Generalized Sphere Packing Bound Arman Arman Fazeli Fazeli (1) , - - PowerPoint PPT Presentation

Generalized Sphere Packing Bound

Arman Arman Fazeli Fazeli(1), Alexander , Alexander Vardy Vardy(1), and , and Eitan Eitan Yaakobi Yaakobi(2)

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(1) - University of California San Diego (2) - Technion Israel Institute of Technology

The Sphere Packing Bound

• Upper bound on a code C with min dist 2r+1

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• This bound is valid for other cases as well

where the error graph is regular ( ) what happens if the graph is not regular?

– Naïve solution: choose to be the minimum size of a ball in the graph

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What about other bounds?

• The Gilbert-Varshamov lower bound:

There exists a code with min dist r+1 of size

• If the graph is not regular, it is possible to choose

as the average size of a ball

• There exists a code with min dist r+1 of size

Does the same analogy hold for the sphere packing bound? Is an upper bound on a code with min dist 2r+1?

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The Deletion Channel

• An example of non-regular graph

– 10010 -> 0010, 1010, 1000, 1001 – 11100 -> 1100, 1110 – 10101 -> 0101, 1101, 1001, 1011, 1010

• It is not possible to apply the sphere packing bound 
• Previous results

– Levenshtein ‘66: asymptotic upper bound – Kulkarni & Kiyavash ‘12: a method to derive explicit non- asymptotic upper bound using tools from hypergraph theory

• Can this method be generalized for other graphs?

– Grain errors:

• Kashyap & Zemor ‘13; Gabrys, Yaakobi, Dolecek ‘13

– Multi-permutations with the Kendall’s tau dist

• Buzaglo, Yaakobi, Etzion, Bruck ‘13

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Hypergraphs

• Let H=(X,E) be a hypergraph, where

– X={x1,…,xn} – set of vertices, E={E1,…,Em} – set of hyperedges – A is a binary n×m incidence matrix of H

• Matching - a collection of pairwise disjoint hyperedges

– The matching number ν (H) - the size of the largest matching

• Transversal - a vertices subset that intersects every

hyperedge

– The transversal number τ(H) - the size of the smallest transversal

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Hypergraphs

• The matching number
• The transversal number
• These problems satisfy weak duality ν(H) ≤ τ(H)
• The relaxation versions of these problems

satisfy strong duality

• Every vector w in τ*(H) is called a fractional transversal

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The Deletion Channel – KK’12

• Define a hypergraph H(X,E):

– X = {0,1}n-1 , E = {all 2n single-deletion balls}

• Every single-deletion correcting code of length n

is a matching in the hypergraph H

• Find the value of τ*(H) or any

fractional transversal to get an explicit upper bound

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The General Case

• G=(X,E) is a graph describing an error channel graph

– X = the set of all possible words (transmitted and received) – E = the set of vertices pairs of dist one

• The distance d(x,y) b/w x and y is the length of the shortest path from

x to y (not necessarily symmetric)

– Br(x) = {y ∊ X : d(x,y)≤r}; degr(x) = |Br(x)|

• For any r>0, H(G,r)=(Xr,Er) is a hypergraph for G

– Xr=X, Er={Br(x) : x∊X}

• Every code C in G is a matching in H(G,r)
• AG(n,d) - the max size of a code w/ min dist d in G

For every r>0:

• Q: Does the following hold?

is called the Average Sphere Packing Value: ASPV(G,r)

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Example: The Z Channel

• GZ=(XZ,EZ), XZ={0,1}n
• H(GZ,r) = (XZ,r,EZ,r), XZ,r=XZ={0,1}n

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BZ,1(10010)={10010,00010,10000}

Example: The Z Channel

• The average size of a ball with radius r
• The average sphere packing value
• For r=1:

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So what is the problem?

• A channel graph G=(X,E) w/ hypergraph H(G,r)=(Xr,Er)
• A code with min dist 2r+1 in G is a matching in H(G,r)
• An upper bound is given by
• The good news: there is an explicit upper bound!
• The bad news: It is not necessarily easy to calculate it

– Need to solve a linear programming… – Usually the number of variables and constraints in exponential

• Our goal: how to calculate the value of

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Some General Results

• Lemma: If for all x∊X, degr

in(x) ≤ Δ then

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Proof:

• Since degr

in(x) ≤ Δ then ⇒

• For any transversal w,
• Therefore,

Some General Results

• Lemma: If for all x∊X, degr

in(x) ≤ Δ then

• Lemma: If for all x∊X, degr(x) ≥ Δ, then

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Proof:

• The vector w=1/Δ is a fractional transversal
• Thus,

Some General Results

• Lemma: If for all x∊X, degr

in(x) ≤ Δ then

• Lemma: If for all x∊X, degr(x) ≥ Δ, then
• Corollary: If G is symmetric and regular then the

generalized sphere packing bound and the sphere packing bound coincide, and

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Monotonicity and Fractional Transversals

• The vector w is a fractional transversal if w ≥ 0 and
• Lemma: The vector w, given by

is a fractional transversal

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Proof:

• If y∊Br(xi), then xi∊Br

in(y) and

• Therefore,

Monotonicity and Fractional Transversals

• The vector w is a fractional transversal if w ≥ 0 and
• Lemma: The vector w, given by

is a fractional transversal

• Def: G is called monotone if for all x∊X and y∊Br(x)
• Lemma: If G is monotone then the vector

is a fractional transversal

• Corollary: If G is monotone an upper bound on AG(n,2r+1)

is called the monotonicity upper bound MB(G,r)

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• Cont. Ex: The Z Channel
• x,y∊{0,1}n, if y∊BZ,r(x), wH(y)≤wH(x) and degr(y)≤degr(x)
• The vector

given by is a fractional transversal

• The monotonicity upper bound for the Z channel:
• For r=1:
• The average sphere packing bound
• Is it possible to do better…?

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Can we find the optimal transversal?

• w is a fractional transversal if w ≥ 0 and
• For the Z channel:

– 2n constraints: Ex, n=3:

• 111: w111+w110+w101+w011 ≥ 1
• 110: w110+w100+w010 ≥ 1, 101: w101+w100+w100 ≥ 1, 011: w011+w010+w001 ≥ 1
• 100: w100+w000 ≥ 1, 010: w010+w000 ≥ 1, 001: w001+w000 ≥ 1
• 000: w000 ≥ 1

– Probably vectors w/ the same weight will have the same value – If so, only n+1 constraints:

• 3: w3+3w2 ≥ 1
• 2: w2+2w1 ≥ 1
• 1: w1+w0 ≥ 1
• 0: w0 ≥ 1
• Is it still possible to find the value of ?

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Automorphisms on Graphs

• Given a graph G=(X,E), an automorphism is a

permutation that preserves adjacency π:X->X s.t. (x,y)∊E iff (π(x),π(y))∊E

• The automorphisms set

Aut(G)={π∊Sn : π is an automorphism in G} is a subgroup of Sn under composition

• Aut(G) induces an equivalence order R on X:

(x,y)∊R iff there exists π∊ Aut(G) and π(x)=y and partitions X into n(G) equivalence classes

• For a vector w and automorphism π, the vector wπ is

(wπ)i = wπ(i)

• Lemma: If w is a transversal and π an automorphism

then wπ is a transversal as well

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Automorphisms on Graphs

• Given a graph G=(X,E), an automorphism is a

permutation that preserves adjacency π:X->X s.t. (x,y)∊E iff (π(x),π(y))∊E

• The automorphisms set

Aut(G)={π∊Sn : π is an automorphism in G} is a subgroup of Sn under composition

• Aut(G) induces an equivalence order R on X:

(x,y)∊R iff there exists π∊ Aut(G) and π(x)=y and partitions X into n(G) equivalence classes

• For a vector w and automorphism π, the vector wπ is

(wπ)i = wπ(i)

• Lemma: If w is a transversal and π an automorphism

then wπ is a transversal as well

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Proof:

• Need to show: for 1 ≤ i ≤n
• y ∊ Br(xi) iff π(y) ∊ Br(π(xi))
• Therefore,

Automorphisms on Graphs

• Aut(G) = {π∊Sn : π is an automorphism in G}
• An equivalence order R on X:

(x,y)∊R iff there exists π∊ Aut(G) and π(x)=y

• Lemma: If w is a transversal and π an automorphism then

wπ is a transversal as well

• Wc = {w : w is a transversal and Σwi=c}
• Theorem: If Wc≠∅ then Wc contains a transversal which

assigns the same weight to the equivalence classes of R

• This result holds also for any subgroup of Aut(G)

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• Cont. Ex: The Z Channel
• For every σ∊Sn define a permutation πσ:{0,1}n->{0,1}n

for all x∊{0,1}n, (πσ(x))i=xσ(i)

• The set K={πσ : σ∊Sn} is a subgroup of Aut(GZ) and

partitions {0,1}n into n+1 equivalence classes XK(GZ) = {X0,X1,…,Xn}, Xi=all vectors of weight i

• The generalized sphere packing bound now becomes:

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• Cont. Ex: The Z Channel
• The generalized sphere packing bound now becomes:
• For r=1:

where , and

– For example, n=3: w*

3=0, w* 2= (1-0)/3=1/3, w* 1= (1-1/3)/2=1/3, w* 0= 1

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Best known upper bound by Weber, De Vroedt, and Boekee ‘88

• Cont. Ex: The Z Channel
• For arbitrary r, the optimal solution is given by

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Comparison for r=2 Comparison for r=3

Limited Magnitude Channels

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• Asymmetric errors GA,q=(X,E): X=[q]n
• The graph is monotone
• The monotonicity upper bound for r=1 is
• The average sphere packing bound

The Asymmetric Channel

• The linear programming problem to find is given by
• By automorphisms as in the Z channel we can divide to

the equivalence classes

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The Asymmetric Channel

• Theorem: the vector given by

is a fractional transversal

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Results for q=3

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The Symmetric Channel

• No longer monotonicity because the channel is

symmetric

• Can follow similar steps…

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Comparison for q=3 Comparison for q=4

Does the average sphere packing hold?

• The average size of the ball is (1∙5+4∙1)/5=9/5
• The average sphere packing value

5/(9/5)=25/9

• But the min dist of the code {x2,x3,x4,x5} is infinity…
• Another example

– n=k2 vertices into two groups k and n-k – Every vertex from the 1st group is connected to exactly k-1 vertices from the 2nd group – All n-k vertices in the second group are connected – But there is a code with k vertices…

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Back to the Deletion Channel…

• GD=(XD,ED), XD={0,1}nU{0,1}n-1 and
• For x∊{0,1}n,
• The hypergraph H(GD,1)=(XD,1,ED,1),

XD,1={0,1}n-1, ED,1 = {BD,1(x) : x∊{0,1}n }

• The generalized sphere packing bound

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Back to the Deletion Channel…

• For x∊{0,1}n, ρ(x)= the number of runs in x

– x=001010010, ρ(x)=7

• degD,1(x) = ρ(x)
• For every y∊BD,1(x), ρ(y)≤ρ(x)=degD,1(x)
• This graph satisfies a similar property to monotonicity
• The vector given by

is a fractional transversal (KK’12)

• The corresponding upper bound is

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Is it possible to do better…?

• A transversal is given by wx=1/ρ(x)
• A vector x has ρ(x) neighbors

– If they all have ρ(x) runs then we can’t do better

• x=001100: BD(x)={01100(w=1/3), 00100(w=1/3), 00110(w=1/3)}

– But if many neighbors have less than ρ(x) runs…?

• x=000010: BD(x)={000010(w=1/3), 00000(w=1), 00001(w=1/2)}
• For x, μ(x) = # of middle runs of length 1

– x=001010010, μ(x)=4 – 0≤μ(x) ≤ρ(x)-2

• Nn(ρ, μ) = # of vectors with ρ runs and μ middle-1 runs

– Nn(1,0)=2

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Is it possible to do better…?

• For x, μ(x) = # of middle runs of length 1
• Nn(ρ, μ) = # of vectors with ρ runs and μ middle-1 runs
• Lemma: For 2≤ρ≤n and 0≤μ≤ρ-2,
• Theorem: The vector

defined by is a fractional transversal

• Theorem: The value

satisfies

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Comparison of the Different Bounds

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Conclusion

• A method to generalize the sphere packing bound
• The Deletion channel – Kulkarni & Kiyavash ’12
• General Results
• Some specific channels

– The Z channel – Asymmetric errors – Deletion channel – Grain errors – Projective space

Thank You!

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