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prelim2_sol

# prelim2_sol - Brief solutions to Prelim 2 1 A binary code...

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Brief solutions to Prelim 2 1. A binary code of dimension k has 2 k codewords. So the answer is 2. 2. The multiplicative inverse of x 3 + x + 1 mod x 4 + x 3 + 1 in F 2 [ x ] is x 3 + x (by the Euclidean algorithm, for instance). 3. Computing powers of 2 mod 17 we find that H consists of the 8 elements ± 1 , ± 2 , ± 4 , ± 8. So the index of H is 16 / 8 = 2. The two cosets are H and 3 H = 3 , ± 6 , ± 5 , ± 7 } . 4. (a) Since G is in the standard form G = ( I | A ), we know that H := ( - A T | I ) is a parity check matrix: H = 2 2 1 1 0 1 0 2 0 1 . This has no zero column, but it does have two columns that are scalar multiples of one another. So d = 2. (b) Using the parity check matrix in (a), we can check that the given vector x satisfies H x T 6 = 0 . So it is not a codeword. Alternatively, just check directly that x is not a linear combination of the 3 rows of G . [This involves a system of 5 equations in 3 unknowns, which one checks to be inconsistent.] 5. (a) φ (51) = (3 - 1)(17 - 1) = 32. The decoding exponent is the multiplicative inverse of 11 mod 32, which is 3.
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