MAP D in .NET Generating QR Code 2d barcode in .NET MAP D

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MAP D using barcode printer for .net vs 2010 control to generate, create qr-code image in .net vs 2010 applications. EAN-14 Consider block-wise maxim Visual Studio .NET qr bidimensional barcode um a posteriori (MAP) decoding; i.e.

, the decoding rule is ( . ) xMAP (y) = argmaxx C pX Y (x y)..

T T Write the de ning equ ation Hx T = 0T in the form HE xE + HE xE = 0T , which, since we are working over F2 , is equivalent to. ( . ). T T HE xE = HE xE . T Note that sT = HE xE , .net vs 2010 qr-codes the right-hand side of ( . ), is known to the receiver since T xE = yE .

Consider the equation HE xE = sT . Since, by assumption, the transmitted word is a valid codeword, we know that this equation has at least one solution. In particular, rank(HE ) E .

If rank(HE ) = E , then block-wise MAP decoding can T be accomplished by solving HE xE = sT . On the other hand, there are multiple solutions (i.e.

, the MAP decoder is not able to recover the codeword uniquely) if and only if rank(HE ) < E . More formally, let T T X MAP (y) = x C HE xE = HE yE ; xE = yE , . i.e., X MAP (y) is the se t of all codewords compatible with the received word y.

Since the prior is uniform, ( . ) becomes xMAP (y) = argmaxx C pY. X (y Now, for any codeword x, if xE yE , then pY X (y x) = 0 and if xE = yE , then n E E MAP pY X (y x) = (1 ) . us, all elements of X (y) are equally likely and the transmitted vector x is either uniquely determined by y or there are multiple solutions. erefore, we say xMAP (y) = x X MAP (y), if X MAP (y) = 1, , otherwise.

e correct solution x is an. We remark that a (erasure) is not the same as an error. element of X MAP (y). . . .B MAP D Y ( . Consider now the bit-wise MAP decoder that uses the decoding rule ( . ) MAP xi (y) = argmax If i E , when can xi be r ecovered Intuitively, we expect that xi can be recovered if and only if all elements of X MAP (y) have the same value for the i-th bit. is is in fact correct. More speci cally, we claim that xi can not be recovered if and only if H i is an element of the space spanned by the columns of HE i .

is is equivalent to the statement that HwT = 0T has a solution with wi = 1 and wE = 0. Now, if there is such a solution then for every element x of X MAP (y) we also have x +w X MAP (y). It follows that exactly half the elements x X MAP (y) have xi = 0 and half have xi = 1.

Conversely, if we can nd two elements x and x of X MAP (y) with xi xi , then w = x + x solves HwT = 0T and has wi = 1 and wE = 0. Proceeding formally, we get MAP xi (y) = argmax = argmax . Y ( . y) = argmax 0,1 x 0,1. pX Y (x y). xi = pX Y (x y) =. x C xi = , if x X MAP (y), xi = , , otherwise. We conclude that optimal QR-Code for .NET (block or bit) decoding for the BEC can be accomplished in complexity at most O(n3 ) by solving a linear system of equations (e.g.

, by Gaussian elimination). Further, we have a characterization of decoding failures of a MAP decoder for both the block and the bit erasure case in terms of rank conditions. E .

(P H(n, k)). In Problem . you are asked to show that the average block erasure probability of Gallager s parity-check ensemble (see De nition .

) satis es EH(n,k) [PB (H, )]. n k e=0 n n e n e e n+k n e n e ( ) 2 + ( ) e e=n k+1 e n k e=0 = 2k n ( )n 2 .
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