Connectivity of Fibonacci cubes, Lucas cubes and generalized cubes

If f is a binary word and d a positive integer, then the generalized Fibonacci cube Q d ( f ) is the graph obtained from the d -cube Q d by removing all the vertices that contain f as a factor, while the generalized Lucas cube Q d (↽— f ) is the graph obtained from Q d by removing all the vertices that have a circulation containing f as a factor. The Fibonacci cube Γ d and the Lucas cube Λ d are the graphs Q d ( 11 ) and Q d (↽— 11 ) , respectively. It is proved that the connectivity and the edge-connectivity of Γ d as well as of Λ d are equal to ⌊ d + 23 ⌋ . Connected generalized Lucas cubes are characterized and generalized Fibonacci cubes are proved to be 2-connected. It is asked whether the connectivity equals minimum degree also for all generalized Fibonacci/Lucas cubes. It was checked by computer that the answer is positive for all f and all d ≤ 9 .


Introduction
Fibonacci cubes [4] and Lucas cubes [16] form hypercube-like classes of graphs that have found several applications and were extensively studied so far, see the recent survey [9].The topics studied include different metric aspects [1,10,11,13], a number of computer science issues [3,18,20,21], applications in chemistry [23,24,25], and a variety of additional topics [2,15,17].It is hence quite surprising, that, to the best of our knowledge, the connectivity of these cubes has not yet been established.This is even more surprising after recalling that Fibonacci cubes were originally introduced as a model for interconnection networks.Actually, in the seminal paper on the Fibonacci cubes it was stated without a proof that if d ≥ 3, Theorem 3].Moreover, an exact value for the connectivity of Fibonacci cubes (in a more general framework) was asserted in [5, Theorem 2, point 3], but no proof was provided and the stated result also does not appear to be correct (at least for Fibonacci cubes).
In this paper we fill this gap by determining the vertex-and the edge-connectivity of Fibonacci cubes and of Lucas cubes, see Section 2. In the subsequent section we prove that the generalized Fibonacci cubes are always 2-connected while in Section 4 we characterize connected generalized Lucas cubes.In the final section we ask whether the connectivity of all generalized Fibonacci/Lucas cubes equals the minimum degree.Using a computer, the answer turned out to be positive for all strings f and all dimensions d ≤ 9.
In the rest of this section we recall the basic concepts needed and notation used in this paper.The d-cube Q d , d ≥ 0, is the graph whose vertices are the binary words (alias strings) of length d, two vertices are adjacent if they differ in exactly one bit.In particular, Q 0 = K 1 and Q 1 = K 2 .The vertex deleted d-cube, that is, the graph obtained from Q d by removing one of its vertices, will be denoted by Q − d .The Fibonacci cube Γ d , d ≥ 0, is the graph obtained from Q d by removing all vertices that contain two consecutive 1s, while the Lucas cube Λ d , d ≥ 0, is obtained from Γ d by further removing the vertices that start and end with 1.We will use the concatenation notation, in particular, for a string u, the notation u = 1u ′ means that u starts with 1, and 1 d denotes the string consisting of d 1s.As usual, for a graph G, its (vertex-)connectivity and edge-connectivity will be denoted by κ(G) and κ ′ (G), respectively, and its minimum degree with δ(G).We will write G ≅ H to denote that G and H are isomorphic graphs.Finally, the Fibonacci numbers are defined by F 0 = 0, F 1 = 1, and Recall (cf.[9]) that V (Γ d ) = F d+2 .From [12] (cf.Corollary 3.3 and the last remark in Section 5) we also recall:

Fibonacci cubes and Lucas cubes
3 ⌋.Our first main result now reads as follows.
Proof: The result can be checked for d ≤ 5 by inspection.Suppose now that the result is true for d ≤ 3k+2, k ≥ 1. Continuing by induction we are going to prove the result for d = 3k + 3, d = 3k + 4, and d = 3k + 5.
Let d = 3k + 4. We need to show that κ(Γ 3k+4 ) = k + 2 and using Lemma 2.1 we only need to prove that κ(Γ 3k+4 ) ≥ k + 2. Let X 3k+2 ≅ Γ 3k+2 and X 3k+3 ≅ Γ 3k+3 be the subgraphs of Γ 3k+4 induced by the fundamental decomposition A 3k+4 and B 3k+4 , respectively.By the induction hypothesis and the already proved case where d = 3k + 3, κ(X 3k+2 ) = k + 1 and κ(X 3k+3 ) = k + 1. Suppose that Γ 3k+4 contains a separating set S with S = k + 1.If S ⊆ A 3k+4 , then since any vertex of A 3k+4 ∖ S has a neighbor in B 3k+4 , the graph Γ 3k+4 ∖ S is connected.Similarly, if both A 3k+4 and B 3k+4 contain some vertices of S, then X 3k+2 ∖ S and X 3k+3 ∖ S are both connected and so is Γ 3k+4 ∖ S. Assume finally that S ⊆ B 3k+4 and consider the fundamental decomposition of X 3k+3 ≅ Γ 3k+3 .It decomposes into Y ≅ Γ 3k+1 and Z ≅ Γ 3k+2 that are (by induction) both of connectivity k + 1.Since S disconnects X 3k+3 and S = k + 1, we infer that S ⊆ V (Z), that is, every vertex of S, considered as a vertex of Z, begins with 0. It follows that any vertex of S is considered as a vertex of X 3k+3 and starts with 00.Now, the subgraph of X 3k+3 induced by the vertices starting with 01 is connected.Moreover, since there are V (Γ 3k+1 ) = F 3k+3 > k + 1 independent edges between the vertices of X 3k+3 starting with 010 and the vertices starting with 000, there is a vertex of X 3k+3 starting with 01 that has a neighbor starting with 00 in the graph Γ 3k+4 ∖ S.This vertex has in turn a neighbor in X 3k+2 .Recalling that A 3k+4 ∩ S = ∅ and using the fact that any vertex of X 3k+2 ∖ S has a neighbor in X 3k+3 , we conclude that Γ 3k+4 ∖ S is connected.
The last case to consider is , and X 3k+4 ≅ Γ 3k+4 have the same meaning as in the previous cases.By the already proved previous cases, κ(X 3k+3 ) = k + 1 and κ(X 3k+4 ) = k + 2. We need to show that κ(Γ 3k+5 ) ≥ k + 2 and assume that Γ 3k+5 contains a separating set S with S = k + 1.But now S cannot lie completely in X 3k+4 (as κ(X 3k+4 ) = k + 2), while in the other cases we can argue again as we did in the first paragraph to conclude that S cannot be a separating set.
Hence For the Lucas cubes we have a result parallel to Theorem 2.2 with a single exception: Proof: We proceed similarly as in the proof of Theorem 2.2, however not all arguments will be parallel.Moreover, in the proof we will apply Theorem 2.2.First, we have checked the result for d ≤ 8 by computer.
Let d = 3k + 4. We need to show that κ(Λ 3k+4 ) = k + 2. By Lemma 2.1 we only need to prove that κ(Λ 3k+4 ) ≥ k + 2. Let X 3k+1 ≅ Γ 3k+1 and X 3k+3 ≅ Γ 3k+3 be the subgraphs of Λ 3k+4 induced on the fundamental decomposition, A 3k+4 and B 3k+4 , respectively.By Theorem 2.2, κ(X 3k+1 ) = k + 1 and κ(X 3k+3 ) = k + 1. Suppose that Λ 3k+4 contains a separating set S with S = k + 1.If S ⊆ A 3k+4 , then since any vertex in A 3k+4 ∖ S has a neighbor in B 3k+4 , the graph Λ 3k+4 ∖ S is connected.Similarly, if both A 3k+4 and B 3k+4 contain some vertices of S, then X 3k+1 ∖ S and X 3k+3 ∖ S are both connected and so is Λ 3k+4 ∖ S. Assume finally that S ⊆ B 3k+4 and consider the decomposition of X 3k+3 into the subgraphs Y and Z, whose vertices start with 01 and 00, respectively where each vertex in V (Y ) has a neighbor in V (Z).Then Y ≅ Γ 3k+1 and Z ≅ Γ 3k+2 and they are both of connectivity k + 1 by Theorem 2.2.If S ⊆ V (Y ), then as any vertex in V (Y ∖ S) has a neighbor in V (Z), Λ 3k+4 ∖ S is connected.If both Y and Z contain some vertices of S, then Y ∖ S and Z ∖ S are both connected.Considering that S < V (Y ) < V (Z) , we conclude that X 3k+3 ∖ S is connected and so is Λ 3k+4 ∖ S. Assume S ⊆ V (Z).Decompose Z into the subgraphs C and D, whose vertices end with 0 and 1, respectively.Then C ≅ Γ 3k+1 and D ≅ Γ 3k that are of connectivity k +1 and k, respectively by Theorem 2.2.Also decompose Y into the subgraphs E and H, whose vertices end with 0 and 1, respectively.Then E ≅ Γ 3k and H ≅ Γ 3k−1 are of connectivity k by Theorem 2.2.Note that every vertex in V (Y ) has a neighbor in V (Z).Also each vertex in V (D) has a neighbor in V (C) and each vertex in V (H) has a neighbor in V (E).As S < V (E) , there is a vertex x ∈ V (C ∖ S) which has a neighbor x ′ in V (E) ⊆ V (Y ).Also x has a neighbor x ′′ in A 3k+4 .Considering that X 3k+1 and Y are both connected, is connected, we only need to show that all the vertices in V (D ∖ S) are connected to some vertex in V (Y ) ∪ V (C ∖ S).As S < V (H) , there is a vertex y in V (D ∖ S) which has a neighbor in V (H) ⊆ V (Y ).As long as S ∩ V (C) ≥ 2, we have S ∩ V (D) ≤ k − 1 and hence D ∖ S is connected.Therefore in this case all the vertices in V (D ∖ S) are connected to y and hence to the vertices in Finally let d = 3k +5, and let A 3k+5 , B 3k+5 , X 3k+2 ≅ Γ 3k+2 , and X 3k+4 ≅ Γ 3k+4 have the same meaning as before.By the already proved previous cases, κ(X 3k+2 ) = k + 1 and κ(X 3k+4 ) = k + 2. We can now proceed as in the last part of the proof of Theorem 2.2.◻

Generalized Fibonacci cubes
Fibonacci cubes and Lucas cubes were recently extended to generalized Fibonacci cubes [6] and to generalized Lucas cubes [7] as follows.If f is an arbitrary binary word and d is a positive integer, then the generalized Fibonacci cube Q d (f ) is the graph obtained from Q d by removing all the vertices that contain f as a factor.We should point out that earlier the term generalized Fibonacci cubes was used in [14,22] for the special classes Q d (1 n ).Similarly, the generalized Lucas cube Q d ( ↽ f ) is the graph obtained from Q d by removing all the vertices that have a circulation containing f as a factor.Using these notations, It was observed in [8, p.2] that every generalized Fibonacci cube is connected.The case Q d (10) ≅ P d+1 is not interesting and we have treated the case Q d (11) = Γ d in the previous section.For any other forbidden string f we have the following general result.
Proof: Due to Lemma 3.1 we may assume throughout the proof that f begins with 0. In addition, we may also assume that d > f , where f denotes the cardinality of f .Indeed, if Let now u be an arbitrary vertex of Q d (f ) containing at least one 0. Let P 1 (u) be the path in Q d between u and 1 d that is obtained by changing from left to right one by one the bits 0 of u.This path lies completely in Q d (f ).Indeed, if f would be a substring of a vertex w that lies on P 1 (u), then since f starts with 0, only the bits left of this 0 would be changed.This would in turn imply that f would already be a substring of u ∈ V (Q d (f )).We now distinguish the following cases.
Let P 2 (u) be the path between u and 1 d that is obtained by changing from right to left one by one the bits 0 of u.Since f ends with 0, an argument parallel to the above implies that P 2 (u) lies completely in Q d (f ).By the construction, P 1 (u) and P 2 (u) are different, internally disjoint paths.Subcase 1.2: u = 1 r 01 s , r, s ≥ 0, r + s = d − 1.Since d ≥ 4, we may assume by the symmetry that s ≥ 2. Let now P 2 (u) be the path u → 1 r 001 s−1 → 1 r+1 01 s−1 → 1 d .Note that P 2 (u) lies in Q d (f ) and that it is internally disjoint from P 1 (u); the latter path in this case is u → 1 d .
In any of the two subcases, for any vertex We still need to consider the forbidden strings f that finish with 1, and first consider the strings that start with two zeros.
) holds for any d ≥ 1 by Lemma 4.1(i), we may without loss of generality Hence assume in the rest that 1 < w(f ) ≤ f 2. Then since w(f ) ≥ 2 and w(f ) ≤ f 2 we infer that f ≥ 4. We now distinguish the following cases.
Case 1: f has two blocks.We may without loss of generality assume that f = Consider the following path from 0 d to v which we obtain by changing from left to right the bits at even coordinates one by one: Each vertex of this path does not have a circulation which contains 11 as a substring and hence it does not have a circulation which contains f as a substring either.In other words, this path lies completely in Next consider the path from v to 1 d which we obtain by changing from left to right the 0 bits to 1 bits one by one.Certainly each vertex of this path does not have a circulation which contains 00 as a substring and hence it does not have a circulation which contains f as a substring either.Hence also this path lies in Q d ( ↽ f ).Therefore by concatenating the above two paths we obtain a path from In the second subcase let f = 1 p 0 q 1 r , where q ≥ 2 and p, r ≥ 1.Then first construct the path Since d − q − 1 ≥ 1 we can change in v all 0 bits one by one (in any order) to complete the above path from 0 d to 1 d .
Case 3: f has more than three blocks.Consider the following path 0 d → 10 d−1 → ⋯ → 1 d obtained by changing from left to right the 0 bits to 1 bits one by one.Each vertex in this path has only two blocks and hence it does not have a circulation which contains f as a substring.◻ We can now characterize the connected generalized Lucas cubes as follows.
Theorem 4.3 Let f be a binary string and d an integer such that d ) consists of an isolated vertex and a connected component containing all the other vertices.
Proof: Suppose first that w(f ) = 0, that is, f = 0 k .Let u be an arbitrary vertex of Q d ( ↽ 0 k ).Then changing one by one the 0 bits of u to 1, we stay in Q d ( ↽ 0 k ) and reach the vertex 1 d .Hence every vertex is connected with a path to 1 d and so Q d ( ↽ 0 k ) is connected.The case when w(f ) = 1 was done in [7, Proposition 10].Hence, having in mind Lemma 4.1(i), we can assume in the rest that 2 ≤ w(f ) ≤ f − 2.
Using Lemma 4.1 again we may without loss of generality assume that f starts with 0. For a string u let b(u) denote the length of its longest block of 1's in a circular manner.For instance, b(101110) = 3 and b(1100111011) = 4.We are going to show that each vertex in We distinguish the following two cases.
Suppose on the contrary that it contains a copy f of f as a (circular) substring.Then, as f starts with 0, f is contained in v i+b(v)+1 v i+b(v)+2 . . .v i+b(v)−1 .But this means that f is contained in a circulation of v as a substring also, a contradiction.Hence the claim is proved.We now proceed by changing one by one the bits 0 that appear after the position i + b(v).Again, all the obtained vertices lie in Q d ( ↽ f ).Indeed, if at some point the bit v j , j ≥ i + b(v) + 1, was changed, and if the obtained word would contain a copy f of f , then, using the fact that b(f ) ≤ b(v), f would be contained in v j+1 v j+2 . . .v i+b(v)−1 .So again v would contain f , the final contradiction.◻ The same kind of question can be asked for generalized Fibonacci cubes.The answers to the above questions are likely to be positive.In both cases, using computer, the connectivity was confirmed to be equal to the minimum degree for all forbidden strings f and for dimensions 4 ≤ d ≤ 9.The computations were performed using the Sage program [19].The same program was used for the computations to verify the base cases of Theorem 2.3.

Concluding remarks
If the answers to the above questions are indeed positive, an approach different from the one that we used for Fibonacci cubes and for Lucas cubes (that is, using their fundamental decompositions) will be needed in order to prove the corresponding theorems.
To determine the connectivity of Fibonacci cubes, we first recall the fundamental decomposition of Γ d .If d ≥ 1, then the vertex set of Γ d naturally partitions into the sets A d = {b 1 . . .b d b 1 = 1} and B d = {b 1 . . .b d b 1 = 0} .Since a string of A d , d ≥ 2, necessarily starts with 10, the set A d induces a subgraph of Γ d isomorphic to Γ d−2 .Similarly, B d induces a subgraph of Γ d isomorphic to Γ d−1 .Moreover, each vertex 1u of A d has exactly one neighbor in B d , namely the vertex 0u.In other words, the edges between A d and B d form a matching from A d to B d .
we have proved that κ(Γ d ) = δ(Γ d ).As for any graph G, the inequalities κ(G) ≤ κ ′ (G) ≤ δ(G) hold, the result follows.◻ Lucas cubes also admit a fundamental decomposition as follows.The vertex set of Λ d , d ≥ 1, partitions into the sets A d = {b 1 . . .b d b 1 = 1} and B d = {b 1 . . .b d b 1 = 0} .Since a string of A d , d ≥ 3, necessarily starts with 10 and ends with 0, the set A d induces a subgraph of Λ d isomorphic to Γ d−3 .Similarly, B d induces a subgraph of Λ d isomorphic to Γ d−1 .Moreover, each vertex 1u of A d has exactly one neighbor in B d , namely the vertex 0u.Thus the edges between A d and B d form a matching from A d to B d .
).For a binary string b = b 1 . . .b d , let b be the binary complement of b and let b R = b d . . .b 1 be the reverse of b.The following basic result helps to significantly reduce the number of cases to be considered.Lemma 3.1 [6, Lemmas 2.2 and 2.3] If f be a binary string and d ≥ 1, then Q

Lemma 4 . 2
Let f be a binary string with w(f ) ∈ [ f − 1], and d an integer such that d ≥ f ≥ 2. Then the vertices 0 d and 1 d of Q d ( ↽ f ) are in the same connected component if and only if w

Case 1 :
b(v) < b(f ).Consider a path from v to 0 d we obtain by changing one by one the bits 1 to 0. For any vertex u in this path we have b(u) ≤ b(v) < b(f ) and hence it does not have a circulation which contains f as a substring.We have therefore shown that any vertex of Q d ( ↽ f ) is connected with a path to 1 d .By Lemma 4.2 this case is done.

Theorem 4 .Question 5 . 1
3 thus asserts that for all practical purposes, each generalized cube can be considered as connected (by neglecting an isolated vertex).Confronting Theorem 4.3 with Theorem 3.2, a question arises whether all connected generalized Lucas cubes are actually 2-connected.More generally, we pose the following question:Is it true that κ(Q d ( ↽ f )) = δ(Q d ( ↽ f )) holds for all f and d, except for Q 4 ( ↽11))?

Question 5 . 2
Is it true that κ(Q d (f )) = δ(Q d (f ))for all f and d?