On a conjecture on bidimensional words - CiteSeerX
On a conjecture on bidimensional words Chiara Epifanio
Michel Koskas y
Filippo Mignosiz
Abstract
We prove that, given a double sequence w over the alphabet A (i.e. a mapping from ZZ 2 to A), if there exists a pair (n0; m0) 2 ZZ 2 such that 1 pw (n0 ; m0) < 100 n0 m0, then w has a periodicity vector, where pw of w is the complexity function of w.
1 Introduction In combinatorics on words the notions of complexity and periodicity are of fundamental importance. The complexity function of a formal language counts, for any natural number n, the number of words in the language of length n. The complexity function of a word ( nite, in nite, biin nite) is the complexity function of the formal language whose elements are all the factors (or blocks, or also subwords) of the word. The Morse-Hedlund Theorem states that there exists an important relationship between periodicity and complexity. In particular it states that for any biin nite word w if the number of its di erent factors of length n is less
Dipartimento di Matematica ed Applicazioni, Universita degli Studi di Palermo,
[email protected] y
Faculte de Mathematiques et d'Informatique, Universite de Picardie Jules Verne,
[email protected] z
Dipartimento di Matematica ed Applicazioni, Universita degli Studi di Palermo,
[email protected]
1
than or equal to n, then the word is periodic. Moreover the period of w is smaller than or equal to n. Several generalizations of the complexity function exist in the literature. One of the most known is the complexity function pw (n; m) that counts the number of di erent rectangles of size n m that are \factors" of the double sequence w. This is the one that is considered and studied in this paper. Its rigorous de nition will be given in next section. In their seminal and fundamental work (cf. [1], [2], [3]) Amir and Benson introduced the notion of bidimensional periodicity (in particular the notions of simmetry and periodicity vector) and proved theorems analogous to the \periodicity lemma" (cf. [23]). After them, many researchers have worked on bidimensional periodicity and its applications (cf., for instance, [4], [5], [6], [16], [17], [18], [27]). In this paper we prove an extension of the Morse-Hedlund Theorem to the bidimensional case. On this subject let us state a conjecture (cf. [25]), known among the researchers in the eld of Combinatorics on Words as the Nivat's conjecture on bisequences. The rigorous de nitions of the notations used in it will be given in next section. Conjecture: If there exists a pair (n; m) such that the complexity function pw (n; m) of the double sequence w veri es pw (n; m) nm, then w has at least a periodicity vector. Many researchers have worked on this conjecture and on a related one of L. Vuillon that is presented in this paper in Example 2.1. Some partial results on this conjecture have been proved for small values of m or n (cf. [31], [32], [33]). The diculty of this conjecture is related also to the fact that all the known proofs of the Morse-Hedlund Theorem are intrinsecally \unidimensional", in the sense that they make use of properties of \unidimensional" words, such as the possibility of concatenating one letter to the right or to the left of a word. For such properties there are, to our knowledge, no generalizations that it is possible to use in the bidimensional case. Recently, we have been able to give a new proof of the Morse-Hedlund Theorem by using the periodicity Theorem of Fine and Wilf. For this theorem, which is the tight version of the periodicity lemma, there exist generalizations to the bidimensional case. This fact has suggested us to try to 2
extend our proof in order to settle above conjecture. Indeed, in this paper we prove a weak version of above conjecture, making use of the tools and techniques developed by the researchers in the eld of bidimensional periodicity. In particular we make use of some bidimensional generalizations of the periodicity lemma. Since there exist also bidimensional generalizations of the Theorem of Fine and Wilf (cf. [18]), we hope that this approach will allow to settle the original Nivat's conjecture. The remainder of this paper is organized as follows. In the second section we give the new proof of the Morse-Hedlund Theorem, de ne basic notations, discuss some preliminaries and give examples. In the third section we state and demonstrate some preliminaries lemmas. In the fourth section we prove the main theorem after having demonstrated two fundamental lemmas. Finally, in the last two sections we give a family of examples that shows some deep di erences between the unidimensional and the bidimensional case, and make some concluding remarks.
2 Preliminaries For any notation not explicitely de ned in this paper we refer to [18], [20] and [22]. A word w = a a an has period p if ai = aj for any 1 i; j n such that i j (mod p). Note that, following this de nition, any natural number p > jwj turns out to be a period of w, and, in this case, it is called an improper period of w. The smallest period of w is called the period of w. Notice that classically (cf. [20]) improper periods are not considered as periods. 1 2
We start with an easy proposition.
Proposition 2.1 Let w be a word of length n and period p n. If there exists a factor u of w of length juj p and period q where q divides p, then w has period q.
3
Proof
Let w = a an and u = ah ak , where 1 h < k n; j u j= k ?h p. Let i and j be two integers lying between 1 and n such that i j (mod q). There exist two integers i0 and j 0 such that i i0 (mod p), j j 0 (mod p) and h i0; j 0 k. Since i0 j 0 (mod q) and 1 i0; j 0 k, ai = aj . Since i i0 (mod p) and j j 0 (mod p), ai = ai and aj = aj and the conclusion holds. 2 1
0
0
0
0
The following theorem is a classical result in combinatorics on words (cf.[14], [20] and [21]).
Theorem 2.1 (Fine and Wilf, 1965) Let w be a p-periodic and q-periodic
word. If jwj p + q ? gcd(p; q) then w is gcd(p; q)-periodic.
Above theorem is also tight. Indeed for any p, q, p < q such that p does not divide q, there exists a word w; jwj = p + q ? gcd(p; q) ? 1, that is p-periodic and q-periodic but it is not gcd(p; q)-periodic. Let us consider an (unidimensional) word w and a positive integer n. We de ne the complexity function of w
pw (n) =j fu j u is a factor of w of length ng j The Morse-Hedlund Theorem (cf. [24]) states that there exists an important relationship between periodicity and complexity. Here we state the version relative to biin nite (two sides) words. The proof of the theorem that we give is a new one and it makes use of the Theorem of Fine and Wilf. The main idea of this proof is that the hypothesis implies the existence of a \local" period in any position of the biin nite word. The Theorem of Fine and Wilf, together with the Proposition 2.1, forces all these \local" periods to be the same all over the biin nite word. This is also the general idea of the proof of our main theorem in Section 4. Moreover the same \unidimensional" technique used here will be used also in both the proofs of our two main lemmas. We can now state and give our new proof of the Morse-Hedlund Theorem. 4
Theorem 2.2 (Morse and Hedlund) Let w be a biin nite word. If there exists an integer n such that pw (n) n, then w is periodic. Moreover the
period of w is smaller than or equal to n.
Proof
Let n be the smallest integer such that pw (n ) n . Let us consider in w the n + 1 positions 0; 1; ; n . Each position determines uniquely the factor of length n which starts in this position. 0
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a0 a1 0
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n -2
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By the pigeon's holes principle two of these n + 1 positions must determine the same factor of length n (by hypothesis pw (n ) n ). Let i < j be two of such positions such that j is the smallest position greater than i that determines the same factor of length n determined by i. Hence, for h = 0; ; n ? 1, one has that ai h = aj h. This is equivalent to say that the word v = ai aj n ? has period p = j ? i. The word v has length n + p . Moreover, by the choice of j , p n is the minimal period of v. 0
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+
+ 0
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+
j
n0
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i+n 0-1
p =j?i
v = a i aj
1
j+n 0-1
n ?1
+ 0
We want to prove that p is a period for the whole word w. Let us suppose, by contradiction, that p is not a period for w. Therefore there exists a position H such that aH 6= aH ?p . In what follows we suppose that H > j + n ? 1. The proof in the case where H < i + p is analogous. 1
1
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5
Let us suppose that H is the smallest integer greater than j + n ? 1 such that aH 6= aH ?p . Let u = aH ?n ?p aH be the word of length (n + p ) which last letter is aH and let u? be its pre x of length j u? j=j u j ?1. Since H is the smallest integer greater than j + n ? 1 such that aH 6= aH ?p and j u? j= (n + p ? 1), then u? has period p . Moreover, since p is the minimal period of the word v = ai aj n ? , then it is also the minimal period of the word v = ai aH ? whose sux of length (n + p ? 1) is the pre x u? of u of the same length. Therefore the word b = ai ai p ? is a primitive one. By a lemma of Aldo de Luca all the p words conjugates to b are di erent. This fact implies that if we consider the rst p positions of the word u, since they determine all the p conjugates of the word b of length p < n , they determine di erent words of length n . If we consider the rst (n + 1) positions of the word u (i.e. H ? n ? p + 1; ; H ? p + 1), we obtain, again by the hypothesis that pw (n ) n and by the pigeon's holes lemma, that two among these must determine the same factor of length n , i0 and j 0, i0 < j 0. By previous result, j 0 is not one of the rst p positions of the word u, i.e. j 0 (H ? n ? p +1)+ p : Moreover the word v0 which starts at the position i0 and whose length is n + (j 0 ? i0), has period j 0 ? i0 = p n . Since j 0 (H ? n ? p + 1) + p , the word v0 contains position H and, since aH 6= aH ?p , the word v0 cannot have period p. 0
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1 +1
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+ 0
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+ 1
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H-n 0-p1 +1
i’
j’
H
H+1
p 1
The pre x of v0 that has both periods p and p is the word u0 = ai aH ? . Its length is j u0 j= H ? 1 ? i0 + 1 = H ? i0 = H ? j 0 + p . But j 0 is one of the rst (n + 1) positions of the word u, i.e. j 0 H ? p + 1. Hence ju0j = H ? j 0 + p p + p ? 1. Therefore we can apply the Theorem of Fine and Wilf to u0 and obtain that u0 has period d, where d is the greatest common divisor between p and 1
2
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1
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2
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6
p . Since v0 has a pre x u0 of length greater than or equal to p that has a period d that divides p , and since p is a period of v0 then, by Proposition 2.1, v0 has also period d. This period d is also a divisor of p , consequently v0 has also period p , that contradicts the fact that aH 6= aH ?p . 2 2
2
2
2
1
1
1
Let us consider, now, the bidimensional case. A double sequence w over the alphabet A is a mapping from ZZ to A. A factor u of a bidimensional word w is the restriction of w to a subset S ZZ , which is called the shape of u and it is denoted by sh(u). A factor u is a factor of u if the shape of u is contained in the shape of u . Notice that u is a factor of itself as well w is a factor of w. Two factors u and u are equal if there exists a translation of the plane such that (S ) = S , where S and S are respectively the shapes of u and of u . Moreover this translation must be compatible with the word w, i.e. for any (x; y) 2 S , w(x; y) = w( (x; y)). A rectangle R is a subset fi ; i + 1; ; i g fj ; j + 1; ; j g of ZZ . For brevity we denote R = [i ; i ] [j ; j ]. The numbers i ? i + 1 = n and j ? j + 1 = m are the lengths of the sides of R. A rectangle R having sides of lengths n and m is called an n m rectangle. If the shape of a factor u of w is a n m rectangle R then u is a n m factor and n; m are also called the lengths of the sides of the factor. Clearly, if (n; m) 6= (n0; m0) then every n m factor is di erent from every n0 m0 factor. We de ne the complexity function pw (n; m) of w as the numbers of different n m factors. 2
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De nition! 2.1 Let u be a factor of w and let R be its shape. A vector 6 0 is a symmetry vector for u if w((i; j )) = w((i + v ; j + v )) v = vv = 1
1
2
2
for any element (i; j ) in R such that (i + v1 ; j + v2) is still in R. Clearly, if u is the whole w then v is a symmetry vector for w if, for any (i; j ), w((i; j )) = w((i + v1; j + v2)). We say that v is a periodicity vector for u if for any integer z the vector zv is a symmetry vector for u. In this case we will also say that v is a period for u (for w resp.) or also that u (w resp.) has period v.
7
This de nition of symmetry vector is analogous to the de nition of symmetry vector given by Amir and Benson in [1], [2], [3]. The de nition of periodicity vector is analogous to the one of L =< v >-periodicity given by Giancarlo and Mignosi in [18] or by Koskas in [19]. We can here state the Nivat's conjecture. Conjecture: If there exists a pair (n; m) such that the complexity function pw (n; m) of the double sequence w veri es pw (n; m) nm, then w has at least a periodicity vector. Let us state two easy propositions without proof.
Proposition 2.2 A vector v is a symmetry vector for u (resp. w) if and only if (?v) is still a symmetry vector for u (resp. w). Moreover, if v is a periodicity vector then 8z 2ZZ ; zv is still a periodicity vector. In IR the de nitions of convex set and convex hull are well known. The next de nition concerns \discrete" convexity. 2
De nition 2.2 A subset S of ZZ is convex if its convex hull in IR does not contain any point of (ZZ n S ): Proposition 2.3 Any symmetry vector for an n m factor u or for the whole double sequence w is also a periodicity vector. 2
2
2
More generally, any symmetry vector for any factor whose shape is a convex subset of ZZ 2, is also a periodicity vector.
Given two vectors v, v0 2ZZ we denote by < v; v0 > the subgroup of ZZ generated by v and v0, i.e. the set of all vectors v^ of the form v^ = iv + jv0 with i; j 2 ZZ . It is known that if v and v0 are not parallel, the subgroup < v; v0 > has nite index, i.e. there exists a nite subset T of ZZ such that for each point (x; y) 2 ZZ there exists an unique element t 2 T such that (x; y) ? t 2< v; v0 >. 2
2
0
2
2
Remark 2.1 If v and v0 are parallel, then the subgroup < v; v0 >=< v00 >
is generated by only one element v 00 whose length divides the lengths of v and v 0.
8
De nition 2.3 If v and v0 are not parallel, a factor u is full-periodic (or lattice-periodic) with respect to v and v 0 if for all v^ 2< v; v 0 > v^ is a periodicity vector for u. In this case we also say that u is < v; v 0 >-periodic. Notice that if we require that there exists n such that pw (n; n) n, then every restriction of w to unidimensional lines parallel to the axes is, by the Morse-Hedlund theorem, periodic with period smaller than or equal to n. Hence w is! lattice-periodic,!where the lattice is generated by the two vectors v = n0! and v0 = n0! .
Consequently, this form of generalization of the Morse-Hedlund Theorem is relatively simple and can be extended to any dimension. Moreover J. Cassaigne has proved (cf. [10]) that the lengths of v and v0 can be bounded by n. Another similar generalization of the Morse-Hedlund Theorem is stated and proved in [13] (cf. also [12]) for Delone (or Delaunay) sets, where factors are \replaced" by neighborhoods. One of the diculties in this new setting comes from the fact that two neighborhoods are \equal" if they are isometric and not just equal by a translation as in the case of double sequences (cf. also the pinwheel tilings of the plane in [26] and the Delone sets that can be obtained by the Conway tasselation, as explained in [13]).
Examples:
Example 2.1 In the following example w((i; j )) = 0 for any point (i; j )
except for one point, where it has value 1. The complexity function is equal to pw (n; m) = nm +1 and w has no periodicity vectors. J. Cassaigne has proved [8] that there are no recurrent double sequences having the same complexity pw (n; m) = nm + 1 of w. Recall that a double sequence w is recurrent if for any couple of integers (n; m) there exists a couple (n0; m0) such that any n m factor u1 of w is equal to a factor of any n0 m0 factor u2 . A conjecture on the complexity functions of recurrent double sequences was stated by L. Vuillon (cf. [7], [34], [35], [36]). This conjecture states that, for any recurrent double sequence w that has no periodicity vector, there exist n^ , m ^ such that for any n > n^ , m > m^ one has pw (n; m) nm + min(n; m).
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Example 2.2 The following example is due to Cassaigne and Bernardi (cf. [9]). Let us consider a Christo el approximation of a line in the plane having irrational slope. This line de nes two distinct half-planes HP1 and HP2 . We de ne a word w such that for any point (i; j ) in HP1 w(i; j ) = 1 and for any point (i; j ) in HP2 w(i; j ) = 0. In this case it is possible to prove that pw (n; m) = nm + 1. The idea of the proof is the following. Let us assume to write the letter w(i; j ) inside the unit square that has (i; j ) as the bottom-left corner. If we slide an n m rectangle from HP2 to HP1 , the labels w(i; j ) associated to the points (i; j ) of the rectangle change from 0's in 1's always in the same order. Since the slope is irrational, it is impossible that two of them could change their value at the same moment. This remark allows us to numerate these rectangles, i.e. the shapes of all the factors of w. Hence the complexity is pw (n; m) = nm + 1. 1
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3 Preliminaries Lemmas Before proving the main theorem, we will prove or recall some preliminaries lemmas. As remarked in the previous section, the proof of our main theorem is based on the fact that the hypothesis implies the existence of a \local" period in every position of the double sequence. This is proved in next lemma. Afterwards, we state two lemmas that are analogous to the periodicity lemma 10
and that will be used in next section to force all \local" periods to be \almost the same" all over the double sequence.
Lemma 3.1 (local 1-periodicity) Let w be a double sequence whose complexity function pw is such that pw (n ; m ) < n m for some (n ; m ) 2 ZZ . Then any factor u of w of shape [i; i + h] [j; j + k] with i; j 2 ZZ and! 0 h < n ? d pn e, 0 k < m ? d pm e has a periodicity vector v = vv such that 0 v d pn e ? 1 and 0 jv j d pm e ? 1. Proof Let u be the factor of w of shape [i; i + h] [j; j + k] with i; j 2 ZZ and 0 h < n ? d pn e, 0 k < m ? d pm e. For any (s; t) 2 [0; d pn e ? 1] [0; d pm e ? 1] let us;t be the factor of w of shape [i ?d pn e + s; i ?d pn e + s + n ? 1] [j ?d pm e + t; j ?d pm e + t + m ? 1]. Since h n ? d pn e ? 1 and k m ? d pm e ? 1, u is a factor of us;t . Moreover, since pw (n ; m ) < n m and since us;t is a n m -factor, there exist two di erent couples (s ; t ); (s ; t ) in [0; d pn e] [0; d pm e] such that us ;t = us ;t . Indeed, there are at least d pn ed pm e n m positions for the factors us;t but less than n m di erent factors. 0
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u s ,t
0 0
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s Since u is a factor of us ;t and a factor of us ;t , v = st ? ? t is a periodicity vector for u. We may of course assume that s ? s 0 because if v is a periodicity vector, so is ?v. It remains to notice that by construction we have s ? s d pn e ? 1 and jt ? t j d pm e ? 1, and the lemma is proved. 2 0
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The following lemma has been proved by Amir and Benson (cf. [3], Lemma 3.6)
Lemma 3.2 Let u be an n m factor of w. If v = vv
are two periodicity vectors of u such that
! 1 2
and v 0 =
v0 v0
!
1
2
v 0; v0 0; v 0; v0 0, v + v0 < n; jv j + v0 < m: Then, for all v^ 2< v; v 0 >; v^ is a periodicity vector for u. Therefore, if v and v 0 are not parallel, u is lattice-periodic, where the lattice is generated by 1
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v and v0.
Next lemma is an easy consequence of Lemma 7 in [16]. A similar result can be found also in [2]. !
!
0 De nition 3.1 Let v = vv and v0 = vv0 be two vectors such that v 0; v0 0; v 0; v0 0; j v j< n ; j v0 j< n ; j v j< m ; j v0 j< m : Let R be an n m-rectangle, R = [i; i + n ? 1] [j; j + m ? 1]. The intersection of ZZ and the convex hull of the six points f(i; j ); (i; j + m ? 1 ? v ? v0 ); (i + v + v0 ; j + m ? 1); (i + n ? 1; j + m ? 1); (i + n ? 1; j + v + v0 ); (i + n ? 1 ? v ? v0 ; j )g will be called the (v; v0)-hexagon of R. 1
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Lemma 3.3! Let u be an n! m factor of w and let R be its shape. If 0 v = vv and v0 = vv0 are two periodicity vectors of u such that 1
1
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2
v 0; v0 0; v 0; v0 0; j v j< n ; j v0 j< n ; j v j< m ; j v0 j< m 1
2
1
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then the (v; v 0)-hexagon of R is periodic with period generated by both vectors. In particular, if v and v 0 are not parallel, this hexagon is lattice periodic.
A direct consequence of these two lemmas is the following theorem.
Theorem 3.1!(Lattice periodicity) ! If a rectangle R has two periodicity vecv v
tors v =
1 2
v0 such that v0 j v j< n4 ; j v0 j< n4 ; j v j< m4 ; j v0 j< m4 ;
and v2 =
1
2
1
2
1
2
then the rectangle centered in the original one and having side lengths n ? 2vmax1 and m ? 2vmax2 (where vmax1 = max(v1; v10 ) and vmax2 = max(v2; v20 )) is periodic with period generated by both vectors. In particular, if v and v 0 are not parallel, this centered rectangle is lattice periodic, where the lattice is < v; v0 >.
4 The main theorem We now state and prove the main theorem of this paper.
Theorem 4.1 Let w : ZZ ! A be a double sequence and let pw (n; m) be its complexity function. Let us suppose that there exist two natural numbers n ; m such that pw (n ; m ) < n m , where = 100. Then w has a 2
0
0
periodicity vector.
0
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In order to prove this theorem we need two fundamental lemmas, to each of those we will dedicate a subsection. Each of them will be devoted to extend local periodicity obtained by Lemma 3.1 to \larger" portions of the plane. In this paper we will prove the main theorem with = 144, because we make use of Theorem 3.1 that synthesize Lemmas 3.2 and 3.3 with loss of information. A proof based on the use of the two lemmas reaches a constant = 100. Since the proof is already involved and hard to follow, we have chosen to let it this way, easier to read, even if the constant is higher. Let us give some de nitions and notations that will be useful in the following. ! ! v 0 De nition 4.1 Given a vector v = v 6= 0 2 ZZ , and a point P = (i; j ) 2 ZZ , let l(v; P ) be the line in IR that contains P and that has the same direction of v, i.e. the line of equation v y ? v x = v j ? v i. We de ne the v-Christo el line C (v; P ) through the point P to be the set of points (r; s) 2 ZZ that are, roughly speaking, immediately below l(v; P ). More precisely if v = 0 then C (v; P ) = f(i; s) j s 2 ZZg. If v 6= 0 then rstly de ne C (v) = f(r; s) 2 ZZ j b vv (r ? 1)c s b vv rcg: De ne now C (v; P ) = C (v) + P . We also de ne, given a real number h > 0, the line-segment S (v; h; P ) of length hjvj starting in P to be if v = 0 then S (v; h; P ) = C (v; P ) \ fv + P j 0 < hg. If v 6= 0 then rstly de ne S (v; h) = f(i; j ) 2 ZZ j 0 < i hv and b vv (i ? 1)c j < b vv icg[f(0; 0)g. De ne now S (v; h; P ) = S (v; h)+ P . 1
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Remark 4.1 Remark that our de nition of Christo el line implies that for
every point (x; y) in any v-line there are exactly two points between the four points of the form (x 1; y 1) that belong also to the same v-line. In other words the line is 4-connected and not 8-connected as the usual de nition (cf. [11], [21], [28], [29]). Indeed all the arguments and proofs in this paper can be easily modi ed to work also with the usual de nition. The fact that, by using this de nition, the number of points in any v-segment of length jvj is exactly jv1j + jv2j instead of max(jv1j; jv2j) will simplify some notations and proofs. !
De nition 4.2 A vector v = vv 2 ZZ is an -vector if m e ? 1: 0 jv j d pn e ? 1; 0 jv j d p Since the number of ?vectors is nite, we can nd the longest ?!vector having the same direction of v. Let us denote this vector vv = vv ;v . 1
2
2
0
1
0
2
1
;v
2
Let now R be an h k rectangle, h 2jv;vj, k ! 2jv;v j,!whose bottom left corner is the point P = (x; y) and let v = vv 6= 00 be an -vector. 1
2
1 2
We consider two cases. 1. (v 0 and v 0) or (v 0 and v 0). In this case set P = (x; y + k ? 2jv;v j) and P = (x + h ? 2jv;v j; y) (see next gure). 2. (v > 0 and v < 0) or (v < 0 and v > 0). In this case set P = (x ; y ) = (x; y + 2jv;v j) and P = (x ; y ) = (x + h ? 2jv;v j; y + k). 1
2
1
2
2
1
1
2
1
2
1
1
2
1
2
2
2
2
1
1
De nition 4.3 The (R; v)-Christo el stripe is the set of points that lie be-
tween the v-Christo el line through P1 = (x1; y1) and the v-Christo el line through P2 = (x2; y2), the points of these two lines included.
15
The intersection of R and the (R; v)-Christo el stripe is called the fundamental (R; v) discrete hexagon H (see next gure). P1
H
R
P2
We say that a vector v is a periodicity vector for a Christo el stripe S (or that S has period v) if v is a periodicity vector for the factor that has S as its shape.
Notice that the fundamental (R; v)-discrete hexagon H contains properly an h0 k0 rectangle R where h0 = h ? 2jv;v j, k0 = k ? 2jv;vj and the bottom left corner is (x ; y ) = (x; y), the same of R. Indeed the (R; v)-Christo el stripe contains a bisequence (Rz )z2ZZ of translates of R (cf. next gure) such that, if (xz ; yz ) is the bottom-left corner of Rz , then jxz ? x j + jyz ? y j = jzj (we may assume, thanks to the symmetry, that xz is increasing with z). These rectangles Rz are called the (R; v)-stripe rectangles, or (R; v)rectangles. Moreover the (R; v)-Christo el stripe is the union of this sequence of rectangles and the (R; v) fundamental discrete hexagon is the union of the rectangles Rz with 0 z < 2jv;vj + 2jv;v j. Notice also that the line-segments in the direction of v starting in P and P that are subsets of the discrete hexagon both contain at least 2jv;vj+2jv;vj points (in the case of v having the same direction of an axis, this number is strictly greater than 2jv;v j + 2jv;vj). 0
0
1
2
0
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1
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2
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1
2
16
2
R3 R2 R0 R1
R3 R2 R0 R1
Notice that an (R; v)-Christo el stripe does not contain R and, moreover, its intersection with R gives rise to the discrete fundamental hexagon, that is obtained as the set di erence between R and two \angles". Note also that the (R; v)-rectangles have side sizes strictly smaller than the ones of the original rectangle. The reasons of this de nition are two. The rst and more important is that this de nition is necessary if we want to apply the same unidimensional argument used in the proof of Theorem 2.2 in our two main lemmas of this section. The second reason is that with this de nition we can use in next main lemma the following proposition. Proposition 4.1 (Propagation of lattice periodicity) Suppose that a factor u of shape an h k rectangle R, h 4d pn e , k 4d pm e, has v and v0 as periodicity vectors, where v and v 0 are two -vectors. Let us also suppose that the union of the rst (l+1) (R; v)-stripe rectangles [ss l Rs is v0-periodic, for some natural number l. 0
= =0
17
0
Therefore, for any rectangle Rs; s = 0; : : :; l, there exists a factor u0s having as shape a rectangle R0s Rs that is < v; v 0 >-lattice periodic. This factor is an ^h k^ factor where h^ h ? 3d pn e + 3 and k^ k ? m 3d p e + 3. 0
0
Proof
By Theorem 3.1, the centered factor having sides h ? 2d pm e + 2 and k ? 2d pm e + 2, is lattice periodic where the lattice is generated by v and v0. Let us call Cs the shape of this centered rectangle. Let us consider now R0s Cs the (h ? 3d pm e +3; k ? 3d pm e +3)-rectangle having the same bottom-left point of Cs if the coordinates of v0 are both non negative or both non positive, and the same top-left point otherwise. Let us nally consider U = [ls R0s . Let (i; j ) and (i0; j 0) be two points in the shape of u0 that are in the same coset of < v; v0 >, i.e. (i ? i0; j ? j 0) = v^ 2< v; v0 >. All we have to prove is that w((i; j )) = w((i0; j 0)). Since U is composed by construction by at least jv0 j + jv0 j consecutive rectangles, there exist two integers n and n0 such that (i; j ) + nv0 2 U and (i0; j 0)+n0v0 2 U . Hence (i; j )+nv0 ?((i0; j 0)+n0v0) = v^+(n?n0 )v0 2< v; v0 >, i.e. (i; j ) + nv0 and (i0; j 0) + n0v0 are in the same coset. Since u is < v; v0 >periodic and U is contained in its shape, w((i; j ) + nv0) = w((i0; j 0) + n0v0). Since u0 has period v, w((i; j ) + nv) = w((i; j )) and w((i0; j 0) + n0v) = w((i0; j 0)). By the transitivity of equality the proposition is proved. 2 0
0
0
0
=0
1
4.1 First Lemma
2
This section is devoted to extend a local periodicity, given by Lemma 3.1, to a periodicity in a Christo el stripe. The main idea is to try to extend the local periodicity in the direction of the period. If it is possible to cover all a \large enough" Christo el stripe associated to this periodicity vector, then we obtain the thesis. If this is not possible, then at a certain point in this direction there must exist an \error" in the periodicity and, by Lemma 3.1, there must exist another periodicity. This new periodicity cannot be parallel to the original one by the unidimensional argument used in the proof of Theorem 2.2. Then we make one \turn" and try to extend this new local periodicity in its own direction. The bidimensional generalizations of the periodicity lemma allow us to prove that 18
it is possible to extend this local periodicity to a \large enough" Christo el stripe. The reason of that relies on the fact that, by extending, we bring with us the informations of the previous periodicity and of the \error". Roughly speaking, we are allowed to make one \turn" but not two \turns". We will need, in the proof of the rst lemma, the following results. Covering Lemma: Let v0 and v00 be two non parallel -vectors and R be an h k-rectangle with h 2d pn e ? 2 and k 2d pm e ? 2. Then there exists a vector v 2< v 0; v 00 > belonging to R. The proof of this lemma is left to the reader. 0
0
Corollary 4.1 Let v0 and v00 be two non parallel -vectors and R be an h krectangle with h 2d pn e ? 2 and k 2d pm e ? 2. Then, for any (x; y) 2 ZZ there exists (x0; y 0) 2 R such that (x; y) ? (x0 ; y 0) 2< v 0; v 00 >. Proof Consider the rectangle R0 = R ? (x; y). By previous lemma there exists a vector v 2< v0; v00 > belonging to R0. Hence the point (x0; y0) = (x; y) + v belongs to R and, so, (x; y) ? (x0; y0) 2< v0; v00 >. 2 0
2
0
Lemma 4.1 (Contradictory conditions) Let u be a factor having shape R . The following properties cannot hold together.
1
1
It has two non parallel periodicity vectors v0 and v00. There exists in R a factor uL having shape a rectangle RL that is 1
< v; v0 >-lattice periodic and it is \centered enough" in R , i.e. for any PL = (iL; jL ) 2 RL iL ? i d pn e ? 1; jL ? j d pm e ? 1 for any (i ; j ) belonging to the sides of R . This factor is an h^ k^ factor where ^h 3d pn e ? 3 and k^ 3d pm e ? 3. There exists a couple of points P^ and P^ = P^ + v such that w(P^ ) 6= w(P^ ). Moreover these points are \centered enough", i.e. j P^ ? ^ P^ g and (i ; j ) (i ; j ) j (d pn e ? 1; d pm e ? 1), for any P^ 2 fP; belonging to the sides of R . 1
0
1
1
1
0
1
1
0
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1
1
1
1
0
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19
1
1
Proof
Let R^ be the rectangle centered in R , i.e. the rectangle whose points P^ = (^i; ^j) are such that ^i ? i d pn e ? 1; ^j ? j d pm e ? 1 for any (i ; j ) belonging to the sides of R . By Theorem 3.1, the factor having shape the rectangle R^ centered in R is L =< v0; v00 >-periodic. Moreover it contains P^ and P^ and also it has as factor uL that has shape RL. 1
1
0
1
0
1
1
1
1
1
such that P^ + v^ is in RL(RL ? v). Consequently P^ + v^ belongs to both. Since P^ + v^ belongs to RL ? v, P^ + v + v^ = P^ + v^ belongs to RL. By the < v0; v00 >-periodicity w(P^ ) = w(P^ + v^) = b, and w(P^ ) = w(P^ + v^) = a. But P^ + v^ and P^ + v^ belong to RL and P^ + v + v^ = P^ + v^. Since uL is < v; v0 >-periodic we have a contradiction. 2 1
1
1
1
1
In the next lemma we will set the constant to be = 144. As we have 20
said previously, with a more involved proof we may reduce this constant to = 100.
Lemma 4.2 (First Lemma) Let w : ZZ ! A be a double sequence whose n m for some n 4p ; m complexity function veri es p ( n ; m ) < w p 4 2 IN . 2
0
0
0
Then there exists an -vector v = vv
1 2
0
0
0
! , and a (R; v){Christo el stripe
S such that R = [i; i + h] [j; j + k] with h 3 pn ; k 3 pm and v is a periodicity vector of S . 0
0
Before giving the complete proof of this lemma, we want to give a detailed sketch of it, in order to allow an easier reading. The \ingredients" of this proof are 1. an unidimensional argument similar to the one used in the proof of Theorem 2.2; 2. Lemma 3.1 (local 1-periodicity); 3. Theorem 3.1 (lattice periodicity); 4. Proposition 4.1 (propagation of lattice periodicity); 5. Lemma 4.1 (contradictory conditions). By Lemma 3.1 we start with an h k-factor u that admits a periodicity vector v. Let us consider the sh(u) v-Christo el stripe S , where sh(u) is the h k-rectangle that is the shape of u. If S is v-periodic the statement of the lemma is true. If S is not v-periodic, then there exists a rectangle Rt where there is an \error" in this periodicity, i.e. there exist two points (i; j ) and (i0; j 0) = (i; j )+ v in Rt such that w((i; j )) = b and w((i0; j 0)) = a; a 6= b. Again for Lemma 3.1 there must exist another periodicity vector v0. A claim that uses an unidimensional argument similar to the one used in the proof of Theorem 2.2 states that v and v0 are not parallel. 1
1
1
21
Theorem 3.1 states that a rectangle centered in Rt is L =< v; v0 >-lattice periodic. Now we consider the (R0; v0)-Christo el stripe S , where R0 is a rectangle that is v0-periodic and that contains a rectangle L-lattice periodic. If the whole stripe S is v0-periodic, then the lemma is proved. If it is not so, there will exist another rectangle R00 where there is an \error" in v0-periodicity. Moreover in R00 there is a couple of points (i; j ) + v0 and (i0; j 0) + v0 = (i; j ) + v + v0 such that w((i; j ) + v0) = b and w((i0; j 0) + v0) = a. These two points represent a kind of \memory" of the error in the v-periodicity of stripe S . By Lemma 3.1 there must exist another periodicity vector v00. By using again an unidimensional argument similar to the one used in the proof of Theorem 2.2 we get that v0 and v00 are not parallel. By Proposition 4.1, R00 contains also a rectangle that is L-lattice periodic. We get a contradiction on the existence of R00 from Lemma 4.1. Hence the whole stripe S is v0-periodic. The reason of that relies on the fact that R00 contains some informations of the previous periodicities and of the \error". Roughly speaking, we are allowed to make one \turn" but not two \turns". 2
2
1
2
Proof
Let u be a factor of shape an [0; h] [0; k] rectangle R with h = n ? and k = m ? d pm e ? 1. By Lemma 3.1, u admits a periodicity vector, say v. We may assume that its coordinates are greater than or equal to 0, because of symmetry (the other cases can be proved in the same way). Let S be the (R; v)-Christo el stripe. If S is v-periodic, it is enough to set S = S in order to prove the lemma. Let us suppose it is not the case. Let t be the smallest integer, that we may assume positive without loss of generality, such that the factor ut having shape the union Ut of the rst t + 1 (R; v)-rectangles Rs , 0 s t is not v-periodic. Therefore there exist two points (i; j ) and (i0; j 0) = (i; j ) + v in Ut such that w((i; j )) = b and w((i0; j 0)) = a, as shown in the gure:
d pn0 e ? 1
0
0
0
1
1
1
22
a
b
Recall also that Rt is an h0 k0 rectangle, where h0 = h ? 2jv;v j and k0 = k ? 2jv;v j (for the de nition of v;v and v;v cf. 4.2). Since t is increasing, the error is either on the right or on the top side of the rectangle. Let (x; y) be the bottom-left corner of Rt. Let us distinguish now three cases: 1
2
1
2
1. x i0 < x + 4(d pn e ? 1); j 0 = y + k; 0
x + 4(d pn e ? 1) i0 x + h; j 0 = y + k OR i0 = x + h; y + 4(d pm e ? 1) j 0 y + k; 3. i0 = x + h; y j 0 < y + 4(d pm e ? 1) 2.
0
0
0
23
Case 1
Case 2
Case 3
We may introduce a new point P 0, according to the cases above-stated. In case 1 P 0 = (x ? 3(d pn e ? 1); y + d pm e ? 1), in case 2 P 0 = (x + d pn e ? 1; y + d pm e ? 1) and in case 3 P 0 = (x + d pn e ? 1; y ? 3(d pm e ? 1)). Let us consider the h k rectangle R0 having P 0 as bottom-left corner. This new rectangle has been chosen such that the points (i; j ) and (i0; j 0) are \centered enough" in R0, i.e. (i; j ) ? (i,j) 2(d pn e ? 1; d pm e ? 1) for any (i; j ) 2 f(i; j ); (i0; j 0)g and (i,j) belonging to the sides of R0. Let now u0 be the factor of w having shape R0. By Lemma 3.1, u0 admits a periodicity vector, say v0. 0
0
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Claim: v cannot be parallel to v0.
The technique used here to prove the claim is similar to the one used in the Theorem 2.2. Let us suppose by the absurd that v is parallel to v0. Since we are supposing t to be positive and since all the factors having shape an (R; v)-rectangle inside the fundamental hexagon have period v, then t > 2(jv;v j + jv;vj). Each (R; v)-rectangle Rs is obtained by translating rectangle Rs? of one unit line. Let s be this translation and let Q ; Q ; : : :; Qt be the unique sequence of points such that Q 2 R , Qs = s(Qs? ), 1 s t, and Qt = (i0; j 0). By de nition it is easy to prove that (i; j ) belongs to this 1
2
1
0
0
0
24
1
1
sequence and, more precisely, that (i; j ) = Qt , t0 = t?jv j?jv j, where v and v are the components of vector v. Let u = a a at be the unidimensional word where as = w(Qs ), 0 s t. Since v is a periodicity vector for the factor ut, the pre x u? of u having length juj ? 1 has period jv j + jv j. By the choice of factor u0 all the factors having shape a Rs, t ? (jv j + jv j + jv;v j + jv;vj) s t, have period v0. Consequently, the sux w of the word u of length jwj = jv j + jv j + jv;v j + jv;vj has period jv0 j + jv0 j, where v0 and v0 are the components of vector v0. Hence, the pre x w? of w of length jw? j = jwj ? 1 = jv j + jv j + jv;v j + jv;vj ? 1 has both periods. Since jv;v j = jv;v j and jv;vj = jv;v j, jv j + jv j + jv;vj + jv;v j ? 1 jv j + jv j + jv0 j + jv0 j ? 1. Therefore we can apply the Theorem of Fine and Wilf to w? . We obtain that w is d = gcd(jv j + jv j; jv0 j + jv0 j)-periodic. Let us consider, now, the word w. This unidimensional word is (jv0 j + 0 jv j)-periodic. The factor w? of w has length greater than or equal to jv0 j + jv0 j. Thus it contains surely a factor of length jv0 j + jv0 j that is dperiodic. Therefore, by Proposition 2.1, w is d-periodic, and, consequently, (jv j + jv j)-periodic. This is absurd because at 6= at , t0 = t ? jv j ? jv j, and this concludes the proof of the claim. 1
0
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1
0 1
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2
So the h k factor u0 has period v0 and it has a periodic factor whose shape is an h0 k0 rectangle that is v-periodic where h0 = h ? 3(d pn e ? 1) and k0 = k ? 3(d pm e? 1). This factor has shape the intersection between the shape of u and the shape of u0. Notice that h0 n ? 4d pn e +2 8d pn e? 10 and k0 m ? 4d pm e + 2 8d pm e ? 10. Let us now consider the (R0; v0)-Christo el stripe S . If the whole stripe S is v0-periodic, then the lemma is proved. If it is not the case, let us consider the minimum integer t0 (we may again assume it is positive) such that the factor u0t having shape the union Ut0 of the rst t0 + 1 (R0; v0)-rectangles R0s, 0 s t0, is not v0-periodic. Then there exist two points (i ; j ) and (i0 ; j 0 ) = (i ; j ) + v0 in R0t such that w((i ; j )) = c and w((i0 ; j 0 )) = c0, c 6= c0. By Proposition 4.1 for any rectangle R0s ; s = 0; : : : ; t0, there exists a factor 0 us having as shape a rectangle R0s Rs that is < v; v0 >-lattice periodic. This factor is an ^h k^ factor where ^h h0 ? 3d pn e +3 n ? 7d pn e +5 5d pn e ? 7 and k^ k0 ? 3d pm e + 3 m ? 7d pm e + 5 5d pm e ? 7. 0
0
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25
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By the choice of R0, by the v0-periodicity and by the fact that v0 is an -vector, one has that for any sequence of jv;v j + jv;v j consecutive (R0; v0)rectangles, there exists a rectangle R0s that contains a \translate by v0" of the \initial" couple of points (i; j ) and (i0; j 0) that is \centered enough" in R0s, i.e. (i; j ) ? (i,j) (d pn e? 1; d pm e? 1)) for any (i; j ) 2 f(i; j ); (i0; j 0)g and (i,j) belonging to the sides of R0s . For a \translate by v0" we intend a couple of points of the form (i; j )+ v0 and (i0; j 0) + v0, for an integer (cf. gure below). 0
1
0
2
0
0
a
v
v’
b
a
v’
v
v’
b
a
v’
b
v’
v
a
v’
b
v
!
0 Let us suppose now that v0 = vv0 is such that v0 0 and v0 0. The proof in the other cases is analogous. Let (x0; y0) be the bottom-left corner of R0t . We introduce a new point P 00 = (x0 ? 2v0;v ; y0). Let us consider the h k rectangle R00 having P 00 as bottom-left corner. This new rectangle has been chosen such that it contains a sequence of 2(jv;v j + jv;v j) consecutive (R0; v0)-rectangles. 1
1
2
2
0
1
1
2
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26
0
Let now u00 be the factor of w having shape R00. By Lemma 3.1, u00 admits a periodicity vector, say v00. By using an argument analogous to that used in previous claim, it turns out that v00 is not parallel to v0. Moreover there exists a (R0; v0)-rectangle R R00 that contains a \translate by v0" of the \initial" couple of points (i; j ) and (i0; j 0) that is \centered enough" in R , i.e. (i; j ) ? (i ; j ) (d pn e ? 1; d pm e ? 1)) for any (i; j ) 2 f(i; j ); (i0; j 0)g and (i ; j ) belonging to the sides of R . Recall that by Proposition 4.1 for any (R0; v0)-rectangle R0s ; s = 0; : : : ; t0, there exists a factor u0s having as shape a rectangle R0s Rs that is < v; v0 >lattice periodic. This factor is an ^h k^ factor where ^h h0 ? 3d pn e +3 n ? 7d pn e +5 5d pn e ? 7 and k^ k0 ? 3d pm e + 3 m ? 7d pm e + 5 5d pm e ? 7. This fact means that in R there exists a factor uL having shape a rectangle RL that is < v; v0 >-lattice periodic and it is \centered enough" in R , i.e. for any PL = (iL; jL) 2 RL iL ? i d pn e ? 1; jL ? j d pm e ? 1 for any (i ; j ) belonging to the sides of R . This factor is an h k factor where h 4d pn e ? 6 and k 4d pm e ? 6. p p If n 4 and m 4 then h 3d pn e ? 3 and k 3d pm e ? 3. Let u be the factor having shape R , the Lemma 4.1 gives a contradiction. 2 1
1
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4.2 Second Lemma
In the main result of this section, we extend the local periodicity v of a Christo el stripe obtained in Lemma 4.2 to the whole plane. The main point in this procedure is that, starting from the -vector v, we can obtain other vectors that are parallel to v that are periodicity vectors for stripes parallel to the original one. In other word the whole plane can be subdivided in \large" Christo el stripes all associated to v or to any vector parallel to v, each of them admitting an -vector as periodicity vector having same direction of v. Since the number of -vectors is nite, then the vector v having same direction of v and length the product of the lengths of all -vectors parallel to v is a periodicity vector for the whole plane. Remark that from the proof it turns out that the length of such vector can \explode" with respect to the size of the two numbers (n ; m ). And this 0
27
0
can really happen, as shown in the examples of next section. The techniques used in this lemma are elementary, and we make use of the unidimensional periodicity lemma.
Lemma 4.3 Let S be a Christo el stripe associated to an -vector v. Suppose that S contains a s t rectangle with s > pn and t > pm , v is a periodicity vector of S . then there exists a vector parallel to v that is a periodicity vector for the whole double sequence w. 0
0
Proof
Let us suppose that the two parallel lines r and r associated to the Christo el stripe S (that are also parallel to v) have a slope greater than or equal to 0. The proof in the case in which the slope is smaller than 0 is analogous. Let 1
and
2
S = S + (1; ?1) = fz 2ZZ j z = v + (1; ?1); v 2 S g 1
S = S + (?1; 1) = fz 2ZZ j z = v + (?1; 1); v 2 S g and let us de ne two functions f and g as follows: 2
f (S ) = S [ S; g(S ) = S [ S : 1
2
g(S)
28
f(S)
If for all k 2 IN v is a periodicity vector for both f k (S ) and gk (S ), then v is a periodicity vector for the whole double sequence w. Otherwise, there exists a k 2 IN such that f k (S ) and gk (S ) admit v as a periodicity vector, while either f k (S ) or gk (S ) contain (contains) two points P = (i ; j ) and P = (i ; j ) such that +1
1
1
+1
0
0
0
1
1) P + v = P ; 2) w(P ) = a and w(P ) = b, where a and b are two di erent letters. 0
1
0
1
Let us suppose that these two points are in f k (S ) (the case in which they are in gk (S ) is analogous). Since f k (S ) is still a stripe associated to v then neither P nor P belongs to f k (S ) and, consequently, they belong to the set (f k (S ) ? f k (S )). +1
+1
+1
0
1
v
By Lemma 3.1 the rectangle R centered in P = (i ; j ) of shape [i ? 0
0
0
0
d pn e; i0 + d 5pn 0 e? 2] [j0 ?d 4pm 0 e; j0 + d 5pm 0 e? 2] has a new periodicity vector v0. Notice that this rectangle contains also P1. Let us examine the direction of this new periodicity vector. If v0 is not parallel to v then, since the stripe f k (S ) contains an s t rectangle with s pn0 and t pm 0 , there would exists 4 0
29
2 f+1; ?1g such that both P + v0 and P + v0 are in the stripe f k (S ). In this case a = w(P ) = w(P + v0) = w(P + v0) = w(P ) = b; that is an absurd. Therefore v0 must be parallel to v. Let us consider now the (R; v0)-Christo el stripe. It is an in nite stripe that has v0 as periodicity vector and, by very de nition, it contains an s t rectangle with s pn and t pm . Let us suppose by absurd that v0 is not a periodicity vector for this new stripe. Let i be the smallest integer, that we may assume positive without loss of generality, such that the factor ui, having as shape the union Ui of the rst i +1 (R; v0)-rectangles Rs, 0 s i, is not v0-periodic. Then there exist two points P 0 ; P 0 2 Ui such that 0
0
1
0
1
1
0
0
0
1
1) P 0 + v0 = P 0 ; 2) w(P 0 ) = c and w(P 0 ) = d, with (c 6= d). 0
1
0
1
Let (x; y) be the bottom left corner of rectangle Ri and let R^ i be the h k rectangle having (x ? (d pn e ? 1); y ? (d pm e ? 1)) as bottom left corner, with h = n ? d pn e ? 1 and k = m ? d pm e ? 1. By Lemma 3.1, the factor that has R^i as shape admits a periodicity vector, say v00. An argument analogous to the one used in the proof of the claim of the Lemma 4.2 can be used to prove that v00 cannot be parallel to v0. Therefore v00 incides v0. Notice also that since R^ i contains Ri , also the factor that has shape Ri admits v00 as periodicity vector. Since factor ui? having as shape the union Ui? of the rst i (R; v0)rectangles Rs, 0 s (i ? 1), is v0-periodic and since the sides of R0 have lengths greater than the respective components of v0, there exists 2ZZ such that 0
0
0
0
0
0
1
1. 2. 3. 4.
1
P + v0 = P 2 R0 w(P ) = w(P ) = a (P ) + v0 = P 2 R0 w(P ) = w(P ) = b 0
0
0
1
0
1
1
1
30
By construction and since P and P belong to the set (f k (S ) ? f k (S )), also P and P belong to the set (f k (S ) ? f k (S )). 0
0
+1
1 +1
1
Since the stripe f k (S ) contains an s t rectangle with s pn and t pm , ^ 00 and P + v ^ 00 are there would exists ^ 2 f+1; ?1g such that both P + v ^ 00; P + ^v00 all belong in the stripe f k (S ). In this case, since P ; P ; P + v 0 00 to R that has period v ^ 00) = w(P ) = b 6= a ^ 00) = w(P + v a = w(P ) = w(P + v that is an absurd. Hence there exists another stripe parallel to f k (S ), overlapping f k (S ) and that contains some Christo el lines not contained in f k (S ), which has a periodicity vector parallel to the periodicity vector of S and which contains an s t rectangle with s pn and t pm . Iterating previous argument we can cover the plane with such stripes. Therefore we are able to nd a periodicity vector for the whole double sequence w. This vector is parallel to v and its length is the product of the lengths of the periodicity -vectors of the di erent stripes. Notice that the number of di erent stripes can be in nite but the number of di erent vectors is nite. 2 0
0
0
0
0
0
1
0
1
1
1
1
0
0
4.3 Proof of the Theorem
Proof
p
p
First case n0 4 orpm0 4 . p Let us suppose m0 4 , the proof for n0 4 is analogous.
Let B = Am a new alphabet (i.e. the letters of B are the words of length m over A). Let us de ne a new bidimensional word u in the following way: 0
0
u[i; j ] = b 2 B () b = w[i; m j ]w[i; m j + 1] w[i; m j + m ? 1] In other words we have grouped the rows of w in packets of size m . It 0
0
is not dicult to prove by construction that 31
0
0
0
u has a periodicity vector () w has a periodicity vector. The complexity function pu of u is such that pu (n ; 1) pw (n ; m ) < 1 n m n Therefore, by Theorem 2.2, any line of u is!periodic with period smaller than or equal to n . Consequently u has n0 ! as periodicity vector. Hence also w has a periodicity vector. p p Second case n 4 or m 4 . The proof of this case is a direct consequence of Lemma 4.2 and of Lemma 4.3. 2 0
0
0
0
0
0
0
0
0
0
5 Examples In this section we show that there exists a sequence of bidimensional words (wn )n such that pwn (n; n) = o(n ), even if wn admits a periodicity vector which has length greater than or equal to 2
9
n
n +o( n ) n 2 lg n :
p
p
2 lg
p
For any natural n 9, let us take m = b nc and let x ; x m be the increasing sequence of prime numbers smaller than or equal to m. Let us consider the sequence z ; z m , such that for any i, 1 i (m), zi is the greatest power of xi smaller than or equal to m, i.e. 1
(
1
(
)
)
zi = xih m; xih > m. +1
It is easy to prove that for any i, 1 i (m), zi > pm. Let us now de ne the binary bidimensional word wn as the juxtaposition of the (2(m) + 1) vertical stripes de ned in the following. 32
Each one of these stripes except the rst one and the last one, which have in nite width, have width n . If j is odd, then wn (x; y) = 0 for any (x; y) belonging to the j -th stripe. If j is even, j = 2k for some integer 1 k (m), then wn (x; y) = 0 for any (x; y) belonging to the j -th stripe such that zk is a divisor ! of 0 y, wn (x; y) = 1 elsewhere. In this way this stripe admits z as k periodicity vector. 2
n2
n2
n2
n2
n2
0
1
1
0
0
1
1
0
0
0
0
0
0
1
1
0
0
0
0
0
0
1
1
0
0
0
0
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
0
0
0
0
1
1
0
0
1
1
0
0
1
1
0
0
0
0
0
0
0
0
0
0
1
1
0
0
1
1
0
0
1
1
0
0
0
0
0
0
1
1
0
0
1
1
0
0
1
1
1
1
0
1
1
0
0
1
1
0 0
0
0
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
!
It is easy to see that any periodicity vector v = vv of the whole wn is such that v = 0. Moreover the length of this periodicity vector is the least common multiple of all the natural numbers smaller than or equal to n, that is also equal to the product of all the zi. Let us give now a rough estimate of this number. Since for any i, 1 i m(m), zmi > m = , the product of all the zi is greater than m m = = p o m , by the Prime Number Theorem. Since m = b nc, then we m m obtain that 1 2
1
1 2
(
+ ( 2 lg( ) ) 2 lg( )
33
) 2
m m = = n (
) 2
n +o( n ) n 2 lg n
p
p
2 lg
We want now to prove that pwn (n; n) = o(n ). If a square of side n is completely inside a stripe or a semiplane where the ! word wn has periodicity vector z0 , then the number of di erent factors i is zi. If the square is completely inside a stripe or a semiplane where wn has ! 0 periodicity vector 1 , then there is only a factor of side n. If it is between two consecutive stripes or a stripe and a semiplane, the number of di erent factors is 2(n ? 1)zi. In the rst case, i.e. in the case in which the square is completely inside a stripe or a semiplane, we get at most 1 + i m zi factors, while in the second case we get at most i m 2(n ? 1)zi factors. By the Prime Number Theorem, since for any i, zi m, it follows that 2
( ) =1
( ) =1
1 + i m zi + i m (2(n ? 1)zi) = o(m (n ? 1)) = o(n ) ( ) =1
( ) =1
2
2
Previous example shows that di erent phenomena happen in the one and in the two dimensional case. Indeed, in the one dimensional case, pw (n) < n implies not only that w has a period but also that this period is bounded by n. In the bidimensional case, if pwn (n; n) < n , the bound of the length of the periodicity vector is not even polinomial in n. 2
6 Concluding Remarks In this paper we proved a weak version of the Nivat conjecture. We hope that with this approach it will be possible to settle completely the conjecture. Very recently J. W. Sander and R. Tijdeman (cf. Example 5 of [33]) have shown, with a simple but interesting example, that it is not possible to extend in a completely analogous way the Nivat's conjecture to the k-dimensional case. His example shows that for any n there exists a k-dimensional sequence w that has no periodicity vector and such that the number of factors having shape a full dimensional hypercube of side n is n + 1. 2
34
We now state the following conjecture that somehow extend the Nivat's conjecture to the k-dimensional case. Conjecture: if the number of factors having shape a full dimensional hypercube of side n is smaller than or equal to n , then the k-dimensional sequence w is k ? 1 dimensional-lattice periodic L, i.e. the minimal number of generators of L is k ? 1. 2
Acknowledgements. We thank J. C. Lagarias and D. Bernardi for discussions and comments.
References [1] A. Amir, G.E. Benson: Alphabet Independent Two-Dimensional Pattern Matching. Proc. 24th ACM Symp. Theory on Comp. (1992), pp. 59{68 [2] A. Amir, G. E. Benson: Two-Dimensional Periodicity and its Applications. Proc. 3rd ACM-SIAM Symp. on Discr. Algorithms (1992), pp. 440{452 [3] A. Amir, G. E. Benson: Two-Dimensional Periodicity in Rectangular Arrays. SIAM Journal of Computing, Vol. 27, No. 1 (1998), pp. 90{106 [4] A. Amir, G. E. Benson, M. Farach: An Alphabet Independent Approach to Two Dimensional Pattern Matching. SIAM Journal of Computing Vol. 23, No. 2 (1994) pp. 313-323, (preliminary version appeared in STOC 92). [5] A. Amir, M. Farach: Ecient Matching of Nonrectangular Shapes. Annals of Mathematics and Arti cial Intelligence, special issue on the Foundations of Aarti cial Intelligence, 1991, No. 4, pp. 211-224. [6] A. Amir, M. Farach: Two Dimensional Dictionary Matching. Information Processing Letters, 1992, Vol. 44, pp. 233-239. [7] V. Berthe, L. Vuillon: Tiling and Rotations. Pretirage 97-19, Institut de Mathematiques de Luminy. 35
[8] J. Cassaigne: Communications of the Workshop \Complexite des suites doubles", Marseille, June 2-4 1998. [9] J. Cassaigne, D. Bernardi: Private communication, 1997. [10] J. Cassaigne: Private communication, 1999. [11] I. Debled, J.P. Reveilles: A Linear algorithm for segmentation of digital curves IEEE Int. Journ. P.R.A.I. (1995). [12] B. N. Delone [B. N. Delaunay], N. P. Dolbilin, M. I. Shtogrin, R. V. Galiulin: A local criterion for regularity of a system of points. Sov. Math. Dokl., Vol. 17, No. 2 (1976), pp. 319{322. [13] N. P. Dolbilin, J. C. Lagarias, M. Senechal: Multiregular Point Systems. To appear in Discrete and Comp. Geom. [14] N.J. Fine, H.S. Wilf: Uniqueness Theorem for Periodic Functions. Proc. Am. Math. Soc., Vol. 16 (1965), pp. 109{114. [15] Z. Galil, R. Giancarlo: On the Exact Complexity of String Matching: Upper Bounds. SIAM J. Comp., Vol. 20, No. 6, pp. 1008{1020. [16] Z. Galil, K. Park: Alphabet-Independent Two-Dimensional Witness Computation. Siam J. Comput., Vol. 25, No. 5 (1996), pp. 907{935. [17] Z. Galil, K. Park: Truly Alphabet Independent Two-Dimensional Pattern Matching. Proc. 33th IEEE Symp. on Foundations of Computer Science (1992), pp. 247{256. [18] R. Giancarlo, F. Mignosi: Generalizations of the Periodicity Theorem of Fine and Wilf. CAAP94, Lecture Notes in Computer Science, Vol. 787, pp. 130-141. [19] M. Koskas: Complexites de Suites de Toeplitz. Discr. Math., 83 (1998), pp. 161-183. [20] Lothaire: Combinatorics on Words. Encyclopedia of Mathemathics and its Applications, Vol. 17, Addison-Wesley 1983. 36
[21] Lothaire: Algebraic Combinatorics on Words. Available for the moment at URL http://www-igm.univ-mlv.fr/berstel/Lothaire. [22] R.C. Lyndon, P.E. Schupp: Combinatorial Group Theory. Springer Verlag, 1977. [23] R.C. Lyndon, M.P. Schutzenberger: The Equation am = bncp in a Free Group. Michigan Math. J., Vol. 9, No. 4 (1962), pp. 289{298. [24] M. Morse, G.A. Hedlund: Symbolic dynamics II. Sturmian trajectories. Amer. J. Math., Vol. 62 (1940), pp. 1-42. [25] M. Nivat: Invited talk at ICALP'97. [26] C. Radin: The pinwheel tilings of the plane. Ann. Math., Vol. 139 (1994), pp. 661{702. [27] M. Regnier, L. Rostami: A Unifying Look at d-Dimensional Periodicities and Space Coverings. Proc. 4th Symposium on Combinatorial Pattern Matching (1993), pp. 215{227. [28] J.P. Reveilles: Geometrie Discrete, Calcul en nombres entiers et algorithmique. These D'E tat, Universite Louis Pasteur, Strasbourg, (1991). [29] J.P. Reveilles: Combinatorial pieces in digital lines and planes. Vision Geometry, IV, San Diego, CA, 1995, pp. 23-34. [30] A. Rosenfeld, A.C. Kak: Digital Picture Processing. Academic Press, 1982. [31] J. W. Sander, R. Tijdeman: Low complexity functions and convex sets in ZZ k . To appear in Mathem. Zeitschr. [32] J. W. Sander, R. Tijdeman: The complexity of functions on lattices. To appear in Theoret. Comp. Sci. [33] J. W. Sander, R. Tijdeman: The rectangle complexity of functions on two-dimensional lattices. Theoret. Comp. Sci., to appear. [34] L. Vuillon: Combinatoire des motifs d'une suite sturmienne bidimensionnelle. Theor. Comp. Sci., Vol. 209 (1998), pp. 261-285. 37
[35] L. Vuillon: Contribution a l'etude des pavages et des surfaces discretisees. PhD. thesis, Aix-Marseille II, December 1996. [36] L. Vuillon: Local con gurations in a discrete plane. Bull. Belg. Math. Soc., to appear.
38
Michel Koskas y
Filippo Mignosiz
Abstract
We prove that, given a double sequence w over the alphabet A (i.e. a mapping from ZZ 2 to A), if there exists a pair (n0; m0) 2 ZZ 2 such that 1 pw (n0 ; m0) < 100 n0 m0, then w has a periodicity vector, where pw of w is the complexity function of w.
1 Introduction In combinatorics on words the notions of complexity and periodicity are of fundamental importance. The complexity function of a formal language counts, for any natural number n, the number of words in the language of length n. The complexity function of a word ( nite, in nite, biin nite) is the complexity function of the formal language whose elements are all the factors (or blocks, or also subwords) of the word. The Morse-Hedlund Theorem states that there exists an important relationship between periodicity and complexity. In particular it states that for any biin nite word w if the number of its di erent factors of length n is less
Dipartimento di Matematica ed Applicazioni, Universita degli Studi di Palermo,
[email protected] y
Faculte de Mathematiques et d'Informatique, Universite de Picardie Jules Verne,
[email protected] z
Dipartimento di Matematica ed Applicazioni, Universita degli Studi di Palermo,
[email protected]
1
than or equal to n, then the word is periodic. Moreover the period of w is smaller than or equal to n. Several generalizations of the complexity function exist in the literature. One of the most known is the complexity function pw (n; m) that counts the number of di erent rectangles of size n m that are \factors" of the double sequence w. This is the one that is considered and studied in this paper. Its rigorous de nition will be given in next section. In their seminal and fundamental work (cf. [1], [2], [3]) Amir and Benson introduced the notion of bidimensional periodicity (in particular the notions of simmetry and periodicity vector) and proved theorems analogous to the \periodicity lemma" (cf. [23]). After them, many researchers have worked on bidimensional periodicity and its applications (cf., for instance, [4], [5], [6], [16], [17], [18], [27]). In this paper we prove an extension of the Morse-Hedlund Theorem to the bidimensional case. On this subject let us state a conjecture (cf. [25]), known among the researchers in the eld of Combinatorics on Words as the Nivat's conjecture on bisequences. The rigorous de nitions of the notations used in it will be given in next section. Conjecture: If there exists a pair (n; m) such that the complexity function pw (n; m) of the double sequence w veri es pw (n; m) nm, then w has at least a periodicity vector. Many researchers have worked on this conjecture and on a related one of L. Vuillon that is presented in this paper in Example 2.1. Some partial results on this conjecture have been proved for small values of m or n (cf. [31], [32], [33]). The diculty of this conjecture is related also to the fact that all the known proofs of the Morse-Hedlund Theorem are intrinsecally \unidimensional", in the sense that they make use of properties of \unidimensional" words, such as the possibility of concatenating one letter to the right or to the left of a word. For such properties there are, to our knowledge, no generalizations that it is possible to use in the bidimensional case. Recently, we have been able to give a new proof of the Morse-Hedlund Theorem by using the periodicity Theorem of Fine and Wilf. For this theorem, which is the tight version of the periodicity lemma, there exist generalizations to the bidimensional case. This fact has suggested us to try to 2
extend our proof in order to settle above conjecture. Indeed, in this paper we prove a weak version of above conjecture, making use of the tools and techniques developed by the researchers in the eld of bidimensional periodicity. In particular we make use of some bidimensional generalizations of the periodicity lemma. Since there exist also bidimensional generalizations of the Theorem of Fine and Wilf (cf. [18]), we hope that this approach will allow to settle the original Nivat's conjecture. The remainder of this paper is organized as follows. In the second section we give the new proof of the Morse-Hedlund Theorem, de ne basic notations, discuss some preliminaries and give examples. In the third section we state and demonstrate some preliminaries lemmas. In the fourth section we prove the main theorem after having demonstrated two fundamental lemmas. Finally, in the last two sections we give a family of examples that shows some deep di erences between the unidimensional and the bidimensional case, and make some concluding remarks.
2 Preliminaries For any notation not explicitely de ned in this paper we refer to [18], [20] and [22]. A word w = a a an has period p if ai = aj for any 1 i; j n such that i j (mod p). Note that, following this de nition, any natural number p > jwj turns out to be a period of w, and, in this case, it is called an improper period of w. The smallest period of w is called the period of w. Notice that classically (cf. [20]) improper periods are not considered as periods. 1 2
We start with an easy proposition.
Proposition 2.1 Let w be a word of length n and period p n. If there exists a factor u of w of length juj p and period q where q divides p, then w has period q.
3
Proof
Let w = a an and u = ah ak , where 1 h < k n; j u j= k ?h p. Let i and j be two integers lying between 1 and n such that i j (mod q). There exist two integers i0 and j 0 such that i i0 (mod p), j j 0 (mod p) and h i0; j 0 k. Since i0 j 0 (mod q) and 1 i0; j 0 k, ai = aj . Since i i0 (mod p) and j j 0 (mod p), ai = ai and aj = aj and the conclusion holds. 2 1
0
0
0
0
The following theorem is a classical result in combinatorics on words (cf.[14], [20] and [21]).
Theorem 2.1 (Fine and Wilf, 1965) Let w be a p-periodic and q-periodic
word. If jwj p + q ? gcd(p; q) then w is gcd(p; q)-periodic.
Above theorem is also tight. Indeed for any p, q, p < q such that p does not divide q, there exists a word w; jwj = p + q ? gcd(p; q) ? 1, that is p-periodic and q-periodic but it is not gcd(p; q)-periodic. Let us consider an (unidimensional) word w and a positive integer n. We de ne the complexity function of w
pw (n) =j fu j u is a factor of w of length ng j The Morse-Hedlund Theorem (cf. [24]) states that there exists an important relationship between periodicity and complexity. Here we state the version relative to biin nite (two sides) words. The proof of the theorem that we give is a new one and it makes use of the Theorem of Fine and Wilf. The main idea of this proof is that the hypothesis implies the existence of a \local" period in any position of the biin nite word. The Theorem of Fine and Wilf, together with the Proposition 2.1, forces all these \local" periods to be the same all over the biin nite word. This is also the general idea of the proof of our main theorem in Section 4. Moreover the same \unidimensional" technique used here will be used also in both the proofs of our two main lemmas. We can now state and give our new proof of the Morse-Hedlund Theorem. 4
Theorem 2.2 (Morse and Hedlund) Let w be a biin nite word. If there exists an integer n such that pw (n) n, then w is periodic. Moreover the
period of w is smaller than or equal to n.
Proof
Let n be the smallest integer such that pw (n ) n . Let us consider in w the n + 1 positions 0; 1; ; n . Each position determines uniquely the factor of length n which starts in this position. 0
0
0
0
0
0
a0 a1 0
1
an
0
2
an
-2
n -2
0
n -1
0
an
-1
0
n
0
0
By the pigeon's holes principle two of these n + 1 positions must determine the same factor of length n (by hypothesis pw (n ) n ). Let i < j be two of such positions such that j is the smallest position greater than i that determines the same factor of length n determined by i. Hence, for h = 0; ; n ? 1, one has that ai h = aj h. This is equivalent to say that the word v = ai aj n ? has period p = j ? i. The word v has length n + p . Moreover, by the choice of j , p n is the minimal period of v. 0
0
0
0
0
0
+
+ 0
0
1
1
1
1
0
i
+
j
n0
0
i+n 0-1
p =j?i
v = a i aj
1
j+n 0-1
n ?1
+ 0
We want to prove that p is a period for the whole word w. Let us suppose, by contradiction, that p is not a period for w. Therefore there exists a position H such that aH 6= aH ?p . In what follows we suppose that H > j + n ? 1. The proof in the case where H < i + p is analogous. 1
1
1
0
1
5
Let us suppose that H is the smallest integer greater than j + n ? 1 such that aH 6= aH ?p . Let u = aH ?n ?p aH be the word of length (n + p ) which last letter is aH and let u? be its pre x of length j u? j=j u j ?1. Since H is the smallest integer greater than j + n ? 1 such that aH 6= aH ?p and j u? j= (n + p ? 1), then u? has period p . Moreover, since p is the minimal period of the word v = ai aj n ? , then it is also the minimal period of the word v = ai aH ? whose sux of length (n + p ? 1) is the pre x u? of u of the same length. Therefore the word b = ai ai p ? is a primitive one. By a lemma of Aldo de Luca all the p words conjugates to b are di erent. This fact implies that if we consider the rst p positions of the word u, since they determine all the p conjugates of the word b of length p < n , they determine di erent words of length n . If we consider the rst (n + 1) positions of the word u (i.e. H ? n ? p + 1; ; H ? p + 1), we obtain, again by the hypothesis that pw (n ) n and by the pigeon's holes lemma, that two among these must determine the same factor of length n , i0 and j 0, i0 < j 0. By previous result, j 0 is not one of the rst p positions of the word u, i.e. j 0 (H ? n ? p +1)+ p : Moreover the word v0 which starts at the position i0 and whose length is n + (j 0 ? i0), has period j 0 ? i0 = p n . Since j 0 (H ? n ? p + 1) + p , the word v0 contains position H and, since aH 6= aH ?p , the word v0 cannot have period p. 0
1
1 +1
0
0
1
1
1
0
1
0
1
1
1
1
+ 0
1
1
1
0
1
1
+ 1
1
1
1
1
1
0
0
0
1
0
1
0
0
0
1
0
1
1
0
2
0
0
1
1
1
1
H-n 0-p1 +1
i’
j’
H
H+1
p 1
The pre x of v0 that has both periods p and p is the word u0 = ai aH ? . Its length is j u0 j= H ? 1 ? i0 + 1 = H ? i0 = H ? j 0 + p . But j 0 is one of the rst (n + 1) positions of the word u, i.e. j 0 H ? p + 1. Hence ju0j = H ? j 0 + p p + p ? 1. Therefore we can apply the Theorem of Fine and Wilf to u0 and obtain that u0 has period d, where d is the greatest common divisor between p and 1
2
1
0
2
0
1
2
1
2
1
6
p . Since v0 has a pre x u0 of length greater than or equal to p that has a period d that divides p , and since p is a period of v0 then, by Proposition 2.1, v0 has also period d. This period d is also a divisor of p , consequently v0 has also period p , that contradicts the fact that aH 6= aH ?p . 2 2
2
2
2
1
1
1
Let us consider, now, the bidimensional case. A double sequence w over the alphabet A is a mapping from ZZ to A. A factor u of a bidimensional word w is the restriction of w to a subset S ZZ , which is called the shape of u and it is denoted by sh(u). A factor u is a factor of u if the shape of u is contained in the shape of u . Notice that u is a factor of itself as well w is a factor of w. Two factors u and u are equal if there exists a translation of the plane such that (S ) = S , where S and S are respectively the shapes of u and of u . Moreover this translation must be compatible with the word w, i.e. for any (x; y) 2 S , w(x; y) = w( (x; y)). A rectangle R is a subset fi ; i + 1; ; i g fj ; j + 1; ; j g of ZZ . For brevity we denote R = [i ; i ] [j ; j ]. The numbers i ? i + 1 = n and j ? j + 1 = m are the lengths of the sides of R. A rectangle R having sides of lengths n and m is called an n m rectangle. If the shape of a factor u of w is a n m rectangle R then u is a n m factor and n; m are also called the lengths of the sides of the factor. Clearly, if (n; m) 6= (n0; m0) then every n m factor is di erent from every n0 m0 factor. We de ne the complexity function pw (n; m) of w as the numbers of different n m factors. 2
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De nition! 2.1 Let u be a factor of w and let R be its shape. A vector 6 0 is a symmetry vector for u if w((i; j )) = w((i + v ; j + v )) v = vv = 1
1
2
2
for any element (i; j ) in R such that (i + v1 ; j + v2) is still in R. Clearly, if u is the whole w then v is a symmetry vector for w if, for any (i; j ), w((i; j )) = w((i + v1; j + v2)). We say that v is a periodicity vector for u if for any integer z the vector zv is a symmetry vector for u. In this case we will also say that v is a period for u (for w resp.) or also that u (w resp.) has period v.
7
This de nition of symmetry vector is analogous to the de nition of symmetry vector given by Amir and Benson in [1], [2], [3]. The de nition of periodicity vector is analogous to the one of L =< v >-periodicity given by Giancarlo and Mignosi in [18] or by Koskas in [19]. We can here state the Nivat's conjecture. Conjecture: If there exists a pair (n; m) such that the complexity function pw (n; m) of the double sequence w veri es pw (n; m) nm, then w has at least a periodicity vector. Let us state two easy propositions without proof.
Proposition 2.2 A vector v is a symmetry vector for u (resp. w) if and only if (?v) is still a symmetry vector for u (resp. w). Moreover, if v is a periodicity vector then 8z 2ZZ ; zv is still a periodicity vector. In IR the de nitions of convex set and convex hull are well known. The next de nition concerns \discrete" convexity. 2
De nition 2.2 A subset S of ZZ is convex if its convex hull in IR does not contain any point of (ZZ n S ): Proposition 2.3 Any symmetry vector for an n m factor u or for the whole double sequence w is also a periodicity vector. 2
2
2
More generally, any symmetry vector for any factor whose shape is a convex subset of ZZ 2, is also a periodicity vector.
Given two vectors v, v0 2ZZ we denote by < v; v0 > the subgroup of ZZ generated by v and v0, i.e. the set of all vectors v^ of the form v^ = iv + jv0 with i; j 2 ZZ . It is known that if v and v0 are not parallel, the subgroup < v; v0 > has nite index, i.e. there exists a nite subset T of ZZ such that for each point (x; y) 2 ZZ there exists an unique element t 2 T such that (x; y) ? t 2< v; v0 >. 2
2
0
2
2
Remark 2.1 If v and v0 are parallel, then the subgroup < v; v0 >=< v00 >
is generated by only one element v 00 whose length divides the lengths of v and v 0.
8
De nition 2.3 If v and v0 are not parallel, a factor u is full-periodic (or lattice-periodic) with respect to v and v 0 if for all v^ 2< v; v 0 > v^ is a periodicity vector for u. In this case we also say that u is < v; v 0 >-periodic. Notice that if we require that there exists n such that pw (n; n) n, then every restriction of w to unidimensional lines parallel to the axes is, by the Morse-Hedlund theorem, periodic with period smaller than or equal to n. Hence w is! lattice-periodic,!where the lattice is generated by the two vectors v = n0! and v0 = n0! .
Consequently, this form of generalization of the Morse-Hedlund Theorem is relatively simple and can be extended to any dimension. Moreover J. Cassaigne has proved (cf. [10]) that the lengths of v and v0 can be bounded by n. Another similar generalization of the Morse-Hedlund Theorem is stated and proved in [13] (cf. also [12]) for Delone (or Delaunay) sets, where factors are \replaced" by neighborhoods. One of the diculties in this new setting comes from the fact that two neighborhoods are \equal" if they are isometric and not just equal by a translation as in the case of double sequences (cf. also the pinwheel tilings of the plane in [26] and the Delone sets that can be obtained by the Conway tasselation, as explained in [13]).
Examples:
Example 2.1 In the following example w((i; j )) = 0 for any point (i; j )
except for one point, where it has value 1. The complexity function is equal to pw (n; m) = nm +1 and w has no periodicity vectors. J. Cassaigne has proved [8] that there are no recurrent double sequences having the same complexity pw (n; m) = nm + 1 of w. Recall that a double sequence w is recurrent if for any couple of integers (n; m) there exists a couple (n0; m0) such that any n m factor u1 of w is equal to a factor of any n0 m0 factor u2 . A conjecture on the complexity functions of recurrent double sequences was stated by L. Vuillon (cf. [7], [34], [35], [36]). This conjecture states that, for any recurrent double sequence w that has no periodicity vector, there exist n^ , m ^ such that for any n > n^ , m > m^ one has pw (n; m) nm + min(n; m).
9
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Example 2.2 The following example is due to Cassaigne and Bernardi (cf. [9]). Let us consider a Christo el approximation of a line in the plane having irrational slope. This line de nes two distinct half-planes HP1 and HP2 . We de ne a word w such that for any point (i; j ) in HP1 w(i; j ) = 1 and for any point (i; j ) in HP2 w(i; j ) = 0. In this case it is possible to prove that pw (n; m) = nm + 1. The idea of the proof is the following. Let us assume to write the letter w(i; j ) inside the unit square that has (i; j ) as the bottom-left corner. If we slide an n m rectangle from HP2 to HP1 , the labels w(i; j ) associated to the points (i; j ) of the rectangle change from 0's in 1's always in the same order. Since the slope is irrational, it is impossible that two of them could change their value at the same moment. This remark allows us to numerate these rectangles, i.e. the shapes of all the factors of w. Hence the complexity is pw (n; m) = nm + 1. 1
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3 Preliminaries Lemmas Before proving the main theorem, we will prove or recall some preliminaries lemmas. As remarked in the previous section, the proof of our main theorem is based on the fact that the hypothesis implies the existence of a \local" period in every position of the double sequence. This is proved in next lemma. Afterwards, we state two lemmas that are analogous to the periodicity lemma 10
and that will be used in next section to force all \local" periods to be \almost the same" all over the double sequence.
Lemma 3.1 (local 1-periodicity) Let w be a double sequence whose complexity function pw is such that pw (n ; m ) < n m for some (n ; m ) 2 ZZ . Then any factor u of w of shape [i; i + h] [j; j + k] with i; j 2 ZZ and! 0 h < n ? d pn e, 0 k < m ? d pm e has a periodicity vector v = vv such that 0 v d pn e ? 1 and 0 jv j d pm e ? 1. Proof Let u be the factor of w of shape [i; i + h] [j; j + k] with i; j 2 ZZ and 0 h < n ? d pn e, 0 k < m ? d pm e. For any (s; t) 2 [0; d pn e ? 1] [0; d pm e ? 1] let us;t be the factor of w of shape [i ?d pn e + s; i ?d pn e + s + n ? 1] [j ?d pm e + t; j ?d pm e + t + m ? 1]. Since h n ? d pn e ? 1 and k m ? d pm e ? 1, u is a factor of us;t . Moreover, since pw (n ; m ) < n m and since us;t is a n m -factor, there exist two di erent couples (s ; t ); (s ; t ) in [0; d pn e] [0; d pm e] such that us ;t = us ;t . Indeed, there are at least d pn ed pm e n m positions for the factors us;t but less than n m di erent factors. 0
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u s ,t
1 1
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u s ,t
0 0
11
0
0 0
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!
s Since u is a factor of us ;t and a factor of us ;t , v = st ? ? t is a periodicity vector for u. We may of course assume that s ? s 0 because if v is a periodicity vector, so is ?v. It remains to notice that by construction we have s ? s d pn e ? 1 and jt ? t j d pm e ? 1, and the lemma is proved. 2 0
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The following lemma has been proved by Amir and Benson (cf. [3], Lemma 3.6)
Lemma 3.2 Let u be an n m factor of w. If v = vv
are two periodicity vectors of u such that
! 1 2
and v 0 =
v0 v0
!
1
2
v 0; v0 0; v 0; v0 0, v + v0 < n; jv j + v0 < m: Then, for all v^ 2< v; v 0 >; v^ is a periodicity vector for u. Therefore, if v and v 0 are not parallel, u is lattice-periodic, where the lattice is generated by 1
2
1
1
2
2
1
2
v and v0.
Next lemma is an easy consequence of Lemma 7 in [16]. A similar result can be found also in [2]. !
!
0 De nition 3.1 Let v = vv and v0 = vv0 be two vectors such that v 0; v0 0; v 0; v0 0; j v j< n ; j v0 j< n ; j v j< m ; j v0 j< m : Let R be an n m-rectangle, R = [i; i + n ? 1] [j; j + m ? 1]. The intersection of ZZ and the convex hull of the six points f(i; j ); (i; j + m ? 1 ? v ? v0 ); (i + v + v0 ; j + m ? 1); (i + n ? 1; j + m ? 1); (i + n ? 1; j + v + v0 ); (i + n ? 1 ? v ? v0 ; j )g will be called the (v; v0)-hexagon of R. 1
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Lemma 3.3! Let u be an n! m factor of w and let R be its shape. If 0 v = vv and v0 = vv0 are two periodicity vectors of u such that 1
1
2
2
v 0; v0 0; v 0; v0 0; j v j< n ; j v0 j< n ; j v j< m ; j v0 j< m 1
2
1
1
4
1
2
2
4
4
2
4
then the (v; v 0)-hexagon of R is periodic with period generated by both vectors. In particular, if v and v 0 are not parallel, this hexagon is lattice periodic.
A direct consequence of these two lemmas is the following theorem.
Theorem 3.1!(Lattice periodicity) ! If a rectangle R has two periodicity vecv v
tors v =
1 2
v0 such that v0 j v j< n4 ; j v0 j< n4 ; j v j< m4 ; j v0 j< m4 ;
and v2 =
1
2
1
2
1
2
then the rectangle centered in the original one and having side lengths n ? 2vmax1 and m ? 2vmax2 (where vmax1 = max(v1; v10 ) and vmax2 = max(v2; v20 )) is periodic with period generated by both vectors. In particular, if v and v 0 are not parallel, this centered rectangle is lattice periodic, where the lattice is < v; v0 >.
4 The main theorem We now state and prove the main theorem of this paper.
Theorem 4.1 Let w : ZZ ! A be a double sequence and let pw (n; m) be its complexity function. Let us suppose that there exist two natural numbers n ; m such that pw (n ; m ) < n m , where = 100. Then w has a 2
0
0
periodicity vector.
0
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13
0
In order to prove this theorem we need two fundamental lemmas, to each of those we will dedicate a subsection. Each of them will be devoted to extend local periodicity obtained by Lemma 3.1 to \larger" portions of the plane. In this paper we will prove the main theorem with = 144, because we make use of Theorem 3.1 that synthesize Lemmas 3.2 and 3.3 with loss of information. A proof based on the use of the two lemmas reaches a constant = 100. Since the proof is already involved and hard to follow, we have chosen to let it this way, easier to read, even if the constant is higher. Let us give some de nitions and notations that will be useful in the following. ! ! v 0 De nition 4.1 Given a vector v = v 6= 0 2 ZZ , and a point P = (i; j ) 2 ZZ , let l(v; P ) be the line in IR that contains P and that has the same direction of v, i.e. the line of equation v y ? v x = v j ? v i. We de ne the v-Christo el line C (v; P ) through the point P to be the set of points (r; s) 2 ZZ that are, roughly speaking, immediately below l(v; P ). More precisely if v = 0 then C (v; P ) = f(i; s) j s 2 ZZg. If v 6= 0 then rstly de ne C (v) = f(r; s) 2 ZZ j b vv (r ? 1)c s b vv rcg: De ne now C (v; P ) = C (v) + P . We also de ne, given a real number h > 0, the line-segment S (v; h; P ) of length hjvj starting in P to be if v = 0 then S (v; h; P ) = C (v; P ) \ fv + P j 0 < hg. If v 6= 0 then rstly de ne S (v; h) = f(i; j ) 2 ZZ j 0 < i hv and b vv (i ? 1)c j < b vv icg[f(0; 0)g. De ne now S (v; h; P ) = S (v; h)+ P . 1
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v
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Remark 4.1 Remark that our de nition of Christo el line implies that for
every point (x; y) in any v-line there are exactly two points between the four points of the form (x 1; y 1) that belong also to the same v-line. In other words the line is 4-connected and not 8-connected as the usual de nition (cf. [11], [21], [28], [29]). Indeed all the arguments and proofs in this paper can be easily modi ed to work also with the usual de nition. The fact that, by using this de nition, the number of points in any v-segment of length jvj is exactly jv1j + jv2j instead of max(jv1j; jv2j) will simplify some notations and proofs. !
De nition 4.2 A vector v = vv 2 ZZ is an -vector if m e ? 1: 0 jv j d pn e ? 1; 0 jv j d p Since the number of ?vectors is nite, we can nd the longest ?!vector having the same direction of v. Let us denote this vector vv = vv ;v . 1
2
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2
1
;v
2
Let now R be an h k rectangle, h 2jv;vj, k ! 2jv;v j,!whose bottom left corner is the point P = (x; y) and let v = vv 6= 00 be an -vector. 1
2
1 2
We consider two cases. 1. (v 0 and v 0) or (v 0 and v 0). In this case set P = (x; y + k ? 2jv;v j) and P = (x + h ? 2jv;v j; y) (see next gure). 2. (v > 0 and v < 0) or (v < 0 and v > 0). In this case set P = (x ; y ) = (x; y + 2jv;v j) and P = (x ; y ) = (x + h ? 2jv;v j; y + k). 1
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De nition 4.3 The (R; v)-Christo el stripe is the set of points that lie be-
tween the v-Christo el line through P1 = (x1; y1) and the v-Christo el line through P2 = (x2; y2), the points of these two lines included.
15
The intersection of R and the (R; v)-Christo el stripe is called the fundamental (R; v) discrete hexagon H (see next gure). P1
H
R
P2
We say that a vector v is a periodicity vector for a Christo el stripe S (or that S has period v) if v is a periodicity vector for the factor that has S as its shape.
Notice that the fundamental (R; v)-discrete hexagon H contains properly an h0 k0 rectangle R where h0 = h ? 2jv;v j, k0 = k ? 2jv;vj and the bottom left corner is (x ; y ) = (x; y), the same of R. Indeed the (R; v)-Christo el stripe contains a bisequence (Rz )z2ZZ of translates of R (cf. next gure) such that, if (xz ; yz ) is the bottom-left corner of Rz , then jxz ? x j + jyz ? y j = jzj (we may assume, thanks to the symmetry, that xz is increasing with z). These rectangles Rz are called the (R; v)-stripe rectangles, or (R; v)rectangles. Moreover the (R; v)-Christo el stripe is the union of this sequence of rectangles and the (R; v) fundamental discrete hexagon is the union of the rectangles Rz with 0 z < 2jv;vj + 2jv;v j. Notice also that the line-segments in the direction of v starting in P and P that are subsets of the discrete hexagon both contain at least 2jv;vj+2jv;vj points (in the case of v having the same direction of an axis, this number is strictly greater than 2jv;v j + 2jv;vj). 0
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2
R3 R2 R0 R1
R3 R2 R0 R1
Notice that an (R; v)-Christo el stripe does not contain R and, moreover, its intersection with R gives rise to the discrete fundamental hexagon, that is obtained as the set di erence between R and two \angles". Note also that the (R; v)-rectangles have side sizes strictly smaller than the ones of the original rectangle. The reasons of this de nition are two. The rst and more important is that this de nition is necessary if we want to apply the same unidimensional argument used in the proof of Theorem 2.2 in our two main lemmas of this section. The second reason is that with this de nition we can use in next main lemma the following proposition. Proposition 4.1 (Propagation of lattice periodicity) Suppose that a factor u of shape an h k rectangle R, h 4d pn e , k 4d pm e, has v and v0 as periodicity vectors, where v and v 0 are two -vectors. Let us also suppose that the union of the rst (l+1) (R; v)-stripe rectangles [ss l Rs is v0-periodic, for some natural number l. 0
= =0
17
0
Therefore, for any rectangle Rs; s = 0; : : :; l, there exists a factor u0s having as shape a rectangle R0s Rs that is < v; v 0 >-lattice periodic. This factor is an ^h k^ factor where h^ h ? 3d pn e + 3 and k^ k ? m 3d p e + 3. 0
0
Proof
By Theorem 3.1, the centered factor having sides h ? 2d pm e + 2 and k ? 2d pm e + 2, is lattice periodic where the lattice is generated by v and v0. Let us call Cs the shape of this centered rectangle. Let us consider now R0s Cs the (h ? 3d pm e +3; k ? 3d pm e +3)-rectangle having the same bottom-left point of Cs if the coordinates of v0 are both non negative or both non positive, and the same top-left point otherwise. Let us nally consider U = [ls R0s . Let (i; j ) and (i0; j 0) be two points in the shape of u0 that are in the same coset of < v; v0 >, i.e. (i ? i0; j ? j 0) = v^ 2< v; v0 >. All we have to prove is that w((i; j )) = w((i0; j 0)). Since U is composed by construction by at least jv0 j + jv0 j consecutive rectangles, there exist two integers n and n0 such that (i; j ) + nv0 2 U and (i0; j 0)+n0v0 2 U . Hence (i; j )+nv0 ?((i0; j 0)+n0v0) = v^+(n?n0 )v0 2< v; v0 >, i.e. (i; j ) + nv0 and (i0; j 0) + n0v0 are in the same coset. Since u is < v; v0 >periodic and U is contained in its shape, w((i; j ) + nv0) = w((i0; j 0) + n0v0). Since u0 has period v, w((i; j ) + nv) = w((i; j )) and w((i0; j 0) + n0v) = w((i0; j 0)). By the transitivity of equality the proposition is proved. 2 0
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=0
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4.1 First Lemma
2
This section is devoted to extend a local periodicity, given by Lemma 3.1, to a periodicity in a Christo el stripe. The main idea is to try to extend the local periodicity in the direction of the period. If it is possible to cover all a \large enough" Christo el stripe associated to this periodicity vector, then we obtain the thesis. If this is not possible, then at a certain point in this direction there must exist an \error" in the periodicity and, by Lemma 3.1, there must exist another periodicity. This new periodicity cannot be parallel to the original one by the unidimensional argument used in the proof of Theorem 2.2. Then we make one \turn" and try to extend this new local periodicity in its own direction. The bidimensional generalizations of the periodicity lemma allow us to prove that 18
it is possible to extend this local periodicity to a \large enough" Christo el stripe. The reason of that relies on the fact that, by extending, we bring with us the informations of the previous periodicity and of the \error". Roughly speaking, we are allowed to make one \turn" but not two \turns". We will need, in the proof of the rst lemma, the following results. Covering Lemma: Let v0 and v00 be two non parallel -vectors and R be an h k-rectangle with h 2d pn e ? 2 and k 2d pm e ? 2. Then there exists a vector v 2< v 0; v 00 > belonging to R. The proof of this lemma is left to the reader. 0
0
Corollary 4.1 Let v0 and v00 be two non parallel -vectors and R be an h krectangle with h 2d pn e ? 2 and k 2d pm e ? 2. Then, for any (x; y) 2 ZZ there exists (x0; y 0) 2 R such that (x; y) ? (x0 ; y 0) 2< v 0; v 00 >. Proof Consider the rectangle R0 = R ? (x; y). By previous lemma there exists a vector v 2< v0; v00 > belonging to R0. Hence the point (x0; y0) = (x; y) + v belongs to R and, so, (x; y) ? (x0; y0) 2< v0; v00 >. 2 0
2
0
Lemma 4.1 (Contradictory conditions) Let u be a factor having shape R . The following properties cannot hold together.
1
1
It has two non parallel periodicity vectors v0 and v00. There exists in R a factor uL having shape a rectangle RL that is 1
< v; v0 >-lattice periodic and it is \centered enough" in R , i.e. for any PL = (iL; jL ) 2 RL iL ? i d pn e ? 1; jL ? j d pm e ? 1 for any (i ; j ) belonging to the sides of R . This factor is an h^ k^ factor where ^h 3d pn e ? 3 and k^ 3d pm e ? 3. There exists a couple of points P^ and P^ = P^ + v such that w(P^ ) 6= w(P^ ). Moreover these points are \centered enough", i.e. j P^ ? ^ P^ g and (i ; j ) (i ; j ) j (d pn e ? 1; d pm e ? 1), for any P^ 2 fP; belonging to the sides of R . 1
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Proof
Let R^ be the rectangle centered in R , i.e. the rectangle whose points P^ = (^i; ^j) are such that ^i ? i d pn e ? 1; ^j ? j d pm e ? 1 for any (i ; j ) belonging to the sides of R . By Theorem 3.1, the factor having shape the rectangle R^ centered in R is L =< v0; v00 >-periodic. Moreover it contains P^ and P^ and also it has as factor uL that has shape RL. 1
1
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1
such that P^ + v^ is in RL(RL ? v). Consequently P^ + v^ belongs to both. Since P^ + v^ belongs to RL ? v, P^ + v + v^ = P^ + v^ belongs to RL. By the < v0; v00 >-periodicity w(P^ ) = w(P^ + v^) = b, and w(P^ ) = w(P^ + v^) = a. But P^ + v^ and P^ + v^ belong to RL and P^ + v + v^ = P^ + v^. Since uL is < v; v0 >-periodic we have a contradiction. 2 1
1
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1
1
In the next lemma we will set the constant to be = 144. As we have 20
said previously, with a more involved proof we may reduce this constant to = 100.
Lemma 4.2 (First Lemma) Let w : ZZ ! A be a double sequence whose n m for some n 4p ; m complexity function veri es p ( n ; m ) < w p 4 2 IN . 2
0
0
0
Then there exists an -vector v = vv
1 2
0
0
0
! , and a (R; v){Christo el stripe
S such that R = [i; i + h] [j; j + k] with h 3 pn ; k 3 pm and v is a periodicity vector of S . 0
0
Before giving the complete proof of this lemma, we want to give a detailed sketch of it, in order to allow an easier reading. The \ingredients" of this proof are 1. an unidimensional argument similar to the one used in the proof of Theorem 2.2; 2. Lemma 3.1 (local 1-periodicity); 3. Theorem 3.1 (lattice periodicity); 4. Proposition 4.1 (propagation of lattice periodicity); 5. Lemma 4.1 (contradictory conditions). By Lemma 3.1 we start with an h k-factor u that admits a periodicity vector v. Let us consider the sh(u) v-Christo el stripe S , where sh(u) is the h k-rectangle that is the shape of u. If S is v-periodic the statement of the lemma is true. If S is not v-periodic, then there exists a rectangle Rt where there is an \error" in this periodicity, i.e. there exist two points (i; j ) and (i0; j 0) = (i; j )+ v in Rt such that w((i; j )) = b and w((i0; j 0)) = a; a 6= b. Again for Lemma 3.1 there must exist another periodicity vector v0. A claim that uses an unidimensional argument similar to the one used in the proof of Theorem 2.2 states that v and v0 are not parallel. 1
1
1
21
Theorem 3.1 states that a rectangle centered in Rt is L =< v; v0 >-lattice periodic. Now we consider the (R0; v0)-Christo el stripe S , where R0 is a rectangle that is v0-periodic and that contains a rectangle L-lattice periodic. If the whole stripe S is v0-periodic, then the lemma is proved. If it is not so, there will exist another rectangle R00 where there is an \error" in v0-periodicity. Moreover in R00 there is a couple of points (i; j ) + v0 and (i0; j 0) + v0 = (i; j ) + v + v0 such that w((i; j ) + v0) = b and w((i0; j 0) + v0) = a. These two points represent a kind of \memory" of the error in the v-periodicity of stripe S . By Lemma 3.1 there must exist another periodicity vector v00. By using again an unidimensional argument similar to the one used in the proof of Theorem 2.2 we get that v0 and v00 are not parallel. By Proposition 4.1, R00 contains also a rectangle that is L-lattice periodic. We get a contradiction on the existence of R00 from Lemma 4.1. Hence the whole stripe S is v0-periodic. The reason of that relies on the fact that R00 contains some informations of the previous periodicities and of the \error". Roughly speaking, we are allowed to make one \turn" but not two \turns". 2
2
1
2
Proof
Let u be a factor of shape an [0; h] [0; k] rectangle R with h = n ? and k = m ? d pm e ? 1. By Lemma 3.1, u admits a periodicity vector, say v. We may assume that its coordinates are greater than or equal to 0, because of symmetry (the other cases can be proved in the same way). Let S be the (R; v)-Christo el stripe. If S is v-periodic, it is enough to set S = S in order to prove the lemma. Let us suppose it is not the case. Let t be the smallest integer, that we may assume positive without loss of generality, such that the factor ut having shape the union Ut of the rst t + 1 (R; v)-rectangles Rs , 0 s t is not v-periodic. Therefore there exist two points (i; j ) and (i0; j 0) = (i; j ) + v in Ut such that w((i; j )) = b and w((i0; j 0)) = a, as shown in the gure:
d pn0 e ? 1
0
0
0
1
1
1
22
a
b
Recall also that Rt is an h0 k0 rectangle, where h0 = h ? 2jv;v j and k0 = k ? 2jv;v j (for the de nition of v;v and v;v cf. 4.2). Since t is increasing, the error is either on the right or on the top side of the rectangle. Let (x; y) be the bottom-left corner of Rt. Let us distinguish now three cases: 1
2
1
2
1. x i0 < x + 4(d pn e ? 1); j 0 = y + k; 0
x + 4(d pn e ? 1) i0 x + h; j 0 = y + k OR i0 = x + h; y + 4(d pm e ? 1) j 0 y + k; 3. i0 = x + h; y j 0 < y + 4(d pm e ? 1) 2.
0
0
0
23
Case 1
Case 2
Case 3
We may introduce a new point P 0, according to the cases above-stated. In case 1 P 0 = (x ? 3(d pn e ? 1); y + d pm e ? 1), in case 2 P 0 = (x + d pn e ? 1; y + d pm e ? 1) and in case 3 P 0 = (x + d pn e ? 1; y ? 3(d pm e ? 1)). Let us consider the h k rectangle R0 having P 0 as bottom-left corner. This new rectangle has been chosen such that the points (i; j ) and (i0; j 0) are \centered enough" in R0, i.e. (i; j ) ? (i,j) 2(d pn e ? 1; d pm e ? 1) for any (i; j ) 2 f(i; j ); (i0; j 0)g and (i,j) belonging to the sides of R0. Let now u0 be the factor of w having shape R0. By Lemma 3.1, u0 admits a periodicity vector, say v0. 0
0
0
0
0
0
0
0
Claim: v cannot be parallel to v0.
The technique used here to prove the claim is similar to the one used in the Theorem 2.2. Let us suppose by the absurd that v is parallel to v0. Since we are supposing t to be positive and since all the factors having shape an (R; v)-rectangle inside the fundamental hexagon have period v, then t > 2(jv;v j + jv;vj). Each (R; v)-rectangle Rs is obtained by translating rectangle Rs? of one unit line. Let s be this translation and let Q ; Q ; : : :; Qt be the unique sequence of points such that Q 2 R , Qs = s(Qs? ), 1 s t, and Qt = (i0; j 0). By de nition it is easy to prove that (i; j ) belongs to this 1
2
1
0
0
0
24
1
1
sequence and, more precisely, that (i; j ) = Qt , t0 = t?jv j?jv j, where v and v are the components of vector v. Let u = a a at be the unidimensional word where as = w(Qs ), 0 s t. Since v is a periodicity vector for the factor ut, the pre x u? of u having length juj ? 1 has period jv j + jv j. By the choice of factor u0 all the factors having shape a Rs, t ? (jv j + jv j + jv;v j + jv;vj) s t, have period v0. Consequently, the sux w of the word u of length jwj = jv j + jv j + jv;v j + jv;vj has period jv0 j + jv0 j, where v0 and v0 are the components of vector v0. Hence, the pre x w? of w of length jw? j = jwj ? 1 = jv j + jv j + jv;v j + jv;vj ? 1 has both periods. Since jv;v j = jv;v j and jv;vj = jv;v j, jv j + jv j + jv;vj + jv;v j ? 1 jv j + jv j + jv0 j + jv0 j ? 1. Therefore we can apply the Theorem of Fine and Wilf to w? . We obtain that w is d = gcd(jv j + jv j; jv0 j + jv0 j)-periodic. Let us consider, now, the word w. This unidimensional word is (jv0 j + 0 jv j)-periodic. The factor w? of w has length greater than or equal to jv0 j + jv0 j. Thus it contains surely a factor of length jv0 j + jv0 j that is dperiodic. Therefore, by Proposition 2.1, w is d-periodic, and, consequently, (jv j + jv j)-periodic. This is absurd because at 6= at , t0 = t ? jv j ? jv j, and this concludes the proof of the claim. 1
0
2
2
1
0 1
1
1
2
1
2
1
2
1
1
2
1
2
1
1
1
1
1
2
2
1
2
1
1
2
1
2
0
2
0
2
1
2
1
2
1
2
2
1
1
1
2
1
1
2
1
2
1
1
2
2
1
0
2
So the h k factor u0 has period v0 and it has a periodic factor whose shape is an h0 k0 rectangle that is v-periodic where h0 = h ? 3(d pn e ? 1) and k0 = k ? 3(d pm e? 1). This factor has shape the intersection between the shape of u and the shape of u0. Notice that h0 n ? 4d pn e +2 8d pn e? 10 and k0 m ? 4d pm e + 2 8d pm e ? 10. Let us now consider the (R0; v0)-Christo el stripe S . If the whole stripe S is v0-periodic, then the lemma is proved. If it is not the case, let us consider the minimum integer t0 (we may again assume it is positive) such that the factor u0t having shape the union Ut0 of the rst t0 + 1 (R0; v0)-rectangles R0s, 0 s t0, is not v0-periodic. Then there exist two points (i ; j ) and (i0 ; j 0 ) = (i ; j ) + v0 in R0t such that w((i ; j )) = c and w((i0 ; j 0 )) = c0, c 6= c0. By Proposition 4.1 for any rectangle R0s ; s = 0; : : : ; t0, there exists a factor 0 us having as shape a rectangle R0s Rs that is < v; v0 >-lattice periodic. This factor is an ^h k^ factor where ^h h0 ? 3d pn e +3 n ? 7d pn e +5 5d pn e ? 7 and k^ k0 ? 3d pm e + 3 m ? 7d pm e + 5 5d pm e ? 7. 0
0
0
0
0
0
0
0
2
2
0
0
0
0
0
0
0
0
0
0
0
0
0
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0
0
25
0
0
0
0
By the choice of R0, by the v0-periodicity and by the fact that v0 is an -vector, one has that for any sequence of jv;v j + jv;v j consecutive (R0; v0)rectangles, there exists a rectangle R0s that contains a \translate by v0" of the \initial" couple of points (i; j ) and (i0; j 0) that is \centered enough" in R0s, i.e. (i; j ) ? (i,j) (d pn e? 1; d pm e? 1)) for any (i; j ) 2 f(i; j ); (i0; j 0)g and (i,j) belonging to the sides of R0s . For a \translate by v0" we intend a couple of points of the form (i; j )+ v0 and (i0; j 0) + v0, for an integer (cf. gure below). 0
1
0
2
0
0
a
v
v’
b
a
v’
v
v’
b
a
v’
b
v’
v
a
v’
b
v
!
0 Let us suppose now that v0 = vv0 is such that v0 0 and v0 0. The proof in the other cases is analogous. Let (x0; y0) be the bottom-left corner of R0t . We introduce a new point P 00 = (x0 ? 2v0;v ; y0). Let us consider the h k rectangle R00 having P 00 as bottom-left corner. This new rectangle has been chosen such that it contains a sequence of 2(jv;v j + jv;v j) consecutive (R0; v0)-rectangles. 1
1
2
2
0
1
1
2
0
26
0
Let now u00 be the factor of w having shape R00. By Lemma 3.1, u00 admits a periodicity vector, say v00. By using an argument analogous to that used in previous claim, it turns out that v00 is not parallel to v0. Moreover there exists a (R0; v0)-rectangle R R00 that contains a \translate by v0" of the \initial" couple of points (i; j ) and (i0; j 0) that is \centered enough" in R , i.e. (i; j ) ? (i ; j ) (d pn e ? 1; d pm e ? 1)) for any (i; j ) 2 f(i; j ); (i0; j 0)g and (i ; j ) belonging to the sides of R . Recall that by Proposition 4.1 for any (R0; v0)-rectangle R0s ; s = 0; : : : ; t0, there exists a factor u0s having as shape a rectangle R0s Rs that is < v; v0 >lattice periodic. This factor is an ^h k^ factor where ^h h0 ? 3d pn e +3 n ? 7d pn e +5 5d pn e ? 7 and k^ k0 ? 3d pm e + 3 m ? 7d pm e + 5 5d pm e ? 7. This fact means that in R there exists a factor uL having shape a rectangle RL that is < v; v0 >-lattice periodic and it is \centered enough" in R , i.e. for any PL = (iL; jL) 2 RL iL ? i d pn e ? 1; jL ? j d pm e ? 1 for any (i ; j ) belonging to the sides of R . This factor is an h k factor where h 4d pn e ? 6 and k 4d pm e ? 6. p p If n 4 and m 4 then h 3d pn e ? 3 and k 3d pm e ? 3. Let u be the factor having shape R , the Lemma 4.1 gives a contradiction. 2 1
1
1
1
0
0
1
1
1
0
0
0
0
0
0
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1
1
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1
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1
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1
1
4.2 Second Lemma
In the main result of this section, we extend the local periodicity v of a Christo el stripe obtained in Lemma 4.2 to the whole plane. The main point in this procedure is that, starting from the -vector v, we can obtain other vectors that are parallel to v that are periodicity vectors for stripes parallel to the original one. In other word the whole plane can be subdivided in \large" Christo el stripes all associated to v or to any vector parallel to v, each of them admitting an -vector as periodicity vector having same direction of v. Since the number of -vectors is nite, then the vector v having same direction of v and length the product of the lengths of all -vectors parallel to v is a periodicity vector for the whole plane. Remark that from the proof it turns out that the length of such vector can \explode" with respect to the size of the two numbers (n ; m ). And this 0
27
0
can really happen, as shown in the examples of next section. The techniques used in this lemma are elementary, and we make use of the unidimensional periodicity lemma.
Lemma 4.3 Let S be a Christo el stripe associated to an -vector v. Suppose that S contains a s t rectangle with s > pn and t > pm , v is a periodicity vector of S . then there exists a vector parallel to v that is a periodicity vector for the whole double sequence w. 0
0
Proof
Let us suppose that the two parallel lines r and r associated to the Christo el stripe S (that are also parallel to v) have a slope greater than or equal to 0. The proof in the case in which the slope is smaller than 0 is analogous. Let 1
and
2
S = S + (1; ?1) = fz 2ZZ j z = v + (1; ?1); v 2 S g 1
S = S + (?1; 1) = fz 2ZZ j z = v + (?1; 1); v 2 S g and let us de ne two functions f and g as follows: 2
f (S ) = S [ S; g(S ) = S [ S : 1
2
g(S)
28
f(S)
If for all k 2 IN v is a periodicity vector for both f k (S ) and gk (S ), then v is a periodicity vector for the whole double sequence w. Otherwise, there exists a k 2 IN such that f k (S ) and gk (S ) admit v as a periodicity vector, while either f k (S ) or gk (S ) contain (contains) two points P = (i ; j ) and P = (i ; j ) such that +1
1
1
+1
0
0
0
1
1) P + v = P ; 2) w(P ) = a and w(P ) = b, where a and b are two di erent letters. 0
1
0
1
Let us suppose that these two points are in f k (S ) (the case in which they are in gk (S ) is analogous). Since f k (S ) is still a stripe associated to v then neither P nor P belongs to f k (S ) and, consequently, they belong to the set (f k (S ) ? f k (S )). +1
+1
+1
0
1
v
By Lemma 3.1 the rectangle R centered in P = (i ; j ) of shape [i ? 0
0
0
0
d pn e; i0 + d 5pn 0 e? 2] [j0 ?d 4pm 0 e; j0 + d 5pm 0 e? 2] has a new periodicity vector v0. Notice that this rectangle contains also P1. Let us examine the direction of this new periodicity vector. If v0 is not parallel to v then, since the stripe f k (S ) contains an s t rectangle with s pn0 and t pm 0 , there would exists 4 0
29
2 f+1; ?1g such that both P + v0 and P + v0 are in the stripe f k (S ). In this case a = w(P ) = w(P + v0) = w(P + v0) = w(P ) = b; that is an absurd. Therefore v0 must be parallel to v. Let us consider now the (R; v0)-Christo el stripe. It is an in nite stripe that has v0 as periodicity vector and, by very de nition, it contains an s t rectangle with s pn and t pm . Let us suppose by absurd that v0 is not a periodicity vector for this new stripe. Let i be the smallest integer, that we may assume positive without loss of generality, such that the factor ui, having as shape the union Ui of the rst i +1 (R; v0)-rectangles Rs, 0 s i, is not v0-periodic. Then there exist two points P 0 ; P 0 2 Ui such that 0
0
1
0
1
1
0
0
0
1
1) P 0 + v0 = P 0 ; 2) w(P 0 ) = c and w(P 0 ) = d, with (c 6= d). 0
1
0
1
Let (x; y) be the bottom left corner of rectangle Ri and let R^ i be the h k rectangle having (x ? (d pn e ? 1); y ? (d pm e ? 1)) as bottom left corner, with h = n ? d pn e ? 1 and k = m ? d pm e ? 1. By Lemma 3.1, the factor that has R^i as shape admits a periodicity vector, say v00. An argument analogous to the one used in the proof of the claim of the Lemma 4.2 can be used to prove that v00 cannot be parallel to v0. Therefore v00 incides v0. Notice also that since R^ i contains Ri , also the factor that has shape Ri admits v00 as periodicity vector. Since factor ui? having as shape the union Ui? of the rst i (R; v0)rectangles Rs, 0 s (i ? 1), is v0-periodic and since the sides of R0 have lengths greater than the respective components of v0, there exists 2ZZ such that 0
0
0
0
0
0
1
1. 2. 3. 4.
1
P + v0 = P 2 R0 w(P ) = w(P ) = a (P ) + v0 = P 2 R0 w(P ) = w(P ) = b 0
0
0
1
0
1
1
1
30
By construction and since P and P belong to the set (f k (S ) ? f k (S )), also P and P belong to the set (f k (S ) ? f k (S )). 0
0
+1
1 +1
1
Since the stripe f k (S ) contains an s t rectangle with s pn and t pm , ^ 00 and P + v ^ 00 are there would exists ^ 2 f+1; ?1g such that both P + v ^ 00; P + ^v00 all belong in the stripe f k (S ). In this case, since P ; P ; P + v 0 00 to R that has period v ^ 00) = w(P ) = b 6= a ^ 00) = w(P + v a = w(P ) = w(P + v that is an absurd. Hence there exists another stripe parallel to f k (S ), overlapping f k (S ) and that contains some Christo el lines not contained in f k (S ), which has a periodicity vector parallel to the periodicity vector of S and which contains an s t rectangle with s pn and t pm . Iterating previous argument we can cover the plane with such stripes. Therefore we are able to nd a periodicity vector for the whole double sequence w. This vector is parallel to v and its length is the product of the lengths of the periodicity -vectors of the di erent stripes. Notice that the number of di erent stripes can be in nite but the number of di erent vectors is nite. 2 0
0
0
0
0
0
1
0
1
1
1
1
0
0
4.3 Proof of the Theorem
Proof
p
p
First case n0 4 orpm0 4 . p Let us suppose m0 4 , the proof for n0 4 is analogous.
Let B = Am a new alphabet (i.e. the letters of B are the words of length m over A). Let us de ne a new bidimensional word u in the following way: 0
0
u[i; j ] = b 2 B () b = w[i; m j ]w[i; m j + 1] w[i; m j + m ? 1] In other words we have grouped the rows of w in packets of size m . It 0
0
is not dicult to prove by construction that 31
0
0
0
u has a periodicity vector () w has a periodicity vector. The complexity function pu of u is such that pu (n ; 1) pw (n ; m ) < 1 n m n Therefore, by Theorem 2.2, any line of u is!periodic with period smaller than or equal to n . Consequently u has n0 ! as periodicity vector. Hence also w has a periodicity vector. p p Second case n 4 or m 4 . The proof of this case is a direct consequence of Lemma 4.2 and of Lemma 4.3. 2 0
0
0
0
0
0
0
0
0
0
5 Examples In this section we show that there exists a sequence of bidimensional words (wn )n such that pwn (n; n) = o(n ), even if wn admits a periodicity vector which has length greater than or equal to 2
9
n
n +o( n ) n 2 lg n :
p
p
2 lg
p
For any natural n 9, let us take m = b nc and let x ; x m be the increasing sequence of prime numbers smaller than or equal to m. Let us consider the sequence z ; z m , such that for any i, 1 i (m), zi is the greatest power of xi smaller than or equal to m, i.e. 1
(
1
(
)
)
zi = xih m; xih > m. +1
It is easy to prove that for any i, 1 i (m), zi > pm. Let us now de ne the binary bidimensional word wn as the juxtaposition of the (2(m) + 1) vertical stripes de ned in the following. 32
Each one of these stripes except the rst one and the last one, which have in nite width, have width n . If j is odd, then wn (x; y) = 0 for any (x; y) belonging to the j -th stripe. If j is even, j = 2k for some integer 1 k (m), then wn (x; y) = 0 for any (x; y) belonging to the j -th stripe such that zk is a divisor ! of 0 y, wn (x; y) = 1 elsewhere. In this way this stripe admits z as k periodicity vector. 2
n2
n2
n2
n2
n2
0
1
1
0
0
1
1
0
0
0
0
0
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1
1
0
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0
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1
0
0
0
0
0
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1
0
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1
1
0
0
1
1
0
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1
1
0
0
1
1
0
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1
0
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0
0
0
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1
1
0
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1
1
0
0
1
1
0
0
0
0
0
0
0
0
0
0
1
1
0
0
1
1
0
0
1
1
0
0
0
0
0
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1
1
0
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1
1
0
0
1
1
1
1
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1
1
0
0
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1
0 0
0
0
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
!
It is easy to see that any periodicity vector v = vv of the whole wn is such that v = 0. Moreover the length of this periodicity vector is the least common multiple of all the natural numbers smaller than or equal to n, that is also equal to the product of all the zi. Let us give now a rough estimate of this number. Since for any i, 1 i m(m), zmi > m = , the product of all the zi is greater than m m = = p o m , by the Prime Number Theorem. Since m = b nc, then we m m obtain that 1 2
1
1 2
(
+ ( 2 lg( ) ) 2 lg( )
33
) 2
m m = = n (
) 2
n +o( n ) n 2 lg n
p
p
2 lg
We want now to prove that pwn (n; n) = o(n ). If a square of side n is completely inside a stripe or a semiplane where the ! word wn has periodicity vector z0 , then the number of di erent factors i is zi. If the square is completely inside a stripe or a semiplane where wn has ! 0 periodicity vector 1 , then there is only a factor of side n. If it is between two consecutive stripes or a stripe and a semiplane, the number of di erent factors is 2(n ? 1)zi. In the rst case, i.e. in the case in which the square is completely inside a stripe or a semiplane, we get at most 1 + i m zi factors, while in the second case we get at most i m 2(n ? 1)zi factors. By the Prime Number Theorem, since for any i, zi m, it follows that 2
( ) =1
( ) =1
1 + i m zi + i m (2(n ? 1)zi) = o(m (n ? 1)) = o(n ) ( ) =1
( ) =1
2
2
Previous example shows that di erent phenomena happen in the one and in the two dimensional case. Indeed, in the one dimensional case, pw (n) < n implies not only that w has a period but also that this period is bounded by n. In the bidimensional case, if pwn (n; n) < n , the bound of the length of the periodicity vector is not even polinomial in n. 2
6 Concluding Remarks In this paper we proved a weak version of the Nivat conjecture. We hope that with this approach it will be possible to settle completely the conjecture. Very recently J. W. Sander and R. Tijdeman (cf. Example 5 of [33]) have shown, with a simple but interesting example, that it is not possible to extend in a completely analogous way the Nivat's conjecture to the k-dimensional case. His example shows that for any n there exists a k-dimensional sequence w that has no periodicity vector and such that the number of factors having shape a full dimensional hypercube of side n is n + 1. 2
34
We now state the following conjecture that somehow extend the Nivat's conjecture to the k-dimensional case. Conjecture: if the number of factors having shape a full dimensional hypercube of side n is smaller than or equal to n , then the k-dimensional sequence w is k ? 1 dimensional-lattice periodic L, i.e. the minimal number of generators of L is k ? 1. 2
Acknowledgements. We thank J. C. Lagarias and D. Bernardi for discussions and comments.
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