Distributing labels on infinite trees

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4.3.1 Phases of a treeLet us call Φ v the set of numbers that can be phases of a node v and Φ, the set of the possiblephases of a tree is the union of all possible phases of its nodes: Φ = ∪ b Φ v . The set Φ may becountable or uncountable. Countable for example when α/(d − 1) is normal since there are atmost as many phases as nodes. Uncountable for example for the tree with all label 0, for whichfor each node, all phases in [0; 1) work. Nevertheless, the set of possible phases is dense is [0; 1).Indeed, at least all phases defined by the relation (10) are in Φ. If φ is the phase of the root,then all nodes at level k have a phase which is the fractional part of:φ+α+i kd +α+i k−1d+ · · · + α + i 1= α( 1 dd k . . . 1 d ) + φ d k + i kd k + · · · + i 1d 1 , (16)with 0 ≤ i j < d for all j. Conversely all of these numbers are the phases of some node at level k.As k tends to infinity, by a proper choice of i 1 , . . . , i k the fractional part of this number canbe as close as possible to any number in [0; 1]. Thus the set of phases of the tree is dense in [0; 1].If the density isp(d−1)d n+k −d k(with n+k minimal) one can show that the set of all possible phasesfor a given node is [ dm −1d−1 α; min( dm+1 −1d−1α, 1)) for some m ∈ 0, . . . , n + k − 1. As Φ is dense in[0; 1), it contains all of these intervals. Therefore, Φ = [0; 1) and the tree has exactly n + kdifferent factors of size greater than n + k. Hence its minimal graph has exactly n + k nodes.4.4 Equivalence between strongly balanced and mechanical treesAs we have seen in section 4.1, there are strong relations between balanced and mechanicalwords. In this part, we will see that we can prove the same results between strongly balancedand mechanical trees. This result is formally stated in the following theorem.A tree is ultimately mechanical if all nodes (except finitely many) are mechanical (i.e. satisfiesthe equations 8 or 9).Theorem 4.1. The following statements are true.(i) A mechanical tree is strongly balanced.(ii) An irrational strongly balanced tree is mechanical.(iii) A rational strongly balanced tree is ultimately mechanical.This theorem is the analog of the theorem linking balanced and mechanical words. We haveseen that the word 0 k 10 ∞ is balanced but not mechanical, only ultimately mechanical. Itscounterpart for trees would be a tree with all label equal to 0 except for one node which has alabel 1. The number 1 can be chosen as deep as desired, which shows that we can not boundthe size of the “non-mechanical” beginning of the tree. A more complicated example is drawnFigure 8.Let us begin by the proof the first part of the theorem:Lemma 4.2.1. A mechanical tree is strongly balanced.Proof. Let n, k ∈ N. For all node v, h v (n, k) is the number of 1 in the factor of size n and widthk rooted in v. We want to prove that for all pairs of nodes v and v ′ : |h v (n, k) − h v ′(n, k)| ≤ 1.We assume that the nodes v and v ′ are inferior of phase φ and φ ′ (the proof with superiorphases is similar).h v (n, k) − h v ′(n, k) = ⌊ dn+k − 1d − 1 α + φ⌋ − ⌊dk − 1d − 1 α + φ⌋ − ⌊dn+k − 1d − 1 α + φ′ ⌋ + ⌊ dk − 1d − 1 α + φ′ ⌋.16
Using the well-known inequality x − x ′ − 1 < ⌊x⌋ − ⌊x ′ ⌋ < x − x ′ + 1, one can show that−2 < h v (n, k) − h v ′(n, k) < 2.As h v (n, k) and h v ′(n, k) are integers, we have −1 ≤ h v (n, k) − h v ′(n, )k ≤ 1 which ends theproof of the lemma.We will see in the next section 4.5 that a tree is rational if and only if its density can bepwritten asS(n,k)(p, k, n ∈ N), therefore we will do the proof of theorem 4.1 distinguishingstrongly balanced tree with density of this form or not.Lemma 4.2.2. If A is a strongly balanced tree of density α which can not be written as(p, k, n ∈ N) then A is mechanical.pS(n,k)Proof. let τ be a real number and v a node. At least one of the two following properties is true:∀n ≥ 1 : h v (n) ≤ ⌊S(n)α + τ⌋, (17)∀n ≥ 1 : h v (n) ≥ ⌊S(n)α + τ⌋. (18)To prove this, assume that it is not true. Then there exists k, n such that h v (n) < ⌊S(n)α + τ⌋and h v (k) > ⌊S(k)α + τ⌋. In that case the number of 1 in the factor of size n, n − k (or k, k − nif k > n) is h v (n) − h v (k) ≤ ⌊S(n)α + φ⌋ − ⌊S(k)α + φ⌋ − 2 < dn −d kd−1α − 1 which violates theformula (7).Let us define now the number φ as the minimum τ that satisfies (17)φ = infτ{For all n : h v (n) ≤ ⌊S(n)α + τ⌋}.For all τ > φ, the equation (17) is true, while for all τ ′ < φ, the equation (18) is true. Thismeans that for all ɛ > 0 and all n:S(n)α + φ − ɛ − 1 ≤ ⌊S(n)α + φ − ɛ⌋ ≤ h v (n) ≤ ⌊S(n)α + φ + ɛ⌋ ≤ S(n)α + φ + ɛ. (19)Taking the limit when ɛ tends to 0 shows that:S(n)α + φ − 1 ≤ h v (n) ≤ S(n)α + φ. (20)Therefore, unless S(n)α + φ ∈ N, h v (n) = ⌊S(n)α + φ⌋ = ⌈S(n)α + φ − 1⌉.If there exists n ∈ N such that S(n)α + φ ∈ N, then there are no other k ∈ N (k ≠ n)psuch that S(k)α + φ ∈ N – otherwise that would violate the condition α /∈ {S(n,k), p, k, q ∈ N}.Therefore, for this particular n, either h v (n) = S(n)α + φ = ⌊S(n)α + φ⌋ – in that case the nodeis inferior of phase φ – or h v (n) = S(n)α + φ − 1 = ⌈S(n)α + φ − 1⌉ – in that case the node issuperior of phase 1 − φ.Lemma 4.2.3. Let A be a tree such that there exist n and k such that all factors of size (n, k)have the same number of nodes with label 1. Then the tree is mechanical.Proof. Let us take n and k satisfying the property, such that n + k is minimal and let us callp is the common number of ones in the factors of size (n, k). Obviously, the tree as a densityα =p(d−1)d k (d n −1) .Let v be the root of the tree. The same proof as in the irrational case can be used to establishthat there exists φ such thatS(n)α + φ − 1 ≤ h v (n) ≤ S(n)α + φ,17
Page 1 and 2: Distributing <stro
Page 3 and 4: 2 Infinite TreesFor ordered infinit
Page 5 and 6: 2.1 Irreducibility and periodicityB
Page 7 and 8: Figure 4: The first tree has a dens
Page 9 and 10: Let us now consider the case when t
Page 11 and 12: If x is a factor of w, its height h
Page 13 and 14: Definition 4.2 (Mechanical tree). A
Page 15: Let x 1 , . . . , x k , . . . (resp
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Page 23 and 24: Algorithm 1 Testing if a rational t
Page 25 and 26: C v (n) def= g(d v (n)).We can defi
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Page 29 and 30: Looking at the extremal cases for x

Distributing labels on infinite trees

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