CODES FOR HIGH-SPEED MEMORIES III: BIT / BYTE ERROR CONTROL CODES in .NET

Implement QR Code JIS X 0510 in .NET CODES FOR HIGH-SPEED MEMORIES III: BIT / BYTE ERROR CONTROL CODES
CODES FOR HIGH-SPEED MEMORIES III: BIT / BYTE ERROR CONTROL CODES
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Step 2. Let Fq denote a column vector having R=2 binary elements. The weight of Fq must be even for an odd weight of G and must be odd for an even weight of G. The number of Fq s is 2R=2 1 , where q 0; 1; . . . ; 2R=2 1 1. Step 3. Two column vectors Fi and Fj are selected arbitrarily from the Fq s, where 0 i 6 j 2R=2 1 1. Using Fi and Fj , de ne the following R 4 matrix: " Hi; j G Fi Fj Fi G Fi Fj Fj Fi G Fi Fj Fj G Fi Fj # :
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Here the plus sign represents the mod-2 sum of the column vectors. Step 4. Construct the H matrix as follows: p 2R=2 1 1 H H0;1 H0;2 H0;3 . . . Hi; j . . . Hp 1;p ; Construction 2 De ne
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Let W be a row vector of N 0 s followed by N 1 s, where N n b. ! W ; HR HR
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HR 1
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where HR is the parity-check matrix of the code C R; b with code length n bytes and check-bit length R. Then the code C R 1; b de ned by the parity-check matrix HR 1 is an SEC-DED-SbED code with a code length of 2n bytes. Here C R; b denotes an SECDED-SbED code with R check bits and b bits in a byte. This is the same code extension method as that shown in Theorem 5.8. Theorem 6.10 The code designed by construction 1 is an odd-weight-column SECDED-S4ED code with an even number of check bits larger than 3 and a code length in bytes n 2R 3 2R=2 2 . Proof Obviously, the column vectors in H shown in Eq. (6.7) have odd weight. Now let us assume that Fi has odd weight. In this case G must have even weight. When Fi has even weight, the proof is the same, as will be shown. 1. Proof of single-error correction capability. The weight of Fi is odd and the weight of G Fi Fj is even. Then we assume that the following equations are satis ed for the column vectors: Fi Fi0 ; G Fi Fj G Fi0 Fj0 : If Fi Fi0 , then by construction 1, Fj should not be equal to Fj0 . However, we can derive Fi Fi0 and Fj Fj0 from the equations above, and this is a contradiction. Therefore the column vectors in H shown in Eq. (6.7) are distinct. Consequently the codes are SEC-DED codes. 2. Proof of single 4-bit byte error detection capability. Double-bit errors within each byte can be detected through the DED capability. We should verify the error detection capability of triple-bit errors and quadruple-bit errors within a byte.
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SINGLE-BYTE / BURST ERROR DETECTING SEC-DED CODES
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a. Triple-bit errors. In the H shown in Eq. (6.7), replace arbitrarily three column vectors in the byte corresponding to the erroneous triple-bit errors in this byte, and then do a mod-2 sum of these vectors. The resultant column vector should be equal to the syndrome caused by the triple-bit errors. It can be easily shown that either the lower-half R=2 bits or the upper-half R=2 bits in the syndrome are equal to G. From construction 1, Fi is not equal to Fj . Therefore G Fi Fj cannot be equal to G. So triple-bit errors within a byte can be detected. b. Quadruple-bit errors. In the H shown in Eq. (6.7), take a mod-2 sum of the four column vectors in a byte corresponding to the erroneous quadruple-bit errors in this byte. Obviously, the resultant column vector has even weight and is not equal to the all-0 vector. Thus the quadruple-bit errors within a byte can be detected. Note that the number of Fq s is 2R=2 1 . In addition the number of Hi; j s is the combination number of two elements from 2R=2 1 elements. Hence the code length in Q.E.D. bytes is equal to n 2R=2 2 2R=2 1 1 2R 3 2R=2 2 . If an odd number of check bits is required, construction 2 can be applied to the code in Theorem 6.10. The codes obtained have the code length of 2R 2 2R=2 1 bytes for R ( odd) check bits. Example 6.6 [KANE84]
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For b 4 and R 8, let G be an all-1 column vector: 2 3 1 617 G 6 7: 415 1 This G is even weight; therefore, Fq must be odd weight. There are eight Fq s: 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 1 1 1 0 1 1 0 1 1 0 1 1 0 1 1 1
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and   these Fq s, two columns are chosen. Then the Hij can be constructed. There among 8 are 28 Hij s. Examples for H01 and H02 are indicated as follows: 2 2 1 1 0 1 0 0 0 0 1 0 1 0 1 1 0 0 0 0 3 2 1 1 0 1 0 0 1 0 0 0 0 3
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