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RS] p. 14). However the latter was given two different solutions some twelve years ago (cf. [CF] and [ERI]) while the former remains open. In fact, when considering the number of iterations on which two non equivalent DOL-sequences agree there exists a huge gap between the double exponential of the so far best upper bound (cf. [ER2]) and the linear lower bound (cf. [Ni]). Yet, in the case of binary alphabets a sharp result can be stated since it is known that exactly 4 iterations are sufficient for deciding that two DOL-sequences are equivalent (cf.

The L ambiguity sets L1 (h,B) and Amb 1 (h,B) are now defined as 27 follows: L 1 (h,B) and {(b1 ,bi)··· (bm,b~) (bm+l'~)'" (bm+n'~) I bi,bj e B, h(b 1 ; ••. • ,b~ e B with (b 1 , ••• ,bn ) ¢ (bi, ••• ,b~), h(b 1 ;···;bn ) = h(bi;···;b~)}. We show that L1 (h,B) and Amb 1 (h,B) are effectively regular, if every word of B contains a growing letter and h is nonerasing. If some word of B does not contain a growing letter, the languages L1 (h,B) and Amb 1 (h,B) are not, in general, regular. This is partly due to the possible presence of arbitrarily long overflows consisting of bounded letters only.

Otherwise, we have bp(u) P (u), u'(p(u) 1\ p(v» or -1 bp(u) = u'[v' b']p(v) according to whether p(u) 1\ p(v) '" p(v) or p(u) 1\ p(v) = p(v). -1 • Here v, b' belongs to Y • Suppose now that the length of u is at least 3N. Then the length of u' is greater than N. By Lemmas 10 and 9 the suffix of u' of length N contains a growing letter. Hence, if u contains 3N consecutive bounded letters, so does u'. Therefore the claim follows inductively. c Lemma 12. If no letter of Y is semibounded, the state set of A2 (p,B) is finite.

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