Global Journal of Science Frontier Research, F: Mathematics and Decision Science, Volume 21 Issue 4

Maximum Distance Separable Codes to Order The first matrix takes the first 6 rows of F 12 ; the second matrix takes rows { e 1 , e 6 , e 11 , e 4 , e 9 , e 2 } which are 6 rows in sequence with arithmetic difference 5, gcd(12 , 5) = 1, starting with the second row. These are generator matrices for (12 , 6 , 7) codes over GF (13) = Z 13 and each can correct 3 errors. Efficient decoding algorithms for the codes are established in [6]. Here is an example to show how the algorithms work in practice. The matrix K as above, formed from the first 6 rows of Fourier matrix F 12 , is the generator matrix of a (12 , 6 , 7) code. Apply Algorithm 6.1 from [6] to correct up to 3 errors of the code as follows. Note the work is done in Z 13 = GF (13) using modular arithmetic. 1. The word w = (8 , 9 , 2 , 6 , 3 , 3 , 10 , 8 , 4 , 1 , 5 , 7) is received. 2. Apply check matrix to w and get e = (2 , 9 , 12 , 10 , 11 , 11). Thus there are errors and w is not a codeword. (The check matrix ( e 1 T , e 2 T , e 3 T , e 4 T , e 5 T , e 6 T ) is immediate, see section 2.) 3. Find a non-zero element of the kernel of   2 9 12 10 9 12 10 11 12 10 11 11   . This is a 3 × 4 Hankel matrix, formed from e ; the first row consists of elements (1 − 4) of e , the second row consists of elements (2 − 5) of e , and the third row consists of elements (3 − 6) of e . A non-zero element of the kernel is x = (7 , 1 , 7 , 1) T . 4. Now a = ( e 1 , e 2 , e 3 , e 4 ) ∗ x = (3 , 12 , 7 , 0 , 1 , 0 , 1 , 2 , 4 , 0 , 10 , 12). Thus errors occur at 4 th , 6 th , 10 th positions (which are the positions of the zeros of a ). 5. Solve (from 4 th , 6 th , 10 th columns of (2 − 7) rows of F 12 ):   8 6 5 12 10 12 5 8 8 1 9 1 8 2 5 12 12 12     x 1 x 2 x 3   =   2 9 12 10 11 11   In fact only the first three equations need be solved; answer is (10 , 1 , 4) T . Thus error vector is k = (0 , 0 , 0 , 10 , 0 , 1 , 0 , 0 , 0 , 4 , 0 , 0). 6. Correct codeword is c = w − k = (8 , 9 , 2 , 9 , 3 , 2 , 10 , 8 , 4 , 10 , 5 , 7). 7. If required, the original data word can be obtained directly by multiplying by the right inverse of the generator matrix; the right inverse is read off as K = ( e 0 , e 11 , e 10 , e 9 , e 8 , e 7 ) T ∗ 12. Then c ∗ K = (1 , 2 , 3 , 4 , 5 , 6) which is the original data word to be safely transmitted. The equations to be solved are Hankel matrices of size the order of t × t where t is the error-correcting capability. 2 3 − 1 = 7 , 2 4 − 1 = 15 , 2 5 − 1 = 31 , 2 6 − 1 = 63 , . . . . In general consider the characteristic 2 field GF (2 q ). In this field acquire an element of order n = (2 q − 1) and construct the Fourier n × n matrix over GF (2 q ). From this, MDS ( n, r ) codes are constructed for 1 ≤ r ≤ n . It is better to take odd r from consideration of the error-correcting capability. 1. 2 3 − 1 = 7. From the Fourier 7 × 7 matrix over GF (2 3 ) construct the MDS { (7 , 5 , 3) , (7 , 3 , 5) , (7 , 1 , 7) } codes which can correct respectively { 1 , 2 , 3 } errors. 2. 2 4 − 1 = 15. From the Fourier 15 × 15 matrix over GF (2 4 ) construct { (15 , 13 , 3) , (15 , 11 , 5) , (15 , 9 , 7) , (15 , 7 , 9) , (15 , 5 , 11) , (15 , 3 , 13) , (15 , 1 , 15) } MDS codes over GF (2 4 ) which can correct respectively { 1 , 2 , 3 , 4 , 5 , 6 , 7 } errors. © 2021 Global Journals 1 Global Journal of Science Frontier Research Volume XXI Issue IV Year 2021 8 ( F ) Version I i. Correcting errors sample g) Length (2 q − 1) MDS codes in GF(2 q ) 6. Ted Hurley and Donny Hurley, “ Coding theory: the unit-derived methodology ” , Int. J. Information and Coding Theory, Vol. 5, no.1, 55-80, 2018. R ef