Tighter Cut-BasedBounds for k-pairs

Communication Problems

Nick HarveyRobert Kleinberg

Overview Definitions

Sparsity and Meagerness Bounds Show these bounds very loose Define Informational Meagerness

Based on Informational Dominance Show that it can be slightly loose

M1M2

M1⊕M2

S(1) S(2)

T(2) T(1)

k-pairs Communication Problem

Concurrent Rate

Source i desires communication rate di. Rate r is achievable if rate vector

[ rd1, rd2, …, rdk ] is achievable Rate region interval of R+

Def: “Network coding rate” (or NCR) := sup { r : r is achievable }

M1M2

M1⊕M2

S(1) S(2)

T(2) T(1)

k-pairs Communication Problem

d1 = d2 = 1ce = 1 eERate 1 achievable

Upper bounds on rate

[Classical]: Sparsity bound for multicommodity flows

[CT91]: General bound for multi-commodity information networks

[B02]: Application of CT91 to directed network coding instances; equivalent to sparsity.

[KS03]: Bound for undirected networks with arbitrarytwo-way channels

[HKL04]: Meagerness

[SYC03], [HKL05]: LP bound

[KS05]: Bound based on iterative d-separation

Vertex-Sparsity

Def: For U V,

VS (G) := minUV VS (U)

Claim: NCR VS (G)

Capacity of edges crossing between U and U

Demand of commodities separated by UVS

(U) :=

Edge-Sparsity Def: For A E,

ES (G) = minAE ES (A)

Claim: Max-Flow ES (G)

But: Sometimes NCR > ES (G)

Capacity of edges in A

Demand of commodities separated in G\AES (A) :=

NCR > Edge-Sparsity

S(1) S(2)

T(2) T(1)

Cut {e} separates S(1) and S(2)

ES ({e}) = 1/2 But rate 1 achievable!

e

Meagerness Def: For A E and P [k],

A isolates P if for all i,j P,S(i) and T(j) disconnected in G\A.

M (G) := minAE M (A)

Claim: NCR M (G)

Capacity of edges in A

Demand of commodities in PM (A) := minP isolated by A

Meagerness & Vtx-Sparsity are weak

Thm: M (Gn) = VS (Gn) = (1),but NCR 1/n.

S(3) S(2)S(n) S(n-1) f2fn-1 f3 S(1)f1

T(1)T(n-1)T(n) T(3)hn-1 h1h3 T(2)h2

g2g3 g1gn-1gn

Gn :=

A Proof Tool

Def: Let A,B E. B is downstream of Aif B disconnected from sources in G\A.Notation: A B.

Claim: If A B then H(A) H(A,B).

Pf: Because S A B form Markov chain.

Proof:

{gn} {gn,T(1),h1}

S(3) S(2)S(n) S(n-1) f2fn-1 f3 S(1)f1

T(1)T(n-1)T(n) T(3)hn-1 h1h3 T(2)h2

g2g3 g1gn-1gn

Gn :=

Lemma: NCR 1/n

Proof:

{gn} {gn,T(1),h1} {S(1),f1,g1,h1}

S(3) S(2)S(n) S(n-1) f2fn-1 f3 S(1)f1

T(1)T(n-1)T(n) T(3)hn-1 h1h3 T(2)h2

g2g3 g1gn-1gn

Gn :=

Lemma: NCR 1/n

Proof:

{gn} {gn,T(1),h1} {S(1),f1,g1,h1}

{S(1),f1,T(2),h2}

S(3) S(2)S(n) S(n-1) f2fn-1 f3 S(1)f1

T(1)T(n-1)T(n) T(3)hn-1 h1h3 T(2)h2

g2g3 g1gn-1gn

Gn :=

Lemma: NCR 1/n

Proof:

{gn} {gn,T(1),h1} {S(1),f1,g1,h1}

{S(1),f1,T(2),h2} {S(1),S(2),f2,g2,h2}

S(3) S(2)S(n) S(n-1) f2fn-1 f3 S(1)f1

T(1)T(n-1)T(n) T(3)hn-1 h1h3 T(2)h2

g2g3 g1gn-1gn

Gn :=

Lemma: NCR 1/n

h3

Proof:

{gn} {gn,T(1),h1} {S(1),f1,g1,h1}

{S(1),f1,T(2),h2} {S(1),S(2),f2,g2,h2}

{S(1),S(2),f2,T(3),h3}

S(3) S(2)S(n) S(n-1) f2fn-1 f3 S(1)f1

T(1)T(n-1)T(n) T(3)hn-1 h1T(2)h2

g2g3 g1gn-1gn

Gn :=

Lemma: NCR 1/n

Proof:

{gn} … {S(1),S(2),…,S(n)}

Thus 1 H(gn) H(S(1),…,S(n)) = n∙r

So 1/n r

S(3) S(2)S(n) S(n-1) f2fn-1 f3 S(1)f1

T(1)T(n-1)T(n) T(3)hn-1 h1h3 T(2)h2

g2g3 g1gn-1gn

Gn :=

Lemma: NCR 1/n

Towards a stronger bound Our focus: cut-based bounds

Given A E, we want to infer thatH(A) H(A,P) where P{S(1),…,S(k)}

Meagerness uses Markovicity:(sources in P) A (sinks in P)

Markovicity sometimes not enough…

Informational Dominance

Def: A dominates B if information in A determines information in Bin every network coding solution.Denoted A B.

Trivially implies H(A) H(A,B) How to determine if A dominates B?

[HKL05] give combinatorial characterization and efficient algorithm to test if A dominates B.

i

Informational Meagerness Def: For A E and P {S(1),…,S(k)},

A informationally isolates P ifAP P.

iM (A) = minP

for P informationally isolated by A

iM (G) = minA E iM (A)

Claim: NCR iM (G).

iCapacity of edges in A

Demand of commodities in P

iMeagerness Example

“Obviously” NCR = 1. But no two edges disconnect t1 and t2 from both

sources!

s1 s2

t1

t2

iMeagerness Example

After removing A, still a path from s2 to t1!

Cut A

s1 s2

t1

t2

Informational Dominance Examples1 s2

t1

t2

Our characterization shows A {t1,t2}

H(A) H(t1,t2) and iM (G) = 1

Cut A

i

A bad example: Hn

Thm: iMeagerness gap of Hn is (log |V|)

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

q(00) q(01) q(10) q(11)r(00) r(01) r(10) r(11)

t(00)t(0)

t(01) t(10) t(11)t(1)

t(ε)

Capacity 2-nH2

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

Tn = Binary tree of depth n

Source S(i) iTn

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

Tn = Binary tree of depth n

Source S(i) iTn

Sink T(i) iTn

t(00)t(0)

t(01) t(10) t(11)t(1)

t(ε)

r(00) r(01) r(10) r(11)

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

t(00)t(0)

t(01) t(10) t(11)t(1)

t(ε)

q(00) q(01) q(10) q(11)

Nodes q(i) and r(i) for every leaf i of Tn

r(00) r(01) r(10) r(11)

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

t(00)t(0)

t(01) t(10) t(11)t(1)

t(ε)

q(00) q(01) q(10) q(11)

Complete bip. graph between sources and q’s

r(00) r(01) r(10) r(11)

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

t(00)t(0)

t(01) t(10) t(11)t(1)

t(ε)

q(00) q(01) q(10) q(11)

(r(a),t(b)) if b ancestor of a in Tn

r(00) r(01) r(10) r(11)

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

q(00) q(01) q(10) q(11)

t(00)t(0)

t(01) t(10) t(11)t(1)

t(ε)

(s(a),t(b)) if a and b cousins in Tn

r(00) r(01) r(10) r(11)

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

q(00) q(01) q(10) q(11)

t(00)t(0)

t(01) t(10) t(11)t(1)

t(ε)

Capacity 2-n

All edges have capacity except (q(i),r(i))

r(00) r(01) r(10) r(11)

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

q(00) q(01) q(10) q(11)

t(00)t(0)

t(01) t(10) t(11)t(1)

t(ε)

Capacity 2-n

Demand of source at depth i is 2-i

Properties of Hn

Lemma: iM (Hn) = (1)

Lemma: NCR < 1/n

Corollary: iMeagerness gap is n=(log |V|)

Properties of Hn

Lemma: iM (Hn) = (1)

Lemma: NCR < 1/n

Corollary: iMeagerness gap is n=O(log |V|)

We will prove this

Entropy moneybags i.e., sets of RVs

Entropy investments Buying sources and edges, putting into moneybag Loans may be necessary

Profit Via Downstreamness or Info. Dominance Earn new sources or edges for moneybag

Corporate mergers Via Submodularity New Investment Opportunities and Debt Consolidation

Debt repayment

Proof Ingredients

Submodularity of Entropy

Claim: Let A and B be sets of RVs.Then H(A)+H(B) H(AB)+H(AB)

Pf: Equivalent to I( X; Y | Z ) 0.

Proof: Two entropy moneybags:

F(a) = { S(b) : b not an ancestor of a }E(a) = F(a) { (q(b),r(b)) : b is descendant of a }

Lemma: NCR < 1/n

Entropy Investment

Let a be a leaf of Tn

Take a loan and buy E(a).

r(00) r(01) r(10) r(11)

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

q(00) q(01) q(10) q(11)

a

t(00)

Earning Profit

Claim: E(a) T(a)

Pf: Cousin-edges not from ancestors.Vertex r(00) blocked by E(a).

r(00) r(01) r(10) r(11)

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

q(00) q(01) q(10) q(11)

a

Earning Profit

Claim: E(a) T(a)

Result:E(a) gives free upgrade to E(a){S(a)}.Profit = S(a).

r(00) r(01) r(10) r(11)

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

q(00) q(01) q(10) q(11)

a

t(00)

q(00) q(01)

r(00) r(01) r(10) r(11)

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

q(00) q(01) q(10) q(11)

E(aL){S(aL)}

r(00) r(01) r(10) r(11)

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

q(10) q(11)

E(aR){S(aR)}

aL

aR

q(00) q(01)

r(00) r(01) r(10) r(11)

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

q(00) q(01) q(10) q(11)

(E(aL){S(aL)})

(E(aR){S(aR)})

r(00) r(01) r(10) r(11)

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

q(10) q(11)

(E(aL){S(aL)})

(E(aR){S(aR)})

Applying submodularity

r(00) r(01) r(10) r(11)

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

q(00) q(01) q(10) q(11)

(E(aL){S(aL)})

(E(aR){S(aR)})

New Investment

Union term has more edges

Can use downstreamnessor informational dominance again!

(E(aL){S(aL)}) (E(aR){S(aR)}) = E(a)

a

Debt Consolidation Intersection term has only sources

Cannot earn new profit.

Used for later “debt repayment” (E(aL){S(aL)}) (E(aR){S(aR)}) = F(a)

q(00) q(01)r(00) r(01) r(10) r(11)

s(00)s(0)

s(01) s(10) s(11)s(1)

s(ε)

q(10) q(11)

(E(aL){S(aL)})

(E(aR){S(aR)})

a

What have we shown? Let aL,aR be sibling leaves; a is their parent.

H(E(aL)) + H(E(aR)) H(E(a)) + H(F(a)) Iterate and sum over all nodes in tree

where r is the root. Note: E(v) = F(v) {(q(v),r(v))} when v is a leaf

vl

vFHrEHlEH nonleaf leaf

))(())(())((

v

l l

vFHrEH

HlFH lrlq

nonleaf

leaf leaf

))(())((

)())(( ))(),((

Debt Repayment

Claim:

Pf: Simple counting argument.

vl

vFHlFH nonleaf leaf

))(())((

v

l l

vFHrEH

HlFH lrlq

nonleaf

leaf leaf

))(())((

)())(( ))(),((

))(()( leaf

))(),(( rEHHl

lrlq

Finishing up

l lrlqc

leaf ))(),((

))(()( leaf

))(),(( rEHHl

lrlq

i

iSH ))((=

1

=

Rate < 1/n

nv

vdepth )(2

=

(where α = rate of solution)