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Proof: The uniqueness of the factorizations of conditions (i) and (ii) is trivial. To prove the existence of such a factorization, let be the longest prefix of such that , where is the corresponding suffix with . We denote by the first symbol of . The two possible values for are and . One has . Since is maximal, one has , but the rhythmic oddity property implies , thus . - If
, being even implies that is even, thus equal to zero, and so that condition (i) is satisfied by the pair and condition (ii) by the pair . - If
, being even implies that is odd, thus equal to one. This gives two remaining cases. - If
, the equality implies that so that condition (i) of the property is satisfied by the pair . - If
, the factorization is such that , so that condition (ii) of the property is satisfied by the pair .![[]]() Proposition 1 implies that (with being even) so that and , thus expressing the »half minus one / half plus one« characterization of these patterns given in Arom (1991). We introduce the notion of asymmetric pair that will be the key of our construction. Let , be words over . We say that is an asymmetric pairs if no pair of prefixes of and respectively exist such that . For instance, is an asymmetric pair, but is not. Proof: Let or with . The existence of a pair of prefixes of and such that and with is equivalent to the existence of a cyclic shift of (whatever being the
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