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A  Reduc(on  Approach  to  the  Mul(ple-­‐Unicast  Conjecture  in  Network  Coding

Zongpeng  Li

What  is  Network  Coding?

•  Encoding  data  during  a  mul(-­‐hop  transmission  – mul(ple  unicasts  – mul(cast

Coding  Advantage

•  Improve  throughput  for  mul(cast

Coding  Advantage

•  Improve  throughput  for  mul(ple-­‐unicast

t2 t1

s1 s2 a b

a+b a b

Coding  Advantage

•  Save  bandwidth  –  Network  Coding:  9  bits  –  Rou(ng:  10  bits

t2 t1

s1 s2 a b

a+b a b

Network  Models

Directed  Networks •  Not  necessarily  bidirec(onal  •  A  pair  of  reverse  links  each  

has  its  own  capacity

Undirected  Networks •  Bidirec(onal  •  Capacity  can  be  freely  

allocated  to  two  direc(ons

2

3 6 4

4

5

6 4

4

Coding  Advantage  in  Undirected  Networks

•  Improve  throughput  for  mul(cast  – Up  to  a  bounded  factor  

•  Network  Coding:  2  bps  •  Rou(ng:                1.875  bps  

LeVer:  0.25bps;  Number:  0.125bps

Coding  Advantage  in  Undirected  Networks

•  Reduce  cost  for  mul(cast

Rou(ng:    4.64 Network  Coding:    4.57

Mul(ple-­‐Unicast  +  Undirected  Networks?

       Coding  advantage  vanishes!

a

a

b

ba+b

a+ba+b

s1

t2

a1

b1

a1

b1

b2a1

a2

b2

a2

b2

a2 b1

s1

t2

2

t1

s2

t1

s

Another  Example

a

a b b c

ca+b

a+b

a+b

b+c

b+ca+ca+b+c

a

a

b c

c b

b+c

a b c

ac b

a2

c1

a1

c1

b1c1

b2

c1

c1

b2

c2

b2

a1 b1

a1

c2

a1b2

c2b1

c2b1

c2

a2 a2b1 a2

b2

a1a2

The  Conjecture

In  terms  of  improving  throughput  or  saving  bandwidth,  Network  coding  has  no  advantage  over  rou(ng  for  mul(ple  unicast  sessions  in  undirected  networks.  [Li  and  Li  2004]

Comments

•  Mitzenmacher  :  No.1  of  seven  open  problems  in  network  coding  (2007)  

•  Chekuri  :  “bold  conjecture”,  the  problem  of  fully  understanding  network  coding  for  mul(ple  unicast  sessions  is  s(ll  “wild  open”.  

•  Adler  :  “arguably  the    most  important  open  problem  in  the  field  of  network  coding”  (2006)  

•  The  conjecture  implies  an  affirma(ve  answer  to  a  28-­‐year-­‐old  open  problem.

Verified  Cases

•  2  unicast  sessions  •  Terminal  co-­‐face  planar  networks  •  Complete  networks  with  uniform  link  length  •  Grid  networks  with  uniform  link  length  and  aligned  source-­‐receivers  

•  Each  source  is  closer  to  its  receiver  than  other  receivers

Verified  Cases

•  Okamura-­‐Seymour  Network  (K3,2)  

•  Hu’s  3-­‐commodity  network  

•  Complete  bipar(te  networks  with  uniform  link  length

s1 t1

t3

s3

s2

t2

t4

s4

s1 t1 t3 s3

s2

t2

Overview  of  our  reduc(on  approach

Undirected)Networks�

)�

Atom)Networks�

…�…�

Decompose�

Cut6set)Bound:)No� Theorem)1:)No�

) �

Require)Coding?�

Assemble� Theorem)3:)No)need)to)code)in)networks)that)can)be)decomposed)into)these)atoms)networks.�

Theorem)2:))when)&)how)to)decompose�

?�

Highlights  of  our  results

•  Generalize  proofs  of  verified  cases  •  Prove  the  conjecture  for  up  to  6  nodes  &  most  7-­‐node  networks  

•  Find  an  interes(ng  example  where  new  techniques  may  be  necessary  

Cost  Domain

•  Link  capacity  is  ignored  •  Each  link  is  assigned  with  a  non-­‐nega(ve  length  le  

•  Let  fe  denote  the  amount  of  informa(on  transmiVed  on  link  e  

•  Cost:  Σe  fe  le

Rela(ons  Between  Cost  Domain  and  Throughput  Domain

The  conjecture  in  Cost  Domain

Basic  Techniques  -­‐-­‐  inequali(es

•  Cut-­‐set:  a  set  of  edges  dividing  nodes  into  two  parts  

•  Cut-­‐set  bound:     F fe

e∈F∑ ≥ H (Xi )

i∈Sep(F )∑

Example  for  the  cut-­‐set  bound

•  Unit  link  length  •  For  each  cut-­‐set  Fj:  

•  Sum  up:  

t2 t1

s1 s2

fee∈Fj

∑ ≥ H (X1)+H (X2 )

F1

F2 F3

fee∈E∑ ≥ 3H (X1)+3H (X2 )

Overview  of  our  reduc(on  approach

Undirected)Networks�

)�

Atom)Networks�

…�…�

Decompose�

Cut6set)Bound:)No� Theorem)1:)No�

) �

Require)Coding?�

Assemble� Theorem)3:)No)need)to)code)in)networks)that)can)be)decomposed)into)these)atoms)networks.�

Theorem)2:))when)&)how)to)decompose�

?�

An  observa(on

•  Under  the  condi(on  Network  coding  is  necessary  in  G1  iff  it  is  necessary  in  G2  

t2 t1

s1 s2

t2 t1

s1 s2

F2

fee∈F2

∑ ≥ H (X1)+H (X2 )

Contract  edges  in  F2

G1 G2

Generalize  the  idea

•  Cut-­‐set  à  Arbitrary  edge  set  F  •  The  problem  is  about  the  condi(on:

fee∈F2

∑ ≥ H (X1)+H (X2 )

fee∈F∑ ≥ ?

i∑ H (Xi )

An  Equivalent  form  of  the  conjecture

Explana(on

•  An  edge  set  F  decomposes  G  in  to  G/F  and  G/F.  

t2 t1

s1 s2

t2 t1

s1 s2

F2

Decompose

t2 t1

s1 s2

•  As  long  as  the  decomposi(on  preserves  the  distance  between  each  pair  of  source-­‐receiver:  

•  Network  coding  is  unnecessary  in  G/F  and  G/F            è  it  is  unnecessary  in  G.  –  Cost  of  Network  Coding:    –  Cost  of  Rou(ng:    

fee∈E∑ = fe

e∈F∑ + fe

e∈F∑

dG (si, ti )H (Xi )i∑ = dG/F (si, ti )H (Xi )

i∑ + dG/F (si, ti )H (Xi )

i∑

When  there  exists  a  decomposi(on

•  An  example  – dG(s,t)  =  2  – dG/F(s,t)  =dG/F(s,t)  =  0  

•  A  path  p  in  G  à                  length  |p      F|  in  G/F                  length  |p      F|  in  G/F  •  There  exist  two  shortest  paths  p1,p2  in  G:  

|p1        F|≠|p2        F|  

s

t

F

∩

∩

∩ ∩

When  there  exists  a  decomposi(on

•  Another  example  •  Observa(on:  – Non-­‐shortest  paths  have  some  redundancy  

– Shortest  paths  intersect  F  the  minimum  (me  

s

t F

1  >  0  +  0

2  =  2  +  0

scale  up  link  length

When  there  exists  a  decomposi(on

Theorem  2  If  there  is  an  edge  set  F  that  is  compa(ble  with  all  sessions,  there  exists  a  decomposi(on.  

Overview  of  our  reduc(on  approach

Undirected)Networks�

)�

Atom)Networks�

…�…�

Decompose�

Cut6set)Bound:)No� Theorem)1:)No�

) �

Require)Coding?�

Assemble� Theorem)3:)No)need)to)code)in)networks)that)can)be)decomposed)into)these)atoms)networks.�

Theorem)2:))when)&)how)to)decompose�

?�

When  the  cut-­‐set  bound  is  insufficient

•  Intui(vely,  we  need  to  combine  several  fe  to  show  that  their  sum  is  no  less  than  some  H(Xi).  

•  Consider  the  following  solu(on:  

•  LHS:              

a d s1

b Xab

c

Xba

Xca

Xac

t1

s2

t2

fab + fac ≥ H (X1)

fab = H (Xab )+H (Xba )

fac = H (Xac )+H (Xca )

Xab = Xbd = X1 Xba = Xac = X2

fab + fac = H (X1)+ 2H (X2 )Loss!

A  Finer  Technique  -­‐-­‐  Informa(on  Inequality

•  Use                                                    instead  of  the  combined  version    

•  Submodularity  

– Here  A,B  are  sets  of  variables  Xi  ,  Xuv  

fuv

H (Xuv ),H (Xvu )

H (A)+H (B) ≥ H (A∪B)+H (A∩B)

Flexible!

Might  save  some  loss!

•  If  messages  B  are  determined  by  messages  A  –   H(A)  ≥  H(B)  

•  Input-­‐output  Inequality  – The  messages  leaving  node  set  U  are  determined  by  the  messages  entering  U  

•  Crypto  Inequality  – A  source  message  is  determined  by  the  messages  transmiVed  through  a  cut-­‐set  separa(ng  the  source  and  the  receiver.  

Example  using  informa(on  inequali(es

s1 t1

t3

s3

s2

t2

t4

s4

d a b

c

e

H (Xac )+H (Xbc )+H (X2 )≥ H (Xac,Xbc,X2 )≥ H (Xac,Xbc,X2,X4,Xca,Xcb )

brought  in  by    the  Input-­‐output  Inequality

Similarly,  combine  messages  enters  d  and  e,  respec(vely.  We  obtain

H (Xad,Xbd,X3,X2,Xda,Xdb )H (Xae,Xbe,X4,X3,Xea,Xeb )

Borrowed

Example  using  informa(on  inequali(es

s1 t1

t3

s3

s2

t2

t4

s4

d a b

c

e

H (Xac,Xbc,X2,X4,Xca,Xcb )+H (Xad,Xbd,X3,X2,Xda,Xdb )≥ H (...,X2,X3,X4 )+H (X2 )

Then  combine  the  3  resul(ng  entropies:

H (...,X2,X3,X4 )+H (Xae,Xbe,X4,X3,Xea,Xeb )≥ H (XE,X2,X3,X4 )+H (X3,X4 )

returned

set  of  messages  on  every  link

Example  using  informa(on  inequali(es

s1 t1

t3

s3

s2

t2

t4

s4

d a b

c

e

To  sum  up,  

H (XE,X2,X3,X4 )≥ H (XE,X2,X3,X4,X1)≥ H (X1)+H (X2 )+H (X3)+H (X4 )

crypto  inequality

source  independent

H (Xuv )u=a,b; v=c,d,e∑ ≥ H (Xi )

i=1,2,3,4∑

Similarly,  we  can  derive   H (Xvu )

u=a,b; v=c,d,e∑ ≥ H (Xi )

i=1,2,3,4∑

Lessons  Learned  from  the  example

•  Splixng  fe  into  H(Xuv)  and  H(Xvu)  is  helpful.  •  Entropy  terms  H(A)  can  be  combined  in  a  cascade  way.  – we  first  combine  the  entropies  of  messages  entering  each  node,  then  combine  the  resul(ng  entropies.  

•  Borrowing  source  messages  to  trigger  the  input-­‐output  inequality  is  OK.  – what  actually  maVers  is  the  number  of  source  messages  brought  into  the  deriva(on  by  the  input-­‐output/crypto  inequality.

•  We  study  the  edge  sets  that  are  a  liVle  bit  more  complicate  than  cut-­‐sets  –  the  union  of  two  cut-­‐sets  

•  For  such  an  edge  set  F,  we  find  a  way  to  combine  the  entropy  terms  to  derive  that      

     where  zi  equals  dG/F(si,ti)  for  one  session  and          min{2,  dG/F(si,ti)}  for  the  other  sessions.  

H (Xuv )+H (Xvu )e=uv∈F∑ ≥ ziH (Xi )

i∑

s1 t1

t3

s3

s2

t2

t4

s4 F1 F2

s1 t1 t3 s3

s2

t2 F1 F2

Proof  of  Theorem  1

•  Each  component  is  labeled  according  to  its  distance  to  s  in  G/F.

s t

F1

F2

Connected  Component

U0

U1

U’1

U3

U2 U4

Proof  of  Theorem  1  (cont.)

•  Step  1:  combine  the  entropies  of  messages  entering  each  component  Ui;  

•  Step  2:  combine  the  resul(ng  entropies  of  U1  and  U’1  

•  Step  3:  similarly,  combine  U1  ,U’1,U3;  combine  U0,  U2,  U4.  

Combine  results  together

A  cut-­‐set  F  is  orthogonal  to  session  i,  if  each  shortest  si-­‐ti  path  crosses  F  at  most  once.

Remarks

•  Condi(ons  P1  and  P2  only  relate  to  cut-­‐sets  and  shortest  paths.    

•  Can  be  verified  in  (me  O(2^|V|),  in  contrast  to  O(2^|E|)  for  the  state-­‐of-­‐art  LP  outer-­‐bound.

The  Next  Atom  Network

Examine  the  Next  Atom  Network

s1

t1

t3

s3

s2

t2

s1

t1

t3

s3

s2

t2

Conclusion

•  A  Reduc(on  Approach  – brings  the  abstract  conjecture  to  concrete  small  networks  

•  Prove  the  conjecture  for  up  to  6  nodes  •  An  interes(ng  example  for  future  research

Q&A

•  Thanks  for  your  (me!