Edges in Fibonacci Cubes, Lucas Cubes and Complements

Abstract

The Fibonacci cube of dimension n, denoted as $\Gamma _n$, is the subgraph of the hypercube induced by vertices with no consecutive 1s. The Lucas cube $\Lambda _n$ is the cyclic version of $\Gamma _n$. The irregularity of a graph G is the sum of $|d(x)-d(y)|$ over all edges $\{x,y\}$ of G. In two recent papers based on the recursive structure of $\Gamma _n$, it is proved that the irregularity of $\Gamma _n$ and $\Lambda _n$ is two times the number of edges of $\Gamma _{n-1}$ and 2n times the number of vertices of $\Gamma _{n-4}$, respectively. Using an interpretation of the irregularity in terms of couples of incident edges of a special kind, we give a bijective proof of both results. For these two graphs, we deduce also a constant time algorithm for computing the imbalance of an edge. In the last section using the same approach, we determine the number of edges and the sequence of degrees of the cube complement of $\Gamma _n$.

1 Introduction

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 popular interconnection network because of its structural properties.

The Fibonacci cube of dimension n, denoted as $\Gamma _n$, is the subgraph of the hypercube induced by vertices with no consecutive 1s. This graph was introduced in [7] as a new interconnection network.

$\Gamma _n$ is an isometric subgraph of the hypercube which is inspired by the Fibonacci numbers. It has attractive recurrent structures such as its decomposition into two subgraphs which are also Fibonacci cubes themselves. Structural properties of these graphs were more extensively studied afterwards; see for example the survey [9].

Lucas cubes $\Lambda _n$, introduced in [13], have attracted attention as well due to the fact that these cubes are the cyclic version of Fibonacci cubes. They have also been widely studied [3,4,5, 10, 12, 14]. The determination of degree sequence [12] is one of the first enumerative results about Fibonacci cubes.

Let $G=(V(G),E(G))$ 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 $e=\{x,y\}\in E(G)$ is defined by $imb_G(e)=|d_G(x)-d_G(y)|$. The irregularity of a graph G is

$$\begin{aligned} irr(G)=\sum _{e\in E(G)}{imb_G(e)}. \end{aligned}$$

The irregularity of a regular graph is evidently 0 and this concept of irregularity was introduced in [1] as a measure of graph’s global non-regularity.

Using the inductive structure of Fibonacci cubes, it is proved in two recent papers [2, 6] that $irr(\Gamma _n)=2|E(\Gamma _{n-1})|$ and $irr(\Lambda _n)=2n|V(\Gamma _{n-4})|$. Our main motivation is to give direct bijective proofs of these remarkable properties.

The generalized Fibonacci cube $\Gamma _n(s)$ is the graph obtained from $Q_n$ by removing all vertices that contain a given binary string s as a substring. For example $\Gamma _n(11)=\Gamma _n$. Daisy cubes are an other kind of generalization of Fibonacci cubes introduced in [11].

For an induced subgraph G of $Q_n$, the cube-complement of G is the graph induced by the vertices of $Q_n$ which are not in G. In [16], the questions of whether the cube complement of generalized Fibonacci cube is connected, an isometric subgraph of a hypercube or a median graph is studied. It is also proved in the same paper that the cube-complement of a daisy cube is a daisy cube. We consider in the last section ${\overline{\Gamma }}_n$ the cube complement of $\Gamma _n$.

We give the number of edges of ${\overline{\Gamma }}_n$ and determine, using the main lemma of the first section, the degree sequence of ${\overline{\Gamma }}_n$. We will also study the embedding of $\Gamma _n$ in ${\overline{\Gamma }}_n$.

2 Preliminaries

We will next give some concepts and notations needed in this paper. We denote by [1, n] the set of integers i such that $1\le i \le n$. The vertex set of the hypercube of dimension n $Q_n$ is the set ${{\mathcal {B}}}_n$ of binary strings of length n, two vertices being adjacent iff they differ in precisely one position. We will denote by $\overline{x_i}$ the binary complement of $x_i$.

Let $x=x_1\ldots x_n$ be a binary string and $i\in [1,n]$ we will denote by $x+\delta _i$ the string $x'_1\ldots x_n'$ where $x'_j=\overline{x_j}$ for $j=i$ and $x'_j=x_j$ otherwise. We will say that the edge $\{x,x+\delta _i\}$ uses the direction i. The endpoint x such that $x_i=1$ of an edge using the direction i will be called upper endpoint and $y=x+\delta _i$ the lower endpoint. In the Hasse diagram of the Boolean algebra, encoded as binary strings of length n for vertices, “upper” corresponds to the position above in the covering relation.

A Fibonacci string of length n is a binary string $b=b_1 b_2\ldots b_n$ with $b_i\cdot b_{i+1}=0$ for $1\le i<n$. In other words, a Fibonacci string is a binary string without 11 as substring.

The Fibonacci cube $\Gamma _n$ ($n\ge 1$) is the subgraph of $Q_n$ induced by the Fibonacci strings of length n. Because of the empty string $\epsilon$, $\Gamma _0 = K_1$.

A Fibonacci string b of length n is a Lucas string if $b_1 \, \cdot b_n \ne 1$. That is, a Lucas string has no two consecutive 1s including the first and the last elements of the string. The Lucas cube $\Lambda _n$ is the subgraph of $Q_n$ induced by the Lucas strings of length n. We have $\Lambda _0 = \Lambda _1 = K_1$.

Fig. 1

$\Gamma _2=\Lambda _2$, $\Gamma _3$, $\Lambda _3$, $\Gamma _4$, ${\overline{\Gamma }}_3$ and ${\overline{\Gamma }}_4$

Let $F_n$ be the nth Fibonacci number where $F_0=0$, $F_1=1$, $F_n=F_{n-1}+F_{n-2}$ for $n\ge 2$.

Denote by ${{\mathcal {F}}}_n$ and ${{\mathcal {L}}}_n$ the sets of strings of Fibonacci strings and Lucas strings of length n. Let ${{\mathcal {F}}}_n^{1.}$ and ${{\mathcal {F}}}_n^{0.}$ be the set of strings of ${{\mathcal {F}}}_n$ that begin with 1 and that do not begin with 1, respectively. Note that with this definition ${{\mathcal {F}}}_0^{0.}=\{\epsilon \}$ and ${{\mathcal {F}}}_0^{1.}=\emptyset$. Let ${{\mathcal {F}}}_n^{.0}$ be the set of strings of ${{\mathcal {F}}}_n$ that do not end with 1. Thus, $|{{\mathcal {F}}}_n^{.0}|=|{{\mathcal {F}}}_n^{0.}|$. Let ${{\mathcal {F}}}_n^{00}$ be the set of strings of ${{\mathcal {F}}}_n^{0.}$ that do not end with 1. With this definition ${{\mathcal {F}}}_0^{00}=\{\epsilon \}$, ${{\mathcal {F}}}_1^{00}=\{0\}$ and ${{\mathcal {F}}}_2^{00}=\{00\}$.

From ${{\mathcal {F}}}_{n+2}=\{0\varvec{s}; \varvec{s}\in {{\mathcal {F}}}_{n+1}\}\cup \{10\varvec{s}; \varvec{s}\in {{\mathcal {F}}}_{n}\},{{\mathcal {F}}}_{n+1}^{0.}=\{0\varvec{s}; \varvec{s}\in {{\mathcal {F}}}_{n}\} \text {and }{{\mathcal {F}}}_{n+1}^{1.}=\{1\varvec{s}; \varvec{s}\in {{\mathcal {F}}}_{n}^{0.}\}$, we obtain the following classical result.

Proposition 2.1

Let $n \ge 0$. The numbers of Fibonacci strings in ${{\mathcal {F}}}_{n}$, ${{\mathcal {F}}}_{n}^{0.}$ and ${{\mathcal {F}}}_{n}^{1.}$ are $|{{\mathcal {F}}}_{n}|=F_{n+2}$, $|{{\mathcal {F}}}_{n}^{0.}|=F_{n+1}$ and $|{{\mathcal {F}}}_{n}^{1.}|=F_{n}$, respectively. For $n \ge 1$, the number of Fibonacci strings in ${{\mathcal {F}}}_n^{00}$ is $|{{\mathcal {F}}}_n^{00}|=F_{n}$.

The following expressions for the number of edges in $\Gamma _{n}$ are obtained in [8] and [13].

Proposition 2.2

Let $n \ge 0$. The number of edges in $\Gamma _{n}$ is $|E(\Gamma _{n})|=\sum _{i=1}^n{F_{i}F_{n-i+1}}=\frac{nF_{n+1}+2(n+1)F_n}{5}$ and satisfies the recursion $|E(\Gamma _{n+2})|=|E(\Gamma _{n+1})|+|E(\Gamma _{n})|+|V(\Gamma _{n})|$ .

Remark 2.3

Let $\{x,x+\delta _i\}$ be an edge and $\theta (x)=((x_{1}x_{2}\ldots x_{i-1}), (x_{i+1} x_{i+2}\ldots x_{n}))$. A combinatorial interpretation of $|E(\Gamma _{n})|=\sum _{i=1}^n{F_{i}F_{n-i+1}}$ is that for any $i\in [1,n]$, $\theta$ is a one to one mapping between the set of edges using the direction i and the Cartesian product ${{\mathcal {F}}}_{i-1}^{.0}\times {{\mathcal {F}}}_{n-i}^{0.}$.

Let G be an induced subgraph of $Q_n$. Let $e=\{x,y\}$ be an edge of G where y is the lower endpoint of e and $x=y+\delta _i$. An edge $e'=\{y,y+\delta _j\}$ of G will be called an imbalanced edge for e if $x+\delta _j \notin V(G)$ (and thus $\{x,x+\delta _j\}\notin E(G)$). Note that such couple of edges does not exist for $G= Q_n$. We will prove in the next two sections that for $G=\Gamma _n$ and $G=\Lambda _n$, the irregularity of G is the number of such couples of edges (Fig. 2).

Fig. 2

$irr(\Gamma _n)$ and $irr(\Lambda _n)$ count the couples of edges $(e,e')$ of the right kind

3 Edges in Fibonacci Cube

Lemma 3.1

Let x, y be two strings in ${{\mathcal {F}}}_n$ with $y=x+\delta _i$ and $x_i=1$. Then for all $j\in [1,n]$, we have

$$\begin{aligned} x+\delta _j\in {{\mathcal {F}}}_n \text{ implies } y+\delta _j\in {{\mathcal {F}}}_n. \end{aligned}$$

Proof

Assume $y+\delta _j\notin {{\mathcal {F}}}_n$ then $y_k=1$ for some k in $\{j-1,j+1\}\cap [1,n]$. But for all $p\in [1,n]$ $x_p=0$ implies $y_p=0$. Thus, $x_k=1$ and $x+\delta _j\notin {{\mathcal {F}}}_n$. $\square$

Lemma 3.2

Let x, y be two strings in ${{\mathcal {F}}}_n$ with $y=x+\delta _i$. Then for all $j\in [1,n]$ with $|i-j|>1$, we have

$$\begin{aligned} x+\delta _j\in {{\mathcal {F}}}_n \text{ if } \text{ and } \text{ only } \text{ if } y+\delta _j\in {{\mathcal {F}}}_n. \end{aligned}$$

Proof

• If $x_j=1$ then $y_j=x_j=1$ and both $x+\delta _j$ and $y+\delta _j$ belong to ${{\mathcal {F}}}_n$.

• Assume $x_j=0$ thus $y_j=0$. We have

  $$\begin{aligned} x+\delta _j\in {{\mathcal {F}}}_n\text { if and only if } x_{k}=0 \text { for all } k\in \{j-1,j+1\}\cap [1,n] \end{aligned}$$

  and

  $$\begin{aligned} y+\delta _j\in {{\mathcal {F}}}_n\text { if and only if } y_{k}=0 \text { for all } k\in \{j-1,j+1\}\cap [1,n]. \end{aligned}$$

  But $i\notin \{j-1,j+1\}\cap [1,n]$ thus $x_k=y_k$ for all k in this set and the two conditions are equivalent.

$\square$

Corollary 3.3

Let $n\ge 2$. Then, $irr(\Gamma _n)$ is the number of couples $(e,e')\in E(\Gamma _n)^2$ where $e'$ is an imbalanced edge for e.

Proof

By Lemma 3.1, if $e=\{x,y\}$ is an edge using the direction i with upper endpoint x then $d(y)\ge d(x)$ and imb(e) is the number of imbalanced edges for e. The conclusion follows. $\square$

Furthermore assume that $e=\{x,y\}$ uses the direction i with $x_i=1$ and let $e'=\{y,y+\delta _j\}$ be an imbalanced edge for e. Then by Lemma 3.2, we have $j=i+1$ or $j=i-1$. Note that for $i=1$ or $i=n$, only one case is possible.

We will call $e'$ a right or left imbalanced edge for e accordingly. Note that for $i=1$ or $i=n$, only one case is possible. Let $R_{\Gamma _n}$ and $L_{\Gamma _n}$ be the sets of couples $(e,e')$ where $e'$ is a right imbalanced edge for e and a left imbalanced edge for e, respectively, where e goes through $E(\Gamma _n$).

Theorem 3.4

Let $n\ge 2$. There exists a one to one mapping between $R_{\Gamma _n}$ or $L_{\Gamma _n}$ and $E(\Gamma _{n-1})$.

Proof

Let $(e,e')\in R_{\Gamma _n}$. Assume that x is the upper endpoint of $e=\{x,y\}$. We have thus $y=x+\delta _i$ and $x_i=1$ for some $i\in [1,n-1]$.

Let $\theta ((e,e'))= \{x_1 x_2 \dots x_{i-1}1x_{i+2}x_{i+3}\dots x_{n},x_1 x_2 \dots x_{i-1}0x_{i+2}x_{i+3}\dots x_{n}\}$. Since x and y belong to ${{\mathcal {F}}}_n$ and the edges e, $e'$ use the direction i, $i+1$ we have $x_k=y_k=0$ for k in $\{i-1,i+2\}\cap [1,n]$. Therefore, $x_1 x_2 \dots x_{i-1}1x_{i+2}x_{i+3}\dots x_{n}$ is a Fibonacci string and $\theta ((e,e'))$ belongs to $E(\Gamma _{n-1})$.

Conversely, let $f=\{z_1 z_2 \dots z_{i-1}0z_{i+1}z_{i+2}\dots z_{n-1},z_1 z_2 \dots z_{i-1}1z_{i+1}z_{i+2}\dots z_{n-1}\}$ be an arbitrary edge of $\Gamma _{n-1}$ then $z_k=0$ for k in $\{i-1,i+1\}\cap [1,n-1]$. Thus, $x=z_1 z_2 \dots z_{i-1}10z_{i+1}z_{i+2}\dots z_{n-1}$ and $t=z_1 z_2 \dots z_{i-1}01z_{i+1}z_{i+2}\dots z_{n-1}$ are in ${{\mathcal {F}}}_n$. The edge $\{t,t+\delta _{i+1}\}$ is a right imbalanced edge for the edge $\{x,x+\delta _{i}\}$. Furthermore, $\theta (\{x,x+\delta _i\},\{t,t+\delta _{i+1}\})=f$ and $\theta$ is a bijection.

Similarly, let $\phi ((e,e'))= \{x_1 x_2 \dots x_{i-2}1x_{i+1}x_{i+2}\dots x_{n},x_1 x_2 \dots x_{i-2}0x_{i+1}x_{i+2}\dots x_{n}\}$ where x is the upper end point of an edge e using the direction i and such that $(e,e')\in L_{\Gamma _n}$. Then, $\phi$ is a one to one mapping between $L_{\Gamma _n}$ and $E(\Gamma _{n-1})$. $\square$

As an immediate corollary, we deduce the result of Alizadeh and his co-authors [2].

Corollary 3.5

$$\begin{aligned} irr(\Gamma _n)=2|E(\Gamma _{n-1})|. \end{aligned}$$

An other consequence of Lemma 3.2 is the following classification of the edges according to their imbalance. Note that from this classification we obtain a constant time algorithm for computing the imbalance of an edge of $\Gamma _n$.

Theorem 3.6

Let $n\ge 4$ and $e=\{x,y\}$ be an edge of $\Gamma _n$ using direction i. Then, $imb(\{x,y\})$ follows Table 1.

Table 1 imb(e) in $\Gamma _n$

Proof

Assume that x is the upper endpoint of the edge $e=\{x,y\}$. There exists an edge $e'$ such that $e'$ is a right imbalanced edge for e if and only if $i\in [1,n-1]$ and $e'=\{y,y+\delta _{i+1}\}$ thus if $y+\delta _{i+1}\in {{\mathcal {F}}}_n$. Since $y_i=0$, $y+\delta _{i+1}$ is a Fibonacci string if and only if $i=n-1$ or if $y_{i+2}=x_{i+2}=0$ in the general case $i\in [1,n-2]$.

Similarly, there exists an edge $e'$ such that $e'$ is a left imbalanced edge for e if and only if $i\in [2,n]$ and $e'=\{y,y+\delta _{i-1}\}$ thus if $y+\delta _{i-1}\in {{\mathcal {F}}}_n$. Since $y_i=0$, $y+\delta _{i-1}$ is a Fibonacci string if and only if $i=2$ or if $y_{i-2}=x_{i-2}=0$ when $i\in [3,n]$.

Therefore, imb(e) is completely determined by the values of $x_{i+2},x_{{i-2}}$ according to Table 1. $\square$

Let e be an edge of $\Gamma _n$ then by Lemma 3.2$imb(e)\le 2$. Let A, B, C be the sets of edges with $imb(e)=0$, $imb(e)=1$ and $imb(e)=2$, respectively.

Theorem 3.7

Let $n\ge 2$. The numbers of edges of $\Gamma _n$ with imbalance 0,1 and 2 are, respectively,

$$\begin{aligned} |A|= & {} \sum _{i=3}^{n-2}{F_{i-2}F_{n-i-1}}+2F_{n-2} \\ |B|= & {} 2\sum _{i=1}^{n-3}{F_{i}F_{n-i-2}}+2F_{n-1} \\ |C|= & {} \sum _{i=2}^{n-1}{F_{i-1}F_{n-i}}. \end{aligned}$$

Remark 3.8

Note that we can deduce immediately these three equalities from equation (19) in [6], and since $|B|+2|C|=2|E(\Gamma _{n-1})|$ we obtain again the result of Alizadeh and his co-authors.

Proof

The case $n\le 3$ is obtained by direct inspection.

Assume $n\ge 4$.

For $i\in [1,n]$, let $E_i$ be the set of edges $\{x,y\}$ of $\Gamma _i$ with $y=x+\delta _i$. Let $A_i=A\cap E_i$, $B_i=B\cap E_i$ and $C_i=C\cap E_i$. Let $e=\{x,y\}$ be an edge of $\Gamma _n$.

• If $e\in A_i$ then by Table 1 we have $i\in [3,n-2]$ or $i\in \{1,n\}$. If $i\in [3,n-2]$ then $\theta (e)=(x_{1}x_{2}\ldots x_{i-2}, x_{i+2} x_{i+3} \ldots x_{n})$ is a one to one mapping between $A_i$ and ${{\mathcal {F}}}_{i-2}^{.1}\times {{\mathcal {F}}}_{n-i-1}^{1.}$. If $i=1$ then $\phi (e)=x_{3} x_{4} \ldots x_{n}$ is a one to one mapping between $A_1$ and ${{\mathcal {F}}}_{n-2}^{1.}$. Similarly, $\Psi (e)=x_{1} x_{2} \ldots x_{n-2}$ is a one to one mapping between $A_n$ and ${{\mathcal {F}}}_{n-2}^{.1}$. By Proposition 2.1, we obtain $|A|=\sum _{i=3}^{n-2}{F_{i-2}F_{n-i-1}}+2F_{n-2}$.

• If $e\in C_i$ then by Table 1 we have $i\in [2,n-1]$. Let $\theta (e)=(x_{1}x_{2}\ldots x_{i-2},x_{i+2} x_{i+3} \ldots x_{n})$. Then, $\theta$ is a one to one mapping between $C_i$ and ${{\mathcal {F}}}_{i-2}^{.0}\times {{\mathcal {F}}}_{n-i-1}^{0.}$. The expression of |C| follows.

• Assume $e\in B_i$ and that there exists a right imbalanced edge for e therefore no left imbalanced edge. We have thus $i\in [1,n-1]$ and $i\ne 2$. If $i\in [3,n-1]$ then $\theta (e)=(x_{1}x_{2}\ldots x_{i-2}, x_{i+2} x_{i+3} \ldots x_{n})$ is a one to one mapping this kind of edges and and ${{\mathcal {F}}}_{i-2}^{.1}\times {{\mathcal {F}}}_{n-i-1}^{0.}$. If $i=1$ then $\phi (e)=x_{3} x_{4} \ldots x_{n}$ is a one to one mapping between this kind of edges and ${{\mathcal {F}}}_{n-2}^{0.}$. Thus, this case contributes $\sum _{i=3}^{n-1}{F_{i-2}F_{n-i}}+F_{n-1}=\sum _{i=1}^{n-3}{F_{i}F_{n-i-2}}+F_{n-1}$ to B.

• Assume $e\in B_i$ and that there exists a left imbalanced edge for e thus no right imbalanced edge. By a similar construction, this case contributes also $\sum _{i=1}^{n-3}{F_{i}F_{n-i-2}}+F_{n-1}$ to B. The expression of |B| follows.

4 Edges in Lucas Cube

For any integer, i let $\varvec{i}=((i-1)\mod n) +1$. Thus, $\varvec{i}=i$ for $i\in [1,n]$ and $\varvec{n+1}=1$, $\varvec{0}=n$. With this notation, i and $\varvec{i+1}$ are cyclically consecutive in [1, n]. Therefore for $x\in {{\mathcal {L}}}_n$ with $x_i=0$, the string $x+\delta _i$ belongs to ${{\mathcal {L}}}_n$ if and only if $x_k=0$ for all $k\in \{\varvec{i-1},\varvec{i+1}\}$. Note also that $k\in \{\varvec{i-1},\varvec{i+1}\}$ if and only if $i\in \{\varvec{k-1},\varvec{k+1}\}$

Lemma 4.1

Let x, y be two strings in ${{\mathcal {L}}}_n$ with $y=x+\delta _i$ and $x_i=1$. Then for all $j\in [1,n]$, we have

$$\begin{aligned} x+\delta _j\in {{\mathcal {L}}}_n \text{ implies } y+\delta _j\in {{\mathcal {L}}}_n. \end{aligned}$$

Proof

Assume $y+\delta _j\notin {{\mathcal {L}}}_n$ then $y_k=1$ for some k in $\{\varvec{j-1},\varvec{j+1}\}$. But for all $p\in [1,n]$ $x_p=0$ implies $y_p=0$. Thus, $x_k=1$ and $x+\delta _j\notin {{\mathcal {L}}}_n$. $\square$

Lemma 4.2

Let x, y two strings in ${{\mathcal {L}}}_n$ with $y=x+\delta _i$. Then, for all $j\in [1,n]$ with $j\notin \{\varvec{i-1},\varvec{i+1}\}$ we have

$$\begin{aligned} x+\delta _j\in {{\mathcal {L}}}_n \text{ if } \text{ and } \text{ only } \text{ if } y+\delta _j\in {{\mathcal {L}}}_n. \end{aligned}$$

Proof

This is true for $j=i$ thus assume $j\ne i$.

• If $x_j=1$ then $y_j=x_j=1$ and both $x+\delta _j$ and $y+\delta _j$ belong to ${{\mathcal {L}}}_n$.

• Assume $x_j=0$ thus $y_j=0$. We have

  $$\begin{aligned} x+\delta _j\in {{\mathcal {L}}}_n\text { if and only if } x_{k}=0 \text { for all } k\in \{\varvec{j-1},\varvec{j+1}\} \end{aligned}$$

  and

  $$\begin{aligned} y+\delta _j\in {{\mathcal {L}}}_n\text { if and only if } y_{k}=0 \text { for all } k\in \{\varvec{j-1},\varvec{j+1}\}. \end{aligned}$$

  But $i\notin \{\varvec{j-1},\varvec{j+1}\}$ thus $x_k=y_k$ for all k in this set and the two conditions are equivalent.

$\square$

From these two lemmas, we deduce the equivalent for Lucas cube of Corollary 3.3.

Corollary 4.3

Let $n\ge 2$ then $irr(\Lambda _n)$ is the number of couples $(e,e')\in E(\Lambda _n)^2$ where $e'$ is an imbalanced edge for e.

Let $e'=\{y,y+\delta _j\}$ be an imbalanced edge for e then by Lemma 4.2 we have $j=\varvec{i+1}$ or $j=\varvec{i-1}$. We will call $e'$ a cyclically right or cyclically left imbalanced edge for e accordingly. Let $R^i_{\Lambda _n}$ be the set of $(e,e')$ where $e'$ is a cyclically right imbalanced edge for e and e uses the direction i. Similarly, let $L^i_{\Lambda _n}$ be the equivalent set for cyclically left imbalanced edges.

Theorem 4.4

Let $n\ge 4$ and $i\in [1,n]$. There exists a one to one mapping between $R^i_{\Lambda _n}$ or $L^i_{\Lambda _n}$ and ${{\mathcal {F}}}_{n-4}$.

Proof

Since $x_1 x_2\dots x_n\mapsto x_i x_{i+1}\dots x_n x_1 x_2 \dots x_{i-1}$ is an automorphism of $\Lambda _n$ we can assume without loss of generality that $i=1$. Let $(e,e')$ in $R^1_{\Lambda _n}$. Assume that x is the upper endpoint of $e=\{x,y\}$. We have thus $y=x+\delta _1$ and $x_1=1$. Let $\theta ((e,e'))= x_4 x_5 \dots x_{n-1}$. As a substring of x, the string $x_4 x_5 \dots x_{n-1}$ belongs to ${{\mathcal {F}}}_{n-4}$. Furthermore, since e and $e'$ use the directions 1 and 2 we have $x_n=x_2=x_3=0$. Therefore, $\theta ((e,e'))= x_4 x_5 \dots x_{n-1}$ defines x thus defines $(e,e')$ and $\theta$ is injective. Conversely, let $z_1 z_2 \dots z_{n-4}$ be an arbitrary string of ${{\mathcal {F}}}_{n-4}$. Let $x=100z_1 z_2\dots z_{n-4}0$, $t=010z_1 z_2\dots z_{n-4}0$, $e=\{x,x+\delta _1\}$ and $e'=\{t,t+\delta _2\}$. Note that $t+\delta _2=x+\delta _1$ and $x+\delta _2\notin {{\mathcal {L}}}_n$ thus by Lemma 4.2$(e,e')\in R^1_{\Lambda _n}$. Therefore, $\theta$ is surjective. The proof that $\phi ((e,e'))= x_3 x_4 \dots x_{n-2}$ where $e=\{x,x+\delta _1\}$ defines a one to one mapping between $L^1_{\Lambda _n}$ and ${{\mathcal {F}}}_{n-4}$ is similar. $\square$

As an immediate corollary, we deduce the result obtained in [6]:

Corollary 4.5

For all $n\ge 3$ $irr(\Lambda _n)= 2n|F_{n-2}|$.

Like in $\Gamma _n$, it is not necessary to know the degree of endpoints for computing the imbalance of an edge in $\Lambda _n$.

Theorem 4.6

Let $n\ge 4$ and $e=\{x,y\}$ be an edge of $\Lambda _n$ with $y= x+\delta _i$. Then, imb(e) follows Table 2 where their indices $\varvec{i-2}$ and $\varvec{i+2}$ are taken cyclically in [1, n].

Table 2 imb(e) in $\Lambda _n$

Proof

Assume that x is the upper endpoint of the edge. Since $x_{\varvec{i+1}}=y_{\varvec{i+1}}=0$ there exists a couple $(e,e')$ in $R^i_{\Lambda _n}$ if and only if $e'=\{y,y+\delta _{\varvec{i+1}}\}$ and $x_{\varvec{i+2}}=0$. Since $x_{\varvec{i-1}}=y_{\varvec{i-1}}=0$ there exists a couple $(e,e')$ in $L^i_{\Lambda _n}$ if and only if $e'=\{y,y+\delta _{\varvec{i-1}}\}$ and $x_{\varvec{i-2}}=0$. Therefore, imb(e) is completely determined by the values of $x_{\varvec{i+2}},x_{\varvec{i-2}}$ according to Table 2. $\square$

Let $e=\{x,x+\delta _i\}$ be an edge of $\Lambda _n$ and $\theta (e)=x_{i+1} x_{i+2}\ldots x_{n} x_{1} x_{2}\ldots x_{i-1}$. Note that $\theta$ is a one to one mapping between the set of edges using the direction i and ${{\mathcal {F}}}_{n-1}^{00}$. This remark gives a combinatorial interpretation of the well-known result $|E(\Lambda _n)|=nF_{n-1}$ [13]. We will use the same idea for a combinatorial interpretation for the number edges with a given imbalance already given by equation (26) in [6].

Corollary 4.7

Let $n\ge 5$ then the imbalance of any edge $e=\{x,y\}$ in $\Lambda _n$ is at most 2. Furthermore, if A, B and C are the sets of edges with imbalance 0,1 and 2, respectively, then $|A|=nF_{n-5}$, $|B|=2nF_{n-4}$ and $|C|=nF_{n-3}$.

Proof

For $i\in [1,n]$, let $E_i$ be the set of edges $\{x,y\}$ using direction i. Since the number of edges in $E_i$ with a given imbalance is independent of i we can assume without loss of generality that $i=1$ and consider $A_1=A\cap E_1$, $B_1=B\cap E_1$ and $C_1=C\cap C_1$. Let x be the end point such that $x_1=1$. We have thus $x_2=x_n=0$, $x+\delta _2\notin {{\mathcal {L}}}_n$ and $x+\delta _n\notin {{\mathcal {L}}}_n$.

• Assume $x_3=x_{n-1}=0$. Then, $y+\delta _2\in {{\mathcal {L}}}_n$, $y+\delta _n\in {{\mathcal {L}}}_n$, and the edge $\{x,y\}$ belongs to $C_1$. Furthermore, $\theta (x)= x_{3} x_{4}\ldots x_{n-1}$ is one to one mapping between the set of this kind of edges and ${{\mathcal {F}}}_{n-3}^{00}$. The contribution of this case to $C_1$ is $F_{n-3}$.

• Assume $x_3=x_{n-1}=1$. Then, $y+\delta _2\notin {{\mathcal {L}}}_n$, $y+\delta _n\notin {{\mathcal {L}}}_n$, and the edge $\{x,y\}$ belongs to $A_1$. Since $x_4=x_{n-2}=0,$ $\theta (x)= x_{4} x_{5}\ldots x_{n-2}$ is one to one mapping between the set of this kind of edges and ${{\mathcal {F}}}_{n-5}^{00}$. The contribution of this case to $A_1$ is $F_{n-5}.$

• Assume $x_3=1$ and $x_{n-1}=0$. Then, $y+\delta _2\notin {{\mathcal {L}}}_n$, $y+\delta _n\in {{\mathcal {L}}}_n$. The edge $\{x,y\}$ belongs to $B_1$. Furthermore, $\theta (x)= x_{4} x_{5}\ldots x_{n-1}$ is one to one mapping between the set of this kind of edges and ${{\mathcal {F}}}_{n-4}^{00}$. The contribution of this case to $B_1$ is $F_{n-4}$.

• The case $x_3=0$ and $x_{n-1}=1$ is similar and thus contributes also $F_{n-4}$ to $B_1$.

$\square$

5 Cube-Complement of Fibonacci Cube

Let ${\overline{{{\mathcal {F}}}}}_n$ be the set of binary strings of length n with 11 as substring. We will call the strings in ${\overline{{{\mathcal {F}}}}}_n$ non-Fibonacci strings of length n. The cube complement of $\Gamma _n$ is ${\overline{\Gamma }}_n$ the subgraph of $Q_n$ induced by ${\overline{{{\mathcal {F}}}}}_n$.

Note that ${\overline{{{\mathcal {F}}}}}_n$ is connected since there is always a path between any vertex $x\in V({{\overline{\Gamma }}_n})$ and $1^n$. Furthermore, $|V({{\overline{\Gamma }}_n})|=2^n-F_{n+2}$.

Let $A_n$,$B_n$,$C_n$ be the sets of edges of $Q_n$ incident to 0,1 and 2, respectively, strings of ${{\mathcal {F}}}_n$. We have thus $A_n=E({{\overline{\Gamma }}_n})$ and $C_n=E(\Gamma _n)$.

Proposition 5.1

$|E(Q_n)|=|E({\overline{\Gamma }}_n)|+|B_n|+|E(\Gamma _n)|$ is the total number of 0s in binary strings of length n.

$|B_n|+|E(\Gamma _n)|$ is the total number of 0s in Fibonacci strings of length n.

$|B_n|+|E({\overline{\Gamma }}_n)|$ is the total number of 1s in non-Fibonacci strings of length n.

$|E({\overline{\Gamma }}_n)|$ is the total number of 0s in non-Fibonacci strings of length n.

$|E(\Gamma _n)|$ is the total number of 1s 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 $\phi (\{x,y\})= (x,i)$ and $\psi (\{x,y\})= (y,i)$. Note that $\phi$ is a one to one mapping between $E(Q_n)$ and $\{(s,i); s\in {{\mathcal {B}}}_n,s_i=0\}$ the set of 0s appearing in strings of ${{\mathcal {B}}}_n$. Likewise, $\psi$ is a one to one mapping between $E(Q_n)$ and $\{(s,i); s\in {{\mathcal {B}}}_n,s_i=1\}$ the set of 1s appearing in strings of ${{\mathcal {B}}}_n$.

Furthermore, an edge incident to exactly one Fibonacci string z is mapped by $\phi$ to (z, i) and by $\psi$ to $(z+\delta _i,i)$. Therefore, the restriction of the reverse mapping $\phi ^{-1}$ to $\{(s,i); s\in {{\mathcal {F}}}_n,s_i=0\}$ the set of 0s appearing in strings of ${{\mathcal {F}}}_n$, is a one to one mapping to the edges of $E(\Gamma _n)\cup B_n$. For the same reason, the restriction of the reverse mapping $\psi ^{-1}$ to $\{(s,i); s\in {{\mathcal {F}}}_n,s_i=1\}$, the set of 1s appearing in strings of ${{\mathcal {F}}}_n$, is a one to one mapping to the edges of $E ({\overline{\Gamma }}_n)\cup B_n$.

Since a binary string is a Fibonacci string or a non-Fibonacci string we can deduce the last two affirmations from the previous. We can also give a direct proof. Indeed, the restriction of $\phi$ to edges of ${\overline{\Gamma }}_n$ defines a one to one mapping between $E({\overline{\Gamma }}_n)$ and $\{(s,i); s\in {\overline{{{\mathcal {F}}}}}_n,s_i=0\}$. Likewise, the restriction of $\psi$ to edges of $\Gamma _n$ defines a one to one mapping between $E(\Gamma _n)$ and $\{(s,i); s\in {{\mathcal {F}}}_n,s_i=1\}$. $\square$

Proposition 5.2

The total number of 0s in Fibonacci strings of length n is $\sum _{i=1}^{n}F_{i+1}F_{n-i+2}$.

Proof

Let s be a Fibonacci string of length n and $i\in [1,n]$ then $s_1s_2\dots s_{i-1}$ and $s_{i+1}s_{i+2}\dots s_{n}$ are Fibonacci strings. Reciprocally if u and v are Fibonacci strings then u0v is also a Fibonacci string. Therefore, the mapping defined by $\theta (s,i)=(s_{1}s_{2}\dots s_{i-1}, s_{i+1}s_{i+2}\dots s_{n})$ is a one to one mapping between $\{(s,i); s\in {{\mathcal {F}}}_n,s_i=0 \}$ and the Cartesian product ${{\mathcal {F}}}_{i-1}\times {{\mathcal {F}}}_{n-i}$. The identity follows. $\square$

Theorem 5.3

The number of edges of ${\overline{\Gamma }}_n$ is given by the equivalent expressions:

• (i) $|E({\overline{\Gamma }}_n)|=n2^{n-1}-\sum _{i=1}^{n}{F_{i+1}F_{n-i+2}}.$

• (ii) $|E({\overline{\Gamma }}_n)|=n2^{n-1}-\frac{4nF_{n+1}+(3n-2)F_n}{5}.$

Proof

Combining the first two identities in Proposition 5.1 together with Proposition 5.2, we obtain the first expression.

For the second expression, note first that the n edges of $Q_n$ incident to a vertex of ${{\mathcal {F}}}_n$ belongs to $E(\Gamma _n)$ or $B_n$. Making the sum over all vertices of ${{\mathcal {F}}}_n$, the edges of $E(\Gamma _n)$ are obtained two times therefore $nF_{n+2}=|B_n|+2|E(\Gamma _n)|$. By Proposition 5.1$|E({\overline{\Gamma }}_n)|=|E(Q_n)|-|B_n|-|E(\Gamma _n)|=n2^{n-1}-nF_{n+2}+|E(\Gamma _n)|$. Using the expression of $|E(\Gamma _n)|$ given by Proposition 2.2, we obtain the final result. $\square$

The sequence $(|E({\overline{\Gamma }}_n)|,n\ge 1)=0,0,2,10,35,104,\dots$ can also be obtained by an inductive relation:

Proposition 5.4

The number of edges of ${\overline{\Gamma }}_n$ is the sequence defined by

$|E({\overline{\Gamma }}_n)|=|E({\overline{\Gamma }}_{n-1})|+|E({\overline{\Gamma }}_{n-2})|+(n+4)2^{n-3}-F_{n+2} \ \ (n \ge 3)$

$|E({\overline{\Gamma }}_1)|=|E({\overline{\Gamma }}_2)|=0.$

Proof

Let $n\ge 3$. Let $\overline{{{\mathcal {F}}}_n}^{1.}$ be the set of strings of ${\overline{{{\mathcal {F}}}}}_n$ that begin with 1. Since $\overline{{{\mathcal {F}}}_n}^{1.}=\{10\varvec{s}; \varvec{s}\in {\overline{{{\mathcal {F}}}}}_{n-2}\}\cup \{11\varvec{s}; \varvec{s}\in {{\mathcal {B}}}_{n-2}\}$ we have $|\overline{{{\mathcal {F}}}_n}^{1.}|=2^{n-1}-F_n$. This identity is also valid for $n=1$ or $n=2$.

Consider the following partition of the set of vertices of ${\overline{\Gamma }}_n$: ${\overline{{{\mathcal {F}}}}}_{n}=\{0\varvec{s}; \varvec{s}\in {\overline{{{\mathcal {F}}}}}_{n-1}\}\cup \{10\varvec{s}; \varvec{s}\in {\overline{{{\mathcal {F}}}}}_{n-2}\}\cup \{11\varvec{s}; \varvec{s}\in {{\mathcal {B}}}_{n-2}\}$.

From this decomposition, the sequence of vertices of ${\overline{\Gamma }}_n$ follows the induction

$$\begin{aligned} |V({\overline{\Gamma }}_{n})|=|V({\overline{\Gamma }}_{n-1})|+|V({\overline{\Gamma }}_{n-2})|+2^{n-2}. \end{aligned}$$

We deduce also a partition of the edges ${\overline{\Gamma }}_n$ in six sets:

-$|E({\overline{\Gamma }}_{n-1})|$ edges between vertices of $\{0\varvec{s}; \varvec{s}\in {\overline{{{\mathcal {F}}}}}_{n-1}\}$.

-$|E({\overline{\Gamma }}_{n-2})|$ edges between vertices of $\{10\varvec{s}; \varvec{s}\in {\overline{{{\mathcal {F}}}}}_{n-2}\}$.

-$|E(Q_{n-2}|$ edges between vertices of $\{11\varvec{s}; \varvec{s}\in {{\mathcal {B}}}_{n-2}\}$.

-Edges between vertices of $\{0\varvec{s}; \varvec{s}\in {\overline{{{\mathcal {F}}}}}_{n-1}\}$ and $\{10\varvec{s}; \varvec{s}\in {\overline{{{\mathcal {F}}}}}_{n-2}\}$. Those edges are the $|V({\overline{\Gamma }}_{n-2})|$ edges $\{00s,10s\}$ where $s\in {\overline{{{\mathcal {F}}}}}_{n-2}$.

-Edges between vertices of $\{10\varvec{s}; \varvec{s}\in {\overline{{{\mathcal {F}}}}}_{n-2}\}$ and $\{11\varvec{s}; \varvec{s}\in {{\mathcal {B}}}_{n-2}\}$. Those edges are the $|V({\overline{\Gamma }}_{n-2})|$ edges $\{10s,11s\}$ where $s\in {\overline{{{\mathcal {F}}}}}_{n-2}$.

-Edges between vertices of $\{0\varvec{s}; \varvec{s}\in {\overline{{{\mathcal {F}}}}}_{n-1}\}$ and $\{11\varvec{s}; \varvec{s}\in {{\mathcal {B}}}_{n-2}\}$. Those edges are the $2^{n-2}-F_{n-1}$ edges $\{0s,1s\}$ where s is a string of ${\overline{{{\mathcal {F}}}}}_{n-1}^{1.}$ .

Therefore $|E({\overline{\Gamma }}_n)|=|E({\overline{\Gamma }}_{n-1})|+|E({\overline{\Gamma }}_{n-2})|+(n-2)2^{n-3}+2(2^{n-2}-F_{n})+2^{n-2}-F_{n-1}$. $\square$

We will call block of a binary string s a maximal substring of consecutive 1s. Therefore, a string in ${\overline{{{\mathcal {F}}}}}_n$ is as string with a least one block of length greater that 1. The degree of a vertex of $\Gamma _{n}$ lies between $\lfloor (n+2)/3\rfloor$ and n. The number of vertices of a given degree is determined in [12].

We will now give a similar result for ${\overline{\Gamma }}_{n}$.

Theorem 5.5

The degree of a vertex in ${\overline{\Gamma }}_{n}$ is n, $n-1$ or $n-2$ and the number of vertices of a given degree are:

$|E(\Gamma _{n-1})|$ vertices of degree $n-2$ $|E(\Gamma _{n-2})|$ vertices of degree $n-1$ $\sum _{k=0}^{n-4}{2^k|E(\Gamma _{n-k-3})|}$ vertices of degree n.

Remark 5.6

Using Proposition 2.2 these numbers can be rewritten as, respectively, $\frac{(n-1)F_{n}+2n F_{n-1}}{5}$, $\frac{(n-2)F_{n-1}+(2n-2)F_{n-2}}{5}$ and $2^{n}-\frac{(3n+7)F_{n}+(n+5)F_{n-1}}{5}$.

Proof

This is true for $n\le 3$ thus assume $n\ge 4$. Let x be a vertex of ${\overline{\Gamma }}_{n}$ 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 1s. Thus, $i_l=min\{i/x_ix_{i+1}=11\}$ and $i_r=max\{i/x_ix_{i+1}=11\}$. 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, $x_1\dots x_{i_l-1}\in {{\mathcal {F}}}_{i_l-1}^{.0}$ and $x_{i_l+2}\dots x_n\in {{\mathcal {F}}}_{n-i_l-1}^{0.}$. For $i=i_l$ or $i=i_l+1$, the string $x+\delta _i$ is a Fibonacci string. For i distinct of $i_l$ and $i_l+1$ then $x+\delta _i$ is a string of ${\overline{{{\mathcal {F}}}}}_{n}$. Therefore, $d(x)=n-2$. Since $x_1\dots x_{i_l-1}$ and $x_{i_l+2}\dots x_n$ are arbitrary strings of ${{\mathcal {F}}}_{i_l-1}^{.0}$ and ${{\mathcal {F}}}_{n-i_l-1}^{0.}$ the number of vertices of this kind is by Proposition 2.1$\sum _{i_l=1}^{n-1}{F_{i_l}F_{n-i_l}}=|E(\Gamma _{n-1})|$.

• $i_r=i_l+1$. Then, there exists a unique block of length at least 2 and this block $x_{i_l}x_{i_l+1}x_{i_l+2}$ is of length 3. Thus, $x= x_1\dots x_{i_l-1}111x_{i_l+3}\dots x_n$ where $x_1\dots x_{i_l-1}\in {{\mathcal {F}}}_{i_l-1}^{.0}$ and $x_{i_l+3}\dots x_n\in {{\mathcal {F}}}_{n-i_l-2}^{0.}$. For $i=i_l+1$, the string $x+\delta _i$ is a Fibonacci string. For i distinct of $i_l+1$, then $x+\delta _i$ is a string of ${\overline{{{\mathcal {F}}}}}_{n}$. Therefore, $d(x)=n-1$. Since $x_1\dots x_{i_l-1}$ and $x_{i_l+3}\dots x_n$ are arbitrary strings of ${{\mathcal {F}}}_{i_l-1}^{.0}$ and ${{\mathcal {F}}}_{n-i_l-2}^{0.}$ the number of vertices of this kind is $\sum _{i_l=1}^{n-2}{F_{i_l}F_{n-i_l-1}}=|E(\Gamma _{n-2})|$.

• $i_r\ge i_l+2$. Then, there exists a unique block of length at least 4 or there exist at least two blocks of length at least 2. In both cases for any $i\in [1,n]$ $x+\delta _i$ is a string of ${\overline{{{\mathcal {F}}}}}_{n}$. Therefore, $d(x)=n$. Let $k=i_r-i_l-2$. Note that $k\in [0,n-4]$ and k fixed $i_l\in [1,n-k-3]$. The strings $x_1x_2\dots x_{i_l-1}$ and $x_{i_l+k+4}x_{i_l+k+5}\dots x_n$ are arbitrary strings in ${{\mathcal {F}}}_{i_l-1}^{.0}$ and ${{\mathcal {F}}}_{n-k-i_l-3}^{0.}$. Since $x_{i_l+2}\dots x_{i_r-1}$ is an arbitrary string in ${{\mathcal {B}}}_{k}$ the number of vertices of this kind is $\sum _{k=0}^{n-4}{\sum _{i_l=1}^{n-k-3}{2^k F_{i_l}F_{n-k-i_l-2}}}=\sum _{k=0}^{n-4}{2^k|E(\Gamma _{n-k-3})|}$.

$\square$

The sequence $0,0,0,1,4,13,36,\dots$ formed by the numbers of vertices on degree n in ${\overline{\Gamma }}_n, (n\ge 1)$ already appears in OEIS [15] as sequence A235996 of the number of length n binary words that contain at least one pair of consecutive 0s followed by (at some point in the word) at least one pair of consecutive 1s. This is clearly the same sequence.

As noticed in Figure 1, ${\overline{\Gamma }}_{3}$ and ${\overline{\Gamma }}_{4}$ are isomorphic to $\Gamma _2$ and $\Gamma _4$, respectively. Our last result completes this observation.

Theorem 5.7

For $n\ge 4$ $\Gamma _n$ is isomorphic to an induced subgraph of ${\overline{\Gamma }}_{n}$.

Proof

Let $n\ge 4$ and define a mapping between binary strings of length n by $\theta (x)=\theta (x_1x_2\dots x_n)={\overline{x}}_4{\overline{x}}_2{\overline{x}}_3{\overline{x}}_1{\overline{x}}_5{\overline{x}}_6\dots {\overline{x}}_n$. Let $\sigma$ be the permutation on $\{1,2,\dots ,n\}$ defined by $\sigma (1)=4$, $\sigma (4)=1$ and $\sigma (i)=i$ for $i\notin \{1,4\}$.

Note first that $x\in {{\mathcal {F}}}_n$ implies $\theta (x)\in {\overline{{{\mathcal {F}}}}}_n$. Indeed, since $x_2x_3\ne 11$ we have three cases

• $x_2x_3=00$ then ${\overline{x}}_2{\overline{x}}_3=11$ is a substring of $\theta (x)$

• $x_2x_3=10$ then $x_1=0$ and ${\overline{x}}_3{\overline{x}}_1=11$ is a substring of $\theta (x)$

• $x_2x_3=01$ then $x_4=0$ and ${\overline{x}}_4{\overline{x}}_2=11$ is a substring of $\theta (x)$.

Therefore, $\theta$ maps vertices of $\Gamma _n$ to vertices in ${\overline{\Gamma }}_n$.

Let $\{x,x+\delta _i\}$ be an edge of $\Gamma _n$ then by construction we have $\theta (x+\delta _{i})=\theta (x)+\delta _{\sigma (i)}$, and therefore, $\theta (x)$ and $\theta (x+\delta _i)$ are adjacent in ${\overline{\Gamma }}_n$.

Since $\theta$ is a transposition we have also for all $i\in [1,n]$ that $\theta (x)+\delta _i=\theta (x+\delta _{\sigma (i)})$. Therefore, if $\{\theta (x),\theta (y)\}$ is an edge in the subgraph induced by $\theta (\Gamma _n)$ then $\theta (y)=\theta (x)+\delta _i$ for some i, $y=x+\delta _{\sigma (i)}$ and thus $\{x,y\}\in E (\Gamma _n)$. $\square$

Since $0^n$ is a vertex of degree n in $\Gamma _n$ this graph cannot be a subgraph of ${\overline{\Gamma }}_m$ for $m<n$; thus, this mapping in ${\overline{\Gamma }}_n$ is optimal. Conversely, it might be interesting to determine the minimum m such that ${\overline{\Gamma }}_{n}$ is isomorphic to an induced subgraph of $\Gamma _{m}$. We already know that $m\le 2n-1$ since the hypercube $Q_n$ is an induced subgraph of $\Gamma _{2n-1}$ [10].