Panchromatic patterns by paths

TL;DR

This paper characterizes certain classes of digraphs related to $H$-paths, kernels, and absorbent sets in arc-colored digraphs, advancing understanding of path-based structures in directed graphs.

Contribution

It provides a complete characterization of the class $ ilde{ ext{B}}_2$ and structural insights into $ ilde{ ext{B}}_3$, except for one specific three-vertex digraph.

Findings

Characterization of $ ilde{ ext{B}}_2$ digraphs.

Structural properties of $ ilde{ ext{B}}_3$ digraphs.

Remaining open case of a three-vertex digraph.

Abstract

Let $H=(V_H,A_H)$ be a digraph, possibly with loops, and let $D=(V_D, A_D)$ be a loopless multidigraph with a colouring of its arcs $c: A_D \rightarrow V_H$. An $H$-path of $D$ is a path $(v_0, \dots, v_n)$ of $D$ such that $(c(v_{i-1}, v_i), c(v_i,v_{i+1}))$ is an arc of $H$ for every $1 \le i \le n-1$. For $u, v \in V_D$, we say that $u$ reaches $v$ by $H$-paths if there exists an $H$-path from $u$ to $v$ in $D$. A subset $S \subseteq V_D$ is $H$-absorbent of $D$ if every vertex in $V_D-S$ reaches by $H$-paths some vertex in $S$, and it is $H$-independent if no vertex in $S$ can reach another (different) vertex in $S$ by $H$-pahts. An $H$-kernel is an independent by $H$-paths and absorbent by $H$-paths subset of $V_D$. We define $\tilde{\mathscr{B}}_1$ as the set of digraphs $H$ such that any $H$-arc-coloured tournament has an $H$-absorbent by paths vertex; the set $\tilde{\mathscr{B}}_2$ consists of the digraphs $H$ such that any $H$-arc-coloured digraph $D$ has an independent, $H$-absorbent by paths set; analogously, the set $\tilde{\mathscr{B}}_3$ is the set of digraphs $H$ such that every $H$-arc-coloured digraph $D$ contains an $H$-kernel by paths. In this work, we present a characterization of $\tilde{\mathscr{B}}_2$, and provide structural properties of the digraphs in $\tilde{\mathscr{B}}_3$ which settle up its characterization except for the analysis of a single digraph on three vertices.

Figures

Equations

$$c(a)=\left\{\begin{array}[]{ll}c_{1}(a)&\text{ if }a\in A_{D_{1}}\\ c_{2}(a)&\text{ if }a\in A_{D_{2}}\\ c_{0}&\text{ if }a=(u,v)\text{ with }u\in V_{D_{1}}\text{ and }v\in V_{D_{2}}\text{ where }c_{0}\text{ may or not be in }H.\end{array}\right.$$

$$c_{xy}((u,v))=\left\{\begin{array}[]{ll}c((u,v))&\text{ if }c((u,v))\neq x\text{ and }c((u,v))\neq y\\ v_{xy}&\text{ if }c((u,v))=x\text{ or }c((u,v))=y.\end{array}\right.$$

$$K=(K^{\prime }\cup \left\{s\in V_D:\hat{s}\in K^{\prime }\right\})-\hat{S}.$$

Full text

We define $\tilde{\mathscr{B}}_{1}$ as the set of digraphs $H$ such that any $H$-arc-coloured tournament has an $H$-absorbent by paths vertex; the set $\tilde{\mathscr{B}}_{2}$ consists of the digraphs $H$ such that any $H$-arc-coloured digraph $D$ has an independent, $H$-absorbent by paths set; analogously, the set $\tilde{\mathscr{B}}_{3}$ is the set of digraphs $H$ such that every $H$-arc-coloured digraph $D$ contains an $H$-kernel by paths.

In this work, we present a characterization of $\tilde{\mathscr{B}}_{2}$, and provide structural properties of the digraphs in $\tilde{\mathscr{B}}_{3}$ which settle up its characterization except for the analysis of a single digraph on three vertices.

Keywords: digraph kernel, $H$-kernel, kernel by monochromatic paths, panchromatic pattern.

1 Introduction

A kernel in a digraph $D$ is a subset $K$ of $V_{D}$ which is independent and absorbent111The set $K$ is absorbent if for every vertex $u\in V_{D}-K$ there exists a vertex $v$ in $K$ such that $(u,v)$ is an arc of $D$.. Numerous generalizations of this concept have been studied in the literature, but one that has received particular attention is the notion of a kernel by monochromatic walks. Given a digraph $D$ whose arc set is coloured with a set of colours $C$, we say that a vertex $u$ reaches a vertex $v$ by monochromatic walks in $D$ if there is a (directed) walk from $u$ to $v$ such that all its arcs have the same colour. With this generalized reachability concept, the notions of independence and absorbance by monochromatic walks come naturally; as one would expect, a subset $S$ of $V_{D}$ is independent by monochromatic walks if no vertex in $S$ can reach another vertex in $S$ by monochromatic walks, and it is absorbent by monochromatic walks if every vertex in $V_{D}-S$ reaches some vertex in $S$ by monochromatic walks. Evidently, a kernel by monochromatic walks is a subset $K$ of $V_{D}$ which is independent and absorbent by monochromatic walks. Sands, Sauer and Woodrow proved in [12] that every digraph whose arc set is coloured with two colours has a kernel by monochromatic walks.

In [11], Linek and Sands further generalized kernels by monochromatic walks by considering a broader notion of reachability. Instead of colouring the arcs of the digraph $D$ with an arbitrary set, they used the vertex set of another digraph $H$; with this setting, instead of only considering monochromatic walks, the arcs of $H$ could be used to codify permitted colour changes in the walks of $D$ to define reachability. Formally, a digraph $H$, possibly with loops will be called a pattern of colours, or pattern for short, and $D$ is an $H$-arc-coloured digraph if it is an irreflexive multidigraph together with a colouring $c$ of its arcs, $c\colon A_{D}\to V_{H}$. An $H$-walk $W$ in $D$ is a walk $W=(x_{0},\dots,x_{k})$ such that $(c(x_{0},x_{1}),\dots,c(x_{k-1},x_{k}))$ is a walk in $H$. For $u,v\in V_{D}$, we say that $u$ reaches $v$ by $H$-walks if there exists an $H$-walk from $u$ to $v$ in $D$. Naturally, this notion of reachability allows us to define independence and absorbance by $H$ walks. A susbset $S$ of $V_{D}$ is independent by $H$-walks if no vertex in $S$ can reach by $H$-walks another vertex in $S$, and it is absorbent by $H$-walks if every vertex in $V_{D}-S$ can reach by $H$-walks some vertex in $S$; is $S$ is both independent by $H$-walks and absorbent by $H$-walks, we say that $S$ is a kernel by $H$-walks. Notice that if the only arcs of $H$ are loops, then the only possible $H$-walks are precisely monochromatic walks, and hence, a kernel by $H$-walks is just a kernel by monochromatic walks.

In this context, Arpin and Linek [1] introduced three interesting classes of patterns. The family $\mathscr{B}_{1}$ is the class of patterns $H$ with the property that for every $H$-arc-coloured tournament $T$ there exists a vertex which is $H$-absorbent by $H$-walks. Noticing that every independent set in a tournament has a single vertex, we may consider a special subclass of $\mathscr{B}_{1}$; the family $\mathscr{B}_{2}$ which consist of the patterns $H$ with the property that for every $H$-arc-coloured digraph $D$ there exists an $H$-absorbent set by walks which is an independent set. Further noticing that every independent by $H$-walks set is independent, the family $\mathscr{B}_{3}$ of the patterns $H$ such that every $H$-arc-coloured digraph contains a kernel by $H$-walks, seems to be a nice subfamily of $\mathscr{B}_{2}$ to consider. The patterns in $\mathscr{B}_{3}$ are known as panchromatic patterns (by walks). In the same article, Arpin and Linek characterized the class $\mathscr{B}_{2}$; it consists of those patterns $H$ with no odd directed cycle in their complement.

In [8] Galeana-Sánchez and Strausz presented a characterization of the class $\mathscr{B}_{3}$. Nonetheless, the authors of the present work recently found that the proof of a crucial lemma in the aforementioned work is incorrect, and thus, the characterization of this class remains as an open problem. Galeana-Sánchez and Hernández-Cruz used this flawed characterization in [7] to prove that every pattern $H$ is either a panchromatic pattern, or the problem of determining whether an $H$-arc-coloured digraph has a kernel by $H$-walks is $NP$-complete; luckily, the main argument is robust enough to remain true even if the family of panchromatic patterns by $H$-walks turns out to be a family different from the one proposed in [8].

All the developements mentioned so far use a notion of reachability that relies on the concept of $H$-walks. It is easy to notice that every monochromatic walk contains a monochromatic path, so it comes to mind to consider reachability by $H$-paths instead of reachability by $H$-walks. Of course, this new notion of reachability yields new notions of independence, absorbance and kernels, but also, analogous to the families $\mathscr{B}_{i}$ might be defined for $i\in\left\{1,2,3\right\}$ using reachability by $H$-paths. So, the aim of the present work is to study these concepts, similarities and differences with the reachability by $H$-walks case. To avoid the cumbersome nomenclature, a kernel by $H$-walks will be simply called an $H$-kernel, and, analogously, $H$-independence and $H$-absorbance will refer to independence by $H$-paths and absorbance by $H$-paths, respectively.

It is worth noticing that reachability problems in arc-coloured digraphs (and tournaments in particular) have received a a lot of attention and many variants are being studied, consider, for an instance, [3, 5, 6, 9].

We refer to [2] and [4] for general concepts. A digraph is reflexive if every vertex has a loop. Let $H$ be a reflexive pattern. Two different vertices $x$ and $y$ of $H$ are true twins if $N^{+}(x)=N^{+}(y)$ and $N^{-}(x)=N^{-}(y)$. Notice that if follows from the previous definition that, if $x$ and $y$ are true twins, then we obtain isomorphic patterns if we delete either of them or we contract the set $\{x,y\}$ and delete any parallel arcs that were created from this contraction. We will denote this kind of contraction of the set $\{x,y\}$ (without parallel arcs) as $H_{xy}$. From the previous observation it is clear that $H_{xy}$ is isomorphic to an induced subdigraph of $H$.

The rest of the paper is organized as follows. In Section 2 some basic results regarding the differences and similarities between reachability by $H$-walks and by $H$-paths are presented, in particular, for a pattern $H$ we propose constructions to find infinite families of $H$-coloured digraphs having $H$-kernel but no kernel by $H$-walks, and vice versa. Section 3 is devoted to prove that the class $\tilde{\mathscr{B}_{2}}$, the $H$-path reachability analogue to $\mathscr{B}_{2}$, is actually equal to $\mathscr{B}_{2}$. In Section 4, we study the patterns $H$ such that for every $H$-arc-coloured digraph, every $H$-walk between two vertices of $D$ contains an $H$-path with the same endpoints. Sections 5 and 6 are devoted to panchromatic patterns by paths; in the former, we analyze thirteen of the patterns on three vertices, and in the latter we deal with two of the remaining three. In Section 7 we point out some properties of the pattern $F_{1}$, which is the only one we were unable to decide whether it is as a panchromatic pattern by paths. Section 8 closes this work by presenting conclusions and some open problems.

2 Basic results

As we mentioned before, our main goal is to obtain a development parallel to that made in [1], but considering reachability by $H$-paths instead of reachability by $H$-walks. Hence, we start defining the following classes of digraphs.

• •

Let $\tilde{\mathscr{B}_{1}}$ the class of patterns $H$ such that every $H$-arc-colourerd tournament contains an $H$-kernel.

• •

Let $\tilde{\mathscr{B}_{2}}$ the class of patterns $H$ such that every $H$-arc-coloured digraph $D$, contains an independent set of $V_{D}$ which is also $H$-absorbent.

• •

Let $\tilde{\mathscr{B}_{3}}$ the class of patterns $H$ such that for every $H$-arc-coloured digraph $D$ contains an $H$-kernel.

By simply considering the definitions it is easy to observe that $\tilde{\mathscr{B}_{3}}\subseteq\tilde{\mathscr{B}_{2}}\subseteq\tilde{\mathscr{B}_{1}}$. Notice that if $H$ is a complete reflexive digraph, then $H\in\tilde{\mathscr{B}_{3}}$. Sands, Sauer and Woodrow proved in [12] that if $H$ consists of two reflexive isolated vertices, then $H\in\tilde{\mathscr{B}_{3}}$, hence $\tilde{\mathscr{B}_{i}}$ is nonempty for $i\in\left\{1,2,3\right\}$. Naturally, we are interested in a characterization of these classes; our following result states some simple observations about them.

Lemma 1**.**

Let $H$ be a pattern.

If $H\in\tilde{\mathscr{B}}_{i}$, then every vertex of $H$ is reflexive, for $i\in\left\{1,2,3\right\}$. 2. 2.

If $H\in\tilde{\mathscr{B}}_{i}$ and $H_{1}$ is an induced subdigraph of $H$ then $H_{1}\in\tilde{\mathscr{B}}_{i}$, for $i\in\left\{1,2,3\right\}$. 3. 3.

If $H\in\tilde{\mathscr{B}}_{i}$ and $H_{1}$ is a spanning superdigraph of $H$, then $H_{1}\in\tilde{\mathscr{B}}_{i}$, for $i\in\left\{1,2\right\}$.

Proof.

For the first item, consider a pattern $H$ with a vertex $x$ such that $(x,x)\notin A_{H}$, then the $H$-arc-coloured digraph $D$ obtained by colouring each arc of $C_{3}$ with $x$ does not have an $H$-kernel, and thus, $H\notin\tilde{\mathscr{B}}_{i}$ for $i\in\left\{1,2,3\right\}$.

For (2), since every $H_{1}$-arc-colouring is also an $H$-arc-colouring, then any example showing $H_{1}\notin\tilde{\mathscr{B}}_{i}$ also shows that $H\notin\tilde{\mathscr{B}}_{i}$, for $i\in\left\{1,2,3\right\}$.

For (3) we have that any $H$-path is also an $H_{1}$-path, thus $H_{1}\in\tilde{\mathscr{B}}_{i}$, for $i\in\left\{1,2\right\}$. ∎

Since the existence of an $H$-walk between two vertices does not guarantee the existence of an $H$-path between those vertices and the concatenation of two $H$-paths is not always an $H$-walk, we claim that if $D$ has a kernel by $H$-walks, then $D$ not necessarily has an $H$-kernel, as the example in Figure 1 shows. In Figure 1 we have that $\left\{w\right\}$ is a kernel by $H$-walks of $D$, because $(x,y,z,y,w)$ is a spanning $H$-walk in $D$ that finishes in $w$. It is easy to check that $D$ has no $H$-kernel (notice that every $H$-independent set by paths of $D$ has cardinality one).

We also claim that if $D$ has an $H$-kernel, then $D$ not necessarily has a kernel by $H$-walks as the example in Figure 2 shows. In Figure 2 we have that $\left\{u,v\right\}$ is an $H$-kernel of $D$. It is easy to see that $D$ has no $H$-kernel by walks (notice that every independent by $H$-walks set in $D$ has cardinality one because $(u,y,z,y,v)$ is an $H$-walk between in $D$).

To convince the reader that these are not isolated cases, we now exhibit inifinite families with the same behaviour as the previous two examples.

Given two multidigraphs $D_{1}$ and $D_{2}$, we define the multidigraph $D_{1}\bullet D_{2}$, which we call the linear sum of $D_{1}$ and $D_{2}$, to have vertex set $V_{D_{1}\bullet D_{2}}=V_{D_{1}}\cup V_{D_{2}}$ (disjoint union) and arc set $A_{D_{1}\bullet D_{2}}=A_{D_{1}}\cup A_{D_{2}}\cup V_{D_{1}}\times V_{D_{2}}$.

Let $H$ be a pattern, and let $D_{1}$ and $D_{2}$ be two $H$-arc-coloured digraphs with colourings $c_{1}$ and $c_{2}$, respectively. Consider the linear sum of $D_{1}$ and $D_{2}$. We define an $H$-arc-colouring $c$ of $D_{1}\bullet D_{2}$ as follows, $c:A_{D_{1}\bullet D_{2}}\to V_{H}$ is such that

[TABLE]

Based on the previous $H$-arc-colouring we can obtain the following lemmas.

Lemma 2**.**

Let $H$ be a pattern. If $D_{1}$ and $D_{2}$ are $H$-arc-coloured digraphs, then $D_{1}\bullet D_{2}$ has an $H$-kernel and has no kernel by $H$-walks if and only if $D_{2}$ has $H$-kernel and has no kernel by $H$-walks.

Proof.

Suppose first that $D_{1}\bullet D_{2}$ has an $H$-kernel. We will show that $D_{2}$ has an $H$-kernel and has no kernel by $H$-walks. Let $K$ be an $H$-kernel of $D_{1}\bullet D_{2}$. We will prove that $K$ is an $H$-kernel of $D_{2}$.

Since every vertex in $V_{D_{1}}$ dominates every vertex in $V_{D_{2}}$, it is not hard to notice that $K\subseteq V_{D_{2}}$. On the other hand, by the definition of linear sum and its associated colouring $c$, it follows that any $H$-path in $D_{1}\bullet D_{2}$ between two vertices $x$ and $y$ in $V_{D_{2}}$, is also an $H$-path in $D_{2}$. Hence, since $K$ is an $H$-kernel of $D_{1}\bullet D_{2}$, then $K$ is $H$-kernel of $D_{2}$.

We will prove that if $D_{1}\bullet D_{2}$ does not have a kernel by $H$-walks, then neither does $D_{2}$ by contrapositive. Again, by the definition of $D_{1}\bullet D_{2}$, every vertex in $V_{D_{1}}$ dominates every vertex in $V_{D_{2}}$ and no vertex in $V_{D_{2}}$ dominates a vertex in $V_{D_{1}}$. Also, the definition of $c$ guarantees that an $H$-walk in $D_{2}$ is also an $H$-path in $D_{1}\bullet D_{2}$. It follows that a kernel by $H$-walks in $D_{2}$ is also kernel by $H$-walks in $D_{1}\bullet D_{2}$.

Analogously, an $H$-kernel of $D_{2}$ is also an $H$-kernel of $D_{1}\bullet D_{2}$. So, proceeding again by contrapositive, assume that $D_{1}\bullet D_{2}$ has a kernel by $H$-walks $K$. Again, by the structure of $D_{1}\bullet D_{2}$, it must be the case that $K\subseteq V_{D_{2}}$. Also, every $H$-walk in $D_{1}\bullet D_{2}$ between vertices of $D_{2}$ is an $H$-walk in $D_{2}$. Thus, $K$ is a kernel by $H$-walks in $D_{2}$. ∎

The proof of our following lemma is analogous to the proof of Lemma 2, so we omit it.

Lemma 3**.**

Let $H$ be a pattern. If $D_{1}$ and $D_{2}$ are $H$-arc-coloured digraphs, then $D_{1}\bullet D_{2}$ has a kernel by $H$-walks and has no $H$-kernel if and only if $D_{2}$ has a kernel by $H$-walks and has no $H$-kernel.

As noted above, using the example in Figure 2 and Lemma 2 we can recursively construct an infinite family of digraphs having an $H$-kernel and no kernel by $H$-walks as follows. Let $H_{2}$ and $D_{2}$ be digraphs as in Figure 2. Define

• •

$D^{1}=D_{2}$.

• •

$D^{j+1}=K_{1}\bullet D^{j}$.

The resulting family of digraphs clearly has the desired properties.

Similarly, using the example in 1 and Lemma 3 an infinite family of digraphs having a kernel by $H$-walks and has no $H$-kernel can be recursively defined as follows. Let $H_{1}$ and $D_{1}$ be digraphs as in Figure 2.

• •

$E^{1}=D_{1}$.

• •

$E^{j+1}=K_{1}\bullet E^{j}$.

Hence, the problem of determining the existence of an $H$-kernel and the problem of determining the existence of a kernel by $H$-walks are indeed different problems. In view of these examples, the results of the following sections come a bit as a surprise.

3 The $\tilde{\mathscr{B}}_{2}$ class

In this section we characterize the patterns that belong to $\tilde{\mathscr{B}}_{2}$ as the reflexive digraphs such that their complement do not contain odd directed cycles, i.e., $\tilde{\mathscr{B}}_{2}=\mathscr{B}_{2}$.

Lemma 4**.**

Let $H_{1}$ and $H_{2}$ be reflexive patterns. If $H_{1}$, $H_{2}\in\tilde{\mathscr{B}}_{2}$, then $H_{1}\bullet H_{2}\in\tilde{\mathscr{B}}_{2}$.

Proof.

Let $D$ be an $H_{1}\bullet H_{2}$-arc-colured multidigraph. For $i\in\{1,2\}$, let $D_{i}$ be the spanning subdigraph of $D$ with arc set consisting of precisely those arcs of $D$ coloured with the vertices of $H_{i}$. Let $K$ be an independent $H_{1}$-absorbent set of $D_{1}$, and let $K^{\prime}$ be an independent $H_{2}$-absorbent set of $D_{2}[K]$, the subdigraph induced by $K$ in $D_{2}$. Clearly, $K^{\prime}$ is independent in $D$; to verify that $K^{\prime}$ is $(H_{1}\bullet H_{2})$-absorbent, let $x$ be a vertex of $D$ not in $K^{\prime}$. If $x\in K$, then the choice of $K^{\prime}$ guarantees that $x$ is $H_{2}$-absorbed by it. If $x\notin K$, then there is an $H_{1}$-path $P_{1}$ in $D_{1}$ (and thus also in $D$) from $x$ to some vertex $y\in K$. If $y\notin K^{\prime}$, then, as in the previous case, there is an $H_{2}$-path $P_{2}$ from $y$ to some vertex $z\in K^{\prime}$ in $D_{1}[K]$. Let $s$ be the last vertex in $P_{2}$ which is also in $P_{1}$, then, by the structure of $H_{1}\bullet H_{2}$, we have that $xP_{1}sP_{2}z$ is an $(H_{1}\bullet H_{2})$-path in $D$. Therefore, $K^{\prime}$ is an independent $(H_{1}\bullet H_{2})$-absorbent set. ∎

Theorem 5**.**

If $H$ is a reflexive pattern, then $H\in\tilde{\mathscr{B}}_{2}$ if and only if $H^{c}$ does not contain any odd cycles.

Proof.

For the necessity, we prove the contrapositive. If $H^{c}$ contains the odd cycle $(0,1,2,\dots,2k,0)$ then let $D$ be the odd cycle $D=(x_{0},x_{1},\dots,x_{2k},x_{0})$ and $c(x_{i},x_{i+1})=i$, for $0\leq i\leq 2k-1$, and $c(x_{2k},x_{0})=2k$. Every $H$-path in $D$ is a single arc, hence $D$ has no independent $H$-absorbent set and $D\notin\tilde{\mathscr{B}}_{2}$.

For the sufficiency, let us suppose that $H^{c}$ has no odd cycle. Let $H_{1},\dots,H_{k}$ be an ordering of the strong components of $H^{c}$, such that $H_{i}$ is an initial component of $H^{c}-(\bigcup_{j=1}^{i-1}H_{j})$. Since every $H_{i}$ is strongly connected and has no odd cycles, then it is a bipartite digraph with bipartition $(V_{i1},V_{i2})$, for $1\leq i\leq k$. It follows that in $H$ every $V_{ij}$ induces a complete reflexive digraph, for $1\leq i\leq k$ and for $1\leq j\leq 2$. Then, by Lemma 1.3, $H_{i}\in\tilde{\mathscr{B}}_{2}$.

By definition of $H_{i}$ in $H$, $H^{\prime}=H_{k}\bullet(\cdots\bullet(H_{3}\bullet(H_{2}\bullet H_{1}))\cdots)$ is a spanning subdigraph of $H$. Then, it follows from the previous lemma that $H\in\tilde{\mathscr{B}}_{2}$. ∎

4 Transitive patterns

In this section we present a characterization of the patterns $H$, such that for every digraph $D$ and every $H$-arc-colouring of $D$, every $H$-walk between two vertices in $D$ contains an $H$-path with the same endpoints.

Given a set of digraphs $\mathcal{S}$, we say that a digraph is $\mathcal{S}$-free if it does not contain any digraph in $\mathcal{S}$ as an induced subdigraph. Recall that if a property $\mathcal{P}$ is closed under taking induced subdigraphs, then there exists a family of forbidden digraphs, $\mathcal{F}$, not necessarily finite, that characterize the digraphs with that property. In other words, a digraph satisfies the property $\mathcal{P}$ if and only if it is $\mathcal{F}$-free. As an example, consider the following theorem.

Theorem 6**.**

If $H$ is a reflexive digraph, then $H$ is a transitive digraph if and only if it is $\mathcal{T}$-free, where $\mathcal{T}$ is the family of reflexive digraphs depicted in Figure 3.

Proof.

The necessity is trivial. The sufficiency follows from a simple analysis of cases considering two consecutive arcs in a $\mathcal{T}$-free digraph $D$. ∎

If we consider the elements of $\mathcal{T}$ as colour patterns $H$, each of them is an example showing that not every $H$-walk contains an $H$-path with the same endpoints; see Figure 3. Our following result states that these are the only (minimal) patterns with this property.

Theorem 7**.**

Let $H$ be a reflexive digraph. The following statements are equivalent.

$H$* is transitive.* 2. 2.

$H$* is $\mathcal{T}$-free.* 3. 3.

For every $H$-arc-coloured digraph $D$, and every pair of different vertices $x,y$ of $D$, every $H$-walk from $x$ to $y$ in $D$ contains an $H$-path from $x$ to $y$ in $D$.

Proof.

The equivalence of (1) and (2) has been already proved in Theorem 6.

We will prove (3) implies (2) by contrapositive. Suppose that $H$ is a reflexive digraph containing one of the patterns in Figure 3 as an induced subdigraph. Hence, the digraph at the bottom right of the same figure is an $H$-arc-coloured digraph with vertices $x$ and $w$ such that an $H$-walk exists from $x$ to $w$, but no $H$-path exists with the same endpoints.

For (1) implies (3), suppose that $H$ is transitive reflexive digraph. Let $D$ be an $H$-arc-coloured digraph, let $x$ and $y$ be different vertices in $D$ and let $\gamma=(x=x_{0},\dots,x_{n}=y)$ be an $H$-walk from $x$ to $y$ in $D$. Proceeding by induction on the length of $\gamma$, $\ell(\gamma)$, we will prove that $\gamma$ contains an $H$-path from $x$ to $y$ in $D$.

If $\ell(\gamma)=1$, then $\gamma$ is an $H$-path from $x$ to $y$ in $D$. Let $\gamma=(x=x_{0},x_{1},\dots,x_{n}=y)$ be an $H$-walk from $x$ to $y$ in $D$ of length $n$. If $x_{i}\neq x_{j}$ for each $i\neq j$, then $\gamma$ is an $H$-path from $x$ to $y$ in $D$. Else, consider $i,j\in\{0,\dots,n\}$, such that $x_{i}=x_{j}$ and $i<j$. Since $\gamma$ is an $H$-walk in $D$, we have that $(c(x_{k-1},x_{k}),c(x_{k},x_{k+1}))\in A_{H}$ for each $k\in\{1,\dots,n-1\}$. Moreover, as $(c(x_{i-1},x_{i}),c(x_{i},x_{i+1}),\dots,c(x_{j-1},x_{j}),c(x_{j},x_{j+1}))$ is a directed walk in $H$, then $(c(x_{i-1},x_{i}),c(x_{j},x_{j+1}))\in A_{H}$ by the transitivity of $H$. Therefore $\gamma^{\prime}=(x=x_{0},x_{1},\dots,x_{i}=x_{j},x_{j+1},\dots,x_{n}=y)$ is an $H$-walk form $x$ to $y$ in $D$ of length less than $n$. By induction hypothesis, $\gamma^{\prime}$ contains an $H$-path from $x$ to $y$ in $D$, say $T$ which is also contained in $\gamma$. ∎

The above result allows us to establish the following theorem.

Theorem 8**.**

Let $H$ be a transitive reflexive digraph and $D$ an $H$-arc-coloured digraph. The subset $K$ of $V_{D}$ is a kernel by $H$-walks of $D$ if and only if it is an $H$-kernel of $D$.

Proof.

Let $K\subseteq V_{D}$ be a kernel by $H$-walks of $D$. Since $K$ is independent by $H$-walks in $D$ and every $H$-path is an $H$-walk in $D$ then, in particular, $K$ is an $H$-independent set in $D$. To show that $K$ is an $H$-absorbent set in $D$, consider a vertex $x$ in $V_{D}-K$. Since $K$ is absorbent by $H$-walks in $D$, then there is a $w$ in $K$ such that there is an $H$-walk from $x$ to $w$ in $D$, say $\gamma$. By Theorem 7 $\gamma$ contains an $H$-path from $x$ to $w$ in $D$. Therefore $K$ is an $H$-kernel of $D$.

Conversely, let $K\subseteq V_{D}$ be an $H$-kernel of $D$. Every $H$-path is an $H$-walk in $D$, so we have that $K$ is absorbent by $H$-walks. It follows from Theorem 7 that every $H$-walk contains an $H$-path; since there are no $H$-paths between vertices in $K$, it follows that there are neither $H$-walks between vertices of $K$, and thus, $K$ is independent by $H$-walks. ∎

5 Panchromatic patterns by paths

We now turn our attention to $\tilde{\mathscr{B}}_{3}$. In [7], a characterization of panchromatic patterns (by walks) was given in terms of forbidden subdigraphs and, although it is based on the characterization of $\mathscr{B}_{3}$ found in [8] (which we mentioned earlier is flawed), a similar approach can be used to describe the structure of the panchromatic patterns by paths. The idea is to classify all the patterns on three vertices; knowing which of them are panchromatic patterns by paths gives us enough information to describe the general structure of patterns in this family.

We begin our analysis with a simple observation relating $H$-kernels with $H$-kernels by walks. It is a direct consequence of Theorem 8, so its proof will be omitted.

Proposition 9**.**

If $H$ is a transitive reflexive digraph, then $H$ is a panchromatic pattern by paths if and only if $H$ is a panchromatic pattern (by walks).

Arpin and Linek showed in [1] that every reflexive digraph with order one or two is a panchromatic pattern. These patterns are trivially transitive, so by Proposition 9 we have that the panchromatic patterns with order one or two are also panchromatic patterns by paths.

Using Proposition 9 we can analyze nine of the patterns on three vertices depicted in Figure 4. The patterns $K_{3}$, $K_{1}\bullet K_{2}$, $K_{2}\bullet K_{1}$ and $K_{2}+K_{1}$ are transitive and belong to $\mathscr{B}_{3}$, and hence, they also belong to $\tilde{\mathscr{B}}_{3}$. Analogously, the patterns $2K_{1}\bullet K_{1}$, $T_{3}$, $(K_{1}\bullet K_{1})+K_{1}$, $K_{1}\bullet 2K_{1}$ and $3K_{1}$ do not belong to $\tilde{\mathscr{B}}_{3}$.

The following result was proved by Arpin and Linek in [1]; although they stated it for panchromatic patterns, it is easy to notice that the same proof works for panchromatic patterns by paths.

Lemma 10**.**

Let $H$ be a pattern. If $H$ contains a walk $W$ such that $W=(x_{0},x_{1},\dots,w_{k})$ and

for all $x_{j}$, $0\leq j\leq k-1$, there is a colour $c_{j}\in V_{H}$ such that $(x_{j},c_{j})\notin A_{H}$, 2. 2.

$(x_{k},x_{0})\notin A_{H}$,

then $H\notin\tilde{\mathscr{B}_{3}}$.

As in [1], Lemma 10 has the following simple consequence dealing with four of the remaining non-transitive patterns on three vertices.

Proposition 11**.**

None of the patterns in Figure 5 is in a panchromatic pattern by paths.

Proof.

The digraph $D$ In Figure 5, does not contain an $H$-kernel for any of the four patterns $H$ on three vertices found in the same figure. It is worth noticing that the digraph $D$ was obtained using the construction given by the proof of Lemma 10. ∎

Our following lemma states that contraction of true twins and its inverse operation, blowing up vertices to complete reflexive digraphs, preserve panchromatic patterns by paths.

Lemma 12**.**

Let $H$ be a reflexive pattern, and let $x$ and $y$ be a pair of true twins in $H$. Then, $H$ is a panchromatic pattern by paths if and only if $H_{xy}$ is.

Proof.

Since $H_{xy}$ is isomorphic to $H-y$, an induced subdigraph of $H$, it follows from Lemma 1 that if $H\in\tilde{\mathscr{B}_{3}}$, then, also $H_{xy}\in\tilde{\mathscr{B}}_{3}$.

So, assume that $H_{xy}$ is a panchromatic pattern by paths and let $D$ be an $H$-arc-coloured digraph with colouring $c$. We will abuse notation to consider the digraph $D$ with two different colourings of its arc set, $c$ and $c_{xy}$, where the latter is an $H_{xy}$-arc-colouring of $D$ defined as follows:

[TABLE]

By hypothesis, $D$ has an $H_{xy}$-kernel, $K$. We will prove that $K$ is also an $H$-kernel of $D$.

To verify the $H$-independence, let $u$ and $v$ be different vertices in $K$. Proceeding by contradiction, suppose that there is an $H$-path from $u$ to $v$ in $D$, say $P=(u=w_{0},w_{1},\dots,w_{n}=v)$. Since $K$ is an $H_{xy}$-kernel of $D$, then $P$ is not an $H_{xy}$-path in $D$. So, it follows that there is an arc $(w_{j},w_{j+1})$ in $A(P)$ such that $c((w_{j},w_{j+1}))=x$ or $c((w_{j},w_{j+1}))=y$; assume without loss of generality that $c((w_{j},w_{j+1}))=x$. By definition of $c_{xy}$, we have that $c_{xy}((w_{j},w_{j+1}))=v_{xy}$. Since $H_{xy}$ is reflexive, the length of $P$ must be greater than one, otherwise $P$ would be an $H_{xy}$-path in $D$. If $j>0$, then there is a vertex $z$ in $V_{H}$ such that $(z,x)\in A_{H}$ and $(c((w_{j-1},w_{j})),c((w_{j},w_{j+1})))=(z,x)$. So, by the definition of $H_{xy}$ we have that $(c_{xy}((w_{j-1},w_{j})),c_{xy}((w_{j},w_{j+1})))=(z,v_{xy})\in A(H_{xy})$. Analogously, if $j<n-1$, there is $z^{\prime}$ in $V_{H}$ such that $(c_{xy}((w_{j},w_{j+1})),c_{xy}((w_{j+1},w_{j+2})))=(v_{xy},z^{\prime})\in A(H_{xy})$. Notice that either $z$ or $z^{\prime}$ could be $x$ or $y$, but $(v_{xy},v_{xy})$ is also an arc of $H_{xy}$. Therefore, we conclude that $P$ is an $H_{xy}$-path in $D$, contradicting that $K$ is an $H_{xy}$-kernel of $D$. Therefore, $K$ is $H$-independent.

To prove that $K$ is $H$-absorbent, let $u$ be a vertex in $V_{D}-K$. Since $K$ is an $H_{xy}$-kernel of $D$, then there exists $v\in K$ such that there is an $H_{xy}$-path from $u$ to $v$ in $D$, say $P=(u=w_{0},w_{1},\dots,w_{n}=v)$. We will show that $P$ is also an $H$-path of $D$. If $c_{xy}((w_{j},w_{j+1}))\neq v_{xy}$, for $0\leq j\leq n-1$, then $P$ is trivially an $H$-path of $D$. Suppose there exists $(w_{i},w_{i+1})\in A(T)$ such that $c_{xy}((w_{i},w_{i+1}))=v_{xy}$. By definition of $c_{xy}$, we have that $c((w_{i},w_{i+1}))=x$ or $c((w_{i},w_{i+1}))=y$, but since $x$ and $y$ are true twins, we can assume without loss of generality that $c((w_{i},w_{i+1}))=x$. Since $P$ is an $H_{xy}$-path of $D$ and by the definition of $H_{xy}$, it follows that, if $i>0$, then $(c((w_{i-1},w_{i})),c((w_{i},w_{i+1})=x))$ is an arc of $H$ and if $i<n-1$, then $(x=c((w_{i},w_{i+1})),c((w_{i+1},w_{i+2})))$ is an arc of $H$. Therefore, $P$ is an $H$-path of $D$ and hence, $K$ is an $H$-absorbent set of $D$.

We conclude that $K$ is an $H$-kernel of $D$. ∎

So far, we have dealt with thirteen of the sixteen reflexive patterns on three vertices. The remaining three patterns are harder to deal with. The following two sections are devoted to patterns $F_{4}$, $F_{5}$ and $F_{1}$.

6 The patterns $F_{4}$ and $F_{5}$

In [8], it is proved that patterns $F_{4}$ and $F_{5}$ are not panchromatic patterns. Nonetheless, while adapting the proof for the $H$-kernel case, we noticed that the proof of the key lemma to prove this fact in [8] is flawed. At this moment it is neither known whether $F_{4}\in\mathscr{B}_{3}$ nor whether $F_{5}\in\mathscr{B}_{3}$. Despite this fact, a techique similar to the one proposed in [8] can be used to prove that $F_{4},F_{5}\notin\tilde{\mathscr{B}_{3}}$.

Lemma 13**.**

The pattern $H$ in Figure 6 is not a panchromatic pattern by paths.

Proof.

Consider the digraphs $H$ and $H^{\prime}$ as in Figure 6. We start by showing that for every $H^{\prime}$-coloured digraph $D$ there is an $H$-coloured multidigraph $D^{\prime}$, such that if $D^{\prime}$ has an $H$-kernel, then $D$ has a kernel by $H^{\prime}$-paths. Let $D$ be an $H^{\prime}$-coloured digraph and construct $D^{\prime}$ from $D$ as follows. For each path $(r,s,t)$ in $D$ with arcs $(r,s)$ and $(s,t)$ coloured by $x$ and $w$, respectively, we add a new vertex $\hat{s}$, with two arcs from $s$ to $\hat{s}$, one of them coloured $y$ and the other coloured $w$, also, an arc $(\hat{s},s)$ coloured $y$ and an arc $(\hat{s},t)$ coloured $w$. Let $\hat{S}$ be the set of all those new vertices $\hat{s}$ in $D^{\prime}$. (See Figure 6.)

Suppose $K^{\prime}\subseteq V_{D^{\prime}}$ is a kernel by $H$-paths in $D^{\prime}$. Consider the set $K$ defined as

[TABLE]

We affirm that $K$ is an $H^{\prime}$-kernel in $D$. For the $H^{\prime}$-independence in $D$, consider vertices $u$ and $v$ in $K$. Proceeding by contradiction suppose that there is an $H^{\prime}$-path $P=(u=u_{0},u_{1},\ldots,u_{n}=v)$ from $u$ to $v$ in $D$. Note that $P$ is a path in $D^{\prime}$. If $P$ is not an $H$-path in $D^{\prime}$, then by construction there is at least one subpath $(r,s,t)$ of $P$ such that $(r,s)$ has colour $x$ and $(s,t)$ has colour $w$. Consider the $H$-path $(r,s,\hat{s},t)$ with $(r,s)$ coloured $x$, $(s,\hat{s})$ coloured by $y$ and $(\hat{s},t)$ coloured by $w$. Let $P^{\prime}$ be the path that results from making such a replacement for each path with form $(r,s,t)$ with the respective $(r,s,\hat{s},t)$. By construction $P^{\prime}$ is an $H$-path from $u$ to $v$ in $D^{\prime}$. We have 4 cases.

Case 1. $\{u,v\}\subseteq K^{\prime}-\left\{s\in V_{D}\colon\ \hat{s}\in K^{\prime}\right\}$.

It follows that $P$ or $P^{\prime}$ is an $H$-path between two vertices of $K^{\prime}$ in $D^{\prime}$, contradicting the independence by $H$-paths of $K^{\prime}$.

Case 2. $u\in\left\{s\in V_{D}\colon\ \hat{s}\in K^{\prime}\right\}$ and $v\in K^{\prime}-\left\{s\in V_{D}\colon\ \hat{s}\in K^{\prime}\right\}$.

As $N^{+}_{H}(y)=V_{H}$, then adding the arc $(\hat{s},s=u)$ with colour $y$ at the beginning of $P$ (or $P^{\prime}$) results in an $H$-path between two vertices of $K^{\prime}$ in $D^{\prime}$, contradicting the independence by $H$-paths of $K^{\prime}$.

Case 3. $u\in K^{\prime}-\left\{s\in V_{D}\colon\ \hat{s}\in K^{\prime}\right\}$ and $v\in\left\{s\in V_{D}\colon\ \hat{s}\in K^{\prime}\right\}$.

As $N^{-}_{H}(y)=\left\{x,y\right\}$ and $N^{-}_{H}(w)=\left\{z,w\right\}$, then adding the arc $(s,\hat{s})$ with colour $y$, if $(u_{n-1},u_{n}=v)$ is coloured $x$ or $y$, at the end of $P$ (or $P^{\prime}$), or adding the arc $(s,\hat{s})$ with colour $w$, if $(u_{n-1},u_{n}=v)$ is coloured $z$ or $w$, at the end of $P$ (or $P^{\prime}$) results in an $H$-path between two vertices of $K^{\prime}$ in $D^{\prime}$, contradicting the $H$-independence of $K^{\prime}$.

Case 4. $\left\{u,v\right\}\subseteq\left\{s\in V_{D}\colon\ \hat{s}\in K^{\prime}\right\}$.

As $N^{+}_{H}(y)=V_{H}$, $N^{-}_{H}(y)=\left\{x,y\right\}$ and $N^{-}_{H}(w)=\left\{z,w\right\}$, then adding the arc $(\hat{s},s=u)$ coloured $y$ at the beginning of $P$ (or $P^{\prime}$) and adding the arc $(s,\hat{s})$ with colour $y$, if $(u_{n-1},u_{n}=v)$ is coloured $x$ or $y$, at the end of $P$ (or $P^{\prime}$), or adding the arc $(s,\hat{s})$ with colour $w$, if $(u_{n-1},u_{n}=v)$ is coloured $z$ or $w$, at the end of $P$ (or $P^{\prime}$) results in an $H$-path between two vertices of $K^{\prime}$ in $D^{\prime}$, contradicting the $H$-independence $K^{\prime}$.

To show that $K$ is $H^{\prime}$-absorbent in $D$. Consider $u\in V_{D}-K$. By definition of $K$, we have $u\in V_{D^{\prime}}-K^{\prime}$. Since $K^{\prime}$ is an $H$-kernel in $D^{\prime}$ there is $v\in K^{\prime}$ such that $u$ reaches $v$ by $H$-paths in $D^{\prime}$. Let $P=(u=u_{0},u_{1},\ldots,u_{n}=v)$ be an $H$-path in $D^{\prime}$. Suppose that $v\notin\hat{S}$. If $\hat{s}\in V(P)$, then $\hat{s}=u_{i}$ for some $i\in\left\{1,\ldots,n-1\right\}$, by construction of $D^{\prime}$, it follows that $s=u_{i-1}$, then $(u_{i-1},u_{i+1})\in A_{D}$ with colour $w$, even more, $(u_{i},u_{i+1})$ has colour $w$, thus, since $N^{-}_{H^{\prime}}(w)=V_{H^{\prime}}$, $(u_{i-2},u_{i-1},u_{i+1},u_{i+2})$ is an $H^{\prime}$-path. It follows that removing all $\hat{s}$ of $P$ an $H^{\prime}$-path from $u$ to $v$ in $D$ is obtained. If $v\in\hat{S}$, then $u_{n-1}\in K$ hence the above procedure gives us an $H^{\prime}$-path from $u$ to $K$ in $D$. Thus, $K$ is absorbent by $H^{\prime}$-paths in $D$ and therefore $K$ is a kernel by $H^{\prime}$-paths in $D$.

Notice that the subdigraph of $H^{\prime}$ induced by $\left\{x,z,w\right\}$ does not belong to $\hat{\mathscr{B}}_{3}$, and hence, $H^{\prime}\notin\hat{\mathscr{B}}_{3}$. Therefore, there is an $H^{\prime}$-coloured digraph $D$ such that $D$ has no $H^{\prime}$-kernel. Let $D^{\prime}$ be constructed as before. If $H\in\hat{\mathscr{B}}_{3}$, then $D^{\prime}$ has a kernel by $H$-paths, thus by the previous argument $D$ has a kernel by $H^{\prime}$-paths, which cannot happen. Therefore $H\notin\hat{\mathscr{B}}_{3}$. ∎

We will say that a vertex $z\in V_{H}$ is universal if and only if $N^{-}_{H}(z)=N^{+}_{H}(z)=V_{H}$.

The following lemma is very similar in statement and proof to the previous one.

Lemma 14**.**

The pattern $H$ in Figure 7 is not a panchromatic pattern by paths.

Proof.

Let $H$ and $H^{\prime}$ be the digraphs depicted in Figure 7. As in the previous lemma, we will show that for every $H^{\prime}$-coloured digraph $D$ there is an $H$-coloured multidigraph $D^{\prime}$, such that if $D^{\prime}$ has an $H$-kernel, then $D$ has an $H^{\prime}$-kernel. For an $H^{\prime}$-coloured digraph $D$, construct $D^{\prime}$ from $D$ as follows. For each path $(r,s,t)$ in $D$ with arcs $(r,s)$ and $(s,t)$ coloured $x$ and $w$, respectively, we add a new vertex $\hat{s}$, the arcs $(s,\hat{s})$, $(\hat{s},s)$ coloured $y$ and the arc $(\hat{s},t)$ coloured $w$. Let $\hat{S}$ be the set of all those new vertices $\hat{s}$ in $D^{\prime}$. (See Figure 7.)

Suppose $K^{\prime}\subseteq V_{D^{\prime}}$ is an $H$-kernel in $D^{\prime}$, and consider the set $K$ defined as

[TABLE]

We will show that $K$ is an $H^{\prime}$-kernel in $D$. For the $H^{\prime}$-independence in $D$, consider two different vertices $u$ and $v$ in $K$. Proceeding by contradiction suppose that there is an $H^{\prime}$-path $P=(u=u_{0},u_{1},\ldots,u_{n}=v)$ from $u$ to $v$ in $D$. Note that $P$ is a path in $D^{\prime}$. If $P$ is not an $H$-path in $D^{\prime}$, by construction there is at least on subpath $(r,s,t)$ of $P$ such that $(r,s)$ with colour $x$ and $(s,t)$ with colour $w$. Consider the $H$-path $(r,s,\hat{s},t)$ with $(r,s)$ coloured $x$, $(s,\hat{s})$ coloured $y$ and $(\hat{s},t)$ coloured $w$. Let $P^{\prime}$ be the path that results from replacing each of these $(r,s,t)$ paths by the respective $(r,s,\hat{s},t)$. By construction $P^{\prime}$ is an $H$-path from $u$ to $v$ in $D^{\prime}$. We have 4 cases.

Case 1. $\left\{u,v\right\}\subseteq K^{\prime}-\left\{s\in V_{D}\colon\ \hat{s}\in K^{\prime}\right\}$.

It follows that $P$ or $P^{\prime}$ is an $H$-path between two vertices of $K^{\prime}$ in $D^{\prime}$, contradicting the independence by $H$-paths of $K^{\prime}$.

Case 2. $u\in\left\{s\in V_{D}\colon\ \hat{s}\in K^{\prime}\right\}$ and $v\in K^{\prime}-\left\{s\in V_{D}\colon\ \hat{s}\in K^{\prime}\right\}$.

Since $y$ is a universal vertex, then adding the arc $(\hat{s},s=u)$ with colour $y$ at the beginning of $P$ (or $P^{\prime}$) results in an $H$-path between two vertices of $K^{\prime}$ in $D^{\prime}$, contradicting the $H$-independence of $K^{\prime}$.

Case 3. $u\in K^{\prime}-\left\{s\in V_{D}\colon\ \hat{s}\in K^{\prime}\right\}$ and $v\in\left\{s\in V_{D}\colon\ \hat{s}\in K^{\prime}\right\}$.

As in the previous case, $y$ is a universal vertex, and thus, adding the arc $(s,\hat{s})$ with colour $y$ at the end of $P$ (or $P^{\prime}$) results in an $H$-path between two vertices of $K^{\prime}$ in $D^{\prime}$, a contradiction.

Case 4. $\left\{u,v\right\}\subseteq\left\{s\in V_{D}\colon\ \hat{s}\in K^{\prime}\right\}$.

Again, $y$ is a universal vertex, so adding the arc $(\hat{s},s=u)$ coloured $y$ at the beginning of $P$ (or $P^{\prime}$) and adding the arc $(s,\hat{s})$ coloured $y$ at the end of $P$ (or $P^{\prime}$) results in an $H$-path between two vertices of $K^{\prime}$ in $D^{\prime}$, a contradiction.

To show that $K$ is $H^{\prime}$-absorbent in $D$, consider $u\in V_{D}-K$. By definition of $K$, we have $u\in V_{D^{\prime}}-K^{\prime}$. Since $K^{\prime}$ is an $H$-kernel in $D^{\prime}$ there exists $v\in K^{\prime}$ such that $u$ reaches $v$ by an $H$-path in $D^{\prime}$. Let $P=(u=u_{0},u_{1},\ldots,u_{n}=v)$ be an $H$-path in $D^{\prime}$. Suppose that $v\notin\hat{S}$. If there is some vertex $s$ such that $\hat{s}\in V(P)$, then $\hat{s}=u_{i}$ for some $i\in\left\{1,\ldots,n-1\right\}$ and by the construction of $D^{\prime}$, it follows that $s=u_{i-1}$, so $(u_{i-1},u_{i+1})$ is an arc of $D$ coloured $w$. Moreover, $(u_{i},u_{i+1})$ has colour $w$ and thus, since $N^{-}_{H^{\prime}}(w)=V_{H^{\prime}}$, $(u_{i-2},u_{i-1},u_{i+1},u_{i+2})$ is an $H^{\prime}$-path. It follows that by removing all $\hat{s}$ vertices of $P$ in this way, an $H^{\prime}$-path from $u$ to $v$ in $D$ is obtained. If $v\in\hat{S}$, then $u_{n-1}\in K$ hence the above procedure results in an $H^{\prime}$-path from $u$ to $K$ in $D$. Thus, $K$ is absorbent by $H^{\prime}$-paths in $D$ and therefore $K$ is an $H^{\prime}$-kernel in $D$.

Finally, proceeding by contradiction, suppose that $H\in\tilde{\mathscr{B}}_{3}$. Since $H\notin\tilde{\mathscr{B}}_{3}$, then there exists an $H^{\prime}$-coloured digraph $D$ which does not have an $H^{\prime}$-kernel. Let $D^{\prime}$ be constructed and coloured as before. Since $H\in\tilde{\mathscr{B}}_{3}$ then $D^{\prime}$ has an $H$-kernel. Therefore, by the previous argument, $D$ would have an ¡$H^{\prime}$-kernel contradicting the second sentence of this paragraph.

∎

Lemma 15**.**

The patterns $F_{4}$ and $F_{5}$ (Figure 4) are not panchromatic patterns by paths.

Proof.

Note that $z$ and $w$ are true twins in the pattern $H$ in Lemmas 13 and 14. Then, by Lemma 12, the pattern $H$ is a panchromatic pattern by paths if and only if the pattern obtained from $H$ by contracting $z$ and $w$ is. But the pattern obtained by contracting $z$ and $w$ is $F_{4}$ for $H$ in Lemma 13 and $F_{5}$ for $H$ in Lemma 14. It follows from Lemmas 13 and 14 that $F_{4}$ and $F_{5}$ are not panchromatic patterns by paths. ∎

7 The pattern $F_{1}$

So far, for every pattern on three vertices we know whether it is a panchromatic pattern by paths, except for the pattern $F_{1}$. Arpin and Linek proved in [1] that $F_{1}$ is a panchromatic pattern (by walks) using the concept of closure of a coloured digraph. Unfortunately, their technique is neither directly applicable in the path case, nor seems to be modifiable to suit it. One interesting fact that should be pointed ot from Arpin and Linek’s technique is that they construct a multidigraph (parallel arcs are present) having the same reachability of a given $F_{1}$-arc-coloured digraph, and where some substructures always exist. From here, one may ask whether there might be an $F_{1}$-arc-coloured multidigraph without an $F_{1}$-kernel, while every $F_{1}$-arc-coloured digraph (without parallel arcs) has an $F_{1}$-kernel. If so, then looking for a counterexample would imply looking at multidigraphs, which represents a broader search space than just considering digraphs. Fortunately, it is not the case, the main result of this section implies that we can only consider digraphs.

Theorem 16**.**

If every $F_{1}$-arc-coloured digraph has an $F_{1}$-kernel, then every $F_{1}$-arc-coloured multidigraph has an $F_{1}$-kernel.

Proof.

We will prove that for every $F_{1}$-arc-coloured multidigraph $D$ there is an $F_{1}$-arc-coloured digraph $\widehat{D}$ such that $D$ has an $F_{1}$-kernel if and only if $\widehat{D}$ has an $F_{1}$-kernel. Let be $r,g$ and $b$ the vertices of $F_{1}$, and assume that the only missing arc in $F_{1}$ is $(b,g)$.

Let $A_{D}[x,y]=\left\{f\in A_{H}\colon\ f\text{ has head }x\text{ and tail }y\right\}$. We construct $\widehat{D}$, an $F_{1}$-arc-coloured digraph obtained from $D$ through the following modifications. Whenever there are vertices $u$ and $v$ in $V_{D}$ with $|A_{D}[u,v]|\geq 2$, then

If there is $f\in A_{D}[u,v]$ with colour $r$, we replace $A_{D}[u,v]$ by only one arc from $u$ to $v$, $e_{uv}$, coloured $r$. 2. 2.

If $A_{D}[u,v]=\left\{f,f^{\prime}\right\}$ is such that $f$ is coloured $g$ and $f^{\prime}$ is coloured $b$, then we create new vertices $z_{1}$ and $z_{2}$, add them to $V_{D}$, and create new arcs $(u,v)$, $(u,z_{1})$, $(z_{1},z_{2})$ and $(z_{1},v)$, with colours $g$, $b$, $g$ and $b$, respectively, add them to $A_{D}$ and delete $f$ and $f^{\prime}$ from $A_{D}$.

Let $Z_{i}$ be the set of all new vertices $z_{i}$ in $\widehat{D}$, $i\in\left\{1,2\right\}$.

Claim 1. Let $x,y\in V_{D}$. There is an $F_{1}$-path from $x$ to $y$ in $D$ if and only if there is an $F_{1}$-path from $x$ to $y$ in $\widehat{D}$.

Proof of Claim 1. Let $W$ be the $F_{1}$ path $W=(x=x_{0},x_{1},\ldots,x_{n}=y)$ in $D$. Construct $\widehat{W}$, an $F_{1}$-path from $x$ to $y$ in $\widehat{D}$, by performing the following modifications on $W$. For each $i\in\left\{0,1,\ldots,n-1\right\}$ such that $|A_{D}[x_{i},x_{i+1}]|\geq 2$, we replace the arc from $x_{i}$ to $x_{i+1}$ used in $W$ with

$e_{x_{i}x_{i+1}}$ coloured $r$, if there is $f\in A_{D}[x_{i},x_{i+1}]$ coloured $r$, 2. 2.

$(x_{i},x_{i+1})$ coloured $g$, if there is no $f\in A_{D}[x_{i},x_{i+1}]$ coloured $r$ and the arc from $x_{i}$ to $x_{i+1}$ in $W$ is coloured $g$, 3. 3.

$(x_{i},z_{1},x_{i+1})$ the monochromatic path coloured $b$, if there is no $f\in A_{D}[x_{i},x_{i+1}]$ coloured $r$ and the arc from $x_{i}$ to $x_{i+1}$ in $W$ is coloured $b$.

Since $W$ is an $F_{1}$-path in $D$, it is enough to check that for each arc substitution in $W$, the sequence of colours of the new arcs are still a walk in $F_{1}$. For each substitution of type 1, $(x_{i-1},x_{i},x_{i+1},x_{i+2})$ is an $F_{1}$-path in $\widehat{D}$ because $N^{+}(r)=N^{-}(r)=V(F_{1})$ and $(x_{i},x_{i+1})$ is coloured $r$ in $\widehat{W}$. For each substitution of type 2, $(x_{i-1},x_{i},x_{i+1},x_{i+2})$ is an $F_{1}$-path in $\widehat{D}$ because the arc from $x_{i}$ to $x_{i+1}$ in $W$ is coloured $g$ and the new arc $(x_{i},x_{i+1})$ in $\widehat{W}$ is coloured $g$. For each substitution of type 3, $(x_{i-1},x_{i},z_{1},x_{i+1},x_{i+2})$ is an $F_{1}$-path because the arc from $x_{i}$ to $x_{i+1}$ in $W$ is coloured $b$ and the path $(x_{i},z_{1},x_{i+1})$ is monochromatic with colour $b$ in $\widehat{D}$. Hence, $\widehat{W}$ is an $F_{1}$-path from $x$ to $y$ in $\widehat{D}$.

Conversely, let $\widehat{W}=(x=x_{0},x_{1},\ldots,x_{n}=y)$ be an $F_{1}$-path from $x$ to $y$ in $\widehat{D}$ and construct an $F_{1}$-path $W$ from $x$ to $y$ in $D$ by modifying $\widehat{W}$ in the following way. For each $i\in\left\{1,\ldots,n-1\right\}$ such that there is $x_{i}\in Z_{1}$, we replace the monochromatic path $(x_{i-1},z_{1}=x_{i},x_{i+1})$ coloured $b$ by the arc from $x_{i-1}$ to $x_{i+1}$ coloured $b$ in $D$, and for each $i\in\left\{0,1,\ldots,n-1\right\}$ such that $|A_{D}[x_{i},x_{i+1}]|\geq 2$ different from those obtained by the first modifications, we replace the arc from $x_{i}$ to $x_{i+1}$ in $\widehat{W}$ with

$f\in A_{D}[x_{i},x_{i+1}]$ coloured $r$ if the arc $x_{i}$ to $x_{i+1}$ in $\widehat{W}$ is $e_{x_{i}x_{i+1}}$ coloured $r$, 2. 2.

the arc from $x_{i}$ to $x_{i+1}$ coloured $g$ if there is no $f\in A_{D}[x_{i},x_{i+1}]$ coloured $r$ and the arc from $x_{i}$ to $x_{i+1}$ in $\widehat{W}$ is coloured $g$.

Note that the above modifications change a monochromatic path of colour $b$ into an arc coloured $b$ and replace arcs by other arcs with the same colour. Therefore, it follows that $W$ is an $F_{1}$-path from $x$ to $y$ in $D$. This ends the proof of Claim 1.

Let $K$ be an $F_{1}$-kernel in $D$. We consider $\widehat{K}=K\cup Z_{2}$, we will prove that $\widehat{K}$ is an $F_{1}$-kernel in $\widehat{D}$. Notice that $\widehat{K}\cap Z_{1}=\varnothing$ and $Z_{2}$ is reached by $F_{1}$-paths only by the vertices in $Z_{1}$. Also, each vertex in $Z_{2}$ is a sink in $\widehat{D}$, so there are no $F_{1}$-paths that start in $Z_{2}$. By the $F_{1}$-independence of $K$ and Claim 1, there is no $F_{1}$-path in $\widehat{D}$ between vertices of $K$. Since $Z_{1}\cap K=\varnothing$ and the observations at the beginning of this paragraph, there are neither $F_{1}$-paths in $\widehat{D}$ between vertices of $K$ and vertices of $Z_{2}$, nor between vertices of $Z_{2}$. Hence $\widehat{K}$ is $F_{1}$-independent in $\widehat{D}$.

Let $x$ be a vertex in $V_{\widehat{D}}-\widehat{K}$. If $x\in Z_{1}$, then there is $z_{2}\in Z_{2}$ such that $(x,z_{2})\in A_{\widehat{D}}$. If $x\in V_{D}-K$, then there is $y\in K$ such that there exist an $F_{1}$-path from $x$ to $y$ in $D$, by Claim 1 there is an $F_{1}$-path from $x$ to $y$ in $\widehat{D}$. Hence, $\widehat{K}$ is absorbent by $F_{1}$-paths in $\widehat{D}$, and therefore $\widehat{K}$ is an $F_{1}$-kernel in $\widehat{D}$.

Let $\tilde{K}$ be an $F_{1}$-kernel in $\widehat{D}$. Consider $K=\tilde{K}\cap V_{D}$, we will prove that $K$ is an $F_{1}$-kernel in $D$. Since every vertex in $Z_{2}$ is a sink in $\widehat{D}$, then $Z_{2}\subset\tilde{K}$ and consequently $Z_{1}\cap\tilde{K}=\varnothing$, moreover, the only vertices that reach $Z_{2}$ by $F_{1}$-paths in $\widehat{D}$ are the vertices in $Z_{1}$. Since $\tilde{K}$ is $F_{1}$-independent in $\widehat{D}$, then by Claim 1, $K=\tilde{K}\cap V_{D}$ is an $F_{1}$-independent set in $D$. Hence, if $x\in V_{D}-K$, then by the $F_{1}$-absorbency of $\tilde{K}$ in $\widehat{D}$, we have that $K$ is an $F_{1}$-absorbent set in $D$. Therefore $K$ is an $F_{1}$-kernel in $D$.

The contrapositive of the statement of the theorem now follows directly using the previous construction. ∎

The following corollary is immediate.

Corollary 17**.**

If every $F_{1}$-arc-coloured digraph (without parallel arcs) has an $F_{1}$-kernel, then $F_{1}$ is a panchromatic pattern by paths.

8 Conclusions

To round up our comparisson between reachability by $H$-paths and reachability by $H$-walks, let us recall that Arpin and Linek proved in [1] that there is a pattern in $\mathscr{B}_{2}$ which is not in $\mathscr{B}_{3}$. It is easy to see that the proof of this result is analogous for paths (see Figure 9). So we obtain the following result.

Lemma 18**.**

The digraph $H$ such that $V_{H}=\left\{u,u^{\prime},b,g\right\}$ and $A(H_{2})\subseteq V(H_{2})-\left\{(b,u),(g,u^{\prime}),(b,g),(g,b)\right\}$ is not a panchromatic pattern by paths.

Observe that $H_{2}^{c}$ has no odd cycles, then $H_{2}\in\tilde{\mathscr{B}}_{2}-\tilde{\mathscr{B}}_{3}$. Therefore $\tilde{\mathscr{B}}_{3}\subset\tilde{\mathscr{B}}_{2}$.

So, we know exactly which of the reflexive patterns on three vertices are panchromatic patterns by paths, except for $F_{1}$. Despite the missing pattern, we have enough information to state the following theorem.

Theorem 19**.**

Let $H$ be a pattern. Then $H$ is a panchromatic pattern by paths if and only if $V_{H}$ admits a partition $V_{H}=(V_{1},V_{2})$ such that $V_{1}$ and $V_{2}$ induce reflexive complete digraphs (every symmetric arc is present) and exactly one of the following holds

• •

There are no arcs between vertices in $V_{1}$ and vertices in $V_{2}$.

• •

Every vertex in $V_{1}$ dominates every vertex in $V_{2}$, and no vertex in $V_{2}$ dominates a vertex in $V_{1}$ (if $F_{1}$ is not a panchromatic pattern by paths).

• •

Every vertex in $V_{1}$ dominates every vertex in $V_{2}$ and vertices in $V_{2}$ may dominate vertices in $V_{1}$ (if $F_{1}$ is a panchromatic pattern by paths).

Proof.

If $F_{1}$ is a panchromatic pattern by paths, the proof of this result is found in [7]. Otherwise, the proof is analogous (cf. [10]). ∎

We would like to point out two open problems regarding panchromatic patterns by paths.

Problem 20**.**

Determine whether $F_{1}$ is a panchromatic pattern by paths.

Problem 21**.**

Find an $H$-arc-coloured digraph without an $H$-kernel for $H\in\left\{F_{4},F_{5}\right\}$.

A solution to problem 20 would settle the characterization of panchromatic patterns by paths, and Corollary 17 guarantees that we can restrict either a proof or the search for an example to $F_{1}$-arc-coloured digraphs (without parallel arcs). Regarding problem, 21, recall that the proofs of Lemmas 13 and 14 are not constructive, so, we do not have a single example of an $H$-arc-coloured with the desired property. It would be really nice to see such and example, for the sake of completness of the subject.