Cyclic Codes for Error Detection

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Cyclic Codes for Error Detection

• W. W. Peterson and D. T. Brown

Presented by Ian Strascina 11/7/03
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Why do we need Error Detection/Correction???

• We want messages/values to remain correct if
  something corrupts them, or at least know there
  was an error to try to fix it
• The following is a practical example

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X
BAD!!!
Shuttle

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v
GOOD
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Error Detecting vs. Error Correcting

• Error Detecting no correction
• appropriate for communication-like environments
• data can be resent if necessary
• more efficient, useful for serial
• Error Correcting corrects corrupted data
• appropriate for memory-like environments
• less efficient, useful for parallel

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Relationship between them

• Error detection is used for error correction
• A code can detect 2t errors ? It can correct t
  errors
• Example Take t1 (can detect 2 errors, can
  correct 1 error)
• ? Assume code can detect 2 errors
• To correct one error, it could just try flipping
  bits until it was found If any bit other than
  the actual error was flipped, it would create a
  2nd error bit, which it can detect by assumption
• ? Assume code can correct 1 error
• To be shown, but not at all based on the
  assumption (theorem 3)

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Polynomial representation of bit strings

• Coding a k-bit message by appending n-k check
  bits and transmitting all n of them together
• Convenient to think of a bit string as a
  polynomial of the dummy variable X (even though
  we know X2)
• Example
• The bit string
• can be represented
• by the polynomial

110101
101011
1XX3X5
X5X3X1
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Polynomial representation of bit strings

• Normal rules of algebra apply to polynomials,
  except we will use modulo 2 addition (? addition
  modulo 2)
• addition/subtraction w/o carry/borrow (resp.)
• Thus
• 1Xa 1Xa 0Xa 1Xa 0Xa 1Xa 0Xa 1Xa
• 0Xa 0Xa 0Xa -1Xa 1Xa

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Polynomial representation of bit strings

• Simple examples of addition and multiplication
  (resp.)
• X3 X1 X1
• X2X1 X1
• X3X2 X1
• X2X___
• X2 1
• This corresponds to the binary operation
• Associativity, Distributivity, Commutativity, and
  Unique factorization of primes also hold

XOR!!!
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A little math review/preview

• The degree of polynomial is the largest exponent
  which appears in the polynomial with a non-zero
  coefficient
• a b read a divides b if there exists an
  integer q s.t. b aq
• Thus 6 12 (for q2)
• However, 12 does not divide 6 (there is no q s.t.
  6 12q), but 12 is divisible by 6. Confused
  yet?
• Let ngt0, a?Z, then ?! q,r?Z s.t. aqnr, 0rltn
  (division algorithm)
• (G, ?) is a group (in the domain G) under the
  operation ? if
• G is closed under ?
• ? is associative
• identity inverse exist in domain
• (Z,) is a group, but (Z,) is not
• More on groups??? Take Math 331

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Cyclic Codes

• Cyclic code defined in terms of a generator
  polynomial P(X) of degree n-k
• A polynomial of degree n is a code ? it is
  divisible by P(X)
• Cyclic codes form a group with an added property
  the cyclic shift of a code vector is also a
  code vector
• Group? Let C(X)D(X)P(X) and F(X)G(X)P(X)
  W(X)Z(X)P(X) be codes
• C(X)F(X) (D(X)G(X))P(X) still a code
• (C(X)F(X))W(X) ((D(X)G(X))Z(X))P(X)
  (D(X)(G(X)Z(X)) )P(X) C(X)(F(X)W(X))
• Identity is 0 A codes inverse is itself
• Consider only codes for which P(X) is not
  divisible by X (important for rest of paper!!!)
• if it is, P(X) (and all codes) will have X0 0
• This (coefficient) is useless as it gives no
  information

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Cyclic Codes

• Could form a code poly. by multiplying any poly.
  C(X) (s.t. degree(C(X)) lt k) by P(X)
• More advantageous multiplying message poly.
  G(X) by Xn-k, divide this by P(X) and
  add(subtract) the remainder R(X) to the result of
  the multiplication
• Xn-kG(X) Q(X)P(X) R(X) or
• F(X) Xn-kG(X) R(X) Q(X)P(X)
• F(X) code poly. Q(X) quotient of division
• Since degree(P(X)) n-k, degree(R(X)) lt n-k (by
  division alg.)
• Xn-kG(X) has zeros in n-k lowest order positions
• So F(X) has coefficients of R(X) in the low order
  n-k positions (check bits) and coefficients of
  G(X) in the k highest-order positions (message
  bits)

F(X) G(X) R(X), degree(F(X))n k positions
n-k positions
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Cyclic Codes

• Example take n8, k5, n-k3
• Take P(X) X3X1 (1011) and G(X) X4X2X1
  (10111)
• Xn-kG(X) X3(X4X2X1) X7X5X4X3
• Xn-kG(X) / P(X) (X41) (X1) (long division
  omitted)
• Xn-kG(X) (X41)(X3X1) (X1)
• Q(X) P(X) R(X)
• F(X) Xn-kG(X) R(X) X7X5X4X3X1
  (10111011)
• message check

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Principles of Error Detection/Correction

• Can represent an encoded message containing
  errors by
• H(X) F(X) E(X)
• Where F(X) is the correct message and E(X) has a
  term for each erroneous position
• As previously stated, a code poly. is one that is
  divisible by P(X)
• Thus, if P(X) does not divide H(X), then clearly
  H(X) has one or more errors
• Since we already know P(X) divides F(X), it is
  enough to show that P(X) does not divide E(X)
• Good checks will make sure that no E(X) we wish
  to detect is divisible by P(X)

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Principles of Error Detection/Correction

• So to look for errors, we can just divide H(X) by
  P(X)
• if remainder is not zero, we found an error
• if it is zero, then either no errors or
  undetectable errors exist
• Example Take P(X) X3X1 from last example,
  let
• F(X) X5X2X1 (X21)P(X) 100111
• E(X) X4 010000
• H(X) X5 X4X2X1 110111
• We get the remainder of H(X)/P(X) is X2X
• This is the same result as E(X)/P(X)

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Error Detection Schemes

• Single-Error Detection
• Double- and Triple-Error Detection
• Burst-Error Detection
• Detecting Two Bursts of Errors
• Implementation

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Detection of Single Errors

• Theorem 1 A cyclic code generated by a
  polynomial P(X) with more than one term detects
  all single errors.
• Proof An error at the ith position of an encoded
  message it represented by the error poly. E(X)
  Xi. To detect this, we must only show the P(X)
  does not divide Xi or Xi ? P(X)Q(X) for some
  poly. Q(X). Clearly, we can find no such Q(X) if
  P(X) has more than one term
• The simplest such P(X) is X1.

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Detection of Single Errors

• Theorem 2 Every polynomial divisible by X1 has
  an even number of terms
• Proof Let F(X) XaXbXn (X1)Q(X)
• If we evaluate F with X1, we get
• F(X) 111 (11)Q(1) 0
• Since the result is 0, then there must have been
  an even number of terms for the additions (xors)
  to equal 0.
• Corollary 1 The term X1 can detect any odd
  number of errors, acting as a parity check to
  ensure an even number of ones
• Corollary 2 If a polynomial has a term Xc1 then
  it can detect any odd number of errors, since
  Xc1 (X1)(Xc-1Xc-21).

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Detection of Double and Triple Errors

• Definition A poly. P(X) is said to belong to an
  exponent e if e is the least positive integer
  s.t. P(X) divides Xe1.
• Theorem 3 A code generated by the poly. P(X)
  detects all single and double errors if the
  length n of the code is no greater than the
  exponent e to which P(X) belongs
• Proof Two errors are represented by the error
  poly. E(X) Xj Xi (iltj). So we must show P(X)
  does not divide Xj Xi. We can factor Xj Xi
  into Xi(Xj-i1). P(X) is assumed not to be
  divisible by X (and thus Xi), so we only need to
  show P(X) does not divide Xj-i1. Since j-iltne,
  cleary P(X) cannot divide Xj-i1. So this code
  can detect 2 errors. Since P(X) not divisible by
  X and certainly not just be the constant 1, it
  has more than one term. Thus, by theorem 1 it
  detects single errors also.

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Detection of Double and Triple Errors

• Definition It can be shown (dont know how???)
  that for any m there exists at least one poly.
  P(X) of degree m that belongs to e2m-1. This is
  the maximum possible value for e. These type of
  polys. are called primitive polys. and are always
  irreducible.

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Detection of Double and Triple Errors

• For any m, there is a double-error detecting code
  of length n2m-1 generated by a poly. P(X) of
  degree m. The code thus has m check bits and
  2m-1-m message bits. These can be shown (not by
  me) to be equivalent to single-error correcting
  Hamming codes

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Detection of Double and Triple Errors

• Theorem 4 A code generated by P(X) (X1)P1(X)
  detects all single, double, and triple errors if
  the length n of the code is no greater than the
  exponent e to which P1(X) belongs.
• Proof The factor X1 detects all single and
  triple errors by theorem 2. The double errors
  are caught because P1(X) belongs to e and ne, as
  we just saw in theorem 3.
• If P1(X) is a primitive poly., a code of maximal
  length is produced. These are equivalent to
  single-error correcting, double-error detecting
  (SECDED) Hamming codes.

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Detection of Burst-Errors

• Definition A burst-error of length b is any
  pattern of errors for which the number of symbols
  between the first and last errors (inclusive) is
  b.
• Thus for E(X)
• X31 (1001), X3X1 (1011),
• X3X21 (1101), X3X2X1 (1111)
• all have burst-errors of length 4, even though
  they contain a different number of errors.
• For bgt3, there are 2b-2 different possible error
  polys. of burst-length b.
• It also follows that for any b, there are a total
  of 2b-1 different error polys. of burst-length b

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Detection of Burst-Errors

• Theorem 5 Any cyclic code generated by a poly.
  P(X) of degree n-k detects any burst-error of
  length b n-k.
• Proof We can factor any burst-error into XiE1(X)
  and degree(E1(X)) is b-1. This burst can be
  detected if P(X) does not divide E(X). P(X) is
  already assumed to not divisible by X (and
  therefore Xi), so we must show P(X) does not
  divide E1(X). We know degree(E1(X)) is b-1,
  which is at most n-k-1. P(X) is assumed to be of
  degree n-k. Since P(X) is of a higher degree
  than E1(X), P(X) clearly cannot divide it.
  
• Next, we show that some bursts of length bgtn-k
  can even be detected

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Detection of Burst-Errors

• Theorem 6 The fraction of bursts of length bgtn-k
  that are undetectable is
• 2-(n-k) if bgtn-k1, 2-(n-k-1) if bn-k1
• Proof Error poly. is E(X) XiE1(X) where
  degree(E1(X)) b-1. We know E1(X) has the terms
  Xb-1 and X0 with b-2 terms in between. Each of
  these could have a coefficient of either 0 or 1,
  so there are 2b-2 different polys. for E1(X). We
  can only detect this error if P(X) divides E1(X)
  (E1(X) P(X)Q(X) for some poly. Q(X)). Since
  P(X) has degree n-k then Q(X) must have degree
• b-1-(n-k). So if bn-k1 then Q(X) is 1 (degree
  0), and the only possible error is E1(X) P(X).
  So 1 undetectable error out of the total number
  of possible bursts is 1/2b-2 2-(n-k-1). On the
  other hand, if bgtn-k1, Q(X) has terms X0,
  Xb-1-(n-k), and b-2-(n-k) other coefficients. So
  there are 2b-2-(n-k) possibilities for Q(X). So
  now that ratio of undetectable to total patterns
  is 2b-2-(n-k)/2b-2 or 2-(n-k).

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Detection of Two Bursts of Errors

• Theorem 7 The cyclic code generate by P(X)
  (X1)P1(X) detects any combination of two
  burst-errors of length 2 or less if the length of
  the code n is no greater than e, the exponent to
  which P1(X) belongs.
• Proof We can see four patterns of two bursts,
  each length 2 or less
• 1) E(X) Xj Xi 2) E(X) Xj (Xi1 Xi)
• 3) E(X) (Xj1 Xj) Xi 4) E(X) (Xj1 Xj)
  (Xi1 Xi)
• Cases 2 and 3 are automatically covered by the
  factor X1 since they have an odd number of
  errors (Corollary 2 of theorem 2). For case 4,
  we can rewrite E(X) to be (X1)(XjXi). The X1
  terms cancel out, so we must show that P1(X) does
  not divide XjXi (the same for case 1). We
  already showed this by theorem 3. So all 4 cases
  are not divisible by P(X) and are thus
  detectable.

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Detection of Two Bursts of Errors

• These are equivalent to Abramson codes, which
  correct single and double adjacent errors. They
  are also the same as the SECDED Hamming codes of
  theorem 6.
• Theorem 8 The cyclic code generated by
• P(X) (Xc1)P1(X)
• will detect any combination of two bursts
• E(X) XjE2(X) XiE1(X)
• provided
• c1 the sum of the lengths of both bursts
• P1(X) is irreducible and of degree length of
  the shorter burst
• length of the code the least common multiple of
  c and the exponent e to which P1(X) belongs
• Proof (next slide)

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Implementation

• Recap
• To encode a message G(X), we multiply it by Xn-k,
  divide this by P(X), and take the remainder R(X)
  and add it to Xn-kG(X).
• The only thing yet unseen polynomial division
  (modulo 2)
• Example X5X4X3X / X21
• X3 X2 0X 1
• X20X1 / X5 X4 X3 0X2 X 0 lt
  dividend
• divisor X5 0X4 X3
• 0 X4 0X3 0X2 X 0
• X4 0X3 X2
• 0 0 X2 X 0
• X2 0X 1
• X 1

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Implementation

• Quotient is unnecessary for us just need the
  remainder
• The variables are also unnecessary just need
  the coefficients
• Algorithm for modulo 2 division to find remainder
• Line up the highest degree term of the dividend
  with the highest degree term of the divisor and
  add/subtract (xor) each coefficient
• Line up the highest degree term of the
  addition/subtraction with the highest degree term
  of the divisor and subtract again
• Repeat until the difference has a lower degree
  than divisor

________________ 1 0 1 1 1 1 0 1 0
1 0 1 1 0 0 1
0 1 1 1 0
1 0 1 1 1 1 1 corresponds
to X1
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Implementation

• This can be implemented by a shift register and
  XOR gates
• Number of registers is equal to the degree of
  P(X)
• Place XOR gates at positions where P(X) has a
  coefficient of one for the corresponding term

dividend
?
?
?
Shift register for dividing by X5X4X21
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Implementation

• Can modify initial design to be able to check for
  errors or encode a message

dividend
OR
Delay of n-k shifts
G1
?
?
?
G2
One method of encoding on detecting errors (P(X)
X5X4X21)
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Implementation

• Can modify again to avoid the delay of n-k shifts.

OR
dividend
G1
?
?
G2
?
More efficient circuit for encoding and error
detection (P(X) X5X4X21)
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A little bit on Error Correction

• More difficult than detection
• Can be shown that each different correctable
  error pattern corresponds to a different
  remainder when dividing by P(X)
• Can be achieved in the follow manner
• divide received message H(X)F(X)E(X) by P(X) to
  get remainder
• get error E(X) corresponding to the remainder
  from a lookup table or some calculation
• Subtract E(X) from H(X) to obtain the correct
  trasmitted message F(X)
• More complex equipment for error
  lookup/computation
• Entire H(X) must be temporarily stored while
  computing the remainder and then E(X)
• Step 2 calculation can be done with a shift
  register for burst- or single-error correcting
  codes

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Conclusion

• Messages can be corrupted and we would at least
  like to be able to detect them
• Can represent a message bit string in terms of a
  polynomial
• Encoded message F(X) Xn-kG(X) R(X)
  Q(X)P(X)
• The methods of detection are highly efficient and
  easily implementable with shift registers
• We want our astronauts to return safely!!!

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