25 quantization and_compression

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.

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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.

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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.

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•  Sampling is typically done nowadays during signal acquisition (e.g., digital camera for pictures or audio recording equipment for music and speech).

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•  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.

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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.

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•  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.

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since it can be any number between 0 and 255. •  It would take 25 bytes to store a 5x5 image.

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since it can be any number between 0 and 255. •  It would take 25 bytes to store a 5x5 image. •  Can we do better?

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255 255 255 255 255

200 200 200 200 200

200 200 200 200 100

Can we do better than 25 bytes?

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Two key ideas •  Idea #1:

–  Transform the data to create lots of zeros.

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–  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]

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–  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

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–  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!

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–  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.

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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

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Entropy coding

value of X 0 255 −55 −100 codeword 00 01 10 11

Consider the following fixed-length encoder:

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Entropy coding

For a file with 25 numbers, E[file size] = 25*2*(22/25+1/25+1/25+1/25) = 50 bits

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Entropy coding

Now consider the following encoder:

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Entropy coding

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!

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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.

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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.

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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?

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Can we do even better than 30 bits? What about this alternative encoder?

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E[file size] = 25(22/25 + 2/25 + 1/25+2/25) = 27 bits

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E[file size] = 25(22/25 + 2/25 + 1/25+2/25) = 27 bits Is there anything wrong with this encoder?

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The second encoding is not uniquely decodable!

Encoded string ‘01’ could either be 255 or 0 followed by −55

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The second encoding is not uniquely decodable!

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.

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What kinds of distributions are amenable to entropy coding?

a b c d0

a b c d

Cannot do better than two bits per symbol

Can do a lot better than two bits per symbol

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a b c d0

a b c d

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).

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a b c d0

a b c d

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.

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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

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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

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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}.

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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.

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•  Discretizing the argument values t, x, and y (or sampling), is studied in ECE 301, 438, and 440.

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•  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.

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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.

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How to compare two quantizers?

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•  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

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[From Mannos-Sakrison IEEE-IT 1974]

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High and low frequencies may be quantized more coarsely

[From Mannos-Sakrison IEEE-IT 1974]

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But there are many other intricacies in the way human

visual system computes similarity…

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Are these two images similar?

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What about these two?

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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.

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Scalar vs Vector Quantization

r 0 127

•  quantize each value separately •  simple thresholding

•  quantize several values jointly •  more complex

r 0 95 255

95 127

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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.

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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.

r 0 95 255

95 127

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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.

r 0 95 255

95 127

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.

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•  Does it make sense to do scalar quantization with different quantization bins for different variables?

More on Scalar Quantization

255 r 0 127

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•  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.

255 r 0 127

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•  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.

255 r 0 127

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Structure of a Typical Lossy Compression Algorithm for Audio,

Images, or Video

transform quantization entropy coding

compressed bitstream data

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Structure of a Typical Lossy Compression Algorithm for Audio,

Images, or Video

Let’s more closely consider quantization and entropy coding. (Various transforms are considered in ECE 301 and ECE 438.)

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Quantization: problem statement

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Source (e.g., image, video, speech signal)

Sequence of discrete or continuous random variables X(1),…,X(N) (e.g., transformed image pixel values).

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Source (e.g., image, video, speech signal) Quantizer

Sequence of discrete random variables Y(1),…,Y(N), each distributed over a finite set of values (quantization levels)

Errors: D(1),…,D(N) where D(n) = X(n) − Y(n)

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Source (e.g., image, video, speech signal) Quantizer

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

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•  Suppose data X(1),…,X(N) are quantized, to result in Y(1),…,Y(N).

E X(n) −Y (n)( )2

∑⎡⎣⎢

⎤⎦⎥= E D(n)( )2

∑⎡⎣⎢

⎤⎦⎥

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.

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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.

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Uniform vs non-uniform quantization

•  Uniform quantization is not a good strategy for distributions which significantly differ from uniform.

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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.

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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.

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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.

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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 = ∞ – 

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−∞ < q1 < x2 ≤ q2 < x3 ≤ q3 <… ≤ qL−1 < xL ≤ qL < +∞I.e., qk ∈k-th quantization interval

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

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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].

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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.

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Scalar Lloyd-Max quantizer: derivation

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E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞

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E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞

∫ = y(x) − x( )2 fX (x)dxxk

∫k=1

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E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞

∫ = y(x) − x( )2 fX (x)dxxk

∫k=1

∑ = qk − x( )2 fX (x)dxxk

∫k=1

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E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞

∫ = y(x) − x( )2 fX (x)dxxk

∫k=1

Minimize w.r.t. qk : ∂∂qk

E Y − X( )2⎡⎣ ⎤⎦ = 2 qk − x( ) fX (x)dxxk

∫ = 0

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E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞

∫ = y(x) − x( )2 fX (x)dxxk

∫k=1

E Y − X( )2⎡⎣ ⎤⎦ = 2 qk − x( ) fX (x)dxxk

∫ = 0

qk fX (x)dxxk

∫ = xfX (x)dxxk

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E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞

∫ = y(x) − x( )2 fX (x)dxxk

∫k=1

E Y − X( )2⎡⎣ ⎤⎦ = 2 qk − x( ) fX (x)dxxk

∫ = 0

qk fX (x)dxxk

∫ = xfX (x)dxxk

∫ , therefore qk =xfX (x)dx

fX (x)dxxk

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E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞

∫ = y(x) − x( )2 fX (x)dxxk

∫k=1

E Y − X( )2⎡⎣ ⎤⎦ = 2 qk − x( ) fX (x)dxxk

∫ = 0

qk fX (x)dxxk

∫ = xfX (x)dxxk

fX (x)dxxk

∫= E X | X ∈k-th quantization interval[ ]

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E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞

∫ = y(x) − x( )2 fX (x)dxxk

∫k=1

E Y − X( )2⎡⎣ ⎤⎦ = 2 qk − x( ) fX (x)dxxk

∫ = 0

qk fX (x)dxxk

∫ = xfX (x)dxxk

fX (x)dxxk

∫= E X | X ∈k-th quantization interval[ ]

This is a minimum, since ∂2

∂qk2 E Y − X( )2⎡⎣ ⎤⎦ = 2 fX (x)dx

∫ > 0.

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E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞

∫ = y(x) − x( )2 fX (x)dxxk

∫k=1

Minimize w.r.t. xk , for k = 2,…,L

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E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞

∫ = y(x) − x( )2 fX (x)dxxk

∫k=1

Minimize w.r.t. xk , for k = 2,…,L:

∂∂xk

E Y − X( )2⎡⎣ ⎤⎦ =∂∂xk

qk−1 − x( )2 fX (x)dxxk−1

∫ + qk − x( )2 fX (x)dxxk

∫⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

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E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞

∫ = y(x) − x( )2 fX (x)dxxk

∫k=1

∂∂xk

E Y − X( )2⎡⎣ ⎤⎦ =∂∂xk

∫⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

= qk−1 − xk( )2 fX (xk ) − qk − xk( )2 fX (xk )

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E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞

∫ = y(x) − x( )2 fX (x)dxxk

∫k=1

∂∂xk

E Y − X( )2⎡⎣ ⎤⎦ =∂∂xk

∫⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

= 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 .

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E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞

∫ = y(x) − x( )2 fX (x)dxxk

∫k=1

∂∂xk

E Y − X( )2⎡⎣ ⎤⎦ =∂∂xk

∫⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

= 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.

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E Y − X( )2⎡⎣ ⎤⎦ = y(x) − x( )2 fX (x)dx−∞

∫ = y(x) − x( )2 fX (x)dxxk

∫k=1

∂∂xk

E Y − X( )2⎡⎣ ⎤⎦ =∂∂xk

∫⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

= 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.

Nonlinear system to be solved

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qk =xfX (x)dx

fX (x)dxxk

∫= E X | X ∈k-th quantization interval[ ], for k = 1,…,L

xk =qk−1 + qk

2, for k = 2,…,L

⎪⎪⎪⎪

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qk =xfX (x)dx

fX (x)dxxk

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.

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qk =xfX (x)dx

fX (x)dxxk

xk =qk−1 + qk

2, for k = 2,…,L

⎪⎪⎪⎪

•  In general, an approximate solution can be found numerically, via an iterative algorithm (e.g., lloyds command in Matlab).

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qk =xfX (x)dx

fX (x)dxxk

xk =qk−1 + qk

2, for k = 2,…,L

⎪⎪⎪⎪

•  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?

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X = X(1),…,X(N )( ) = source random vector with a given joint distribution.L = a desired number of quantization points.

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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

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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

∑⎡⎣⎢

⎤⎦⎥

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

∑⎡⎣⎢

⎤⎦⎥

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…

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Problem statement

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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

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

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

… …

Unique Decodability

•  How to decode the following string: 0001? •  It could be aaab or aad or acb or cab or cd. •  Not uniquely decodable!

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symbol codeword

A condition that ensures unique decodability

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•  Prefix condition: no codeword in the code is a prefix for any other codeword.

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•  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.

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•  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.

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•  Prefix condition holds if an only if all the codewords are leaves of the binary tree.

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•  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

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symbol codeword

•  Codeword 0 is a prefix for both codeword 00 and codeword 01

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symbol codeword

0 wa=0

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symbol codeword

0 wa=0

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symbol codeword

Example: prefix condition, all words are leaves

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symbol codeword

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symbol codeword

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symbol codeword

wd=001

0 wc=000

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symbol codeword

wd=001

0 wc=000

•  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

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symbol codeword

wd=001

0 wc=000

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.

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wa=1 wb=01

wd=001

0 wc=000

000001101 How to decode the following string?

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wa=1 wb=01

wd=001

0 wc=000

000001101

Ilya Pollak

wa=1 wb=01

wd=001

0 wc=000

000001101

Ilya Pollak

wa=1 wb=01

wd=001

0 wc=000

000001101

Ilya Pollak

wa=1 wb=01

wd=001

0 wc=000

000001101

output: c

Ilya Pollak

wa=1 wb=01

wd=001

0 wc=000

000001101

output: c

Ilya Pollak

wa=1 wb=01

wd=001

0 wc=000

000001101

output: c

Ilya Pollak

wa=1 wb=01

wd=001

0 wc=000

000001101

output: c

Ilya Pollak

wa=1 wb=01

wd=001

0 wc=000

000001101

output: cd

Ilya Pollak

wa=1 wb=01

wd=001

0 wc=000

000001101

output: cd

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wd=001

0 wc=000

000001101

output: cda

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wd=001

0 wc=000

000001101

output: cda

Ilya Pollak

wd=001

0 wc=000

000001101

output: cda

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wd=001

0 wc=000

000001101

output: cdab

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wd=001

0 wc=000

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

•  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}.)

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•  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.

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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.

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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.)

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•  Consider “aggregate outcome” a12 = {a1,a2} and a discrete r.v. X’ such that

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X ' = a12 if X = a1 or X = a2

X otherwise

⎧⎨⎪

⎩⎪

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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

⎧⎨⎪

⎩⎪

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

⎧⎨⎪

⎩⎪

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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

⎧⎨⎪

⎩⎪

Example

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letter pX(letter)

a1 0.10

a2 0.10

a3 0.25

a4 0.25

a5 0.30

Example

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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

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letter pX(letter)

a1 0.10

a2 0.10

a3 0.25

a4 0.25

a5 0.30

a12 0.20

a3 0.25

a4 0.25

a5 0.30

Example

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letter pX(letter)

a1 0.10

a2 0.10

a3 0.25

a4 0.25

a5 0.30

a12 0.20

a3 0.25

a4 0.25

a5 0.30

a12 Tree for X:

Tree for X’ (still to be constructed)

Example

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Step 2: combine the two least likely letters from the new alphabet.

letter pX’’(letter)

a123 0.45

a4 0.25

a5 0.30

a12 0.20

a3 0.25

a4 0.25

a5 0.30

Example

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a123 0.45

a4 0.25

a5 0.30

a12 0.20

a3 0.25

a4 0.25

a5 0.30

Example

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a123 0.45

a4 0.25

a5 0.30

a12 0.20

a3 0.25

a4 0.25

a5 0.30

a12 Tree for X:

Tree for X’’

Example

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a123 0.45

a4 0.25

a5 0.30

a12 0.20

a3 0.25

a4 0.25

a5 0.30

a12 Tree for X:

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

a123 0.45

a4 0.25

a5 0.30

a4 a45

Example

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a123 0.45

a45 0.55

a12 Tree for X:

a123 0.45

a4 0.25

a5 0.30

a4 a45

Tree for X’’’

Example

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a123 0.45

a45 0.55

a12 Tree for X:

a123 0.45

a4 0.25

a5 0.30

a4 a45

Tree for X’’

Tree for X’’’

Example

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a123 0.45

a45 0.55

a12 Tree for X:

Tree for X’

a123 0.45

a4 0.25

a5 0.30

a4 a45

Tree for X’’

Tree for X’’’

Example

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Step 4: combine the last two remaining letters letter pX’’’(letter)

a123 0.45

a45 0.55 Done!

a12 Tree for X:

a4 a45

a12345 1

Example

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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.

Tree for X:

Example

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Tree for X:

letter pX(letter) codeword

a1 0.10 111

a2 0.10

a3 0.25

a4 0.25

a5 0.30

Example

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Tree for X:

a1 0.10 111

a2 0.10 110

a3 0.25

a4 0.25

a5 0.30

Example

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Tree for X:

a1 0.10 111

a2 0.10 110

a3 0.25 10

a4 0.25

a5 0.30

Example

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Tree for X:

a1 0.10 111

a2 0.10 110

a3 0.25 10

a4 0.25 01

a5 0.30

Example

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Tree for X:

a1 0.10 111

a2 0.10 110

a3 0.25 10

a4 0.25 01

a5 0.30 00

Example

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Tree for X:

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.

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outcomes a1,…,aM and with PMF pX. •  Self-information of outcome am is I(am) = −log2 pX(am) bits.

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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.

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•  Self-information of X is I(X) = −log2 pX(X) and is a random variable.

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•  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:

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H (X) = E I(X)[ ] = − pX (am )log2m=1

∑ 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

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•  Suppose that X has M=2K possible outcomes a1,…,aM.

Example

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•  Suppose that X has M=2K possible outcomes a1,…,aM. •  Suppose that X is uniform, i.e., pX (a1) = … = pX (aM) = 2−K.

Example

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•  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

∑ 2−K( ) = 2K −2−K( ) −K( ) = K

Example

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H (X) = E I(X)[ ] = − 2−K log2k=1

∑ 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

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H (X) = E I(X)[ ] = − 2−K log2k=1

∑ 2−K( ) = 2K −2−K( ) −K( ) = 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

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H (X) = E I(X)[ ] = − 2−K log2k=1

∑ 2−K( ) = 2K −2−K( ) −K( ) = 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

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•  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

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If integers d1,…,dM satisfy the inequality

2−dm

∑ ≤ 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

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A full binary tree of depth D has 2D leaves.

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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

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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

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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.

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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 .

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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.

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 .

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

∑ = 2dM

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2dM −dm

∑ = 2dM ⇔ 2−dm

∑ = 1

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2dM −dm

∑ = 2dM ⇔ 2−dm

∑ = 1

Therefore, 2−dm

∑ = 2−dm

∑ + 2−dm

m= r+1

∑ > 1. This violates (1).

Ilya Pollak

2dM −dm

∑ = 2dM ⇔ 2−dm

∑ = 1

Therefore, 2−dm

∑ = 2−dm

∑ + 2−dm

m= r+1

∑ > 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 .

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.

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 .

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 .

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

∑ ≤ 2dM

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

∑ ≤ 2dM ⇔ 2−dm

∑ ≤ 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.

Ilya Pollak

H (X) − E[C] = − pX (am )log2m=1

∑ pX (am ) − pX (am )dmm=1

Ilya Pollak

H (X) − E[C] = − pX (am )log2m=1

∑ = pX (am ) log21

pX (am )⎛⎝⎜

⎞⎠⎟− log2 2dm

⎣⎢

⎦⎥

Ilya Pollak

H (X) − E[C] = − pX (am )log2m=1

∑ = pX (am ) log21

pX (am )⎛⎝⎜

⎞⎠⎟− log2 2dm

⎣⎢

⎦⎥

= pX (am ) log21

pX (am )2dm

⎛⎝⎜

⎞⎠⎟

⎣⎢

⎦⎥

Ilya Pollak

H (X) − E[C] = − pX (am )log2m=1

∑ = pX (am ) log21

pX (am )⎛⎝⎜

⎞⎠⎟− log2 2dm

⎣⎢

⎦⎥

= pX (am ) log21

pX (am )2dm

⎛⎝⎜

⎞⎠⎟

⎣⎢

⎦⎥

≤ pX (am ) 1pX (am )2dm

−1⎛⎝⎜

⎞⎠⎟m=1

∑ log2 e (by Lemma 1)

Ilya Pollak

H (X) − E[C] = − pX (am )log2m=1

∑ = pX (am ) log21

pX (am )⎛⎝⎜

⎞⎠⎟− log2 2dm

⎣⎢

⎦⎥

= pX (am ) log21

pX (am )2dm

⎛⎝⎜

⎞⎠⎟

⎣⎢

⎦⎥

−1⎛⎝⎜

⎞⎠⎟m=1

= 12dm

∑ − pX (am )m=1

∑⎛⎝⎜

⎞⎠⎟

log2 e

Ilya Pollak

H (X) − E[C] = − pX (am )log2m=1

∑ = pX (am ) log21

pX (am )⎛⎝⎜

⎞⎠⎟− log2 2dm

⎣⎢

⎦⎥

= pX (am ) log21

pX (am )2dm

⎛⎝⎜

⎞⎠⎟

⎣⎢

⎦⎥

−1⎛⎝⎜

⎞⎠⎟m=1

= 12dm

∑ − pX (am )m=1

∑⎛⎝⎜

⎞⎠⎟

log2 e

= 2−dm

∑ −1⎛⎝⎜

⎞⎠⎟

log2 e ≤ 0

Ilya Pollak

H (X) − E[C] = − pX (am )log2m=1

∑ = pX (am ) log21

pX (am )⎛⎝⎜

⎞⎠⎟− log2 2dm

⎣⎢

⎦⎥

= pX (am ) log21

pX (am )2dm

⎛⎝⎜

⎞⎠⎟

⎣⎢

⎦⎥

−1⎛⎝⎜

⎞⎠⎟m=1

= 12dm

∑ − pX (am )m=1

∑⎛⎝⎜

⎞⎠⎟

log2 e

= 2−dm

∑ −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 )

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 )

Ilya Pollak

dm ≥ − log2 pX (am ) ⇒ − dm ≤ log2 pX (am ) ⇒ 2−dm ≤ pX (am ) ⇒ 2−dm

∑ ≤ pX (am )m=1

∑ = 1.

Ilya Pollak

∑ ≤ pX (am )m=1

∑ = 1.

Therefore, Kraft inequality is satisfied, and we can construct a prefix condition code with codewordlengths d1,…,dM .

Ilya Pollak

∑ ≤ pX (am )m=1

∑ = 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

Ilya Pollak

∑ ≤ pX (am )m=1

∑ = 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 )

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∑ ≤ pX (am )m=1

∑ = 1.

⇒ pX (am )dmm=1

∑ < − pX (am )log2 pX (am ) + pX (am )( )m=1

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∑ ≤ pX (am )m=1

∑ = 1.

⇒ pX (am )dmm=1

⇒ E[C] < − pX (am )log2 pX (am )( )m=1

∑ + pX (am )m=1

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∑ ≤ pX (am )m=1

∑ = 1.

⇒ pX (am )dmm=1

⇒ E[C] < − pX (am )log2 pX (am )( )m=1

∑ + pX (am )m=1

∑ = H (X) +1

Note: Huffman code may often be very far from the entropy

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•  Let X have two outcomes, a1 and a2, with probabilities 1−2−d

and 2−d, respectively.

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and 2−d, respectively. •  Huffman code: 0 for a1; 1 for a2. •  Expected codeword length: 1.

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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.

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example, if d=20, this is 0.0000204493. •  Problem: no codeword can have fractional numbers of bits!

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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

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•  (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

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•  Huffman code: 0 for (a1,a1); 10 for (a1,a2); 110 for (a2,a1); 111 for (a2,a2).

Example

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•  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

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•  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

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•  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

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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( )

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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( )

∑ = I xn( )n=1

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∑ = I xn( )n=1

Therefore, the entropy of X is

H X( ) = E I X( )⎡⎣ ⎤⎦ = E I Xn( )n=1

∑⎡⎣⎢

⎤⎦⎥= H Xn( )

∑ = NH Xn( ).

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∑ = I xn( )n=1

H X( ) = E I X( )⎡⎣ ⎤⎦ = E I Xn( )n=1

∑⎡⎣⎢

⎤⎦⎥= H Xn( )

∑ = 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

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∑ = I xn( )n=1

H X( ) = E I X( )⎡⎣ ⎤⎦ = E I Xn( )n=1

∑⎡⎣⎢

⎤⎦⎥= H Xn( )

∑ = NH Xn( ).

NH Xn( ) ≤ E CN[ ] < NH Xn( ) +1,

where E CN[ ] is the expected codeword length for the optimal uniquely decodable code for X

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∑ = I xn( )n=1

H X( ) = E I X( )⎡⎣ ⎤⎦ = E I Xn( )n=1

∑⎡⎣⎢

⎤⎦⎥= H Xn( )

∑ = NH Xn( ).

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.

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Arithmetic coding

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•  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

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