2005
DOI: 10.1016/j.tcs.2005.03.022
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An enhanced property of factorizing codes

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Cited by 6 publications
(25 citation statements)
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“…In this regard, it could be interesting to look for a generalization of the constructions given in Sections 7,8. Another related question is to find conditions under which a set of words C 1 satisfying C 1 = a I (a−1)S 1 +P 1 (a−1)a J +a I ba J , where (I, J) is a Krasner pair and P 1 , S 1 are polynomials with coefficients 0, 1, could be embedded in a factorizing code. Some sufficient conditions have been stated in [10,12,14,15].…”
Section: Discussionmentioning
confidence: 99%
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“…In this regard, it could be interesting to look for a generalization of the constructions given in Sections 7,8. Another related question is to find conditions under which a set of words C 1 satisfying C 1 = a I (a−1)S 1 +P 1 (a−1)a J +a I ba J , where (I, J) is a Krasner pair and P 1 , S 1 are polynomials with coefficients 0, 1, could be embedded in a factorizing code. Some sufficient conditions have been stated in [10,12,14,15].…”
Section: Discussionmentioning
confidence: 99%
“…Another objective is the description of the structure of the (positively) factorizing codes, i.e., codes satisfying the factorization conjecture. There are several papers devoted to this problem [8,9,10,11,12,13,14,15,24]. In particular, the structure of m-codes, m ≤ 3, has been characterized, as well as that of codes C such that C = P (A − 1)S + 1, with P ⊆ A * , S ⊆ a * .…”
Section: Introductionmentioning
confidence: 99%
“…Since S , S S ⊆ a * ∪ a * ba * and S ⊆ a * , we must have S ⊆ a * . Furthermore, if a h , a k ∈ S with h < k, looking at equation (8), we have S S ∩a * ba * = a M ba 19 and so, for each a i ba j ∈ S , we have 19 = j + h < j + k = 19, which is a contradiction. Consequently, we have |S| = 1 which, together with 1 ∈ S, yields S = 1.…”
Section: Remark 34 C Is a Factorizing Completion Formentioning
confidence: 94%
“…(I, J) will be called the Krasner companion factorization of (R, T ) with respect to the chain of divisors of n given in equation (2). It is easy to prove that all the Krasner companion factorizations of a given Hajós factorization (R, T ) are exactly the Krasner companion factorizations of (R, T ) obtained in this way (i.e., (I, J) is a Krasner companion factorization of (R, T ) if and only if there exists a chain C of divisors of n defining (R, T ) such that (I, J) is the Krasner companion factorizations of (R, T ) with respect to C) [19].…”
Section: Recursive Constructions Of Hajós Factorizationsmentioning
confidence: 98%
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