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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discretized.
•  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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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.

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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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grayscale images
•  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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grayscale images
•  It would take 25 bytes to store a 5x5 image.

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grayscale images
•  It would take 25 bytes to store a 5x5 image.
•  Can we do better?

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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.

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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]

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Two key ideas
•  Idea #1:
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:
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:
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
Suppose we are encoding realizations of a discrete random variable X such that
value of X

0

255

−55

−100

probability

22/25

1/25

1/25

1/25

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Entropy coding
value of X

0

255

−55

−100

probability

22/25

1/25

1/25

1/25

Consider the following fixed-length encoder:
value of X

0

255

−55

−100

codeword

00

01

10

11

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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

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

Now consider the following encoder:
value of X

0

255

−55

−100

codeword

1

01

000

001

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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

Now consider the following encoder:
value of X

0

255

−55

−100

codeword

1

01

000

001

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

value of X

0

255

−55

−100

probability

22/25

1/25

1/25

1/25

codeword

1

01

000

001

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

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value of X

0

255

−55

−100

probability

22/25

1/25

1/25

1/25

codeword

1

01

000

001

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

value of X

0

255

−55

−100

probability

22/25

1/25

1/25

1/25

codeword

1

01

000

001

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

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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?
0.7
0.6
0.5
0.4
0.3

0.3

0.2

0.2

0.1

0.1

0
a

b

c

d

Can do a lot better than
two bits per symbol

0
a

b

c

d

Cannot do better than
two bits per symbol

Ilya Pollak

0.7
0.6
0.5
0.4
0.3

0.3

0.2

0.2

0.1

0.1

0
a

b

c

d

two bits per symbol

0
a

b

c

d

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

0.7
0.6
0.5
0.4
0.3

0.3

0.2

0.2

0.1

0.1

0
a

b

c

d

two bits per symbol

0
a

b

c

d

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

253.5

253.5

253.5

253.5

253.5

254

254

254

255

254

253.5

253.5

253.5

253.5

253.5

252

255

255

254

252

253.5

253.5

253.5

253.5

253.5

253

253

254

254

254

253.5

253.5

253.5

253.5

253.5

252

255

253

252

253

253.5

253.5

253.5

253.5

253.5

Quantization

Ilya Pollak

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


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

Ilya Pollak


•  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.

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How to compare two quantizers?
•  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.

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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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[From Mannos-Sakrison IEEE-IT 1974]

High and low frequencies may be quantized more coarsely
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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
s

s

255

255

127
95

r
•  quantize each value separately
•  simple thresholding

0

127

255

0

95

255

r

•  quantize several values jointly
•  more complex

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What kinds of joint distributions are
amenable to scalar quantization?
s
255

127

r
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.
0

127

255

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s

s

255

255

127
95

r
0

127

255

r
If (r,s) are jointly uniform over yellow
region, knowing r tells us a lot about s.
0

95

255

decisions jointly.

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s

s

255

255

127
95

r
0

127

255

r
If (r,s) are jointly uniform over yellow
region, knowing r tells us a lot about s.
0

95

255

decisions jointly.

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

More on Scalar Quantization
•  Does it make sense to do scalar
quantization with different
quantization bins for different
variables?

s
255

127

0

127

255

r

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variables?
–  No reason to do this if we are
quantizing grayscale pixel values.

s
255

127

0

127

255

r

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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.

s
255

127

0

127

255

r

Ilya Pollak

Structure of a Typical Lossy
Compression Algorithm for Audio,
Images, or Video
data

transform

quantization

entropy
coding

compressed
bitstream

Ilya Pollak

Structure of a Typical Lossy
Compression Algorithm for Audio,
Images, or Video
data

transform

quantization

entropy
coding

compressed
bitstream

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

Ilya Pollak

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

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

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values).


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

Quantizer

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values).


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

Quantizer

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

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MSE is a widely used measure of
distortion of quantizers
•  Suppose data X(1),…,X(N) are quantized, to result in Y(1),…,Y(N).

⎡N
⎡N
2⎤
2⎤
E ⎢ ∑ ( X(n) − Y (n)) ⎥ = E ⎢ ∑ ( D(n)) ⎥
⎣ n =1
⎦
⎣ n =1
⎦
2
If D(1),..., D(N ) are identically distributed, this is the same as NE ⎡( D(n)) ⎤ , for any n.
⎣
⎦

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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.

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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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–  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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•  Quantization intervals (x1,x2), [x2,x3),…[xL,xL+1) and levels q1, …, qL such that
–  x1 = −∞
–  xL+1 = ∞
–  −∞ < q1 < x2 ≤ q2 < x3 ≤ q3 < … ≤ qL −1 < x L ≤ qL < +∞
I.e., qk ∈k-th quantization interval

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•  Quantization intervals (x1,x2), [x2,x3),…[xL,xL+1) and levels q1, …, qL such that
–  x1 = −∞
–  xL+1 = ∞
–  −∞ < q1 < x2 ≤ q2 < x3 ≤ q3 < … ≤ qL −1 < x L ≤ qL < +∞
I.e., qk ∈k-th quantization interval

•  Y = the result of quantizing X, a discrete random variable with L possible
outcomes, q1, q2,…, qL, defined by
⎧
⎪
⎪
⎪
Y = Y (X) = ⎨
⎪
⎪
⎪
⎩

q1

if X < x2

q2

if x 2 ≤ X < x3



qL −1 if x L −1 ≤ X < x L
qL

X ≥ xL
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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].

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
E ⎡(Y − X ) ⎤ =
⎣
⎦
2

∞

( y(x) − x )2 f X (x)dx
∫

−∞

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E ⎡(Y − X ) ⎤ =
⎣
⎦
2

∞

∫ ( y(x) − x )

−∞

2

L xk+1

f X (x)dx = ∑

( y(x) − x )2 f X (x)dx
∫

k =1 xk

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E ⎡(Y − X ) ⎤ =
⎣
⎦
2

∞

∫ ( y(x) − x )

−∞

2

L xk+1

f X (x)dx = ∑

∫ ( y(x) − x )

k =1 xk

2

L xk+1

f X (x)dx = ∑

( qk − x )2 f X (x)dx
∫

k =1 xk

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E ⎡(Y − X ) ⎤ =
⎣
⎦
2

∞

∫ ( y(x) − x )

−∞

2

L xk+1

f X (x)dx = ∑

∂
2
Minimize w.r.t. qk :
E ⎡(Y − X ) ⎤ =
⎦
∂qk ⎣

∫ ( y(x) − x )

k =1 xk

xk+1

∫ 2 (q

k

2

L xk+1

f X (x)dx = ∑

( qk − x )2 f X (x)dx
∫

k =1 xk

− x ) f X (x)dx = 0

xk

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E ⎡(Y − X ) ⎤ =
⎣
⎦
2

∞

∫ ( y(x) − x )

−∞

2

f X (x)dx = ∑

∂
2
E ⎡(Y − X ) ⎤ =
⎦
∂qk ⎣
xk+1

∫q

xk

L xk+1

∫ ( y(x) − x )

k =1 xk

xk+1

∫ 2 (q

k

2

L xk+1

f X (x)dx = ∑

( qk − x )2 f X (x)dx
∫

k =1 xk

− x ) f X (x)dx = 0

xk

xk+1

f (x)dx =

k X

∫

xf X (x)dx

xk

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E ⎡(Y − X ) ⎤ =
⎣
⎦
2

∞

∫ ( y(x) − x )

−∞

2

L xk+1

f X (x)dx = ∑

∂
2
E ⎡(Y − X ) ⎤ =
⎦
∂qk ⎣

∫ ( y(x) − x )

k =1 xk

xk+1

∫ 2 (q

k

2

L xk+1

f X (x)dx = ∑

( qk − x )2 f X (x)dx
∫

k =1 xk

− x ) f X (x)dx = 0

xk

xk+1
xk+1

∫

xk

xk+1

qk f X (x)dx =

∫

xk

xf X (x)dx, therefore qk =

∫

xf X (x)dx

∫

f X (x)dx

xk
xk+1

xk

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E ⎡(Y − X ) ⎤ =
⎣
⎦
2

∞

∫ ( y(x) − x )

−∞

2

L xk+1

f X (x)dx = ∑

∂
2
E ⎡(Y − X ) ⎤ =
⎦
∂qk ⎣

∫ ( y(x) − x )

k =1 xk

xk+1

∫ 2 (q

k

2

L xk+1

f X (x)dx = ∑

( qk − x )2 f X (x)dx
∫

k =1 xk

− x ) f X (x)dx = 0

xk

xk+1
xk+1

∫

xk

xk+1

qk f X (x)dx =

∫

xk


∫

xf X (x)dx

∫

f X (x)dx

xk
xk+1

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

xk

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E ⎡(Y − X ) ⎤ =
⎣
⎦
2

∞

∫ ( y(x) − x )

−∞

2

L xk+1

f X (x)dx = ∑

∂
2
E ⎡(Y − X ) ⎤ =
⎦
∂qk ⎣

∫ ( y(x) − x )

k =1 xk

xk+1

∫ 2 (q

k

2

L xk+1

f X (x)dx = ∑

( qk − x )2 f X (x)dx
∫

k =1 xk

− x ) f X (x)dx = 0

xk

xk+1
xk+1

∫

xk

xk+1

qk f X (x)dx =

∫

xk


∫

xf X (x)dx

∫

f X (x)dx

xk
xk+1

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

xk

∂2
2
This is a minimum, since 2 E ⎡(Y − X ) ⎤ =
⎦
∂qk ⎣

xk+1

∫ 2f

X

(x)dx > 0.

xk

Ilya Pollak

E ⎡(Y − X ) ⎤ =
⎣
⎦
2

∞

∫ ( y(x) − x )

−∞

2

L xk+1

f X (x)dx = ∑

∫ ( y(x) − x )

k =1 xk

2

L xk+1

f X (x)dx = ∑

( qk − x )2 f X (x)dx
∫

k =1 xk

Minimize w.r.t. xk , for k = 2,…, L

Ilya Pollak

E ⎡(Y − X ) ⎤ =
⎣
⎦
2

∞

∫ ( y(x) − x )

−∞

2

L xk+1

f X (x)dx = ∑

∫ ( y(x) − x )

k =1 xk

2

L xk+1

f X (x)dx = ∑

( qk − x )2 f X (x)dx
∫

k =1 xk

Minimize w.r.t. xk , for k = 2,…, L:
x
xk+1
⎫
∂
∂ ⎧ k
2
2
⎪
⎪
2
⎡(Y − X ) ⎤ =
E⎣
⎨ ∫ ( qk −1 − x ) f X (x)dx + ∫ ( qk − x ) f X (x)dx ⎬
⎦ ∂x
∂xk
k ⎪ xk−1
⎪
xk
⎩
⎭

Ilya Pollak

E ⎡(Y − X ) ⎤ =
⎣
⎦
2

∞

∫ ( y(x) − x )

2

L xk+1

f X (x)dx = ∑

∫ ( y(x) − x )

k =1 xk

−∞

2

L xk+1

f X (x)dx = ∑

( qk − x )2 f X (x)dx
∫

k =1 xk

x
xk+1
⎫
∂
∂ ⎧ k
2
2
⎪
⎪
2
⎡(Y − X ) ⎤ =
E⎣
⎦ ∂x
∂xk
k ⎪ xk−1
⎪
xk
⎩
⎭

= ( qk −1 − xk ) f X (xk ) − ( qk − xk ) f X (xk )
2

2

Ilya Pollak

E ⎡(Y − X ) ⎤ =
⎣
⎦
2

∞

∫ ( y(x) − x )

2

L xk+1

f X (x)dx = ∑

∫ ( y(x) − x )

k =1 xk

−∞

2

L xk+1

f X (x)dx = ∑

( qk − x )2 f X (x)dx
∫

k =1 xk

x
xk+1
⎫
∂
∂ ⎧ k
2
2
⎪
⎪
2
⎡(Y − X ) ⎤ =
E⎣
⎦ ∂x
∂xk
k ⎪ xk−1
⎪
xk
⎩
⎭

= ( qk −1 − xk ) f X (xk ) − ( qk − xk ) f X (xk ) = ( qk −1 − qk ) ( qk −1 + qk − 2xk ) f X (xk ) = 0.
2

2

By assumption, f X (x) ≠ 0 and qk −1 ≠ qk .

Ilya Pollak

E ⎡(Y − X ) ⎤ =
⎣
⎦
2

∞

∫ ( y(x) − x )

2

L xk+1

f X (x)dx = ∑

∫ ( y(x) − x )

k =1 xk

−∞

2

L xk+1

f X (x)dx = ∑

( qk − x )2 f X (x)dx
∫

k =1 xk

x
xk+1
⎫
∂
∂ ⎧ k
2
2
⎪
⎪
2
⎡(Y − X ) ⎤ =
E⎣
⎦ ∂x
∂xk
k ⎪ xk−1
⎪
xk
⎩
⎭

2

2

By assumption, f X (x) ≠ 0 and qk −1 ≠ qk . Therefore,
q + qk
xk = k −1
, for k = 2,…, L.
2

Ilya Pollak

E ⎡(Y − X ) ⎤ =
⎣
⎦
2

∞

∫ ( y(x) − x )

2

L xk+1

f X (x)dx = ∑

∫ ( y(x) − x )

k =1 xk

−∞

2

L xk+1

f X (x)dx = ∑

( qk − x )2 f X (x)dx
∫

k =1 xk

x
xk+1
⎫
∂
∂ ⎧ k
2
2
⎪
⎪
2
⎡(Y − X ) ⎤ =
E⎣
⎦ ∂x
∂xk
k ⎪ xk−1
⎪
xk
⎩
⎭

2

2

By assumption, f X (x) ≠ 0 and qk −1 ≠ qk . Therefore,
q + qk
xk = k −1
, for k = 2,…, L.
2
∂2
2
This is a minimum, since 2 E ⎡(Y − X ) ⎤ = 2 ( qk − qk −1 ) f X (xk ) > 0.
⎦
∂xk ⎣

Ilya Pollak

Nonlinear system to be solved
xk+1
⎧
⎪
∫ xfX (x)dx
xk
⎪
qk = xk+1
= E [ X | X ∈k-th quantization interval], for k = 1,…, L
⎪
⎪
⎨
∫ fX (x)dx
⎪
xk
⎪
⎪ xk = qk −1 + qk , for k = 2,…, L
⎪
2
⎩

Ilya Pollak

xk+1
⎧
⎪
∫ xfX (x)dx
xk
⎪
qk = xk+1
⎪
⎪
⎨
∫ fX (x)dx
⎪
xk
⎪
⎪ xk = qk −1 + qk , for k = 2,…, L
⎪
2
⎩

•  Closed-form solution can be found only for very simple PDFs.
–  E.g., if X is uniform, then Lloyd-Max quantizer = uniform quantizer.

Ilya Pollak

xk+1
⎧
⎪
∫ xfX (x)dx
xk
⎪
qk = xk+1
⎪
⎪
⎨
∫ fX (x)dx
⎪
xk
⎪
⎪ xk = qk −1 + qk , for k = 2,…, L
⎪
2
⎩


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

Ilya Pollak

xk+1
⎧
⎪
∫ xfX (x)dx
xk
⎪
qk = xk+1
⎪
⎪
⎨
∫ fX (x)dx
⎪
xk
⎪
⎪ xk = qk −1 + qk , for k = 2,…, L
⎪
2
⎩


•  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.

Ilya Pollak

Vector Lloyd-Max quantizer?
X = ( X(1),…, X(N )) = source random vector with a given joint distribution.
L = a desired number of quantization points.

Ilya Pollak

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 ,…, q L ∈AL

Ilya Pollak

(2) L quantization points q1 ∈A1 ,…, q L ∈AL ,

such that the quantized random vector, defined by
Y = q k if X ∈Ak , for k = 1,…, L,
minimizes the mean-square error,
⎡N
2⎤
E ⎡ Y − X ⎤ = E ⎢ ∑ (Y (n) − X(n)) ⎥
⎣
⎦
⎣ n =1
⎦
2

Ilya Pollak

(2) L quantization points q1 ∈A1 ,…, q L ∈AL ,

such that the quantized random vector, defined by
Y = q k if X ∈Ak , for k = 1,…, L,
minimizes the mean-square error,
⎡N
2⎤
E ⎡ Y − X ⎤ = E ⎢ ∑ (Y (n) − X(n)) ⎥
⎣
⎦
⎣ n =1
⎦
2

Difficulty: cannot differentiate with respect to a set Ak , and so unless the set of all allowed
partitions is somehow restricted, this cannot be solved.

Ilya Pollak

Hopefully, prior discussion gives
you some idea about various
issues involved in quantization.
And now, on to entropy coding…
data

transform

quantization

entropy
coding

compressed
bitstream

Ilya Pollak

Problem statement
video, speech signal,
or quantizer output)

Sequence of discrete
(e.g., transformed image pixel values),
assumed to be independent and
identically distributed over a finite
alphabet {a1,…,aM}.

Ilya Pollak

Problem statement

Encoder: mapping

between source
symbols and binary
strings (codewords)

Binary string

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

Problem statement

Encoder: mapping

between source
symbols and binary
strings (codewords)

Binary string

•  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.
symbol

codeword

a1

w1

…

…

aM

wM
Ilya Pollak

Unique Decodability
symbol

codeword

a

0

b

1

c

00

d

01

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

Ilya Pollak

A condition that ensures unique
decodability
•  Prefix condition: no codeword in the code is a prefix for
any other codeword.

Ilya Pollak

decodability
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.

Ilya Pollak

decodability
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.

Ilya Pollak

decodability
any other codeword.
node forms w.
•  Prefix condition holds if an only if all the codewords are
leaves of the binary tree.

Ilya Pollak

decodability
any other codeword.
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.

Ilya Pollak

Example: no prefix condition, no unique
decodability, one word is not a leaf
symbol

codeword

a

0

b

1

c

00

d

01

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

Ilya Pollak

symbol

codeword

a

0

b

1

c
d

wa=0
0
1
wb=1
Ilya Pollak

symbol

codeword

a

0

b

1

c

00

d

wc=00

0
wa=0
0
1
wb=1
Ilya Pollak

symbol

codeword

a

0

b

1

c

00

d

01

wc=00

0
wa=0
0
1

wd=01

1
wb=1
Ilya Pollak

Example: prefix condition, all words are
leaves
symbol

codeword

a

1

b
c
d

0
1
wa=1

Ilya Pollak

leaves
symbol

codeword

a

1

b

01

c
d

0
0
1
wb=01

1
wa=1

Ilya Pollak

leaves
symbol
a

1

b

01

c

000

d

wc=000

codeword

001

0
0
1

wd=001

0
1

wb=01

1
wa=1

Ilya Pollak

leaves
symbol
a

1

b

01

c

000

d

wc=000

codeword

001

0
0
1

wd=001

0

•  No path from the root to a
codeword contains another
codeword. This is equivalent
to saying that the prefix
condition holds.

1
wb=01

1
wa=1

Ilya Pollak

leaves => unique decodability
symbol
a

1
wd=001

000

d

0

01

c

0

1

b

wc=000

codeword

001

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.
0

1
wb=01

1
wa=1

Ilya Pollak

How to decode the following string?
wc=000

000001101

0
0
1

0
1

wd=001
wb=01

1
wa=1

Ilya Pollak


wc=000

000001101

0
0
1

0
1

wd=001
wb=01

1
wa=1

Ilya Pollak


wc=000

000001101

0
0
1

output: c
0

1

wd=001
wb=01

1
wa=1

Ilya Pollak


wc=000

000001101

0
0
1

output: cd
0

1

wd=001
wb=01

1
wa=1

Ilya Pollak


wc=000

000001101

0
0
1

wd=001

output: cda
0

1
wb=01

1
wa=1

Ilya Pollak


wc=000

000001101

0
0
1

wd=001

output: cdab
0

1
wb=01

1
wa=1

Ilya Pollak


wc=000

000001101

0
0
1

wd=001

0
1

wb=01

final output:
cdab

1
wa=1

Ilya Pollak

Prefix condition and unique
decodability
•  There are uniquely decodable codes
which do not satisfy the prefix condition
(e.g., {0, 01}).

Ilya Pollak

decodability
(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

decodability
(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 “aggregate outcome” a12 = {a1,a2} and a discrete r.v. X’ such that
⎧ a12
⎪
X' = ⎨
⎪ X
⎩

if X = a1 or X = a2
otherwise

Ilya Pollak

Huffman code
⎧ a12
⎪
X' = ⎨
⎪ X
⎩

if X = a1 or X = a2
otherwise

⎧ p ( a ) + p ( a ) if a = a
⎪ X 1
X
2
12
pX ' ( a ) = ⎨
if a = a3 ,…, aM
⎪ pX ( a )
⎩

Ilya Pollak

Huffman code
⎧ a12
⎪
X' = ⎨
⎪ X
⎩

if X = a1 or X = a2
otherwise

⎧ p ( a ) + p ( a ) if a = a
⎪ X 1
X
2
12
pX ' ( a ) = ⎨
if a = a3 ,…, aM
⎪ pX ( a )
⎩

•  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.

Ilya Pollak

Huffman code
⎧ a12
⎪
X' = ⎨
⎪ X
⎩

if X = a1 or X = a2
otherwise

⎧ p ( a ) + p ( a ) if a = a
⎪ X 1
X
2
12
pX ' ( a ) = ⎨
if a = a3 ,…, aM
⎪ pX ( a )
⎩

•  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.
Ilya Pollak

letter

pX(letter)

a1

0.10

a2

0.10

a3

0.25

a4

0.25

a5

Example

0.30

Ilya Pollak

letter

pX(letter)

a1

0.10

a2

0.10

a3

0.25

a4

0.25

a5

Example

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

Ilya Pollak

letter

pX(letter)

a1

0.10

a2

0.10

a3

0.25

a4

0.25

a5

0.30

Example

a2

1

pX’(letter)

a12

0.20

a3

0.25
0.25

a5

a1

letter

a4

Step 1: combine
letters.

0.30

a12

0

Ilya Pollak

letter

pX(letter)

a1

0.10

a2

0.10

a3

0.25

a4

0.25

a5

0.30

Example

1

Tree for X:
a2

a12

pX’(letter)

a12

0.20

a3

0.25
0.25

a5

a1

letter

a4

Step 1: combine
letters.

0.30

Tree for X’
(still to be
constructed)

0

Ilya Pollak

Example
letter

pX’(letter)

a12

0.20

a3

0.25

a4

0.25

a5

Step 2: combine
letters from the new
alphabet.

0.30

letter

pX’’(letter)

a123

0.45

a4

0.25

a5

0.30

Ilya Pollak

Example
letter

pX’(letter)

a12

0.20

a3

0.25

a4

0.25

a5

Step 2: combine
alphabet.

0.30

1

0

a123

0.45

a4

0.25
0.30

a12
1

a2

pX’’(letter)

a5

a1

letter

a3

a123

0

Ilya Pollak

Example
letter

pX’(letter)

a12

0.20

a3

0.25

a4

0.25

a5

Step 2: combine
alphabet.

0.30

1

Tree for X:
0

a123

0.45

a4

0.25
0.30

a12
1

a2

pX’’(letter)

a5

a1

letter

a3

0

a123
Tree for
X’’

Ilya Pollak

Example
letter

pX’(letter)

a12

0.20

a3

0.25

a4

0.25

a5

Step 2: combine
alphabet.

0.30

1

Tree for X:
0

a12

a3

a123

0.45

a4

0.25
0.30

Tree for X’
1

a2

pX’’(letter)

a5

a1

letter

0

a123
Tree for
X’’

Ilya Pollak

Example
letter

pX’’(letter)

a123

0.45

a4

0.25

a5

0.30

Step 3: again combine
letters

a1

1

0

pX’’’(letter)

a123

0.45

a45

0.55

a12
1

a2

letter

a3
a4
a5

a123

0
1

a45

0

Ilya Pollak

Example
letter

pX’’(letter)

a123

0.45

a4

0.25

a5

0.30

letters

a1

1

Tree for X:
0

pX’’’(letter)

a123

0.45

a45

0.55

a12
1

a2

letter

a3

a123
Tree for X’’’

0

a4
a5

1

a45

0

Ilya Pollak

Example
letter

pX’’(letter)

a123

0.45

a4

0.25

a5

0.30

letters

a1

1

Tree for X:

a12

0

a3

a5

a123

0.45

a45

0.55

Tree for X’’’

0

a4

pX’’’(letter)

Tree for X’’
a123

1
a2

letter

1

a45

0

Ilya Pollak

Example
letter

pX’’(letter)

a123

0.45

a4

0.25

a5

0.30

letters

a1

1

Tree for X:
0

a12

a3

a5

a123

0.45

a45

0.55

Tree for X’’
a123
Tree for X’’’

0

a4

pX’’’(letter)

Tree for X’

1
a2

letter

1

a45

0

Ilya Pollak

Example
letter

pX’’’(letter)

a123

0.45

a45

Step 4: combine the last
two remaining letters

0.55

Done!
a1

1

Tree for X:

a12
1

a2

0

a3
a4
a5

a123
1

0
1a

45

a12345

0

0

Ilya Pollak

Example
letter

pX’’’(letter)

a123

0.45

a45

Step 4: combine the last Done! The codeword
two remaining letters
for each leaf is the sequence

0.55

of 0’1 and 1’s along the path
from the root to that leaf.
a1

1

Tree for X:

1
a2

0

a3
a4
a5

1
0
1

0

0

Ilya Pollak

Example

a1

1

Tree for X:

1
a2

0

a3
a4
a5

letter

0
1

0

0

codeword

a1

0.10

111

a2

1

pX(letter)

0.10

a3

0.25

a4

0.25

a5

0.30

Ilya Pollak

Example

a1

1

Tree for X:

1
a2

0

a3
a4
a5

letter

0
1

0

0

codeword

a1

0.10

111

a2

1

pX(letter)

0.10

110

a3

0.25

a4

0.25

a5

0.30

Ilya Pollak

Example

a1

1

Tree for X:

1
a2

0

a3
a4
a5

letter

0
1

0

0

codeword

a1

0.10

111

a2

1

pX(letter)

0.10

110

a3

0.25

10

a4

0.25

a5

0.30

Ilya Pollak

Example

a1

1

Tree for X:

1
a2

0

a3
a4
a5

letter

0
1

0

0

codeword

a1

0.10

111

a2

1

pX(letter)

0.10

110

a3

0.25

10

a4

0.25

01

a5

0.30

Ilya Pollak

Example

a1

1

Tree for X:

1
a2

0

a3
a4
a5

letter

0
1

0

0

codeword

a1

0.10

111

a2

1

pX(letter)

0.10

110

a3

0.25

10

a4

0.25

01

a5

0.30

00

Ilya Pollak

Example
Expected codeword length: 3(0.1) + 3(0.1) + 2(0.25) + 2(0.25) + 2(0.3) = 2.2 bits

a1

1

Tree for X:

1
a2

0

a3
a4
a5

letter

0
1

0

0

codeword

a1

0.10

111

a2

1

pX(letter)

0.10

110

a3

0.25

10

a4

0.25

01

a5

0.30

00

Ilya Pollak

Self-information
•  Consider again a discrete random variable X with M possible
outcomes a1,…,aM and with PMF pX.

Ilya Pollak

Self-information
•  Self-information of outcome am is I(am) = −log2 pX(am) bits.

Ilya Pollak

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

Ilya Pollak

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:
M

H (X) = E [ I(X)] = − ∑ p X (am )log 2 p X (am )
m =1

Ilya Pollak

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 codeword
length is < H (X) + 1.

Ilya Pollak

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

Ilya Pollak

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

Ilya Pollak

Example
•  Suppose that X is uniform, i.e., pX (a1) = … = pX (aM) = 2−K. Then
2K

( )

(

)

H (X) = E [ I(X)] = − ∑ 2 − K log 2 2 − K = 2 K −2 − K ( −K ) = K
k =1

Ilya Pollak

Example
2K

( )

(

)

H (X) = E [ I(X)] = − ∑ 2 − K log 2 2 − K = 2 K −2 − K ( −K ) = K
k =1

•  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.

Ilya Pollak

Example
2K

( )

(

)

H (X) = E [ I(X)] = − ∑ 2 − K log 2 2 − K = 2 K −2 − K ( −K ) = K
k =1

•  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.

Ilya Pollak

Example
2K

( )

(

)

H (X) = E [ I(X)] = − ∑ 2 − K log 2 2 − K = 2 K −2 − K ( −K ) = K
k =1

•  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.
Ilya Pollak

Lemma 1: An auxiliary result helpful for
proving the source coding theorem
•  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.

Ilya Pollak

Another auxiliary result: Kraft inequality
If integers d1 ,…, d M satisfy the inequality
M

∑2

− dm

≤ 1,

(1)

m =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.

Ilya Pollak

Some useful facts about full binary trees
A full binary tree of depth D has
2D leaves.

Ilya Pollak

Tree depth D = 4

2D leaves. (Here, depth is D=4 and
the number of leaves is 24=16.)

Ilya Pollak

Tree depth D = 4

Depth of red
node = 2

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.)

Ilya Pollak

Kraft inequality: proof of

⇒

Suppose d1 ≤ … ≤ d M satisfy (1). Consider the full binary tree of depth d M , and consider all its
nodes at depth d1 . Assign one of these nodes to symbol a1 .

Ilya Pollak


⇒

nodes at depth d1 . Assign one of these nodes to symbol a1 . Consider all the nodes at depth d2 which
are not a1 and not descendants of a1 . Assign one of them to symbol a2 .

Ilya Pollak


⇒

are not a1 and not descendants of a1 . Assign one of them to symbol a2 . Iterate like this M times.

Ilya Pollak


⇒

If we have run out of tree nodes to assign after r < M iterations, it means that every leaf in the full
binary tree of depth d M is a descendant of one of the first m symbols, a1 ,…, ar .

Ilya Pollak


⇒

binary tree of depth d M is a descendant of one of the first m symbols, a1 ,…, ar . But note that every
node at depth dm has 2 dM − dm descendants. Note also that the full tree has 2 dM leaves. Therefore, if
every leaf in the tree is a descendant of a1 ,…, ar , then
r

∑2

d M − dm

= 2 dM

m =1

Ilya Pollak


⇒

r

∑2
m =1

d M − dm

=2

dM

⇔

r

∑2

− dm

=1

m =1

Ilya Pollak


⇒

r

∑2

d M − dm

=2

r

∑2

⇔

dM

m =1

=1

m =1

M

Therefore,

− dm

∑2
m =1

− dm

r

= ∑2
m =1

− dm

+

M

∑

2 − dm > 1. This violates (1).

m = r +1

Ilya Pollak


⇒

r

∑2

d M − dm

=2

r

∑2

⇔

dM

m =1

=1

m =1

M

Therefore,

− dm

∑2
m =1

− dm

r

= ∑2
m =1

− dm

+

M

∑

2 − dm > 1. This violates (1).

m = r +1

Thus, our procedure can in fact go on for M iterations. After the M -th iteration, we will have
constructed a prefix condition code with codeword lengths d1 ,…, d M .

Ilya Pollak


⇐

Suppose d1 ≤ … ≤ d M , and suppose we have a prefix condition code with there codeword lengths.
Consider the binary tree corresponding to this code.

Ilya Pollak


⇐

Consider the binary tree corresponding to this code. Complete this tree to obtain a full tree of
depth d M .

Ilya Pollak


⇐

depth d M . Again use the following facts:
the full tree has 2 dM leaves;
the number of leaf descendants of the codeword of length dm is 2 dM − dm .

Ilya Pollak


⇐

The combined number of all leaf descendants of all codewords must be less than or equal to
the total number of leaves in the full tree:
M

∑2

d M − dm

≤ 2 dM

m =1

Ilya Pollak


⇐

The combined number of all leaf descendants of all codewords must be less than or equal to
the total number of leaves in the full tree:
M

∑2
m =1

d M − dm

≤2

dM

⇔

M

∑2

− dm

≤ 1.

m =1

Ilya Pollak

Source coding theorem: proof of
H(X)≤E[C]
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]
M

M

m =1

m =1

H (X) − E[C] = − ∑ p X (am )log 2 p X (am ) − ∑ p X (am )dm

Ilya Pollak

H(X)≤E[C]
⎡
⎤
⎛ 1 ⎞
dm
H (X) − E[C] = − ∑ p X (am )log 2 p X (am ) − ∑ p X (am )dm = ∑ p X (am ) ⎢ log 2 ⎜
− log 2 2 ⎥
⎝ p X (am ) ⎟
⎠
m =1
m =1
m =1
⎣
⎦
M

M

M

Ilya Pollak

H(X)≤E[C]
⎡
⎤
⎛ 1 ⎞
dm
− log 2 2 ⎥
⎝ p X (am ) ⎟
⎠
m =1
m =1
m =1
⎣
⎦
M

M

M

⎡
⎛
⎞⎤
1
= ∑ p X (am ) ⎢ log 2 ⎜
⎥
⎝ p X (am )2 dm ⎟ ⎦
⎠
m =1
⎣
M

Ilya Pollak

H(X)≤E[C]
⎡
⎤
⎛ 1 ⎞
dm
− log 2 2 ⎥
⎝ p X (am ) ⎟
⎠
m =1
m =1
m =1
⎣
⎦
M

M

⎡
⎛
⎞⎤
1
= ∑ p X (am ) ⎢ log 2 ⎜
⎥
⎝ p X (am )2 dm ⎟ ⎦
⎠
m =1
⎣
M
⎛
⎞
1
≤ ∑ p X (am ) ⎜
− 1⎟ log 2 e
⎝ p X (am )2 dm
⎠
m =1

M

M

(by Lemma 1)

Ilya Pollak

H(X)≤E[C]
⎡
⎤
⎛ 1 ⎞
dm
− log 2 2 ⎥
⎝ p X (am ) ⎟
⎠
m =1
m =1
m =1
⎣
⎦
M

M

⎡
⎛
⎞⎤
1
= ∑ p X (am ) ⎢ log 2 ⎜
⎥
⎝ p X (am )2 dm ⎟ ⎦
⎠
m =1
⎣
M
⎛
⎞
1
≤ ∑ p X (am ) ⎜
− 1⎟ log 2 e
⎝ p X (am )2 dm
⎠
m =1

M

M

(by Lemma 1)

M
⎛ M 1
⎞
= ⎜ ∑ dm − ∑ p X (am )⎟ log 2 e
⎝ m =1 2
⎠
m =1

Ilya Pollak

H(X)≤E[C]
⎡
⎤
⎛ 1 ⎞
dm
− log 2 2 ⎥
⎝ p X (am ) ⎟
⎠
m =1
m =1
m =1
⎣
⎦
M

M

⎡
⎛
⎞⎤
1
= ∑ p X (am ) ⎢ log 2 ⎜
⎥
⎝ p X (am )2 dm ⎟ ⎦
⎠
m =1
⎣
M
⎛
⎞
1
≤ ∑ p X (am ) ⎜
− 1⎟ log 2 e
⎝ p X (am )2 dm
⎠
m =1

M

M

(by Lemma 1)

M
⎛ M 1
⎞
= ⎜ ∑ dm − ∑ p X (am )⎟ log 2 e
⎝ m =1 2
⎠
m =1

⎛ M − dm
⎞
= ⎜ ∑ 2 − 1⎟ log 2 e ≤ 0
⎝ m =1
⎠

Ilya Pollak

H(X)≤E[C]
⎡
⎤
⎛ 1 ⎞
dm
− log 2 2 ⎥
⎝ p X (am ) ⎟
⎠
m =1
m =1
m =1
⎣
⎦
M

M

⎡
⎛
⎞⎤
1
= ∑ p X (am ) ⎢ log 2 ⎜
⎥
⎝ p X (am )2 dm ⎟ ⎦
⎠
m =1
⎣
M
⎛
⎞
1
≤ ∑ p X (am ) ⎜
− 1⎟ log 2 e
⎝ p X (am )2 dm
⎠
m =1

M

M

(by Lemma 1)

M
⎛ M 1
⎞
= ⎜ ∑ dm − ∑ p X (am )⎟ log 2 e
⎝ m =1 2
⎠
m =1

⎛ M − dm
⎞
= ⎜ ∑ 2 − 1⎟ log 2 e ≤ 0
⎝ m =1
⎠
By Kraft inequality, this holds for any prefix condition code. But it is also true for any uniquely
decodable code.
Ilya Pollak

Source coding theorem: how to satisfy
E[C] < H(X)+1?
Choose dm = − ⎡ log 2 p X (am ) ⎤ (where ⎡ x ⎤ stands for the smallest integer which is ≥ x). Then
⎢ ⎥
⎢
⎥
dm ≥ − log 2 p X (am )

Ilya Pollak

E[C] < H(X)+1?
⎢ ⎥
⎢
⎥
dm ≥ − log 2 p X (am ) ⇒ − dm ≤ log 2 p X (am ) ⇒ 2 − dm ≤ p X (am )

Ilya Pollak

E[C] < H(X)+1?
⎢ ⎥
⎢
⎥

⇒

− dm ≤ log 2 p X (am )

⇒

2

− dm

≤ p X (am )

⇒

M

∑2
m =1

− dm

M

≤ ∑ p X (am ) = 1.
m =1

Ilya Pollak

E[C] < H(X)+1?
⎢ ⎥
⎢
⎥

⇒


⇒

2

− dm

≤ p X (am )

⇒

M

∑2
m =1

− dm

M

≤ ∑ p X (am ) = 1.
m =1

Therefore, Kraft inequality is satisfied, and we can construct a prefix condition code with codeword
lengths d1 ,…, d M .

Ilya Pollak

E[C] < H(X)+1?
⎢ ⎥
⎢
⎥

⇒


⇒

2

− dm

≤ p X (am )

⇒

M

∑2
m =1

− dm

M

≤ ∑ p X (am ) = 1.
m =1

lengths d1 ,…, d M . Also, by construction,
dm − 1 < − log 2 p X (am ) ⇒ dm < − log 2 p X (am ) + 1

Ilya Pollak

E[C] < H(X)+1?
⎢ ⎥
⎢
⎥

⇒


⇒

2

− dm

≤ p X (am )

⇒

M

∑2
m =1

− dm

M

≤ ∑ p X (am ) = 1.
m =1

⇒ p X (am )dm < − p X (am )log 2 p X (am ) + p X (am )

Ilya Pollak

E[C] < H(X)+1?
⎢ ⎥
⎢
⎥

⇒


⇒

2

− dm

≤ p X (am )

⇒

M

∑2
m =1

− dm

M

≤ ∑ p X (am ) = 1.
m =1

⇒

M

∑p
m =1

M

X

(am )dm < ∑ ( − p X (am )log 2 p X (am ) + p X (am ))
m =1

Ilya Pollak

E[C] < H(X)+1?
⎢ ⎥
⎢
⎥

⇒


⇒

2

− dm

≤ p X (am )

⇒

M

∑2
m =1

− dm

M

≤ ∑ p X (am ) = 1.
m =1

⇒

M

∑p
m =1

M

X

m =1

M

M

m =1

m =1

⇒ E[C] < ∑ ( − p X (am )log 2 p X (am )) + ∑ p X (am )

Ilya Pollak

E[C] < H(X)+1?
⎢ ⎥
⎢
⎥

⇒


⇒

2

− dm

≤ p X (am )

⇒

M

∑2
m =1

− dm

M

≤ ∑ p X (am ) = 1.
m =1

⇒

M

∑p
m =1

M

X

m =1

M

M

m =1

m =1

⇒ E[C] < ∑ ( − p X (am )log 2 p X (am )) + ∑ p X (am ) = H (X) + 1

Ilya Pollak

Note: Huffman code may often be very
far from the entropy
•  Let X have two outcomes, a1 and a2, with probabilities 1−2−d
and 2−d, respectively.

Ilya Pollak

•  Huffman code: 0 for a1; 1 for a2.
•  Expected codeword length: 1.

Ilya Pollak

•  Entropy: −(1−2−d) log2(1−2−d) + d2−d ≈ 0 for large d. For
example, if d=20, this is 0.0000204493.

Ilya Pollak

•  Problem: no codeword can have fractional numbers of bits!

Ilya Pollak

•  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.
Ilya Pollak

Example
•  (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.

Ilya Pollak

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

Ilya Pollak

Example
respectively.
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

Ilya Pollak

Example
respectively.
for (a2,a2).
–  [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

Ilya Pollak

Example
respectively.
for (a2,a2).
–  [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.

Ilya Pollak

Source coding theorem for sequences
of independent, identically distributed
random variables
Suppose we are jointly encoding independent, identically distributed discrete
random variables X1 ,…, X N , 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
1
length is < H (Xn ) + .
N

Ilya Pollak

Proof of the source coding theorem
for iid sequences
Consider random vector X = ( X1 ,…, X N ) . The self-information of its outcome x = ( x1 ,…, x N ) is
I(x) = − log 2 p X1 ,…, XN ( x1 ,…, x N )

Ilya Pollak

for iid sequences
N

N

n =1

n =1

I(x) = − log 2 p X1 ,…, XN ( x1 ,…, x N ) = − ∑ log 2 p Xn ( xn ) = ∑ I ( xn ).

Ilya Pollak

for iid sequences
N

N

n =1

n =1

Therefore, the entropy of X is
⎡N
⎤ N
H ( X ) = E ⎡ I ( X ) ⎤ = E ⎢ ∑ I ( Xn ) ⎥ = ∑ H ( Xn ) = NH ( Xn ) .
⎣
⎦
⎣ n =1
⎦ n =1

Ilya Pollak

for iid sequences
N

N

n =1

n =1

⎡N
⎤ N
⎣
⎦
⎣ n =1
⎦ n =1
Therefore, applying the single-symbol source coding theorem to X, we have:
H ( X ) ≤ E [ C N ] < H ( X ) + 1,

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

Ilya Pollak

for iid sequences
N

N

n =1

n =1

⎡N
⎤ N
⎣
⎦
⎣ n =1
⎦ n =1
H ( X ) ≤ E [ C N ] < H ( X ) + 1,

NH ( Xn ) ≤ E [ C N ] < NH ( Xn ) + 1,

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

Ilya Pollak

for iid sequences
N

N

n =1

n =1

⎡N
⎤ N
⎣
⎦
⎣ n =1
⎦ n =1
H ( X ) ≤ E [ C N ] < H ( X ) + 1,

NH ( Xn ) ≤ E [ C N ] < NH ( Xn ) + 1,
1
,
N
is the expected codeword length for the optimal uniquely decodable code for X,

H ( Xn ) ≤ E [C ] < H ( Xn ) +
where E [ C N ]

E [CN ]
and E [ C ] =
is the corresponding expected codeword length per symbol.
N
Ilya Pollak

Arithmetic coding
•  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

Ilya Pollak

25 quantization and_compression

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