1

Network Coding and its Applications in Communication

Networks

Alex Sprintson

Computer Engineering GroupDepartment of Electrical and

Computer EngineeringTexas A&M University

NP-hardness result

Lemma (Rasala et. al. 2003). Deciding whether there exists a linear network code with alphabet size q for a multicast network coding instance is NP-hard when q is a prime power.

NP-hardness result

Proof: Reduction from the chromatic number problem

s

s1 s2 s3 s4

T3T1T4 T5T2

S1

S2 S3

S4

qmin= least prime power > X(G)-1

y

x+1*y

yx y

x+y S1

S2

S4

S3

x+0*y

Cyclic networks

In certain settings, cycles are necessary

s s

t1 t2

Acyclic Network Networks with Cycles

t1 t2

s

t1 t2

Quiz

Given a network (G,s,T) with minimum cut h, is it possible to transmit h packets to all T terminals if each link can only transmit one packet?

Always yes, if the network is acyclics

a b

a

a

b

aÅb

t1 t2

Quiz

Sometimes, if the network has cycles

t1 t2

a

a

a

b

b

b

t1 t2

a b

a

a b

ba

aÅb

aÅb

aÅbaÅb

Quiz

These networks are not really cyclic:

t1 t2

s

a b

a

a b

ba

aÅb

aÅb

aÅbaÅb

v

t1 t2

s

a b

a

a b

ba

aÅb

aÅb

aÅb

aÅb

v

Quiz

A example of a “truly” cyclic network We prove: it is impossible to send two

packets in one round to all destinations

t1 t2

s

a b

a

a b

bx y

x

xy

y

Proof: By way of contradiction

Quiz

Let e be the first link of the cycle v1->v2->v3->v4->v1 that transmits a packet.

Case study. If e=(v1,v2) then t1 gets no information about b

t1 t2

s

a b

a

a b

bx y

x

xy

y

v1

v2

v3

v4

Solution

We have shown that it is not possible to send two packets in one round to both terminals

However, it is possible to send 2n packets in n+1 rounds

Asymptotically, the rate is two packets per round

Use convolution codes

Convolution Codes

Idea: mix messages from different rounds

t1 t2

s

a b

a

a b

bx y

x

xy

xi=aiÅyi-1=aiÅbi-1Åxi-2=aiÅbi-1Åai-2Åyi-3=aiÅbi-1Åai-2Åbi-3 . . . Åa2Åb1

yi=biÅxi-1=biÅai-1Åyi-2=biÅai-1Åbi-2Åxi-3=biÅai-1Åbi-2Åai-3 . . . Åb2Åa1

Å

otherwise

1 if

1

1

ii

i ya

iax

Å

otherwise

1 if

1

1

ii

i xb

iby

Convolution Codes

Recovery at terminal t1

t1

x

a

Å

otherwise

1 if

1

1

ii

i ya

iax

a1

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

otherwise...

even is if...

123321

1234321

abababa

ibabababax

iiii

iiii

i

1 2 3 4

a2

b1

a3 a4

b2 b3

5

a5

b4

13

Algebraic framework

t1 t2 t3 t4 t5 t6

z1 z2 z3 z4

s x2x1

=

z1

z2

z3

z4

a11 a12

a22a21

a31 a32x2x1

a41 a42

Ax=z

Observation: Any two rows of matrix A must be linearly independent

aij,xi,ziGF(2n)

14

Algebraic framework (cont.)

A4x=z

Observation: Matrices A1, A2,…,Ak must be of full rank (invertible)

t3

=

z1

z2

z3

a11a21a31

x2

x1

G

s

t1

t2t4

t5

z82

z64

z54

a12a22a32

a13a23a33

x3

y11

y12y21

y22

y23

z31y41

z42 y43

y44

y51

y52

y61

y62 y71

z73

z74

y75

y81z82

z83z84

z91

y24

y11=b1x3+b2y24

y4 y1y2

x2 x3x1

- the collection of () packets generated at node

- Packets transmitted on link e

- Packets received by the destination node

15

Notation

All operations in the network are linear

Y(e)

16

Intermediate source

All operations in the network are linear

Y(e1) Y(e2)

Y(e3)

X(v,l)

At the intermediate nodes

17

Receiver node

Y(e1) Y(e2)

Z(e3)

18

Example 1

Example 2

Transfer matrix

2,1,

2,1,

55

44

00

00

00

ee

ee

B

Transfer matrix

Observation: The information propagates throughout networks on paths with different length.

Length of the path - the number of hops between the source and the tail of the edge.

For example, for edge e5 there are two paths - one through path (e2,e5) and one through path (e1,e3,e5)

Transfer matrix

Let be the vector of packets transmitted over all links

Let be the contribution (gain) of paths on length i:

Note that

0i ,,,,,54321

ie

ie

ie

ie

ie

i YYYYYY

54321

,,,, eeeee YYYYYY

...210 YYYY

Transfer matrix

00010

00001 ,,,, 21

00000

0

54321

XXYYYYY

AXY

eeeee

Contribution of paths of length zero:

Contribution of paths of length one

00000

00000

0000

0000

000

00010

00001

00000

00000

0000

0000

000

,,,,,,,,

53

52

4131

53

52

4131

5432154321

21

0000011111

01

ee

ee

eeee

ee

ee

eeee

eeeeeeeeee

XX

YYYYYYYYYY

FAXFYY

Contribution of paths of length two

00000

00000

00000

00000

0000

00010

00001

00000

00000

0000

0000

000

,,,,,,,,

5331

53

52

4131

5432154321

21

1111122222

220112

eeee

ee

ee

eeee

eeeeeeeeee

XX

YYYYYYYYYY

FAXFYYFYY

Transfer matrix

00010

00001 ,,,, 21

00000

54321

XXYYYYY eeeee

00000

00000

0000

0000

000

00010

00001 ,,,,

53

52

4131

54321 2111111

ee

ee

eeee

eeeee XXYYYYY

00000

00000

00000

00000

0000

00010

00001 ,,,,

5331

54321 2122222

eeee

eeeee XXYYYYY

Transfer matrix

10

01

00

00

00

00000

00000

00000

00000

0000

00000

00000

0000

0000

000

10000

01000

00100

00010

00001

00010

00001

,,,,

5331

53

52

4131

5431

21

2

eeee

ee

ee

eeee

eeeee

XX

YYYYY

Transfer matrix

10

01

00

00

00

,,,,5432121 eeeee YYYYYZZ

10

01

00

00

00

00000

00000

00000

00000

0000

00000

00000

0000

0000

000

10000

01000

00100

00010

00001

00010

00001

5331

53

52

4131

21

21

eeee

ee

ee

eeee

XX

ZZ

Transfer matrix

10

01

00

00

00

10000

01000

0100

0010

01

00010

00001

53

52

53314131

2121 ee

ee

eeeeeeee

XXZZ

2121 0

52

533141 ZZXXee

eeeeee

TBFFIAM ...)( 2

Transfer matrix

Transfer matrix

10

01

00

00

00

10000

01000

0100

0010

01

00010

00001

53

52

53314131

2121 ee

ee

eeeeeeee

XXZZ

0

52

533141

2121

ee

eeeeeeXXZZ

TBFFIAM ...)( 2

Linear network system

34

Example 3

Point-to-point connections

35

Example 3

36

Point-to-Point connections

Define matrices A, B, and M

37

Point-to-Point connections

Idea: Choose parameters in an extension field such that the determinant of

is nonzero over

Then, we can set A to be the identity matrix and B so that M is the identity matrix

38

Critical property

The equation

Admits a choice of variables over so that the polynomial does not evaluate to zero!!

39

Point-to-Point connections

A possible solution:Set

And set to be equal zero.

Equivalent to finding three disjoint paths between s and t!

An algebraic max flow min cut condition

[KM01, 02, 03]

41

Recap (from Wikipedia)

( )

42

Polynomial Ring

Let F be a field. The set of all polynomials with coefficients in the field F, together with the addition and the multiplication forms itself a ring, the polynomial ring over F, which is denoted by F[X].