1 / 15

236601 - Coding and Algorithms for Memories Lecture 8

236601 - Coding and Algorithms for Memories Lecture 8. Relative Vs. Absolute Values. Less errors More retention. 0. 1. Jiang, Mateescu , Schwartz, Bruck , “Rank modulation for Flash Memories”, 2008. The New Paradigm Rank Modulation. Absolute values  Relative values

derron
Download Presentation

236601 - Coding and Algorithms for Memories Lecture 8

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 236601 - Coding and Algorithms for MemoriesLecture 8

  2. Relative Vs. Absolute Values Less errors More retention 0 1 Jiang, Mateescu, Schwartz, Bruck, “Rank modulation for Flash Memories”, 2008

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

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

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

  6. 1 2 3 1 2 3 3 1 2 3 1 2 2 3 1 2 3 1 2 1 3 2 1 3 3 2 1 3 2 1 1 3 2 1 3 2 1 2 3 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 n=3 3 cycles 1 2 3 Transition graph, n=3

  7. 1 2 3 3 1 2 2 3 1 2 1 3 3 2 1 1 3 2 ~ (n-1)! Gray Codes for Arbitrary n • Recursive construction: • Keep bottom cell fixed • (n-1)! transitions with others 1 3 2 3 1 2 2 1 3 2 3 1 3 2 1 1 2 3 444444 4 1 2 1 4 2 2 4 1 2 1 4 1 2 4 4 2 1 333333 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

  8. Rewriting with Rank Modulation • If we represent n! symbols then in the worst case we apply n-1 push-to-the-top operations to transfer from one permutation to another • Problem: Is it possible to use less push-to-the-top operations in case less than n! symbols are represented? • Rank Modulation Rewriting code (RMRC) (n,M) consists of • Update function: E: Sn×[M] -> Sn • Decoding function D: Sn-> [M]

  9. 1 2 3 1 2 3 3 1 2 3 1 2 2 3 1 2 3 1 2 1 3 2 1 3 3 2 1 3 2 1 1 3 2 1 3 2 Rewriting with Rank Modulation • Definition: The cost of changing s1 into s2,α(s1->s2), is the min number of push-to-the-top operations needed to change s1 to s2 • Ex: α([123]->[213]) = 1, α([123]->[321]) = 2 • The rewriting cost of a RMRC is the maximum update cost • The transition graph Gn=(Vn,En) • Vn = Sn, En ={(s1,s2) : α(s1->s2)=1} • The ballor radius r: Br(s)={ s’ : α(s->s’) ≤ r } • The sphereor radius r: Sr(s)={ s’ : α(s->s’) = r } • The balls and the sphere sizes do not depend on rBr,Sr

  10. Rewriting with Rank Modulation • Lemma: Br =n!/(n-r)! • For n,M, define r(n,M) to be the smallest integer such that Br(n,M) ≥ M • Lemma (Lower Bound): For any RMRC (n,M), its rewriting cost is at least r(n,M) • A tight upper bound on the rewriting cost is given by a construction • Theorem: There exists a RMRC with parameters (n,M≤Br) and cost r • Ex. (n,n) with cost 1 • Ex. (n,n(n-1)) with cost 2 • Ex. (n,n!) with cost n-1 • Ex. (n,n!/2) with cost n-2

  11. 3 2 4 1 3 2 1 4 2 3 1 4

  12. 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  andπ = Number of adjacent transpositions to change  to beπ =2413 andπ=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  2134  2143  2314  2143  2134

  13. 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)

  14. 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

  15. How to Construct ECCs for the Kendall’s Tau Distance? • Goal: Construct codes with some prescribed min Kendall’s tau dist d

More Related