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

Alex Dimakis

based on collaborations with Dimitris Papailiopoulos

Viveck Cadambe Kannan Ramchandran

USC

Tutorial on Distributed Storage Problems and Regenerating Codes

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

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?

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

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

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

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

The promise: coding is much more reliable

… 33 encoded packets

… 21 copies

1GB

21 Replication uses 21GB. (33,10) Code uses 33*0.1=3.3GB600% more storage for the same reliability.

… 10 packets

1GB

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

(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

We can reconstruct after any 2 failures

a

b

c

d

a+c

b+d

b+c

a+b+d

1GB

1GB

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)

overview

13






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.

The Repair problem

15

a b c d

e

??

?




• Filesize B, B/k per block.

Do I need to reconstruct the Whole data object to repair one failure?

The Repair problem

16

a b c d

e

??

?




• 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.

The Repair problem

17

a b c d

e

??

?


• If we have 1 failure however, how do we rebuild the lost blocks in a new disk?


• 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)

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

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

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

Proof sketch: Information flow graph

a

e

2GB

a

b b

c c

d d

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

Proof sketch: reduction to multicasting

a

e

a

b b

c

d d

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

The infinite graph for Repair

x1α

αα

α

αβ

d

αβ

d

αβ

d

αβ

d

data collector

k data collector

x2

…

xn

24

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

€

αmin =M /k, γ ∈ [ f (0),∞),

M − g(i)γ

k − i, γ ∈ [ f (i), f (i −1)).

⎧ ⎨ ⎪

⎩ ⎪

€

f (i) :=2Md

(2k − i −1)i + 2k(d − k +1)

g(i) :=(2d − 2k + i +1)i

2d

Storage-Communication tradeoff

25

Storage-Communication tradeoff

Min-Storage Regenerating code

Min-Bandwidth Regenerating code

α

(D, Godfrey, Wu, Wainwright, Ramchandran, IT Transactions (2010) )

γ=βd

overview

26






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? ?

?

?

x1?

28

Repair vs Exact Repair

x1α

αα

α

αβ

d

αβ

d

αβ

d

αβ

d

data collector

k data collector

x2

…

xn• 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

Exact Storage-Communication tradeoff?

αExact repair feasible?

γ=βd

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

• 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

Min-Storage Regenerating code

(no known practical codes for high rates)

Min-Bandwidth Regenerating code (practical)

α

γ=βd

E-MSR PointE-MBR Point

Exact Storage-Communication tradeoff?

overview

33






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).

A1+2A2+ B1+B2=y1

2A1+A2+ B1+B2=y2

B1+B2=y3

Interference alignment

How to form codes that have multiple alignments at the same time?

3535

Exact Repair-(4,2) example

x1 x3

x2 x4

x1+x3

x2+x4

x1+2x3

2x2+3x4

x1?

x2?

x1+x2+x3+x4 2-1x1+2 3-1x2+x3+x4

2-1

3-1

x3+x4

(Wu and D. , ISIT 2009)

11

1 1

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)

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)

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?

overview

38






File is Separated in m blocks

An MDScode produces T blocks.

Each coded block is stored in r nodes.

m

Each storage nodeStores d coded blocks.

n

Simple regenerating codes

Adjacency matrix of an expander graph.

Every k right nodes are adjacent to m left nodes.

The ring coden=5 k=3

Any 3 nodes must suffice to recover the data.

set x5=x1+x2+x3+x

4

The ring code

41

n=5 k=3

Any 3 nodes know m=4 packets.

An MDScode produces T=5 blocks.

Each coded block is stored in r=2 nodes.

The ring code

4242


m=4

n=5




m


n




Claim 1: This code has the (n,k) recovery property.




m


n




Claim 1: This code has the (n,k) recovery property.

Choose k right nodesThey must know

m left nodes




m


n




Claim 2: I can do easy lookup repair.

[Rashmi et al. 2010, El Rouayheb & Ramchandran 2010]

d packets lostBut each packet is replicated r times. Find copy in another node.




m


n




Claim 2: I can do easy lookup repair.

[Rashmi et al. 2010, El Rouayheb & Ramchandran 2010]

d packets lostBut each packet is replicated r times. Find copy in another node.

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.




m


n




Great. Now everything depends on which graph I use and how much expansion it has.


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.

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

Two excellent expanders to try at homeThe 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.

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.


Each coded block is stored in r=2 nodes.

m=20


n=22

Ring RC vs RS

k=19, n=22 Ring RC. Assume B=20MB. Each Node stores d=2 packets. α= 2MB.Total storage =44MB1/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= 23.11/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.

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

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.


A code (possibly MDS code) produces T blocks.

Each coded block is stored in r=1.5 nodes.

m


n



+

+





m


n




Claim: I can still do easy lookup repair.

d packets lost

+

+




m


n

Simple regenerating codes (SRC)



Claim: I can still do easy lookup repair. 2d disk IO and communication

[ Papailipoulos et al. to be submitted]

d packets lost

+

+

High rate SRCs

59

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

overview

61





security through coding

62

Startup Cleversafe is introducing data security through distributed coding.

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

Security during Repair ?

a

b

ce

Incorrect linear equations

d

Repair bandwidth in the presence of byzantine adversaries?

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

Coding for Storage wiki

6767

fin

6868

Exact Repair-(4,2) example

x1 x3

x2 x4

x1+x3

x2+x4

x1+2x3

2x2+3x4

x1?

x2?

x1+x2+x3+x4 2-1x1+2 3-1x2+x3+x4

2-1

3-1

x3+x4

(Wu and D. , ISIT 2009)

11

1 1

69

1 0

0 1

0 0

0 0

0 0

0 0

1 0

0 1

1 0

0 1

1 0

0 1

1 0

0 2

2 0

0 3

1 1

1 1

2-1 3-1

0 0 1 1

1 1 1 1

2-1 23-1 1 1

v2

v3

v4

=

=

=

Exact Repair-interference alignment

70

1 0

0 1

0 0

0 0

0 0

0 0

1 0

0 1

1 0

0 1

1 0

0 1

1 0

0 2

2 0

0 3

1 1

1 1

2-1 3-1


=

=

=

[Cadambe-Jafar 2008, Cadambe-Jafar-Maleki-2010]

We want this full rank

71

1 0

0 1

0 0

0 0

0 0

0 0

1 0

0 1

1 0

0 1

1 0

0 1

1 0

0 2

2 0

0 3

1 1

1 1

2-1 3-1


=

=

=

Choose same V’ and V

Make all A diagonal iid

Want this in the span of V’

72


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


Ok. Lets start by choosing V’ to be one vector w

Must be in the rowspan of


And fold it back in…


And fold it back in…

And again fold it back in…. And again fold it back in….

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.

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

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