥ x). Then
dm ≥ − log2 pX (am ) ⇒ − dm ≤ log2 pX (am ) ⇒ 2−dm ≤ pX (am ) ⇒ 2−dm
m=1
M
∑ ≤ pX (am )m=1
M
∑ = 1.
Therefore, Kraft inequality is satisfied, and we can construct a prefix condition code with codewordlengths d1,…,dM . Also, by construction,dm −1 < − log2 pX (am ) ⇒ dm < − log2 pX (am ) +1 ⇒ pX (am )dm < − pX (am )log2 pX (am ) + pX (am )
⇒ pX (am )dmm=1
M
∑ < − pX (am )log2 pX (am ) + pX (am )( )m=1
M
∑
⇒ E[C] < − pX (am )log2 pX (am )( )m=1
M
∑ + pX (am )m=1
M
∑ = H (X) +1
Note: Huffman code may often be very far from the entropy
Ilya Pollak
• Let X have two outcomes, a1 and a2, with probabilities 1−2−d
and 2−d, respectively.
Note: Huffman code may often be very far from the entropy
Ilya Pollak
• Let X have two outcomes, a1 and a2, with probabilities 1−2−d
and 2−d, respectively. • Huffman code: 0 for a1; 1 for a2. • Expected codeword length: 1.
Note: Huffman code may often be very far from the entropy
Ilya Pollak
• Let X have two outcomes, a1 and a2, with probabilities 1−2−d
and 2−d, respectively. • Huffman code: 0 for a1; 1 for a2. • Expected codeword length: 1. • Entropy: −(1−2−d) log2(1−2−d) + d2−d ≈ 0 for large d. For
example, if d=20, this is 0.0000204493.
Note: Huffman code may often be very far from the entropy
Ilya Pollak
• Let X have two outcomes, a1 and a2, with probabilities 1−2−d
and 2−d, respectively. • Huffman code: 0 for a1; 1 for a2. • Expected codeword length: 1. • Entropy: −(1−2−d) log2(1−2−d) + d2−d ≈ 0 for large d. For
example, if d=20, this is 0.0000204493. • Problem: no codeword can have fractional numbers of bits!
Note: Huffman code may often be very far from the entropy
Ilya Pollak
• Let X have two outcomes, a1 and a2, with probabilities 1−2−d
and 2−d, respectively. • Huffman code: 0 for a1; 1 for a2. • Expected codeword length: 1. • Entropy: −(1−2−d) log2(1−2−d) + d2−d ≈ 0 for large d. For
example, if d=20, this is 0.0000204493. • Problem: no codeword can have fractional numbers of bits! • If we have a source which produces independent random
variables X1, X2 , …, all identically distributed to X, a single Huffman code can be constructed for several of them, effectively resulting in fractional numbers of bits per random variable.
Example
Ilya Pollak
• (X1,X2) will have four outcomes, (a1,a1), (a1,a2), (a2,a1), (a2,a2), with probabilities 1−2−d+1+2−2d, 2−d−2−2d, 2−d−2−2d, and 2−2d, respectively.
Example
Ilya Pollak
• (X1,X2) will have four outcomes, (a1,a1), (a1,a2), (a2,a1), (a2,a2), with probabilities 1−2−d+1+2−2d, 2−d−2−2d, 2−d−2−2d, and 2−2d, respectively.
• Huffman code: 0 for (a1,a1); 10 for (a1,a2); 110 for (a2,a1); 111 for (a2,a2).
Example
Ilya Pollak
• (X1,X2) will have four outcomes, (a1,a1), (a1,a2), (a2,a1), (a2,a2), with probabilities 1−2−d+1+2−2d, 2−d−2−2d, 2−d−2−2d, and 2−2d, respectively.
• Huffman code: 0 for (a1,a1); 10 for (a1,a2); 110 for (a2,a1); 111 for (a2,a2).
• Expected codeword length per random variable: – [1−2−d+1+2−2d + 2(2−d−2−2d) + 3(2−d−2−2d)+ 3(2−2d)]/2
Example
Ilya Pollak
• (X1,X2) will have four outcomes, (a1,a1), (a1,a2), (a2,a1), (a2,a2), with probabilities 1−2−d+1+2−2d, 2−d−2−2d, 2−d−2−2d, and 2−2d, respectively.
• Huffman code: 0 for (a1,a1); 10 for (a1,a2); 110 for (a2,a1); 111 for (a2,a2).
• Expected codeword length per random variable: – [1−2−d+1+2−2d + 2(2−d−2−2d) + 3(2−d−2−2d)+ 3(2−2d)]/2 – This is 0.500001 for d=20
Example
Ilya Pollak
• (X1,X2) will have four outcomes, (a1,a1), (a1,a2), (a2,a1), (a2,a2), with probabilities 1−2−d+1+2−2d, 2−d−2−2d, 2−d−2−2d, and 2−2d, respectively.
• Huffman code: 0 for (a1,a1); 10 for (a1,a2); 110 for (a2,a1); 111 for (a2,a2).
• Expected codeword length per random variable: – [1−2−d+1+2−2d + 2(2−d−2−2d) + 3(2−d−2−2d)+ 3(2−2d)]/2 – This is 0.500001 for d=20
• Can get arbitrarily close to entropy by encoding longer sequences of Xk’s.
Source coding theorem for sequences of independent, identically distributed
random variables
Suppose we are jointly encoding independent, identically distributed discreterandom variables X1,…,XN , each taking values in {a1,…,aN}.For any uniquely decodable code, the expected codeword length is ≥ H (Xn ).Moreover, there exists a prefix condition code for which the expected codeword
length is < H (Xn ) + 1N
.
Ilya Pollak
Proof of the source coding theorem for iid sequences
Consider random vector X = X1,…,XN( ). The self-information of its outcome x = x1,…, xN( ) isI(x) = − log2 pX1 ,…,XN
x1,…, xN( )
Ilya Pollak
Proof of the source coding theorem for iid sequences
Consider random vector X = X1,…,XN( ). The self-information of its outcome x = x1,…, xN( ) is
I(x) = − log2 pX1 ,…,XNx1,…, xN( ) = − log2 pXn xn( )
n=1
N
∑ = I xn( )n=1
N
∑ .
Ilya Pollak
Proof of the source coding theorem for iid sequences
Consider random vector X = X1,…,XN( ). The self-information of its outcome x = x1,…, xN( ) is
I(x) = − log2 pX1 ,…,XNx1,…, xN( ) = − log2 pXn xn( )
n=1
N
∑ = I xn( )n=1
N
∑ .
Therefore, the entropy of X is
H X( ) = E I X( )⎡⎣ ⎤⎦ = E I Xn( )n=1
N
∑⎡⎣⎢
⎤⎦⎥= H Xn( )
n=1
N
∑ = NH Xn( ).
Ilya Pollak
Proof of the source coding theorem for iid sequences
Consider random vector X = X1,…,XN( ). The self-information of its outcome x = x1,…, xN( ) is
I(x) = − log2 pX1 ,…,XNx1,…, xN( ) = − log2 pXn xn( )
n=1
N
∑ = I xn( )n=1
N
∑ .
Therefore, the entropy of X is
H X( ) = E I X( )⎡⎣ ⎤⎦ = E I Xn( )n=1
N
∑⎡⎣⎢
⎤⎦⎥= H Xn( )
n=1
N
∑ = NH Xn( ).
Therefore, applying the single-symbol source coding theorem to X, we have:H X( ) ≤ E CN[ ] < H X( ) +1,
where E CN[ ] is the expected codeword length for the optimal uniquely decodable code for X
Ilya Pollak
Proof of the source coding theorem for iid sequences
Consider random vector X = X1,…,XN( ). The self-information of its outcome x = x1,…, xN( ) is
I(x) = − log2 pX1 ,…,XNx1,…, xN( ) = − log2 pXn xn( )
n=1
N
∑ = I xn( )n=1
N
∑ .
Therefore, the entropy of X is
H X( ) = E I X( )⎡⎣ ⎤⎦ = E I Xn( )n=1
N
∑⎡⎣⎢
⎤⎦⎥= H Xn( )
n=1
N
∑ = NH Xn( ).
Therefore, applying the single-symbol source coding theorem to X, we have:H X( ) ≤ E CN[ ] < H X( ) +1,
NH Xn( ) ≤ E CN[ ] < NH Xn( ) +1,
where E CN[ ] is the expected codeword length for the optimal uniquely decodable code for X
Ilya Pollak
Proof of the source coding theorem for iid sequences
Consider random vector X = X1,…,XN( ). The self-information of its outcome x = x1,…, xN( ) is
I(x) = − log2 pX1 ,…,XNx1,…, xN( ) = − log2 pXn xn( )
n=1
N
∑ = I xn( )n=1
N
∑ .
Therefore, the entropy of X is
H X( ) = E I X( )⎡⎣ ⎤⎦ = E I Xn( )n=1
N
∑⎡⎣⎢
⎤⎦⎥= H Xn( )
n=1
N
∑ = NH Xn( ).
Therefore, applying the single-symbol source coding theorem to X, we have:H X( ) ≤ E CN[ ] < H X( ) +1,
NH Xn( ) ≤ E CN[ ] < NH Xn( ) +1,
H Xn( ) ≤ E C[ ] < H Xn( ) + 1N
,
where E CN[ ] is the expected codeword length for the optimal uniquely decodable code for X,
and E C[ ] = E CN[ ]N
is the corresponding expected codeword length per symbol.
Ilya Pollak
Arithmetic coding
Ilya Pollak
• Another form of entropy coding. • More amenable to coding long sequences of symbols than
Huffman coding. • Can be used in conjunction with on-line learning of conditional
probabilities to encode dependent sequences of symbols: – Q-coder in JPEG (JPEG also has a Huffman coding option) – QM-coder in JBIG – MQ-coder in JPEG-2000 – CABAC coder in H.264/MPEG-4 AVC