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Error Detection and Correction - PowerPoint - PowerPoint

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  • pg 1
									01010000001001
by Colin Kriwox

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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
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Detect Error On Credit Card

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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)

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Detect Error On Credit Card

n=3

d1 d2 d3 … d15 d16
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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

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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 start with an account number that is 15 digits long and 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).
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What is a code?
A code is defined as an n-tuple of q elements. Where q is any alphabet.

Ex. 1001 Ex. 2389047298738904 Ex. (a,b,c,d,e)

n=4, q={1,0} n=16, q={0,1,2,3,4,5,6,7,8,9} 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
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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

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Send a message

Message 10

Encoder 101010

Channel noise

Decoder 001010

Message 10

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Encoding
Naïve approach
Hamming codes

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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
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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

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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 G 0  0

0 0 0 0 1 1 1 0 0 1 0 1  0 1 0 1 1 0  0 0 1 1 1 1

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Hamming [7,4] Codes
1000011 0100101 0010110 0001111 1100110 1010101 1001100 0110011 0101010 0011001 1101001 1001010 1111111 0111100 0011001 0000000

2  16
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Codes

2  128
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Possible codes

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Minimum Weight Theorem
Definitions

Proof of Theorem

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

e.g.

d(G)= 3
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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).

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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)

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Definitions
The sphere of radius r about a vector u is defined as:

S r (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)
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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.

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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  S t (u )  S t ( w)

u

v

w

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Proof
By triangle inequality

dist (u, w)  dist (u, v)  dist (v, w)  2t

v u w

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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

We have a contradiction. Showing the sphere of radius t about codes are disjoint.

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

Since d(G) = 3 then for t = [(3 – 1)/2] = 1 or fewer errors, the received code is in a disjoint sphere about a unique code word.

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Decoding
list all possible messages

using vectors
syndrome

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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 Other Received Words 1000011 0000011 1000001 0010011 1100011 … 0100101 0000101 0100111 0001101 1100101 … 0010110 … 0000110 0010100 1010110 0110110 …

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

Code words Other Received Words

1000011 0000011 1000001 0010011 1100011 …

0100101 0000101 0100111 0001101 1100101 …

0010110 … 0000110 0010100 1010110 0110110 …

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Vector Decoding
1 0 G 0  0 0 0 0 0 1 1 1 0 0 1 0 1  0 1 0 1 1 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).
If x : ( x1 ,..., x7 )
y : ( y1 ,..., y 7 ) then inner product =

x y
i 1 i

7

i

mod( 2)

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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 1001 Encoder 1001100 Channel noise Decoder 1000100 Message ?

a u  1 b u  0 c u  0

Error at 100 = digit 4. Decode to 1001100 and message equals 1001
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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

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syndrome
The cosets of C are determined by {a  c | c  C} Some facts about cosets: (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

q nk (iv) C has

cosets
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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

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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.

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

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

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

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Syndrome example
Let x:= 1001100 be the original message
Message 1001 Encoder 1001100 Channel noise Decoder 1000100 Message ?

syn ( y )  Hy T Compute the syndrome of the received code
0 1 1 1 1 0 0   1    H  1 0 1 1 0 1 0  0    1 1 0 1 0 0 1  0     
0  1    0  1   

1  1  1 

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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 of radius one about each code-word, times sixteen unique codes. Therefore, the Hamming [7,4] code with minimum weight 3 is perfect since all the vectors (128) are contained in the sphere of radius 1.

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