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BASIC ELEMENTS in STORAGE

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BASIC ELEMENTSin STORAGE

A.J. Han Vinck

March 2003

We consider the properties of simple elements in storage

Write Once Memory

Write unidirectional

Memory with defects

CODING is MORE THAN ERROR-CORRECTION !

Example: IBM punchcardpunching 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 ?

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!!

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 ?

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 -

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-

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

Assumptions:Cell stuck-at with probability p

0

0

0

1

1

1

stuck-at 0 stuck-at 1

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)

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!)

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

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:

log27

4

ONE APPROACH:

- LOOK at PRESENT WORD or STATE S
- CHOOSEWRITE 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

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!

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)