01010000001001 by Colin Kriwox

1. 01010000001001by Colin Kriwox

2. 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

3. Detect Error On Credit Card

4. 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)

5. Detect Error On Credit Card n = 3 d1 d2 d3 … d15 d16

6. 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

7. 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).

8. 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.

9. The purpose A message can become distorted through a wide range of unpredictable errors. • Humans • Equipment failure • Lighting interference • Scratches in a magnetic tape

10. 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

11. Send a message Message Encoder Channel Decoder Message 10 101010 noise 001010 10

12. Encoding Naïve approach Hamming codes

13. 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

14. 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

15. 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:

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

17. Minimum Weight Theorem Definitions Proof of Theorem

18. 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

19. 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).

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

21. Definitions The sphere of radius r about a vector u is defined as: (0,0,0,0,0,1,1) e.g. u=(1,0,0,0,0,1,1) (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)

22. 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.

23. 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 u v w

24. Proof By triangle inequality v u w

25. Proof 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.

26. 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.

27. Decoding list all possible messages using vectors syndrome

28. 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 Received 0010011 0001101 1010110 Words 1100011 1100101 0110110 … … …

29. 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 Received 0010011 0001101 1010110 Words 1100011 1100101 0110110 … … …

30. 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). If then inner product =

31. Vector Decoding 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 ? Error at 100 = digit 4. Decode to 1001100 and message equals 1001

32. 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

33. syndrome The cosetsof C are determined by • Some facts about cosets: • Every coset of C has the same number of elements as C does • Any two cosets are either disjoint or identical • V is the union of all cosets of C • C has cosets

34. 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

35. syndrome The first step is to create a list of syndromes corresponding the coset leaders. The syndrome of each vector y is found by 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.

36. Syndrome example Note that G=(I | A) and H = ( | I).

37. 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

38. 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.