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236608 - Coding and Algorithms for Memories Lecture 3 – WOM Codes

236608 - Coding and Algorithms for Memories Lecture 3 – WOM Codes. Write-Once Memories (WOM). Introduced by Rivest and Shamir , “ How to reuse a write-once memory ” , 1982

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236608 - Coding and Algorithms for Memories Lecture 3 – WOM Codes

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  1. 236608 - Coding and Algorithms for MemoriesLecture 3 – WOM Codes

  2. Write-Once Memories (WOM) • Introduced by Rivest and Shamir, “How to reuse a write-once memory”, 1982 • The memory elements represent bits (2 levels) and are irreversibly programmed from ‘0’ to ‘1’ Q:How many cells are required to write 100 bits twice? P1: Is it possible to do better…? P2: How many cells to write kbits twice? P3: How many cells to write kbits ttimes? P3’: What is the total number of bits that is possible to write in n cells in t writes? 1st Write 2nd Write

  3. Binary WOM Codes • k1,…,kt:the number of bits on each write • n cells and t writes • The sum-rate of the WOM code is R = ()/n • Rivest Shamir: R = (2+2)/3 = 4/3

  4. Definition: WOM Codes • Definition: An [n,t;M1,…,Mt] t-write WOM code is a coding scheme which consists of n cells and guarantees any t writes of alphabet size M1,…,Mt by programming cells from zero to one • A WOM code consists of t encoding and decoding maps Ei, Di, 1 ≤i≤ t • E1: {1,…,M1}  {0,1}n • For 2 ≤i≤ t, Ei: {1,…,Mi}×Im(Ei-1)  {0,1}nsuch that for all (m,c)∊{1,…,Mi}×Im(Ei-1), Ei(m,c) ≥ c • For 1 ≤i≤ t, Di: {0,1}n {1,…,Mi}such that for Di(Ei(m,c)) = m for all (m,c)∊{1,…,Mi}×Im(Ei-1) The sum-rate of the WOM code is R = ()/n Rivest Shamir: [3,2;4,4], R = (log4+log4)/3=4/3

  5. The Coset Coding Scheme • C[n,n-r] is an ECC with an r×nparity check matrix H • Write r bits: Given a syndrome s of r bits, find a length-n vector e such that H⋅eT = s • Example: H is aparity check matrix of a Hamming code • s=100, v1 = 0000100: c = 0000100 • s=000, v2 = 1001000: c = 1001100 • s=111, v3 = 0100010: c = 1101110 • s=010, …  can’t write! • This matrix gives a [7,3:8,8,8] WOM code • The Golay(23,12,7) code: [23,3; 211,211,211], R=33/23=1.43 • The Hamming code: r bits, 2r-2+2 times, 2r–1 cells: R=r(2r-2+2)/(2r –1) • Improved my Godlewski (1987) to 2r-2+2r-4+2 times: R=r(2r-2+2r-4+2)/(2r –1)

  6. The Coset Coding Scheme • Cohen, Godlewski, and Merkx ‘86 – The coset coding scheme • Use Error Correcting Codes (ECC) in order to construct WOM-codes • Let C[n,n-r] be an ECC with parity check matrix H of size r×n • Write r bits: Given a syndrome s of r bits, find a length-n vector e such that H⋅eT = s • Use ECC’s that guarantee on successive writes to find vectors that do not overlap with the previously programmed cells • The goal is to find a vector e of minimum weight such that only 0s flip to 1s

  7. Binary Two-Write WOM-Codes • C[n,n-r] is a linear code w/ parity check matrix H of size r×n • For a vector v ∊ {0,1}n, Hv is the matrix H with 0’s in the columns that correspond to the positions of the 1’s in v v1= (0 1 0 1 1 00)

  8. Binary Two-Write WOM-Codes • First Write: program only vectors v such that rank(Hv) = rVC = { v ∊ {0,1}n | rank(Hv) = r} • For H we get |VC| = 92 - we can write 92 messages • Assume we write v1= 0 1 0 1 1 0 0 v1= (0 1 0 1 1 00)

  9. Binary Two-Write WOM-Codes • First Write: program only vectors v such that rank(Hv) = r, VC = { v ∊ {0,1}n | rank(Hv) = r} • Second Write Encoding: • Second Write Decoding: Multiply the received word by H: H⋅(v1 + v2) = H⋅v1 + H⋅v2 = s1+ (s1 + s2) = s2 • s2 = 001 • s1 = H⋅v1 = 010 • Hv1⋅v2 = s1+s2 = 011a • v2 = 0 0 0 0 0 1 1 • v1+v2 = 0 1 0 1 1 1 1 • Write a vector s2 of r bits • Calculates1=H⋅v1 • Find v2 such that Hv1⋅v2= s1+s2a • v2 exists since rank(Hv1) = ra • Write v1+v2 to memory v1= (0 1 0 1 1 00)

  10. Example Summary 1 1 1 0 1 0 0 1 0 1 1 0 1 0 1 1 0 1 0 0 1 H = • Let H be the parity check matrix of the [7,4] Hamming code • First write: program only vectors v such that rank(Hv) = 3 VC = { v ϵ {0,1}n | rank(Hv) = 3} • For H we get |VC| = 92 - we can write 92 messages • Assume we write v1= 0 1 0 1 1 0 0 • Write 0’s in the columns of H corresponding to 1’s in v1: Hv1d • Second write: write r = 3 bits, for example: s2 = 0 0 1 • Calculate s1 = H⋅v1 = 0 1 0 • Solve: find a vector v2 such that Hv1⋅v2 = s1 + s2= 0 1 1d • Choose v2 = 0 0 0 0 0 1 1 • Finally, writev1 + v2 = 0 1 0 1 1 1 1 • Decoding: 1 0 1 0 0 0 0 Hv1 = 1 0 1 0 0 1 0 1 0 0 0 0 0 1 1 1 1 0 1 0 0 1 0 1 1 0 1 0 1 1 0 1 0 0 1 . [0 1 0 1 1 1 1]T = [0 0 1]

  11. Sum-rate Results • The construction works for any linear code C • For any C[n,n-r] with parity check matrix H, VC = { v ∊ {0,1}n | rank(Hv) = r} • The rate of the first write is: R1(C) = (log2|VC|)/n • The rate of the second write is: R2(C) = r/n • Thus, the sum-rate is: R(C) = (log2|VC| + r)/n • In the last example: • R1= log(92)/7=6.52/7=0.93, R2=3/7=0.42, R=1.35 • Goal: Choose a code C with parity check matrix H that maximizes the sum-rate

  12. Sum-rate Results • The (23,11,8)Golay code: (0.9415,0.5217), R = 1.4632 • The (16,5,8) Reed-Muller (4,2) code: (0.7691, 0.6875), R = 1.4566 • We can limit the number of messages available for the first write so that both writes have the same rate, R1 = R2 = 0.6875, andR = 1.375 • Computer search results • Best code with rate 1.4928 • For fixed rate on both writes 1.4546

  13. Capacity Region and Achievable Rates of Two-Write WOM codes

  14. Capacity Achieving Results • The Capacity region C2-WOM={(R1,R2)|∃p∊[0,0.5],R1≤h(p), R2≤1-p} • Theorem: For any (R1, R2)∊C2-WOM and 𝛆>0, there exists a linear code C satisfying R1(C) ≥ R1-𝛆and R2(C) ≥ R2–𝛆

  15. The Capacity of WOM Codes • The Capacity Region for two writes C2-WOM={(R1,R2)|∃p∊[0,0.5],R1≤h(p), R2≤1-p} h(p) – the entropy function h(p) = -plog(p)-(1-p)log(1-p) • The Capacity Region for t writes: Ct-WOM={(R1,…,Rt)| ∃p1,p2,…pt-1∊[0,0.5], R1≤ h(p1), R2 ≤ (1–p1)h(p2),…, Rt-1≤ (1–p1)(1–pt–2)h(pt–1) Rt≤ (1–p1)(1–pt–2)(1–pt–1)} • p1 - prob to prog. a cell on the 1st write: R1≤ h(p1) • p2 - prob to prog. a cell on the 2nd write (from the remainder): R2≤(1-p1)h(p2) • pt-1 - prob to prog. a cell on the (t-1)th write (from the remainder): Rt-1≤(1–p1)(1–pt–2)h(pt–1) • Rt≤ (1–p1)(1–pt–2)(1–pt–1) because(1–p1)(1–pt–2)(1–pt–1) cells weren’t programmed • The maximum achievable sum-rate is log(t+1)

  16. The Capacity for Fixed Rate • The capacity region for two writes C2-WOM={(R1,R2)|∃p∊[0,0.5],R1≤h(p), R2≤1-p} • When forcing R1=R2 we get h(p) = 1-p • The (numerical) solution is p = 0.2271, the sum-rate is 1.54 • Multiple writes: A recursive formula to calculate the maximum achievable sum-rateRF(1)=1RF(t+1) = (t+1)root{h(zt/RF(t))-z}where root{f(z)} is the min positive value zs.t.f(z)=0 • For example:RF(2) = 2root{h(z)-z} = 2 0.7729=1.5458RF(3) = 3root{h(2z/1.54)-z}=3 0.6437=1.9311

  17. More Constructions • Shpilka, “New constructions of WOM codes using the Wozencraft ensemble” • An efficient capacity-achieving two-write construction • 1st write – program any vector of weight at most m (fixed) • 2nd write – instead of using one matrix, use a set of matrices such that at least one of them succeeds on the second write • Need to index the matrix for the 2nd write – negligible if the number of matrices is small • Use the Wozencraft ensemble of linear codes to construct a good set of matrices

  18. Polar WOM Codes • A probabilistic approach to construct WOM codes which works with high probability • On each write, encode more bits and write a vector that matches the bits which were already programmed • Can combine with ECC so the redundancy is used both for rewriting and error correction • Another recent construction using LDPC codes

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