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BASIC ELEMENTS in STORAGE

BASIC ELEMENTS in STORAGE. A.J. Han Vinck March 2003. BASIC ELEMENTS in STORAGE. We consider the properties of simple elements in storage. Write Once Memory Write unidirectional Memory with defects. CODING is MORE THAN ERROR- CORRECTION !. WRITE ONCE MEMORY.

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BASIC ELEMENTS in STORAGE

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  1. BASIC ELEMENTSin STORAGE A.J. Han Vinck March 2003

  2. BASIC ELEMENTS in STORAGE We consider the properties of simple elements in storage Write Once Memory Write unidirectional Memory with defects CODING is MORE THAN ERROR-CORRECTION !

  3. WRITE ONCE MEMORY Example: IBM punchcard punching a hole is destructive Obvious method: Use card only once Efficiency: 1 bit/cell hole or not More complicated method:Use card T times Efficiency: log2 (T+1) bits per cell WHY ?

  4. EXAMPLE: Card with 3 positions FIRSTTRANSMISSION PUNCH 1 hole  log23 bits SECONDTRANSMISSION PUNCH 2 bits 00 01 10 11 TOGETHER:(log23+2)/3 = 1.2 bits per position > 1!!

  5. Memories with known defects (ROM-type) Problem:output fixedand cell is useless! correct stuck-at 0 stuck-at 1 Assumptions:Cell stuck-at with probability p READER knows? Storage Capacity per cell WRITER knows? WHY? Yes Yes 1-p defect cells are not used RUSSIAN INVENTION Kuznetsov/Tsybakov (1970) Yes No ?

  6. 1 N EXAMPLE:maximum of 1 defect in a word of length 3 defect 1 defect 0 STORE:for defect 1 00 or 00 for a 1-defect: 01 10 11 for a 0-defect similar as for 00 In general:N-1 bits in N positions  Efficiency = 1 -

  7. n-k t k n n n SOLUTION:Yes-No situation Construct matrix C n -k k 0000000 0000001 … 1111111 CODE C X’ 0000000 INFORMATION • PROPERTIES: • Any t pattern is present in some row of C • Rows uniquely represented by n-k first digits RESULT: for t  n-k defects; R = = 1- = 1-

  8. Memories with known defects (ROM-type) 4 situations WRITER knows? READER knows? Storage Capacity per cell WHY? Yes Yes 1-p defect cells are not used No Yes 1-p defect found as erasure probability(e) = p Yes No ? Additive Coding invented by: Kuznetsov/Tsybakov No No 1-h(p/2) defect is random error; probability(error) = p/2 the result is a BSC

  9. Problem:output fixed and cell is useless! Assumptions:Cell stuck-at with probability p 0 0 0 1 1 1 stuck-at 0 stuck-at 1

  10. ENCODING: EXAMPLE:We store 3 bits of information in 6 locations • info X written as X’ (0 ,0 , 0 ,X1,X2,X3) • add modulo-2 the code vector C(d) = (C1,C2,C3,C4,C5,C6) • STOREX’ C(d) =    R(d,X) = (C1,C2,C3,S4,S5,S6)

  11. PROPERTY: The components of R(d,X) areequal to the 2 given defects at the defect location for any defect pair (condition 1 on code C, covering) DECODING: CalculateC(d) R(d,X) = C(d) C(d) X’  we obtain X (condition 2 on code, uniqueness!)

  12. 1 2 3 4 5 6 0 0 0 0 0 1 1 1 0 0 1 0 1 0 1 1 1 1 0 0 0 0 1 1 0 0 1 1 1 0 1 0 0 1 1 1 Efficiency 1/2 < 1 - 2/6 C = X’ = _ _ _ 1 0 0 R(d,X) = (1 1 0 , 1 0 1) or ? d = 1 _ _ _ 0 _ we decide to add row 3 In GENERAL CODES CAN BE CONSTRUCTED with EFFICIENCY 1-2/n

  13. MAGNETO-OPTICAL MEMORY WRITING PROCESS: first ERASE then WRITE erase write erase write    EFFICIENCY: .5 bit per cycle/cell QUESTIONS: Can we do better? How? How much? APPLICATION: MINI DISK IMPROVEMENT:CHANGEWRITING STRATEGY:

  14. log27 4 ONE APPROACH: • LOOK at PRESENT WORD or STATE S • CHOOSE WRITE or ERASE Example: words of length N = 4, # of messages M = 7 a = 0 0 0 0 b = 0 0 0 or 0 c = 0 0 0 or 0 d = 0 0 0 or 0 e = 0 0 0 or 0 f = 0 0 or 0 0 g = S SUPPOSE S = 0 0 Check that we can write the strings 0 0 0 0 0 0 0 0 0 0 0 0 For n  Capacity = 0.69 bits/cell/cycle < 1 ! STORAGE CAPACITY = bits/cell

  15. Example: 6 messages, word length N = 5 Messages present at ERASE WRITE EXAMPLE: write erase PROPERTY: From ANY word(message) at erasewe may write ANY messsage(word) and vice versa (n = 11 gives .53 b/c and M=58) Efficiency is log6/5 = .517 bits/cell!

  16. PIM PAM PET? WOM write once memory (Rivest, 1983) WUM write unidirectional (Willems Vinck, 1986) WIM write inhibited memory (Cohen, 1998) WEM write efficient memory (Ahlswede, 1990) WAM write address fault memory (Fuja, 1995)

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