About Fibonacci trees III: multiple Fibonacci trees
aa r X i v : . [ c s . D M ] S e p About Fibonacci trees. − III − :multiple Fibonacci trees Maurice M argenstern
September 5, 2019
Abstract
In this third paper, we revisit the question to which extent the propertiesof the trees associated to the tilings { p , } of the hyperbolic plane are stilltrue if we consider a finitely generated tree by the same rules but rooted ata black node? What happens if, considering the same distinction betweenblack and white nodes but changing the place of the black son in the rules.What happens if we change the representation of the numbers by anotherset of digits? We tackle all of these questions in the paper. Section 1 indicateswhich Section or Sub-section is devoted to which problem. Section 6.4.2concludes the paper. The present paper is an extension of the previouspapers [5, 6]. Paper [5] investigated the question what happens if the rules generating thestandard Fibonacci tree are applied to a tree whose root is a black node. Thequestion was investigated with what is called in this paper the leftmost assign-ment: in the generating rules, the black son is always the first one. In that papertoo, the question was raised of what happens if instead the standard Fibonaccisequence we consider what was called the golden sequence which is associatedwith the square of the golden number while the standard Fibonacci sequenceis associated with the golden number itself. The question was considered forboth the white and the black Fibonacci trees.Paper [6] generalizes the context of the same questions. Instead of consider-ing trees more or less connected with the tilings { , } and { , } of the hyperbolicplane, that paper considers the trees which span the tilings { p , } and { p + , } of the same plane. Those trees are finitely generated by rules which generalizethe rules of the case p = p =
5. In the present paper, we perform twonew steps in the generalization. On one hand, we consider the definition ofthe generating rules themselves. Instead of considering that the black son isthe first one in each rule, we consider various possibilities if the position ofthat son is changed, whether the change is always the same or if the change isalso submitted to variations. On another hand, we consider the representationof the numbers. In paper [6], the metallic sequences are defined with digitsin { .. p − } . What if we impose the digits to be in { .. p − } , considering onlypositive integers which is the case for the numbers we attach to the tiles?Section 2 recalls definitions about trees and about the trees we considerin the present paper. Section 3 recalls the results about the metallic numbersand the standard representation of positive numbers we can infer from them.The section also considers the representation where the set of digits is { .. p − } .Section 4 defines what we call an assignment, a way of applying rules inthe construction process of a tree and the section studies the properties ofthe representations studies in Section 3 with respect to various assignments.Section 6.4.2 investigates the contribution of the paper and the problems whichremain open. In this section, Sub-section 2.1 recalls the vocabulary we shall use in the consid-erations of the trees which appear in the paper and in the properties connectedwith them, considering, in particular, the numbering we may attach to thenodes of a finitely generated tree. In Sub-section 2.2, we consider the metallictrees which we shall study in this paper.
Consider an infinite tree T with finite branching at each node. Number thenodes from the root which receives 1, then, level by level and, on each level,from left to right with the conditions that for each node, the numbers of itssons are consecutive numbers. We then say that T is numbered or that itis endowed with its natural numbering . In what follows, we shall considernumbered trees only. Clearly, a sub-tree S of T can also be numbered in the justabove described way but it can also be numbered by the numbers of its nodesin T . In that case, a node ν may receive two numbers: n S , the number definedin S as a numbered tree and n T , its number as a node of T . A node may haveno son, it is then called a leaf . A path from µ to ν is a finite sequence of nodes { λ i } i ∈ [0 .. k ] , if it exists, such that λ = µ , λ k = ν and, for all i with i ∈ [0 .. k − λ i + isa son of λ i . A branch of T is a maximal finite or infinite sequence of paths { π i } from the root of T to nodes of that tree such that for all i , j , π i ⊆ π j or π j ⊆ π i .Accordingly, a branch connects the root to a leaf or it is infinite. It is clear thatfor any node, they are connected to the root by a unique path. The length of2he path from a node to one of its son is always 1. If the length of a path from µ to ν is k , the length of the path from µ to any son of ν , assuming that ν is nota leaf, is k +
1. The length of the path leading from the root to a node ν of T is called the distance of ν to the root ρ and it is denoted by dist { ρ, ν } . We alsodefine dist { ρ, ρ } =
0. The level k of T is the set of its nodes which are at thedistance k from its root. Denote it by L k , T . Define T n as the set of levels k of T with k ≤ n . Say that the height of T n is n . By definition, T n is a sub-tree of T .For each node ν of T , λ T ( ν ) is its level in T , i.e. its distance from the root, and σ T ( ν ) is the number of its sons. Clearly, if ν ∈ T n and if λ T ( ν ) = n , then σ ( ν ) = S is a sub-tree of T , denote it by S ⊳ T , and if ν ∈ S , then λ S ( ν ) ≤ λ T ( ν ) andthe numbers may be not equal.Consider two infinite numbered trees T and T . Say that T and T are isomorphic if there is a bijection β from T onto T such that: f ( n T ) = n T for any n ∈ N . λ T ( f ( n T )) = λ T ( n ). σ T ( f ( n T )) = σ T ( n ). (0)Clearly, if T and T are two infinite numbered trees, they are isomorphic ifand only if there is a bijection from the nodes of T into those of T such thata node of T and its image in T have the same number, they are on the samelevel of their respective trees and they have the same number of sons. We call metallic tree an infinite tree constituted by two kind of nodes, b - and w -ones called black and white respectively, finitely generated by the followingrules: b → bw p − and w → bw p − . (1)with p ≥ literal status . Wealso associate to the node its numerical status : 0 or 1 depending on whetherthe node is white or black respectively. If it is not specified, status will refer tothe literal one.We shall mainly investigate two kinds of infinite metallic trees. When theroot of the tree is a white, black node, we call such a metallic tree a white , blackmetallic tree respectively. We denote the infinite white metallic tree by W andwe endow it with its natural numbering. We do the same with the infinite blackmetallic tree B . Note that we can construct a bijective morphism between B and a part B of W as follows. The morphism is the identity on B and we fixthe following conditions: σ B (1) = σ W (1) − σ B ( n ) = σ W ( n ), for all positive integer n .Moreover, the nodes numbered by n ∈ [1 .. p −
2] in W also belong to B andreceive the same numbers in the natural numbering of B . This morphism3llows us to identify B with B , so that in our sequel, we shall speak of B only. From what we just said, it is plain that for a node ν ∈ B , if ν B > p − ν B < ν W . We shall look closer to the connection between ν B and ν W inSection ?? . Later on, we shall use ⇋ to introduce a notation for an expression.We may wonder whether the simplicity of the rules (1) allow us to givea precise connection between the number of a node and those of its sons,whether in W or in B . These questions were partially studied in [6]. We shallturn to them in Section 4. But before, we need to recall the introduction ofan appropriate representation of the numbers used to number the nodes of ametallic tree. It is the goal of Section 3 to which we new turn. These numbers are introduced by the computation of the number of nodeswhich lie on a given level of a metallic tree. We consider that point in Sub-section 3.1. The sequence allow us to represent the numbers. We considersome basic properties of the standard representation in Sub-section 3.2. Westudy the same properties in the representation where the digits are restrictedto { .. p − } in Sub-section 3.3. Let m n , b n be the number of nodes on L n , W and L n , B respectively. We also define M n , and B n as the number of nodes of W n and B n respectively. It appears thatthese numbers are defined by a simple induction equation as stated in thefollowing statement:T heorem [2, 6] Consider the numbers m n defined as the number of nodes on L n , W ,where W is the white metallic tree. The numbers m n satisfy the following inductionequation: m n + = ( p − m n + − m n with m = and m − = . (2) We call white metallic sequence the sequence { m n } n ∈ N . See the proof in [6] for instance.As the black metallic tree is defined by the same rules, we may concludethat the same equation rules the sequence { b n } n ∈ N :T heorem The sequence { b n } n ∈ N of the number of nodes on L n , B satisfies the equation:b n + = ( p − b n + − b n with b = p − and b = . (3) We call black metallic sequence the sequence { b n } n ∈ N . Note that we could define the white metallic sequence by the initial con-ditions m = p − m =
1. In our sequel we shall say metallic sequence white metallic sequence for a reason which will be made more clearin a while.Before turning to the properties of the integers with respect to the metallicnumbers, we have to consider the numbers M n and B n already introduced withrespect to the finite trees W n and B n .T heorem (see [3]) On the level k of W , with non-negative k, the rightmost nodehas the number M k , so that the leftmost node on the level k + has the number M k + .On the level k of B with non-negative k, the rightmost node has the number m k , sothat the leftmost node on the level k + has the number m k + .The sequence { M n } n ∈ N satisfies the following induction equation:M n + = ( p − M n + − M n + , (4) with the initial conditions M = and M − = , while the sequence { B n } n ∈ N satisfythe equation (2) with the same initial conditions, which means that B n = m n for anynon-negative n. We also have:M n + = B n + + M n and m n + = b n + + m n (5)See the proof in [6]. . Let us go back to the sequence { m n } n ∈ N of metallic numbers. It is clear thatthe sequence defined by (2) is increasing starting from m : from (2), we getthat m n + > ( p − m n + if we assume that m n < m n + . As p ≥
5, we get that thesequence is increasing starting from m . Now, as the sequence is increasing, itis known that any positive integer n can be written as a sum of distinct metallicnumbers whose terms are defined by Theorem 1: n = k X i = a i m i with a i ∈ { .. p − } . (6)The sum of a i m i ’s in (6) is called the metallic representation of n and the m i ’sin (6) are the metallic components of n .From now on, we use bold characters for the digits of a metallic representa-tion of a number. In particular, we define d to represent p − c to represent p − e to represent p − p >
5. Of course, , , and represent 0, 1, 2 and3 respectively.First, note that the representation (6) is not unique.L emma [7, 3, 6] For any integers n and h with ≤ h ≤ n, we have: ( p − m n + + n X k = h + ( p − m k + ( p − m h = ( p − m n + + n X k = h + ( p − m k + ( p − m h + − m h + m h − (7)5 orollary [7, 3] For any positive integer n, we have: ( p − m n + + n X k = ( p − m k + ( p − m = m n + (8)See the proofs in [6] for instance.Let us write the a i ’s of (5) as a word a k ..a a which we call a metallicword for n as the digits a i which occur in (5) are not necessarily unique fora given n . They can be made unique by adding the following condition onthe corresponding metallic word for n : the pattern dc ∗ d is ruled out from thatword. It is called the forbidden pattern . Lemma 1 proves that property whichis also proved in [7, 3]. We reproduced it here for the reader’s convenience.When a metallic representation for n does not contain the forbidden patternit is called the metallic code of n which we denote by [ n ]. We shall write ν = ([ ν ]) when we wish to restore the number from its metallic code. Let us call signature of ν the rightmost digit of [ ν ] = a k .. a a and denote it by sg . Let σ , σ , ..., σ k with k = p − k = p − ν . We call sons signature of ν the word s ...s k , where s i = sg ( σ i ). We shall denote the literal status of ν by ℓ s ( ν )and its numerical one by sn ( ν ). It is known that given a basis b with b ≥ n can be written n = k X i = a i b i with a i ∈ { .. b } (9)Let a + = a + a − = a − a . The representation (9)was used by Quine in order to encode any finite sequence of natural numbers:writing n as in (9), b is used as a separator and the other digits which lie in[1 .. b − ] can be interpreted as the representations of positive numbers in the base b − which requires b ≥
2. Smullyan, see [ ? ] makes use of such a representationin order to prove G ¨odel’s theorem on the incompleteness of Peano arithmetics.The question is: taking the metallic numbers m n as a basis, is it possible tohave a representation which rules out all 0’s in the representation of a positivenumber? The answer is given by the following proposition:T heorem Let n be a positive natural number. Then it is possible to write n as:n = k X i = a i m i with a i ∈ { .. b } , (10) where b = p − . Proof. As all digits of n in (10) should be not smaller than , we consider k de-fined by the unique value such that M k ≤ n < M k + . We then define n ⇋ n − M k .6onsider the digits a i of [ n ], the metallic code of n . We have n = k X i = a i m i and M k = k X i = m i , so that we get n = k X i = s i m i where s i = a i + i ∈ { .. k } . As0 ≤ a i < b for all i in { .. k } we get 1 ≤ s i ≤ b for the same indices. (cid:3) When we use the representation (10) to write a positive integer, we use x todenote the digit whose value is b = p −
2. We know that the representation withthe metallic numbers of a positive integer is not necessarily unique. This is whywe needed to select a criterion in order to ensure the uniqueness of the metalliccode. It is also the case that the representation (10) is not unique, despite thefact that it satisfies the constraint of no among the digits. To see that point,we need the following result which enlarges a lemma from [6]:L emma For any integers n and h with ≤ h ≤ n, we have: ( p − m n + + n X k = h + ( p − m k + ( p − m h = ( p − m n + + n X k = h + ( p − m k + ( p − m h + + m h − (11)C orollary For any positive integer n, it has a unique representation (10) providedthat the pattern xd ∗ x is ruled out and is called a forbidden pattern . A metallic codewhere the -digit is ruled out and where the forbidden pattern does not occur is called a non-zero metallic code , nzm -code for short. The nzm -code of n is denoted by [ ν ] nz .In order to get a [ ν ] nz from (10) , we apply the conversion rules: nxd k xm = n + k + m + and n + k + m + = ndc k dm (12) where n , m are non-zero digits in { .. x } , n + = n ⊕ and m + = m ⊕ , with a ⊕ = a + if a + < p − and a ⊕ = if a = p − . We call the forbidden pattern defined in Corollary 2 the nzm -forbiddenpattern in order to distinguish it from that of Lemma 1. Note that the forbiddenpattern of Lemma 1 is no more forbidden in an nzm -code. Of course, theapplication of (12) may be repeated in (10) as long as all occurrences of the nzm -forbidden pattern are replaced by their permitted equivalent expressiongiven in (12).It is not di ffi cult to adapt the algorithms of [6] to operations on nzm -codes.We just mention the change about the incrementation and decrementation al-gorithms.We take the notations given in the caption of Algorithm 1. The algorithmfirst detects whether the nzm -code of ν has a su ffi x xd h . If it is the case,appending 1 to a would create a nzm -forbidden pattern. And so, in that case, xd h is replaced by h + according to (12) and 1 is added to the digit which isto the left of x and which is less than p −
2. If appending 1 would not raise aforbidden pattern, the algorithm looks at whether a is x or not. If it is x , a is7eplaced by and 1 is added to a which is not x as a was x . If a is is not x , 1is added to it. Algorithm 1
Algorithm giving [ ν + nz from that of [ ν ] nz . Recall that [ ν ] nz does notcontain a nzm -forbidden pattern. We assume that [ ν ] nz = a k .. a .i : = while a i = d loop i : = i + end loop ; if i > then if a i = xthen for j in [0 .. i ] loop a j : = ; end loop ; a i + : = a i + + ; else a : = x ; end if ; else if a = xthen a : = ; a : = a + else a : = a + ; end if ; end if ;The decrementation algorithm works exactly in the opposite way: if a is not , we can replace it by a − . Otherwise, we look at the position h of the leftmostitem of consecutive ’s. We replace a h + by a h + − , we replace a h by x and thenwe replace all a i from 0 up to h − d : see Algorithm 2. Note that when a i is found di ff erent from in the while -loop, i >
0, so that x is always writtenin place of the digit at i −
1. If i = while -loopwe write x instead of a and the range of the for -loop is empty. Accordingly,Algorithm 2 works in all cases for a positive integer ν . Algorithm 2
Algorithm giving the [ ν − nz from [ ν ] nz ,provided that ν is positive. Re-call that [ ν ] nz does not contain an nzm -forbidden pattern. Assume that [ ν ] nz = a k .. a . if a , : = a − ; else i : = while a i = i : = i + end loop ; a i : = a i − ; i : = i − a i : = x ; for j in [0 .. i − in reverseloop a j : = d ; end loop ; end if ;It can be noted that if we provide Algorithms 1 and 2 with nzm -codes, the8esult is again an nzm -code in both cases. In this section, we consider the notion of assignment which we define in Sub-section 4.1 where we consider a tool to compare the assignments. In Sub-section 4.2, we focus our attention on a particular assignment, which we callthe penultimate one. In Sub-section 4.3, we characterise a property shared bythe assignments, a property which we shall discover with the penultimate one.In Sub-section 4.4, we look at another assignment, the mid-assignment whichallows us to have a new look on the particular assignments investigated in theprevious sections. In Sub-section 4.5, we shall investigate the properties of theassignments in the frame of the nzm -codes.
In Sub-section 2.1, we recalled the definition of the natural numbering of W and B . Consider those trees. We can see each of them as an infinite sequence L n of finite sequences of numbers defined by L n + = { U n + .. U n + } , where U n = M n for all n or U n = B n for all n , depending on whether we consider W or B . Inboth cases, we can see the application of the rules (1) as an application α which,to each node ν of the level n associates three numbers ℓ ν , s ν and b ν such that s ν isthe numeral status of ν under α , ℓ ν is the leftmost node of an interval I ν of L n + with the conditions:for all ν , I ν ∩ I n + = ∅ and X ν ∈L n | I ν | = m n + , (13)and b ν is the position of the black node associated to ν among the nodes of theinterval I ν , the leftmost position being 1. The nodes belonging to I ν are calledthe sons of ν under α , for short they are called sons only when it is clear whichassignment is considered. For short they are also called α -sons . The conditions(13) can equivalently be stated as:for all ν with ν ∈ L n , α ( ν ) = ( α ℓ ( ν ) , α b ( ν ) , α s ( ν )),with α ℓ ( ν ) ∈ L n + , α s ( ν ) ∈ { , } , α b ( ν ) ∈ { .. p − − s ν } ,for all ν , α ℓ ( ν + − X k = α ℓ ( ν ) α s ( ν ) = ν ∈ [ M n − + .. M n ],for any positive ν , α s ( α b ( ν )) = α ℓ ( ν + = α ℓ ( ν ) + p − − α s ( ν ), and α ℓ ( M n ) = M n + − p + + α s ( M n ). (14)We call assignment an application α which satisfies (14). We denote by W α the white metallic tree W dotted with the assignment α : it means that, startingfrom the root, the status of each node ν under α is defined by α s ( ν ) and that theposition of the black son of ν among its α -sons is defined by α b ( ν ). When α isassociated with the rules (1), we additionally have that α b ( ν ) = ν .9hat assignation is called the leftmost assignment . It was called the standardassignment in [2] which considers white metallic trees only in the case when p =
5. In [7], another assignment was considered, defined by: α ( ν ) = ( ℓ ν , p − − sn ( ν ) , sn ( ν )) for all ν . (15)It is not di ffi cult to see that whether ℓ s ( ν ) is b or w , the black son is thepenultimate son of ν . For this reason, we call (15) the penultimate assignment .Similarly, we define the rightmost assignment by α ( ν ) = ( ℓ ν , p − − sn ( ν ) , sn ( ν )) for all ν . (16)Say that an assignment α is an a-assignment if and only if for any node ν ,one of its sons exactly has a as its signature. It means that for one son of ν and forone of them only, its code has a among its su ffi xes. We say that an assignment α has the preferred son property if, for any node ν , exactly one of its sons has thecode [ ν ]0 . Note that an assignment which possesses the preferred son propertyis also a -assignment. Accordingly, the preferred son property assumes thatwe consider the representation of the numbers by their metallic codes.Say that an assignment α is a b-a-assignment if and only if all black nodesof the tree and only them have a as their signature. We can note that the notionof b - a -assignment is meaningful also in the case of the representation of thenumbers by their nzm -code when it exists. We shall see a bit later that thereare many values of a for which the b - a -assignment exists.In order to establish the property characterised in Sub-section 4.3, we con-sider the following tool which measures the distance between two assign-ments as follows. Let α and β be two assignments of the white metallictree. Call apartness between α and β denoted by δ αβ the function definedby δ αβ ( ν ) = β ℓ ( ν ) − α ℓ ( ν ) for any node ν of W . We have the easy property:L emma Let α , β and γ be three assignments on the white metallic tree. For anynode ν of W we have: δ αβ ( ν ) = δ γβ ( ν ) − δ γα ( ν ) . (17)Consider the metallic codes which are associated to the numbers by (6),see Sub-section 3.2. Consider the metallic codes of the nodes which lie on L .One of them only has the signature : it is the node numbered by p − . Consider the nodes on L . Their numbers grow from M + M and the metallic codes go from to . The nodes whose signatureis are: , , ..., c0 , d0 , , . We can see that the distance between twoconsecutive such nodes is p − d0 with whose distance is p − any node of W whose signature is . On L n + , the -nodes run from n − up to n + . Note that if we erase the last digit of thosecodes, we get the codes from n − up to n + , i.e. the metallic codes of the nodeson L n . Accordingly, the number of -nodes on L n + is the number of nodeson L n . Moreover, we can observe that the distance between two consecutive -nodes on a level is p − p −
3. When it is p −
3? On the level L , the distance p − d0 and . Indeed, ⊖ = dc and the distance between d0 and dc is p −
4. More generally, we can state:10 emma On the level L n let µ and ν be two consecutive -nodes, with µ < ν . Then ν − µ = p − if and only if [ ν ] = [ ω ]0 k with k ≥ . When it is not the case, ν − µ = p − . Proof. The decrementation algorithm tells us that [[ ω ]0 k ⊖ = [[ ω ] ⊖ k − andthe distance between that latter metallic code and [[ ω ] ⊖ k − , the metalliccode of the previous -node is p − L n between [ ω ]0 k andthe previous -node is p −
3. If [ ν ] = [ ω ]0 , assume that [ ω ] = [ ω ] dc k with k > ω dc k ] ⊖ = [ ω ] dc k − c - d , where c - = c - . That latter metallic code doesnot contain the forbidden pattern. And so, the distance from that node to theprevious -node is p −
3, so that ν − µ = p −
2. Now, if [ ω ] does not contain thesu ffi x dc + , the last digit of ω which is greater than can be reduced by so thatin that case too ν − µ = p − (cid:3) Is there a connection between this two values between two consecutive -nodes and the smaller occurrence of the smaller distance with the distinctionbetween white and black nodes which have p − p − Say that an assignment α possesses the preferred son property if and only if forany node ν of W , the signature of one of its sons under α and one of them onlyis and if the metallic code of that son is [ ν ]0 . When an assignment α possessesthe preferred son property, for any node ν , the node whose signature is [ ν ] iscalled its preferred son under α . Note that if α and β are two assignments whichpossess the preferred son property, for each node, the preferred son under α and that under β coincide.We can state:T heorem The penultimate assignment possesses the preferred son property and it isthe b - -assignment. Proof. It is based on the following property on the signatures and on the metalliccodes of the sons with respect to those of the node.L emma Let ν be a node of W equipped with the penultimate assignment π . Thesignatures of its sons under π is defined by the following rules: b0 → w2 ( wa ) p − wc . b0 . w1 , wa → w2 ( wa ) p − wd . b0 . w1 , (18) The metallic codes of the sons of ν under π are given by the following table, where a k .. a a ⇋ [ ν ] and b k .. b ⇋ [[ a k .. a ] ⊖ ] and a is in { .. d } in lines to . ν range son metallic code ref. b0 .. p − h b k .. b h + p − a k .. a p − a k .. a wa .. p − h b k .. b h + p − a k .. a p − a k .. a (19)11learly, Theorem 5 follows from Lemma 5.Proof of Lemma 5. We perform it by complete induction on ν . The lemma isclearly true for the root. It is applied the rule w1 of (18) and its sons satisfiesthe lines , and of (19) as here b = . Assume that the lemma is provedup to ν which is on the level n +
1. Let λ k + be the leftmost node of L k + ,so that λ k + = M k +
1. Clearly, [ λ k + ] = k , so that recursively applying theincrementation algorithm to the code n + , which is [ λ n + ] , the leftmost son of λ n + , the sons of λ n + have the metallic codes given by the lines , and of (19).Consider that ν is a white node. If its signature is a with a < c , then theinduction hypothesis entails that the metallic code of the leftmost son of ν + ν ] . Accordingly, the incrementation algorithm ensures that the metallic codesof the sons of ν + , and of (19). If the signature of ν is c andif that of ν + d , we can apply the same argument as when a < c . Assumethat the signature of ν + . We know from Lemma 4 that [ ν ] = [ ω ] dc k + ,so that its rightmost son ρ satisfies [ ρ ] = [ ν ] . Accordingly, the leftmost son λ of ν + ρ + ν ]. Applying consecutively p − λ + p − = [ ν ] c = [ ω ] dc k + . Accordingly,the metallic code of the next two sons of ν + ν + and [ ν + . Now, dueto the signature of ν , ν and ν + L n sothat the rules of (18) apply which means that ν + ν +
1, we get that we have p − ν + , and of (19).We remain with the case when ν is a black node. We have that [ ν ] = [ ω ] , sothat the signature of ν + as required. From the induction hypothesis, themetallic code of the rightmost son of ν is [ ω ] , so that the metallic code of theleftmost son of ν + ω ] . Iterating p − p − ν + ω ] = [ ν + . (cid:3) .We can now establish another property. Consider the metallic tree W anda tree T . Assume that both trees are isomorphic. If we equip W with theassignment α , we can endow T with a way of defining the branchings of thetree in accordance with the rules entailed by α on W by putting the samebranchings on T by the isomorphism. Let α be an assignment which possessesthe preferred son property. Consider the set T of the -nodes of W . We definea tree structure on T as follows. Take as root the node whose metallic codeis . Assuming that we defined the level n of T , we define the level n + ν be a node on the level n of T . Let µ be the father of ν in W . Wesay that the status of ν is that of µ . Moreover, each son of µ has a preferredson π on the level n +
1: it is a -node which, by definition belongs to T . Wedefine those nodes π as the sons of ν . Note that the nodes π and ν belongto the subtree of W rooted at µ . It is not di ffi cult to see that in this way weconstruct an isomorphism from W onto T and that isomorphism transports α as an assignment on T which coincide with the definition of the status of thenodes of T which we above indicated. We proved:12 heorem Consider the application ϕ such that ϕ ([ ν ] ) = ([ ν ]) , defining a bijectionof W on the set T of the -nodes of W . Define the sons of a node ν of T as the -nodesof W which are in the subtree S of W rooted at ([ ν ] ) and which are on the level of S . Define the status of ν in T as the status of ([ ν ] ) . Then, the bijection ϕ definesan isomorphism of W onto T which transports the assignment α onto T whatever α . In this subsection, we shall see that all assignments on the white metallic treepossess the preferred son property.To that purpose, we compare the assignments to the same one: the leftmostassignment which we denote by λ . The reason of this choice lies in the followingproperty:L emma For any node ν of W , we have: ≤ δ λα ( ν ) ≤ st α ( ν ) is the status of the node ν under α . It will be easier toprove the lemma, once we have proved the following one.L emma Let λ be the leftmost assignment on W and let α be an assignmenton W . If st λ ( ν ) = st α ( ν ) then, δ λα ( ν + = δ λα ( ν ) . Otherwise, if st λ ( ν ) = w then δ λα ( ν + = δ λα ( ν ) − and if st λ ( ν ) = b , then δ λα ( ν + = δ λα ( ν ) + . Proof of Lemma 7. We set ℓ s α ( ν ) = st α ( ν ) = b and ℓ s α ( ν ) = st α ( ν ) = w .Then α ℓ ( ν + = α ℓ ( ν ) + p − − ℓ s α ( ν ). Considering the four cases raised by thedi ff erent distributions between λ and α for defining the leftmost son of ν andof ν + (cid:3) Proof of Lemma 6. The lemma is of course true for the root. However, we shallprove a stronger property. For any node ν of W , we have: δ λα ( α ℓ )( ν ) = δ λα ( α ℓ )( ν + = ν , not about ν itself. We prove the property by induction on ν . First, we have to prove theproperty for the root, which means that we have to compute δ λα for ν = upto ν = . Consider the node which is λ -black. Its λ -sons are , , up to included: we check that there are p − is λ -black. If is α -black too δ λα ( ) =
0, from Lemma 7. Now, as a node has a single black node, whateverthe assignment we can see that on W , the other sons of the root up to arewhite nodes under both λ and σ so that δ λα ( ν ) = ν up to included.If is an α -white node, δ λα ( ) = δ λα ( ν ) = ν is α -white from until we meet the α -black son of the root, say k . Accordingly, δ λα ( k ) = k is λ -white, we get that δ λα ( k + = δ λα ( ν ) = ν until if any. Note that if k = , we already know that δ λα ( ) = ν by an α -black node assuming that δ λα ( λ ℓ ( ν )) = ν is α -white. From the induction hypothesisapplied to ν −
1, we know that δ λα ( λ ℓ ( ν )) =
0: the leftmost son of ν is the sameunder whatever λ or α . If ν is α -white too, the above argument can be repeated:if κ is the α -black son of ν , we necessarily have that κ < λ ℓ ( ν + κ = λ ℓ ( ν ),there is nothing to prove and δ λα ( σ ) = ν under both λ and α .If κ , λ ℓ ( ν ), as κ < λ ℓ ( ν + δ λα ( σ ) takes thevalue 1 on σ = λ ℓ ( ν ) +
1, keeping that value until σ = κ and then, takes back thevalue 0 from κ + ν .If ν is a λ -white node and an α -black one, as κ ≤ α ℓ ( ν ) < λ ℓ ( ν ), the same ar-gument holds proving that δ λα ( α ℓ ( ν ) + =
0. Accordingly, the proof of Lemma 7is completed. (cid:3) T heorem Any assignment α on W do possess the preferred son property. In W α ,whatever α , we have that m n + is the preferred son of m n . Call the sequence of nodes { m n } n ∈ N the . Proof. First, we prove that λ , the leftmost assignment, possesses the preferredson property, by proving an analog of the rules (18) and Table (19) for λ . Wehave:L emma Let ν be a node of W equipped with the leftmost assignment λ . Thesignatures of its sons under λ is defined by the following rules: b1 , b2 → b2 ( wa ) p − wd . w0 , wa → b1 ( wa ) p − wd . w0 , w0 → b1 ( wa ) p − wc . w0 . w1 , w1 → b2 ( wa ) p − wd . w0 . w1 . (22) The metallic codes of the λ -sons of ν are given by the following table, where a k .. a a ⇋ [ ν ] and b k .. b ⇋ [[ a k .. a ] ⊖ ] , and in lines and , a is in { .. d } . ν range son metallic code ref. b1 , b2 .. p − h b k .. b h + p − a k .. a w0 .. p − h b k .. b h p − a k .. a p − a k .. a w1 .. p − h b k .. b h + p − a k .. a p − a k .. a wa .. p − h b k .. b h p − a k .. a (23)Proof of the lemma. We know from Theorem 5 that the -nodes coincide withthe black nodes under π , the penultimate assignment. Clearly, under λ , the rootobeys the rule w1 of (22) and the metallic codes of its λ -sons satisfy the lines , and of (23). Consider the nodes on the level n of W under λ and comparethe intervals assigned by λ with those assigned by π . The lefmost node of thelevel is a λ -black node while it is an α -white one. Accordingly, the first -nodeof the level n + λ -son of the first node on the level n . From our14revious studies about the positions of the -nodes on a level, we have that thenext -nodes on the level n + λ -rightmost sons of the first nodes ν of thelevel n whose signatures run from until d . The signature of the next node is so that the rule b2 is observed by the λ -sons of ν . The incrementation algorithmapplied to the λ -rightmost son of the leftmost node on the level n so that therules wa apply to ν +
1, and the lines and of (23) are observed by the metalliccodes of the λ -sons of ν +
1. Clearly, what the argument can be repeated whengoing from a λ -white node whose signature is a with < a < d , to the new nodewhose signature is a ⊕ . Applying the induction hypothesis on a node whosesignature is d and iteratively applying the incrementation algorithm, we obtainthat λ -white nodes whose signature is and are applied the rules w0 and w1 respectively and that the metallic codes of their λ -sons are those indicated bythe lines up to included of Table (23). The proof of Lemma 8 is completed. (cid:3) Note that the rules can be identified by their left-hand side part, which weshall do later on. We also have:L emma Let ν be a node of W and let α be an assignment among the leftmost, thepenultimate and the rightmost ones. Let σ be the leftmost son of ν under α . If [ ω ] a is thecode of ν , then we have that [ σ ] = [ ω ] a − u , where ω = ω if a > and [ ω ] = [ ω ] ⊖ otherwise and u ∈ { , , } . More exactly, u = when α is the rightmost assignmentand ν is black or is a w1 -node; u = when ν is the root; when α is the penultimateassignment; when α is the leftmost assignment and then ν is black or is a w1 -node;when α is the rightmost assignment and then ν is a w0 -node or a wa -node with a , . Proof of the lemma. Lemma 9 is a reformulation of Lemmas 5 and 8 for whatare the penultimate and the leftmost assignments respectively. We consideralso the rightmost assignment. For that assignment, we have the followingrules and metallic codes for the sons of a node, with the same notations as inLemma 8: b1 , b2 → w3 ( wa ) p − wd . w0 . b1 , wa → w2 ( wa ) p − wd . w0 . b1 , w0 → w2 ( wa ) p − wc . w0 . w1 . b2 , w1 → w3 ( wa ) p − wd . w0 . w1b2 . (24) ν range son metallic code ref. b1 , b2 .. p − h b k .. b h ++ p − a k .. a p − a k .. a w0 .. p − h b k .. b h + p − a k .. a p − a k .. a p − a k .. a w1 .. p − h b k .. b h ++ p − a k .. a p − a k .. a p − a k .. a wa .. p − h b k .. b h + p − a k .. a p − a k .. a (25)15ote that a is in { .. d } in lines to .Let ρ denote the rightmost assignment. From the definition, it seems closeto the penultimate one: the ρ -black son is always the last one among the ρ -sons of a node. This explains that the rule wa of (18) is transformed tothat of (24) by exchanging the status of the nodes with signatures and accordingly. The change in the other rules is a bit more complex. However,the proof of (24) and (25) is very similar to those of Lemmas 5 and 8: it isbased on the consideration of the penultimate assignment, the -nodes and theiterated applications of the incrementation algorithm. This completes the proofof Lemma 9 which synthesizes the Tables (19), (23) and (25). (cid:3) Proof of Theorem 7. Consider an assignment α on W . Let ν be a node of W .If δ λα ( ν ) =
0, then, as the λ and α -leftmost and rightmost sons of ν coincide,and as ν has a preferred son under λ , its -son is also its α -preferred son. If δ λα ( ν ) =
1, clearly ϕ ≤ ρ λ ≤ ρ λ where ρ α , ρ λ denote the α -, λ -rightmost son of ν respectively and ϕ denotes the λ -preferred son of ν .We remain with the proof that m n + is the preferred son of m n . The proofproceeds by induction and on the following remark. By definition of M n , wehave that m n < M n . Note that m > M = m . Assume that M n < m n + . Then, M n + = M n + m n + < m n + < ( p − m n + < m n + . So that, by induction, we getthat M n < m n + < M n + . We can write that m n + = M n + − M n = M n + − n X k = m k .This means that if we assume that m n + is the penultimate λ -son of m n , we getthat m n + = M n + − M n + = M n + − n + X k = m k . Now, if we interpret the sum n X k = m k as the trace on the level of m n + of white trees T h of heights h = h = n , thesum n + X k = m k can be interpreted as the trace on the next level of the same trees,so the height of T h is now increased by 1, plus one node which is the rightmost λ -son of m n + . Together with the application of the rule w0 of (22) together withthe metallic code of the penultimate son of a -node, what we just remarkedproves that m n is the preferred son of m n + .Accordingly, Theorem 7 is proved. (cid:3) Figures 1 and 2 illustrate the property proved in Theorem 7 for the leftmostand the rightmost assignments respectively. In the figures, the red colour isused to mark the black nodes, while the white ones have a blue and a greencolour. The blue and the green colours are used to distinguish between thedi ff erent kinds of white nodes which appear in Tables (19), (23) and (25). Theblue nodes correspond to the nodes marked by wa . The w0 -nodes are indicatedby a green disk with a red circle while the w1 -nodes are indicated by a greendisk with a darker green circle. The numbers in red, above the nodes, indicatethe natural numbering of the tree. The metallic code is mentioned vertically,below each node. In these illustrations, p =
9. However, in order to indicate thegeneral form of the properties, in the metallic codes, and are replaced by c d respectively. Indeed, it corresponds for p = c and to d respectively. In order to make easier the reading of the figures,not all nodes are mentioned. We just mention those which allow us to see theapplication of the rules (22) and to check Table (23). d c1 c2 cd d0 d1 d2 dc dc1 dc2 dcc Figure 1
The white metallic tree. Partial representation of the first three levels ofthe tree when p = with the conventions mentioned in the text. d c2 cd d0 d1 d2 dc dc2 dcc Figure 2
The white metallic tree under the rightmost assignment. Partial rep-resentation of the first three levels of the tree when p = with the conventionsmentioned in the text. We defined the -branch which connects the -nodes we obtain which arethe -son of the previous one except the first one which is the root. We noted thatthe -branch does not depend on the assignment α with which we equipped W .Now, if we take a node ν whose signature is not . It has a unique α -son σ whichis a -node, and we know that the position of σ in W does not depend on α .What depends on α is the position of σ among the α -sons of ν . As an example, m n + is the penultimate λ -son of m n while it is its ante-penultimate ρ -son. Fromwhat we just mentioned, we can construct a sequence { ϕ n } n ∈ N of nodes suchthat ϕ = ν and ϕ n + is the -son of ϕ n for any n . From Theorem 7, we knowthat ϕ n + is always an α -son of ϕ n and again, its position does not depend on α . Call the sequence { ϕ n } n ∈ N the ν . From Lemma 4,we can17tate:T heorem For any assignment α , the -paths indicate the nodes in W α at whichthe application of the incrementation algorithm necessitates a carry, which produce the -signature of the metallic code of the node. Before turning to the connections between the assignments on W and the nzm -codes, we deal with a particular fixed assignment which, in some sense,synthesizes the properties we observed on the leftmost, the penultimate and therightmost assignments. We say that an assignment is fixed if the black nodesare always applied the same rule and if it is the same for the white nodes. In thissub section,we consider what we call a mid-assignment . A mid-assignmentis defined by a constant k with k ∈ { .. p − } which defines the position of theblack son among the sons of a node, avoiding the positions we already studied.Denote by W µ, k the white metallic tree equipped with such an assignment. Itis illustrated by Figure 3 in the case when p = k = k c x c1 d1 dk dc dc1 dcc Figure 3
The white metallic tree. Partial representation of the first three levels ofthe tree when p = with the conventions mentioned in the text. The rules for the nodes are given by (26) and the sons of a node by (27).We note that the root obeys the rule wa of (26). From the metallic code of theleftmost son and from Lemma 4, we can see that the rule wa is applied until wemeet the sons of the first black node on the father level. We also can check thatlines to of Table (27) are observed. bk → w2 .. wk - .. bk . wk + .. wd . w0 , wa → w2 .. wk - bk . wk + .. wd . w0 . w1 , with < a < k , wb → w1 .. wk - bk . wk + .. wd . w0 , with k < b ≤ d , w0 → w1 .. wk - bk . wk + .. wc . w0 . w1 . (26)Then, it is easy to see that the rule bk is applied and that the corresponding18ines and of (27) are observed too by the metallic codes of the sons of the node.As a black node has p − p − , so that the signature of the leftmost son of the following whitenode on the father level is and not as required by the rule wa . Accordinglythe rule wb is applied until the -node is met on the father level and we can seethat the metallic sons of the corresponding nodes of the father level obey lines and of (27). Then, we apply the rule w0 : the signature of the leftmost nodeis as the signature of the rightmost node on the father level was . Lines is applied to get the metallic codes of the sons. Now, as the father node ν is a -node, the signature of the previous node on the level is either d or c but, inthat latter case, the metallic code of ν − ffi x dc ∗ . It is the reason whythe metallic code of the rightmost son of ν is [ ν ] . ν range son nzm -code ref. bk .. p − j a h .. a a - j + p − a k .. a a wa .. p − j a h .. a a - j + p − a k .. a a p − a k .. a a wb .. p − j a h .. a a - j p − a k .. a a w0 .. p − j a h .. a a - j p − a k .. a a p − a k .. a a (27)It is interesting to note that the rules corresponding to the leftmost, thepenultimate and the rightmost assignments can be derived from the rules (26).As an example, consider the leftmost assignment: the rule b1 , b2 of (22) comesfrom the rule wa of (26) noticing that the leftmost signature w2 becomes b2 because of the leftmost position of the black nodes in W λ . Due to the leftmostposition of the black nodes, the rule wa of (22) comes from the rule wb of (26)up to the change in the positions of the black node. The rule w0 of (22) is thesame of that of (26) up to the change in the black nodes and the rule w1 of (22)comes from the rule wa of (26) up to the change due to the position of the blacknode. W and the nzm-codes As the -signature has no more any meaning in nzm -codes, the property of thepreferred son can be reformulated as follows: is there a value a such that eachnode ν has among its α -sons exactly one of them whose nzm -code is [ ν ]a for atleast one assignment α ?Before addressing that issue, note that we can easily characterise in nzm -terms the nodes whose signature is in the metallic code. As far as the nodesdo not change but their sons according to the assignment α we set on W , let us19till call those nodes -nodes even in that context. We have:L emma Let ν be a -node and let [ ν ] = [ ω ]0 k be its metallic code, where k > andthe signature of [ ω ] is not . Then we have: [ ω ]0 = [ ω − ]x and, when k ≥ , [ ω ] k = [ ω − dc k − d (28)The lemma is an immediate application of (12).The lemma tells us that the su ffi xes x and dc ∗ d cannot be used for replacingthe notion of preferred son in the context of the metallic codes: the nzm -codeof the -son of ν contains [ ω −
1] and not [ ω ].Now, it is not di ffi cult to see that occurs among the sons of the root ,whatever the assignment. Also, the first nodes on the level L are , ..., , and is the p − th node. Accordingly, if is α -white, occurs as its rightmost α -son. Let us call the nodes of W whose signature of their nzm -codeis . We may wonder what is the distribution of the -nodes in W ? In fact,what we already said is a valuable hint to the solution: from Lemma 2 and itsCorollary 2, we know that [ ω − ]x ⊕ = [ ω ]1 and that [ ω − ]xd k + = [ ω ]1 k + . Thisallows us to prove:L emma Let µ and ν be two consecutive -nodes of the level L n with µ < ν . Then ν − µ = p − if and only if is a su ffi x of ν and, when it is not the case, ν − µ = p − . Proof. When is a su ffi x of ν , we can write [ ν ] nz = [ ω ] nz k + , with k anatural integer. Using Algorithm 2, we get that [ ν − ] nz = [ ω − ] nz xd k + , sothat [ µ ] nz = [ ω − ] nz xd k whose distance to ν − p − ν − ν = p − [ ν ] nz = [ ω ] nz , with the signature of [ ω ] nz beinggreater than . Then, [ µ ] = [ ω − ] nz , which can be checked by iterated applica-tions of Algorithm 2. Accordingly, ν − µ = p −
2, which completes the proof ofLemma 11. (cid:3)
For any node ν , call successor of ν , denoted by succ ( ν ), the node whose nzm -code is [ ν ] nz . Lemma 11 and our study of the penultimate assignment on W with respect to the metallic codes suggests to state:T heorem Let W equipped with the rightmost assignment ρ and consider the nzm -codes of its nodes. Then, for any node ν , its successor occurs among its ρ -sons and noother ρ -sons of ν is a -node, so that we can call [ ν ] nz the nzm -preferred son of ν .Moreover, ρ is the unique assignment α such that for any node, its successor occursamong its α -sons. Proof. By induction, we prove that W ρ can also be the defined by application ofthe rules (29) and that the nzm -codes of the sons of ν are defined by Table (30). b1 → w2 . w3 .. wd . b1 , wa → w2 .. wd . wx . b1 , (29)20 range son nzm -code ref. b1 .. p − h a k .. a a - h + p − a k .. a a wa .. p − h a k .. a a - h + p − a k .. a a (30)We can see that the root is applied the rule wa with the exceptional value a = which is used for the root only. We already seen, that the rule w2 appliesto , the leftmost node of L . Then, by induction and applying p − wa apply up to the node d . Now, thenext node on L after d is x . From (30) and Algorithm 2, we can see that theleftmost ρ -node of x has d2 as nzm -code. The p − th son of ν is dd , which is avalid nzm -code, so that the next son is dx and the last one is x1 as required bythe rule wx . Accordingly, the leftmost node of is x2 . Iteratively applyingAlgorithm 1, we get that the p − th node of is xd , so that by Corollary 2, therightmost node is . Consequently, the rule b1 applies to . By the way, wecan check that the ρ -sons of satisfy the lines and of (30).We can repeat this progression on the trace of each sub-tree rooted at a nodeof the level L on the level L starting from the nodes of the level L in orderto transport (29) and (30) on L . We note that when the root is a black node, wehave just one missing white node which changes nothing in the application ofthe rules wa , so that the rule b1 applies as we have seen: the rule b1 applieswhen the distance from an -node to the previous one is p −
3. The occurrenceof the pattern xd ∗ allows us to spare one node, avoiding the signature x , so thatthe following -node is applied the rule b1 .Presently, consider an assignment α . Let ν be the first node of W such that st λ ( ν. ) , st α ( ν. )If ν is the rightmost node ρ n of the level n , it is α white, so thatit contains the rightmost son of ρ − -node. And so, as an α -node, ν contains two -sons. If ν is not the rightmost node of a level, it is necessarily α -black, so that it does not contains the successor of ν which is the rightmost ρ -son of the node which is ρ -white. But this discrepancy induces a shift of theleftmost α -son with respect with the ρ -one: for each node µ after ν on the samelevel, the -son is the successor of ν −
1. Accordingly, the proof of Theorem 9 iscompleted. (cid:3) d x c2 cd cx d1 d2 dd dx x1 x2 xc xd dc2 dcd dcx dd1 dx2 dxd x11 xd2 xdc xdd Figure 4
The white metallic tree and the rightmost assignment under the nzm -codes for the nodes. Partial representation of the first three levels of the tree whenp = with the conventions mentioned in the text. Figure 5 illustrates W λ . The colours are again those of Figures 1 and 2. Asin those latter figures, the blue colour indicates an application of the rules wa . d x c1 c2 cd cx d1 d2 dc dd dx x1 x2 xc xd dc1 dc2 dcc dcd dcx xd1 xd2 xdc xdd Figure 5
The white metallic tree and the leftmost assignment under the nzm -codes for the nodes. Partial representation of the first three levels of the tree whenp = with the conventions mentioned in the text. Figure 4 illustrates W ρ . The convention on the colours are the same as forFigures 1 and 2. We can see on the figure that two colours only occur: blue andred. This corresponds to the fact that (29) mentions two rules only, the rule b1 and the rules wa with a ∈ { .. x } . We can check on the figure that the nzm -codeswhich are displayed in the figure in the same way as for the metallic codes inFigures 1 and 2 satisfy Table (30). The green colour with a red circle indicatesthe nodes which are before the -nodes on the same level and the colour greenwith a green circle indicate the -nodes of the rightmost branch of the tree. Theblack nodes lie on the leftmost branch of the tree and also in the -nodes whichare not on that branch. On Figure 5, we can see that the λ -assignment does notpossess the nzm -preferred son property. For all nodes ν except those which22ie on the rightmost branch of the tree, the successor of ν is the leftmost λ -sonof ν + W ρ . We define the as the sequence of nodes { ω n } n ∈ N ,where ω is the root and ω n + is the successor of ω n . The -branch is theanalog in the nzm -codes context of the -branch in the context of the metalliccodes. Similarly, we define the ν in W ρ . Note thedi ff erence with the previous situation: a -path is a path whose terms exceptthe first one are sons of the previous term, whatever the assignment given to W . In the context of the nzm -codes, a -path is a true path in W ρ and it is nota path in any other W α as established in the proof of Theorem 9. We can state:T heorem Let V be the set of -nodes of W ρ , equipped with the rightmost assign-ment ρ , being excepted. Define the mapping ϕ from V onto W ρ by ϕ (([ ν ] nz )) ⇋ ([ ν ] nz ) .Define the sons of ( [ ν ] nz ) as the -sons of the ρ -sons of ( [ ν ] nz ). Then ϕ definesan isomorphism between V equipped with its natural numbering and W ρ and ϕ − transports the ρ -assignment onto V . As defined in Subsection 2.2, the black metallic tree B is defined by the samerules as the white one, the di ff erence being that the root of B is a black node.We know that the number of nodes on the level n of B is b n which satisfies (3).We also know that B n = m n . Accordingly, the nodes of the rightmost branch of B are numbered by m n , as known from Theorem 3, and we get from (19) thattheir nzm -code is dc n − d .In [6], we proved the properties of the sons signatures of the nodes in theblack metallic tree under the leftmost assignment and when the nodes arefitted with their metallic code. The properties are di ff erent from those we havenoted in the white one in the similar context. In Sub section 5.1 we considerthe properties of B when its nodes are fitted with the metallic codes. Subsubsection 5.1.1 studies the case of B λ when B is constructed under the λ -assignment. Figure 6 illustrates the black metallic tree in that context for p = W and B . We recall the results in Sub-subsection 5.1.1. A moredetailed study of that comparison can be found in [6]. In the present section,we shall stress on the the comparison with the situation of the black metallictree under the rightmost assignment and also, in both the leftmost and therightmost assignments when the nodes are equipped with their nzm -codes.We shall write B λ , B ρ for B equipped with the λ -, ρ -assignments respectively.The study of B λ with the nzm -codes is dealt with in Sub-subsection 5.2.1, whilethe similar study with B ρ is the goal of Sub-subsection 5.2.2.23 .1 The black metallic tree and the metallic codes We now turn to the black metallic tree B and we look at properties, similar tothose which hold for the white metallic tree, which are still valid in that tree andwe try to see the reason why for those which are not valid. Sub subsection 5.1.1looks at the situation for B λ , the tree B when it is fitted with the leftmostassignment λ . Sub subsection 5.1.2 deals with the situation for B ρ , the tree B when it is fitted with the rightmost assignment ρ . We recall the reader that inthis subsection, we consider the metallic codes for the representations of thenumbers attached to the nodes. Figure 6 shows us that the preferred is not true in B λ . The leftmost son of alevel, a black node, has no son whose signature is . All other nodes have a sonwhose signature is , and among them, the last node of a level has two sonswhose signature is , so that the leftmost assignment is not even a -assignmentfor the leftmost one. Now, for a node ν which has a unique son whose signatureis , the metallic code of that node is not [ ν ]0 but it is [ ν − . d c0 c1 cc cd d0 d1 dc cc0 ccd cd0 cdc d00 dc0 dcc Figure 6
The black metallic tree with the leftmost assignment and with the metalliccode of the nodes. The same convention about colours of the nodes and of the edgesbetween nodes as in Figure is used. We can see that the preferred son property isnot true in the present setting. Here too, call successor of the node ν , the node whose metallic code is [ ν ]0 .We can state:T heorem see [6] . In B λ , the nodes are applied the rules of (31) and the metalliccodes of the λ -sons of a node ν are given by Table (32) , the root being excepted. Theroot is applied the rule b1 → b2w3 .. wdw0 . For the other nodes there are two kindsof black nodes, the nodes b1 and the nodes b0 which follow the rules for black nodesin (31) . The rule b1 is also followed by the node whose signature is . The nodes b1 are present on the leftmost branch of B , the node being excepted, and only on those laces. The other black nodes, the leftmost one of the λ -sons of a node are b0 nodes.There are two types of white nodes, w0 , and wa with a > . The metallic codes of thesons of a node are given in Table (32) in terms of b k .. b ⇋ [ ν ] − . The w0 -nodes areexactly the nodes of the rightmost branch of the tree, the root being excepted.The nodes of the rightmost branch of the tree being excepted, the successor of ν isthe leftmost λ -son of ν . The tree B λ does not observe the preferred son property. Thenodes on the extremal branches of the tree being excepted but the root being included,any other node has a -node among its λ -sons which is not its successor. The root isthe single node of the tree which has a preferred son. b0 → b0 , w1 , .., wc b1 → b1 , w2 , .., wdw0 → b0 , w1 , .., wc , w0 wa → b0 , w1 , .., wd . (31) For any node ν which is not the rightmost one on a level, the successor of ν isthe leftmost node of ν + . For the rightmost node on the level n, its successor is therightmost node on the level n + . We also have that the type b1 occurs for the leftmostnode of a level only and that the type w0 occurs for the rightmost node of a level only.Table (32) gives the metallic code of a node ν in terms of [ ν ] . ν range son metallic code ref. b0 .. p − h [ b k .. b ] h - b1 .. p − h k − h wa .. p − h [( b k .. b ) − h - w0 .. p − h dc k − h - p − k + (32)Proof. As usual, we proceed by induction on the level n and, on each level, byinduction on ν from the leftmost node of the level to its rightmost one. Theleftmost node λ n on the level n is m n − + when n = n − when n >
1. Accordingly, as the leftmost son of λ n is λ n + whosemetallic code is n , so that we easily obtain the line of Table (32). Accordingly,that line is proved which also proves the rule b1 . Starting from λ n +
1, we havethat the distance between two consecutive black nodes on a level which havethe same grand-father ϕ is p −
2. This shows us that the rules of (31) apply to the λ -sons of ϕ . We can note that the property is true whether ϕ is black or white.The distance between the rightmost node of a level, which is a -node and thefirst -node of the next level is p −
2. Now, as far as the signature of the leftmostson of a level is consequently and as far as that node is black under λ , we getthat the first -node of a level is the second black node on the level. From thatsituation, we have that the case when the distance between two -nodes is p − w0 -nodes which have two -nodes among their λ -sons. Thefact that the rightmost son of the level n is m n , as proved in Theorem 3, explainsthe line of Table (32). Presently, consider the rightmost son ρ n of the level n .We know that its number is m n so that its metallic code is n . Applying theAlgorithm for decreminting a metallic code, see [6], we get that [ m n − = dc n − .Applying again that algorithm which here consists in decrementing the lastdigit only, we get the lines and of Table (32). Consequently, Theorem 11 is25roved. (cid:3) Note that, par abus de langage , we can also say for the black metallic treeequipped with the leftmost assignment that m k + is the preferred son of m k . Figure 7 illustrates B ρ . The figure can be compared with Figure 2. The coloursindicates that the rules in the case of B ρ seems to be simpler than the rulesfor W ρ , see (24). If we compare Figure 7 with Figure 6, we can see that thepreferred son property which is not observed in B λ as stated in Theorem 11seems to be satisfied in B ρ .Indeed, we have a stronger property tightly connected with Lemma 4 andit reminds what we noted in the case of the nzm -codes: d c1 cc cd d0 d1 dc cc1 cd0 cd1 d00 dc1 dcc Figure 7
The black metallic tree with the rightmost assignment and with themetallic code of the nodes. The same convention about colours of the nodes and ofthe edges between nodes as in Figure is used. We can see that the preferred sonproperty is true in the present setting. T heorem Let B ρ be the black metallic tree equipped with the rightmost assignment.Consider the metallic representations of its nodes. The rules which may be used forconstructing the tree are given by (33) and the metallic codes of the ρ -sons of a node ν are given by Table (34) in terms of [ ν ] and of [ ν ] − . Equipped with the rightmostassignment, B possesses the preferred son property. But the tree does not possess thatproperty if it is fitted with another assignment. Proof. Our first remark is that Lemma 4 is also true for B . The reason isthat the property given on the lemma follows from the order on the numbersthemselves and on the properties of the incrementation and not on the fact thatwe may use the numbers to identify the nodes of an infinite finitely generatedtree.From that remark, we note that the first -node on level 2 in B is the p − th ofthe level so that it is the rightmost ρ -son of the node 2 of B which is the leftmost26ne on level 1. Accordingly, all the ρ -white nodes on level 1 have the successorof their metallic code as the metallic code of their rightmost ρ -son. Accordingly,the argument performed for the analysis of the rightmost assignment on W canbe repeated for those white nodes. Now, the rightmost ρ -son of the penultimatenode d on level 1 is d0 , so that the leftmost ρ -son of the penultimate node onlevel 1 is d1 and the penultimate node on level 2 is then dc , so that the last nodeis . Lemma 4 tells that from d1 up to there are p − d0 to be a ρ -black node. Next, the argument goeson as in the proof of the rules (24). We can see that the rules are exactly thoseof (33). This also proves the codes given in Table (34) which also proves thatthe preferred son property is true in B ρ with respect to the metallic code. Notethat in (34), a k .. a ⇋ [ ν ] and that b k .. b ⇋ [ ν − b0 → w1 .. wc . b0 , wa → w1 .. wd . b0 , with a > . (33) ν range son metallic code ref. b0 .. p − h [ b k .. b ] h p − a k .. a wa .. p − h [ b k .. b ] h p − a k .. a (34)Now, we can repeat the argument of Theorem 9 as the - ρ -son is the right-most son whatever the node. Indeed, considering another assignment α , takethe first node ν whose status is not the same under α and under ρ . If ν is a ρ black node, as its α -leftmost son is the rightmost ρ -son of ν −
1, it contains two -nodes. If ν is a ρ white node, as it is the first node where the statuses aredi ff erent, the rightmost α -son of ν is a node whose signature is d , so that no α -son of ν is a -node.The proof of Theorem 12 is now completed. (cid:3) We may repeat the proof of Theorem 10 and prove the following result:T heorem Let V be the set of -nodes of B ρ , equipped with the rightmost assign-ment ρ . Define the mapping ϕ from V onto B ρ by ϕ (([ ν ] )) ⇋ ([ ν ]) . Define the sonsof ( [ ν ] ) as the -sons of the ρ -sons of ( [ ν ] ). Then ϕ defines an isomorphism between V equipped with its natural numbering and B ρ and ϕ − transports the ρ -assignmentonto V . The proof combines the argument of the proof of Theorem 6 and that of Theo-rem 10. It is left to the reader as an exercise.
We now turn to the study of B when the numbers of its nodes are written as nzm -codes. In Sub subsection 5.2.1 we investigate the properties for B λ whileSub subsection 5.2.2 is devoted to those of B ρ .27 .2.1 The black metallic tree under the leftmost assignment and the nzm-codes Figure 8 illustrates B λ when the nodes are fitted with their nzm -codes. At firstglance, whatever the digit a , no λ -son of a node ν has the nzm -code [ ν ] nz a .Accordingly, the preferred son cannot be defined for B λ , a situation whichreminds us that of the same tree when we consider the metallic codes of thenodes. d x c1 cc cd cx d1 d2 dc dd dx x1 xc xd c4x ccd ccx cdc cdd cxd d1d d4x dcc dcd Figure 8
The black metallic tree still with the leftmost assignment but with the nzm -codes of the nodes. This time it seems that we have five types of rules forthe nodes in order to define the sons signature. We can see that the preferred sonproperty is not true in the present setting.
In this situation we can state:T heorem In B λ , the black metallic tree dotted with the leftmost assignment, therules giving the status and the nzm -signatures of the sons of a node are given in (35) ,the root being excepted.and the nzm -codes of the λ -sons of a node ν are given byTable (36) . The tree B λ under the leftmost assignment has no preferred son property interm of the nzm -codes of its nodes. Proof. Indeed, it is immediate to see that the root of B ρ is applied the rule r1 of (35) and that its leftmost son, , is applied the rule b2 of (35). The nzm -codesof the respective sons are given in Table (34), on lines and for r1 , online for b2 . Lemma 11 and Algorithm 1 show us that the rule wa applies to thewhite sons of a node, provided that the rule b holds for the black nodes. Therightmost node on level 1 is x and it is dc k − d on the level k , starting from k = and the leftmost node ofthe further levels is dc k − x which explains the rules b1 and bx . Those latterrules are contained in the rule b of (35) which stands for any black node, and being applied the rules r1 and b2 of (35) as already noticed. r1 → b2 , w3 , .., wx , b2 → b1 , w2 , .., wd , b → bx , w1 , .., wc . w1 → bd , wx , .., wc , w2 → bd , w1 , .., wd , wa → bx , w1 , .., wd . with a > . (35)28he particular forms of the nzm -codes of the leftmost nodes of a leveltogether with Algorithm 1 explain lines and of Table (36). The nzm -codesof the rightmost node of a level and Algorithm 2 explain lines and of thetable.In Table (36), we need to consider b k .. b ⇋ [ ν −
1] together with f k .. f ⇋ [ ν − ν range son metallic code ref. r1 .. p − h h + p − x b2 .. p − h b f k .. f ] x .. p − h [ b k .. b ] h - w1 f k .. f ] d f k .. f ] x .. p − h [ b k .. b ] h -- w2 f k .. f ] d .. p − h [ b k .. b ] h - wa f k .. f ] x .. p − h [ b k .. b ] h - (36)Let us see that the table is relevant for the other nodes. Starting from theleftmost node, we can see that Algorithm 1 and Lemma11 show us that for awhite sons of a node ν , the first one has the signature x and the nzm -code isbased on [ ν − λ -son of the second node ona level. It explains the lines and as long as a black son is not met. Theblack son interrupts the sequence x1 ... d of the nzm -signatures on c . So that for w1 , the signature of its leftmost λ -son is d . Accordingly, the sequence of nzm -signatures becomes d1 .. d , so that the sequence of nzm -signatures defined forthe rule wa may again apply. This corresponds with the occurrence of a pattern xd ∗ which is a reason why after d we have in the signatures, but it comesfrom the nzm -signature of ν which is then : but necessarily, the nzm -signatureof nu − d . This also explains the nzm -codes given in Table (36). The lackof preferred son property is also a consequence of the table. Accordingly, theproof of Theorem 14 is completed. (cid:3) Figure 9 illustrates B ρ . The conventions for the representation are the same asfor Figure 8. At first glance, the structure seems to be more regular than in thecase of the leftmost assignment. However, it also seems to do not observe thepreferred son property, whatever the digit a chosen in ,.., d , x . The rules for thenodes are given in (35) and the nzm -codes for the ρ -sons of a node are givenby Table (36). 29e can see that the rule is applied the rule r1 of (37) and that is applied therule wa of (37). Lemma 11 explains that the nzm -signatures .. dx appearing inthe rule wa is repeated as long as we are in the white ρ -sons of a node. Whena black son occurs, the sequence stops at d as indicated in the rule b of (37).Let ν be the ρ -black node for which the sons nzm -signature is thus .. d , wehave that the signature of the leftmost ρ -son of ν + x . Accordingly, the sons nzm -signature for ν + x1 .. d which appears in the rule w1 of (37). In order tosee why the nzm -signature of the rightmost ρ -son of ν +
1, namely d , is followedby the nzm -signature , we have to look at Table (38). d x c1 cc cd cx d1 d2 dc dd dx x1 xc xd cd1 cdd cdx cxd d11 d1x dc1 dcc dcd Figure 9
The black metallic tree still with the rightmost assignment but with the nzm -codes of the nodes. This time it seems that we have three types of rules forthe nodes in order to define the sons signature. We can see that the preferred sonproperty is not true in the present setting. r1 → w2 , .., wd , bx , b → w1 , .., wc , bd . wa → w1 , .., wd , bx , with < a < x , w1 , wx → wx , w1 , .., wc , bd . (37)The particular forms of the nzm -codes of the leftmost nodes of a leveltogether with Algorithm 1 explain lines and of Table (38). ν range son metallic code ref. r1 .. p − h h + p − x b .. p − h [ b k .. b ] h w1 , wx f k .. f ] x .. p − h [ b k .. b ] h - wa .. p − h [ b k .. b ] h p − b k .. b ] x (38)Let us see that the table is relevant for the other nodes. Starting from theleftmost node on a level, we can see that lines and apply: from our studyof Sub-subsection 5.2.1, the leftmost node on level n + n ∈ N is dc n x .This proves lines and for the leftmost node λ n + of the level n +
2. Now, as30he nzm -signatures of λ n + are x1 .. , the nzm -code of the leftmost ρ -node of λ n + + dc n − dx , so that line and again apply but this time, the nzm -codeof the rightmost ρ -son of λ n + + dc n − xd , so that the leftmost ρ -son of λ n + + dc n − d11 : accordingly, the rule wa apply to λ n + + nzm -codes ofits ρ -sons are those which are indicated by lines and of Table (38). As inthe case of Table (36), the fact that a white node w1 whose rightmost son hasthe signature d is followed by a white node w2 whose leftmost son has thesignature comes from two features. The first one is that, according to ourstudy, the successor of ν occurs in as the leftmost or the second ρ -son of ν + nzm -signature x is that of the leftmost ρ -son of a w1 - or wx -nodeor that of the rightmost ρ -son of a wa -node with < a < x . Let ν be such a node.Accordingly, the su ffi x xd occurs in the nzm -code of ν + ρ -son if ν is a wa -node, so that the -node is the next son of ν +
1. If ν is a w1 -node, thesu ffi x xd ∗ is the rightmost ρ -son of ν , so that the the nzm -code of the leftmost ρ -son of ν + ν . If ν is a wx -node, the su ffi x xd ∗ is the rightmost ρ -son of ν + ρ -son of ν + ν +
1. This canbe checked by induction on the rules of Table (38). This allows us to state:T heorem Let B ρ be B , the black metallic tree, equipped with the rightmost assign-ment ρ . The rules which allow us to construct the tree under that assignment are givenin (37) and the nzm -codes of the ρ -sons of a node ν are given in (38) in terms of the nzm -codes of ν − and of ν − . Under that assignment, the tree does not observe thepreferred son property, whatever the digit chosen for that purpose. The successor of thenode ν is a ρ -son of ν + : its leftmost ρ -son or the next ρ -son of ν + . No assignmentallows to establish any preferred son property on B . Proof. The largest part of the proof is given with the proof of (37) and ofTable (38). For what is the assignment and a preferred son property, the factthat the successor of ν is a son of ν + ν ] nz a , whatever the digit a as far as ≤ a : therightmost assignment allows us to grasp as far as possible a complete set of nzm -signatures, some circular permutation on { ,.., d , x } . If we change the the ρ -status of the black son of the node ν to an α -white one, we must change the ρ -status of a white son of ν to an α -black one, which reduces the range graspedby the α -white nodes which lies after the α -black son. Accordingly, no preferredson property can be observed in B α for all the nodes of the tree. The proof ofTheorem 15 is now completed. (cid:3) . { p , } and { p + , } of the hyperbolic plane With the previous sections, we established the properties of the metallictrees. As already mentioned in [6], the metallic trees are connected with twofamilies of tilings of the hyperbolic plane: the tilings { p , } and the tilings31 p + , } . The first tiling is generated by the regular convex polygon with p sidesand the right angle as interior angle at each vertex by reflections in its sidesand, recursively, by the reflections of the images in their sides. The second oneis generated in the same way from the regular convex polygon with p + π { p , } and that three of them do the samein { p + , } . There is another way to generate those tilings which rely on themetallic trees. Figure 10 illustrates the considered tilings in the case when p = Figure 10
The tilings generated by the white metallic tree with p = . To left, thethe tiling { , } to right, the tiling { , } In Sub section 6.1, we define the regions of the tilings which are associatedwith the metallic trees and in Sub section 6.2, we explain the correspondencebetween the trees and the regions. In Sub section 6.3 we look carefully at thecase of the white metallic tree and in Sub section 6.4, we study the case of theblack one. { p , } and { p + , } of thehyperbolic plane The metallic trees are associated with two kinds of regions of the consideredtilings. The sub section is devoted to the definition of those regions.The regions addressed by the white metallic tree is called a sector . In { p , } a sector of the tiling is defined by two rays u and v issued from a vertex V of atile T , u and v being supported by the sides of T which meet at V . The sectordefined by u and v is the set of tiles whose center is contained in the right angledefined by those rays. The left hand-side picture of Figure 11 illustrates how p sectors can be displayed around a once and for all fixed tile which we call the central tile , say T . The sectors and the central tile cover the hyperbolic planewith no hole and their interiors do not intersect.The right hand-side picture of the figure illustrates the same display of32ectors around T in the tiling { p + , } with, this time, p + { p + , } , the definition of a sector is more complicate.It is again defined by two rays u and v . Consider a tile T , a vertex V of T . Twosides of T meet at V , say a and b , and a third side c , belonging to the other tilessharing V with T , also meets V . Then u and v are issued from the midpointof c , u and v passing through the midpoints of a and b respectively. The sectordefined by u and v is the set of tiles whose center lies in the acute angle definedby u and v . The p + T in { p + , } and T cover the hyperbolicplane with no hole and their interiors do not intersect. Figure 11
The sectors around the central tile fixed once and for all.
In both tilings, the tile T we above considered to define a sector is called the head of the sector or, also, its leading tile .Presently, let us define the strips in those tilings. Figure 12
The strips around the central tile fixed once and for all. In { p , } , a strip is defined by two rays u and v together with a side a of atile T , u and v being issued from the ends of a and being supported by the sidesof T which meet a . The left hand-side of Figure 12 illustrates the strip in { p , } .The strip, in the tiling, is the set of tiles whose centre lies in the intersection33f the three closed half-planes defined by u , v and a which contain a and therays. We can see in the figure that a strip is, in some sense, smaller than asector. As the figure points at that, the p strips displayed around the central tileand T itself do not cover the hyperbolic plane. As can be seen on the figure, inbetween two strips associated by two consecutive sides of T , there is a sector.In { p + , } , a strip is also defined by two rays u and v together with a side a of a tile T . Let b and c be the sides of T which share a vertex with a . Then, u , v isthe ray issued from the foot of the perpendicular to a issued from the midpointof b , c respectively which pass through the midpoint by which it is defined,also see the right hand-side picture of Figure 14. In the tiling, the strip is theset of tiles whose centre lies in the intersection of the three closed half-planesdefined by u , v and the line supporting a which contains that side and therays. On the right-hand side picture of Figure 12, we can see that the stripsaround T together with that tile do not cover the hyperbolic plane. Applyingthe definition of a sector in that context, we can see that in between the stripsdefined by two consecutive sides of T , there is a sector.Here too, in both tilings, the tile T we considered for defining the strip iscalled the head of the strip or also, its leading tile . It is the time to precisely describe the connection between the metallic treesand the regions defined in Sub section 6.1. Figures 13 and 14 illustrate theseconnections. As shown in [7, 3], there is a bijection between the white metallictree and a sector of both { p , } and { p + , } for the same value of p used fordefining the tree.From now on, if T is a tile of the tiling, we number its side starting from up to h with h = p or h = p +
2, depending on whether T belongs to { p , } orto { p + , } respectively. Once side is fixed, the other sides are increasinglynumbered from 1 while counterclockwise turning around the tile starting fromside . Denote by ( T ) i , with i ∈ { .. h } the tile which shares the side i of T withthat latter tile. In a tiling, a tile which shares a side with T is called a neighbour of T .The comparison between Figure 11 and 13 allows us to better see the treestructure in a sector. The idea is to associate white nodes to the head of a sectorand black nodes to the head of a strip.First, consider the case of the tiling { p , } . The root of the white metallic treeis associated with the head T of a sector S . Let u and v be the rays defining S .We fix number in such a way that side 1 is supported by u , so that side p issupported by v , exchanging the names of u and v if necessary for the numberingof the sides of T . From that numbering and the definition of a sector, ( T ) and( T ) p are outside S . From [7, 3], we know that the neighbours ( T ) i of T with i ∈ { .. p − } are in S . We precisely associate the λ -sons of the root in the orderof their numbers to the ( T ) i ’s inside S in the order of their numbers too. Next,34onsider a tile τ already associated with a node ν of W . We number the sidesof τ as already mentioned, the number being given to the side shared with thetile associated to the father of ν . If ν is white, we associate its λ -sons in the orderof their numbers to the neighbours ( τ ) i of τ with i in { .. p − } in that order. If ν is black, we associate its λ -sons in the order of their numbers to the neighbours( τ ) j of τ with j in { .. p − } in that order too. From [7, 3], it is known that thisprocess establishes a bijection between the nodes of W and the tiles of S .Similarly, consider a strip S defined by the rays u , v and the side a of T , itsleading tile. Fix a as side of T and let side be supported by u and side p besupported by v , exchanging the names of u and v if needed by the numberingof the sides of T . Then ( T ) , ( T ) and ( T ) p are outside S while the neighbours( T ) j of T with j in { .. p − } are in the strip, see [7, 3]. We can repeat the aboveprocess, considering the head of S as associated to the root of B as there areexactly p − T inside S . It is not di ffi cult to prove from that thatthe same process as for S starting from the head T of S establishes a bijectionbetween the nodes of B and the tiles of S . The reason is that B can be obtainedfrom W λ by removing the sub tree rooted at the rightmost son of the root of W , and that subtree is isomorphic to W . Now, it is proved in [7, 3], that a strip R can be obtained from a sector S with head T by removing the image of thesector defined by the sides and p of ( T ) p − , the last neighbour of T in S , seealso Figure 14, and the head of R is T too. Figure 13
How the white metallic tree generates the tilings { , } and { , } : thesectors are delimited by colours, each sector being associated with three colours whichare attached to the status of the nodes. Each sector in the above figures is spannedby the white metallic tree. Secondly, consider the case of the tiling { p + , } . Again, we associate theroot of W with the head T of a sector S . Let u and v be the rays defining S andlet a be the side of another tile which meets T at the vertex belonging to theconsecutive sides of T met by u and v at their midpoints. Let the side met by u while the side p + is met by v , exchanging the names of u and v if needed inorder to be coherent with the numbering of the sides of T . We can see that thetiles ( T ) , ( T ) and ( T ) p + have their centre outside S . It is proved in [3] that the35eighbours ( T ) i of T with i in { .. p } have their centre in S . We apply the sameprocess as in the case of the tiling { p , } with this di ff erence that to the λ -sons ofa node ν associated to the tile τ , we associate in the order of the numbers of thesons the neighbours ( τ ) i with i in { .. p } in this order if ν is white and if ν is black,we associate the the neighbours ( τ ) j with j in { .. p } . It is proved in [3] that thejust described process establishes a bijection between S and the tiles of a sectorin the tiling { p + , } . The right hand-side of Figure 14 illustrates the structureof the tree in S . It also illustrates the fact that the same process establishes abijection between B and the tiles of a strip S in the tiling { p + , } . Figure 14
The decomposition of a sector spanned by the white metallic tree intoa tile, then two copies of the same sector and a strip spanned by the black metallictree. To left: the decomposition in the tiling { p , } ; to right, the decomposition in thetiling { p + , } . In both cases, the dark blue colour indicates the black nodes whilethe white ones are indicated in dark yellow, in green and in purple. We note that the same tree is in bijection of a sector both in { p , } and { p + , } .The di ff erence of two sides for the regular convex polygons generating thosetilings lies in the fact that as three tiles meet at a vertex instead of four of them,the number of neighbours of the head which are outside the sector is biggerin { p + , } than in { p , } . It is also the same situation for a strip. Let u , v bethe rays and a be the side of its leading tile T which define a strip S . Take theside of T as a and Number the other sides as already indicated, exchangingthe names of u and v if needed for side p + to be identified with the side whichis crossed by v and which shares a vertex with a . Then, it is not di ffi cult to seethat the centres of the neighbours ( T ) , ( T ) , ( T ) and ( T ) p + are outside S . Theother neighbours have their centres inside S and there are p − B .We close this sub section by reminding something we already mentionedin [6]. Indeed, we indicated there a property mentioned too in [4]: a sector S can be spit into a sequence { S n } n ∈ N . The first term of the sequence is the strip S whose head is the head too of S . Note that according to our conventions,the side of T as the head of S is the side p of T as the head of S . The ray u defining S is the ray u defining S . The ray v defining S passes through themidpoint of the side p of T as head of S . We take this occasion to note that36he same side of a tile may receive di ff erent numbers depending on the contextwhich defines the choice of the side which may di ff er from one situation toanother one. The head of S n + is the neighbour ( τ n ) p of the head τ n of S n . Theray u n + which defines S n + is the ray v n which defines S n , and the ray v n + defining S n + passes through the midpoint of the side p + of ( τ n ) p , the side of that neighbour being the side it shares with τ n . The construction of the firstelements of that sequence is illustrated by Figure 14, for the tiling { p , } by theleft hand-side picture and by the right hand-side one for the tiling { p + , } .The following sub sections, 6.3 and 6.4 study the applications of the num-bering and their representations to location problems of the tiles in a sectorand in a strip. Such problems are at the basis of an implementation of cellularautomata in the settings of those hyperbolic tilings. As explained in Sub section 6.2, the white metallic tree is connected with thetilings { p , } and { p + , } of the hyperbolic plane with p ≥
5. Recall that Figure 10illustrates the tiling { , } , left hand side, and the tiling { , } , right hand side,associated to p =
7. In the present sub section, we take use of the studies ofSections 4 and 5 in order to solve two location problems of the tiles in a sectorof those tilings. The first problem which we address is to find an algorithmcomputing the path from a tile to the head of a sector. The problem is addressedby Sub subsection 6.3.1. The second problem is to compute the codes of theneighbours of a tile, which is solved in Sub subsection 6.3.2.
In [6], we provided an algorithm to compute the path from a tile τ of a sector S tothe head of S which was based on the metallic code of τ , and W was supposedto be fitted with the leftmost assignment. Here, we revisit the algorithm,assuming that W is fitted with the rightmost assignment. In [6] two algorithmswere provided, the first one reading the digits of [ τ ] from the lowest to thehighest and the second one performs the same in the reverse order. In thatsecond algorithm two paths are constructed, one to right, the second to leftand, eventually the expected path is the to left one. The second algorithm of [6]has a decisive advantage: its complexity is linear in the size of the metallic codeof τ . Accordingly, we provide a similar algorithm based on the metallic codesas codes for the nodes of W ρ .To that purpose, let us have a look on Figure 2 which we reproduce asFigure 15 for the convenience of the reader. Note that for a node ν of the level n such that ν < m n , the metallic code has n digits and when, on the same level, ν ≥ m n , the metallic code has n + . If we look at the highest digit of the metallic codes of the nodes on level 2,we note it is a for the last two sons of the node a and for all sons of the node( a ) +
1, its last two sons being excepted. At this level, there is an exception when a is : the last two sons of , the sons of and those of , its last two sons37eing excepted. And so, in that case, three sons of three nodes are concernedwith as the highest digit. If we look at the nodes of level 3, the second highestdigit of their metallic code is connected with the last digit in the metallic codeof nodes of level 2.More generally. Assume that ν = a k .. a a . Table (25) that for most nodes,the metallic code of their sons but the last two ones are based on [ ν − ν − ν , we know that the nodewhose metallic code is [ ν ] b h .. b with b i in { , .. d } is either in the sub tree issuedfrom [ ν b h ] or [([ ν b h ]) +
1] at a time when we know [ ν ] b h without knowing thedigits b i with i < h . Algorithm 3 answers allows us to compute the path fromthe head of S . The path is given as a table whose length is that of the metalliccode of ν . Each entry of the table contains the indication of the son σ of a node ν as the rank of σ among the sons of ν , the leftmost son being given rank 1, andit also contains the status of ν . What just mentioned and table (25) allows us todevise the algorithm. d c2 cd d0 d1 d2 dc dc2 dcc Figure 15
The white metallic tree under the rightmost assignment. We are hereinterested in the metallic codes.
Let us know see the details of that computation. according to this generalprinciple. Nevertheless, it is needed to lightly tune the computation. Indeed, ifthe highest digit is , we need to know the next one: if the next digit is , thenthe left hand-side path goes through the preferred son and the right hand-sidepath goes through the rightmost son. If the digit is or greater than , the lefthand-side path goes through the rightmost son and the right hand-side pathgoes through the leftmost son. Then, the for -loop deals with the other digitsfrom high ones to low ones.In the working of the algorithm, it is assumed that when we examine thecurrent digit a , the left hand-side list ℓ path goes through a node ν and the righthand-side path list r goes through ν +
1. The last registered digit b occurs in thesignature of ν . Let a be the digit we examine: it is the signature of a son of ν or of ν +
1. If ν is white and if b = , then if a = , list ℓ goes on the rightmostbranch of the tree rooted at ν and list r goes through the leftmost branch of the38 lgorithm 3 The path from the root to the node ν in a sector S in bijection with W ρ .We set a k .. a ⇋ [ ν ] . The lists register the status of the current node and its rank amongthe sons of its father, the leftmost son having rank . proc update (side; from; upto) isbegin if side = left then for j in { from..upto } loop list r ( j ) : = list ℓ ( j ); end loop ; else for j in { from..upto } loop list ℓ ( j ) : = list r ( j ); end loop ; end if ; restart : = upto +
1; handside : = side; end proc ;restart : =
0; handside : = left ; if ( a k = ) and ( a k − = ) then list ℓ (0) : = w . p −
3; list r (0) : = b . p − else if ( a k = ) then list ℓ (0) : = b . p −
2; list r (0) : = w .1; else list ℓ : = w .( a i ) −
1; list r : = w .( a i ); end if ; end if ; for i in { .. k − } in reverseloop if (status(list ℓ ( k − i − = w ) and then ( a i + = ) then case a i iswhen 2 ⇒ list ℓ ( k − i ) = b . p −
2; list r ( k − i ) = w .1; when 0 | ⇒ update (side: left , from: restart, upto: k − i − ℓ ( k − i ) = w . p − + ( a i ); list r ( k − i ) = w . p − + ( a i ); when others ⇒ update (side: right , from: restart, upto: k − i − ℓ ( k − i ) = w .( a i ) −
2; list r ( k − i ) = w .( a i ) − end case ; else last : = p −
2; place : = ( a i ) − if status(list ℓ ( k − i − = bthen last : = last −
1; place : = place − end if ; case a i iswhen 1 ⇒ list ℓ ( k − i ) = b .last; list r ( k − i ) = w .1; when 0 ⇒ update (side: left , from: restart, upto: k − i − ℓ ( k − i ) = w .last −
1; list r ( k − i ) = b .last; when others ⇒ update (side: right , from: restart, upto: k − i − ℓ ( k − i ) = w .place; list r ( k − i ) = w .place + end case ; end if ; end loop ; if handside = right then update(side: right , from: restart, upto: k ); end if ; 39ree rooted at ν + σ is the new end of list ℓ , σ + r . The samesituation occurs for ν if its signature is not and if a = . In the other cases,the number to be remembered is ( a ) − ν to ν for the left hand-side path and it is ( a ) for the edge to ν + ν + for -loop in such a way that theresult is the left hand-side path. Algorithm 4
Algorithm for constructing the path from a tile ν to the leading tile ofits sector from [ ν ] nz = a k .. a a . In the first if , means that the path remains on theroot. The result is in list r .j : = k ; update : = if a k = list ℓ (0) : = r . ; list r (0) : = w . ; else list ℓ (0) : = w . (a k ) − ; list r (0) : = w . (a k ) ; end if ; for j in [0.. k − in reverseloop if a j = status( π ℓ ( j + in { w , r } then list ℓ ( k − j ) : = b . p − ; else list ℓ ( k − j ) : = b . p − ; end if ;list r ( k − j ) : = w . ; else for i in [update + k − j − loop list ℓ ( i ) : = list r ( i ); end loop ;update : = k − j − ℓ ( k − j ) : = w . (a j ) − ; list r ( k − j ) : = w . (a j ) ; end if ; end loop ; if a = i in [update + k ] loop list r ( i ) : = list ℓ ( i ); end loop ; end if ;Algorithm 4 does a similar computation when the nzm -codes are used ascoordinates of the nodes of W ρ . As there are only two rules for the rightmostassignment, two rules which are very similar due to the fact that a black nodeoccurs when the signature d is followed by in the nzm -code of the next value40f the number, the algorithm is simpler. The path goes to left when the currentdigit is , in case the next digit will be again . Otherwise, the path goes toright. Accordingly, the updating occurs only for following the right hand-sidepath. As mentioned in the caption of the algorithm, the result is in the righthand-side path. In the present sub subsection, we turn to another problem. Knowing thecoordinate [ ν ] or [ ν ] nz of a tile, how to get the coordinates of the same typefor its neighbours? The answer is given by Table 2 for the metallic codes andby Table 1 for the nzm -ones. Both tables consider W ρ , i.e. the metallic treeequipped with the rightmost assignment. The tables give the neighbours bothin { p , } and { p + , } .Table 1 is shorter as there are only two rules for W ρ with the nzm -codes. Table 1
Table of the neighbours of ν : to left, in the tiling { p , } , to right, in the tiling { p + , } . In the table, the black son of ( ω ) is denoted by ( ω ) b . in { p , } in { p + , } w -node rep. tile nzm -code ( ν ) [( a k .. a ) − ( ν − b [( a k .. a a ) − j w -sons [( a k .. a a ) − i i ∈ { .. p -2 } , j = i + p ( ν ) b a k .. a a rep. tile nzm -code ( ν ) [( a k .. a ) − ν − a k .. a a ) − ( ν − b [( a k .. a a ) − j w -sons [( a k .. a a ) − i i ∈ { .. p -2 } , j = i + p + ( ν ) b a k .. a a p + ν + a k .. a a ) + b -node rep. tile nzm -code ( ν ) a k .. a ( ν − b [( a k .. a a ) − j w -sons [( a k .. a a ) − i i ∈ { .. p -3 } , j = i + p − ( ν ) b a k .. a a p ( ν ) + a k .. a ) + rep. tile nzm -code ( ν ) a k .. a ⊖ ν − a k .. a a ) − ( ν − b [( a k .. a a ) − j w -sons [( a k .. a a ) − i i ∈ { .. p -3 } , j = i + p ( ν ) b a k .. a a p + ν + a k .. a a ) + p + ( ν ) + a k .. a ) + ν ) j for j ∈ { .. p } for a white node and j ∈ { .. p − } for a black one. In both cases, ( ν ) isthe father and ( ν ) is the rightmost son of ν − nzm -codes can easily be41erived from Table (30). In the case of the black node, ( ν ) p is ( ν ) +
1, the nodewhich lies just after the father of ν on the level of ( ν ) . The part of the tablesdevoted to { p + , } involves two specific neighbours: ν − ν +
1. For a whitenode they are ( ν ) and ( ν ) p + respectively. For a black node, they are ( ν ) and( ν ) p + respectively as far as in that case ( ν ) p + is ( ν ) + Table 2
Table of the neighbours of ν : to left, in the tiling { p , } , to right, in the tiling { p + , } . In the table, the black son of ( ω ) is denoted by ( ω ) b , its preferred son by ( ω ) π . in { p , } in { p + , } wa -node rep. tile metalic code ( ν ) [( a k .. a ) − ( ν − b [( a k .. a a ) − j w -sons [( a k .. a a ) − i i ∈ { .. p -3 } , j = i + p − ( ν ) π a k .. a a p ( ν ) b a k .. a a rep. tile metalic code ( ν ) [( a k .. a ) − ν − a k .. a a ) − ( ν − b [( a k .. a a ) − j w -sons [( a k .. a a ) − i i ∈ { .. p -3 } , j = i + p ( ν ) π a k .. a a p + ( ν ) b a k .. a a p + ν + a k .. a a ) + w0 -node rep. tile metallic code p − ( ν ) π a k .. a a p − ( ν ) p − a k .. a a p ( ν ) b a k .. a a rep. tile metallic code p − ( ν ) π a k .. a a p ( ν ) p − a k .. a a p + ( ν ) b a k .. a a -node rep. tile metallic code ( ν ) [( a k .. a ) − ( ν − b [( a k .. a a ) − j w -sons [( a k .. a a ) − i i ∈ { .. p -3 } , j = i p − ( ν ) π a k .. a a p − ( ν ) p − a k .. a a p ( ν ) b a k .. a a rep. tile metallic code ( ν ) [( a k .. a ) − ν − a k .. a a ) − ( ν − b [( a k .. a a ) − j w -sons [( a k .. a a ) − i i ∈ { .. p -3 } , j = i + p − ( ν ) π a k .. a a p ( ν ) p a k .. a a p + ( ν ) b a k .. a a p + ν + a k .. a a ) + b -node p ( ν ) + a k .. a p + ν + a k .. a a ) + p + ( ν ) + a k .. a W ρ whenthe metallic codes are used. In order to reduce the number of repetitions, thelines for w0 -nodes again takes the lines for wa -nodes except for the lines p-2 , p-1 and p in the case of { p , } and the lines p-1 , p and p + in the case of { p + , } .The same thing was done in the case of the b -nodes which takes the lines ofthe w1 -nodes except the line p for { p , } and the lines p + and p + for { p + , } .The table rewrites the exceptional lines accordingly. The reason of the changesis clear. The codes for the first sons of a wa -node and for a w0 -one are builtone the same way from the metallic code of the node. For a b -node, a similarremark is relevant. In this section, we consider the same problems for the black metallic tree.Sub-subsection 6.4.1 deals with the path from a node to the root of B ρ , whileSub-subsection 6.4.2 computes the codes of the neighbours of a tile in a strip. Algorithm 5 constructs the path from a node ν in B ρ to the root of the tree. Itmakes use of the same procedure ’update’ as in Algorithm 3. Algorithm 5
The path from the root to the node ν in B ρ . Here, a k .. a ⇋ [ ν ] . Theinput is [ ν ] . The result is in list ℓ . restart : =
0; handside : = left ; if a k = list ℓ (0) : = r .0; list r (0) : = w .1; else if a k = list ℓ (0) : = b . p −
3; list r (0) : = w .1; else list ℓ : = w .( a i ); list r : = w .( a i ) + end if ; end if ; for i in { .. k − } in reverseloop if status(list ℓ ( k − i − = wthen last : = p − else last : = p − end if ; if a i = handside = right then update (side: left , from: restart, upto: k − i − end if ;list ℓ ( k − i ) : = b .last; list r ( k − i ) : = w .1; else update (side: right , from: restart, upto: k − i − ℓ ( k − i ) : = w . a i ; list r ( k − i ) : = w .( a i ) + end if ; end loop ; 43he algorithm makes use of the metallic code of ν . The lists register thestatus of the current node and its rank among the sons of its father, the leftmostson having rank 1.The justification of Algorithm ?? is straightforward. It is similar to the caseof Algorithm 4. It is based on the fact that outside the case when the currentdigit a is , the path necessarily goes within the tree rooted at the node whosesignature is a + a th node among the sons of its father. When a = ,except the case when the path has to move to right, it is needed to go on onboth paths at a distance 1 from each other as at that moment, the next digit isnot known. Algorithm 6
The path from the root to the node ν in a strip S in bijection with B ρ .We set a k .. a ⇋ [ ν ] . The lists register the status of the current node and its rank amongthe sons of its father, the leftmost son having rank . restart : =
0; handside : = left ; last : = k − if a k = list ℓ (0) : = b . p −
3; list r (0) : = w . ; else if a k = xthen list ℓ (0) : = w .1; list r (0) : = w .1;list ℓ (1) : = w .1; list r (1) : = w .2; last : = last − else if a i < dthen list ℓ (0) : = w .( a i ) +
1; list r (0) : = w .( a i ) + else list ℓ (0) : = r .0; list r (0) : = w . ;list ℓ (1) : = b . p −
3; list r (0) : = w . ; last : = last − end if ; end if ; end if ; for i in { .. final } in reverseloop last : = p − if status(list ℓ ( k − i − = b then last : = p − end if ; case a i iswhen x ⇒ update (side: right , from: restart, upto: k − i − ℓ ( k − i ) = w .1; list r ( k − i ) = w .2; when d | c ⇒ update (side: right , from: restart, upto: k − i − ℓ ( k − i ) = b .last; list r ( k − i ) = w .1; when others ⇒ if a i in { c , d } then update (side: right , from: restart, upto: k − i − else update (side: left , from: restart, upto: k − i − end if ;list ℓ ( k − i ) = w .( a i ) +
1; list r ( k − i ) = w .( a i ) + end case ; end loop ; 44lgorithm 6 addresses the same issue when the coordinates of the nodes aregiven through their nzm -code. We can see that its structure is more complexthan that of Algorithm 5.The reason is not only the fact this time we have three rules for the nodesinstead of two for Algorithm 5, it is also due to the fact that the occurrence ofthe pattern dc ∗ d entails that when the pattern is followed by a digit a with a < c ,the appropriate node is in the node pointed at by the right hand-side path: itfollows from Table (38). We now turn to the computation of the coordinates of the neighbours of a tile ν which lies in a strip S in bijection with B ρ . we first study that computationwhen it is based on [ ν ]. The computation should be easier as there are two rulesonly for the sons of a node. Table 3
Table of the metallic codes of the neighbours of a tile ν in both tilings { p , } and { p + , } in the black metallic tree under the rightmost assignment. We assumethat a k .. a a ⇋ [ ν ] and ( ν ) i indicates the neighbour i. in { p , } in { p + , } wa -node rep. tile metallic code ( ν ) [( a k .. a ) + ( ν − p [ ν − j w -sons [ ν − i i ∈ { .. p − } , j = i + p b -son [ ν ] rep. tile metallic code ( ν ) [( a k .. a ) + ν − ν − ( ν -1) p + [ ν − j w -sons [ ν − i i ∈ { .. p − } , j = i + p + b -son [ ν ] p + ν + ν + b0 -node rep. tile metallic code ( ν ) [ a k .. a ] ( ν − p [ ν − j w -sons [ ν − i i ∈ { .. p − } , j = i + p b -son [ ν ] rep. tile metallic code ( ν ) [ a k .. a ] ν − ν − ( ν -1) p + [ ν − j w -sons [ ν − i i ∈ { .. p − } , j = i + p b -son [ ν ] p + ν + ν + p + ( ν ) + a k .. a ) + ν in { p + , } are the nodes ν − ν + ν in B ρ . This introduce a small change in thenumbering of the sons of the node compared with their numbering in { p , } . Table 4
Table of the nzm -codes of the neighbours of a tile ν in both tilings { p , } and { p + , } in the black metallic tree under the rightmost assignment. We assume that a k .. a a ⇋ [ ν ] nz and ( ν ) i indicates the neighbour i. When references of neighbours fora type of nodes are missing, they have to be seen in the same column at the previoustype and, again at the previous one if they are still missing. in { p , } in { p + , } wa -node rep. tile nzm -code ( ν ) [( a k .. a ) + nz ( ν − p [ ν − nz x j w -sons [ ν − nz i i ∈ { .. p -3 } , j = i + p b -son [( ν − nz x rep. tile nzm -code ( ν ) [( a k .. a ) + nz ν − ν − nz ( ν -1) p + [ ν − nz x j w -sons [ ν − nz i i ∈ { .. p -3 } , j = i + p + b -son [( ν − nz x p + ν + ν + nz w1 -node rep. tile nzm -code ( ν − p [ ν − nz d ( ν ) [ ν − nz x j w -sons [ ν − nz i i ∈ { .. p -4 } , j = i + p b -son [( ν − nz d rep. tile nzm -code ( ν − p [ ν − nz d ( ν ) [ ν − nz x j w -sons [ ν − nz i i ∈ { .. p -4 } , j = i + p + b -son [( ν − nz dwx -node rep. tile nzm -code ( ν ) [( a k .. a ) + nz rep. tile nzm -code ( ν ) [( a k .. a ) + nz bd , bx -nodes rep. tile metallic code ( ν ) [( a k .. a ) + nz ( ν − p [ ν − nz x j w -sons [ ν − nz i i ∈ { .. p -3 } , j = i + p ( ν ) p [ ν + nz rep. tile metallic code ( ν ) [( a k .. a ) + nz ν − ν − nz ( ν − p [ ν − nz x j w -sons [ ν − nz i i ∈ { .. p -3 } , j = i + p + ( ν ) p + [ ν + nz p + ( ν ) + a k .. a ) + nz nzm -codes of the neighbours ( ν ) i of ν in B ρ . The tablefollows from the rules (37) and (38). It should be remarked that the nzm -codesof the neighbours of a wx -node are very similar to those of a w1 -one. Thedi ff erence is in computation of the nzm -code of the father. For the wx -node,the nzm -code of the father is given by [( a k .. a ) + nz and not by [( a k .. a ) + nz asit the case for a w1 -node. Conclusion
We can conclude the paper with several remarks.The present paper deepens the research started with [5] and which wascontinued by [6].The extension addressed both the connection of the trees with tilings of thehyperbolic plane and the generalization of the golden sequence to the metallicones performed in [6]. A comparison was made in that latter paper between thewhite metallic tree and the black one, finding an explanation of the surprisingresults obtained in the case of the black metallic tree were the preferred sonproperty was no more true.The present paper was an occasion to revisit the previous results with twonew features: the nzm -codes and the notion of assignment, a notion that wasalready introduced in [2] in the case of the Fibonacci trees. A key result hereis Theorem 7 which partially solves a question raised in [2] as far as any as-signment possesses the preferred son property in the frame of the metalliccodes applied to white metallic trees. The introduction of the nzm -codes radi-cally changed the situation even for the white metallic trees: the preferred sonproperty is true for a single assignment as shown by Theorem 9. A uniqueassignment also possesses the preferred son property for the black metallic treein the frame of the metallic codes. But no assignment possess that propertywhen the black metallic tree is fitted with the nzm -codes.The present paper also computes the metallic and the nzm -codes of theneighbours of a tile, taking advantage of the isomorphism established betweenthe metallic trees and particular regions of the tilings { p , } and { p + , } of thehyperbolic plane. The paper also investigates algorithms to compute the pathfrom a node to the root of its tree using metallic codes and also using nzm -ones.All these algorithms are linear in the size of the code which is used;The paper is more conclusive than [6] claimed for itself. Other problemsare probably still open: the present author does not pretend to have solved anypossible problems in these settings, just to have closed one or two issues. References [1] C. Iwamoto, T. Andou, K. Morita, K. Imai, Computational Complexityin the Hyperbolic Plane,
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