Codingrotations on intervals

Transcription

1 Theoretical Computer Science 28 (22) Codingrotations on intervals Jean Berstel a, Laurent Vuillon b; a Institut Gaspard Monge (IGM), Universite de Marne-la-Vallee, 5, boulevard Descartes, Marne-la-Vallee Cedex2, France b Laboratoire d informatique algorithmique: fondements et applications (LIAFA) and CNRS, Universite Denis-Diderot (Paris VII), 2, place Jussieu, 7525 Paris Cedex5, France Abstract We show that the codingof a rotation by on m intervals can be recoded over m Sturmian words of angle. c 22 Elsevier Science B.V. All rights reserved.. Introduction The codingof rotations is a tool for the construction of innite words over a nite alphabet. Consider a rotation R, given by an angle, and dened for a point x by R (x)={x + }, where {y} denotes the fractional part of y. Consider next a partition of the unit circle in m + half open intervals {B ;B ;:::;B m }. For any startingpoint x with 6x, one gets an innite word u by I(x)I(R (x))i(r(x)) 2 I(R(x)) n :::; where I(y)=i if y B i. In the special case where is irrational and the partition is B =[;[ and B =[; [, this construction produces exactly the well-known Sturmian words. These words appear in various domains as computer sciences [2], Physics, Mathematical optimization and play a crucial role in this article. It is remarkable that Sturmian words have a combinatorial characterization. Thus, they are exactly aperiodic words with (subword) complexity p(n)=n + where the complexity function p : N N counts the number of distinct factors of length n in the innite word u [2]. The same general construction allows also to compute Rote words with complexity p(n)=2n by usingan irrational rotation and the partition B =[; 2 [ and B =[ 2 ; [ (see [7]). More generally, one can Correspondingauthor. address: laurent.vuillon@liafa.jussieu.fr (L. Vuillon) /2/$ - see front matter c 22 Elsevier Science B.V. All rights reserved. PII: S (2)9-9

2 J. Berstel, L. Vuillon / Theoretical Computer Science 28 (22) Fig.. Automata for m =; 2; 3. obtain innite words with complexity p(n)=an + b; where a and b are real, by coding of rotation [,3]. In addition, codings of rotation with an irrational value of and the partition B =[;[ and B =[; [ are intimately related to Sturmian words. Indeed, the rst sequence is the dierence term by term of two Sturmian words [6]. Didier gives a characterization of the codingof rotation with a partition of m intervals of length greater than by usingsturmian words and cellular automata [5]. Finally, Blanchard and Kurka study the complexity of formal languages that are generated by coding of rotation [4]. The goal of this article is to show that the coding of a rotation by on m + intervals can be recoded over m + Sturmian words of angle by the use of an automaton. Let the partition of [; [ be given for j =;:::;m, by [ j ; j+ [, where = 2 j m+ =, then the kth Sturmian word is given by J k (x)j k (R (x))j k (R 2 (x)) J k (R n (x)) ; where J k (y)= if y I k =[ k ; k + [ and otherwise (thus the m + Sturmian words have the same language). The automaton for the recodingis a universal automaton (i.e. it is the same for all (m + ) tuples of Sturmian words). It has m + states. Its edges are (m + ) tuples of and. The edge labeled by the vector composed of the ith letters of each Sturmian word gives the ith letter of the codingof rotation (see Fig. ).

3 J. Berstel, L. Vuillon / Theoretical Computer Science 28 (22) 99 7 Fig. 2. Partition of the unit circle. 2. Examples Fig. 2 shows a partition of the unit circle into 4 intervals of the form [ j ; j+ [. It shows also the positions of i + and the codingby 8 intervals associated with binary vectors called X K in Section 3. We can nd the codingof the interval [ j ; j+ [bythe automaton for m=3 applied to the binary vector values. As an example, usingthe universal automaton for m=3, the four followingsturmian words can be recoded on a word on a four letter alphabet: is recoded on the followingword: : 3. Notation We will consider subsets of [; [ that we call intervals. Let x; y be in [; [. Then we set {z x 6 z y} if x y; [x; y[= if x = y; {z x 6 z } {z 6 z y} if x y: In particular, [x; y[=[;y[ [x; [ if x y. This is precisely the notion of an interval on the torus T=R=Z. Let ; ;:::; m be numbers in the interval ]; [, with m. It will be convenient to set = and m+ =. The m + intervals B k =[ k ; k+ [; k =;:::;m

4 2 J. Berstel, L. Vuillon / Theoretical Computer Science 28 (22) 99 7 are a partition of [; [. We consider the rotation of angle dened by R (x)=x + mod. Dene intervals I k by (all values are computed modulo ) I k =[ k ; k + [; k =;:::;m: We will be specially interested in the nonempty intervals X K = I k I k : k K k= K Here, K is a subset of M ={;:::;m}, and I k =[; [\I k is the complement of I k. Observe that, for any nonempty interval I =[x; y[, one gets I =[y; x[. 4. Circular order We want to compute intersections of intervals. Although the geometric approach is easy to understand, it is error prone because points are usually not in general position. Therefore, we consider a more combinatoric approach. Given numbers x ;:::;x n [; [, the sequence (x ;:::;x n )iscircularly ordered, or c-ordered for short, if there exists an integer h with 6h6n such that 6 x h 6 x h+ 6 6 x n 6 x 6 6 x h : () If () holds, then either x = =x n, or the integer h is unique. Also, if (x ;:::;x n ) is c-ordered, then clearly (x 2 ;:::;x n ;x )isc-ordered. Any subsequence of a c-ordered sequence is c-ordered. Observe also that if (x ;:::;x n )isc-ordered and x x n then x 6 6x n. Indeed, if () holds for h, then x n 6x. Two rules are useful. Lemma 4.. (i) Translation Rule If (x ;:::;x n ) is c-ordered and y i x i + mod, then (y ;:::;y n ) is c-ordered. (ii) Insertion Rule If (x ;:::;x n ) and (y ;:::;y m ) are c-ordered, if furthermore y y m and x i =y ; x i+ =y m, then (x ;:::;x i ;y 2 ;:::;y m ;x i+ ;:::;x n ) is c-ordered. Proof. (i) We may assume 6x 6 6x n. The real numbers x i + satisfy x + 6 6x n + +x +. Ifx n +, then y i =x i + and (y ;:::;y n )isc-ordered. Otherwise, let h be the smallest integer such that x h +. Then x x h + 6 x h x n + : If h=, one gets x + 6 6x n + 2 and clearly (y ;:::;y n )isc-ordered. If h, then x n + x + implies y h 6 6 y n y 6 6 y h : (ii) There are two cases. If x i = max{x ;:::;x n }, then x i+ 6 6x n 6x 6 6x i. From x i =y ; x i+ =y m, it follows that y m y. Let h be the integer such that 6y h 6 6y m y 6 6y h. Then 6 y h 6 6 y m = x i+ 6 6 x n 6 x 6 6 x i = y 6 6 y h :

5 J. Berstel, L. Vuillon / Theoretical Computer Science 28 (22) If x i max{x ;:::;x n }, then x i =y y m =x i+ and consequently x i =y 6y 2 6 6y m =x i+. We observe that the insertion rule does not hold if y =y m. Consider the two c-ordered sequences (x; x; y) and (x; y; x), where x y. Insertingthe second into the rst give the sequence (x; y; x; y) which is not c-ordered. We prove another useful formula. Lemma 4.2. Let 2. If (x; y; x+) is c-ordered, then (x; y; x+; y+) is c-ordered. Proof. The condition 2 implies that (x; x+; x+2) isc-ordered. By the translation rule, we get that (x + ; y + ; x + 2) isc-ordered. The insertion rule shows that (x; x + ; y + ; x + 2) isc-ordered and, again by the insertion rule, one gets that (x; y; x + ; y + ) isc-ordered. 5. Intersection Circular order is useful in consideringintersections of intervals. Let I =[x; y[ bea nonempty interval. Then x [x; y[ i(x; x ;y) is ordered. Let I =[x; y[ and I =[x ;y [ be nonempty intervals. Then x I i (x; x ;y)isc-ordered. Since I I i x I or x I, the intervals I and I are disjoint i (x; y; x ) and (x ;y ;x) are c-ordered. Consequently, we have shown Lemma 5.. Let I =[x; y[ and I =[x ;y [ be nonempty intervals. Then I I = if and only if (x; y; x ;y ) is c-ordered. The length s of an interval I =[x; y[ is the number s=y x if x6y, and is s= (x y) ify x. In both cases, y x + s mod so that, knowingthe length, we may write I =[x; x + s[. Lemma 5.2. Let I =[x; y[ and I =[x ;y [ be intervals of the same length 2. If I and I intersect, then (x; x ;y;y ) or (x ;x;y ;y) is c-ordered. In the rst case, I I =[x ;y[, in the second case, I I =[x; y [. Observe that if the length of I and I is greater than =2, then the intersection needs not to be an interval. Proof. The discussion before Lemma 5. shows that I and I intersect if and only if (x; x ;y)or(x ;x;y ) are c-ordered. From Lemma 4.2, it follows that (x; x ;y;y ) or (x ;x;y ;y)isc-ordered. Moreover, y x and x y since otherwise (x; y; x ;y ) is c-ordered and the intervals are disjoint by Lemma 4.2. If x=x (or equivalently if y=y ), then I =I. Thus, we may assume that the numbers x; y; x ;y are distinct. Assume the rst orderingholds. The formula for the intersection is straightforward if 6x x y y. If 6x y y x, then I =[;y[ [x; [ and I I =[x ;y[. The two other cases are proved in the same way.

6 4 J. Berstel, L. Vuillon / Theoretical Computer Science 28 (22) 99 7 The previous lemma will be applied to the intervals I k =[ k ; k + [. They all have same length. We write the conclusion for further reference. Lemma 5.3. Let 2. Let I k =[ k ; k + [ and I =[ ; + [ be two intervals. If I k and I intersect then ( k ; ; k + ; + ) or ( ; k ; + ; k + ) is c-ordered. Moreover, I k I =[ ; k + [ in the rst case, and I k I =[ k ; + [ in the second case. The followingobservation is the basic step for analyzingthe codinginduced by a rotation. Recall that for K {;:::;m}, X K = I k I k : k K k= K We assume from now on that 2. Proposition 5.4. Assume X K for some K {;:::;m} and assume ( i ; i2 ; i3 ; i4 ) is a c-ordered sequence. If i ;i 3 K, then i 2 K or i 4 K. Proof. Arguing by contradiction, suppose that i 2 ;i 4 = K. Since X K, the interval I i I i3 is not empty, therefore by Lemma 5.3 ( i ; i3 ; i + ; i3 + ) or( i3 ; i ; i3 + ; i + ) isc-ordered (or the sequence obtained by exchanging i and i 3 ). Consider the rst case, the second is the same by exchanging i 2 and i 4. Since ( i ; i2 ; i3 )is c-ordered, the translation rule shows that ( i + ; i2 + ; i3 + ) isc-ordered which gives, applying twice the insertion rule, that ( i ; i2 ; i3 ; i + ; i2 + ; i3 + ) is c-ordered. From this, we get that ( i ; i2 ; i +; i2 +) isc-ordered. From Lemma 5., we know that I i I i3 =[ i3 ; i + [, and this is then disjoint from I i2 =[ i2 + ; i2 [. Proposition 5.5. If X K is not empty, then there exist integers k; with 6k 6m such that {K; M\K}={{k;:::; }; { ;:::;m;;:::;k }}. Proof. This is a direct consequence of the precedingdiscussion. It follows that there are only (m + )(m + 2) intervals X K to be considered. In fact, consider the numbers ; ;:::; m ; and ; + ;:::; m +. They partition [; [ into exactly 2m + 2 intervals. Each of these intervals is contained in one and only one of the X K (but X may be scattered over several of the small intervals). This means that, amongthe (m + )(m + 2) possible intervals X K, there are only 2m + 2 that are used in a particular settingof the values of ; ;:::; m. Theorem 5.6. Assume K ;M, and X K. Then X K =[ ; k + [ [ k + ; [ where k and are dened in Proposition 5:5. If K ={k} is a singleton, then the formula still holds with =k. Proof. Suppose that K ={k;:::; } with k. The other case is symmetric. We rst prove that n K I n =[ ; k + [: Set Y K = n K I n.

7 J. Berstel, L. Vuillon / Theoretical Computer Science 28 (22) Since X K, the interval I k I is not empty. By Lemma 4.3 there are two cases: either ( k ; ; k +; +) isc-ordered, or ( ; k ; +; k +) isc-ordered. We show that this second case cannot happen. Indeed in this case, Y K I k I = [ k ; + [: Moreover for each n M\K, the sequence ( ; n ; k ) is c-ordered. By translation and insertion, the sequence ( ; n ; k ; +; n +; k +) is c-ordered. This shows that I n I k I, and consequently I k I I n = for each n in M\K, contradictingthe assumption that X K : Thus, ( k ; ; k + ; + ) is c-ordered. This implies that I k I =[ ; k +[: If i K then ( k ; i ; )isc-ordered. By translation, ( k +; i +; +) is c-ordered. By insertion of ( k ; i ; ) into ( k ; ; k + ; + ) one gets ( k ; i ; ; k + ; + ) isc-ordered. Again by insertion of ( k + ; i + ; + ), the sequence ( k ; i ; ; k + ; i + ; + ) isc-ordered. Thus Y K =[ ; k + [: The second part of the proof deals with n M\K I n: In this intersection the index n runs through the set {;:::;k ; ;:::;m}. The set M\K is partitioned into three possibly empty subsets as follows: n N i I n Y K ; n P i I n Y K =[ ; n [ and nally n Q i I n Y K =[ n + ; k + [: Of course, X K = (I n Y K ) (I n Y K ) (I n Y K ): n N n P n Q If one of the sets N;P;Q is empty it does not contribute to the intersection. Clearly, n N (I n Y K )=Y K. Next n P (I n Y K )= n P [ ; n [. If P is not empty then is in P and n P (I n Y K )=[ ; [. Finally, n Q (I n Y K )= n Q [ n +; k +[. If Q is not empty then k isinq and n Q (I n Y K )=[ k +; k +[. To nish the proof, we just have to verify that in each case, X K = n N (I n Y K ) n P (I n Y K ) n Q (I n Y K ) is equal to [ ; k + [ [ k + ; [. If P then n P (I n Y K )=[ ; [ and the sequence ( ; ; k + ) is c-ordered (case P ). If P = then the sequence ( ; k +; )isc-ordered (case P ). If Q then n Q (I n Y K )=[ k +; k +[ and the sequence ( ; k +; k +) is c-ordered (case Q ). If Q = then the sequence ( ; k + ; k + ) isc-ordered (case Q ). Case (P Q ): If P and Q are nonempty then X K =[ ; k +[ [ ; [ [ k + ; k + [. As the sequences ( ; ; k + ) and ( ; k + ; k + ) are c-ordered, by the insertion rule either the sequence ( ; ; k + ; k + ) or( ; k + ; ; k + ) isc-ordered. The rst case is impossible because X K is not empty. The second case implies that X K =[ ; k + [ [ k + ; [. Case (P Q ): If P = and Q, then X K =[ ; k + [ [ k + ; k + [ and the sequences ( ; k + ; ), ( ; k + ; k + ) are c-ordered. By insertion the sequence ( ; k +; k +; )isc-ordered. Thus, X K =[ ; k +[ [ k +; [. Case (P Q ): is symmetric to case (P Q ). Case (P Q ): If P = and Q =, then X K =[ ; k + [ [ k + ; [ and the sequences ( k + ; k + ; ), ( k + ; ; ) are c-ordered. By insertion rule either the sequence ( k +; k +; ; ), or ( k +; k +; ; )isc-ordered. The rst case is impossible because X K is non empty. The second case implies that X K =[ ; k + [ [ k + ; [.

8 6 J. Berstel, L. Vuillon / Theoretical Computer Science 28 (22) 99 7 Remark. As an additional property, the precedingproof shows that X K and the interior of X K does not contain any i or { i + }: is an interval 6. Main result Proposition 6.. If x B i and x + B j X K, then j i + K mod m +. Proof. If K or K M then X K =[ ; k + [ [ k + ; [: As x [ i ; i+ [; by translation rule we have y=x + [ i + ; i+ + [. Furthermore, y [ j ; j+ [: By the precedingremark and by identication, the only possibility is j = and i=k : It follows that j =i + K mod m +: If K = then X K = n M I n: By hypothesis we have y =x+ B j : But y X K implies that y= [ j ; j + [= I j. If I j B j then B j X K should be empty in contradiction with the hypothesis. Thus I j B j. The interval B j is equal to I j [ j + ; j+ [: That is x + [ j + ; j+ [ and x + [ i + ; i+ + [: By identication, we have i=j: If K =M then X K = n M I n: As x I n implies x+ = I n, by contraposition x+ X K implies x= I n for all n. Thus (x; n ;x+) isc-ordered for all n M. Asx B i and x is not in I i the sequence ( i ; i + ; x; i+ )isc-ordered. Thus (x; i+ ;:::; i+m ; i ;x+ ) is c-ordered. Consequently, X K is equal to [ i ; i+ + [: As x + [ j ; j+ [ by identication we nd j =i. From this proposition, we get the following automaton A (Fig. gives the automata for m=; 2; 3). Its set of states is the set M in bijection with the intervals B k. The alphabet is the set of subsets of M correspondingto the nonempty intervals X K.As already mentioned, there are (m+)(m+2) of them. The transitions or edges are given by the proposition: (i; K; j) is a transition if j i + K mod m +. Observe that the automaton is deterministic. Also, it is universal in the following sense: for a particular settingof ; ;:::; m,ifthe i and the j + are two by two distinct, there are only 2m+2 of the edges that are used. Indeed they are exactly 2m+2 intervals in the partition and between i and i+ the codingis uniquely determined for all i. Moreover, the automaton has the noteworthy property that each label provided it is neither (;:::;) nor (;:::;) appears at only one edge. Consequently, both the starting state and the arrival state of such an edge is determined by is label. We denote by A(v ;:::;v m ) the arrival state of the edge with label (v ;:::;v m ). We now explain the result in terms of innite words. Recall that B i =[ i ; i+ [ and I i =[ i ; i + [. Let u be the innite word dened by I(x)I(R (x))i(r 2 (x)) I(R n (x)) ;

9 J. Berstel, L. Vuillon / Theoretical Computer Science 28 (22) where I(y)=i if y B i. Let s k be the kth Sturmian word dened by J k (x)j k (R (x))j k (R 2 (x)) J k (R n (x)) ; where J k (y)= if y I k =[ k ; k + [ and otherwise. Theorem 6.2. The innite word u and the m + Sturmian words s i are related by A(s (n);:::;s m (n)) = u(n) for all n n, where n is the smallest integer such that {s (n );:::;s m (n )} contains both and. For n n the value u(n) is the starting state of the edge labeled {s (n );:::;s m (n )}. Remark that the result is written for irrational in order to manipulate Sturmian words. The construction of the automaton is independent of the rationality of (this case leads to innite periodic words). References [] P. Alessandri, V. Berthe, Three distance theorems and combinatorics on words, Enseig. Math. 44 (998) [2] J. Berstel, P. Seebold, Sturmian words, in: M. Lothaire (Ed.), Algebraic Combinatorics on Words, Chap. 2, Cambridge University Press, Cambridge, to appear. [3] V. Berthe, Etude arithmetique et dynamique de suites algorithmiques, Habilitation a diriger des reserches, Universite d Aix-Marseille II, 999. [4] F. Blanchard, P. Kurka, Language complexity of rotations and Sturmian sequences, Theoret. Comput. Sci. 29 (998) [5] G. Didier, Combinatoire des codages de rotations, Acta Arith. 85 (998) [6] P. Hubert, Proprietes combinatoires des suites denes par billard dans les triangles pavants, Theoret. Comput. Sci. 64 (996) [7] G. Rote, Sequences with subword complexity 2n, J. Number Theory 46 (994)