Information theory

Multi-user information theory

Part 7: A special matrix application

A.J. Han VinckEssen, 2002

content

a special rank k x n natrix

the application in Broadcast channel Switching channel Coding for memories with defects

existance proof

Switching channel

X2={0,1}

X1={0,1}

Y ={,0,1}

Definition: of a uniformuniform rank k matrix

a binary uniformuniform rank k, k x n matrix U

- has rank k

- when deleting (n-k) columns the rank of the remaining matrix may stay = k

n

k

deleted

Application (1): X2 * U = Y

1 1 1 0 1 0 1 1 0 1 1 = X1X2 = ( 0 1...1 ) = Y

U

Y = ( . . . . . . . . )

Result: Y = X2 x U with positions erased () by X1

Sum Rate: ?

Continuation: sum rate ?

X2 can be retrieved from the remaining part

if rank = k i.e. an inverse exists

transmitted k bits

X1 specifies ~ 2nh([ n-k)/n]) = 2nh(1-k)n) sequences

transmitted nh(1-k/n) = nh(k/n) bits

Problem left

Find matrix U with

maximum number of sequences X1

with remaining rank k matrix

Sum Rate: k/n + nh(k/n)

Excercise:

Give the matrix U and efficiency for

k = 1

k = 2

k = n-1

Existance (1)

Ingredients: specify (n-k) erased columns

Property: remaining part of G has rank k

Existance (2)

Y = # different patterns of (n-k) erased columns

X = # of possible rank k matrices for a specific pattern k n-k

ky X

total number of matrices = 2kn

One matrix must have more than entries kn2

XY

Existance (3)

1. # different patterns of column erasure Y ~

2. # of invertible k x k matrices F= (2k–1)(2k–2)•••(2k–2k-1)

3. A specified pattern allows X = 2(n-k)k F matrices G

4. 2(n-k)k F cF 2nk where cF = 0.28

)n

k(nh

2kn

n

Average # of allowed patterns per matrix

Conclusion: there exists at least one (k x n) matrix

for which different patterns of up to (n-k)

column erasures leave a matrix of rank k

)nk(nh

Fnk

)kn(k)nk(nh

2c2

F22

kn2

XY=

Existance (4)

)nk(nh

F 2c

Extension

Ingredients: specify any k‘ ≤ k columns

Property: the specified matrix has rank k‘

Wish: k‘ = k for optimum performance!

Ik

Application (2): the broadcast channel

Z X Y

0 0 0

1 0 1

2 1 1

Step 1: encode information for y

Y has a maximum of k zeros

Y = ( 1 0 1 0 1 1)

C(y)= (1/2, 0, 1/2, 0, 1/2, 1/2)

Application (2): the broadcast channel

K‘-zeros Y = ( 1 0 1 1 0 1 1 )

X=(X1, X2 , Xn-k) C(X) = ( 0 0 0 1 1 0 1 )

C(X,Y) = ( 1 0 0 0 1 0 0 )

C(X) C(X,Y) = ( 1 0 0 1 0 0 1 )

Z = ( 2 0 1 2 0 1 2 )

Property: Z has the same zeros as C(y)

Application (2): the broadcast channel

Z = ( 2 0 1 2 0 1 2 )

y = ( 1 0 1 1 0 1 1 )

C(X) C(X,Y) = ( 1 0 0 1 0 0 1 )

C(X,Y) = ( 1 0 0 0 1 0 0 )

C(X) = ( 0 0 0 1 1 0 1 )

Continuation: Why does it work?

U( ? ? ? ) = C(X,Y)

C(X,Y) C(X)

First k bits of C(X,Y)

uniquely determine C(X,Y)

Any pattern of k‘ bits can be constructed

s.t. C(X) C(X,Y) has zeros where Y has

C(X) = 00000 X1X2 Xn-k

no influence first k bits

Transmitted information:

n-k bits with C(X)

n h( k‘/n)= nh((n-k‘)/n) bits with Y

Hence: efficiency per transmission

(n-k)/n + h((n-k‘)/n)

Memory with defects:

Y specifies a vector with k‘ k defects

Y = ( **0**0*1**1****1*)

C(X) = ( 000000 X1X2 Xn-k )

Store:

C(X) C(X,Y) matches the defects in Y

Read:

C(X) C(X,Y) errorfree and add C(X,Y) to get C(X)

Efficiency: = 1 - k/n !