Fully bordered words
aa r X i v : . [ c s . F L ] J un FULLY BORDERED WORDS ˇSTˇEP ´AN HOLUB AND MIKE M ¨ULLER
Abstract.
We characterize binary words that have exactly two unborderedconjugates and show that they can be expressed as a product of two palin-dromes.
It is well known that each primitive word has an unbordered conjugate. Thetypical example of such a conjugate is the Lyndon conjugate, that is, the minimalone in a lexicographic ordering. Using different lexicographic orders, we can obtainseveral unbordered conjugates. In particular, any primitive binary word has at leasttwo unbordered conjugates. This leads to a natural question about the structureof words that do not have any other unbordered conjugate apart from these oblig-atory two. We shall call them fully bordered . In this paper we give an inductivecharacterization of fully bordered words, which shows that they have some kindof fractal structure. One consequence of this characterization is that each fullybordered word is a product of two palindromes. This result can be also interpretedin terms of palindromic length, introduced in [1], saying that fully bordered wordshave the palindromic length two. The property was conjectured in 2012 by LucaZamboni (personal communication), and it was the motivation for our research.The other extremal case, that is, binary words with as many unbordered conju-gates as possible, was studied in [3]. The present work is also related to researchon critical points of a word, which is a stronger concept than unbordered conju-gate: the conjugate in the critical point is always unbordered, but not vice versa.Some results related to our words can be found in [2], where authors investigatewords with many and few critical points. In particular, they study words with aunique critical point. A related topic is also Duval’s problem and the Ehrenfeucht-Silberger problem, solved in [4, 5, 6] using similar techniques as those we employin this paper. 1.
Preliminaries
We first review basic concepts and facts we use in this paper. Let Σ = { , } bea binary alphabet and Σ ∗ the free monoid generated by Σ using the concatenationoperation with the empty word as the unit. We refer to elements of Σ ∗ as binarywords . The concatenation of two words u, v is denoted by u · v , but we will omitthe operator most of the times and simply write uv . We say that a word w = w w · · · w n , where w i ∈ Σ for all 1 ≤ i ≤ n , is of length n , and we denote this as | w | = n . Mathematics Subject Classification.
Key words and phrases. palindromes, unbordered words, Lyndon words.Supported by the Czech Science Foundation grant number 13-01832S.
The two possible lexicographic orders on Σ ∗ are ⊳ , defined by ⊳ , and ◭ ,where ◭ . The reversal of a word w = w w · · · w n of length n is the word w R = w n · · · w w . A word w is called a palindrome if w = w R .Two words w and w ′ are conjugates if there exist words u and v , such that w = uv and w ′ = vu . The set of all conjugates of w is denoted by [ w ]. For a word w = uvz , we say that u is a prefix of w , v is a factor of w , and z is a suffix of w .We denote these prefix- and suffix-relations by u ≤ p w and z ≤ s w , respectively.Furthermore, we write vz = u − w and uv = wz − in this case. A prefix (suffix) of w is proper if u = w ( z = w ); then we write u < p w ( z < s w ). A cyclic occurrence of u in w is a number 0 ≤ m < | w | such that u is a prefix of w w where w w = w and m = | w | .A word is primitive if w = t k implies k = 1 and thus w = t . It is a basic fact thatif uv = vu then u = t i and v = t j for some t and some i, j ≥ w is bordered , if there is a nonempty word u = w that is both a suffixand a prefix of w . Any such u is called a border of w . Note that a border may be oflength greater than | w | /
2. However, it is easy to see that any bordered word has aborder u such that w = uvu for some (possibly empty) word v . Naturally, a wordis unbordered , if it is not bordered.If w = w w · · · w n and w i + p = w i holds for some p ≥ ≤ i ≤ n − p ,then p is a period of w . The smallest period of w is called the period of w . Theprefix u of w such that | u | is the period of w is called the periodic root of w . Notethat the periodic root is always primitive. Note also that a word w is unbordered ifand only if its smallest period is | w | . In particular, an unbordered word w cannotbe a factor of a word with a period smaller than | w | .A Lyndon word is a primitive word w that is the lexicographically smallestelement (with respect to some order) of [ w ]. To make it clear which order is con-sidered when we compare words or talk about minimal words, we will speak about ⊳ -Lyndon words or ◭ -Lyndon words.Let z , | z | < | w | , be a prefix of a word w and let r be the shortest border of theword w z := z − wz if w z is bordered, and let r = w z otherwise. We say that r is the local periodic root of w at the point m = | z | and | r | is the local period at the point m . We note that our concept of the local period is consistent with the terminologyused in the literature if w is considered as a cyclic word. Accordingly, the localperiodic root r is the shortest nonempty word for which rr is centered at the givenpoint | z | in a cyclically understood word w .We stress that the local periodic root is always unbordered (being defined as theshortest border). This implies that the local period of a primitive word w at anypoint is either less than | w | / | w | . In thelatter case we say that the local period is trivial .We now prove several basic and useful properties of Lyndon words. For conve-nience, we formulate them for the order ⊳ . Lemma 1.
Any Lyndon word is unbordered.Proof.
Let w be a ⊳ -Lyndon word. Assume that there is nonempty u such that w = uvu . Then uvu ⊳ uuv implies vu ⊳ uv , which in turn yields vuu ⊳ uvu , acontradiction. (cid:3) ULLY BORDERED WORDS 3
Lemma 2.
Let w = uv be a ⊳ -Lyndon word and let z be the periodic root of u .Then z is a ⊳ -Lyndon word.Proof. Let u = ( z z ) k z ′ where z = z z is the periodic root of u , z < p z so that z z is the ⊳ -Lyndon conjugate of z , and z ′ < p z . Let ˜ z be the prefix of u of length | z − u | . Since uv is ⊳ -Lyndon, we have ˜ z ⊳ z − u , and since z z is ⊳ -Lyndon, wehave z − u ⊳ ˜ z . Therefore ˜ z = z − u , which implies that z is empty, otherwise | z | is a period of u . (cid:3) Lemma 3.
Let w be a ⊳ -Lyndon word and let z k z ′ · ≤ p w where z is the periodicroot of z k z ′ , and z ′ · ≤ p z . Then z k z ′ · is a ⊳ -Lyndon word.Proof. Suppose that r · is a border of z k z ′ · . By Lemma 2 and Lemma 1, z isunbordered, which implies that r · is a suffix of z ′ · . Therefore r · is a factor of z . Since r · is a prefix of w , we have a contradiction with w being ⊳ -Lyndon. | | | z z . . . z ′ z ′ z ′ r r r We have shown that z k z ′ · is unbordered, that is, it is its own periodic root. Theclaim now follows from Lemma 2. (cid:3) Lemma 4.
Let u and v be nonempty words such that both uv and vu are Lyndon.Then zero is the only cyclic occurrence of u in uv .Proof. Without loss of generality, let uv be ⊳ -Lyndon word. Suppose that uv ′ is a conjugate of uv . From uv ⊳ uv ′ we deduce v ⊳ v ′ , which is equivalent to v ′ ◭ v since | v | = | v ′ | . From vu being ◭ -Lyndon, we have v ◭ v ′ , and therefore v = v ′ . The claim now follows from the fact that a primitive word is not a nontrivialconjugate of itself. (cid:3) Fully bordered words
We say that a binary word w is fully bordered if | w | >
1, and there are exactly twounbordered conjugates of w . If w is fully bordered and uv, vu are the unborderedconjugates of w , then we will slightly abuse terminology and say that ( u, v ) is a fullybordered pair . Note the obvious but important fact that ( u, v ) is a fully borderedpair if and only if the local period of uv is less than | uv | at all points except 0 and | u | .We adopt the following notation. Let z be the periodic root of a nonempty word w . Then w = z k z ′ , where z ′ is a nonempty prefix of z , and we define s w := z ′ , t w := ( z ′ ) − z and k w := k , which yields w = ( s w t w ) k w s w . Note that we assumethat s w is not empty. This means that if w = z k , then s w = z , t w is empty, and k w = k −
1. In particular, k w = 0 if and only if w is unbordered. Lemma 5.
Let ( u, v ) be a fully bordered pair. Then: (a) ( v, u ) is a fully bordered pair. (b) uv , vu , u R v R and v R u R are Lyndon words. (c) s u t u and ( t u s u ) R are Lyndon words. (d) If k u > , let u ′ = ( s u t u ) k u − s u . Then s u ′ = s u and t u ′ = t u . ˇSTˇEP ´AN HOLUB AND MIKE M¨ULLER Proof. (a) Follows directly from the definition.(b) The definition of a fully bordered word implies that uv is primitive and con-tains both letters and . Therefore, there are at least two unbordered conjugatesof uv , namely the Lyndon conjugates with respect to ⊳ and ◭ . This implies that uv and vu are Lyndon words. A word is bordered if and only if its reversal is. Hence u R v R and v R u R are the only unbordered conjugates of ( uv ) R , which means thatthey are its two Lyndon conjugates.(c) Follows from (b) by Lemma 2.(d) The word u ′ has a period | s u t u | . It is the least period since s u t u is unbordered. (cid:3) The key for a characterization of fully bordered words is the following theorem.
Theorem 1.
Let ( u, v ) be a fully bordered pair such that | v | ≤ | u | and | uv | > .Let u ′ = ( s u t u ) − u . Then either • ( u ′ , v ) is a fully bordered pair, or • v = v ′ u ′ v ′ and ( u, v ′ ) is a fully bordered pair.Proof. Suppose, without loss of generality, that is the first letter of u . Then uv and u R v R are ⊳ -Lyndon words, and vu and v R u R are ◭ -Lyndon words. Inparticular, is a suffix of s u , and since | u | ≥
2, it follows that k u ≥ p be the longest common prefix of v and t u s u . Then p is a proper prefix of v since | t u s u | is not a period of uv . It is also a proper prefix of t u s u since vu isunbordered. This implies, since vu is ◭ -Lyndon, that p · is a prefix of t u s u and p · is a prefix of v . Symmetrically, we define q as the longest common suffix of v and s u t u and observe that · q is a suffix of v while · q is a suffix of s u t u since v R u R is ◭ -Lyndon. Claim 1.
We first show that t u s u is a ◭ -Lyndon word. This is trivial if u is a powerof . Let u contain and let z be the ◭ -Lyndon conjugate of s u t u . Consider thelast occurrence of z in up . More precisely, let up = u zu with | u | < | z | . Let r bethe local periodic root of uv at the point | u z | . Since z is unbordered, we deducethat | r | > | u | and therefore u · is a prefix of r . If | r | ≤ | u z | , then u · is afactor of u z . This is a contradiction with z being ◭ -Lyndon, since u · is a prefixof z . Therefore, | r | > | u z | and up · is a factor of rr . Since up · is unborderedby Lemma 3, we obtain that | r | ≥ | up · | and | rr | > | uv | . Therefore, uv is a factorof rr . Since uv is unbordered, we obtain that | uv | ≤ | r | . The definition of the localperiod now yields | r | = | uv | , and since ( u, v ) is fully bordered, this implies u z = u ,and z = t u s u . Claim 2.
We now show that t u is a prefix of v . Suppose, for the sake of contra-diction, that p < p t u . Let r be the local periodic root of uv at the point ( s u t u ) k u .Since s u t u is unbordered, we have s u p · ≤ p r . If | r | < (cid:12)(cid:12) ( s u t u ) k u (cid:12)(cid:12) , then p · is afactor of ( s u t u ) k u , a contradiction with p · being a prefix of the ◭ -Lyndon word t u s u . If | r | > (cid:12)(cid:12) ( s u t u ) k u (cid:12)(cid:12) , then rr contains up · . Since up · is unbordered, weobtain that | r | ≥ | up · | and | rr | > | uv | . As above, this implies | r | = | uv | and thelocal period of uv at the point ( s u t u ) k u is | uv | , a contradiction. Claim 3.
By symmetry we deduce from previous claims that t u is a suffix of v and( s u t u ) R is the Lyndon conjugate of ( t u s u ) R . Claim 4.
We are going to show that the local periodic root of s u t u at a point0 < m < | s u | is short. To be precise, we claim that if s u = s s , with nonempty s and s , then there is a word r which is both a prefix of s p and a suffix of qs . Toprove this, let r be the local periodic root of s u t u at the point | s | . Suppose thatthe claim does not hold. Then the local periodic root r ′ of uv at the point (cid:12)(cid:12) us − (cid:12)(cid:12) contains either p · or · q as a factor. Indeed, if r ′ is both a prefix of s p and asuffix of qs , then r ′ = r and the claim holds. Note that p · is not a factor of qup ,since p · is a prefix of the ◭ -Lyndon word t u s u . Similarly, · q is not a factor of qup , since ( · q ) R is a prefix of the ◭ -Lyndon word ( s u t u ) R . Therefore · qup · is afactor of r ′ r ′ , which once more implies | r ′ r ′ | > | uv | and | r ′ | = | uv | , a contradiction.The proof that r is a suffix of qs is similar, using the point | s | of uv .Consider now two cases corresponding to the two possibilities in the theorem. Case 1.
Suppose that ( s u t u ) k u is not a factor of v . We show that the pair ( u ′ , v )is fully bordered by comparing the local periodic roots of uv , s u t u and u ′ v atcorresponding points.The word uv contains exactly two occurrences of ( s u t u ) k u , namely 0 and | s u t u | .This follows from the assumption of Case 1, from the fact that t u is a prefix of v while t u s u is not, and from the fact that s u t u is unbordered, that is, not overlappingwith itself. Therefore ( s u t u ) k u is not a factor of any nontrivial local periodic rootof uv , since otherwise it would have two non overlapping occurrences.Consequently, the local periodic root of uv at any point | ut u | < m < | uv | is alsoa nontrivial local periodic root of u ′ v at the point m − | s u t u | .For | u | < m ≤ | ut u | , observe that the local periodic root r of uv at m is either afactor of qut u , and then | r | is at most the period of u , that is | s u t u | , or | r | > | qu | ,which implies | rr | > | uv | and | r | = | uv | , a contradiction.Claim 2, Claim 3 and Claim 4 implies that the local periodic root of u ′ v at apoint 0 < m ′ < | u | − | s u t u | is the same as the local periodic root of uv at the point m ′ + | s u t u | , namely the same as the local periodic root of s u t u at the point m ′ mod | s u t u | .To complete the proof for this case, it remains to note that u ′ v and vu ′ areunbordered: if the shortest border of u ′ v or vu ′ is shorter that ( s u t u ) k u , then it isalso a border of uv or vu ; if it is longer, then v contains ( s u t u ) k u . Case 2.
Suppose, now, that v contains a factor ( s u t u ) k u . Then it can be writtenas v ′ u ′ v ′′ , where v ′′ = t u v for some v . By | u | ≥ | v | , we have that | v ′ v | ≤ | s u | .Since s u t u is unbordered and t u is a prefix of v , it follows that uv contains exactly2 k u + 1 occurrences of s u t u , those visible in the factorization uv = ( s u t u ) k u +1 v ( s u t u ) k u v , where v ′ = t u v , and v is a nonempty word since t u s u is not a prefix of v . Thisimplies that uv ′ = ( s u t u ) k u +1 v is the periodic root of uv ′ ( s u t u ) k u . By Lemma2, it is a Lyndon word. Therefore, the local periodic root r of uv at the point (cid:12)(cid:12) ( s u t u ) k u +1 v (cid:12)(cid:12) has a proper prefix ( s u t u ) k u , otherwise r is a border of the Lyndonword ( s u t u ) k u +1 v . Since we know all occurrences of s u t u in uv , we deduce that r = ( s u t u ) k u v = u ′ v ′ .If | s u t u rr | > | uv | , then the second occurrence of r overlaps with s u t u . Then r is bordered, a contradiction. Let uv = s u t u rrz , where | z | < | s u | for length reasons.Suppose that z is not empty. Since s u t u is unbordered, the word s u t u is not a prefixof zs u t u . Therefore we deduce zs u t u ◭ s u t u from s u t u ⊳ zs u t u , which follows from uv being ⊳ -Lyndon. We obtain a contradiction with vu being ◭ -Lyndon, since v ′ s u t u is a prefix of v and v ′ zs u t u is a factor of vu . ˇSTˇEP ´AN HOLUB AND MIKE M¨ULLER | | p q p q m ↓↑ r ru u ′ u uv ′ Figure 1.
Therefore, z is empty, and uv = s u t u rr = uv ′ u ′ v ′ as claimed.We show that ( u, v ′ ) is fully bordered. We have seen that uv ′ is a Lyndon word, inparticular unbordered, which implies that p is a proper prefix of v ′ . Symmetrically,we can show that q is a suffix of v ′ . Claim 4 now implies that the local periodicroot of uv ′ at a point 0 < m < | u | is the same as the local periodic root of uv atthe point m .Let | u | < m ≤ | up | . The argument is similar as in Case 1: The local periodicroot r of uv at the point m has length at most | u | , since | u | ≥ | v | . Therefore, r has a period | s u t u | , which implies | r | ≤ | s u t u | since r is unbordered. Therefore, r is also the local periodic root of uv ′ at the point m . Similarly, we can show thatthe local periodic root of uv ′ at a point | uv ′ | − m , where 0 < m ≤ | q | , is the sameas the local periodic root of uv at the point | uv | − m .Let r be the local periodic root of uv at a point | up | < m < | uv ′ | − | q | . If | r | ≤ | uv ′ u ′ p | − m , then r is also a local periodic root of uv ′ at the point m (seeFigure 1). Suppose that | r | is greater than that. Then the word y = · qu ′ p · is a factor of r . We check all the three cyclic occurrences of u ′ in uv (namely 0, | s u t u | and | uv ′ | ), and verify that there is only one cyclic occurrence of y in uv ; acontradiction. (cid:3) Theorem 1 motivates the following inductive definition. Let F be the smallestsubset of { , } ∗ × { , } ∗ satisfying the following conditions:(1) ( , ) ∈ F (2) If ( u, v ) ∈ F , then ( v, u ) ∈ F .(3) Let ( u, v ) ∈ F . Then ( s u t u u, v ) ∈ F .(4) Let ( u, v ) ∈ F , and let y be a border of u such that | t v | < | y | . Then( u, vyv ) ∈ F .Pairs in F have the following properties. Lemma 6.
Let ( u, v ) ∈ F . (A) Both uv and vu are unbordered. (B) t u is both a prefix and a suffix of v . (C) If y is a border of u such that | t v | < | y | , then ( y, v ) ∈ F . (D) If t u is empty, then s u ∈ { , } ; otherwise ( s u , t u ) ∈ F .Proof. We proceed by induction on the structure of F . All claims are true for ( , )and ( , ).Suppose now that ( u, v ) , ( v, u ) ∈ F . We shall prove the theorem for the fourpairs ( s u t u u, v ), ( v, s u t u u ), ( u, vyv ) and ( vyv, u ) where y is a border of u such that | t v | < | y | . Observe that s s u t u u = s u and t s u t u u = t u . By (C) for ( u, v ) and (A) for ULLY BORDERED WORDS 7 ( y, v ) we have that vy is unbordered which implies that | vy | is the least period of vyv . Therefore s vyv = v and t vyv = y .(A) For all four pairs, the claim follows easily from uv and vu being unbordered.Consider for example vyvu and suppose it has the shortest border r . If | r | ≤ | vy | ,then r is also a border of vu . If | vy | < r ≤ | vyv | , then r is not unbordered. If r = vyvu ′ with u ′ ≤ p u , then vu ′ is a border of vu .(B) For all four pairs, the claim follows directly from (B) for ( u, v ) and ( v, u ).(C) For ( s u t u u, v ), note that the only border of s u t u u which is not also a borderof u is u itself (a border longer than u would imply a period smaller than | s u t u | ).The claim now follows from (C) for ( u, v ).For ( v, s u t u u ), let y ′ be a border of v such that | t u | < | y ′ | . From (C) for ( v, u ),we obtain ( y ′ , u ) ∈ F . Therefore also ( y ′ , s u t u u ) ∈ F by the definition of F .For ( u, vyv ), let y ′ be a border of u such that | y | < | y ′ | . Then ( y ′ , v ) ∈ F followsfrom (C) for ( u, v ). Since y is a border of y ′ , and | t v | < | y | , we have ( y ′ , vyv ) ∈ F from the definition of F .For ( vyv, u ), let y ′ be a border of vyv such that | t u | < | y ′ | . From (C) for ( u, v ),we deduce that vy is unbordered. This implies that | y ′ | ≤ | v | . If y ′ = v , then theclaim hold since ( v, u ) ∈ F . If | y ′ | < | v | , then y ′ is a border of v such that | t u | < | y ′ | and the claim follows from (C) for ( v, u ).(D) Straightforward for pairs ( s u t u u, v ), ( v, s u t u u ) and ( u, vyv ). For ( vyv, u ),we have that t vyv = y is not empty, and we want to show that ( v, y ) ∈ F , since( v, y ) = ( s vyv , t vyv ). This follows from (C) for ( u, v ) and from item (2) of thedefinition of F . (cid:3) We are now ready for the main result.
Theorem 2.
The pair ( u, v ) ∈ { , } ∗ × { , } ∗ is fully bordered if and only if ( u, v ) ∈ F .Proof. The “only if” part of Theorem 2 follows from Theorem 1 by induction onthe length of the pair (that is, on | uv | ) as follows. Both fully bordered pairs ( u, v )with | uv | ≤ F . Consider a fully bordered pair ( u, v ) of length | uv | > | uv | be in F . By symmetry expressedin Lemma 5(a), we can suppose that | v | ≤ | u | .Suppose first that ( u ′ , v ) is fully bordered, where u ′ = ( s u t u ) − u . Then ( u ′ , v ) , ( v, u ′ ) ∈F . If k u >
1, then s u ′ = s u and t u ′ = t u by Lemma 5(d), which implies ( u, v ) ∈ F by item (3) of the definition of F . If k u = 1, then u ′ = s u . Lemma 6(B) impliesthat t u ′ is a prefix of v . Also t u is a prefix of v by Claim 2 in the proof of Theorem 1.Therefore, t u and t u ′ are prefix comparable. The word s u t u is Lyndon by Lemma5(c), and t u s u is Lyndon by Claim 1 in the proof of Theorem 1. Since t u ′ is a factorof s u , Lemma 4 now implies that t u is not a prefix of t u ′ . Therefore, t u ′ is a properprefix of t u . Item (4) of the definition of F applied to the pair ( v, u ′ ) with y = t u yields ( v, u ′ t u u ′ ) = ( v, u ) ∈ F . Therefore also ( u, v ) ∈ F .Suppose now that v = v ′ ( s u t u ) − uv ′ and ( u, v ′ ) is fully bordered. Then ( u, v ′ ) ∈F . From Lemma 6(B) we have | t u | < | v ′ | , which together with | v | ≤ | u | implies | v ′ | < | s u | . Therefore also | t v ′ | < | y | , where y = ( s u t u ) − u . Item (4) of thedefinition of F again implies ( u, v ′ yv ′ ) = ( u, v ) ∈ F .The pairs ( , ) and ( , ) are fully bordered. To prove the “if” part by induction,let ( u, v ) ∈ F , and let all pairs in F shorter than ( u, v ) be fully bordered. ˇSTˇEP ´AN HOLUB AND MIKE M¨ULLER Consider the local periodic root r of s u t u at a point 0 < m < | s u | . If t u isempty, then s u ∈ { , } by Lemma 6(D) and r = s u . If t u is nonempty, thenLemma 6(D) and the induction assumption imply that ( s u , t u ) is fully bordered.Therefore | r | < | s u t u | , and Lemma 4 implies that t u is not a factor of r . This yieldsa factorization t u s u t u = u rru where | u r | = | t u | + m . Since t u is both a prefixand a suffix of v by Lemma 6(B), we deduce that r is the local periodic root of uv at any point 0 < m ′ < | u | such that m = m ′ mod | s u t u | . We also easily observethat the local periodic root of uv at a point | s u | ≤ m ≤ | u | − | s u | is the same as thelocal periodic root of | s u t u | at the point m mod | s u t u | . We have shown that localperiodic roots of uv at all points 0 < m < | u | are short.Repeating the same argument for the word v completes the proof that ( u, v ) isa fully bordered pair. (cid:3) The definition of F and Lemma 6 yield a good insight into the structure of fullybordered words. One of straightforward corollaries is the answer to the questionthat served as a motivation for our research. Theorem 3.
Let ( u, v ) be a fully bordered pair. Then both u and v are palindromes.Proof. Follows directly by induction from the definition of F . It is enough toobserve that s u , t u , and any border of u are palindromes if u is a palindrome. (cid:3) Acknowledgements
We want to thank Luca Zamboni for introducing us to this problem. Further-more, the second author expresses his gratitude to Tero Harju and Luca Zambonifor numerous interesting discussions on the subject. We also thank Jana Hadravov´afor many suggestions improving the manuscript.
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Department of Algebra, Charles University, Sokolovsk´a 83, 175 86 Praha, CzechRepublic
E-mail address : [email protected] Institut f¨ur Informatik, Christian-Albrechts-Universit¨at zu Kiel, D-24098 Kiel, Ger-many
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