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Problem set 6 for the joint course ece 556/cs 577/math 579 at the university of illinois, fall 2005. The due date is october 4, 2005. Students are required to read chapters 4 and 5 of blahut's 'algebraic codes for data transmission'. The problem set includes four problems related to cyclic codes, such as finding the rightmost column of a generator matrix, intersecting two cyclic codes, and investigating properties of reversible codes.
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University of Illinois Fall 2005
Due: October 4, 8:30 a.m. Reading: Blahut, Algebraic Codes for Data Transmission, Chapters 4 and 5. Reminder: Class will not meet on Thursday September 29. Instead, you are urged to attend the 43rd Annual Allerton Conference on Communication, Control and Computing, Allerton House, Monticello IL, September 28 - 30, 2005. http://www.csl.uiuc.edu/allerton/
This Problem Set contains four problems
∑k i=0 hix
i. Show that the rightmost column of P is (−hk− 1 , −hk− 2 ,... , −h 1 , −h 0 )T^.
C(3)^ = C(1)^ ∩ C(2)^ and C(4)^ = {a + b : a ∈ C(^1 ), b ∈ C(^2 )}
are also linear cyclic codes. Find their generator polynomials. (b) Let Ai denote the number of codewords of Hamming weight i in a linear binary cyclic code. Show that i · Ai is a multiple of n. (Note: In Problem 1(c) of Problem Set 3, you proved that, for any code,
i · Ai us a multiple of n. This result is stronger... ) (c) Show that if the generator polynomial g(x) of a linear binary cyclic code is not divisible by (x − 1), then Ai = An−i for 0 ≤ i ≤ n.
(a) Show that g(x) = xn−kg(x−^1 ). (b) What is the corresponding result for nonbinary codes? (c) Prove that g 0 = ±1. (d) Show that if there is an integer m such that qm^ ≡ −1 mod n, then every cyclic code of length n over GF(q) is reversible.
(a) Show that if c(x) is a codeword, then so is [c(x)]^2 mod (xn^ − 1). (b) Let gcd(n, L) = 1. A decimation by L operation on a sequence ζ of period n results in another sequence ζˆ of period n where ζˆi = ζLi = ζLi mod n. Regarding c(x) as one period of a periodic sequence, show that the codeword corresponding to decimation by 2m−^1 is [c(x)]^2 mod (xn^ − 1), the codeword you found in part (a).