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Unformatted text preview: Stony Broo University MAT 312/AMS 351 Fall 2010 Notes on Hamming Codes 1. Errordetecting matrices. Suppose a group code is generated by a matrix G = 1 0 0 0 1 0 0 0 1 0 0 1 1 1 0 0 1 0 0 1 1 0 0 0 1 0 0 1 . We can write this matrix as G = ( I 4  A ) where A is the 4 3 matrix encoding the check bits. Let us construct another matrix using A : H = parenleftBigg A I 3 parenrightBigg . In this case, H = 1 0 0 1 1 1 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1 . In general if our code is f : B m B n , then G is of the form G = ( I m  A ) and H of the form H = parenleftBigg A I n m parenrightBigg . Remark: The product GH is a 4 3 (in general, m ( n m )) matrix of zeroes. What happens is that during the multiplication each element in A gets added to itself, giving 0. Schematically GH = ( I m  A ) parenleftBigg A I n m parenrightBigg = I m A + AI n m = A + A = 1 where here represents the m ( n m ) matrix of zeroes. This means that H , applied on the right to any one of the codewords, must give the length n m 0vector: (0 ,..., 0), since a codeword is of the form wG for some w B m , and ( wG ) H...
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 Fall '10
 AnthonyPhillips

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