C. Choffrut, J. Karhum/iki, N.Ollinger, The commutation of finite sets: a challenging problem, Theoret. Cornput. Sci., 273 (1-2): 69-79, 2002.  Home/Search   Document Details and Download   Summary   Related Articles   Check

....L and regular codes R to show that it is decidable whether L can be expressed as L = i2I R i for some set I N . In contrast to this result, we show that the above question is undecidable for context free languages and two element codes. From this later result it follows, by a result of [ChKO99], that if K is a given two element code, then it is undecidable whether a given context free language L commutes with K , that is, whether the language equation LK = KL holds true. We also present several open problems that involve the mentioned questions on semilinearity, language equivalence ....

....Problem 4 is decidable for deterministic finite turn 2CAs. We shall now show that Problem 4 has a negative answer for context free languages. Indeed, we prove that the commutation problem is undecidable for K = fa; bg and context free languages L. For the proof we need the following lemma from [ChKO99]. Lemma 6.2. Let K be a two element code and L be any language. Then LK = KL if and only if there exists a subset I N such that L = i2I K i : 1) 12 We note that if (1) holds for a code K and a language L, then the set I is uniquely determined, since K i K j = for all i 6= j. ....

C. Choffrut, J. Karhumaki, and N. Ollinger. The commutation of finite sets: challenging problem. Theoret. Comput. Sci., to appear; TUCS Technical Report 303, http://www.tucs.fi, 1999.

....of fact, one can define similarly as above a notion of a semigroup centralizer of a set L. Indeed, observe that for any set of words L, there is a unique maximal element of the set S C E ILS = SL . We call this element the semigroup centralizer of L and we denote it as es (L) We refer to [13] for details. Clearly, for any L, we have es(L) U 1 C e(L) However, this is a strict inclusion in general: it is not true that for any L, es(L) and e(L) coincide modulo the empty word. The reason for this is that e(L) 1 does not commute with L for all sets of words L. Example 4.1.2. i) ....

....shown by the following result. Theorem 4.1.3 ( 27] Let C be a fixed two element code. It is undecidable for context free languages L whether or not CL LC. Several very different approaches have been taken to attack Conway s Problem: combinatorial properties of finite and infinite words ([13], 53] 66] equations on languages ( 37] 38] algebraic results on the commutation of formal power series ( 29] the fixed point approach ( 18] 36] and the branching point approach ( 40] We discuss in this chapter some of these approaches and their outcome up to date. We prove that ....

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C. Choffrut, J. Karhum/iki, N.Ollinger, The commutation of finite sets: a challenging problem, Theoret. Cornput. Sci., 273 (1-2): 69-79, 2002.

....if they are powers of a common word. In the monoid of languages over a finite alphabet, even if the languages are only finite, the situation changes drastically: it is not likely that a complete solution can be found. This view is supported by the fact that the equation have solutions like, cf. [3], X = a ab ba bb and Y = X X bab bbb Part of this work was done while the author visited LIAFA at Denis Diderot University. Supported also by the grant 44087 of the Academy of Finland. 1 X = a aa aaa ab aba b ba and Y = X n faag: Moreover, from Theorem 3 it follows ....

.... and hence Problem 1 is equivalent to a question whether the centralizer is recursively enumerable, cf. 11] As positive results we have: Theorem 1 Conway s Problem is known to have an affirmative answer in the following cases: i) X is a rational prefix code, cf. 14] ii) card(X) 2, cf. [3]; iii) card(X) 3, cf. 11] iv) X is a finite code which is elementary, synchronizing or contains a word of length 1, cf. 3] v) X is a rational code, cf. 8] The proofs of these results use quite different techniques. Historically the oldest result is that of (i) Its proof is ....

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C. Choffrut, J. Karhumaki and N. Ollinger, The commutation of finite sets: a challenging problem, Theoret. Comput. Sci., to appear.

.... due to Bergman [1] Two polynomials p(x) and q(x) commute if and only if they are linear combinations of powers of a common polynomial t(x) A similar result holds also for noncommutative formal power series, see [5] Recently, both of the above problems have been solved for two element sets in [4]. In this case, Conway s problem has an affirmative answer and moreover, the binary sets possess a Bergman type of characterization: Any set commuting with a two element set X is a union of powers of X (or just a union of powers of a primitive word t, if X t , for some word t) On the other ....

.... has an affirmative answer and moreover, the binary sets possess a Bergman type of characterization: Any set commuting with a two element set X is a union of powers of X (or just a union of powers of a primitive word t, if X t , for some word t) On the other hand, as was pointed out also in [4], no similar characterization can be achieved for four element sets, in general. These problems were also considered in the case of codes, in [13] They have been completely and affirmatively answered if X is a prefix code, i.e. no word is a prefix of another. Moreover, it was proved that for a ....

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C. Choffrut, J. Karhumaki, N. Ollinger, The commutation of finite sets: a challenging problem, TUCS Technical Report 303, http://www.tucs.fi/, 1999, to appear in a special issue of Theoret. Comput. Sci. on Words.

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