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1
01010000001001
by Colin Kriwox
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Contents
Introductioncredit card error checkingwhat is a codepurpose of error-correction codes
Encodingnaïve approachhamming codes
Minimum Weight Theoremdefinitionsproof of single error-correction
Decodinglist all possible messagesusing vectorssyndrome
Conclusionperfect 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
d1
d2 d3 … d15 d16
n = 3
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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 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.
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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 Encoder Channel Decoder Message
10 101010 noise 001010 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:
1111000
0110100
1010010
1100001
G
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Hamming [7,4] Codes
1624 1000011010010100101100001111110011010101011001100011001101010100011001110100110010101111111011110000110010000000
12827
Codes
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.
e.g. d(G)= 3
1111000
0110100
1010010
1100001
G
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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
),(),(),( wvdistvudistwudist
),(),( uvdistvudist
0),( uudist
For any u, v, and w in a space V, the following three conditions hold:
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Definitions
The sphere of radius r about a vector u is defined as:
}),(|{)( rvudistVvuSr
e.g. u=(1,0,0,0,0,1,1) (0,0,0,0,0,1,1)
(1,1,0,0,0,1,1)
(1,0,0,0,0,0,1)
(1,0,0,0,0,0,1)
(1,0,1,0,0,1,1)
(1,0,0,1,0,1,1) (1,0,0,0,1,1,1)
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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 )()( wSuSv tt
u wv
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Proof
By triangle inequality
twvdistvudistwudist 2),(),(),(
u wv
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Proof
12 dt12 dtdist(u,w)dist(u,v)dist(u,w)
Since spheres of radius t = [(d – 1)/2] so and this gives
But since
We have a contradiction. Showing the sphere of radius t about codes are disjoint.
dwuwtwudist )(),(
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Result of Theorem
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.
1111000
0110100
1010010
1100001
G
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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 1000011 0100101 0010110 … 0000011 0000101 0000110
Other 1000001 0100111 0010100Received 0010011 0001101 1010110Words 1100011 1100101 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 1000011 0100101 0010110 … 0000011 0000101 0000110
Other 1000001 0100111 0010100Received 0010011 0001101 1010110Words 1100011 1100101 0110110
… … …
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Vector Decoding
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).
1111000
0110100
1010010
1100001
G
),...,(: 71 xxx ),...,(: 71 yyy
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1
)2mod(i
ii yxIf then inner product =
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Vector Decoding
cubuau ,,
1ua0ub
Correct errors by taking inner product of received vector u by a, b, c. We get
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 ?
0uc
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 }|{ Ccca
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
(iv) C has cosetsknq
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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 0TGH
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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 THyysyn )(
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
1111000
0110100
1010010
1100001
G
1001011
0101101
0011110
H
Note that G=(I | A) and H = ( | I). TA
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Syndrome example
Let x:= 1001100 be the original message
Message Encoder Channel Decoder Message
1001 1001100 noise 1000100 ?
Compute the syndrome of the received codeTHyysyn )(
1001011
0101101
0011110
H
1
0
1
0
0
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.