aa r X i v : . [ m a t h . C O ] F e b Edges in Fibonacci cubes, Lucas cubes and complements
Michel Mollard ∗ February 9, 2021
Abstract
The
Fibonacci cube of dimension n , denoted as Γ n , is the subgraph of the hypercubeinduced by vertices with no consecutive 1’s. The irregularity of a graph G is thesum of | d ( x ) − d ( y ) | over all edges { x, y } of G . In two recent paper based on therecursive structure of Γ n it is proved that the irregularity of Γ n and Λ n are twotimes the number of edges of Γ n − and 2 n times the number of vertices of Γ n − ,respectively. Using an interpretation of the irregularity in terms of couples ofincident edges of a special kind (Figure 2) we give a bijective proof of both results.For these two graphs we deduce also a constant time algorithm for computing theimbalance of an edge. In the last section using the same approach we determinethe number of edges and the sequence of degrees of the cube complement of Γ n . Keywords:
Irregularity of graph, Fibonacci cube, Lucas cube, cube-complement, daisycube.
AMS Subj. Class. : 05C07,05C35
An interconnection topology can be represented by a graph G = ( V, E ), where V denotes the processors and E the communication links. The hypercube Q n is a popularinterconnection network because of its structural properties.The Fibonacci cube of dimension n , denoted as Γ n , is the subgraph of the hypercubeinduced by vertices with no consecutive 1’s. This graph was introduced in [7] as a newinterconnection network.Γ n is an isometric subgraph of the hypercube which is inspired in the Fibonaccinumbers. It has attractive recurrent structures such as its decomposition into twosubgraphs which are also Fibonacci cubes by themselves. Structural properties of thesegraphs were more extensively studied afterwards. See [9] for a survey.Lucas cubes, introduced in [13], have attracted the attention as well due to the fact ∗ Institut Fourier, CNRS, Universit´e Grenoble Alpes, France email: [email protected] G = ( V ( G ) , E ( G )) be a connected graph. The degree of a vertex x is denoted by d G ( x ) or d ( x ) when there is no ambiguity. The imbalance of an edge e = { x, y } ∈ E ( G )is defined by imb G ( e ) = | d G ( x ) − d G ( y ) | . The irregularity of a non regular graph G is irr ( G ) = X e ∈ E ( G ) imb G ( e ) . This concept of irregularity was introduced in [1] as a measure of graph’s global non-regularity.In two recent papers [2, 6] using the inductive structure of Fibonacci cubes it isproved that irr (Γ n ) = 2 | E (Γ n − ) | and irr (Λ n ) = 2 n | V (Γ n − ) | . One of our motivationis to give direct bijective proofs of these remarkable properties.The generalized Fibonacci cube Γ n ( s ) is the graph obtained from Q n by removing allvertices that contain a given binary string s as a substring. For example Γ n (11) = Γ n . Daisy cubes are an other kind of generalization of Fibonacci cubes introduced in [11].For G an induced subgraph of Q n , the cube-complement of G is the graph inducedby the vertices of Q n which are not in G . In [16] the questions whether the cubecomplement of generalized Fibonacci cube is connected, an isometric subgraph of ahypercube or a median graph are studied. It is also proved in the same paper that thecube-complement of a daisy cube is a daisy cube. We consider in the last section Γ n the cube complement of Γ n .We give the number of edges of Γ n and determine, using the main lemma of thefirst section, the degree sequence of Γ n . We will also study the embedding of Γ n in Γ n .We will next give some concepts and notations needed in this paper. We note by[1 , n ] the set of integers i such that 1 ≤ i ≤ n . The vertex set of the hypercube ofdimension n Q n is the set B n of binary strings of length n , two vertices being adjacentif they differ in precisely one position. We will note x i the binary complement of x i .Let x = x . . . x n be a binary string and i ∈ [1 , n ] we will denote by x + δ i the string x ′ . . . x ′ n where x ′ j = x j for j = i and x ′ j = x j otherwise. We will say that the edge { x, x + δ i } uses the direction i . The endpoint x such that x i = 1 of an edge using thedirection i will be called upper endpoint and y the lower endpoint .A Fibonacci string of length n is a binary string b = b b . . . b n with b i · b i +1 = 0 for1 ≤ i < n . In other words a Fibonacci string is a binary string without 11 as substring.The Fibonacci cube Γ n ( n ≥
1) is the subgraph of Q n induced by the Fibonacci stringsof length n . Because of the empty string ǫ , Γ = K .A Fibonacci string b of length n is a Lucas string if b · b n = 1. That is, a Lucasstring has no two consecutive 1’s including the first and the last elements of the string.The Lucas cube Λ n is the subgraph of Q n induced by the Lucas strings of length n . Wehave Λ = Λ = K . 2 ✉✉ ✉✉✉ ✉✉ ✉✉✉ ✉ ✉✉ ✉✉✉ ✉✉✉ ✉✉✉ ✉✉ ✉✉✉ ✉✉✉ = Λ , Γ , Λ , Γ , Γ and Γ .Let F n be the n th Fibonacci number: F = 0, F = 1, F n = F n − + F n − for n ≥ F n and L n be the sets of strings of Fibonacci strings and Lucas strings of length n . Let F .n and F .n be the set of strings of F n that begin with 1 and that do not beginwith 1, respectively. Note that with this definition F . = { ǫ } and F . = ∅ . Let F . n bethe set of strings of F n that do not end with 1. Thus |F . n | = |F .n | . Let F n be the setof strings of F .n that do not end with 1. With this definition F = { ǫ } , F = { } and F = { } .From F n +2 = { s ; s ∈ F n +1 } ∪ { s ; s ∈ F n } , F .n +1 = { s ; s ∈ F n } and F .n +1 = { s ; s ∈ F .n } we obtain the following classical result. Proposition 1.1
Let n ≥ . The numbers of Fibonacci strings in F n , F .n and F .n are |F n | = F n +2 , |F .n | = F n +1 and |F .n | = F n respectively. Let n ≥ . The number ofFibonacci strings in F n is |F n | = F n . The following expressions for the number of edges in Γ n are obtained in [8] and [13]. Proposition 1.2
Let n ≥ . The number of edges in Γ n is | E (Γ n ) | = P ni =1 F i F n − i +1 = nF n +1 +2( n +1) F n and satisfies the induction formula | E (Γ n +2 ) | = | E (Γ n +1 ) | + | E (Γ n ) | + | V (Γ n ) | . Remark 1.3
Let { x, x + δ i } be an edge and θ ( x ) = (( x x . . . x i − ) , ( x i +1 x i +2 . . . x n )) .A combinatorial interpretation of | E (Γ n ) | = P ni =1 F i F n − i +1 is that for any i ∈ [1 , n ] θ is a one to one mapping between the set of edges using the direction i and the Cartesianproduct F . i − × F .n − i Let G be an induced subgraph of Q n . Let e = { x, y } be an edge of G where y isthe lower endpoint of e and x = y + δ i . An edge e ′ = { y, y + δ j } of G will be called an imbalanced edge for e if x + δ j / ∈ V ( G ) and thus { x, x + δ j } / ∈ E ( G ). Note that suchcouple of edges does not exist for G = Q n . We will prove in the next to sections thatfor G = Γ n and G = Λ n the irregularity of G is the number of such couples of edges(Figure 2). 3 ✉ ✉✉✉ (cid:0)(cid:0)(cid:0)❅❅❅ ❅❅❅ (cid:0)(cid:0)(cid:0) ... y y + δ i e y + δ j e ′ y + δ i + δ j ✉ ✉ ✉✉✉ (cid:0)(cid:0)(cid:0)❅❅❅ ❅❅❅ (cid:0)(cid:0)(cid:0) ... y y + δ i e y + δ j e ′ ❍❍❍❍✟✟✟✟ Figure 2: irr (Γ n ) and irr (Λ n ) count the couples of edges ( e, e ′ ) of the right kind. Lemma 2.1
Let x, y be two strings in F n with y = x + δ i and x i = 1 . Then for all j ∈ [1 , n ] we have x + δ j ∈ F n implies y + δ j ∈ F n . Proof.
Assume y + δ j / ∈ F n then y k = 1 for some k in { j − , j + 1 } ∩ [1 , n ]. But forall p ∈ [1 , n ] x p = 0 implies y p = 0. Thus x k = 1 and x + δ j / ∈ F n . (cid:3) Lemma 2.2
Let x, y two strings in F n with y = x + δ i . Then for all j ∈ [1 , n ] with | i − j | > we have x + δ j ∈ F n if and only if y + δ j ∈ F n . Proof. • If x j = 1 then y j = x j = 1 and both x + δ j and y + δ j belong to F n . • Assume x j = 0 thus y j = 0. We have x + δ j ∈ F n if and only if x k = 0 for all k ∈ { j − , j + 1 } ∩ [1 , n ]and y + δ j ∈ F n if and only if y k = 0 for all k ∈ { j − , j + 1 } ∩ [1 , n ] . But i / ∈ { j − , j + 1 } ∩ [1 , n ] thus x k = y k for all k in this set and the twoconditions are equivalent. (cid:3) Corollary 2.3
Let n ≥ then irr (Γ n ) is the number of couples ( e, e ′ ) ∈ E (Γ n ) where e ′ is an imbalanced edge for e . roof. By Lemma 2.1 if e = { x, y } is an edge using the direction i with upperendpoint x then d ( y ) ≥ d ( x ) and imb ( e ) is the number of imbalanced edges for e . Theconclusion follows. (cid:3) Furthermore assume that e = { x, y } uses the direction i with x i = 1 and let e ′ = { y, y + δ j } be an imbalanced edge for e . Then by Lemma 2.2 we have j = i + 1 or j = i − e ′ a right or left imbalanced edge for e accordingly. Let R Γ n and L Γ n bethe sets of couples ( e, e ′ ) where e ′ is a right imbalanced edge for e and a left imbalancededge for e , respectively, where e goes through E (Γ n ). Theorem 2.4
Let n ≥ . There exists a one to one mapping between R Γ n or L Γ n and E (Γ n − ) . Proof.
Let ( e, e ′ ) ∈ R Γ n . Assume that x is the upper endpoint of e = { x, y } . Wehave thus y = x + δ i and x i = 1 for some i ∈ [1 , n − θ (( e, e ′ )) = { x x . . . x i − x i +2 x i +3 . . . x n , x x . . . x i − x i +2 x i +3 . . . x n } . Since x and y belong to F n and the edges e , e ′ use the direction i , i + 1 we have x k = y k = 0for k in { i − , i + 2 } ∩ [1 , n ]. Therefore x x . . . x i − x i +2 x i +3 . . . x n is a Fibonaccistring and θ (( e, e ′ )) belongs to E (Γ n − ).Conversely let f = { z z . . . z i − z i +1 z i +2 . . . z n − , z z . . . z i − z i +1 z i +2 . . . z n − } be anarbitrary edge of Γ n − then z k = 0 for k in { i − , i + 1 } ∩ [1 , n − x = z z . . . z i − z i +1 z i +2 . . . z n − and t = z z . . . z i − z i +1 z i +2 . . . z n − are in F n . Theedge { t, t + δ i +1 } is a right imbalanced edge for the edge { x, x + δ i } . Furthermore θ ( { x, x + δ i } , { t, t + δ i +1 } ) = f and θ is a bijection.Similarly let φ (( e, e ′ )) = { x x . . . x i − x i +1 x i +2 . . . x n , x x . . . x i − x i +1 x i +2 . . . x n } where x is the upper end point of an edge e using the direction i and such that( e, e ′ ) ∈ L Γ n . Then φ is a one to one mapping between L Γ n and E (Γ n − ). (cid:3) As an immediate corollary we deduce the result of Alizadeh and his co-authors [2]
Corollary 2.5 irr (Γ n ) = 2 | E (Γ n − ) | . An other consequence of Lemma 2.2 is the following classification of the edges accord-ing to their imbalance. Note that from this classification we obtain a constant timealgorithm for computing the imbalance of an edge of Γ n . Theorem 2.6
Let n ≥ and e = { x, y } be an edge of Γ n using direction i . Then imb ( { x, y } ) follows Table 1. Proof.
Assume that x is the upper endpoint of the edge e = { x, y } . There exists anedge e ′ such that e ′ is a right imbalanced edge for e if and only if i ∈ [1 , n −
1] and5 mb ( { x, x + δ i } ) i = 1 i = 2 3 ≤ i ≤ n − i = n − i = nx x x i − x i +2 x n − x n − imb ( e ) in Γ n e ′ = { y, y + δ i +1 } thus if y + δ i +1 ∈ F n . Since y i = 0, y + δ i +1 is a Fibonacci string ifand only if i = n − y i +2 = x i +2 = 0 in the general case i ∈ [1 , n − e ′ such that e ′ is a left imbalanced edge for e if andonly if i ∈ [2 , n ] and e ′ = { y, y + δ i − } thus if y + δ i − ∈ F n . Since y i = 0, y + δ i − is aFibonacci string if and only if i = 2 or if y i − = x i − = 0 when i ∈ [3 , n ].Therefore imb ( e ) is completely determined by the values of x i +2 , x i − according toTable 1. (cid:3) Let e be an edge of Γ n then by Lemma 2.2 imb ( e ) ≤
2. Let A , B , C be the sets ofedges with imb ( e ) = 0, imb ( e ) = 1 and imb ( e ) = 2, respectively. Theorem 2.7
Let n ≥ . The numbers of edges of Γ n with imbalance 0,1 and 2 arerespectively | A | = n − X i =3 F i − F n − i − + 2 F n − | B | = 2 n − X i =1 F i F n − i − + 2 F n − | C | = n − X i =2 F i − F n − i . Remark 2.8
Note that | B | + 2 | C | = 2 | E (Γ n − ) | and we obtain again the result ofAlizadeh and his co-authors. Proof.
The case n ≤ n ≥ i ∈ [1 , n ] let E i be the set of edges { x, y } of Γ i with y = x + δ i . Let A i = A ∩ E i , B i = B ∩ E i and C i = C ∩ E i . Let e = { x, y } be an edge of Γ n . • If e ∈ A i then by Table 1 we have i ∈ [3 , n −
2] or i ∈ { , n } . If i ∈ [3 , n − θ ( e ) = ( x x . . . x i − , x i +2 x i +3 . . . x n ) is a one to one mapping between A i and F . i − × F .n − i − . 6f i = 1 then φ ( e ) = x x . . . x n is a one to one mapping between A and F .n − .Similarly Ψ( e ) = x x . . . x n − is a one to one mapping between A n and F . n − .By Proposition 1.1 we obtain | A | = P n − i =3 F i − F n − i − + 2 F n − . • If e ∈ C i then by Table 1 we have i ∈ [2 , n − θ ( e ) = ( x x . . . x i − , x i +2 x i +3 . . . x n ).Then θ is a one to one mapping between C i and F . i − × F .n − i − . The expressionof | C | follows. • Assume e ∈ B i and that there exists a right imbalanced edge for e therefore noleft imbalanced edge. We have thus i ∈ [1 , n −
1] and i = 2. If i ∈ [3 , n −
1] then θ ( e ) = ( x x . . . x i − , x i +2 x i +3 . . . x n ) is a one to one mapping this kind of edgesand and F . i − × F .n − i − .If i = 1 then φ ( e ) = x x . . . x n is a one to one mapping between this kind of edgesand F .n − . Thus this case contributes P n − i =3 F i − F n − i + F n − = P n − i =1 F i F n − i − + F n − to B . • Assume e ∈ B i and that there exists a left imbalanced edge for e thus noright imbalanced edge. By a similar construction this case contributes also P n − i =1 F i F n − i − + F n − to B . The expression of | B | follows. For any integer i let i = (( i −
1) mod n ) + 1. Thus i = i for i ∈ [1 , n ] and n + 1 = 1, = n . With this notation i and i + 1 are cyclically consecutive in [1 , n ]. Thereforefor x ∈ L n with x i = 0 the string x + δ i belongs to L n if and only if x k = 0 for all k ∈ { i − , i + 1 } . Note also that k ∈ { i − , i + 1 } if and only if i ∈ { k − , k + 1 } Lemma 3.1
Let x, y be two strings in L n with y = x + δ i and x i = 1 . Then for all j ∈ [1 , n ] we have x + δ j ∈ L n implies y + δ j ∈ L n . Proof.
Assume y + δ j / ∈ L n then y k = 1 for some k in { j − , j + 1 } . But for all p ∈ [1 , n ] x p = 0 implies y p = 0. Thus x k = 1 and x + δ j / ∈ L n . (cid:3) Lemma 3.2
Let x, y two strings in L n with y = x + δ i . Then for all j ∈ [1 , n ] with j / ∈ { i − , i + 1 } we have x + δ j ∈ L n if and only if y + δ j ∈ L n . Proof.
This is true for j = i thus assume j = i . • If x j = 1 then y j = x j = 1 and both x + δ j and y + δ j belong to L n .7 Assume x j = 0 thus y j = 0. We have x + δ j ∈ L n if and only if x k = 0 for all k ∈ { j − , j + 1 } and y + δ j ∈ L n if and only if y k = 0 for all k ∈ { j − , j + 1 } . But i / ∈ { j − , j + 1 } thus x k = y k for all k in this set and the two conditionsare equivalent. (cid:3) From this two lemmas we deduce the equivalent for Lucas cube of Corollary 2.3.
Corollary 3.3
Let n ≥ then irr (Λ n ) is the number of couples ( e, e ′ ) ∈ E (Λ n ) where e ′ is an imbalanced edge for e . Let e ′ = { y, y + δ j } be an imbalanced edge for e then by Lemma 3.2 we have j = i +1 or j = i − . We will call e ′ a cyclically right or cyclically left imbalanced edge for e accordingly. Let R i Λ n be the set of ( e, e ′ ) where e ′ is a cyclically right imbalanced edgefor e and e uses the direction i . Similarly let L i Λ n be the equivalent set for cyclicallyleft imbalanced edges. Theorem 3.4
Let n ≥ and i ∈ [1 , n ] . There exists a one to one mapping between R i Λ n or L iλ n and F n − . Proof.
Since x x . . . x n x i x i +1 . . . x n x x . . . x i − is an automorphism of Λ n wecan assume without loss of generality that i = 1. Let ( e, e ′ ) in R n . Assume that x is the upper endpoint of e = { x, y } . We have thus y = x + δ and x = 1. Let θ (( e, e ′ )) = x x . . . x n − . As a substring of x the string x x . . . x n − belongs to F n − .Furthermore since e and e ′ use the directions 1 and 2 we have x n = x = x = 0.Therefore θ (( e, e ′ )) = x x . . . x n − defines x thus defines ( e, e ′ ) and θ is injective.Conversely let z z . . . z n − be an arbitrary string of F n − . Let x = 100 z z . . . z n − t = 010 z z . . . z n − e = { x, x + δ } and e ′ = { t, t + δ } . Note that t + δ = x + δ and x + δ / ∈ L n thus by Lemma 3.2 ( e, e ′ ) ∈ R n . Therefore θ is surjective. The proof that φ (( e, e ′ )) = x x . . . x n − where e = { x, x + δ } defines a one to one mapping between L λ n and F n − is similar. (cid:3) As an immediate corollary we deduce the result obtained in [6]
Corollary 3.5
For all n ≥ irr (Λ n ) = 2 n | F n − | . Like in Γ n it is not necessary to know the degree of his endpoints for computing theimbalance of an edge in Λ n . 8 mb ( { x, x + δ i } ) x i − x i +2 imb ( e ) in Λ n Theorem 3.6
Let n ≥ and e = { x, y } be an edge of Λ n with y = x + δ i . Then imb ( e ) follows Table 2 where their indices i − and i + 2 are taken cyclically in [1 , n ] . Proof.
Assume that x is the upper endpoint of the edge. Since x i +1 = y i +1 = 0there exists a couple ( e, e ′ ) in R i Λ n if and only if e ′ = { y, y + δ i +1 } and x i +2 = 0. Since x i − = y i − = 0 there exists a couple ( e, e ′ ) in L i Λ n if and only if e ′ = { y, y + δ i − } and x i − = 0. Therefore imb ( e ) is completely determined by the values of x i +2 , x i − according to Table 2. (cid:3) Let e = { x, x + δ i } be an edge of Λ n and θ ( e ) = x i +1 x i +2 . . . x n x x . . . x i − . Notethat θ is a one to one mapping between the set of edges using the direction i and F n − .This remark gives a combinatorial interpretation of the well known result | E (Λ n ) | = nF n − [13]. We will use the same idea for the number edges with a given imbalance. Corollary 3.7
Let n ≥ then the imbalance of any edge e = { x, y } in Λ n is at most 2.Furthermore if A , B and C are the sets of edges with imbalance 0,1 and 2 respectivelythen | A | = nF n − , | B | = 2 nF n − and | C | = nF n − . Proof.
For i ∈ [1 , n ] let E i be the set of edges { x, y } using direction i . Since thenumber of edges in E i with a given imbalance is independent of i we can assume withoutloss of generality that i = 1 and consider A = A ∩ E , B = B ∩ E and C = C ∩ C .Let x be the end point such that x = 1. We have thus x = x n = 0, x + δ / ∈ L n and x + δ n / ∈ L n . • Assume x = x n − = 0. Then y + δ ∈ L n , y + δ n ∈ L n and the edge { x, y } belongs to C . Furthermore θ ( x ) = x x . . . x n − is one to one mapping betweenthe set of this kind of edges and F n − . The contribution of this case to C is F n − . • Assume x = x n − = 1. Then y + δ / ∈ L n , y + δ n / ∈ L n and the edge { x, y } belongs to A . Since x = x n − = 0 , θ ( x ) = x x . . . x n − is one to one mappingbetween the set of this kind of edges and F n − . The contribution of this case to A is F n − . • Assume x = 1 and x n − = 0. Then y + δ / ∈ L n , y + δ n ∈ L n . The edge { x, y } belongs to B . Furthermore θ ( x ) = x x . . . x n − is one to one mapping between9he set of this kind of edges and F n − . The contribution of this case to B is F n − . • The case x = 0 and x n − = 1 is similar and thus contributes also F n − to B (cid:3) Let F n be the set of binary strings of length n with 11 as substring. We will call thestrings in F n non-Fibonacci strings of length n . The cube complement of Γ n is Γ n thesubgraph of Q n induced by F n .Note that F n is connected since there is always a path between any vertex x ∈ V (Γ n )and 1 n . Furthermore | V (Γ n ) | = 2 n − F n +2 .Let A n , B n , C n be the sets of edges of Q n incident to 0,1 and 2, respectively, strings of F n . We have thus A n = E (Γ n ) and C n = E (Γ n ). Proposition 4.1 | E ( Q n ) | = | E (Γ n ) | + | B n | + | E (Γ n ) | is the total number of ’s inbinary strings of length n . | B n | + | E (Γ n ) | is the total number of ’s in Fibonacci strings of length n . | B n | + | E (Γ n ) | is the total number of ’s in non-Fibonacci strings of length n . | E (Γ n ) | is the total number of ’s in non-Fibonacci strings of length n . | E (Γ n ) | is the total number of ’s in Fibonacci strings of length n . Proof.
Let e be an edge of Q n and let x, y such that e = { x, y } with x i = 0 and y i = 1. Define the mappings φ ( { x, y } ) = ( x, i ) and ψ ( { x, y } ) = ( y, i ). Note that φ is a one to one mapping between E ( Q n ) and { ( s, i ); s ∈ B n , s i = 0 } the set of 0’sappearing in strings of B n . Likewise ψ is a one to one mapping between E ( Q n ) and { ( s, i ); s ∈ B n , s i = 1 } the set of 1’s appearing in strings of B n .Furthermore an edge incident to exactly one Fibonacci string z is mapped by φ to( z, i ) and by ψ to ( z + δ i , i ). Therefore the restriction of the reverse mapping φ − to { ( s, i ); s ∈ F n , s i = 0 } the set of 0’s appearing in strings of F n , is a one to one mappingto the edges of E (Γ n ) ∪ B n . For the same reason the restriction of the reverse mapping ψ − to { ( s, i ); s ∈ F n , s i = 1 } , the set of 1’s appearing in strings of F n , is a one to onemapping to the edges of E (Γ n ) ∪ B n .Since a binary string is a Fibonacci string or a non-Fibonacci string we can deducethe last two affirmations from the previous. We can also give a direct proof. Indeedthe restriction of φ to edges of Γ n define a one to one mapping between E (Γ n ) and { ( s, i ); s ∈ F n , s i = 0 } . Likewise the restriction of ψ to edges of Γ n define a one to onemapping between E (Γ n ) and { ( s, i ); s ∈ F n , s i = 1 } . (cid:3) roposition 4.2 The total number of number of ’s in Fibonacci strings of length n is P ni =1 F i +1 F n − i +2 . Proof.
Let s be a Fibonacci string of length n and i ∈ [1 , n ] then s s . . . s i − and s i +1 s i +2 . . . s n are Fibonacci strings. Reciprocally if u and v are Fibonacci strings then u v is also a Fibonacci string. Therefore the mapping define by θ ( s, i ) = ( s s . . . s i − , s i +1 s i +2 . . . s n )is a one to one mapping between { ( s, i ); s ∈ F n , s i = 0 } and the Cartesian product F i − × F n − i . The identity follows. (cid:3) Theorem 4.3
The number of edges of Γ n is given by the equivalent expressions:(i) | E (Γ n ) | = n n − − P ni =1 F i +1 F n − i +2 . (ii) | E (Γ n ) | = n n − − nF n +1 +(3 n − F n . Proof.
Combining the first two identities in Proposition 4.1 together with Proposition4.2 we obtain the first expression.For the second expression note first that the n edges of Q n incident to a vertex of F n belongs to E (Γ n ) or B n . Making the sum over all vertices of F n the edges of E (Γ n ) are obtained two times therefore nF n +2 = | B n | + 2 | E (Γ n ) | . By Proposition 4.1 | E (Γ n ) | = | E ( Q n ) | − | B n | − | E (Γ n ) | = n n − − nF n +2 + | E (Γ n ) | . Using the expressionof | E (Γ n ) | given by Proposition 1.2 we obtain the final result. (cid:3) The sequence ( | E (Γ n ) | , n ≥
1) = 0 , , , , , , . . . can also be obtain by aninductive relation: Proposition 4.4
The number of edges of Γ n is the sequence defined by | E (Γ n ) | = | E (Γ n − ) | + | E (Γ n − ) | + ( n + 4)2 n − − F n +2 ( n ≥ | E (Γ ) | = | E (Γ ) | = 0 . Proof.
Let n ≥
3. Let F n . be the set of strings of F n that begin with 1. Since F n . = { s ; s ∈ F n − } ∪ { s ; s ∈ B n − } we have |F n . | = 2 n − − F n . This identity isalso valid for n = 1 or n = 2.Consider the following partition of the set of vertices of Γ n : F n = { s ; s ∈ F n − } ∪{ s ; s ∈ F n − } ∪ { s ; s ∈ B n − } .From this decomposition the sequence of vertices of Γ n follows the induction | V (Γ n ) | = | V (Γ n − ) | + | V (Γ n − ) | + 2 n − . We deduce also a partition of the edges Γ n in six sets:- | E (Γ n − ) | edges between vertices of { s ; s ∈ F n − } .- | E (Γ n − ) | edges between vertices of { s ; s ∈ F n − } .11 | E ( Q n − | edges between vertices of { s ; s ∈ B n − } .-Edges between vertices of { s ; s ∈ F n − } and { s ; s ∈ F n − } . Those edges are the | V (Γ n − ) | edges { s, s } where s ∈ F n − .-Edges between vertices of { s ; s ∈ F n − } and { s ; s ∈ B n − } . Those edges are the | V (Γ n − ) | edges { s, s } where s ∈ F n − .-Edges between vertices of { s ; s ∈ F n − } and { s ; s ∈ B n − } . Those edges are the2 n − − F n − edges { s, s } where s is a string of F .n − .Therefore | E (Γ n ) | = | E (Γ n − ) | + | E (Γ n − ) | + ( n − n − + 2(2 n − − F n ) + 2 n − − F n − . (cid:3) We will call block of a binary string s a maximal substring of consecutive 1’s.Therefore a string in F n is as string with a least one block of length greater that 1.The degree of a vertex of Γ n lies between ⌊ ( n + 2) / ⌋ and n . The number of verticesof a given degree is determined in [12].We will now give a similar result for Γ n . Theorem 4.5
The degree of a vertex in Γ n is n , n − or n − and the number ofvertices of a given degree are: | E (Γ n − ) | vertices of degree n − | E (Γ n − ) | vertices of degree n − P n − k =0 k | E (Γ n − k − ) | vertices of degree n . Remark 4.6
Using Proposition 1.2 these numbers can be rewritten as, respectively, ( n − F n +(2 n ) F n − , ( n − F n − +(2 n − F n − and n − (3 n +7) F n +( n +5) F n − . Proof.
This is true for n ≤ n ≥
4. Let x be a vertex of Γ n and consider the indices i l and i r such that x i l x i l +1 and x i r x i r +1 are the leftmost,rightmost respectively, pairs of consecutive 1’s. Thus i l = min { i ; x i x i +1 = 11 } and i r = max { i ; x i x i +1 = 11 } . Consider the three possible cases • i r = i l . Then there exists a unique block of length at least 2 and this block x i l x i l +1 is of length 2. Thus x . . . x i l − ∈ F . i l − and x i l +2 . . . x n ∈ F .n − i l − .For i = i l or i = i l + 1 the string x + δ i is a Fibonacci string. For i distinct of i l and i l + 1 then x + δ i is a string of F n . Therefore d ( x ) = n − x . . . x i l − and x i l +2 . . . x n are arbitrary strings of F . i l − and F .n − i l − thenumber of vertices of this kind is by Proposition 1.1 P n − i l =1 F i l F n − i l = | E (Γ n − ) | . • i r = i l + 1. Then there exists a unique block of length at least 2 and thisblock x i l x i l +1 x i l +2 is of length 3. Thus x = x . . . x i l − x i l +3 . . . x n where x . . . x i l − ∈ F . i l − and x i l +3 . . . x n ∈ F .n − i l − .For i = i l + 1 the string x + δ i is a Fibonacci string. For i distinct of i l + 1 then x + δ i is a string of F n . Therefore d ( x ) = n − x . . . x i l − and x i l +3 . . . x n are arbitrary strings of F . i l − and F .n − i l − thenumber of vertices of this kind is P n − i l =1 F i l F n − i l − = | E (Γ n − ) | . • i r ≥ i l + 2. Then there exists a unique block of length at least 4 or there exist atleast two blocks of length at least 2. In both cases for any i ∈ [1 , n ] x + δ i is astring of F n . Therefore d ( x ) = n .Let k = i r − i l −
2. Note that k ∈ [0 , n −
4] and k fixed i l ∈ [1 , n − k − x x . . . x i l − and x i l + k +4 x i l + k +5 . . . x n are arbitrary strings in F . i l − and F .n − k − i l − . Since x i l +2 . . . x i r − is an arbitrary string in B k the number of verticesof this kind is P n − k =0 P n − k − i l =1 k F i l F n − k − i l − = P n − k =0 k | E (Γ n − k − ) | . (cid:3) The sequence 0 , , , , , , , . . . formed by the numbers of vertices on degree nin Γ n , ( n ≥
1) already appears in OEIS [15] as sequence A235996 of the number oflength n binary words that contain at least one pair of consecutive 0’s followed by (atsome point in the word) at least one pair of consecutive 1’s. This is clearly the samesequence.As noticed in Figure 1 Γ and Γ are isomorphic to Γ and Γ respectively. Ourlast result complete this observation. Theorem 4.7
For n ≥ n is isomorphic to an induced subgraph of Γ n . Proof.
Let n ≥ n by θ ( x ) = θ ( x x . . . x n ) = x x x x x x . . . x n . Let σ be the permutation on { , , . . . , n } define by σ (1) = 4, σ (4) = 1 and σ ( i ) = i for i / ∈ { , } .Note first that x ∈ F n implies θ ( x ) ∈ F n . Indeed since x x = 11 we have threecases • x x = 00 then x x = 11 is a substring of θ ( x ) • x x = 10 then x = 0 and x x = 11 is a substring of θ ( x ) • x x = 01 then x = 0 and x x = 11 is a substring of θ ( x ).Therefore θ maps vertices of Γ n to vertices in Γ n .Let { x, x + δ i } be an edge of Γ n then by construction we have θ ( x + δ i ) = θ ( x ) + δ σ ( i ) and therefore θ ( x ) and θ ( x + δ i ) are adjacent in Γ n .Since θ is a transposition we have also for all i ∈ [1 , n ] that θ ( x ) + δ i = θ ( x + δ σ ( i ) ).Therefore if { θ ( x ) , θ ( y ) } is an edge in the subgraph induced by θ (Γ n ) then θ ( y ) = θ ( x ) + δ i for some i , y = x + δ σ ( i ) and thus { x, y } ∈ E (Γ n ). (cid:3) Since 0 n is a vertex of degree n in Γ n this graph cannot be a subgraph of Γ m for m < n thus this mapping in Γ n is optimal. Conversely it might be interesting to13etermine the minimum m such that Γ n is isomorphic to an induced subgraph of Γ m .We already know that m ≤ n − Q n is an induced subgraph ofΓ n − [10]. References [1] Michael Albertson. The irregularity of a graph.
Ars Comb. , 46, 08 1997.[2] Yaser Alizadeh, Emeric Deutsch, and Sandi Klavˇzar. On the irregularity of π -permutation graphs, fibonacci cubes, and trees. Bulletin of the Malaysian Mathe-matical Sciences Society , 43, 04 2020.[3] Aline Castro, Sandi Klavˇzar, Michel Mollard, and Yoomi Rho. On the dominationnumber and the 2-packing number of fibonacci cubes and lucas cubes.
Computersand Mathematics with Applications , 61(9):2655 – 2660, 2011.[4] Aline Castro and Michel Mollard. The eccentricity sequences of fibonacci and lucascubes.
Discrete Mathematics , 312(5):1025 – 1037, 2012.[5] Ernesto Ded´o, Damiano Torri, and Norma Zagaglia Salvi. The observability of thefibonacci and the lucas cubes.
Discrete Mathematics , 255(1):55 – 63, 2002.[6] ¨Omer Egecioglu, Elif Saygı, and Z¨ulf¨ukar Saygi. The irregularity polynomialsof fibonacci and lucas cubes.
Bulletin of the Malaysian Mathematical SciencesSociety , 07 2020.[7] W.-J. Hsu. Fibonacci cubes-a new interconnection topology.
IEEE Transactionson Parallel and Distributed Systems , 4(1):3–12, 1993.[8] Sandi Klavˇzar. On median nature and enumerative properties of fibonacci-likecubes.
Discrete Mathematics , 299:145–153, 08 2005.[9] Sandi Klavˇzar. Structure of fibonacci cubes: a survey.
Journal of CombinatorialOptimization , 25:505–522, 2013.[10] Sandi Klavˇzar and Michel Mollard. Cube polynomial of fibonacci and lucas cubes.
Acta Applicandae Mathematicae , 117, 02 2012.[11] Sandi Klavˇzar and Michel Mollard. Daisy cubes and distance cube polynomial.
European Journal of Combinatorics , 80, 05 2017.[12] Sandi Klavˇzar, Michel Mollard, and Marko Petkovˇsek. The degree sequence offibonacci and lucas cubes.
Discrete Mathematics , 311(14):1310 – 1322, 2011.[13] Emanuele Munarini, Claudio Perelli Cippo, and Norma Salvi. On the lucas cubes.
The Fibonacci Quarterly , 39, 02 2001.1414] Mark Ramras. Congestion-free routing of linear permutations on fibonacci andlucas cubes.
Australasian Journal of Combinatorics , 60:1–10, 01 2014.[15] Neil J. A. Sloane and The OEIS Foundation Inc. The on-line encyclopedia ofinteger sequences, 2020.[16] Aleksander Vesel. Cube-complements of generalized fibonacci cubes.