# Error Detection and Correction by xiangpeng

VIEWS: 4 PAGES: 38

• pg 1
```									01010000001001

by Colin Kriwox

1
Contents
Introduction
credit card error checking
what is a code
purpose of error-correction codes
Encoding
naïve approach
hamming codes
Minimum Weight Theorem
definitions
proof of single error-correction
Decoding
list all possible messages
using vectors
syndrome
Conclusion
perfect codes
2
Detect Error On Credit Card

3
Formula for detecting error
Let d2, d4, d6, d8, d10, d12, d14, d16 be all the even
values in the credit card number.
Let d1, d3, d5, d7, d9, d11, d13, d15 be all the odd
values in the credit card number.
Let n be the number of all the odd digits which have a
value that exceeds four
Credit card has an error if the following is true:
(d1 + d3 + d5 + d7 + d9 + d11 + d13 + d15) x 2 + n +
(d2 + d4 + d6 + d8 + d10 + d12 + d14 + d16)
 0 mod(10)                                             4
Detect Error On Credit Card

n=3

d1

d2   d3   …   d15   d16

5
Now the test
(4 + 4 + 8 + 1 + 3 + 5 + 7 + 9) = 41
(5 + 2 + 1 + 0 + 3 + 4 + 6 + 8) x 2 + 3 = 61
41 + 61 = 102 mod (10) = 2

3             6
Credit Card Summary
The test performed on the credit card number is called a
parity check equation. The last digit is a function of the
other digits in the credit card. This is how credit card
numbers are generated by Visa and Mastercard. They
use the parity check equation to find the value of the
16th digit.

“This method allows computers to detect 100% of
single-position errors and about 98% of other common
errors” (For All Practical Purposes p. 354).
7
What is a code?

A code is defined as an n-tuple of q elements. Where q is
any alphabet.

Ex. 1001                   n=4, q={1,0}
Ex. 2389047298738904       n=16, q={0,1,2,3,4,5,6,7,8,9}
Ex. (a,b,c,d,e)            n=5, q={a,b,c,d,e,…,y,z}

The most common code is when q={1,0}. This is known as
a binary code.                                         8
The purpose
A message can become distorted through
a wide range of unpredictable errors.

•   Humans
•   Equipment failure
•   Lighting interference
•   Scratches in a magnetic tape

9
Why error-correcting code?

To add redundancy to a message so
the original message can be recovered
if it has been garbled.

e.g. message = 10
code = 1010101010

10
Send a message

Message   Encoder   Channel   Decoder   Message

10       101010     noise    001010      10

11
Encoding

Naïve approach
Hamming codes

12
Take Naïve approach

Append the same message multiple times.
Then take the value with the highest
average.

Message:= 1001
Encode:=   1001100110011001
Channel:= 1001100100011001
Decode: = a1 = Average(1,1,0,1) = 1
a2 = Average(0,0,0,0) = 0 ... (a1,a2,a3,a4)
Message:= 1001
13
Hamming [7,4] Code
The seven is the number of digits that make the code.
E.g. 0100101
The four is the number of information digits in the code.
E.g. 0100101

14
Hamming [7,4] Encoding
Encoded with a generator matrix. All codes can be formed
from row operations on matrix. The code generator matrix
for this presentation is the following:

1    0 0 0 0 1 1
0    1 0 0 1 0 1
G                
0    0 1 0 1 1 0
                
0    0 0 1 1 1 1

15
Hamming [7,4] Codes
1000011
0100101
0010110
0001111
1100110
2  16
4
Codes
1010101

2  128
1001100
0110011    7          Possible codes
0101010
0011001
1101001
1001010
1111111
0111100
0011001
0000000
16
Minimum Weight Theorem
Definitions
Proof of Theorem

17
Definitions
The weight of a code is the number of nonzero
components it contains.
e.g. wt(0010110) = 3

The minimum weight of Hamming codes is the weight of
the smallest nonzero vector in the code.
1   0 0 0 0 1 1
0   1 0 0 1 0 1
e.g.   G                  d(G)= 3
0   0 1 0 1 1 0
               
0   0 0 1 1 1 1
18
Definitions
The distance between two codes u and v is the number of
positions which differ
e.g.   u=(1,0,0,0,0,1,1)
v=(0,1,0,0,1,0,1)
dist(u,v) = 4

Another definition of distance is wt(u – v) = dist(u,v).

19
Definitions
For any u, v, and w in a space V, the following three
conditions hold:

dist(u, u )  0
dist(u, v)  dist(v, u )
dist(u, w)  dist(u, v)  dist(v, w)

20
Definitions
The sphere of radius r about a vector u is defined as:
Sr (u)  {v V | dist(u, v)  r}

e.g. u=(1,0,0,0,0,1,1)           (0,0,0,0,0,1,1)

(1,1,0,0,0,1,1)
(1,0,1,0,0,1,1)
(1,0,0,1,0,1,1)       (1,0,0,0,1,1,1)

(1,0,0,0,0,0,1)

(1,0,0,0,0,0,1)
21
Minimum Weight Theorem
If d is the minimum weight of a code C, then C can
correct t = [(d – 1)/2] or fewer errors, and conversely.

22
Proof
Want to prove that spheres of radius t = [(d – 1)/2] about
codes are disjoint. Suppose for contradiction that they
are not. Let u and w be distinct vectors in C, and
assume that v  St (u )  St ( w)

u     v   w

23
Proof
By triangle inequality
dist(u, w)  dist(u, v)  dist(v, w)  2t

v
u           w

24
Proof

Since spheres of radius t = [(d – 1)/2] so 2t  d  1 and this
gives dist(u,w)  dist(u,v) dist(u,w)  2t  d  1

But since dist(u, w)  wt (u  w)  d

25
Result of Theorem
1    0 0 0 0 1 1
0    1 0 0 1 0 1
G                
0    0 1 0 1 1 0
                
0    0 0 1 1 1 1

Since d(G) = 3 then for t = [(3 – 1)/2] = 1 or fewer
unique code word.

26
Decoding
list all possible messages
using vectors
syndrome

27
List all messages

This is done by generating a list of all the possible
messages. For something small like the Hamming [7,4]
codes the task is feasible, but for codes of greater
length it is not. An example of a list is as follows:

Code words   1000011      0100101      0010110 …
0000011      0000101      0000110
Other        1000001      0100111      0010100
Words        1100011      1100101      0110110
…            …            …

28
List all messages

For example, if the received code was 0001101 then it
would be decoded to 0100101 from the list.

Code words   1000011      0100101      0010110 …
0000011      0000101      0000110
Other        1000001      0100111      0010100
Words        1100011      1100101      0110110
…            …            …

29
Vector Decoding
1          0 0 0 0 1 1
0          1 0 0 1 0 1
G                      
0          0 1 0 1 1 0
                      
0          0 0 1 1 1 1

Let a:=(0,0,0,1,1,1,1), b:=(0,1,1,0,0,1,1), and
c:=(1,0,1,0,1,0,1).
7

If x : ( x1 ,..., x7 )   y : ( y1 ,..., y 7 ) then inner product =   x y
i 1
i   i   mod(2)

30
Vector Decoding
Correct errors by taking inner product of received
vector u by a, b, c. We get u  a, u  b, u  c

e.g. recall: a:=(0,0,0,1,1,1,1), b:=(0,1,1,0,0,1,1), and
c:=(1,0,1,0,1,0,1).

Message     Encoder      Channel       Decoder     Message

1001       1001100      noise        1000100        ?

a u  1
Error at 100 = digit 4. Decode to
b u  0      1001100 and message equals 1001
c u  0                                                     31
syndrome
Decodes without having to derive decoding vectors.
In addition to decoding Hamming [7,4] it can decode other
codes
More feasible than a list of messages

32
syndrome
The cosets of C are determined by {a  c | c  C}

(i) Every coset of C has the same number of elements
as C does
(ii) Any two cosets are either disjoint or identical
(iii) V is the union of all cosets of C
(iv) C has q nk   cosets

33
syndrome
A Coset leader is the vector with the minimum weight in
the coset.
The parity check matrix is found by solving the generator
matrix for GH T  0

34
syndrome
The first step is to create a list of syndromes
corresponding the coset leaders. The syndrome of each
vector y is found by syn( y)  Hy T

When a code is received, the syndrome is computed
and compared to the list of syndromes. Let the coset
leader to the syndrome by e. Finally the code is
decoded to x = y – e.

35
Syndrome example
1   0 0 0 0 1 1             0 1 1 1 1 0 0 
0   1 0 0 1 0 1
G                        H  1 0 1 1 0 1 0
              
0   0 1 0 1 1 0
                            1 1 0 1 0 0 1
              
0   0 0 1 1 1 1

Note that G=(I | A) and H = ( AT | I).

36
Syndrome example
Let x:= 1001100 be the original message

Message    Encoder     Channel       Decoder   Message

1001      1001100      noise       1000100     ?

Compute the syndrome of the received code syn( y)  Hy T
0 1 1 1 1 0 0  1       1
 
H  1 0 1 1 0 1 0 0
                        1

1 1 0 1 0 0 1 0
                       1

0 
1 
 
0 
1                                37
 
Conclusion
A code of minimum weight d is called perfect if all the
vectors in V are contained in the sphere of radius t =
[(d – 1)/2] about the code-word.

The Hamming [7,4] code has eight vectors of sphere