1. Alex Dimakis based on collaborations with Dimitris Papailiopoulos Viveck Cadambe Kannan Ramchandran USC Tutorial on Distributed Storage Problems and Regenerating Codes

2. overview 2 Storing information using codes. The repair problem Exact Repair. The state of the art. The role of Interference Alignment Simple Regenerating Codes Future directions: security through coding

3. 3 Massive distributed data storage Numerous disk failures per day. Failures are the norm rather than the exception Must introduce redundancy for reliability Replication or erasure coding?

4. 44 how to store using erasure codes A B A B A+B B A+2B A A+B A B (3,2) MDS code, (single parity) used in RAID 5 (4,2) MDS code. Tolerates any 2 failures Used in RAID 6 k=2 n=3 n=4 File or data object

5. 55 erasure codes are reliable A B A A B B A+B A+2B (4,2) MDS erasure code (any 2 suffice to recover) A B vs Replication File or data object

6. storing with an (n,k) code An (n,k) erasure code provides a way to: Take k packets and generate n packets of the same size such that Any k out of n suffice to reconstruct the original k Optimal reliability for that given redundancy. Well- known and used frequently, e.g. Reed-Solomon codes, Array codes, LDPC and Turbo codes. Assume that each packet is stored at a different node, distributed in a network. 6

7. how much redundancy is there in current systems? most distributed storage systems use replication gmail uses 21x replication(!) some companies are investigating or using Reed- Solomon and other codes (e.g. NetApp, IBM, Google, MSR, Cleversafe) 7

8. The promise: coding is much more reliable … 33 encoded packets … 21 copies 1GB 21 Replication uses 21GB. (33,10) Code uses 33*0.1=3.3GB 600% more storage for the same reliability. … 10 packets 1GB

9. 99 Coding+Storage Networks = New open problems Issues: Communication Update complexity Repair communication Repair bits Read No of nodes accessed for repair d A B ? Network traffic

10. (4,2) MDS Codes: Evenodd a b c d a+c b+d b+c a+b+d M. Blaum and J. Bruck ( IEEE Trans. Comp., Vol. 44, Feb 95) Total data object size= 4GB k=2 n=4, binary MDS code used in RAID systems

11. We can reconstruct after any 2 failures a b c d a+c b+d b+c a+b+d 1GB

12. We can reconstruct after any 2 failures a b c d a+c b+d b+c a+b+d c = a + (a+c) d = b + (b+d)

13. overview 13 Storing information using codes. The repair problem Exact Repair. The state of the art. The role of Interference Alignment Simple Regenerating Codes Future directions: security through coding

14. The Repair problem 14 a b c d e ? ? ? Ok, great, we can tolerate n-k disk failures without losing data. If we have 1 failure however, how do we rebuild the redundancy in a new disk? Naïve repair: send k blocks. Filesize B, B/k per block.

15. The Repair problem 15 a b c d e ? ? ? Ok, great, we can tolerate n-k disk failures without losing data. If we have 1 failure however, how do we rebuild the redundancy in a new disk? Naïve repair: send k blocks. Filesize B, B/k per block. Do I need to reconstruct the Whole data object to repair one failure?

16. The Repair problem 16 a b c d e ? ? ? Ok, great, we can tolerate n-k disk failures without losing data. If we have 1 failure however, how do we rebuild the redundancy in a new disk? Naïve repair: send k blocks. Filesize B, B/k per block Functional repair : e can be different from a. Maintains the any k out of n reliability property. Exact repair : e is exactly equal to a.

17. The Repair problem 17 a b c d e ? ? ? Ok, great, we can tolerate n-k disk failures without losing data. If we have 1 failure however, how do we rebuild the lost blocks in a new disk? Naïve repair: send k blocks. Filesize B, B/k per block Theorem: It is possible to functionally repair a code by communicating only As opposed to naïve repair cost of B bits. (Regenerating Codes)

18. Exact repair with 3GB a b c d a+c b+d b+c a+b+d a = (b+d) + (a+b+d) b = d + (b+d) a? b? 1GB

19. Systematic repair with 1.5GB a b c d a+c b+d b+c a+b+d a = (b+d) + (a+b+d) b = d + (b+d) a? b? 1GB Reconstructing all the data: 4GB Repairing a single node: 3GB 3 equations were aligned, solvable for a,b Reconstructing all the data: 4GB Repairing a single node: 3GB 3 equations were aligned, solvable for a,b

20. Repairing the last node a b c d a+c b+d b+c a+b+d b+c = (c+d) + (b+d) a+b+d = a + (b+d)

21. 21 Proof sketch: Information flow graph a e 2GB a bb cc dd α =2 GB data collector ∞ ∞ β β β 2+2 β ≥4 GB  β ≥1 GB Total repair comm. ≥3 GB S data collector

22. 22 Proof sketch: reduction to multicasting a e a bb c dd data collector    S data collector data collector data collector Repairing a code = multicasting on the information flow graph. sufficient iff minimum of the min cuts is larger than file size M. (Ahlswede et al. Koetter & Medard, Ho et al.) data collector data collector c

23. 23 The infinite graph for Repair x1x1 α α α α α β d α β d α β d α β d data collector k data collector x2x2 … xnxn

24. 24 Theorem 3 : for any (n,k) code, where each node stores α bits, repairs from d existing nodes and downloads d β=γ bits, the feasible region is piecewise linear function described as follows: Storage-Communication tradeoff

25. 25 Storage-Communication tradeoff Min-Storage Regenerating code Min-Bandwidth Regenerating code α (D, Godfrey, Wu, Wainwright, Ramchandran, IT Transactions (2010) ) γ=βd

26. overview 26 Storing information using codes. The repair problem Exact Repair. The state of the art. The role of Interference Alignment Simple Regenerating Codes Future directions: security through coding

27. 27 Key problem: Exact repair a b c d e =a From Theorem 1, an (n,k) MDS code can be repaired by communicating What if we require perfect reconstruction? ? ? ?

28. x1?x1? 28 Repair vs Exact Repair x1x1 α α α α α β d α β d α β d α β d data collector k data collector x2x2 … xnxn Functional Repair= Multicasting Exact repair= Multicasting with intermediate nodes having (overlapping) requests. Cut set region might not be achievable Linear codes might not suffice (Dougherty et al.) Functional Repair= Multicasting Exact repair= Multicasting with intermediate nodes having (overlapping) requests. Cut set region might not be achievable Linear codes might not suffice (Dougherty et al.)

29. 29 Exact Storage-Communication tradeoff? α Exact repair feasible? γ=βd

30. 30 For (n,k=2) E-MSR repair can match cutset bound. [WD ISIT’09] (n=5,k=3) E-MSR systematic code exists (Cullina,D,Ho, Allerton’09) For k/n <=1/2 E-MSR repair can match cutset bound [Rashmi, Shah, Kumar, Ramchandran (2010)] E-MBR for all n,k, for d=n-1 matches cut-set bound. [Suh, Ramchandran (2010) ] What is known about exact repair

31. 31 What can be done for high rates? Recently the symbol extension technique (Cadambe, Jafar, Maleki) and independently (Suh, Ramchandran) was shown to approach cut-set bound for E-MSR, for all (k,n,d). (However requires enormous field size and sub-packetization.) Shows that linear codes suffice to approach cut-set region for exact repair, for the whole range of parameters. Tamo et al., Papailiopoulos et al. and Cadambe et al. are presenting the first constructions of high rate exact regenerating codes at ISIT 2011. What is known about exact repair

32. 32 Min-Storage Regenerating code (no known practical codes for high rates) Min-Bandwidth Regenerating code (practical) α γ=βd E-MSR Point E-MBR Point Exact Storage-Communication tradeoff?

33. overview 33 Storing information using codes. The repair problem Exact Repair. The state of the art. The role of Interference Alignment Simple Regenerating Codes Future directions: security through coding

34. The coefficients of some variables lie in a lower dimensional subspace and can be canceled out. 34 Imagine getting three linear equations in four variables. In general none of the variables is recoverable. (only a subspace). A 1 +2A 2 + B 1 +B 2 =y 1 2A 1 +A 2 + B 1 +B 2 =y 2 B 1 +B 2 =y 3 Interference alignment How to form codes that have multiple alignments at the same time?

35. 35 Exact Repair-(4,2) example x1 x3 x2 x4 x1+x3 x2+x4 x1+2x3 2x2+3x4 x1? x2? x1+x2+x3+x4 2 -1 x1+2 3 -1 x2+x3+x4 2 -1 3 -1 x3+x4 (Wu and D., ISIT 2009) 1 1 1 1

36. Given an error-correcting code find the repair coefficients that reduce communication (over a field) Given some channel matrices find the beamforming matrices that maximize the DoF (Cadambe and Jafar, Suh and Tse) Given some channel matrices find the beamforming matrices that maximize the DoF (Cadambe and Jafar, Suh and Tse) connecting storage and wireless Both problems reduce to rank minimization subject to full rank constraints. Polynomial reduction from one to the other. (Papailiopoulos & D. Asilomar 2010) Both problems reduce to rank minimization subject to full rank constraints. Polynomial reduction from one to the other. (Papailiopoulos & D. Asilomar 2010)

37. 37 Storage codes through alignment techniques The symbol extension alignment technique of [Cadambe and Jafar] leads to exact regenerating codes Exact repair is a non-multicast problem where cut-set region is achievable but needs alignment. It is an improbable match made in heaven (unfortunately not practical) ergodic alignment should have a storage code equivalent? does real alignment have a finite-field equivalent?

38. overview 38 Storing information using codes. The repair problem Exact Repair. The state of the art. The role of Interference Alignment Simple Regenerating Codes Future directions: security through coding

39. File is Separated in m blocks An MDS code produces T blocks. Each coded block is stored in r nodes. m Each storage node Stores d coded blocks. n Simple regenerating codes Adjacency matrix of an expander graph. Every k right nodes are adjacent to m left nodes.

40. The ring code n=5 k=3 Any 3 nodes must suffice to recover the data. set x 5 =x 1 +x 2 +x 3 +x 4

41. The ring code 41 n=5 k=3 Any 3 nodes know m=4 packets. An MDS code produces T=5 blocks. Each coded block is stored in r=2 nodes.

42. The ring code 42 An MDS code produces T blocks. m=4 n=5

43. File is Separated in m blocks An MDS code produces T blocks. Each coded block is stored in r nodes. m Each storage node Stores d coded blocks. n Simple regenerating codes Adjacency matrix of an expander graph. Every k right nodes are adjacent to m left nodes. Claim 1: This code has the (n,k) recovery property.

44. File is Separated in m blocks An MDS code produces T blocks. Each coded block is stored in r nodes. m Each storage node Stores d coded blocks. n Adjacency matrix of an expander graph. Every k right nodes are adjacent to m left nodes. Simple regenerating codes Claim 1: This code has the (n,k) recovery property. Choose k right nodes They must know m left nodes

45. File is Separated in m blocks An MDS code produces T blocks. Each coded block is stored in r nodes. m Each storage node Stores d coded blocks. n Simple regenerating codes Adjacency matrix of an expander graph. Every k right nodes are adjacent to m left nodes. Claim 2: I can do easy lookup repair. [Rashmi et al. 2010, El Rouayheb & Ramchandran 2010] d packets lost But each packet is replicated r times. Find copy in another node.

46. File is Separated in m blocks An MDS code produces T blocks. Each coded block is stored in r nodes. m Each storage node Stores d coded blocks. n Simple regenerating codes Adjacency matrix of an expander graph. Every k right nodes are adjacent to m left nodes. Claim 2: I can do easy lookup repair. [Rashmi et al. 2010, El Rouayheb & Ramchandran 2010] d packets lost But each packet is replicated r times. Find copy in another node.

47. The ring code: lookup repair n=5 k=3 node 1 fails. just read from d=2 other nodes. Minimizing d is proportional to total disk IO.

48. File is Separated in m blocks An MDS code produces T blocks. Each coded block is stored in r nodes. m Each storage node Stores d coded blocks. n Simple regenerating codes Adjacency matrix of an expander graph. Every k right nodes are adjacent to m left nodes. Great. Now everything depends on which graph I use and how much expansion it has.

49. Simple regenerating codes 49 Rashmi et al. used the edge-vertex bipartite graph of the complete graph. Vertices=storage nodes. Edges= coded packets. d=n-1, r=2 Expansion: Every k nodes are adjacent to m= kd – (k choose 2) edges. Remarkably this matches the cut-set bound for the E-MBR point.

50. Simple regenerating codes In cloud storage practice the number of nodes (d) is more important than number of bits read or transferred. Lookup repair is great. The ring code has the smallest d=2. if we wanted to repair from ANY d, we could not make d smaller than k. 50

51. Two excellent expanders to try at home The Petersen Graph. n=10, T=15 edges. Every k=7 nodes are adjacent to m=13 (or more) edges, i.e. left nodes. The ring. n vertices and edges. Maximum girth. Minimizes d which is important for some applications.

52. Example ring RC 52 Every k nodes adjacent to at least k+1 edges. Example pick k=19, n=22. Use a ring of 22 nodes. An MDS code produces T blocks. Each coded block is stored in r=2 nodes. m=20 Each storage node Stores d coded blocks. n=22

53. Ring RC vs RS k=19, n=22 Ring RC. Assume B=20MB. Each Node stores d=2 packets. α= 2MB.Total storage =44MB 1/rate= 44/20 = 2.2 storage overhead Can tolerate 3 node failures. For one failure. d=2 surviving nodes are used for exact repair. Communication to repair γ= 2MB. Disk IO to repair=2MB. k=19, n=22 Reed Solomon with naïve repair. Assume B=20MB. Each Node stores α= 20MB/ 19 =1.05 MB. Total storage= 23.1 1/rate= 22/19 = 1.15 storage overhead Can tolerate 3 node failures. For one failure. d=19 surviving nodes are used for exact repair. Communication to repair γ= 19 MB. Disk IO to repair=19 MB. Double storage, 10 times less resources to repair.

54. How to get high rate? In cloud storage practice the number of nodes (d) is more important than number of bits read or transferred. Lookup repair is great. We need high rate = low storage overhead There is no fractional repetition code or MBR code that has true rate above ½ 54

55. Extending fractional repetition 55 Lookup repair allows very easy uncoded repair and modular designs. Random matrices and Steiner systems proposed by [El Rouayheb et al.] Note that for d< n-1 it is possible to beat the previous E-MBR bound. This is because lookup repair does not require every set of d surviving nodes to suffice to repair. E-MBR region for lookup repair remains open. r ≥ 2 since two copies of each packet are required for easy repair. In practice higher rates are desirable for cloud storage. This corresponds to a repetition code! Lets replace it with a sparse intermediate code.

56. File is Separated in m blocks A code (possibly MDS code) produces T blocks. Each coded block is stored in r=1.5 nodes. m Each storage node Stores d coded blocks. n Adjacency matrix of an expander graph. Every k right nodes are adjacent to m left nodes. + + Simple regenerating codes

57. File is Separated in m blocks An MDS code produces T blocks. Each coded block is stored in r nodes. m Each storage node Stores d coded blocks. n Simple regenerating codes Adjacency matrix of an expander graph. Every k right nodes are adjacent to m left nodes. Claim: I can still do easy lookup repair. d packets lost + +

58. File is Separated in m blocks An MDS code produces T blocks. Each coded block is stored in r nodes. m Each storage node Stores d coded blocks. n Simple regenerating codes (SRC) Adjacency matrix of an expander graph. Every k right nodes are adjacent to m left nodes. Claim: I can still do easy lookup repair. 2d disk IO and communication [ Papailipoulos et al. to be submitted] d packets lost + +

59. High rate SRCs 59

60. Simple regenerating codes if XORs (forks) of degree 2 are used, these SRCs can have true rate approach 2/3 k/n  f/(f+1) rate can be achieved with higher XORs, but requires more nodes to be accessed. We think this is the minimal d for lookup repair. 60

61. overview 61 Storing information using codes. The repair problem Exact Repair. The state of the art. The role of Interference Alignment Future directions: security through coding

62. security through coding 62 Startup Cleversafe is introducing data security through distributed coding.

63. 63 coding allows secret sharing a b c d Four coded blocks are stored in four different cloud storage providers Any two can be used to recover the data Any cloud storage provider knows nothing about the data. [Shamir, Blakley 1979] Distributed coding theory problems?

64. 64 Security during Repair ? a b c e Incorrect linear equations d Repair bandwidth in the presence of byzantine adversaries?

65. 65 Open Problems in distributed storage Cut-Set region matches exact repair region ? Repairing codes with a small finite field limit ? Dealing with bit-errors (security) and privacy ? (Dikaliotis,D, Ho, ISIT’10) What is the role of (non-trivial) network topologies ? Cooperative repair (Shum et al.) Lookup repair region ? Disk IO region ? What are the limits of interference alignment techniques ? Repairing existing codes used in storage (e.g. EvenOdd, B- Code, Reed-Solomon etc) ? Real world implementation, benefits over HDFS for Mapreduce ? 65

66. 66 Coding for Storage wiki

67. 67 fin

68. 68 Exact Repair-(4,2) example x1 x3 x2 x4 x1+x3 x2+x4 x1+2x3 2x2+3x4 x1? x2? x1+x2+x3+x4 2 -1 x1+2 3 -1 x2+x3+x4 2 -1 3 -1 x3+x4 (Wu and D., ISIT 2009) 1 1 1 1

69. 69 10 01 00 00 00 00 10 01 10 01 10 01 10 02 20 03 11 11 2 -1 3 -1 0011 1111 2 -1 23 -1 11 v2v2 v3v3 v4v4 = = = Exact Repair-interference alignment

70. 70 10 01 00 00 00 00 10 01 10 01 10 01 10 02 20 03 11 11 2 -1 3 -1 Exact Repair-interference alignment = = = [Cadambe-Jafar 2008, Cadambe-Jafar-Maleki-2010]

71. We want this full rank 71 10 01 00 00 00 00 10 01 10 01 10 01 10 02 20 03 11 11 2 -1 3 -1 Exact Repair-interference alignment = = = Choose same V’ and V Make all A diagonal iid Want this in the span of V’

72. 72 Exact Repair-interference alignment We have to choose V, V’ so that all the rows in Are contained in the rowspan of The A matrices assumed iid diagonal, no assumption other than that they commute

73. Exact Repair-interference alignment Ok. Lets start by choosing V’ to be one vector w Must be in the rowspan of

74. Exact Repair-interference alignment And fold it back in…

75. Exact Repair-interference alignment And fold it back in… And again fold it back in….

76. Extending this idea 76 Lookup repair allows very easy uncoded repair and modular designs. Random matrices and Steiner systems proposed by [El Rouayheb et al.] Note that for d< n-1 it is possible to beat the previous E-MBR bound. This is because lookup repair does not require every set of d surviving nodes to suffice to repair. E-MBR region for lookup repair remains open. r ≥ 2 since two copies of each packet are required for easy repair. In practice higher rates are more attractive. This corresponds to a repetition code! Lets replace it with a sparse intermediate code.