236608 - Coding and Algorithms for Memories Lecture 11 – Generalized Sphere Packing Bound

1. 236608 - Coding and Algorithms for MemoriesLecture 11 – Generalized Sphere Packing Bound

2. The Sphere Packing Bound • Upper bound on a code C with min dist2r+1 • This bound is valid for other cases as well where the error graph is regular ( ) • Q: what happens if the graph is not regular? • Naïve solution: choose to be the minimum size of a ball in the graph

3. What about other bounds? • The Gilbert-Varshamovlower bound: There exists a code with min distr+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 • Q: Does the same analogy hold for the sphere packing bound? Is an upper bound on a code with min dist. 2r+1?

4. 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  • Levenshtein’66: asymptotic upper bound

5. 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

6. 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

7. 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 distd in GFor every r>0: • Q: Does the following hold? is called the Average Sphere Packing Value: ASPV(G,r)

8. Example: The Z Channel • GZ=(XZ,EZ), XZ={0,1}n • H(GZ,r) = (XZ,r,EZ,r), XZ,r=XZ={0,1}n BZ,1(10010)={10010,00010,10000}

9. Example: The Z Channel • 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 • Is it possible to find the value of ?

10. Example: The Z Channel • The average size of a ball with radius r • The average sphere packing value • For r=1:

11. So what is the problem? • A channel graph G=(X,E) w/ hypergraph H(G,r)=(Xr,Er) • A code with min dist2r+1 in G is a matchingin 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 • The goal: how to calculate the value of

12. Some General Results • Theorem: If G is symmetric and regular then the generalized sphere packing bound and the sphere packing bound coincide, and

13. Monotonicity and Fractional Transversals • The vector w is a fractional transversal if w ≥ 0 and • Lemma: The vector w, given byis a fractional transversal • Def: G is called monotone if for all x∊X and y∊Br(x) • Lemma: If G is monotone then the vectoris 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) • Proof: • If y∊Br(xi), then xi∊Brin(y) and • Therefore,

14. 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 byis 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…?

15. 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 ?

16. Automorphisms on Graphs • Given a graph G=(X,E), an automorphismis a permutation that preserves adjacencyπ:X->Xs.t.(x,y)∊Eiff(π(x),π(y))∊E • The automorphisms setAut(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)=yand 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 • Proof: • Need to show: for 1 ≤i≤n • y ∊ Br(xi)iffπ(y)∊ Br(π(xi)) • Therefore,

17. 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 andΣwi=c} • Theorem: If Wc≠Ø 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)

18. 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:

19. 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

20. Best known upper bound byWeber, De Vroedt, and Boekee ‘88

21. 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

22. 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

23. 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

24. 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 byis a fractional transversal • Theorem: The value satisfies

25. Comparison of the Different Bounds

26. Does the average sphere packing hold? • The average size of the ball is (1∙5+4∙1)/5=9/5 • The average sphere packing value5/(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…