Chapter 5Markov processes

Run length coding

Gray code

|transition probability

Markov Processes

Let S = {s1, …, sq} be a set of symbols. A jth-order Markov process has probabilities p(si | si1

… sij) associated with it, the conditional

probability of seeing si after seeing si1 … sij. This is said to be a j-

memory source, and there are qj states in the Markov process. Transition Graph

b½

½

c

¼¼a

⅓ ⅓

⅓

¼

¼

sj si

p(si | sj)

Weather Example:

Let (j = 1). Think:

a means “fair”

b means “rain”

c means “snow”

Transition Matrix

p(si | sj) i = column, j = row

next symbol

= M

5.2

currentstate

∑ outgoing edges = 1

⅓ ⅓ ⅓¼ ½ ¼¼ ¼ ½

a b c

a

b

c

Ergodic Equilibriums

Definition: A Markov process M is said to be ergodic if

1. From any state we can eventually get to any other state.

2. The system reaches a limiting distribution.

weather.average overall theiswhich 11

4,

11

4,

11

3example, aboveIn

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. ... thisrepeating),,(),,(

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pMpm solutionequilibriupMp

MM

pMppppMppp

5.2

Predictive Coding

Assume a prediction algorithm for the source which given all prior symbols, predicts the next.

s1 ….. sn1 pn en = pn sn error

input stream prediction

What is transmitted is the error, ei. By knowing just the error, the predictor

also knows the original symbols.

source channel destination

predictor predictor

enensn sn

pnpn

must assume that both predictors are identical, and start in the same state

5.7

Accuracy: The probability of the predictor being correct is p = 1 q; constant (over time) and independent of other prediction errors.

Let the probability of a run of exactly n 0’s, (0n1), be p(n) = pn ∙ q.

The probability of runs of any length n = 0, 1, 2, … is:

111

1

00

q

q

p

q

pqpqqp

n

n

n

n

)(

00

)1()1(run a oflength Expected

pf

n

n

n

n pnqqpn

So, .1

1)1()(

0 0

1

n n

nn cp

pcpdppndppf

qq

qpfq

pp

pp

p

pppppf

1)( So, .

)1(

1

)1(

1

)1(

)1()1()(

2222

Note: alternate method for calculating f(p), look at 2

0

n

np

5.8

Coding of Run Lengths

Send a k-digit binary number to represent a run of zeroes whose length is between 0 and 2k 2. (small runs are in binary)

For run lengths larger than 2k 2, send 2k 1 (k ones) followed by another k-digit binary number, etc. (large runs are in unary)

Let n = run length. Fix k = block length.

Let n = i ∙ m + j 0 ≤ j < m = 2k 1

like “reading” the “matrix” with m cells and ∞ many rows.

nj

ikj

i

kk

n lkiBB

j

)1(1111110 length

binary in code

5.9

Let p(n) = the probability of a run of exactly n 0’s: 0n1. The expected code length is:

0n

nlnp

mmm

i

imm

i

mim

i

m

j

jim

i

m

j

jim

i

m

jjmi

p

k

ppkpipk

q

ppiqkppiqk

kiqpljmip

1)1(

1)1()1()1(

1)1()1(

)1()(

20

00

1

0

0

1

00

1

0

5.9

But every n can be written uniquely as i∙m + j where i ≥ 0, 0 ≤ j < m = 2k 1.

Expected length of run length code

Gray Code

Consider an analog-to-digital “flash” converter consisting of a rotating wheel:

0 0

00

0

0

00 0

0

0

0 01

1

11

11

1 1

1

11

1The maximum error in the scheme is ± ⅛ rotation because …

imagine “brushes” contacting the wheel in each of the three circles

The Hamming Distance between adjacent positions is 1.In ordinary binary, the maximum distance is 3 (the max. possible).

5.15-17

codeGray

bit )1(

1

1

0

0

)1(1

0)1(

0

12

12

0

n

G

G

G

G

nGGn

n

1-bit Gray code

}1,0{}1,0{Let 0101 iniini bbbBgggG

10111 nibbgbg iiinnEncoding

total)running a (keep10Decoding1

nigbn

ijji

5.15-17

12

0

)(

nG

G

nG Let :definition Inductive