Distributing labels on infinite trees

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Proof. Let m n be the minimal number of 1 in all factors of size n. As the tree is balanced, forall nodes v and n ≥ 1:m n ≤ h v (n) ≤ m n + 1 (5)Now let us consider a factor of size n + k and root v. It can be decomposed in a factor of sizek of root v and d k factors of size n at the leaves of the previous factor. The number of ones inthese factors can be bounded by m n and m k , therefore we have:m k + d k m n ≤ m n+k ≤ m k + 1 + d k (m n + 1) (6)The density of a factor of size n is mnS(n) ≤ d v(n) = hv(n)S(n)bound d v (n + k) − d v (n):≤ mn+1S(n). Using these facts, we canm n+kS(n + k) − m n + 1≤ d v (n + k) − d v (n) ≤ m n+k + 1S(n)S(n + k) − m nS(n)Using (6), the left inequality can be lower bounded by(d − 1) ( d k m n + m kd n+k − 1 − m n + 1)d n − 1= (d − 1) ( m n + m k /d kd n − 1/d k − m n + 1)d n − 1)≥ (d − 1) ( m nd n − 1 − m n + 1d n − 1≥ − 1S(n)The same method can be used to prove that d v (n + k) − d v (n) ≤ 1S(n), which shows thatfor n big enough, |d v (n + k) − d v (n)| is smaller than ɛ, regardless of k. Thus d v (n) is a Cauchym nS(n)sequence and has a limit α = lim n→∞ . This limit does not depend on v and the tree has adensity.Lets now prove that d v (n) − ⌊S(n)α⌋| ≤ 1: dividing the inequality (6) by S(n, k) and takingthe limit when k tends to ∞ leads to:(d − 1)m n + αd n≤ α ≤ (d − 1)m n + 1 + αd n .This shows that: S(n)α − 1 ≤ m n ≤ S(n)α, which implies Equation (4).Similar ideas can be used to show that Equation (4) can be improved in the case of stronglybalanced tree: for all width and size k, n ≥ 1, the number of ones h(n, k) in a factor of size nand width k satisfies:∣ h(n, k) − ⌊S(n, k)α⌋∣ ∣ ≤ 1 (7)4.3 Mechanical treesBuilding balanced tree is not that easy. According to formula (4), each factors of size n musthave ⌊αS(n)⌋ or ⌊αS(n)⌋ + 1 nodes one. This leads us to the following construction, inspired bythe construction of mechanical words.12
Definition 4.2 (Mechanical tree). A tree is mechanical of density α ∈ [0; 1] if for all node v,there exists a phase φ v which satisfies one of the two following properties:⌋∀n : h v (n) =⌊S(n)α + φ v , (8)⌉or ∀n : h v (n) =⌈S(n)α − φ v . (9)In the first case, we say that φ v is an inferior phase of v. In the second case, we say that φ vis a superior phase of v.This definition suggests that the phases of all nodes could be arbitrary. In fact, we will seethat there exists a unique mechanical tree with a given phase at the root. The second questionraised by this definition is the existence and uniqueness of the phase: we call φ v “a” phase of anode φ v and not “the” phase of φ v since there may exist several phases leading to the same tree.We begin by a characterization of mechanical trees, given in the following formula:Proposition 4.2.1 (Characterization of mechanical trees). For each α ∈ [0; 1] and φ ∈ [0; 1),there exists a unique mechanical tree of density α such that φ is an inferior (resp. superior)phase of the root.Moreover, if φ is an inferior (resp. superior) phase of a node then φ 0 ≤ · · · ≤ φ d−1 areinferior (resp. superior) phases of its d children, withφ i =α + φ + i − ⌊α + φ⌋d(resp. φ v =)φ + ⌈α − φ⌉ − α + i. (10)dThe proof will be done in two steps. First we will see that if we define the phases as in (10)we have a mechanical tree, then we will see that this is the only way to do so.Proof. Existence. Let α ∈ [0; 1] and φ ∈ [0; 1). We want to build a mechanical tree whichroot has an inferior phase φ ( the case of a superior phase if similar and is not detailed here).Let A be an infinite tree. To each node v, we associate a number φ v defined by:• φ root = φ.• If the phase of a node v is φ v , its d children satisfy Equation (10).Then we build a labeled tree by associating to each node v the label ⌊α + φ v ⌋. Let us prove byinduction on n that the following relation holds.⌋For all v : h v (n) =⌊S(n)α + φ v . (11)By definition of the <strong>labels</strong>, (11) holds when n = 1. Let n ≥ 0 and let us assume that (11) holdsfor n. Let v be a node with phase φ v and let φ 0 . . . φ d−1 be the phases of its children. We assumethat α + φ v < 1 (which means that the label of the node is 0) a similar calculation can be donein the other case, α + φ v > 1.13
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: If x is a factor of w, its height h
Page 15 and 16: Let x 1 , . . . , x k , . . . (resp
Page 17 and 18: Using the well-known inequality x
Page 19 and 20: 13A 1 A 0A 1 A 0A 0024Figure 8: Exa
Page 21 and 22: of a strongly balanced tree can be
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
Page 27 and 28: 7. Irreducible mechanical trees - l
Page 29 and 30: Looking at the extremal cases for x

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