Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 3 3G Solution Created 2026-09-24 Updated 2026-10-05
Identify a binary word with its coefficient polynomial in . A cyclic code is exactly an ideal of : closure under multiplication by is closure under cyclic shift. Its inverse image in the principal ideal domain has a unique monic generator polynomial of a cyclic code dividing . PutThe cubics have no root in , hence are irreducible polynomials. The eight monic divisors yield all binary cyclic codes of length seven. A code generated by has basis and dimension , with the zero code treated separately.
For the dual of a cyclic code, let . Orthogonality of coefficient vectors to all cyclic shifts gives the generator . Here . Thus the complete pairing, with the respective dimensions, isThese are respectively the whole-space/zero, even-weight/repetition, and two Hamming code/binary simplex code pairs. The reciprocal formula follows because the product of a word polynomial and the reversed polynomial of an orthogonal word has zero coefficients in all cyclic positions; the resulting inclusion has the correct complementary dimension, hence is equality.
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 4 3I a Solution Created 2026-09-24 Updated 2026-09-25
For the standard dot product on , the dual code isIt is an intersection of kernels of linear functions, and is therefore a linear code.
Let denote cyclic right shift. If and , thenA cyclic code is closed under both and , so and the right-hand side vanishes. Hence , proving directly that the dual of a cyclic code is cyclic.
Identify words with polynomials in . If the generator polynomial of a cyclic code is the monic divisor andthen the generator polynomial of is the monic reciprocalEquivalently, the parity-check polynomial of one code becomes, after reversal, the generator polynomial of its dual.