Distributing labels on infinite trees

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This shows that the tree is balanced. It is not strongly balanced since there are factors ofsize (1, 1) with 2 nodes labeled by one and others with 0 nodes labeled by one as seen inthe bottom right part of figure 18. Also its minimal graph is not isomorphic to the uniqueminimal graph of a mechanical tree of density 4/7 that has only 3 nodes (see the discussionabout graphs of strongly balanced tree section 5.1.1).1 23 4Figure 18: A Rational Balanced Tree that is not strongly balanced9. Irrational balanced tree that is not strongly balanced – Building an irrational tree notstrongly balanced requires more work. We consider a tree that which has a root r labeledby 0 and two children that are mechanical trees of density α and respective phasesφ and φ + a. We will see that under some conditions on α, φ and a this will give us anexample of an irrational tree that is balanced but not strongly balanced neither rationalnor Sturmian.α, φ α, φ + aThe two children of the root are balanced trees which means that the tree is balanced ifand only if for all n:⌊(2 n+1 − 1)α⌋ ≤ h r (n + 1) ≤ ⌊(2 n+1 − 1)α⌋ + 1 (29)Let us call k = ⌊(2 n − 1)α + φ⌋ and x = frac((2 n − 1)α + φ).h r (n + 1) = ⌊(2 n − 1)α + φ⌋ + ⌊(2 n − 1)α + φ + a⌋= k + ⌊k + x + a⌋As (2 n+1 − 1)α = 2k + 2x + α − 2φ, the equation 29 holds if for all x ∈ [0; 1), we have:which holds if for all x ∈ [0; 1):This equation is satisfied if and only if0 ≤ k + ⌊k + x + a⌋ − ⌊2k + 2x + α − 2φ⌋ ≤ 10 ≤ ⌊x + a⌋ − ⌊2x + α − 2φ⌋ ≤ 1(x + a < 1 and − 1 ≤ 2x − 2φ + α < 1) or (x + a ≥ 1 and 0 ≤ 2x − 2φ + α < 2)28
Looking at the extremal cases for x + a < 1 and x + a ≥ 1 which are x = 0, 1 − a, 1, onegets 4 relations:2(1 − a) − 2φ + α < 1−1 ≤ −2φ + α2 − 2φ + α < 2Therefore the tree is balanced if and only if0 ≤ 2(1 − a) − 2φ + α.α2 < φ ≤ α + 1 < φ + a < 1. (30)2Moreover if α + φ ≥ 1 and 3α + φ < 2, the tree is not strongly balanced since its beginningisThere are lots of triples α, φ, a satisfying conditions (30). For example a tree with α = 1 3 +ɛ,φ = 0.6 and a = 0.2 where ɛ ∈ R \ Q with ɛ small enough (for example ɛ < 0.01 works sinceαα+12≈ 0.21 < 0.6 1, 3α + φ ≈ 1.9 < 2).References[1] A.V. Aho and N.J.A. Sloane. Some doubly exponential sequences. Fibonacci Quarterly,11(4):429–437, 1973.[2] E. Altman, B. Gaujal, and A. Hordijk. Discrete-Event Control of Stochastic Networks:Multimodularity and Regularity. Number 1829 in LNM. Springer-Verlag, 2003.[3] J. Berstel. Sturmian and episturmian words (a survey of some recent result results). In G. RahonisS. Bozapalidis, editor, Conference on Algebraic Informatics, Lecture Notes Comput.Sci. 4728, pages 23–47, 2007.[4] J. Berstel, L. Boasson, O. Carton, and I. Fagnot. A First Investigation of Sturmian Trees.LECTURE NOTES IN COMPUTER SCIENCE, 4393:73, 2007.[5] E. Borel. Lesprobabilites denombrables et leurs applications arithmetiques. Rend. Circ.Mat. Palermo, 27:247–271, 1909.[6] J. Cassaigne. Double sequences with complexity mn+1. J. Autom. Lang. Comb., 4(3):153–170, 1999.[7] R. Durrett. Probability: theory and examples. Wadsworth & Brooks/Cole, 1991.[8] T. Fernique. Pavages, Fractions continues et géométrie discrète. PhD thesis, University ofMontpellier, 2007.[9] Nicolas Gast and Bruno Gaujal. Balanced labeled trees: density, complexity and mechanicity.In Words, 6th international conference on words, Marseille, France, 2007.29
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 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: 7. Irreducible mechanical trees - l

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