Gray Codes for AT-Free Orders

TL;DR

This paper presents a method to generate AT-free vertex orders in graphs efficiently, enabling faster algorithms for processing such graphs.

Contribution

It introduces a novel algorithm that produces AT-free orders in constant amortized time, improving upon previous methods.

Findings

AT-free orders can be generated in constant amortized time

The method improves efficiency for processing AT-free graphs

Potential applications in graph algorithms and optimization

Abstract

AT-free graphs are characterized by vertex elimination orders. We show that these AT-free orders of a graph can be generated in constant amortized time.

Equations

$$z\in I(x,y)\Rightarrow x\notin I(z,y)$$

$$z\in \sigma (x+Y)\Rightarrow x\notin \sigma (z+Y)$$

$$\text{where we write}\sigma (U)=\bigcap \{J|J\in \mathcal{C}\text{and}U\subseteq J\}\text{for a subset U of E.}$$

$$y_{i+1}\in \sigma (y_i+X)\text{if i<k and}{y}_1\in \sigma (y_k+X).$$

$$u\in I(v,x)\text{and}v\in I(u,y)\Rightarrow u\in I(x,y)$$

$$\begin{gathered} a\in I(x,b)\text{and}b\in I(y,z)\text{and}a\notin N[y]\cup N[z]\Rightarrow \\ a\in I(x,y)\text{or}a\in I(x,z)\text{or}a\in I(y,z) \end{gathered}$$

$$Y\subseteq N[y_1]\cup C\text{where C is a component of the graph G−N[y1].}$$

$$\bar{\mathcal{C}}=\{E\setminus C|C\in \mathcal{C}\}.$$

$$z\in I(x,y)\Rightarrow x<z\text{or}y<z$$

$$a=a_1\dots a_n\text{and}b=b_1\dots b_m.$$

$$\beta =a_1\dots a_n{b}_1\dots b_m$$

$$x\in I(\omega ,y)\text{and}{\sigma }_2^{-1}(x)<{\sigma }_2^{-1}(y)\Rightarrow {\sigma }^{-1}(\omega )<{\sigma }^{-1}(x)$$

$$h(x)=\#\{z|z\in C\text{and}x\in I(\omega ,z)\text{and}{\sigma }_2^{-1}(x)<{\sigma }_2^{-1}(z)\}$$

$$C+y\in \mathcal{C}$$

$${\forall }_{y\in Y}{\exists }_{y^{\prime }\in Y}{y}^{\prime }\ne y\text{and}{y}^{\prime }\in \sigma (C+y)$$

$$u\in I(v,x)\text{and}v\in I(u,y)\Rightarrow u\in I(x,y)$$

$$C^u(v)=C^u(y)\text{and}{C}^u(v)\ne C^u(x).$$

$$\begin{gathered} a\in I(x,b)\text{and}b\in I(y,z)\text{and}a\notin N[y]\cup N[z]\Rightarrow \\ a\in I(x,y)\text{or}a\in I(x,z)\text{or}a\in I(y,z) \end{gathered}$$

$$N(C^b(y))\cup N(C^b(z))\subseteq N[x].$$

$$N(C^b(y))\subseteq N[x]$$

$$N(x)\cap N[b]\subseteq N(a).$$

$$Y\subseteq N[y_1]\cup C\text{for a component C of G−N[y1].}$$

$${\forall }_{j>i+1}{\forall }_{x\in \sigma (X)}{y}_j\notin I(y_i,x).$$

$$y_3\in \left(N(x_1)\cup N(y_1)\right)\setminus \left(N(x_1)\cap N(y_1)\right).$$

$$y_i\in N[y_2]\cup C^{y_2}(x_1).$$

$$C^{y_3}(x_1)\ne C^{y_3}(x_2)\text{and}{C}^{y_3}(x_1)\ne C^{y_3}(y_2)$$

$$y_4\in \left(N(x_2)\cup N(y_2)\right)\setminus N[y_3]$$

$${\forall }_{i\ge 4}{y}_i\in C^{y_3}(x_2)\cup C^{y_3}(y_2)\cup N(y_3).$$

$$N(C^{y_3}(x_1))\setminus N[y_1]\ne \varnothing .$$

$${\forall }_{i\ge 2}{y}_i\in \left(C^{y_3}(x_2)\cup C^{y_3}(y_2)\right)\setminus N[y_1]\Rightarrow y_i\in N(u)$$

$${\exists }_{x\in X}{y}_j\in I(y_i,x)\Rightarrow j=i+1,$$

Full text

1 Introduction

$\operatorname{\mathsf{AT}}$-free graphs are those that do not have an asteroidal triple  —  that is  —  $\operatorname{\mathsf{AT}}$-free graphs do  not  have three vertices  of  which  every pair is connected by a path  that  avoids  the  neighborhood  of  the  third.

Broersma  et al. [2]  introduced the following ‘betweenness relation’ to characterize $\operatorname{\mathsf{AT}}$-free graphs. Let $G$ be an $\operatorname{\mathsf{AT}}$-free graph. For a vertex $x$  let $N(x)$ denote its neighborhood and let $N[x]$ denote its closed neighborhood;  $N(x)\cup x$. A vertex $z$ is  between  $x$ and $y$ if there is a path from  $z$  to  $x$   that avoids  $N[y]$  and  similarly   there is a path from  $z$  to  $y$   that avoids  $N[x]$.

Let  $\operatorname{\mathrm{\,I\,}}(x,y)$  stand for the set of vertices that are between $x$ and $y$.  Then  a graph is $\operatorname{\mathsf{AT}}$-free  if and only if  for any three vertices $x$, $y$ and $z$ the following property holds.

[TABLE]

A  set system  is a pair $(E,\mathcal{C})$  where $E$ is a finite set and $\mathcal{C}$ is a collection of subsets of $E$. The elements of $\mathcal{C}$ will be called  convex. The problem to determine  for which set systems a greedy algorithm optimizes linear objective functions  has a long history  —  for a brief overview see  eg  Helman  et al. [7]  For set systems that are convex geometries  Kashiwabara  and  Okamoto [8]  characterize  linear programming problems  for which a greedy algorithm finds an optimum.

A  convex geometry  is a set system $(E,\mathcal{C})$ that satisfies the following properties.

1. (1.)

$E\in\mathcal{C}$ and $\varnothing\in\mathcal{C}$. 2. (2.)

$\mathcal{C}$ is closed under intersections. 3. (3.)

The anti-exchange property holds, that is, for all convex sets $Y\in\mathcal{C}$  and  $x,z\notin Y$,  $x\neq z$

[TABLE]

For  some  interesting  ‘prospective applications’  of convex geometries  in  cloud computing  we  refer  to  Kordecki [9].

Let $G$ be an $\operatorname{\mathsf{AT}}$-free graph. Define a set system  on $V=V(G)$  as the collection of convex sets in $G$  —  where a set $X\subseteq V$  is  convex  if it contains  with any two of its elements  the elements that are between them.

In the following section we show that  the collection of convex sets  in an $\operatorname{\mathsf{AT}}$-free graph  constitutes  a  convex  geometry. This completes the result of Alcón  et al. [1]  who proved a similar result for interval graphs.

2 $\operatorname{\mathsf{AT}}$-free convex geometries

A set system $(E,\mathcal{C})$  which satisfies $E,\varnothing\in\mathcal{C}$  and  which is closed under intersections  is called an alignment  by  Edelman and Jamison [6]. They show that an alignment satisfies the anti-exchange property  if and only if  either one  of the following two properties holds.

1. (1.)

For any $C\in\mathcal{C}$ and $y\notin C$  the element $y$ is  extreme  in $\sigma(C+y)$  —  that is  —  $\sigma(C+y)\setminus y$ is convex. 2. (2.)

Any convex set $X$, $X\neq E$, has an element $y\notin X$ such that $X+y$ is convex.

The following definition allows us to characterize convex geometries with a third property (which is equivalent to the characterization [6, Theorem 2.3]).

Definition 1

Let  $(E,\mathcal{C})$  be an alignment,  let  $X,Y\subseteq E$  and  let  $Y=\{y_{1},\dots,y_{k}\}$.  Let  $k\geq 2$.  The set  $Y$  induces a cycle  on  $X$  if  for  all  $i\in[k]$

[TABLE]

Lemma 1

An alignment  $(E,\mathcal{C})$  is a convex geometry  if and only if  any set  $Y$   which induces a cycle on a set  $X$  is contained in  $\sigma(X)$.

All proofs we skip here can be found in the appendix. We proceed to prove  that the convex sets in an $\operatorname{\mathsf{AT}}$-free graph constitute a convex geometry  via the following three lemmas. (Some easily-made drawings might be helpful to the reader.)

Lemma 2

Let $G$ be an $\operatorname{\mathsf{AT}}$-free graph. Any four vertices satisfy the following property.

[TABLE]

Lemma 3

Let  $G$  be an  $\operatorname{\mathsf{AT}}$-free graph. Any five vertices satisfy the following property.

[TABLE]

(In [3]  Chvatál  describes a subclass of convex geometries by a property similar to Lemma 3.) A component of a graph is a maximal subset of vertices  of  which  every pair is connected by a path.

Lemma 4

Let  $G$  be an  $\operatorname{\mathsf{AT}}$-free graph. Let  $X,Y\subseteq V$  and let  $Y=\{y_{1},\dots,y_{k}\}$. Assume that  $Y\cap\sigma(X)=\varnothing$  and that no subset of  $Y$  induces a cycle on  $X$.  If  —  for all  $i\in[k-1]$  —  there exist $x_{i}\in X$ such that  $y_{i+1}\in\operatorname{\mathrm{\,I\,}}(y_{i},x_{i})$  then

[TABLE]

Theorem 2.1

Let  $G$  be an  $\operatorname{\mathsf{AT}}$-free graph.  The convex sets in  $G$  constitute a convex geometry  on  $V(G)$.

3 Generating $\operatorname{\mathsf{AT}}$-free orders

When $(E,\mathcal{C})$ is a convex geometry then $(E,\bar{\mathcal{C}})$ is an  antimatroid  and this defines all antimatroids  —  here we write

[TABLE]

Crapo [5]  characterizes formal languages that are antimatroids  as  follows.

Definition 2

A language  $L$  is an antimatroid  if  its  words  satisfy  the following properties.

1. (1.)

Every symbol of the alphabet occurs in at least one word. 2. (2.)

Every word of  $L$  contains at most one copy of every symbol in the alphabet. 3. (3.)

Every prefix of a word in  $L$  is  in  $L$. 4. (4.)

If  $s,t\in L$  and  if  $s$ contains at least one symbol that is not in  $t$  then  there is a symbol  $x\in s$ such that  $tx\in L$.

—  Observe that  —  when $L$ is the language  whose words are prefixes of $\operatorname{\mathsf{AT}}$-free orders of a graph  then  $L$  is  an  antimatroid. The basic words of $L$ are those of maximal length  which  are  the  $\operatorname{\mathsf{AT}}$-free orders.

Definition 3

A linear order  $<$  of the vertices of a graph  is an  $\operatorname{\mathsf{AT}}$-free order  if any three vertices satisfy the following property.

[TABLE]

A graph is $\operatorname{\mathsf{AT}}$-free  if and only if  it has an $\operatorname{\mathsf{AT}}$-free order [4].

Pruesse  and  Ruskey [11, 12] considered the problem of producing a Gray code for the basic words of an antimatroid.

Let $L$ be an antimatroid. Consider the graph whose vertices are the basic words of $L$  two vertices being adjacent when one is obtained from the other by a transposition of an adjacent pair. The  prism  is obtained from two copies ($+$ and $-$) of this graph  and the addition of edges joining $+/-$ copies of similar vertices.  Pruesse  and  Ruskey  show that this prism is Hamiltonian  for  all  antimatroids.  Their generic algorithm  generates  all the basic words of $L$  in the order of a Hamiltonian traversal of the prism  —  whilst reporting only the (transpositions in the) $+$ copies.

Assume  that a graph  $G$  is connected and $\operatorname{\mathsf{AT}}$-free. Let $\omega$ be a vertex  such  that the number of vertices  in the largest component of $G-N[\omega]$  is  as  large  as  possible.  Let  $C$ be a largest component of $G-N[\omega]$  and  let  $S=N(C)$ and $\Omega=V\setminus(C\cup S)$.  When $G$ is not a clique then $\{\Omega,S,C\}$ is a partition of $V(G)$.

By our choice of $\omega$  every vertex of $\Omega$ is adjacent to every vertex of $S$  —  that is  —  $\Omega$ is a module,  hence,  convex. Since $G$ is $\operatorname{\mathsf{AT}}$-free  $C\cup S$ is convex as well.

Consider $\operatorname{\mathsf{AT}}$-free orders for $G[\Omega]$ and $G[C\cup S]$  —  say

[TABLE]

Notice that the linear order

[TABLE]

is an $\operatorname{\mathsf{AT}}$-free order for $G$  —  we call  $\beta$  a  canonical order  when $a$ and $b$ are that of $G[\Omega]$ and $G[C\cup S]$.  It is easy to obtain a canonical order in polynomial time.

Consider an arbitrary order $\sigma=v_{1}\,\dots\,v_{n}$ of the vertices of $G$. We write $\sigma_{1}$ for the linear order induced on $\Omega$ and $\sigma_{2}$ for the linear order induced on $C\cup S$. Observe that $\sigma$ is an $\operatorname{\mathsf{AT}}$-free order  if  and  only  if

$\sigma_{1}$ and $\sigma_{2}$ are $\operatorname{\mathsf{AT}}$-free orders of $\Omega$ and $C\cup S$ 2. 2.

for $\omega\in\Omega$ and $x,y\in C\cup S$

[TABLE]

that is  —   all  vertices of $\Omega$ should appear  before  the first element of a pair $x,y\in C\cup S$ that satisfies $x\in\operatorname{\mathrm{\,I\,}}(\omega,y)$  and  that  is  in  a  ‘wrong’  order  —  namely  —  $\sigma^{-1}(y)>\sigma^{-1}(x)$.

In  proving  that  the prism of $G$ is Hamiltonian  we may assume that the prisms of  $\Omega$  and  $C\cup S$  are  that.  —  Furthermore  —  we may assume that $\{+\beta,-\beta\}$ is an edge of  both  Hamiltonian  cycles.  A Hamiltonian cycle in the prism of $G$  that uses the edge $\{+\beta,-\beta\}$  is easily obtained from this [12, Theorem 3.3].

It  follows  that  the $\operatorname{\mathsf{AT}}$-free orders of  $G$  can be generated  such  that  each order differs from its predecessor  by  at  most  one  or  two  adjacent  transpositions.

It remains to establish the timebound.

Theorem 3.1

The  $\operatorname{\mathsf{AT}}$-free orders  of an  $\operatorname{\mathsf{AT}}$-free graph can be generated  in  constant  amortized  time.

Proof

Pruesse and Ruskey developed a generic algorithm to produce all basic words of an antimatroid [11, 12]. The amortized time complexity is determined by an  —  antimatroid specific  —  transposition oracle  which answers whether two adjacent elements in a basic word may  swap  places  to produce another basic word.

We use the notation introduced above. For an $\operatorname{\mathsf{AT}}$-free order $\sigma$  with an induced order $\sigma_{2}$ on $C\cup S$  and $x\in C$  define  $h(x)$   as  follows.

[TABLE]

Then $\sigma$ is an $\operatorname{\mathsf{AT}}$-free order  if  and  only  if  $\sigma_{1}$ and $\sigma_{2}$  are  that  and  $\sigma^{-1}(\omega)<\sigma^{-1}(x)$ when $h(x)\geq 1$. We show that $h$ can be maintained  during a swap of two adjacent elements in $\sigma$.  Notice  that $h$ is easily computable for a canonical order.

Sawada [13, Theorem 15 ff.]  introduces the counter $\operatorname{\mathrm{numBad}}(x,y)$  for ordered pairs $x$ and $y$  as the number of vertices $z$ with $x\in\operatorname{\mathrm{\,I\,}}(y,z)$ and $\sigma^{-1}(z)>\sigma^{-1}(x)$. Two elements  $v_{j}$ and $v_{j+1}$  can be swapped to produce a new $\operatorname{\mathsf{AT}}$-free order  only  if  $\operatorname{\mathrm{numBad}}(v_{j+1},v_{j})=0$.  Sawada  shows that $\operatorname{\mathrm{numBad}}$ can be maintained during a generation of $\operatorname{\mathsf{AT}}$-free orders in constant amortized time [13, Theorem 13 and Observation 1].

Notice that $h(x)=\operatorname{\mathrm{numBad}}(x,\omega)$. This proves the theorem. $\blacksquare$

4 Concluding remark

The family of ideals  in a poset   constitutes  a convex geometry on the elements of the poset. This convex geometry is usually referred to as a poset shelling. A convex geometry is a poset shelling  if and only if  its family of convex sets is closed under unions [10]. (See [8] for other characterizations.)

The family of convex sets of the $\operatorname{\mathsf{AT}}$-free graph  shown in the figure  is not a poset shelling.  To see that  let $A=\{y_{1},z_{2},u\}$ and let $B=\{y_{2},z_{1},u\}$. Then $A$ and $B$ are convex  but  their union is not  since $u^{\prime}\in\operatorname{\mathrm{\,I\,}}(z_{1},z_{2})$.

Appendix: proofs

Lemma 2

An alignment  $(E,\mathcal{C})$  is a convex geometry  if and only if  any set  $Y$  which induces a cycle on a set  $X$  is contained in  $\sigma(X)$.

Proof

When $Y$ is a cycle on $X$ then $\sigma(X+y)\setminus y$ is not convex for any $y\in Y$. When  $Y\setminus\sigma(X)\neq\varnothing$  there exists a vertex $y$ such that $\sigma(X+y)\setminus y$ is not convex. By Edelman and Jamison’s characterization  $(E,\mathcal{C})$  is  not  a  convex  geometry.

Assume that any set  that induces a cycle on $X$  is contained in $\sigma(X)$.

Let $C$ be a convex set and assume that $C\neq E$. We show that there exists an element $y\in E\setminus C$ which satisfies

[TABLE]

(notice that this proves the claim  —  by  Edelman and Jamison’s characterization of convex geometries).

Let  $Y\neq\varnothing$  be an  inclusion-minimal  subset of $E\setminus C$ such that $C\cup Y$ is convex. We claim that  $|Y|=1$.  — Otherwise  —  $Y$ has at least two elements. By the assumption  that $Y$ is  set–inclusion  minimal

[TABLE]

which implies that some nonempty subset $Y^{\prime}\subseteq Y$ induces a cycle on $C$.  By the assumption  —  that any set which induces a cycle on $X$ is contained in $\sigma(X)$  —   $Y^{\prime}\subseteq C$  which is a contradiction. $\blacksquare$

For vertices  $x$ and $y$   that are  not adjacent  we write  $C^{x}(y)$  for the component of $G-N[x]$ that contains the vertex $y$.

Lemma 3

Let $G$ be an $\operatorname{\mathsf{AT}}$-free graph. Any four vertices satisfy the following property.

[TABLE]

Proof

If $x=y$ then the left-hand is only satisfied when $\{u,v,x\}$ is an asteroidal triple.  Otherwise

[TABLE]

The vertices $u$ and $y$ are connected by a path that avoids $N[x]$  since $u\in\operatorname{\mathrm{\,I\,}}(v,x)$.  —  Also  —  the vertices $u$ and $x$ are connected by a path that avoids $N[y]$  since $v\in\operatorname{\mathrm{\,I\,}}(u,y)$.

This proves the lemma. $\blacksquare$

Lemma 4

Let  $G$  be an  $\operatorname{\mathsf{AT}}$-free graph. Any five vertices satisfy the following property.

[TABLE]

Proof

Assume that $C^{b}(a)\neq C^{b}(z)$ and $C^{b}(a)\neq C^{b}(y)$.  Observe that the vertices $a$ and $x$ are connected by a path that avoids $N[y]\cup N[z]$. If there are no paths from $a$ to $z$  nor from $a$ to $y$  that avoid $N[x]$  then

[TABLE]

However, there is a path from $a$ to $b$ that avoids $N[x]$  which implies  that there is a path from $a$ to $b$ that avoids $N[z]$. By concatenation of this path  with a path from $b$ to $y$ that avoids $N[z]$  we find a path from $a$ to $y$ that avoids $N[z]$.  Similarly,  there is a path from $a$ to $z$ that avoids $N[y]$. This proves $a\in\operatorname{\mathrm{\,I\,}}(y,z)$.

Assume  $C^{b}(a)=C^{b}(z)$.  There is a path from $a$ to $x$ that avoids $N[y]$. If there is no path from $a$ to $y$ that avoids $N[x]$  we have

[TABLE]

since there is a path connecting $a$ and $b$ that avoids $N[x]$. Since $a\in\operatorname{\mathrm{\,I\,}}(x,b)$  $N[a]$ separates $x$ and $b$  which implies

[TABLE]

There is a path from $b$ to $y$ that avoids $N[z]$  since $b\in\operatorname{\mathrm{\,I\,}}(y,z)$.  Since  $a\notin N[z]$  there is a path from $a$ to $y$ that avoids $N[z]$.  Since  $C^{b}(a)=C^{b}(z)$  there is a path from $a$ to $z$ that avoids $N[y]$  which implies  $a\in\operatorname{\mathrm{\,I\,}}(y,z)$.

Assume  $C^{b}(a)=C^{b}(y)$.  In an analogous manner it follows that $a\in\operatorname{\mathrm{\,I\,}}(x,z)$ or $a\in\operatorname{\mathrm{\,I\,}}(y,z)$.

This proves the lemma. $\blacksquare$

Note that $a\in\operatorname{\mathrm{\,I\,}}(x,b)\quad\text{\rm{and}}\quad b\in\operatorname{\mathrm{\,I\,}}(y,z)\quad\Rightarrow\quad a\notin N[y]\cap N[z].$

Lemma 5

Let  $G$  be an  $\operatorname{\mathsf{AT}}$-free graph. Let  $X,Y\subseteq V$  and let  $Y=\{y_{1},\dots,y_{k}\}$. Assume that  $Y\cap\sigma(X)=\varnothing$  and that no subset of  $Y$  induces a cycle on  $X$.  If  —  for all  $i\in[k-1]$ —   there exist  $x_{i}\in X$ that satisfy  $y_{i+1}\in\operatorname{\mathrm{\,I\,}}(y_{i},x_{i})$  then

[TABLE]

Proof

Assume that the lemma does not hold. Choose $X$ and $Y$ so that $Y$ intersect  at  least  two  components of $G-N[y_{1}]$  and  $|Y|$  is  minimal. Then $|Y|\geq 3$.

Since $Y$ is minimal

[TABLE]

By  Lemma 3

[TABLE]

Assume $y_{3}\in N(x_{1})$. Consider $G-N[y_{2}]$.  For  $i\geq 2$

[TABLE]

Since  $C^{y_{2}}(x_{1})\neq C^{y_{2}}(y_{1})$  there exists of a path from $x_{1}$ to $y_{2}$ in $G-N[y_{1}]$.  This  contradicts  the  assumption.

Assume $y_{3}\in N(y_{1})$. Consider $G-N[y_{3}]$.  We  have  $x_{2}\in N(y_{1})$  otherwise  $y_{1}\in\operatorname{\mathrm{\,I\,}}(y_{2},x_{2})$.  —  Moreover  —

[TABLE]

since the first equality would imply that $\{x_{1},y_{1},y_{2}\}$ is an asteroidal triple  and  the second equality would imply $y_{2}\in\operatorname{\mathrm{\,I\,}}(x_{1},x_{2})$.

When  $|Y|\geq 4$,

[TABLE]

and thus

[TABLE]

Notice that $y_{2}\in\operatorname{\mathrm{\,I\,}}(x_{1},y_{1})$ implies that $x_{1}$ and $y_{2}$ are connected by a path that avoids $N[y_{1}]$.  This  implies

[TABLE]

Let  $u\in N(C^{y_{3}}(x_{1}))\setminus N[y_{1}]$.  Then

[TABLE]

since  otherwise  $y_{1}\in\operatorname{\mathrm{\,I\,}}(x_{1},y_{i})$.  —  Furthermore  —  for $y_{i}\in N[y_{3}]\setminus N[y_{1}]$ which implies that $N[y_{i}]$ intersects any $x_{1},y_{2}$-path that avoids $N[y_{1}]$. This proves the lemma. $\blacksquare$

Theorem 2.1

Let  $G$  be an  $\operatorname{\mathsf{AT}}$-free graph.  The convex sets in  $G$  constitute a convex geometry  on  $V(G)$.

Proof

Assume there exist sets $X$ and $Y$ that contradict this theorem  —  that is  —  $Y$ induces a cycle on $X$.  Let  $Y=\{y_{1},\dots,y_{k}\}$.  We may assume that $Y$ is minimum  and  that  $Y\cap\sigma(X)=\varnothing$.

—  Also  —

[TABLE]

where all arithmetics here are taken with modulo $k$. By  Lemma 3

[TABLE]

Consider  the  following  two  cases.

Case $1$:

Assume $\forall_{\,i}\>y_{i+2}\in N(y_{i})$.  Then

[TABLE]

since  otherwise  $y_{i}\in\operatorname{\mathrm{\,I\,}}(x_{i-1},x_{i})$. By $y_{i}\notin I(x_{i+1},y_{i+1})$,

[TABLE]

Notice that Eq. (4) holds for all $i\in[k]$. Then, by repeatedly apply Eq. (3) we have $y_{i}\in N(x_{j})$ for all $i$ and $j$. This contradicts the assumption that $Y$ induces a cycle on $X$.

Case $2$:

Assume  $\exists_{\,j}\>y_{j+2}\in N(x_{j})$. Without loss of generality  assume  $j=1$. We show that  $y_{2}\in\operatorname{\mathrm{\,I\,}}(x_{1},y_{1})$  and  $y_{1}\in\operatorname{\mathrm{\,I\,}}(x_{1},y_{2})$. It suffices to show that $x_{1}$ and $y_{1}$ are connected by a path that avoids $N[y_{2}]$.

Consider

[TABLE]

Since  $\{y_{1},y_{3}\}\cap N[y_{2}]=\varnothing$  —  by Lemma 4  —  $y_{1}$ and $y_{3}$ are in the same component of $G-N[y_{2}]$.  Since  $x_{1}\in N[y_{3}]$  this proves that $x_{1}$ and $y_{1}$ are connected by a path that avoids $N[y_{2}]$  —  as claimed.

This proves the theorem. $\blacksquare$