On induced saturation for paths

TL;DR

This paper investigates the existence of graphs that are saturated with respect to induced paths, proving the existence of infinitely many such graphs for paths of lengths multiple of three and analyzing specific cases like Kneser graphs.

Contribution

It establishes the existence of infinitely many $P_{3n}$-induced-saturated graphs and explores the properties of Kneser graphs in this context.

Findings

Existence of $P_{3n}$-induced-saturated graphs for all positive integers n.

Construction of infinitely many such graphs for each $P_{3n}$.

Kneser graphs $K(n,2)$ are $P_6$-induced-saturated for all $n  5.

Abstract

For a graph $H$, a graph $G$ is $H$-induced-saturated if $G$ does not contain an induced copy of $H$, but either removing an edge from $G$ or adding a non-edge to $G$ creates an induced copy of $H$. Depending on the graph $H$, an $H$-induced-saturated graph does not necessarily exist. In fact, Martin and Smith (2012) showed that $P_4$-induced-saturated graphs do not exist, where $P_k$ denotes a path on $k$ vertices. Axenovich and Csik\'{o}s (2019) asked the existence of $P_k$-induced-saturated graphs for $k \ge 5$; it is easy to construct such graphs when $k\in\{2, 3\}$. Recently, R\"{a}ty constructed a graph that is $P_6$-induced-saturated. In this paper, we show that there exists a $P_{k}$-induced-saturated graph for infinitely many values of $k$. To be precise, we find a $P_{3n}$-induced-saturated graph for every positive integer $n$. As a consequence, for each positive integer $n$, we construct infinitely many $P_{3n}$-induced-saturated graphs. We also show that the Kneser graph $K(n,2)$ is $P_6$-induced-saturated for every $n\ge 5$.

Figures

Tables

Table 1. Table 1 : The values f ​ ( v ) 𝑓 𝑣 f(v) , p ​ ( v ) 𝑝 𝑣 p(v) , and q ​ ( v ) 𝑞 𝑣 q(v) for v ∈ N G n ​ ( a ​ b , j ) 𝑣 subscript 𝑁 subscript 𝐺 𝑛 𝑎 𝑏 𝑗 v\in N_{G_{n}}(ab,j) .

$v$	$(\overline{a}\overline{b},j)$	$(a\overline{b},j)$	$(ac,j-1)$	$(ac,j+1)$	$(\overline{a}b,j+2{\left(-1\right)}^a)$
$f\left(v\right)$	$(\overline{a}\overline{b},j+1)$	$(a\overline{b},j+1)$	$(ac,j)$	$(ac,j+2)$	$(\overline{a}b,j+2{\left(-1\right)}^a+1)$
$p\left(v\right)$, $j$: even	$(a\overline{b},-j-1)$	$(\overline{a}\overline{b},-j-1)$	$(\overline{a}\overline{c},-j)$	$(\overline{a}\overline{c},-j-2)$	$(ab,-j-2{\left(-1\right)}^a-1)$
$p\left(v\right)$, $j$: odd	$(ab,-j-1)$	$(\overline{a}b,-j-1)$	$(\overline{a}c,-j)$	$(\overline{a}c,-j-2)$	$(a\overline{b},-j-2{\left(-1\right)}^a-1)$
$q\left(v\right)$	$(\overline{a}c,-j+2\overline{a})$	$(a\overline{c},-j+2a)$	$(ab,-j+1+2a)$	$(ab,-j-1+2a)$	$(\overline{a}\overline{c},-j-2{\left(-1\right)}^a+2\overline{a})$

Table 2. Table 2 : N G n ​ ( f ​ ( a ​ b , j ) ) subscript 𝑁 subscript 𝐺 𝑛 𝑓 𝑎 𝑏 𝑗 N_{G_{n}}(f(ab,j)) , N G n ​ ( p ​ ( a ​ b , j ) ) subscript 𝑁 subscript 𝐺 𝑛 𝑝 𝑎 𝑏 𝑗 N_{G_{n}}(p(ab,j)) , and N G n ​ ( q ​ ( a ​ b , j ) ) subscript 𝑁 subscript 𝐺 𝑛 𝑞 𝑎 𝑏 𝑗 N_{G_{n}}(q(ab,j)) .

$N_{G_n}\left(f(ab,j)\right)$	$(\overline{a}\overline{b},j+1),(a\overline{b},j+1),(ac,j),(ac,j+2),(\overline{a}b,j+1+2{\left(-1\right)}^a)$
$\left(=N_{G_n}(ab,j+1)\right)$
$N_{G_n}\left(p(ab,j)\right)$, $j:$ even	$(a\overline{b},-j-1),(\overline{a}\overline{b},-j-1),(\overline{a}\overline{c},-j-2),(\overline{a}\overline{c},-j),(ab,-j-1+2{\left(-1\right)}^{\overline{a}})$
$\left(=N_{G_n}(\overline{a}b,-j-1)\right)$
$N_{G_n}\left(p(ab,j)\right)$, $j:$ odd	$(ab,-j-1),(\overline{a}b,-j-1),(\overline{a}c,-j-2),(\overline{a}c,-j),(a\overline{b},-j-1+2{\left(-1\right)}^{\overline{a}})$
$\left(=N_{G_n}(\overline{a}\overline{b},-j-1)\right)$
$N_{G_n}\left(q(ab,j)\right)$	$(\overline{a}\overline{c},-j+2a),(a\overline{c},-j+2a),(ab,-j+2a-1),(ab,-j+2a+1),(\overline{a}c,-j+2a+2{\left(-1\right)}^a)$
$\left(=N_{G_n}(ac,-j+2a)\right)$

Table 3. Table 3 : The values r ​ ( v ) 𝑟 𝑣 r(v) and r ​ ( w ) 𝑟 𝑤 r(w) for v ∈ A 𝑣 𝐴 v\in A and w ∈ N G n ​ ( v ) ∖ A 𝑤 subscript 𝑁 subscript 𝐺 𝑛 𝑣 𝐴 w\in N_{G_{n}}(v)\setminus A .

$v\in A$	$w\in N_{G_n}\left(v\right)\setminus A$	$r\left(v\right)=q\left(v\right)$	$r\left(w\right)=f^2\circ p^{-1}\left(w\right)$
$(00,0)$	$(00,-1)$, $(11,0)$	$(00,0)$	$(10,2)$, $(00,1)$
$(01,0)$	$(01,-1)$, $(10,0)$	$(01,0)$	$(11,2)$, $(01,1)$
$(00,1)$	$(00,2)$, $(10,3)$	$(00,-1)$	$(11,-1)$, $(00,-2)$
$(01,1)$	$(01,2)$, $(11,3)$	$(01,-1)$	$(10,-1)$, $(01,-2)$
$(10,1)$	$(11,0)$, $(00,-1)$	$(11,1)$	$(00,1)$, $(10,2)$
$(11,1)$	$(10,0)$, $(01,-1)$	$(10,1)$	$(01,1)$, $(11,2)$
$(10,2)$	$(01,2)$, $(11,3)$	$(11,0)$	$(10,-1)$, $(01,-2)$
$(11,2)$	$(00,2)$, $(10,3)$	$(10,0)$	$(11,-1)$, $(00,-2)$

Equations

$$\{(\bar{a}\bar{b},j),(a\bar{b},j),(ac,j-1),(ac,j+1),(\bar{a}c,j+2(-1{)}^a)\},$$

$$f_n(ab,j)$$

$$p_n(ab,j)$$

$$q_n(ab,j)$$

$$p^{-1}(ab,j)=\{\begin{array}{ll}(\bar{a}\bar{b},-j-1)& \text{if }j\equiv 0(\mathrm{mod}2)\\ (\bar{a}b,-j-1)& \text{if }j\equiv 1(\mathrm{mod}2).\end{array}$$

$$U_n$$

$$L_n$$

$$S=\{\begin{array}{ll}\{(01,2n-3)\}& \text{if n is even}\\ \varnothing & \text{if n is odd.}\end{array}$$

$$f^3\circ p^{-1}(v_0^{\prime })=f^3\circ p^{-1}(11,2)=f^3(00,-3)=(00,0)=v_0$$

$$f^3\circ p^{-1}(v_1^{\prime })=f^3\circ p^{-1}(10,1)=f^3(00,-2)=(00,1)=v_1$$

$$f^3\circ p^{-1}(v_2)=f^3\circ p^{-1}(11,1)=f^3(01,-2)=(01,1)=y_1.$$

$$r(ab,j)=\{\begin{array}{ll}q(ab,j)& \text{if }(ab,j)\in A\\ f^2\circ p^{-1}(ab,j)& \text{otherwise.}\end{array}$$

$$r(ab,j)=\{\begin{array}{ll}(ac,-j+2a)& \text{if }(ab,j)\in A,\\ (\bar{a}\bar{b},-j+1)& \text{if }(ab,j)\notin A\text{ and }j\text{ is even},\\ (\bar{a}b,-j+1)& \text{if }(ab,j)\notin A\text{ and }j\text{ is odd}.\end{array}$$

Full text

In this paper, we show that there exists a $P_{k}$-induced-saturated graph for infinitely many values of $k$. To be precise, we find a $P_{3n}$-induced-saturated graph for every positive integer $n$. As a consequence, for each positive integer $n$, we construct infinitely many $P_{3n}$-induced-saturated graphs. We also show that the Kneser graph $K(n,2)$ is $P_{6}$-induced-saturated for every $n\geq 5$.

1 Introduction

We consider only finite simple graphs. Given a graph $G$, let $V(G)$ and $E(G)$ denote the vertex set and the edge set, respectively, of $G$. A non-edge of $G$ is an unordered pair of vertices not in $E(G)$. For a graph $G$ and an edge $e$ (resp. non-edge $e$) of $G$, let $G-e$ (resp. $G+e$) denote the graph obtained by removing $e$ from $G$ (resp. adding $e$ to $G$). For a vertex $v\in V(G)$, let $N_{G}(v)$ denote the set of neighbors of $v$, and let $N_{G}[v]=N_{G}(v)\cup\{v\}$. We use $K_{n}$, $C_{n}$, and $P_{n}$ to denote the complete graph, a cycle, and a path, respectively, on $n$ vertices.

For a graph $H$, a graph $G$ is $H$-saturated if $G$ has no subgraph isomorphic to $H$, but adding a non-edge to $G$ creates a subgraph isomorphic to $H$. Since an $H$-saturated graph on $n$ vertices always exists for a particular graph $H$ when $n\geq|V(H)|$, it is natural to ask for the maximum number of edges of an $H$-saturated graph on $n$ vertices. This gave birth to a central topic in what we now call extremal graph theory. We hit only the highlights here.

For a graph $H$, the maximum number of edges of an $H$-saturated graph on $n$ vertices is known as the Turán number of $H$. The Turán number of an arbitrary complete graph is determined and is known as Turán’s Theorem [15, 14], which extended the work of Mantel [10] in 1907. The remarkable Erdős-Stone-Simonovits Theorem [6, 5] resolved the Turán number asymptotically for an arbitrary graph $H$ as long as $H$ is non-bipartite. Investigating Turán numbers of bipartite graphs remains to be an important, yet difficult open question; see [8] for a thorough survey. At the other extreme, the saturation number of a graph $H$ is the minimum number of edges of an $H$-saturated graph on $n$ vertices. The saturation number was introduced by Erdős, Hajnal, and Moon in [4], where the saturation number of an arbitrary complete graph was determined. Moreover, it is known that all saturation numbers are at most linear in $n$; see the excellent dynamic survey [7] by Faudree, Faudree, and Schmitt for more results and open questions regarding saturation numbers.

A natural generalization of the extremal questions above is to change the containment relation to induced subgraphs. Utilizing the notion of trigraphs, Prömel and Steger [12] and Martin and Smith [11] defined the induced subgraph version of the Turán number and the saturation number, respectively, of a given graph. We omit the exact definitions here, see [7, 9, 2, 3] for more details and other recent work on Turán-type problems concerning induced subgraphs.

Recently, in 2019, Axenovich and Csikós [1] introduced the notion of an $H$-induced-saturated graph, whose definition emphasizes that a non-edge is as important as an edge when considering induced subgraphs. For a graph $H$, a graph $G$ is $H$-induced-saturated if $G$ does not contain an induced copy of $H$, but either removing an edge from $G$ or adding a non-edge to $G$ creates an induced copy of $H$. In contrast to the fact that an $H$-saturated graph always exists for an arbitrary graph $H$, it is not always the case that an $H$-induced-saturated graph exists. For example, a $K_{n}$-induced-saturated graph cannot exist when $n\geq 3$. This is because removing an edge from a graph without $K_{n}$ as a (induced) subgraph cannot create a copy of $K_{n}$. It is also known that a graph $H$ with induced-saturation number zero guarantees the existence of an $H$-induced-saturated graph. Behrens et al. [2] studied graphs $H$ with induced-saturation number zero. In particular, they revealed that an $H$-induced-saturated graph exists when $H$ is a star, a matching, a 4-cycle, or an odd cycle of length at least 5.

Let us now focus our attention to paths. We can easily see that the empty graph on at least two vertices is $P_{2}$-induced-saturated, and the disjoint union of complete graphs in which each component has at least three vertices is $P_{3}$-induced-saturated. Yet, a result by Martin and Smith [11] implies that there is no $P_{4}$-induced-saturated graph. Recently, a simple proof of the aforementioned result was provided by Axenovich and Csikós [1], where they showed various results regarding induced saturation of a certain family of trees, which unfortunately does not include paths. In particular, they pointed out that it is unknown whether a $P_{k}$-induced-saturated graph exists when $k\geq 5$.

Question 1.1** ([1]).**

Does a $P_{k}$-induced-saturated graph exist when $k\geq 5$?

Very recently, Räty [13] answered Question 1.1 in the affirmative for $k=6$ by using an algebraic construction to generate a graph on 16 vertices that is $P_{6}$-induced-saturated. The author indicated that the construction gives no idea for longer paths, yet, we were able to generalize his construction for infinitely many longer paths. Our main result answers Question 1.1 in the affirmative for all $k$ that is a multiple of 3. We now state our main theorem:

Theorem 1.2**.**

For every positive integer $n$, there exists a $P_{3n}$-induced-saturated graph.

In fact, using the graph construction in Theorem 1.2, we can construct an infinite family of $P_{3n}$-induced-saturated graphs for each positive integer $n$.

Corollary 1.3**.**

For every positive integer $n$, there exist infinitely many $P_{3n}$-induced-saturated graphs.

We also note that the Kneser graph $K(n,2)$ is $P_{6}$-induced-saturated for every $n\geq 5$. Recall that the Kneser graph $K(n,r)$ is a graph whose vertices are the $r$-subsets of $\{1,2,\ldots,n\}$, and two vertices are adjacent if and only if the corresponding $r$-subsets are disjoint.

Theorem 1.4**.**

For every integer $n\geq 5$, the Kneser graph $K(n,2)$ is $P_{6}$-induced-saturated.

In Section 2, we present the proofs of Theorem 1.2 and Corollary 1.3. The section is divided into subsections to improve readability. In Section 3, we prove Theorem 1.4. We end the paper with some questions and further research directions in Section 4. We also have the appendix where we provide proofs of some miscellaneous facts.

2 Proofs of Theorem 1.2 and Corollary 1.3

In this section, we prove Theorem 1.2 and Corollary 1.3. Note that it is sufficient to prove both statements for $n\geq 2$, since the disjoint union of complete graphs in which each component has at least three vertices gives a family of infinitely many $P_{3}$-induced-saturated graphs. We first prove Corollary 1.3 using the graph $G_{n}$ in the proof of Theorem 1.2.

Proof of Corollary 1.3.

Let $G_{n}$ be a $P_{3n}$-induced-saturated graph in the proof of Theorem 1.2. By Lemma 2.2 and Proposition 2.5, $G_{n}$ is vertex-transitive and has an induced path on $3n-1$ vertices. For $k\geq 1$, let $H_{n,k}$ be a graph with $k$ components where each component is isomorphic to $G_{n}$. Thus each component of $H_{n,k}$ is $P_{3n}$-induced-saturated, vertex-transitive, and has an induced path on $3n-1$ vertices.

Since each component is $P_{3n}$-induced-saturated and removing an edge from $H_{n,k}$ is removing an edge from a component of $H_{n,k}$, removing an edge from $H_{n,k}$ creates an induced copy of $P_{3n}$. By the same reason, adding a non-edge joining two vertices in a component of $H_{n,k}$ creates an induced copy of $P_{3n}$. Adding a non-edge joining two vertices in different components of $H_{n,k}$ creates an induced copy of $P_{6n-2}$, since each component of $H_{n,k}$ is vertex-transitive and contains an induced path on $3n-1$ vertices. Hence, $H_{n,k}$ is $P_{3n}$-induced-saturated. ∎

The proof of Theorem 1.2 is split into three subsections. We will first construct a graph $G_{n}$ and lay out some important properties of $G_{n}$ in Subsection 2.1. In Subsection 2.2, we prove that $G_{n}$ does not contain an induced path on $3n$ vertices. In Subsection 2.3, we prove that the graph obtained by either adding an arbitrary non-edge to $G_{n}$ or removing an arbitrary edge of $G_{n}$ contains an induced path on $3n$ vertices. We conclude that $G_{n}$ is a $P_{3n}$-induced-saturated graph.

2.1 The construction of $G_{n}$ and its automorphisms

Let $\mathbb{Z}_{2n}=\{0,1,\ldots,2n-1\}$, so that $\mathbb{Z}_{2}=\{0,1\}$. We will always use either 0 or 1 for an element of $\mathbb{Z}_{2}$. On the other hand, we will use any integer to denote an element of $\mathbb{Z}_{2n}$. Given an element $a\in\mathbb{Z}_{2}$, let $\bar{a}$ denote the element of $\mathbb{Z}_{2}$ satisfying $a+\bar{a}=1$. For simplicity, we use $(ab,j)$ to denote the element $(a,b,j)\in\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2n}$.

For each $n\geq 2$, define a graph $G_{n}$ as follows. The vertex set of $G_{n}$ is $\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2n}$, and the neighborhood of each vertex $(ab,j)\in\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2n}$ is exactly the following set:

[TABLE]

where $c=a+b\in\mathbb{Z}_{2}$. Note that we always set $a,b\in\{0,1\}$, and so the last vertex $(\bar{a}c,j+2(-1)^{a})$ is either $(\bar{a}c,j+2)$ or $(\bar{a}c,j-2)$.

We observe that $G_{2}$ is exactly the graph constructed by Räty in [13], see the appendix for details. See the left figure of Figure 1 for an illustration of $G_{3}$. Regarding figures representing $G_{n}$, we will use dots in a 2-dimensional square array to denote the vertex set of $G_{n}$. See the right figure of Figure 1 for an illustration.

Now we reveal some structural properties of $G_{n}$. For simplicity, we will omit the parentheses of a vertex $(ab,j)$ if doing so improves the readability. For instance, we use $N_{G_{n}}(ab,j)$ and $f_{n}(ab,j)$ instead of $N_{G_{n}}((ab,j))$ and $f_{n}((ab,j))$, respectively.

Define the following functions from $V(G_{n})$ to $V(G_{n})$ such that for $(ab,j)\in V(G_{n})$,

[TABLE]

In the following, if there is no confusion, we denote $f_{n}$, $p_{n}$, and $q_{n}$ by $f$, $p$, and $q$, respectively. It is clear that $f$ is a bijection by definition. Since $q\circ q$ is the identity function, $q$ is also a bijection. Moreover, $p$ is a bijection, since it has an inverse:

[TABLE]

Lemma 2.1**.**

For $n\geq 2$, the functions $f$, $p$, and $q$ are automorphisms of $G_{n}$.

The proof of Lemma 2.1 is given in the appendix.

Lemma 2.2**.**

For $n\geq 2$, $G_{n}$ is vertex-transitive; that is, for every two vertices $v$ and $v^{\prime}$, there exists an automorphism $\varphi$ such that $\varphi(v)=v^{\prime}$.

Proof.

For a positive integer $j$, let $f^{j}$ denote the composition $\underbrace{f\circ\cdots\circ f}_{j\text{ times}}$. Note that $f^{j}$ maps $(00,0)$ to $(00,j)$, $f^{j}\circ p\circ p$ maps $(00,0)$ to $(01,j)$, $f^{j+1}\circ p$ maps $(00,0)$ to $(10,j)$, and $f^{j+1}\circ p^{-1}$ maps $(00,0)$ to $(11,j)$. By Lemma 2.1, the aforementioned four compositions are automorphisms of $G_{n}$. This completes the proof. ∎

Let us define the following sets (see Figure 2):

[TABLE]

Lemma 2.3**.**

For $n\geq 2$, there is an automorphism $r$ of $G_{n}$ satisfying that $r(00,0)=(00,0)$, $r(11,0)=(00,1)$, and $r(U_{n})\subset L_{n}$.

The proof of Lemma 2.3 is given in the appendix.

Lemma 2.4**.**

For every $w,w^{\prime}\in N_{G_{n}}(00,0)\setminus\{(01,0)\}$, there is an automorphism $\varphi$ such that $\varphi(00,0)=(00,0)$ and $\varphi(w)=w^{\prime}$.

Proof.

Note that $N_{G_{n}}(00,0)\setminus\{(01,0)\}$ has four vertices $w_{1}=(11,0)$, $w_{2}=(00,1)$, $w_{3}=(10,2)$, and $w_{4}=(00,2n-1)$. Note that the automorphism $q$ of $G_{n}$ satisfies $q(00,0)=(00,0)$, $q(w_{1})=w_{3}$, and $q(w_{2})=w_{4}$. By Lemma 2.3, there is an automorphism $r$ of $G_{n}$ such that $r(00,0)=(00,0)$ and $r(w_{1})=w_{2}$. By considering compositions and inverses of $q$ and $r$, we can find an automorphism fixing $(00,0)$ and mapping $w$ to $w^{\prime}$ for every $w,w^{\prime}\in N_{G_{n}}(00,0)\setminus\{(01,0)\}$. ∎

2.2 A longest induced path of $G_{n}$

In this subsection, we will show that a longest induced path of $G_{n}$ has exactly $3n-1$ vertices. We first provide a set of vertices that induces a path on $3n-1$ vertices. As one can see in Figure 3, the set $\displaystyle\left(\bigcup_{i=0}^{n-1}\{(00,2i),(01,2i)\}\right)\cup\left(\bigcup_{i=0}^{\lceil\frac{n}{2}\rceil-2}\{(01,4i+1),(00,4i+3)\}\right)\cup S$ induces a path on $3n-1$ vertices, where

[TABLE]

We now show that $G_{n}$ does not have an induced path on $3n$ vertices.

Proposition 2.5**.**

For $n\geq 2$, a longest path of $G_{n}$ has exactly $3n-1$ vertices.

Proof.

It is sufficient to show that $G_{n}$ does not have an induced path on $3n$ vertices. Suppose to the contrary that for some $n$, $G_{n}$ has an induced path $P$ on $3n$ vertices, and let $P:v_{0}v_{1}\ldots v_{3n-1}$. We take the minimum such $n$. Since $G_{2}$ is isomorphic to the graph in [13] which is $P_{6}$-induced-saturated (see also the appendix), we have $n\geq 3$.

Claim 2.6**.**

If there is an induced path $Q$ of $G_{n}$ in the unshaded part of the left figure of Figure 4, then $Q$ has at most $3n-4$ vertices. Moreover, if such $Q$ starts with one vertex in $\{a_{i},b_{i}\}$ and $|Q\cap\{a_{i},b_{i}\}|=1$ for some $i\in\{1,2\}$, then $Q$ has at most $3n-5$ vertices.

Proof.

We use the labels of the vertices in Figure 4. Let $Q$ be an induced path of $G_{n}$ in the unshaded part of the left figure of Figure 4. We consider the graph $H$ that is isomorphic to $G_{n-1}$, and whose vertices are labeled as the right figure of Figure 4. Let $Q^{\prime}$ be the subgraph of $H$, which is the embedding of $Q$ into $H$. Note that $Q^{\prime}$ is an induced path of $H$, isomorphic to $Q$. By the minimality of $n$, $Q^{\prime}$ has at most $3n-4$ vertices, and so does $Q$.

Suppose that $Q$ starts with one vertex in $\{a_{i},b_{i}\}$ and $|Q\cap\{a_{i},b_{i}\}|=1$ for some $i\in\{1,2\}$. Then $Q^{\prime}$ starts with one vertex in $\{a_{i}^{\prime},b_{i}^{\prime}\}$ and $|Q^{\prime}\cap\{a_{i}^{\prime},b_{i}^{\prime}\}|=1$ for some $i\in\{1,2\}$. Then $x_{i}^{\prime}+Q^{\prime}$ is an induced path of $H$, since the neighbors of $x_{i}^{\prime}$ in the unshaded part are $a_{i}^{\prime}$ and $b_{i}^{\prime}$. Hence, if $|V(Q^{\prime})|\geq 3n-4$, then $x_{i}^{\prime}+Q$ is an induced path on at least $3n-3$ vertices, which is a contradiction to the minimality of $n$. ∎

By Lemma 2.2, we may assume $v_{0}=(00,0)$. Furthermore, by Lemma 2.4, we may assume either $v_{1}=(01,0)$ or $v_{1}=(00,1)$.

Suppose $v_{1}=(01,0)$. Let $S=N_{G_{n}}(v_{0})\cup N_{G_{n}}(v_{1})$, which are the vertices marked with an $\times$ in Figure 5. The graph $G_{n}-S$ is the disjoint union of $K_{2}$ and a graph $H$, where $H$ is induced by the unshaded part of Figure 5. We know $S$ contains at most three vertices of $P$, so $H$ must contain an induced path on $3n-3$ vertices. Yet, by the minimality of $n$, $G_{n-1}$ does not contain an induced path on $3n-3$ vertices. Since $H$ is an induced subgraph of $G_{n-1}$, $H$ also does not contain an induced path on $3n-3$ vertices, which is a contradiction. Thus $v_{1}=(00,1)$, and $v_{2}\in\{(01,1),(00,2),(10,3),(11,1)\}$.

See the left figure of Figure 6 and we use the labels of the vertices as in the figure. The vertices marked with an $\times$ do not belong to $V(P)$ as they are in $N_{G_{n}}(v_{0})\setminus\{v_{1}\}$. Note that exactly one of $y_{i}$, that is $v_{2}$, belongs to $V(P)$.

If $v_{3}\in\{a_{1},a_{2},b_{1},b_{2}\}$, then $v_{2}\in\{y_{3},y_{4}\}$ and so $V(P)\setminus\{v_{0},v_{1},v_{2}\}$ are in the unshaded part of the left figure of Figure 6, which is a contradiction to Claim 2.6. Hence, $v_{3}\not\in\{a_{1},a_{2},b_{1},b_{2}\}$.

Let $Q:v_{4}v_{5}\ldots v_{3n-1}$, which is is an induced path of $G_{n}$ on $3n-4$ vertices. Suppose that $v_{2}\in\{y_{1},y_{3},y_{4}\}$. Since $v_{3}\not\in\{a_{1},a_{2},b_{1},b_{2}\}$, $v_{3}=u_{i}$ for some $i\in\{1,2\}$. It is clear that $v_{4}\in\{a_{i},b_{i}\}$, so $|V(Q)\cap\{a_{i},b_{i}\}|=1$. Note that $V(Q)$ is contained in the unshaded part of the left figure of Figure 6. This is a contradiction to the ‘moreover’ part of Claim 2.6. Hence, $v_{2}=y_{2}=(11,1)$.

Now see the right figure of Figure 6. Note that the vertices marked with an $\times$ are in $N_{G_{n}}(v_{0})\cup N_{G_{n}}(v_{1})$ and so the shaded part of the figure does not contain a vertex in $V(P)\setminus\{v_{0},v_{1},v_{2}\}$. We consider a path $P^{\prime}$ obtained from $P$ by replacing $v_{0}$ and $v_{1}$ with $v^{\prime}_{0}=(11,2)$ and $v^{\prime}_{1}=(10,1)$, respectively. Then $P^{\prime}$ is also an induced path of $G_{n}$ on $3n$ vertices, since the neighbors of $v_{0}^{\prime}$ and $v^{\prime}_{1}$ are in the shaded part. Then the image of $P^{\prime}$ under the automorphism $f^{3}\circ p^{-1}$ is an induced path starting with $v_{0},v_{1},y_{1}$, since

[TABLE]

We can reach a contradiction by the same argument of the previous paragraph. ∎

2.3 A longest induced path of the graph obtained from $G_{n}$ by either adding a non-edge or removing an edge

Proposition 2.7**.**

For $n\geq 2$ and every edge $e$ of $G_{n}$, $G_{n}-e$ contains an induced path on $3n$ vertices.

Proof.

Take an edge $e=vw$ of $G_{n}$. By Lemmas 2.2 and 2.4, we may assume that $v=(00,0)$ and $w\in\{(01,0),(00,2n-1)\}$.

Suppose that $w=(01,0)$. See Figure 7 (i). If $n$ is even, then by adding the vertex $(00,2n-1)$ (the circled vertex of the left figure of Figure 7 (i)) to the path in Figure 3 (i), we obtain an induced path of $G-e$ on $3n$ vertices. If $n$ is odd, then the path obtained from the path in Figure 3 (ii) by adding the vertices $(00,2n-1)$ and $(11,0)$ (the circled vertices of the right figure of Figure 7 (i)) and by deleting the vertex $(01,2n-2)$ (the shaded vertex of the right figure of Figure 7 (i)), we obtain an induced path of $G-e$ on $3n$ vertices. Suppose that the endpoints of $w=(00,2n-1)$. See Figure 7 (ii). If $n$ is even, then by adding the vertex $(00,2n-1)$ (the circled vertex of the left figure of Figure 7 (ii)) to the path in Figure 3 (i), we obtain an induced path of $G-e$ on $3n$ vertices. If $n$ is odd, then the path obtained from the path in Figure 3 (ii) by adding the vertices $(00,2n-1)$ and $(11,2n-1)$ (the circled vertices of the right figure of Figure 7 (ii)) and by deleting the vertex $(01,2n-2)$ (the shaded vertex of the right figure of Figure 7 (ii)) is an induced path of $G-e$ on $3n$ vertices. ∎

Proposition 2.8**.**

For $n\geq 2$ and every non-edge $e$ of $G_{n}$, $G_{n}+e$ contains an induced path on $3n$ vertices.

Proof.

Consider a non-edge $e$ of $G_{n}$, which we may assume one endpoint is $(00,0)$ by Lemma 2.2. Note that $U_{n}\cup L_{n}$ is the set of all non-neighbors of $(00,0)$, where $U_{n}$ and $L_{n}$ are the sets defined before Lemma 2.3. By Lemma 2.3, it is sufficient to show that $G_{n}+e$ contains an induced path on $3n$ vertices, where the other endpoint $w$ of $e$ is in $L_{n}$. In the following, we consider five cases according to $w$. For each case, we provide two figures, depending on the parity of $n$. Each figure shows the edges of an induced path $P$ of $G_{n}$ on $3n-1$ vertices such that $(00,0)$ is an end vertex of $P$. Now, adding a non-edge joining $(00,0)$ and a circled vertex $w$ to $G_{n}$ extends $P$ to an induced path on $3n$ vertices, which proves the proposition. We remark that the shaded part of each figure represents a repeated pattern of the path.

• (i)

$w$ is either $(10,4t)$ or $(11,4t+2)$ for some $t$

• (ii)

$w$ is either $(10,4t+1)$ or $(11,4t+3)$ for some $t$

• (iii)

$w$ is either $(10,4t+2)$ or $(11,4t)$ for some $t$

• (iv)

$w$ is either $(10,4t+3)$ or $(11,4t+1)$ for some $t$

• (v)

$w=(01,2n-1)$

∎

Hence, from Propositions 2.5, 2.7, and 2.8, Theorem 1.2 follows.

3 Proof of Theorem 1.4

Recall that $K(n,2)$ denotes the Kneser graph and let $n\geq 5$ be an integer. First, we will show that $K(n,2)$ has no induced path on 6 vertices. Suppose to the contrary that $K(n,2)$ has an induced path $P:v_{1}v_{2}\ldots v_{6}$. Without loss of generality, we may assume $v_{1}=\{1,2\},v_{2}=\{3,4\}$, and $v_{3}=\{1,5\}$. Since $v_{4}\cap v_{3}=\emptyset$ and $v_{4}\cap v_{i}\neq\emptyset$ for every $i\in\{1,2\}$, we have $2\in v_{4}$, and so without loss of generality we may assume $v_{4}=\{2,3\}$. Similarly, since $v_{5}\cap v_{4}=\emptyset$ and $v_{5}\cap v_{i}\neq\emptyset$ for every $i\in\{1,2\}$, we know $v_{5}=\{1,4\}$. Since $v_{6}\cap v_{5}=\emptyset$ and $v_{6}\cap v_{i}\neq\emptyset$ for every $i\in\{1,2\}$, we know $v_{6}=\{2,3\}=v_{4}$, which is a contradiction.

To show that deleting an arbitrary edge creates an induced $P_{6}$, we may assume the edge joining $\{2,3\}$ and $\{4,5\}$ is deleted, since $K(n,2)$ is arc-transitive. Now, $\{4,5\}$, $\{1,2\}$, $\{3,4\}$, $\{1,5\}$, $\{2,3\}$, $\{1,4\}$ form an induced $P_{6}$. To show that adding a non-edge creates an induced $P_{6}$, we may assume that the non-edge joining $\{1,2\}$ and $\{1,3\}$ is added, since $K(n,2)$ is arc-transitive. Then $\{1,3\}$, $\{1,2\}$, $\{3,4\}$, $\{1,5\}$, $\{2,3\}$, $\{1,4\}$ form an induced $P_{6}$. Hence, the Kneser graph $K(n,2)$ is $P_{6}$-induced-saturated for all $n\geq 5$.

4 Remarks and further research directions

In this paper, we found a $P_{k}$-induced-saturated graph for infinitely many values of $k$. Yet, there are still many values of $k$ for which we do not know if a $P_{k}$-induced-saturated graph exists. The ultimate goal is to answer the following question.

Question 4.1**.**

Can we classify all positive integers $k$ for which a $P_{k}$-induced-saturated graph exists?

As we know there exists a $P_{k}$-induced-saturated graph when $k\in\{2,3\}$ and there is no $P_{k}$-induced-saturated graph when $k=4$, the first open case of the above question is when $k=5$.

Question 4.2**.**

Does there exist a $P_{5}$-induced-saturated graph?

The graph $G_{n}$ (in Subsection 2.1) is $5$-regular. In other words, there are infinitely many integers $k$ such that there is a 5-regular $P_{k}$-induced-saturated graph.

The Kneser graph $K(5,2)$ in Theorem 1.4, which is $P_{6}$-induced-saturated, is 3-regular. For many values of $k$, we found a 3-regular $P_{k}$-induced-saturated graph with the aid of a computer. For example, the Heawood graph is $P_{k}$-induced-saturated for $k\in\{8,9\}$, the generalized Petersen graph $GP(10,3)$ is $P_{k}$-induced-saturated for $k\in\{12,13\}$, the Coxeter graph is $P_{19}$-induced-saturated, the Dyck graph is $P_{k}$-induced-saturated for $k\in\{21,22\}$, and the generalized Pertersen graphs $GP(17,3)$ and $GP(17,6)$ are $P_{23}$-induced-saturated. These findings made us wonder if the following question is true.

Question 4.3**.**

Are there infinitely many integers $k$ for which there is a 3-regular $P_{k}$-induced-saturated graph?

Let $Q^{d}_{n}$ be the graph obtained by adding the diagonals to the $n$-dimensional hypercube $Q_{n}$. The graph constructed by Räty in [13] is isomorphic to $Q^{d}_{4}$. A computer check confirms that $Q^{d}_{5}$ is a $P_{k}$-induced-saturated graph for $k\in\{12,13,14\}$. We think $Q^{d}_{n}$ is a good candidate to investigate, and put forward a question to encourage research in this direction.

Question 4.4**.**

For every $n\geq 4$, does there exist a $k$ such that $Q^{d}_{n}$ is a $P_{k}$-induced-saturated graph?

Appendix

The $P_{6}$-induced-saturated graph given by Räty in [13]

Let $\mathbb{F}_{16}=\mathbb{F}_{2}(\alpha)/(\alpha^{4}+\alpha+1)$ be the finite field with 16 elements. Then the multiplicative group $\mathbb{F}_{16}^{\times}$ is generated by $\alpha$, and $(\mathbb{F}_{16}^{\times})^{3}=\{1,\alpha^{3},\alpha^{3}+\alpha,\alpha^{3}+\alpha^{2},\alpha^{3}+\alpha^{2}+\alpha+1\}.$ In [13], Räty provided a $P_{6}$-induced-saturated graph $H$, defined by $V(H)=\mathbb{F}_{16}$ and $E(H)=\{ab\mid a+b\in(\mathbb{F}_{16}^{\times})^{3}\}$. One can check that $H$ is isomorphic to $G_{2}$ by Figure 8, which shows the image of an isomorphism from $G_{2}$ to $H$.

Proof of Lemma 2.1

It is sufficient to check that each of $f$, $p$, and $q$ moves an edge to an edge. Fix a vertex $(ab,j)\in V(G_{n})$ and let $c=a+b\in\mathbb{Z}_{2}$. See Tables 1 and 2.

One can check that the set of the five vertices in the $i$th row of Table 1 is equal to that of the vertices in the $i$th row of Table 2. Especially when one compares the last rows of the two tables, note that $-j-2(-1)^{a}+2\bar{a}=-j+2a$ and $-j+2\bar{a}=-j+2a+2(-1)^{a}$, which implies that the last vertex of one table matches the first vertex of the other table.

Proof of Lemma 2.3

For simplicity, let $A=T_{0}\cup T_{1}$, where $T_{j}=\{(00,j),(01,j),(10,j+1),(11,j+1)\}$ for $j\in\{0,1\}$. We define $r:V(G_{n})\rightarrow V(G_{n})$ by

[TABLE]

It is clear that $r$ is well-defined, and Figure 9 shows the image of $r$. To be precise,

[TABLE]

We will show that $r$ is an automorphism of $G_{n}$, which proves Lemma 2.3.

We first show that $r$ is a bijection. It is sufficient to check that if $(ab,j)\neq(a^{\prime}b^{\prime},j^{\prime})$ then $r(ab,j)\neq r(a^{\prime}b^{\prime},j^{\prime})$. Since we know that $q$ and $f^{2}\circ p^{-1}$ are bijections, it is enough to check the case where $(a^{\prime}b^{\prime},j^{\prime})\in A$ and $(ab,j)\not\in A$. Suppose that $r(a^{\prime}b^{\prime},j)=r(ab,j)$. Then, $q(a^{\prime}b^{\prime},j)=f^{2}\circ p^{-1}(ab,j)$, which implies that $a^{\prime}=\bar{a}$ and $-j^{\prime}+2a^{\prime}=-j+1$. Suppose that $a=0$. Then $a^{\prime}=1$ and $j^{\prime}=j+1$. Since $(a^{\prime}b^{\prime},j^{\prime})\in A$ and $a^{\prime}=1$, $j^{\prime}\in\{1,2\}$. Then we have $j\in\{0,1\}$, which is a contradiction to $(ab,j)\not\in A$. Similarly, for the case where $a=1$, we have $a^{\prime}=0$ and $j^{\prime}=j-1$, which also implies that $j\in\{1,2\}$, a contradiction to $(ab,j)\not\in A$. Hence, $r$ is an injection.

Now we check that $r$ preserves an edge. Since $q$ and $f^{2}\circ p^{-1}$ are automorphisms of $G_{n}$, it is sufficient to check that $r(v)r(w)$ is an edge for every edge $vw$ of $G_{n}$, where $v\in A$ and $w\in N_{G_{n}}(v)\setminus A$. For each vertex $v\in A$, there are two such neighbors $w$ and one can check all cases from Table 3.

It remains to show that $r(U_{n})\subset L_{n}$, that is, $r(v)\in L_{n}$ for all $v\in U_{n}$. Since $U_{n}\cap A$ contains a unique vertex $(01,1)$ and $r(01,1)=(01,2n-1)$ as in Figure 9, we know that $r(v)\in L_{n}$ for $v\in U_{n}\cap A$. For a vertex $v\in U_{n}\setminus A$, say $v=(0b,j)$ for some $b\in\mathbb{Z}_{2}$ and $j\in\mathbb{Z}_{2n}$, by the definition of $r$ we have $r(0b,j)=(1b^{\prime},-j+1)$ for some $b^{\prime}\in\mathbb{Z}_{2}$, and so $r(0b,j)\in L_{n}$. This completes the proof.