Distributed Reed-Solomon Codes

Transcription

1 Distributed Reed-Solomon Codes Farzad Parvaresh University of Isfahan Institute for Network Coding CUHK, Hong Kong August 216

2 Research interests List-decoding of algebraic codes Construction of efficient list-decodable codes over GF(2) Efficient List-decoding of RS codes beyond the Johnson bound Applications of compressed sensing Phase retrieval problem Power optimization over relay network Computing the cut-set bound of a relay network efficiently Computing diversity multiplexing tradeoff of generalized half-duplex relay networks Any fun problem

3 Research interests List-decoding of algebraic codes Construction of efficient list-decodable codes over GF(2) Efficient List-decoding of RS codes beyond the Johnson bound Applications of compressed sensing Phase retrieval problem Power optimization over relay network Computing the cut-set bound of a relay network efficiently Computing diversity multiplexing tradeoff of generalized half-duplex relay networks Any fun problem

4 Research interests List-decoding of algebraic codes Construction of efficient list-decodable codes over GF(2) Efficient List-decoding of RS codes beyond the Johnson bound Applications of compressed sensing Phase retrieval problem Power optimization over relay network Computing the cut-set bound of a relay network efficiently Computing diversity multiplexing tradeoff of generalized half-duplex relay networks

5 Reed-Solomon codes Encoding of RS(n,k,d): Information symbols: (u 1, u 2,, u k ) Fk q f X = u 1 + u 2 X + + u k X k 1 F q [X] Polynomial evaluation: f α 1, f α 2,, f(α n ) α 1,, α n Codeword: (c 1, c 2,, c n ) F q n F q n

6 Reed-Solomon codes Encoding of RS(n,k,d): Information symbols: (u 1, u 2,, u k ) Fk q f X = u 1 + u 2 X + + u k X k 1 F q [X] Polynomial evaluation: f α 1, f α 2,, f(α n ) α 1,, α n Equivalently: (u 1, u 2,, u k ) Codeword: (c 1, c 2,, c n ) F q n 1 1 k 1 α 1 αk 1 n = (c 1, c 2,, c n ) F q n

7 Reed-Solomon codes Encoding of RS(n,k,d): Information symbols: (u 1, u 2,, u k ) Fk q f X = u 1 + u 2 X + + u k X k 1 F q [X] Polynomial evaluation: f α 1, f α 2,, f(α n ) α 1,, α n Equivalently: (u 1, u 2,, u k ) Codeword: (c 1, c 2,, c n ) F q n 1 1 k 1 α 1 αk 1 n Generator matrix, G RS, of RS code. = (c 1, c 2,, c n ) F q n

8 Reed-Solomon Encoding u 1 u 2 u k Encoders: v 1 v 2 v n c 1 =ug 1 c 2 =ug 2 c n =ug n G RS g1 g2 gn G RS is usually a Vandermonde matrix

9 Reed-Solomon Encoding Has access to all data u 1 u 2 u k Encoders: v 1 v 2 v n c 1 =ug 1 c 2 =ug 2 c n =ug n G RS g 1 g 2 g n G RS is usually a Vandermonde matrix

10 Distributed RS Encoding u 1 u 2 u k v 1 v 2 v n c 1 =ug 1 c 2 =ug 2 c n =ug n n RS g g g G 2 1 Encoders:

11 Distributed RS Encoding u 1 u 2 u k v 1 v 2 v n c 1 =ug 1 c 2 =ug 2 c n =ug n n RS g g g G 2 1 Encoders:

12 Distributed RS Encoding u 1 u 2 u k v 1 v 2 v n c 1 =ug 1 c 2 =ug 2 c n =ug n n RS g g g G 2 1 Encoders:

13 Distributed RS Encoding u 1 u 2 u k v 1 v 2 v n c 1 =ug 1 c 2 =ug 2 c n =ug n n RS g g g G 2 1 Encoders:

14 Distributed RS Encoding u 1 u 2 u k v 1 v 2 v n c 1 =ug 1 c 2 =ug 2 c n =ug n n RS g g g G 2 1 Encoders:

15 Distributed RS Encoding u 1 u 2 u k v 1 v 2 v n c 1 =ug 1 c 2 =ug 2 c n =ug n n RS g g g G 2 1 Encoders:

16 Distributed RS Encoding u 1 u 2 u k v 1 v 2 v n c 1 =ug 1 c 2 =ug 2 c n =ug n n RS g g g G 2 1 Encoders:

17 Distributed RS Encoding Doesn t have access to all data Encoders: u 1 u 2 u k v 1 v 2 v n c 1 =ug 1 c 2 =ug 2 c n =ug n G RS g 1 g 2 g n

18 Distributed RS Encoding Doesn t have access to all data Encoders: u 1 u 2 u k v 1 v 2 v n G c 1 =ug 1 c 2 =ug 2 c n =ug n g1 g 2 gn RS

19 Distributed RS Encoding Doesn t have access to all data Encoders: u 1 u 2 u k v 1 v 2 v n G c 1 =ug 1 c 2 =ug 2 c n =ug n Is there any generator matrix g1 for the RS code with the g 2 given constraints? gn RS

20 Simple multiple access network r s s 1 s 2 s s r s r s v 1 v 2 v n W. Halbawi, T. Ho, H. Yao and I. Duursma, Distributed codes for simple multiple access network, arxiv: v1, 213. D

21 Simple multiple access network r s s 1 s 2 s s r s r s v 1 v 2 v n z errors At most z relays are compromised by adversaries. D

22 Simple multiple access network r s s 1 s 2 s s r s r s v 1 v 2 v n z errors D At most z relays are compromised by adversaries. If relay nodes encode a RS[n, k, d=2z+1] code, then destination can recover the data.

23 Constrained MDS generator matrices MDS matrix completion problem: Assume M is a binary n k matrix that satisfy no rectangle condition (it has no all-zero submatrix of total dimension exceeding k). Is there exist an MDS completion for M, i.e. replacing 1 s in M by elements of F q such that the constructed matrix generates an MDS code? Balanced sparsest generator matrix for MDS codes: Constructing a generator matrix, M, for an MDS code such that each row of M has weight n-k+1 and column weights of M differ from each other by at most one. Weakly secure cooperative data exchange problem A group of wireless clients have access to different subsets of n packets and the like to exchange the packets over a lossless broadcast channel secuirly.

24 Related works [Yao, Ho, Nita-Rotaru 11] Key agreement for wireless network in the presence of active adversaries [Halbawi, Ho, Yao, Duursma 13] Distributed Reed-Solomon codes for simple multiple access networks [Dau, Song, Dong, Yuen 13] Balanced sparsest generator matrices for MDS codes [El Rouayheb, Sprintson, Sadeghi 1] On coding for cooperative data exchange [Dau, Song, Sprintson, Yuen 15] Constructions of MDS codes via random Vandermonde and Cauchy matrices over small fields [Dau, Song, Yuen 14]On the existence of MDS codes over small fields with constrained generator matrices [Yan, Sprintson 13] Algorithms for weakly secure data exchange [W. Halbawi, M. Thill and B. Hassibi] Coding with constraints: Minimum distance bounds and systematic constructions [W. Halbawi, Z. Liu and B. Hassibi] Balanced Reed-Solomon codes

25 Distributed RS Encoding u 12 u 13 u 21 u 11 u 22 u k1 u k2 u k3 v 1 v 2 v n G c 1 =ug 1 c 2 =ug 2 c n =ug n Is there any generator matrix g1 for the RS code with the g 2 given constraints? gn RS

26 Distributed RS Encoding u 12 u 13 u 21 u 11 u 22 u k1 u k2 u k3 v 1 v 2 v n G c 1 =ug 1 c 2 =ug 2 c n =ug n Is there any generator matrix g1 for the RS code with the g 2 given constraints? gn RS

27 Distributed RS Encoding r s u 12 u 13 u 21 u 11 u 22 u k1 u k2 u k3 v 1 v 2 v n G c 1 =ug 1 c 2 =ug 2 c n =ug n Is there any generator matrix g1 for the RS code with the g 2 given constraints? gn RS

28 Distributed RS Encoding r s s 1 s 2 s s r2 r s r s v 1 v 2 v n G c 1 =ug 1 c 2 =ug 2 c n =ug n Is there any generator matrix g1 for the RS code with the g 2 given constraints? gn RS

29 Example: Three sources s 1 s 2 s 3 v 1 v 2 v n c 1 =ug 1 c 2 =ug 2 c n =ug n GRS n 1 n 2 n 3 n 12 n 13 n 23 n 123 Z 1 ={positions of zeros of the first group} Z 2 ={positions of zeros of the second group} Z 3 ={positions of zeros of the third group}

30 Simple multiple access network r s s 1 s 2 s s r s r s v 1 v 2 v n W. Halbawi, T. Ho, H. Yao and I. Duursma, Distributed codes for simple multiple access network, arxiv: v1, 213. D

31 Necessary condition (network coding) Network s 1 z links in error s 2 D s s We can only hope to find a distributed RS code for rates (,,, r s ) in the capacity region of the network.

32 Network error correction N. Cai and R. W. Yeung (26) Single source multicast networks D. Silva, F. R. Kschischang, and R. Koetter (28) Rank-metric codes S. Mohajer, M. Jafari, S. Diggavi, and C. Fragouli (29) Two source multicast networks T. Dikaliotis, T. Ho, S. Jaggi, S. Vyetrenko, H. Yao, M. Effros, J. Kliewer, and E. Erez (211) Multisource multicast networks X. Guang and Z. Zhang (214) Linear network error correction coding

33 Network error correction Network s 1 z links in error s 2 D s s Capacity region: i I(S ) r i C S 2z, S S T. Dikaliotis, T. Ho, S. Jaggi, S. Vyetrenko, H. Yao, M. Effros, J. Kliewer, and E. Erez, Multiple access network information-flow and correction codes, IEEE IT, 57(2), 211.

34 Network error correction Network s 1 z links in error s 2 D s s Min-cut capacity between S and D Capacity region: i I(S ) r i C S 2z, S S T. Dikaliotis, T. Ho, S. Jaggi, S. Vyetrenko, H. Yao, M. Effros, J. Kliewer, and E. Erez, Multiple access network information-flow and correction codes, IEEE IT, 57(2), 211.

35 Network error correction Network s 1 z links in error s 2 D s s Min-cut capacity between S and D Capacity region: i I(S ) r i C S 2z, S S Computationally efficient linear network codes with decoding success probability of at least 1- s E /p and complexity O(l m S ) T. Dikaliotis, T. Ho, S. Jaggi, S. Vyetrenko, H. Yao, M. Effros, J. Kliewer, and E. Erez, Multiple access network information-flow and correction codes, IEEE IT, 57(2), 211.

36 Network error correction Network s 1 z links in error s 2 D s s Min-cut capacity between S and D Capacity region: i I(S ) r i C S 2z, S S Exponential in S Computationally efficient linear network codes with decoding success probability of at least 1- s E /p and complexity O(l m S ) T. Dikaliotis, T. Ho, S. Jaggi, S. Vyetrenko, H. Yao, M. Effros, J. Kliewer, and E. Erez, Multiple access network information-flow and correction codes, IEEE IT, 57(2), 211.

37 Necessary condition (network coding) r s s 1 s 2 s s v 1 v 2 v n D Min-cut capacity between S and D Necessary condition: i I(S ) r i C S 2z, S S T. Dikaliotis, T. Ho, S. Jaggi, S. Vyetrenko, H. Yao, M. Effros, J. Kliewer, and E. Erez, Multiple access network information-flow and correction codes, IEEE IT, 57(2), 211.

38 Necessary condition (three sources) GRS n 1 n 2 n 3 n 12 n 13 n 23 n 123 Z 1 ={positions of zeros of the first group} Z 2 ={positions of zeros of the second group} Z 3 ={positions of zeros of the third group} Capacity region: r i k Z i, i {1,2,3} r i + r j k Z i Z j, i, j + + k {1,2,3}

39 Necessary condition (three sources) GRS n 1 n 2 n 3 n 12 n 13 n 23 n 123 Z 1 ={positions of zeros of the first group} Z 2 ={positions of zeros of the second group} Z 3 ={positions of zeros of the third group} Capacity region: r i k Z i, i {1,2,3} r i + r j k Z i Z j, i, j + + k {1,2,3}

40 Necessary condition (three sources) GRS n 1 n 2 n 3 n 12 n 13 n 23 n 123 Z 1 ={positions of zeros of the first group} Z 2 ={positions of zeros of the second group} Z 3 ={positions of zeros of the third group} Capacity region: r i k Z i, i {1,2,3} r i + r j k Z i Z j, i, j + + k {1,2,3}

41 Necessary conditions (three sources) GRS n 1 n 2 n 3 n 12 n 13 n 23 n 123 Z 1 ={positions of zeros of the first group} Z 2 ={positions of zeros of the second group} Z 3 ={positions of zeros of the third group} Capacity region: r i k Z i, i {1,2,3} r i + r j k Z i Z j, i, j + + k {1,2,3} If rates are inside the capacity region, Halbawi et al. (214) showed that for up to three sources one can always find G RS.

42 Necessary conditions (three sources) GRS n 1 n 2 n 3 n 12 n 13 n 23 n 123 Z 1 ={positions of zeros of the first group} We are going to show that for any number of sources if rates are inside the capacity region of SMAN, one can always construct the Z 2 ={positions of zeros of the second group} Z 3 ={positions of zeros of the third group} Capacity region: generator matrix r i k GZ i, i {1,2,3} RS for the distributed RS code over r i + a rfinite j k field Z i of Z j size, i, j at {1,2,3} least n. + + k If rates are inside the capacity region, Halbawi et al. (214) showed that for up to three sources one can always find G RS.

43 Proof (by induction on # sources) The result holds for the case of two sources We assume that the result holds for the case of having less than s sources. We show that it holds for the case of s sources Constraints for the case of s sources: k Z 1 k Z 2 r s k Z s r i1 + r i2 + + r il k Z i1 Z i 2 Z il r s k

44 When rates are inside the boundaries All the constraints are not tight. Constraints for the case of s sources: < k Z 1 < k Z 2 r s < k Z s r i1 + r i2 + + r il < k Z i1 Z i 2 Z il r s < k

45 Rates inside the boundaries All the constraints are not tight. Constraints for the case of s sources: < k Z 1 < k Z 2 r s < k Z s r i1 + r i2 + + r il < k Z i1 Z i 2 Z il r s k Increase r i s till we hit the boundaries

46 Rates inside the boundaries All the constraints are not tight. Constraints for the case of s sources: < k Z 1 < k Z 2 r s < k Z s r i1 + r i2 + + r il < k Z i1 Z i 2 Z il r s k Increase r i s till we hit the boundaries Notice: If we construct G RS for the new rates, then by removing some rows of G RS, we can construct G RS for the original rates.

47 Rates on the boundary = k Z 1 = k Z r s = k l 2 l < s i: r i < k Z i r i1 + + r il = k Z i1 Z i l r s = k Z s r i1 + + r il < k Z i1 Z i l Case I Case II Case III

48 Rates on the boundary (Case III) Constraints: k Z 1 i: r i < k Z i r s k Z s r i1 + r i2 + + r il < k Z i1 Z i 2 Z il r s = k

49 Rates on the boundary (Case III) Constraints: k Z 1 i: r i < k Z i r s k Z s r i1 + r i2 + + r il < k Z i1 Z i 2 Z il r s = k GRS

50 Rates on the boundary (Case III) Constraints: k Z 1 i: r i < k Z i r s k Z s r i1 + r i2 + + r il < k Z i1 Z i 2 Z il r s = k We can always add a column of all zeros to the i-th group of G RS without violating the constraints. GRS

51 Rates on the boundary (Case III) Constraints: k Z 1 i: r i < k Z i r s k Z s r i1 + r i2 + + r il < k Z i1 Z i 2 Z il r s = k We can always add a column of all zeros to the i-th group of G RS without violating the constraints. GRS

52 Rates on the boundary (Case III) Constraints: k Z 1 r i k ( Z i + 1) r s k Z s r i1 + r i2 + + r il k ( Z i1 Z i 2 Z il + 1) r s = k We can always add a column of all zeros to the i-th group of G RS without violating the constraints. GRS

53 Rates on the boundary (Case III) Constraints: k Z 1 r i k ( Z i + 1) r s k Z s r i1 + r i2 + + r il k ( Z i1 Z i 2 Z il + 1) r s = k We can always add a column of all zeros to the i-th group of G RS without violating the constraints. GRS Keep adding zero columns until a set of inequalities becomes tight.

54 Rates on the boundary (Case III) Case III By inserting zeros in the generator matrix Case I Case II

55 Rates on the boundary (Case II) Constraints: k Z 1 r i k Z i r s k Z s r i1 + r i2 + + r il = k Z i1 Z i 2 Z il r s k GRS

56 Rates on the boundary (Case II) Constraints: k Z 1 r i k Z i r s k Z s r i1 + r i2 + + r il = k Z i1 Z i 2 Z il r s k GRS

57 Rates on the boundary (Case II) GRS + = k Z 1 Z 2

58 (1) G RS Rates on the boundary (Case II) GRS k Z 1 k Z 2 Create two new problems Problem 1: Problem 2: + = k Z 1 Z 2 G 2 (2) RS + = k Z 1 Z 2 = 2 = k Z 12 k Z 3 = 2 Z k Z 12 Z 3

59 (1) G RS Rates on the boundary (Case II) GRS Problem 1: Problem 2: k Z 1 k Z 2 + = k Z 1 Z 2 G 2 (2) RS + = k Z 1 Z 2 = 2 = k Z 12 k Z 3 = 2 Z 12 Notice: These subspaces are identical. 2 + k Z 12 Z 3

60 (1) G RS Rates on the boundary (Case II) GRS Problem 1: Problem 2: k Z 1 k Z 2 + = k Z 1 Z 2 G 2 (2) RS By induction, we can solve the sub-problems + = k Z 1 Z 2 = 2 = k Z 12 k Z 3 = 2 Z 12 Notice: These subspaces are identical. 2 + k Z 12 Z 3

61 Rates on the boundary (Case I) Constraints: = k Z 1 r i = k Z i r s = k Z s r i1 + r i2 + + r il k Z i1 Z i 2 Z il r s k

62 Case I: Example Consider we are looking for a distributed RS code with length n=6 and dimension k=3 such that = = =1 and the generator matrix has the following form: Evaluation points: G RS = α 1 α 2 α 3 α 4 α 5 α 6 So r i = k Z i = 3 2 = 1, r i + r j 3, + + 3

63 Case I: Example Consider we are looking for a distributed RS code with length n=6 and dimension k=3 such that = = =1 and the generator matrix has the following form: Evaluation points: G RS = α 1 α 2 α 3 α 4 α 5 α 6 So r i = k Z i = 3 2 = 1, r i + r j 3, General form of RS codewords from the first two rows: c 12 x = f x α 5 x α 6 + g (x α 3 )(x α 4 )

64 Case I: Example Consider we are looking for a distributed RS code with length n=6 and dimension k=3 such that = = =1 and the generator matrix has the following form: Evaluation points: G RS = α 1 α 2 α 3 α 4 α 5 α 6 So r i = k Z i = 3 2 = 1, r i + r j 3, General form of RS codewords from the first two rows: c 12 x = f x α 5 x α 6 + g (x α 3 )(x α 4 )

65 Case I: Example Consider we are looking for a distributed RS code with length n=6 and dimension k=3 such that = = =1 and the generator matrix has the following form: Evaluation points: G RS = α 1 α 2 α 3 α 4 α 5 α 6 So r i = k Z i = 3 2 = 1, r i + r j 3, General form of RS codewords from the first two rows: c 12 x = f x α 5 x α 6 + g (x α 3 )(x α 4 ) If α 1 α 5 α 1 α 6 (α 2 α 5 )(α 2 α 6 ) = α 1 α 3 α 1 α 4 f, g : c 12 α 1 = c 12 α 2 = (α 2 α 3 )(α 3 α 4 )

66 Case I: Example Consider we are looking for a distributed RS code with length n=6 and dimension k=3 such that = = =1 and the generator matrix has the following form: Evaluation points: G RS = So α 1 α 2 α 3 α 4 α 5 α 6 The generator matrix r i = k Z i = 3 2 = 1, r i + r j 3, General form of RS codewords from the first two rows: c 12 x = f x α 5 x α 6 + g (x α 3 )(x α 4 ) is not full rank! If α 1 α 5 α 1 α 6 (α 2 α 5 )(α 2 α 6 ) = α 1 α 3 α 1 α 4 f, g : c 12 α 1 = c 12 α 2 = (α 2 α 3 )(α 3 α 4 )

67 Case I: Example Consider we are looking for a distributed RS code with length n=6 and dimension k=3 such that = = =1 and the generator matrix has the following form: Evaluation points: G RS = α 1 α 2 α 3 α 4 α 5 α 6 To have a counterexample, So evaluation points should satisfy General form specific of RS constraints! codewords from the first two rows: r i = k Z i = 3 2 = 1, r i + r j 3, c 12 x = f x α 5 x α 6 + g (x α 3 )(x α 4 ) The generator matrix is not full rank! If α 1 α 5 α 1 α 6 (α 2 α 5 )(α 2 α 6 ) = α 1 α 3 α 1 α 4 f, g : c 12 α 1 = c 12 α 2 = (α 2 α 3 )(α 3 α 4 )

68 Case I: Example Consider we are looking for a distributed RS code with length n=6 and dimension k=3 such that = = =1 and the generator matrix has the following form: Evaluation points: G RS = α 1 α 2 α 3 α 4 α 5 α 6 Choose evaluation points of the distributed RS code carefully. To have a counterexample, So evaluation points should satisfy General form specific of RS constraints! codewords from the first two rows: r i = k Z i = 3 2 = 1, r i + r j 3, c 12 x = f x α 5 x α 6 + g (x α 3 )(x α 4 ) The generator matrix is not full rank! If α 1 α 5 α 1 α 6 (α 2 α 5 )(α 2 α 6 ) = α 1 α 3 α 1 α 4 f, g : c 12 α 1 = c 12 α 2 = (α 2 α 3 )(α 3 α 4 )

69 Rates on the boundary (Case I) Constraints: = k Z 1 r i = k Z i r s = k Z s GRS Theorem: There is always a set of evaluation points such that one can construct a full rank generator matrix in case I.

70 Rates on the boundary (Case I) Theorem: There is always a set of evaluation points such that one can construct a full rank generator matrix in case I. Proof by induction: GRS P 1 x = P 2 x = i Z 1 (x α i ) i Z 2 (x α i ) c 1 x = f x P 1 (x) deg f x 1 c 2 x = g x P 2 (x) deg g x 1

71 Rates on the boundary (Case I) Theorem: There is always a set of evaluation points such that one can construct a full rank generator matrix in case I. Proof by induction: GRS α i1,, α Z3 Z 3 Z 3 P 1 (α i 1 ) α 1 P1 (α i1 i 1 ) P 1 (α i Z 3 ) α 1 i P1 Z3 (α i ) Z 3 P 1 x = P 2 x = i Z 1 (x α i ) i Z 2 (x α i ) P 2 (α i 1 ) α 1 P2 (α i1 i 1 ) P 2 (α i Z 3 ) α 1 i P2 Z3 (α i Z 3 ) c 1 x = f x P 1 (x) deg f x 1 c 2 x = g x P 2 (x) deg g x 1 f f r1 1 g g r2 1 =

72 Rates on the boundary (Case I) Theorem: There is always a set of evaluation points such that one can always construct a full rank generator matrix in case I. Z 3 Proof by induction: P 1 (α i 1 ) α 1 P1 (α i1 i 1 ) P 1 (α i Z 3 ) α 1 i P1 Z3 (α i ) Z k, = k Z 3 Z 3 + P 2 (α i 1 ) α 1 P2 (α i1 i 1 ) P 2 (α i Z 3 ) α 1 i P2 Z3 (α i Z 3 ) f f r1 1 g g r2 1 =

73 Rates on the boundary (Case I) Theorem: There is always a set of evaluation points such that one can construct a full rank generator matrix in case I. Z 3 Proof by induction: P 1 (α i 1 ) α 1 P1 (α i1 i 1 ) P 1 (α i Z 3 ) α 1 i P1 Z3 (α i ) Z k, = k Z 3 Z 3 + P 2 (α i 1 ) α 1 P2 (α i1 i 1 ) P 2 (α i Z 3 ) α 1 i P2 Z3 (α i Z 3 ) f f r1 1 g g r2 1 = M r1 + + = P 1 (y 1 ) r y P1 (y 1 ) P 1 (y r1 + ) r y 1 1 r1 +r P1 2 (y r1 + ) P 2 (y 1 ) r y P2 (y 1 ) P 2 (y r1 + ) r y 2 1 r1 +r P2 2 (y r1 + )

74 Rates on the boundary (Case I) Theorem: There is always a set of evaluation points such that one can construct a full rank generator matrix in case I. Z 3 Proof by induction: P 1 (α i 1 ) α 1 P1 (α i1 i 1 ) P 1 (α i Z 3 ) α 1 i P1 Z3 (α i ) Z k, = k Z 3 Z 3 + P 2 (α i 1 ) α 1 P2 (α i1 i 1 ) P 2 (α i Z 3 ) α 1 i P2 Z3 (α i Z 3 ) f f r1 1 g g r2 1 = M r1 + + = P 1 (y 1 ) r y P1 (y 1 ) P 1 (y r1 + ) r y 1 1 r1 +r P1 2 (y r1 + ) P 2 (y 1 ) r y P2 (y 1 ) P 2 (y r1 + ) r y 2 1 r1 +r P2 2 (y r1 + ) By induction α 1,, α r1 + : det M α 1,, α r1 + det M y 1,, y r1 +

75 Rates on the boundary (Case I) Theorem: There is always a set of evaluation points such that one can construct a full rank generator matrix in case I. Z 3 Proof by induction: P 1 (α i 1 ) α 1 P1 (α i1 i 1 ) P 1 (α i Z 3 ) α 1 i P1 Z3 (α i ) Z k, = k Z 3 Z 3 + P 2 (α i 1 ) α 1 P2 (α i1 i 1 ) P 2 (α i Z 3 ) α 1 i P2 Z3 (α i Z 3 ) f f r1 1 g g r2 1 = M r1 + + = P 1 (y 1 ) r y P1 (y 1 ) P 1 (y r1 + ) r y 1 1 r1 +r P1 2 (y r1 + ) P 2 (y 1 ) r y P2 (y 1 ) P 2 (y r1 + ) r y 2 1 r1 +r P2 2 (y r1 + ) By induction α 1,, α r1 + : det M α 1,, α r1 + det M y 1,, y r1 + Choose α i1,, α i Z 3 such that det M α i 1,, α ir 1+r2

76 Rates on the boundary (Case I) + + r l f n P 1 (y 1 ) r y P1 (y 1 ) P 1 (y n ) r y 1 1 n P1 (y n ) P 2 (y 1 ) r y P2 (y 1 ) P 2 (y n ) r y 2 1 n P2 (y n ) f r1 1 g g r2 1 = c 1 c n

77 Rates on the boundary (Case I) n M + + r l P 1 (y 1 ) r y P1 (y 1 ) P 1 (y n ) r y 1 1 n P1 (y n ) P 2 (y 1 ) r y P2 (y 1 ) P 2 (y n ) r y 2 1 n P2 (y n ) We just showed that there exist a matrix M such that det M = h y 1,, y n f f r1 1 g g r2 1 = c 1 c n

78 Rates on the boundary (Case I) n M + + r l P 1 (y 1 ) r y P1 (y 1 ) P 1 (y n ) r y 1 1 n P1 (y n ) P 2 (y 1 ) r y P2 (y 1 ) P 2 (y n ) r y 2 1 n P2 (y n ) We just showed that there exist a matrix M such that det M = h y 1,, y n choose evaluation points (α 1,, α n ) such that h α 1,, α n f f r1 1 g g r2 1 = c 1 c n

79 Induction Case II Case II Case II Case II Case II Case II Case II Case II

80 Induction Case II Case II Case II Case II Case II Case II M 1 Case II M 3 Case II M 2 M l

81 Induction Case II Case II choose evaluation points α 1, α 2,, α n Case II such that Case II Case II det M 1 det M 2 det M l α1,,α n Case II M 1 Case II M 3 Case II M 2 M l

82 Size of the required finite-field G RS = f 1 α 1 P 1 (α 1 ) f r1 α 1 P 1 (α 1 ) g 1 α 1 P 2 (α 1 ) g r2 α 1 P 2 (α 1 ) n f 1 α n P 1 (α n ) f r1 α n P 1 (α n ) g 1 α n P 2 (α n ) g r2 α n P 2 (α n )

83 Size of the required finite-field G RS = f 1 α 1 P 1 (α 1 ) f r1 α 1 P 1 (α 1 ) g 1 α 1 P 2 (α 1 ) g r2 α 1 P 2 (α 1 ) Choose evaluation points such that: n f 1 α n P 1 (α n ) f r1 α n P 1 (α n ) g 1 α n P 2 (α n ) g r2 α n P 2 (α n ) M (r1 + +r s ) ( + +r s ) det M Full rank deg det M k k 1, max i deg α i k 1 Extended Schwartz-Zippel Lemma: If size of the finite-field is larger than or equal to n, then there are sets of evaluation points that satisfy the inequality.

84 Future work We can construct a randomized algorithm that finds G RS. Find an efficient determinist algorithm for the problem. Extend the results to general multiple-source networks (Gabidulin codes) Look for applications in storage systems