Code Realizations for Networks Ralf Koetter, Coordinated Science - PDF document

Code Realizations for Networks Ralf Koetter, Coordinated Science Lab., University of Illinois e-mail: koetter@csl.uiuc.edu

2 The network.... . . . ...,Muriel Medard, Tracey Ho, David Karger, Michelle Effros, Gerhard Kramer, Irem Koprulu,... . . . January 17, 2003 Ralf Koetter

3 Networks • What is capacity? • How robustly can we communi- cate? • Do we know the network? • How do we achieve capacity? • ?????? January 17, 2003 Ralf Koetter

4 S R 1 2 S 2 R 3 S 3 R A network 1 Vertices: V Edges: E ⊆ V × V , e = ( v, u ) ∈ E Edge capacity: C ( e ) Network: G = ( V, E ) Source nodes: { v 1 , v 2 , . . . , v N } ⊆ V Sink nodes: { u 1 , u 2 , . . . , u K } ⊆ V January 17, 2003 Ralf Koetter

5 Input random processes at v : ( v ) = { X ( v, 1) , X ( v, 2) , . . . , X ( v, µ ( v ) } Output random processes at u : ( u ) = { Z ( u, 1) , Z ( u, 2) , . . . , Z ( u, ν ( u )) } Random processes on edges: Y ( e ) A connection: ( v, u )) , c = ( v, u, ( v, u ) ⊆ ( v ) A connection is established if ( u ) ⊃ ( v, u ) Set of connections: The pair ( G , ) defines a network problem. January 17, 2003 Ralf Koetter

6 The capacity problem Is the problem ( G , ) solvable? How do we find a solution? Disclaimer: We are not dealing with probabilistic descriptions of channels which is way too hard for us as can be experienced by considering a simple problem like the relay channel. Moreover, we are not (really) dealing with the problem of optimizing routing and flows. Listening to this talk is potentially hazardous and is done according to the respective listeners free will. January 17, 2003 Ralf Koetter

7 Codes for networks The example S S 1 2 R R 1 2 = { ( S i , R j , ( S j )) , i, j ∈ { 1 , 2 }} R. Ahlswede, N. Cai, S.-Y.R. Li, R.W. Yeung, 2000 January 17, 2003 Ralf Koetter

8 Simplyfying Assumptions C ( e ) = 1 (links have the same capacity) H ( X ( v, i )) = 1 (sources have the same rate) The X ( v, i ) are mutually independent. Vector symbols of length m are transmitted and interpreted as el- ements in F 2 m . January 17, 2003 Ralf Koetter

9 Linear network codes All operations at network nodes are linear! X(v,i) Y(e ) Y(e ) 1 2 e e 1 2 e Y(e ) 3 3 , Y ( e 3 ) = � i α i X ( v, i )+ � j =1 , 2 β j Y ( e j ) January 17, 2003 Ralf Koetter

10 Definition (Linear Network Coding) µ ( v ) β e ′ ,e Y ( e ′ ) , � � Y ( e ) = α e,l X ( v, l ) + e ′ : head ( e ′ )= tail ( e ) l =1 α e,l , β e ′ ,e ∈ F 2 m . A consequence: ε e ′ ,j Y ( e ′ ) . � Z ( v, j ) = e ′ : head ( e ′ )= v January 17, 2003 Ralf Koetter

✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ � ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ � ✁ � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ✁ 11 X(v,1) Z(u,1) Z(u,2) A linear network X(w,1) Z(u,3) X(w,2) Z(u’,1) X(v’,1) Input: x = ( X ( v, 1) , X ( v, 2) , . . . , X ( v ′ , µ ( v ′ ))) Output: z = ( Z ( u, 1) , Z ( u, 2) , . . . , Z ( u ′ , ν ( u ′ ))) Transfer matrix M : z = xM ξ = ( ξ 1 , ξ 2 , . . . , ) = ( . . . , α e,l , . . . , β e ′ ,e , . . . , ε e ′ ,j , . . . ) M i,j ∈ F 2 [ ξ ] . January 17, 2003 Ralf Koetter

12 An alg. Min-Cut Max-Flow condition Theorem Let a linear network be given. The following three state- ments are equivalent: 1. A point-to-point connection c = ( v, v ′ , ( v, v ′ )) is possible. 2. The Min-Cut Max-Flow bound) is ( v, v ′ ) | . satisfied for a rate R ( c ) = | 3. The determinant of the R ( c ) × R ( c ) transfer matrix M is nonzero over the ring F 2 [ ξ ] 3. ⇒ We have to study the solu- tion sets of polynomial equations. January 17, 2003 Ralf Koetter

� 13 An Example: v 3 e e 1 5 v e 4 2 e e 6 4 v 1 e 3 v e 2 7 = ( v 1 , v 4 , { X ( v 1 , 1) , X ( v , 2) , X ( v 1 , 3) } )     α e 1 , 1 α e 2 , 1 α e 3 , 1 ε e 5 , 1 ε e 5 , 2 ε e 5 , 3  , B =  . A = α e 1 , 2 α e 2 , 2 α e 3 , 2 ε e 6 , 1 ε e 6 , 2 ε e 6 , 3   α e 1 , 3 α e 2 , 3 α e 3 , 3 ε e 7 , 1 ε e 7 , 2 ε e 7 , 3   β e 1 ,e 5 β e 1 ,e 4 β e 4 ,e 6 β e 1 ,e 4 β e 4 ,e 7  B T . M = A β e 2 ,e 5 β e 2 ,e 4 β e 4 ,e 6 β e 2 ,e 4 β e 4 ,e 7  0 β e 3 ,e 6 β e 3 ,e 6 det ( M ) = det ( A ) det ( B ) ( β e 1 ,e 5 β e 2 ,e 4 − β e 2 ,e 5 β e 1 ,e 5 )( β e 4 ,e 6 β e 3 ,e 7 − β e 4 ,e 7 β e 3 ,e 6 ) Choose the coefficients so that det ( M ) � = 0 ! January 17, 2003 Ralf Koetter

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Multicast: 15 Do we really need coding? We do not only need codes - we need all codes! Sender k bits cut n , Rec. 1 Rec. 2 Rec. l C is a [ n, k ] code with l informa- tion sets. Each receiver picks out one information set. January 17, 2003 Ralf Koetter