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Information theory

Information theory. Multi-user information theory Part 7: A special matrix application A.J. Han Vinck Essen, 2002. content. a special rank k x n natrix the application in Broadcast channel Switching channel Coding for memories with defects existance proof.

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Information theory

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  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. Definition: of a uniform rank k matrix • a binary uniform rank k, k x n matrix U - has rank k - after deleting  (n-k) columns the rank of the remaining matrix = k n k deleted

  4. Application (1): the switching channel X1={0,1} Y ={,0,1} X2={0,1} 1 1 1 0 1 0 1 1 0 1 1 = X1 X2 = (01...1) U Result:Y = X2 x U with positions erased () by X1 Sum Rate: k/n + h(k/n)

  5. Continuation: Why? X2 can be retrieved from the remaining part (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

  6. 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)

  7. Application (2): the broadcast channel k-zeros Y = ( 1 0 1 1 0 1 1 ) X=(X1, X2 , Xn-k) C(X) = ( 0 00 1 10 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)

  8. 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 )

  9. Continuation: Why does it work? First k bits of C(X,Y) uniquely determine C(X,Y) U = C(X,Y)  C(X,Y)  C(X) ( ? ? ? ) C(X) = 00000 X1X2 Xn-k  no influence first k bits Any pattern of k bits can be constructed s.t. C(X)  C(X,Y) has zeros where Y has

  10. 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)

  11. Memory with defects: Y specifies a vector with  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 !

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

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

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

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

  16. Existance (4) 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 =

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