Abstract Algebra Theory and Applications - Computer Science ...

116 CHAPTER 7 ALGEBRAIC CODING THEORYTable 7.2. Distances between 4-bit codewords0000 0011 0101 0110 1001 1010 1100 11110000 0 2 2 2 2 2 2 40011 2 0 2 2 2 2 4 20101 2 2 0 2 2 4 2 20110 2 2 2 0 4 2 2 21001 2 2 2 4 0 2 2 21010 2 2 4 2 2 0 2 21100 2 4 2 2 2 2 0 21111 4 2 2 2 2 2 2 0change x to y. Therefore, if d min = 2, we have the ability to detect singleerrors. However, suppose that d(x, y) = 2, y is sent, and a noncodeword zis received such thatd(x, z) = d(y, z) = 1.Then the decoder cannot decide between x and y. Even though we areaware that an error has occurred, we do not know what the error is.Suppose d min ≥ 3. Then the maximum-likelihood decoding scheme correctsall single errors. Starting with a codeword x, an error in the transmissionof a single bit gives y with d(x, y) = 1, but d(z, y) ≥ 2 for any othercodeword z ≠ x. If we do not require the correction of errors, then we candetect multiple errors when a code has a minimum distance that is greaterthan 3.Theorem 7.3 Let C be a code with d min = 2n + 1. Then C can correctany n or fewer errors. Furthermore, any 2n or fewer errors can be detectedin C.Proof. Suppose that a codeword x is sent and the word y is received withat most n errors. Then d(x, y) ≤ n. If z is any codeword other than x, then2n + 1 ≤ d(x, z) ≤ d(x, y) + d(y, z) ≤ n + d(y, z).Hence, d(y, z) ≥ n + 1 and y will be correctly decoded as x. Now supposethat x is transmitted and y is received and that at least one error hasoccurred, but not more than 2n errors. Then 1 ≤ d(x, y) ≤ 2n. Since theminimum distance between codewords is 2n + 1, y cannot be a codeword.Consequently, the code can detect between 1 and 2n errors.□