exercise in the previous class

p: the probability that symbols are delivered correctlyC:

1

00 → 0000001 → 0101110 → 1010111 → 11110

What is the threshold ( 閾値 ) of pwith which using C is good/bad?

without coding, correct probability = p2 = Awith coding, correct probability = p5 + 5p4(1 – p) = BB > A if p5 + 5p4(1 – p) – p2

= p2(p – 1)(–4p2 + p + 1) > 0(roots = –0.39, 0, 0.64, 1)

Using C is better if 0.64 < p < 1.-0.1

-0.05

0

0.05

0.1

0.15

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9 1

what did we learn?

motivation and models of communication channelssimple examples of linear codes

even parity check codes(a1, …, ak) → (a1, …, ak, p) , p = a1 + … + ak mod 2error detection only

horizontal and vertical parity check (2D) code(a1, a2, a3, a4) →

2

a1 a2

a3 a4

x1

x2

y1 y2 z→ (a1, a2, a3, a4, x1, x2, y1, y2, z)

one-bit error correcting, two-bits error detecting

errata: additional remark (cnt’d)

We expect that 2D codes detect all two-bit errors.If we don’t use the parity of parity, then...

3

0 0 00 0 00 0 0

0 0 00 1 10 1 0

codeword codeword0 0 00 1 00 0 0

0 0 00 1 00 1 0

1-bit err.

to thenearestcodeword

1-bit err.

some two-bit errors are not detected,instead, they are decoded to a wrong codeword.

0 0 00 1 00 0 0

to the nearestcodeword×

additional remark (cnt’d) ， corrected

We expect that 2D codes detect all two-bit errors.If we don’t use the parity of parity, then...

4

0 0 00 0 00 0 0

0 0 00 1 10 1 0

codeword codeword0 0 00 1 00 0 0

0 0 00 1 00 1 0

1-bit err.

to thenearestcodeword

1-bit err.

some two-bit errors are not detected,instead, they are decoded to a wrong codeword.

to the nearestcodeword

today’s class

linear codesdefinitionencodingdecoding (error detection and correction)

5

(one of) definition(s) of linear codes

a binary code C is linear C is a vector space for any u, v C, we have u + v C

parity check code with length 3:

6

000 011 101 110000 000 011 101 110011 011 000 110 101101 101 110 000 011110 110 101 011 000

how about parity checkcode with other length?

linearity of parity check codes

C: parity check code for length k data (x1, ..., xk)

... codewords in C are (x1, ..., xk, p) with p = x1 + ... + xk

Theorem: parity check codes are linearProof: confirm that u + v C for any u, v C.

7

u = (u1, ..., uk, p), (p = u1 + ... + uk),v = (v1, ..., vk, q), (q = v1 + ... + vk),u + v = (u1 + v1, ..., uk + vk, p + q).

because p + q = u1 + ... + uk + v1 + ... + vk = (u1 + v1) + ... + (uk + vk),p + q is a valid parity bit for (u1 + v1, ..., uk + vk). u + v is a codeword

Theorem: 2D codes are linear (proof omitted)

computation of parity bits

parity check codes, 2D codes...for information bits (x1, x2, ..., xk),

parity bits are defined by linear equationsp = a1x1 + a2x2 + ... + akxk (a1, a2, ..., ak {0, 1})∈

8

(x1, ..., xk, p) p = x1 + ... + xk

ab

ii

b

ii

b

ijiaj

a

jjiai xprxqxp

111)1(

1)1( ,,

(x1, ..., xab, p1, ..., pb, q1, ..., qa, r)

In both codes, parity bits are sum of some of information bits.

linear parity bits make the code linear

Lemma: for a linear equation f(x1, x2, ..., xk) = a1x1 + a2x2 +...+ akxk,

f(u1, u2, ..., uk) + f(v1, v2, ..., vk) = f(u1 + v1, u2 + v2, ..., uk + vk)

Theorem:If parity bits are defined by linear equations, then C is linear.

9

codeword 　 (u1, u2, ..., uk, ..., f(u1, u2, ..., uk) , ...)codeword 　 (v1, v2, ..., vk, ..., f(v1, v2, ..., vk) , ...) ＋）

(u1 + v1, u2 + v2, ..., uk + vk, ..., f(u1, u2, ..., uk) + f(v1, v2, ..., vk), ...)

(u1 + v1, u2 + v2, ..., uk + vk, ..., f(u1 + v1, u2 + v2, ..., uk + vk), ...)＝codeword

example of linear code construction

k = 3: information bits are (x1, x2, x3).

determine parity bits as you like...p1 = x1 + x2

p2 = x2 + x3

10

⇒ the codeword is (x1, x2, x3, p1, p2)

00000 00101 01011 01110 10010 10111 11001 1110000000 00000 00101 01011 01110 10010 10111 11001 1110000101 00101 00000 01110 01011 10111 10010 11100 1100101011 01011 01110 00000 00101 11001 11100 10010 1011101110 01110 01011 00101 00000 11100 11001 10111 1001010010 10010 10111 11001 11100 00000 00101 01011 0111010111 10111 10010 11100 11001 00101 00000 01110 0101111001 11001 11100 10010 10111 01011 01110 00000 0010111100 11100 11001 10111 10010 01110 01011 00101 00000

codewords for unit vectors

Assume that...we define m parity bits for k-bit information x1, x2, ..., xk,

with the j-th parity bit defined bypj = aj,1x1 + aj,2x2 + ... + aj,kxk (aj,i = 0 or 1)

the codeword for the unit vector (0, ...,0, 1, 0, ..., 0) is ...ci = (0, ..., 1, ..., 0, a1,i, a2,i, ..., am,i)

11

^^ ^i1 k

c1 = 1 0 0 1 0c2 = 0 1 0 1 1c3 = 0 0 1 0 1

p1 = x1 + x2

p2 = x2 + x3

example in p.10: k = 3, m = 2

(a1,1 a1,2 a1,3) = (1 1 0)(a2,1 a2,2 a2,3) = (0 1 1)

transposition (転置

)

“basis” in a linear code

a vector space has a basis ( 基底 ) b1, b2, ..., bk:

any u in the space is written asu = u1b1 + u2b2 + ... + ukbk (ui {0, 1})∈

12

Theorem:Codewords c1, c2, ..., ck for unit vectors constitute the basis of C.

any codeword c C is written asu = u1c1 + u2c2 + ... + ukck (ui {0, 1})∈

basis of the p.8 example

k = 3: information bits are (x1, x2, x3)p1 = x1 + x2

p2 = x2 + x3

13

c1 = 1 0 0 1 0c2 = 0 1 0 1 1c3 = 0 0 1 0 1

codewords0000000101010110111010010101111100111100

= 0・ 10010 + 0・ 01011 + 0・ 00101= 0・ 10010 + 0・ 01011 + 1・ 00101= 0・ 10010 + 1・ 01011 + 0・ 00101= 0・ 10010 + 1・ 01011 + 1・ 00101= 1・ 10010 + 0・ 01011 + 0・ 00101= 1・ 10010 + 0・ 01011 + 1・ 00101= 1・ 10010 + 1・ 01011 + 0・ 00101= 1・ 10010 + 1・ 01011 + 1・ 00101

generator matrix

if the j-th parity bit is pj = aj,1x1 + aj,2x2 + ... + aj,kxk, then

ci = (0, ..., 1, ..., 0, a1,i, a2,i, ..., am,i)

c1, c2, ..., ck is a basis of C

any codeword u is written as u = u1c1 + u2c2 + ... + ukck

14

kmkk

m

m

k

mk

aaa

aaa

aaa

uuu

pppuuu

,,2,1

2,2,22,1

1,1,21,1

21

2121

100

010

001

generator matrix of C

encoding = multiplication of the generator matrix and a vector

the structure of the generator matrix

15

kmkk

m

m

aaa

aaa

aaa

,,2,1

2,2,22,1

1,1,21,1

100

010

001

the generator matrix

k × k identity matrix

row vectors= codewords for unit vector

transposition of the coefficientsof the linear equations of parity bits

example

16

parity check code p = x1 + x2 + x3

.

1100

1010

1001

)()( 321321

uuupuuu

2D code

.

110101000

101100100

110010010

101010001

p1 = x1 + x2

p2 = x3 + x4

q1 = x1 + x3

q2 = x2 + x4

r = x1 + x2 + x3 + x4

x1 x2

x3 x4

p1

p2

q1 q2 r

encoding

17

encoding of a 2D code

110101000

101100100

110010010

101010001

)(

)(

4321

21214321

uuuu

rqqppuuuu

to encode 0111...

.101011110

110101000

101100100

110010010

101010001

)1110(

the codeword is 011110101

hardware encoder

18

110101000

101100100

110010010

101010001

)(

)(

4321

21214321

uuuu

rqqppuuuu

u1u2u3u4

u1 u2 u3 u4 p1 p2 q1 q2 r

data

codeword

XOR

summary of the first half

A code is linear if and only if... it is a vector spaceparity bits are defined by linear equations

codewords of unit vectors is a basis of the code

the generator matrix ...contains basis codewords as its row vectorsgives mathematical principle for encoding an encoder is realizable by a combinatorial circuit

( 組み合わせ回路 )

19

condition for a vector to be a codeword (1)

consider a 2D code

20

p1 = x1 + x2

p2 = x3 + x4

q1 = x1 + x3

q2 = x2 + x4

r = x1 + x2 + x3 + x4

.

110101000

101100100

110010010

101010001

)()( 432121214321

xxxxrqqppxxxx

a vector (x1 x2 x3 x4 p1 p2 q1 q2 r) is a codeword

if and only if

condition for a vector to be a codeword (2)

21

p1 = x1 + x2

p2 = x3 + x4

q1 = x1 + x3

q2 = x2 + x4

r = x1 + x2 + x3 + x4

x1 + x2 – p1 = 0

x3 + x4 – p2 = 0

x1 + x3 – q1 = 0

x2 + x4 – q2 = 0

x1 + x2 + x3 + x4 – r = 0

in binary world, x – y = x ＋ y

x1 + x2 + p1 = 0 x3 + x4 + p2 = 0x1 + x3 + q1 = 0 x2 + x4 + q2 = 0x1 + x2 + x3 + x4 + r = 0

parity check equations

(x1 x2 x3 x4 p1 p2 q1 q2 r) is a codeword

move the lhs to rhs...

condition for a vector to be a codeword (3)

22

x1 + x2 + p1 = 0 x3 + x4 + p2 = 0x1 + x3 + q1 = 0 x2 + x4 + q2 = 0x1 + x2 + x3 + x4 + r = 0

(x1 x2 x3 x4 p1 p2 q1 q2 r) is a codeword

0

0

0

0

0

100001111

010001010

001000101

000101100

000010011

9

8

7

6

5

4

3

2

1

x

x

x

x

x

x

x

x

x

Is (x1 x2 x3 x4 x5 x6 x7 x8 x9) a codeword?transpose the vector,multiply to this matrix from the right,and see if the result is 0 or not.

zero 　 ⇒ it’s a codewordnonzero it’s not a codeword⇒

parity check matrix

parity check matrix for error detection

consider a 2D codeIs 110101101 a codeword? ... yes

23

Is 011011010 a codeword? ... no

0

0

0

0

0

101101011

100001111

010001010

001000101

000101100

000010011

T

0

0

1

0

0

010110110

100001111

010001010

001000101

000101100

000010011

T

generator and check matrices

24

kmkk

m

m

aaa

aaa

aaa

,,2,1

2,2,22,1

1,1,21,1

100

010

001

definition of parity bitsp1 = a1,1x1 + a1,2x2 + ... + a1,kxk p2 = a2,1x1 + a2,2x2 + ... + a2,kxk

:pm = am,1x1 + am,2x2 + ... + am,kxk

generator matrix

identity coefficients transposed

a1,1x1 + a1,2x2 + ... + a1,kxk + p1 = 0a2,1x1 + a2,2x2 + ... + a2,kxk + p2 = 0

:am,1x1 + am,2x2 + ... + am,kxk + pm = 0

k rows

100

010

001

,2,1,

,22,21,2

,12,11,1

kmmm

k

k

aaa

aaa

aaa

parity check matrix

identitycoefficients

n － k row

s

example of matrices

25

2D code ： n = 9, k = 4, m = n － k = 5.

110101000

101100100

110010010

101010001

x1 x2

x3 x4

p1

p2

q1 q2 r

100001111

010001010

001000101

000101100

000010011

generator matrix parity check matrix

p1 = x1 + x2

p2 = x3 + x4

q1 = x1 + x3

q2 = x2 + x4

r = x1 + x2 + x3 + x4

syndrome

For a parity check matrix H and a vector v,if HvT = 0, then v Cif HvT 0, then v C

the vector HvT is called the syndrome ( シンドローム ) of v:if the syndrome of v is zero, then v Cif the syndrome of v is nonzero, then v C

The syndrome is more useful,because it contains the information of errors.

26

syndrome and error

send a codeword u to a binary symmetric channelan error vector e is added to the codeword in the channelthe received vector is v = u + e

27

ue

v = u + e

noise

codeword received

if e = 0 (no error),then the syndrome of v

is...HvT = HuT = 0

if e 0 (error occurs), then the syndrome of v is...HvT = H(u + e)T = HuT + HeT = HeT

the syndrome is solely determined from e, independently from u

... if you see the syndrome, then you can say what e is.

error patterns determine the syndrome

28

100001111

010001010

001000101

000101100

000010011

H

• 000000000 is sent, 000100000 is received...H(0 0 0 1 0 0 0 0 0)T = (0 1 0 1 1)T.

• 110000110 is sent, 110100110 is received...H(1 1 0 1 0 0 1 1 0)T = (0 1 0 1 1)T.

⇒ if the syndrome is (0 1 0 1 1), then the fourth bit is in error

independent from the sent codeword

error correction

if you know the correspondence betweenerror patterns and syndromes, then you can correct errors.

29

received syndrome

error pattern

decodingresult

v = u + e

computethe syndrome:

s = HvT s

table of errors / syndromes...............

...............e

u

one-bit error

Let n be a codeword, and let hi be the i-th column vector of H:

30

nhhh 21H

.

0

1

0

H 21

in

T hhhhe

the syndrome of a one-bit error e = (0 0 ... 0 1 0 ... 0) is...

only one-bit error at the i-th symbol position⇔ 　 syndrome equals to the i-th vector of H

(the table of errors / syndromes not needed)

example of error correction

2D code

31

000000000,000101011,001001101,001100110,010010011,010111000,011011110,011110101,

100010101,100111110,101011000,101110011,110000110,110101101,111001011,111100000.

codewords

100001111

010001010

001000101

000101100

000010011

H

paritycheckmatrix

• if 101001000 is received...⇒ 　 the syndrome is H(1 0 1 0 0 1 0 0 0)T = (1 0 0 0 0)T

⇒ 　 this is the fifth column of H⇒ 　 the fifth-bit is in error, 101011000 must be sent

• if 101100110 is received...⇒ 　 the syndrome is H(1 0 1 1 0 0 1 1 0)T = (1 0 1 0 1)T

⇒ 　 this is the first column of H⇒ 　 the first-bit is in error, 001100110 must be sent

parity check matrix and the ability of codes (1)

one-bit error at the i-th symbol position⇔ 　 syndrome equals the i-th vector of H

if several column vectors in H are the same, thendifferent error patterns result in the same syndromethe error pattern is not uniquely determined

32

parity check code p = x1 + x2 111H

p1 = x1 + x2

p2 = x2 + x3

the example code in p. 8

10110

01011H

erro

r cor

recti

onN

OT

poss

ible

parity check matrix and the ability of codes (2)

one-bit error at the i-th symbol position⇔ 　 syndrome equals the i-th vector of H

if all column vectors in H are different, thendifferent error patterns result in different syndromesthe error pattern is uniquely determined

33

erro

r cor

recti

onpo

ssib

le

100001111

010001010

001000101

000101100

000010011

H2D code

summary

definition of linear codesvector space, parity bits defined by linear equations

generator matrixmatrix of basis codewordscontributes to the encoding

parity check matrixrepresents constraints among symbolsdetermines syndrome

error correction and error detection

34

exercise

Consider an “odd” parity check code C whose codewords are(x1, …, xk, p) with p = x1+…+xk+1. Is C a linear code?

Construct a 2D code for 6-bit information (a1, ..., a6) as follows.

determine the generator and parity check matricesencode 011001 using the generator matrixcorrect an error in the sequence 110111001010

35

a1 a2 a3a4 a5 a6

p1p2

q1 q2 q3 r

(a1, ..., a6)→ (a1, ..., a6, p1, p2, q1, q2, q3, r)