236601 - Coding and Algorithms for Memories Lecture 8

Flash Memory Cell 1 3 2 Introduce errors Can only add water (charge)

The Leakage Problem 3 2 Error! 1 Get rid of the thresholds

The Overshooting Problem 3 2 1 iterations Need to erase the whole block

Possible Solution – Iterative Programming 3 Slow… 2 1 leakage

Relative Vs. Absolute Values Less errors More retention More cells 1 Jiang, Mateescu, Schwartz, Bruck, “Rank modulation for Flash Memories”, 2008

The New Paradigm Rank Modulation Absolute values  Relative values Single cell  Multiple cells Physical cell  Logical cell

Rank Modulation Ordered set of n cells Assume discrete levels Relative levels define a permutation Basic operation: push-to-the-top 1 2 3 4 1 4 3 2 Overshoot is not a concern Writing is much faster Increased reliability (data retention)

New Number Representation System an-1…a3a2a1 = an-1·(n-1)! + … + a3·3! + a2·2! + a1·1! 0 ≤ ai ≤ i permutation in lexicographical order [Lehmer 1906, Laisant 1888] FACTORADIC decimal 0 1 2 0 2 1 1 0 2 1 2 0 2 0 1 2 1 0 0 0 0 1 1 0 1 1 2 0 2 1 1 2 3 4 5

Gray Codes for Rank Modulation The problem: Is it possible to transition between all permutations? Find cycle through n! states by push-to-the-top transitions 1 2 3 2 3 1 2 3 1 n=3 3 cycles 1 2 3 Transition graph, n=3

Gray Codes for Arbitrary n Recursive construction: Keep bottom cell fixed (n-1)! transitions with others 2 3 1 ~ (n-1)! 1 3 2 3 1 2 2 1 3 2 3 1 3 2 1 1 2 3 4 4 4 4 4 4 4 1 2 1 4 2 2 4 1 2 1 4 1 2 4 4 2 1 3 3 3 3 3 3 3 4 2 4 3 2 2 3 4 2 4 3 4 2 3 3 2 4 1 1 1 1 1 1 1 2 3 4 1 2 3 4

Gray Codes for Arbitrary n 1 3 2 3 1 2 2 1 3 2 3 1 3 2 1 1 2 3 4 4 4 4 4 4 4 1 2 1 4 2 2 4 1 2 1 4 1 2 4 4 2 1 3 3 3 3 3 3 3 4 2 4 3 2 2 3 4 2 4 3 4 2 3 3 2 4 1 1 1 1 1 1 4 1 3 4 3 1 1 4 1 3 4 3 3 3 4 1 1 4 2 2 2 2 2 2

3 2 4 1 3 2 1 4 Discrete model 2 3 1 4

Kendall’s Tau Distance For a permutation  an adjacent transposition is the local exchange of two adjacent elements For ,π∊Sm, dτ(,π) is the Kendall’s tau distance between  and π = Number of adjacent transpositions to change  to be π =2413 and π=2314 2413 dτ(,π) = 3 It is called also the bubble-sort distance Lemma: Kendall’s tau distance induces a metric on Sn The Kendall’s tau distance is the number of pairs that do not agree in their order  2143  2143  2134  2134  2314

Kendall’s Tau Distance Lemma: Kendall’s tau distance induces a metric on Sn The Kendall’s tau distance is the number of pairs that do not agree in their order For a permutation , Wτ() = {(i,j) | i<j, -1(i) > -1(i) } Lemma: dτ(,π) = |Wτ()∆Wτ(π)| = |Wτ()\Wτ(π)| + |Wτ(π)\Wτ()| dτ(,id) = |Wτ()| The maximum Kendall’s tau distance is n(n-1)/2 The inversion vector of  is x =(x(2),…,x(n)) x(i) = # of elements to the right of i and are less then i dτ(,id) = |Wτ()| = Σ2≤i≤nx(i)

Kendall’s Tau Distance The Kendall’s tau ball: Br() = {π|dτ(,π) ≤ r} The Kendall’s tau sphere: Sr() = {π|dτ(,π) = r} They do not depend on the center  |B1()| = n, |S1()| = n-1

How to Construct ECCs for the Kendall’s Tau Distance? Goal: Construct codes with some prescribed minimum Kendall’s tau distance d

ECCs for Kendall’s Tau Distance Goal: Construct codes correcting a single error Assume k or k+1 is prime Encode a permutation in Sk to a permutation in Sk+2 A code over Sk+2 with k! codewords s=(s1,…sk) ϵ Sk is the information permutation set the locations of k+1ϵ Zk+1 and k+2ϵ Zk+2 to be loc(k+1) = Σ1k(2i-1)si (mod m) loc(k+2) = Σ1k(2i-1)2si(mod m) m=k if k is prime and m=k+1 is k+1 is prime Ex: k=7, s=(7613245) loc(8) = 17+36+51+73+92+114+135 = 3 (mod 7) loc(9) = 127+326+521+723+922+1124+1325 = 2 (mod 7) E(s) = (769183245)

ECCs for Kendall’s Tau Distance A code over Sk+2 with k! codewords s=(s1,…sk) ϵ Sk is the information permutation set the locations of k+1ϵ Zk+1 and k+2ϵ Zk+2 to be loc(k+1) = Σ1k(2i-1)si (mod m) loc(k+2) = Σ1k(2i-1)2si(mod m) m=k if k is prime and m=k+1 is k+1 is prime Ex: k=3, m=3 123 => 15423 ; loc(4)=11+32+53=1(mod3), loc(5)=11+92+253=1(mod3) 132 => 13542 ; loc(4)=11+33+52=2(mod3), loc(5)=11+93+252=2(mod3) 213 => 21543 ; loc(4)=12+31+53=2(mod3), loc(5)=12+91+253=2(mod3) 231 => 52431 ; loc(4)=12+33+51=1(mod3), loc(5)=12+93+251=0(mod3) 312 => 34512 ; loc(4)=13+31+52=1(mod3), loc(5)=13+91+252=2(mod3) 321 => 35241 ; loc(4)=13+32+51=2(mod3), loc(5)=13+92+251=1(mod3)

ECCs for Kendall’s Tau Distance A code over Sk+2 with k! codewords s=(s1,…sk) ϵ Sk is the information permutation set the locations of k+1ϵ Zk+1 and k+2ϵ Zk+2 to be loc(k+1) = Σ1k(2i-1)si (mod m); loc(k+2) = Σ1k(2i-1)2si(mod m) Theorem: This code can correct a single Kendall’s tau error. Proof: Enough to show that the Kendall’s tau distance between every two codewords is at least 3 s=(s1,…sk) ϵ Sk, u=E(s); t=(t1,…tk) ϵ Sk, v=E(t) If dτ(s,t)≥3 then dτ(u,v)≥3 ✔ If dτ(s,t)=1, write t=(s1,…si+1,si…sk), let δ = si+1-si locs(k+1)-loct(k+1)=(2i-1)si+(2i+1)si+1–(2i-1)si+1-(2i+1)si=2si+1-2si =2δ(mod k) thus, they are not positioned in the same location. ✔ locs(k+2)-loct(k+2)=(2i-1)2si+(2i+1)2si+1–(2i-1)2si+1-(2i+1)2si =8isi+1-8isi =8iδ(mod k) thus, they are not positioned in the same location either ✔.

ECCs for Kendall’s Tau Distance Proof (cont): s=(s1,…sk) ϵ Sk, u=E(s); t=(t1,…tk) ϵ Sk, v=E(t) If dτ(s,t)=1, write t=(s1,…si+1,si…sk), let δ = si+1-si locs(k+1)-loct(k+1)=(2i-1)si+(2i+1)si+1–(2i-1)si+1-(2i+1)si=2si+1-2si =2δ(mod k) thus, they are not positioned in the same location. ✔ locs(k+2)-loct(k+2)=(2i-1)2si+(2i+1)2si+1–(2i-1)2si+1-(2i+1)2si=8isi+1-8isi =8iδ(mod k) thus, they are not positioned in the same location either ✔. If dτ(s,t)=2, write t=(s1,…,si+1,si,…,si+1,si,…,sk), δ1=si+1-si, δ2=sj+1-sj locs(k+1)-loct(k+1)=2(δ1+δ2)(mod k) locs(k+2)-loct(k+2)=8iδ1 +8jδ2 (mod k) If δ1+δ2 is NOT a multiple of k then k+1 appears in different locations ✔ Otherwise, δ2=-δ1 (mod k) and locs(k+2)-loct(k+2)=8(i–j)δ1 (mod k), so k+2 appears in different locations ✔ If dτ(s,t)=2, write t=(s1,…,si+2,si,si+1,…,sk) locs(k+1)-loct(k+1)=(2i-1)si+(2i+1)si+1+(2i+3)si+2-(2i-1)si+2-(2i+1)si-(2i+3)si+1) =4si+2 -2si+1-2si (mod k) locs(k+2)-loct(k+2)=8((2i+1)si+2-(i+1)si+1-isi)(mod k)