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GF(16) Double-Error Correcting Code: Developing and Error Detection - Prof. Julie M. Clark, Assignments of Mathematics

From a university course on applied algebra: codes & ciphers in spring 2009. It focuses on developing a double-error correcting code using gf(16) and detecting errors using syndromes. Addition in gf(16), using maple to find syndromes, and homework problems.

Typology: Assignments

Pre 2010

Uploaded on 08/18/2009

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koofers-user-i1z 🇺🇸

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Download GF(16) Double-Error Correcting Code: Developing and Error Detection - Prof. Julie M. Clark and more Assignments Mathematics in PDF only on Docsity! Math 350: Applied Algebra: Codes & Ciphers Spring 2009 Developing a double-error correcting Code GF(16) (1100) (1010)  (1011) (1111) (1101)   (0110) (0100)  1 (0101)  (0011) (1110)  5(0011)  Addition in GF(16): + α 1 α 2 α 3 α 4 α 5 α 6 α7 α 8 α 9 α 10 α 11 α 12 α 13 α 14 α 0 α 1 0 α 10 α 4 α 2 0 α 10 α 3 0 α 12 1 α 4 0 α 5 0 α 14 α 6 0 1 α 7 0 α 2 α 9 α 8 0 α 7 α 9 0 α 13 α 10 0 α 11 0 α 12 0 α 13 0 α 14 0 α 3 α 0 0 Page 1 Power Notation Binary Notation 0 0000 α 0 0001 α 1 0010 α 2 0100 α 3 1000 α 4 0011 α 5 0110 α 6 1100 α 7 1011 α 8 0101 α 9 1010 α 10 0111 α 11 1110 α 12 1111 α 13 1101 α 14 1001 Math 350: Applied Algebra: Codes & Ciphers Spring 2009 H1 = [1,  4, 2, 8, 5, 10, 3, 14, 9, 7, 6, 13, 11, 12] 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 H             H2 = [1, , 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14] 2 0 0 0 1 0 0 1 1 0 1 0 1 1 1 1 0 0 1 0 0 1 1 0 1 0 1 1 1 1 0 0 1 0 0 1 1 0 1 0 1 1 1 1 0 0 1 0 0 0 1 0 0 1 1 0 1 0 1 1 1 H             2 12 13 14 3 3 3 2 3 12 3 13 3 14 3 1 1 ( ) ( ) ( ) ( ) H                   2 3 4 5 6 7 8 9 10 11 12 13 14 3 3 6 3 6 9 12 1 1 H                            3 0 0 0 1 0 0 1 1 0 1 0 1 1 1 1 0 0 1 0 0 1 1 0 1 0 1 1 1 1 0 0 1 0 0 1 1 0 1 0 1 1 1 1 0 0 1 0 0 0 1 0 0 1 1 0 1 0 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 H                                                           Page 2
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