Common Information of Random Linear Network Coding Over A 1-Hop Broadcast Packet Erasure Channel

Transcription

1 Wang, ISIT 2011 p. 1/15 Common Information of Random Linear Network Coding Over A 1-Hop Broadcast Packet Erasure Channel Chih-Chun Wang, Jaemin Han Center of Wireless Systems and Applications (CWSA) School of Electrical and Computer Engineering, Purdue University Presented in ISIT, 08/04/2011 Sponsored by NSF CCF and CNS

2 Wang, ISIT 2011 p. 2/15 Motivation The COPE Principle The COPE protocol 2-hop relay networks [Katti et al. 06] 4 transmissions w/o coding vs. 3 transmissions w. coding r sends [X + Y]; d 1 decodes X by subtraction. Empirically, % throughput improvement.

3 Wang, ISIT 2011 p. 2/15 Motivation The COPE Principle The COPE protocol 2-hop relay networks [Katti et al. 06] 4 transmissions w/o coding vs. 3 transmissions w. coding r sends [X + Y]; d 1 decodes X by subtraction. Empirically, % throughput improvement. The capacity can be defined over the corresponding PEC network.

4 Wang, ISIT 2011 p. 3/15 The Capacity Region for M = 2 Assume round-based schemes, and p r;d1 p r;d2 > 0.

5 Wang, ISIT 2011 p. 3/15 The Capacity Region for M = 2 Assume round-based schemes, and p r;d1 p r;d2 > 0. The capacity region for M = 2, [W, Asilomar 09]: R 1 min ( p s1 ;r, p r;1 (R 2 p s2 ;1) +) ( R 2 min p s2 ;r, p r;2 p ) r;2 (R 1 p s1 ;2) + p r;1

6 Wang, ISIT 2011 p. 3/15 The Capacity Region for M = 2 Assume round-based schemes, and p r;d1 p r;d2 > 0. The capacity region for M = 2, [W, Asilomar 09]: s R 1 min ( r p s1 ;r, p r;1 (R 2 p s2 ;1) +) ( R 2 min p s2 ;r, p r;2 p ) r;2 (R 1 p s1 ;2) + p r;1 Transmit all info from s i to r.

7 Wang, ISIT 2011 p. 3/15 The Capacity Region for M = 2 Assume round-based schemes, and p r;d1 p r;d2 > 0. The capacity region for M = 2, [W, Asilomar 09]: s R 1 min ( r avail. slots min inter. p s1 ;r, p r;1 (R 2 p s2 ;1) +) ( R 2 min p s2 ;r, p r;2 p ) r;2 (R 1 p s1 ;2) + p r;1 Transmit all info from s i to r. The cost of carrying the not overheard info for the other session.

8 Wang, ISIT 2011 p. 3/15 The Capacity Region for M = 2 Assume round-based schemes, and p r;d1 p r;d2 > 0. The capacity region for M = 2, [W, Asilomar 09]: s R 1 min ( r avail. slots min inter. p s1 ;r, p r;1 (R 2 p s2 ;1) +) ( R 2 min p s2 ;r, p r;2 p ) r;2 (R 1 p s1 ;2) + p r;1 Transmit all info from s i to r. The cost of carrying the not overheard info for the other session.

9 Wang, ISIT 2011 p. 3/15 The Capacity Region for M = 2 2-Stage Scheme takes the following as input: p r;1, p r;2 : The CH parameters, nr 1, nr 2 : The total # of to-be-sent symbols, np s1 ;d 2, np s2 ;d 1 : The overheard info. Assume round-based schemes, and p r;d1 p r;d2 > 0. The capacity region for M = 2, [W, Asilomar 09]: s R 1 min ( r avail. slots min inter. p s1 ;r, p r;1 (R 2 p s2 ;1) +) ( R 2 min p s2 ;r, p r;2 p ) r;2 (R 1 p s1 ;2) + p r;1 Transmit all info from s i to r. The cost of carrying the not overheard info for the other session.

10 A Competing Technique Wang, ISIT 2011 p. 4/15

11 A Competing Technique Wang, ISIT 2011 p. 4/15

12 Wang, ISIT 2011 p. 4/15 A Competing Technique Opportunistic Routing: Allow d 1 to directly hear from s 1. [Chachulski et al. 07]. No (intersession) coding at relay.

13 Wang, ISIT 2011 p. 4/15 A Competing Technique Opportunistic Routing: Allow d 1 to directly hear from s 1. [Chachulski et al. 07]. No (intersession) coding at relay. Capacity of Opp. Routing is characterized by the min-cut/max-flow theorem [Dana et al. 06]. R 1 min(p s1 ;{r or d 1 }, p s1 ;r + p s1 ;d 1 ).

14 Wang, ISIT 2011 p. 4/15 A Competing Technique Opportunistic Routing: Allow d 1 to directly hear from s 1. [Chachulski et al. 07]. No (intersession) coding at relay. Capacity of Opp. Routing is characterized by the min-cut/max-flow theorem [Dana et al. 06]. R 1 min(p s1 ;{r or d 1 }, p s1 ;r + p s1 ;d 1 ). Can we combine the benefits of Network Coding & Opportunistic Routing?

15 Wang, ISIT 2011 p. 5/15 Initial Thoughts Without direct s i d i communication: 2-Stage Scheme takes the following as input: p r;1, p r;2 : The CH parameters, nr 1, nr 2 : The total # of to-be-sent symbols, np s1 ;d 2, np s2 ;d 1 : The overheard info.

16 Wang, ISIT 2011 p. 5/15 Initial Thoughts Without direct s i d i communication: 2-Stage Scheme takes the following as input: p r;1, p r;2 : The CH parameters, nr 1, nr 2 : The total # of to-be-sent symbols, np s1 ;d 2, np s2 ;d 1 : The overheard info. With direct s i d i communication: We thus need p r;1, p r;2 : The CH parameters, The total # of to-be-sent symbols: The overheard info.:

17 Wang, ISIT 2011 p. 5/15 Initial Thoughts Without direct s i d i communication: 2-Stage Scheme takes the following as input: p r;1, p r;2 : The CH parameters, nr 1, nr 2 : The total # of to-be-sent symbols, np s1 ;d 2, np s2 ;d 1 : The overheard info. With direct s i d i communication: We thus need p r;1, p r;2 : The CH parameters, The total # of to-be-sent symbols: [What r has heard] The overheard info.:

18 Wang, ISIT 2011 p. 5/15 Initial Thoughts Without direct s i d i communication: 2-Stage Scheme takes the following as input: p r;1, p r;2 : The CH parameters, nr 1, nr 2 : The total # of to-be-sent symbols, np s1 ;d 2, np s2 ;d 1 : The overheard info. With direct s i d i communication: We thus need p r;1, p r;2 : The CH parameters, The total # of to-be-sent symbols: [What r has heard] [What r and d 1 both have heard] The overheard info.:

19 Wang, ISIT 2011 p. 5/15 Initial Thoughts Without direct s i d i communication: 2-Stage Scheme takes the following as input: p r;1, p r;2 : The CH parameters, nr 1, nr 2 : The total # of to-be-sent symbols, np s1 ;d 2, np s2 ;d 1 : The overheard info. With direct s i d i communication: We thus need p r;1, p r;2 : The CH parameters, The total # of to-be-sent symbols: [What r has heard] [What r and d 1 both have heard] The overheard info.: [What r and d 2 have heard]

20 Initial Thoughts Without direct s i d i communication: 2-Stage Scheme takes the following as input: p r;1, p r;2 : The CH parameters, nr 1, nr 2 : The total # of to-be-sent symbols, np s1 ;d 2, np s2 ;d 1 : The overheard info. With direct s i d i communication: We thus need p r;1, p r;2 : The CH parameters, The total # of to-be-sent symbols: [What r has heard] [What r and d 1 both have heard] The overheard info.: [What r and d 2 have heard] [What r, d 2, and d 1 all have heard] Wang, ISIT 2011 p. 5/15

21 Initial Thoughts Without direct s i d i communication: 2-Stage Scheme takes the following as input: p r;1, p r;2 : The CH parameters, nr 1, nr 2 : The total # of to-be-sent symbols, np s1 ;d 2, np s2 ;d 1 : The overheard info. With direct s i d i communication: We thus need p r;1, p r;2 : The CH parameters, The total # of to-be-sent symbols: [What r has heard] Common Info. [What r and d 1 both have heard] The overheard info.: Common Info. [What r and d 2 have heard] Common Info. [What r, d 2, and d 1 all have heard] Wang, ISIT 2011 p. 5/15

22 Initial Thoughts Without direct s i d i communication: With this motivation, this work studies the Common Info. of Random Linear Network Coding. With direct s i d i communication: We thus need p r;1, p r;2 : The CH parameters, The total # of to-be-sent symbols: [What r has heard] Common Info. [What r and d 1 both have heard] The overheard info.: Common Info. [What r and d 2 have heard] Common Info. [What r, d 2, and d 1 all have heard] Wang, ISIT 2011 p. 5/15

23 Wang, ISIT 2011 p. 6/15 Settings and Definitions Random Linear Network Coding (RLNC) N packets: W = (W 1,, W N ) (GF(q)) N. Each time t, source sends Y t = v t W T via an erasure CH. R t : The set of destinations receive the packet at time t. v t is randomly generated, but is known to all receivers.

24 Wang, ISIT 2011 p. 6/15 Settings and Definitions Random Linear Network Coding (RLNC) N packets: W = (W 1,, W N ) (GF(q)) N. Each time t, source sends Y t = v t W T via an erasure CH. R t : The set of destinations receive the packet at time t. v t is randomly generated, but is known to all receivers. Knowledge Space: Ω k = span{v t : t such that d k R t }. If Z k denotes the pkts rcv d by d k, then I(W; Z k ) = rank(ω k ).

25 Wang, ISIT 2011 p. 6/15 Settings and Definitions Random Linear Network Coding (RLNC) N packets: W = (W 1,, W N ) (GF(q)) N. Each time t, source sends Y t = v t W T via an erasure CH. R t : The set of destinations receive the packet at time t. v t is randomly generated, but is known to all receivers. Knowledge Space: Ω k = span{v t : t such that d k R t }. If Z k denotes the pkts rcv d by d k, then I(W; Z k ) = rank(ω k ). ( Kk=1 ) Common Information (CI): rank Ω k. Gács-Körner CI among Z 1 to Z K is rank ( Kk=1 Ω k ).

26 Settings and Definitions Random Linear Network Coding (RLNC) N packets: W = (W 1,, W N ) (GF(q)) N. Each time t, source sends Y t = v t W T via an erasure CH. R t : The set of destinations receive the packet at time t. v t is randomly generated, but is known to all receivers. Knowledge Space: Ω k = span{v t : t such that d k R t }. If Z k denotes the pkts rcv d by d k, then I(W; Z k ) = rank(ω k ). ( Kk=1 ) Common Information (CI): rank Ω k. Gács-Körner CI among Z 1 to Z K is rank ( Kk=1 Ω k ). Other notations: T S = { t : S Rt : The amount of time when all d k in S receive the pkt; A B = span(v : v A B). Wang, ISIT 2011 p. 6/15

27 Wang, ISIT 2011 p. 7/15 ( Kk=1 ) Our goal: Quantifying rank Ω k for all combinations of K, N, and {R t : t} assuming large GF(q). The total # of to-be-sent symbols: [What r has heard] [What r and d 1 both have heard] The overheard info.: [What r and d 2 have heard] [What r, d 2, and d 1 all have heard]

28 Wang, ISIT 2011 p. 7/15 ( Kk=1 ) Our goal: Quantifying rank Ω k for all combinations of K, N, and {R t : t} assuming large GF(q). The total (# of to-be-sent ) symbols: rank Ω [s 1] r [What r and d 1 both have heard] The overheard info.: [What r and d 2 have heard] [What r, d 2, and d 1 all have heard]

29 Wang, ISIT 2011 p. 7/15 ( Kk=1 ) Our goal: Quantifying rank Ω k for all combinations of K, N, and {R t : t} assuming large GF(q). The total (# of to-be-sent ) symbols: rank Ω [s 1] ( rank r ) Ω [s 1] r Ω [s 1] d 1 The overheard info.: [What r and d 2 have heard] [What r, d 2, and d 1 all have heard]

30 Wang, ISIT 2011 p. 7/15 ( Kk=1 ) Our goal: Quantifying rank Ω k for all combinations of K, N, and {R t : t} assuming large GF(q). The total (# of to-be-sent ) symbols: rank Ω [s 1] ( rank r ) Ω [s 1] r Ω [s 1] d 1 The overheard info.: ( ) rank Ω [s 1] r Ω [s 1] d 2 [What r, d 2, and d 1 all have heard]

31 Wang, ISIT 2011 p. 7/15 ( Kk=1 ) Our goal: Quantifying rank Ω k for all combinations of K, N, and {R t : t} assuming large GF(q). The total (# of to-be-sent ) symbols: rank Ω [s 1] ( rank r ) Ω [s 1] r Ω [s 1] d 1 The overheard ( info.: ) rank Ω [s 1] r Ω [s 1] ( d 2 rank Ω [s 1] r Ω [s 1] d 2 Ω [s 1] d 1 )

32 Wang, ISIT 2011 p. 8/15 The Main Challenge rank ( Kk=1 Ω k )

33 Wang, ISIT 2011 p. 8/15 The Main Challenge rank ( Kk=1 Ω k ) When K = 1, then rank (Ω 1 ) = min(t 1, N).

34 Wang, ISIT 2011 p. 8/15 The Main Challenge rank ( Kk=1 Ω k ) When K = 1, then rank (Ω 1 ) = min(t 1, N). When K = 2, we have rank (Ω 1 Ω 2 ) =rank (Ω 1 ) + rank (Ω 2 ) rank (Ω 1 Ω 2 ) = min(t 1, N) + min(t 2, N) min(t 1 + T 2 T {1,2}, N)

35 Wang, ISIT 2011 p. 8/15 The Main Challenge rank ( Kk=1 Ω k ) When K = 1, then rank (Ω 1 ) = min(t 1, N). When K = 2, we have rank (Ω 1 Ω 2 ) =rank (Ω 1 ) + rank (Ω 2 ) rank (Ω 1 Ω 2 ) = min(t 1, N) + min(t 2, N) min(t 1 + T 2 T {1,2}, N) Our first thought was when K = 3, we should have rank ( 3 k=1 Ω k ) = K k=1 rank (Ω k ) rank ( ) Ω i Ω j + rank (Ω1 Ω 2 Ω 3 ) i<j

36 Wang, ISIT 2011 p. 8/15 The Main Challenge rank ( Kk=1 Ω k ) When K = 1, then rank (Ω 1 ) = min(t 1, N). When K = 2, we have rank (Ω 1 Ω 2 ) =rank (Ω 1 ) + rank (Ω 2 ) rank (Ω 1 Ω 2 ) = min(t 1, N) + min(t 2, N) min(t 1 + T 2 T {1,2}, N) Our first thought was when K = 3, we should have rank ( 3 k=1 Ω k ) = K k=1 rank (Ω k ) rank ( ) Ω i Ω j + rank (Ω1 Ω 2 Ω 3 ) i<j DOES NOT HOLD!! An example is provided in the paper.

37 Wang, ISIT 2011 p. 9/15 Main Result A partition of {1,, K} is a collection of disjoint subsets {S m } = {S 1, S 2,, S M } such that M m=1 S m = {1,, K}. Theorem 1 Define ( ) + = max(, 0). For any receiving sets {R t : t}, with sufficiently large GF(q) we have ( ) { } K M rank Ω k = max N (N T Sm ) + : partition {S m } m=1 k=1.

38 Wang, ISIT 2011 p. 10/15 Proof of A Simplified Example We prove a degenerate case in this presentation. Assume N max k [K] T k S [K], ( k S ( Kk=1 ) We will prove that rank Ω k T k ) ( S 1)N T S. = K k=1 T k (K 1)N.

39 Proof of A Simplified Example We prove a degenerate case in this presentation. Assume N max k [K] T k S [K], ( k S ( Kk=1 ) We will prove that rank Ω k T k ) ( S 1)N T S. = K k=1 T k (K 1)N. An illustrative example w. K = 3: N = 5 dimensional vector space. 7 vectors with reception status 001, 010, 011, 100, 101, 110, 111. T 1 = T 2 = T 3 = 4, T {1,2} = T {1,3} = T {2,3} = 2, and T {1,2,3} = 1. Wang, ISIT 2011 p. 10/15

40 Proof of A Simplified Example We prove a degenerate case in this presentation. Assume N max k [K] T k S [K], ( k S ( Kk=1 ) We will prove that rank Ω k T k ) ( S 1)N T S. = K k=1 T k (K 1)N. ( 3k=1 ) An illustrative example w. K = 3: We will prove rank Ω k = N = 5 dimensional vector space (3 1) 5 = 2. 7 vectors with reception status 001, 010, 011, 100, 101, 110, 111. T 1 = T 2 = T 3 = 4, T {1,2} = T {1,3} = T {2,3} = 2, and T {1,2,3} = 1. Wang, ISIT 2011 p. 10/15

41 Proof (Cont d) Wang, ISIT 2011 p. 11/15

42 Proof (Cont d) Wang, ISIT 2011 p. 11/15

43 Proof (Cont d) Wang, ISIT 2011 p. 11/15

44 Proof (Cont d) Wang, ISIT 2011 p. 11/15

45 Proof (Cont d) Wang, ISIT 2011 p. 11/15

46 Proof (Cont d) Wang, ISIT 2011 p. 11/15

47 Proof (Cont d) Wang, ISIT 2011 p. 11/15

48 Proof (Cont d) Wang, ISIT 2011 p. 11/15

49 Proof (Cont d) Wang, ISIT 2011 p. 11/15

50 Proof (Cont d) Wang, ISIT 2011 p. 11/15

51 Proof (Cont d) Wang, ISIT 2011 p. 11/15

52 Proof (Cont d) Wang, ISIT 2011 p. 11/15

53 Proof (Cont d) Wang, ISIT 2011 p. 11/15

54 Proof (Cont d) Wang, ISIT 2011 p. 11/15

55 Proof (Cont d) Wang, ISIT 2011 p. 11/15

56 Proof (Cont d) Wang, ISIT 2011 p. 11/15

57 Proof (Cont d) Wang, ISIT 2011 p. 11/15

58 Wang, ISIT 2011 p. 12/15 Proof (Cont d) [Koetter et al. 03, Ho et al. 06]: If we can find deterministically 7 vectors s.t. then for almost all random choices of the 7 vectors, rank(ω 1 Ω 2 ) = 5.

59 Wang, ISIT 2011 p. 12/15 Proof (Cont d) [Koetter et al. 03, Ho et al. 06]: If we can find deterministically 7 vectors s.t. Fully Spanned Cond. then for almost all random choices of the 7 vectors, rank(ω 1 Ω 2 ) = 5.

60 Wang, ISIT 2011 p. 12/15 Proof (Cont d) [Koetter et al. 03, Ho et al. 06]: If we can find deterministically 7 vectors s.t. Fully Spanned Cond. then for almost all random choices of the 7 vectors, rank(ω 1 Ω 2 ) = 5. The rank problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned Cond.

61 Wang, ISIT 2011 p. 12/15 Proof (Cont d) [Koetter et al. 03, Ho et al. 06]: If we can find deterministically 7 vectors s.t. Fully Spanned Cond. then for almost all random choices of the 7 vectors, rank(ω 1 Ω 2 ) = 5. The rank problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned Cond. [New]: If we can find deterministically 10 vectors: v 1, v 2, v 3, plus 7, s.t. Fully Spanned

62 Wang, ISIT 2011 p. 12/15 Proof (Cont d) [Koetter et al. 03, Ho et al. 06]: If we can find deterministically 7 vectors s.t. Fully Spanned Cond. then for almost all random choices of the 7 vectors, rank(ω 1 Ω 2 ) = 5. The rank problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned Cond. [New]: If we can find deterministically 10 vectors: v 1, v 2, v 3, plus 7, s.t. Fully Spanned

63 Wang, ISIT 2011 p. 12/15 Proof (Cont d) [Koetter et al. 03, Ho et al. 06]: If we can find deterministically 7 vectors s.t. Fully Spanned Cond. then for almost all random choices of the 7 vectors, rank(ω 1 Ω 2 ) = 5. The rank problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned Cond. [New]: If we can find deterministically 10 vectors: v 1, v 2, v 3, plus 7, s.t. Fully Spanned Fully Intersected Cond.

64 Wang, ISIT 2011 p. 12/15 Proof (Cont d) [Koetter et al. 03, Ho et al. 06]: If we can find deterministically 7 vectors s.t. Fully Spanned Cond. then for almost all random choices of the 7 vectors, rank(ω 1 Ω 2 ) = 5. The rank problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned Cond. [New]: If we can find deterministically 10 vectors: v 1, v 2, v 3, plus 7, s.t. Fully Spanned Fully Spanned Fully Intersected Cond.

65 Wang, ISIT 2011 p. 12/15 Proof (Cont d) [Koetter et al. 03, Ho et al. 06]: If we can find deterministically 7 vectors s.t. Fully Spanned Cond. then for almost all random choices of the 7 vectors, rank(ω 1 Ω 2 ) = 5. The rank problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned Cond. [New]: If we can find deterministically 10 vectors: v 1, v 2, v 3, plus 7, s.t. Fully Spanned Fully Spanned Fully Intersected Cond.

66 Wang, ISIT 2011 p. 12/15 Proof (Cont d) [Koetter et al. 03, Ho et al. 06]: If we can find deterministically 7 vectors s.t. Fully Spanned Cond. then for almost all random choices of the 7 vectors, rank(ω 1 Ω 2 ) = 5. The rank problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned Cond. [New]: If we can find deterministically 10 vectors: v 1, v 2, v 3, plus 7, s.t. Fully Spanned Fully Spanned Fully Intersected Cond. then for almost all rand. choices of the 7 vectors, rank((ω 1 Ω 2 ) Ω 3 ) = 5.

67 Proof (Cont d) [Koetter et al. 03, Ho et al. 06]: If we can find deterministically 7 vectors s.t. Fully Spanned Cond. then for almost all random choices of the 7 vectors, rank(ω 1 Ω 2 ) = 5. The rank problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned Cond. [New]: If we can find deterministically 10 vectors: v 1, v 2, v 3, plus 7, s.t. Fully Spanned Fully Spanned Fully Intersected Cond. then for almost all rand. choices of the 7 vectors, rank((ω 1 Ω 2 ) Ω 3 ) = 5. The Common Info. problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned and Fully Intersected Conds. Wang, ISIT 2011 p. 12/15

68 Proof (Cont d) [Koetter et al. 03, Ho et al. 06]: If we can find deterministically 7 vectors s.t. It does not matter how we choose v 1, v 2, v 3, plus 7, Fully Spanned Cond. and in what order we construct the vectors. then for almost all random choices of the 7 vectors, rank(ω 1 Ω 2 ) = 5. Any deterministic construction will suffice! The rank problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned Cond. [New]: If we can find deterministically 10 vectors: v 1, v 2, v 3, plus 7, s.t. Fully Spanned Fully Spanned Fully Intersected Cond. then for almost all rand. choices of the 7 vectors, rank((ω 1 Ω 2 ) Ω 3 ) = 5. The Common Info. problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned and Fully Intersected Conds. Wang, ISIT 2011 p. 12/15

69 A Deterministic Assignment Wang, ISIT 2011 p. 13/15

70 A Deterministic Assignment Wang, ISIT 2011 p. 13/15

71 A Deterministic Assignment Wang, ISIT 2011 p. 13/15

72 A Deterministic Assignment Wang, ISIT 2011 p. 13/15

73 A Deterministic Assignment Wang, ISIT 2011 p. 13/15

74 A Deterministic Assignment Wang, ISIT 2011 p. 13/15

75 A Deterministic Assignment Wang, ISIT 2011 p. 13/15

76 A Deterministic Assignment Wang, ISIT 2011 p. 13/15

77 A Deterministic Assignment Wang, ISIT 2011 p. 13/15

78 A Deterministic Assignment Wang, ISIT 2011 p. 13/15

79 A Deterministic Assignment Wang, ISIT 2011 p. 13/15

80 Wang, ISIT 2011 p. 14/15 Intuition of the FS and FI Conds. The Common Info. problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned and Fully Intersected Conds. Let x denote the input coding vectors and the local mixing kernels. Then in RLNC, each message along an edge has the form of M = ( f 1 (x), f 2 (x),, f N (x)) where f i (x) are polynomials of x. [Koetter et al. 03].

81 Wang, ISIT 2011 p. 14/15 Intuition of the FS and FI Conds. The Common Info. problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned and Fully Intersected Conds. Let x denote the input coding vectors and the local mixing kernels. Then in RLNC, each message along an edge has the form of M = ( f 1 (x), f 2 (x),, f N (x)) where f i (x) are polynomials of x. [Koetter et al. 03]. When constructing the basis vectors of k Ω k, we need to solve linear equations (i.e., being in( all marginal spaces), which results in the form f1 (x) v = g 1 (x), f 2(x) g 2 (x),, f ) N(x). g N (x)

82 Wang, ISIT 2011 p. 14/15 Intuition of the FS and FI Conds. The Common Info. problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned and Fully Intersected Conds. Let x denote the input coding vectors and the local mixing kernels. Then in RLNC, each message along an edge has the form of M = ( f 1 (x), f 2 (x),, f N (x)) where f i (x) are polynomials of x. [Koetter et al. 03]. When constructing the basis vectors of k Ω k, we need to solve linear equations (i.e., being in( all marginal spaces), which results in the form f1 (x) v = g 1 (x), f 2(x) g 2 (x),, f ) N(x). g N (x) [Fully Intersected] associates a deterministic x 0 assignment with the corresponding fractional expressions; [Fully Spanned] guarantees both g n (x) and det ([v i ]) are non-zero.

83 Wang, ISIT 2011 p. 15/15 Conclusion & Future Works The Common Info. problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned and Fully Intersected Conds.

84 Wang, ISIT 2011 p. 15/15 Conclusion & Future Works The Common Info. problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned and Fully Intersected Conds. Main Results rank ( K k=1 Ω k ) = max { N } M (N T Sm ) + : partition {S m }. m=1

85 Wang, ISIT 2011 p. 15/15 Conclusion & Future Works The Common Info. problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned and Fully Intersected Conds. Main Results rank ( K k=1 Ω k ) = max { N } M (N T Sm ) + : partition {S m }. m=1 Main motivation: The total (# of to-be-sent ) symbols: rank Ω [s 1] ( rank r ) Ω [s 1] r Ω [s 1] d 1 The overheard ( info.: ) rank Ω [s 1] r Ω [s 1] ( d 2 rank Ω [s 1] r Ω [s 1] d 2 Ω [s 1] d 1 )

86 Conclusion & Future Works The Common Info. problem of RLNC is reduced to finding a deterministic assignment satisfying the Fully Spanned and Fully Intersected Conds. Main Results rank ( K k=1 Ω k ) = max { N } M (N T Sm ) + : partition {S m }. m=1 Main motivation: The total (# of to-be-sent ) symbols: rank Ω [s 1] ( rank r ) Ω [s 1] r Ω [s 1] d 1 The overheard ( info.: ) rank Ω [s 1] r Ω [s 1] ( d 2 rank Ω [s 1] r Ω [s 1] d 2 Ω [s 1] d 1 ) Future works: (i) Arbitrary combinations: Ex: (Ω 1 Ω 2 ) Ω 3. (ii) Common Information of RLNC over multi-hop networks. Wang, ISIT 2011 p. 15/15