Suppose the receiver receives 101 it can tell theres

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Unformatted text preview: essage is either a member of S (no errors), or is a member of some particular Ewi (one error), in which case the receiver can deduce the original message was wi . Here’s another simple example: let S = {000, 111}. So E000 = {001, 010, 100} and E111 = {110, 101, 011} (note that E000 doesn’t intersect E111 ). Suppose the receiver receives 101. It can tell there’s been a single error since 101 ∈ S . Moreover it can deduce that the original message was 111 since 101 ∈ E111 . / We can formally state some properties from the above discussion, and state what the error-correcting power of a code whose minimum Hamming distance is at least D. Theorem 6.2 The Hamming distance between n-bit words satisfies the triangle inequality. That is, HD(x, y ) + HD(y, z ) ≥ HD(x, z ). Theorem 6.3 For a BSC error model with bit error probability < 1/2, the maximum likelihood decoding strategy is to map any received word to the valid code word with smallest Hamming distance from the received one (ties may be broken arbitrarily). Theorem 6.4 A code with a minimum Hamming distance of D can correct any error pattern of 7 SECTION 6.4. LINEAR BLOCK CODES AND PARITY CALCULATIONS − − ￿ D2 1 ￿ or fewer errors. Moreover, there is at least one error pattern with ￿ D2 1 ...
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This document was uploaded on 02/26/2014 for the course CS 6.02 at MIT.

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