1. Information theory Multi-user information theory Part 7: A special matrix application A.J. Han Vinck Essen, 2002

2. content a special rank k x n natrix the application in Broadcast channel Switching channel Coding for memories with defects existance proof

3. Switching channel X2={0,1} X1={0,1} Y ={ ,0,1}

4. uniform Definition: of a uniform rank k matrix uniform a binary uniform rank k, k x n matrix U - has rank k - when deleting  (n-k) columns the rank of the remaining matrix may stay = k nknk deleted

5. Application (1): X2 * U = Y 1 1 1 0 1 0 1 1 0 1 1 = X1 X2 = ( 0 1...1 ) = Y U Y = (... . .. .. ) Result: Y = X2 x U with positions erased (  ) by X1 Sum Rate: ?

6. Continuation: sum rate ? X2 can be retrieved from the remaining part if rank = k i.e. an inverse exists transmitted k bits X1 specifies ~ 2 nh([ n-k)/n]) = 2 nh(1-k)n) sequences transmitted nh(1-k/n) = nh(k/n) bits

7. Problem left Find matrix U with maximum number of sequences X1 with remaining rank k matrix Sum Rate: k/n + nh(k/n)

8. Excercise: Give the matrix U and efficiency for k = 1 k = 2 k = n-1

9. Existance (1) Ingredients: specify (n-k) erased columns Property: remaining part of G has rank k

10. Existance (2) Y = # different patterns of (n-k) erased columns X = # of possible rank k matrices for a specific pattern kn-k k y X   t otal number of matrices = 2 kn One matrix must have more than entries

11. Existance (3) 1. # different patterns of column erasure Y ~ 2.# of invertible k x k matrices F= (2 k –1)(2 k –2)(2 k –2 k-1 ) 3.A specified pattern allows X = 2 (n-k)k F matrices G 4.2 (n-k)k F  c F 2 nk where c F = 0.28

12. 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 = Existance (4)

13. Extension Ingredients: specify any k‘ ≤ k columns Property: the specified matrix has rank k‘ Wish: k‘ = k for optimum performance! IkIk

14. Application (2): the broadcast channel ZXY000101211ZXY000101211 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)

15. Application (2): the broadcast channel K‘-zeros Y = ( 1 0 1 1 0 1 1 ) X=(X 1, X 2,  X n-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)

16. 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 )

17. 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 X 1 X 2 X n-k  no influence first k bits

18. 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)

19. Memory with defects: Y specifies a vector with k‘  k defects Y = ( **0**0*1**1****1*) C(X) = ( 000000 X 1 X 2 X n-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 !