The free daily email that shows people how to change their thoughts, so that life and business becomes easy and fulfilling rather than stressful and overwhelming.
That's the affirmation made again and again by Annette Bening's character – a real estate agent – in the movie American Beauty.
The trouble is, affirmations don't work – unless you say them from a place of alignment.
Here's the thing: You might think you can fool your brain and be all positive, but if your underlying mood isn't so great, you're creating conflict within yourself.
Your thinking is brain is saying "yes" but your heart is saying "no".
And in the end, the heart wins.
So don't try to fool it.
Instead, ease yourself into a better-feeling space using the Feel Better Project process so that positive affirmations come naturally.
Ask: How am I feeling right now?
Feel that feeling – don't deny it.
Then ask: What would make me feel just a little bit better?
Repeat that exercise – trying different things if the first ones don't work – until you feel a shift.
Your true desires bubble up to the surface. Then you can express them, backed by the full force of truly positive emotions.
Alignment first – then heartfelt desire.
P.S. The process is simple – but it takes practice. Click here to join the wait list to be the first to hear about the program I am putting together.


text:
FictionMythology
We, Orpheus
By: Shmeilia Rockie
The Underworld's only hope rests with a group of kids endowed with the powers of ancient gods. But can they stop Gaia and her minions, especially when she's out for their blood?
Rated: Fiction K+ - English - Fantasy/Romance - Chapters: 5 - Words: 9,900 - Reviews: 26 - Favs: 4 - Updated: 9/10/2003 - Published: 7/1/2003 - id: 1345173
1. Reality Is Overrated Anyway2. A Bolt From Out of the Blue3. Message from Zeus4. One More Anguish5. And So It Begins
Chapter 5: And So It Begins
A/N: I publicly retract the flamer comment I made earlier, as it turns out the person was joking. I've changed the ending slightly. Comments are welcomed and appreciated.
"You asked for me, Lord Hades?" Julie's mother appeared before the tall man clad in black.
"Yes," he said curtly and offered her a chair. "I was recently informed that your daughter has her powers now, is that correct?"
She nodded and did not relax in the high backed, velvet-cushioned chair. "She received them about two hours ago. Then she left to join her brother in protecting the girl."
He folded his hands and rested his elbows on the desk in front of him. "I'll be honest with you, Ivy, her powers may not work. As Gaia grows stronger, your daughter's powers will weaken. Soon, there will be nothing she can do to help the others."
Ivy's temper rose, and so did her voice. "Julie's got brains! She'll figure out some other way to help them if it's the last thing she does!"
He held out his hands, easing her to sit back down. She hadn't even been aware of standing up. "It may very well be the last thing she does. It took the entire pantheon of ancient gods to defeat Gaia the first time. They're just kids, naïve and ill prepared. I'm afraid she'll obliterate them, unless they get lucky."
"You underestimate them, surely, Lord Hades. Besides, Tybalt knows exactly what he's doing. He's been preparing for this since he was eight."
"Still, one teenager against Gaia and her minions? If they die, it will mean we will need to choose some other Champions."
This was her children he was talking about, so naturally she became even more upset. "They will not die, Your Highness. I have absolute confidence in them, and I assure you they will not fail!"
Unfazed by her words, he stared at her over his hands and said coldly, "For their sake, I can only hope you are right."
"Next week," the history teacher announced with glee. "We will learn about the Salem Witch Trials. Tybalt, that's your hometown. Care to enlighten us on anything?"
Tybalt smirked and laced his fingers behind his head and smirked. "Yeah. You know, it's funny. They hung so many people, yet they didn't get a single one of the real witches."
This was met with a stunned silence and many incredulous looks. Tessa couldn't tell for the life of her if he was joking or not, and it frightened her. Then the bell rang, and nearly everyone in the class jumped in fright, and tore out of there like their lives depended on it. Tybalt merely laughed. "Gullible! Sheesh! Oh, hey Tessa, it's good to see you back in class." He touched her arm once they were out in the crowded hallway and lowered his voice. "How are you?"
She shrugged. "Okay, I guess. My grandmother… She passed away."
"Oh, I'm so sorry."
This caused her to grimace. "There's nothing anyone could do about it. She was already gone before my brother found her. At least I know she's in a better place."
He smiled sadly. "That's the spirit, Tessa. She wouldn't want you dwelling on it."
A chill ran down Tessa's spine, accompanied by that familiar nagging feeling of déjà vu. She felt like she'd had this conversation with him before, and it didn't help that he sounded as though he'd known her grandmother personally. But as she opened her mouth to ask him about it, Ryan approached them, clutching his books as though they'd protect him from the strange stares kids were giving him. He threw Tybalt a suspicious glance and asked Tessa, "Who's this?"
"I'm Tybalt. You must be her best friend, right?" He held out his hand, and Ryan had to shift the books to the other hand before giving him a firm shake.
"Yeah. Name's Ryan."
The three of them stood in awkward silence before Tybalt leaned in toward Ryan and whispered, "Forgive my curiosity, but what was it like to be hit with lightning?"
Ryan rolled his eyes. As if I haven't been asked that already today. "Painful," he admitted in a dead, dull sort of voice. He recited it as though it were a speech he'd memorized for class. "My necklace and class ring melted to my skin. It was a miracle that I didn't need my finger amputated. They said I'll have the scars forever. But you know, I'm alive, so I guess I'm okay." The finality in his voice made it clear that he didn't want to talk about it any more, so Tybalt held out his hands defensively.
"Sorry, dude that was way out of line of me."
"Yes, it was."
Sensing that his presence was no longer appreciated, Tybalt adjusted the weight of his backpack on his shoulder and took a deep breath. "Well, I'll see you around."
Tessa looked hurt. "B-bye, Tybalt." He walked off, and immediately she rounded on Ryan. "What was that for?"
His eyes bugged out momentarily. "What are talking about, he was prying into my personal life! I have every right to be angry, he doesn't even know me. YOU haven't even asked me that."
"You disliked him from the moment you lay eyes on him, you didn't even give him a chance!"
He rubbed his eyes and pinched the bridge of his nose. "Whatever, Tessa. I don't want to argue with you right now."
After a beat she asked, "What's gotten into you lately? You haven't been yourself at all."
"Oh, gee, I don't know. Let me just touch your face to remind you what's gotten into me. Not to mention the fact that I almost died last week. But you know me; I'll bounce back just like that. Come on, I'm only human." He seemed almost to be talking to himself more than to Tessa. "I didn't ask for any of this."
She took his free hand and held it with both of her own. "I understand why you're bitter, Ryan. I wish it had been me instead."
He squeezed his eyes shut. "Don't you dare say that."
"But it's true. I don't like to see you hurting."
He looked into her eyes, now brimming with tears, and felt his heart warm slightly. "But I would gladly suffer in your place, Tessa. I care too much about you to let you take my burden. I'm sorry I've been so nasty lately."
The bell rang, and the hallway began to empty. Still the two stood there. Finally, Tessa threw her arms around his neck, causing him to drop his books. "You're my best friend, Ryan. I don't know what I'd do without you!"
He patted her back cautiously, making sure to keep her away from his exposed skin. "And vice versa, Tessa."
The rest of the day passed with the two of them pretending that the previous conversation hadn't happened. Neither one knew what to make of it, but they didn't really want to find out. So they walked home in an uncomfortable silence. Sunshine drenched the two of them in its golden glow, but still the air felt somewhat chilly. A nervous tension hung in the silence, and it didn't really have anything to do with their awkwardness. Something was going to happen, and soon.
"So…" Ryan started.
"So?" A pink flower petal fell off the tree overhead and landed on her shoulder. She picked it off as an excuse not to look him in the eye. Some part of her mind recalled that cherry blossom petals were used as a sign of romance in anime, but she shoved the thought away. This was Ryan she was talking about, after all… Desperately, she tried to change the conversation. "Did you get harassed all day long?"
He sighed. "Yeah, I did. Suddenly I'm some freak of nature, and everyone I meet feels the need to remind me of that. I just want them to leave me alone, but no, that's just too much to ask."
Okay, the choice of topics probably wasn't the wisest. She smiled at him as best she could. "Well, just think: in a couple of weeks, they'll have forgotten all about it."
"I hope so." He laughed, without a hint of humor. "Then maybe I can pretend that I'm normal again." They walked on, both feeling they should say something, but not knowing what to say. This was so different than what they were used to; they could normally discuss anything without feeling uncomfortable. Something had happened between them, besides the obvious. And neither one liked it a bit. Ryan took a deep breath. "I'm sorry."
She furrowed her eyebrows. "What for?"
"I've been acting like jerk. I just…" He made a random motion with his hands. "I don't know. I feel so different now. I want to go back to the way things were before, but… I don't think I can."
"What you went through was horrible, Ryan." She retorted. "Don't try to down-play it. It's only natural for you to feel differently now. You know, I read somewhere that people who live through traumatic often come away with a whole new look on life."
"Yeah, but they normally appreciate life more, don't they?" He stared at her, daring her to answer. She shrugged and averted her eyes. "I almost wish I hadn't survived."
"DON'T SAY THAT!" Tessa stopped dead in her tracks. "Have you no concern for those around you? Your poor mother would have to burry her son! No one wants to outlive their child, Ryan. Not to mention everybody else-"
"You mean you." Both had their voices raised now.
"Yes, ME! I would miss you, okay? I mean, heck, my grandmother just died! If my best friend had too, I don't know what I would do with myself. I… I don't think I could go on without you two. You and she are the two most important people in my life. I didn't want to loose her, and I don't want to loose you either." Her throat tightened, and she choked on her next words. "I care about you and believe it or not, I AM concerned about you."
The comment took him by total surprise. They stood staring at each other, the tension dangerously high. The air crackled with it, as though it could spark and ignite at any moment. He expected Tessa to dissolve into tears again, as she seemed to be doing a lot of it lately. However, she did not, but stared at him ruthlessly in a way that really didn't suit her. He couldn't stand up to that stare, he realized, and so gave up fighting it. He was just so tired… "Tessa-"
Before he could even finish forming the thought in his head, he stopped. Something in the back of his head screamed a brief warning, manifested by a white-hot burst of pain behind his eyes. He squeezed his eyes shut and when he opened them again, the look in Tessa's eyes told him she'd felt it too. The semi-fight flew from their minds with this new whatever it was. His senses amplified drastically, to almost painful levels and he felt adrenalin poor into his veins. Time seemed to slow down by a fraction, and a wave of power washed over him that if he hadn't been feeling the Fight or Flight Response, he would have been drunk with it.
Then the rumbling started.
The ground shook in a mini earthquake, threatening to throw them off their feet. Tessa grabbed his arm for balance and he nearly jumped out of his skin. All his senses were screaming at him "Something's terribly wrong!" He felt his heart would explode from the pressure.
Just when he thought it couldn't possibly get any worse, the ground opened up and regurgitated something. The thing was tall, standing at about six or so feet and looking like a cross between a small T-Rex and a Venus Fly Trap, neither truly plant nor animal. It had black eyes that radiated an insane hunger. It turned that gaze on Tessa and her blood turned to ice.
Ryan instinctively put himself between her and the monster, but then froze. How could he stop something like that? The monster took advantage of his slight hesitation and merely shoved him out of the way. Tessa screamed. It was upon her-
A strong gust of wind blew the creature off its feet and into the trunk of a near-by tree, stunned. Tybalt ran up to Tessa. "Are you okay?"
"Huh?" Her eyes widened even farther with the shock of it all. "Tybalt? What's going on?"
He didn't answer but rounded on Ryan, who was still lying face down in the grass. He pointed to the creature, which by now was getting to its feet again. "Don't just lie there; use your power on it!"
"How do you-?"
"NOW! Before it recovers!"
Ryan leapt to his feet and removed his gloves. He wouldn't hesitate again, not when she had almost been killed. In a feat of daring he didn't realize he was capable of, he tackled the monster. He put his hands over the creature's eyes and let loose all the energy swimming in his system. It flowed out of his fingertips, into the monster, and down into the ground where it could no longer be conducted. The creature went into a fit of convulsions, and then… It was gone. A cloud of smoke hung in the air, giving him the distinct impression he must have vaporized the thing.
Tessa's eyes were as wide as saucers. "R-Ryan!"
He opened his mouth to explain, but couldn't find the strength to. She turned to Tybalt, who still hovered at her elbow. "Who ARE you? WHAT are you?"
"Listen, Tessa…"
"No!" She began backing away from him. "No, you just stay away from me!" And then she was gone, running away as fast as her legs could carry her. He started after her.
"Leave her alone!" Ryan swayed on his feet, exhausted but in no way defeated. He picked up a glove and tugged it on. "If I see you near her again, I will hurt you."
Tybalt laughed, a harsh barking noise coming from deep within him. "You can't hurt a fly now." He waved his hand and a gust of hot wind toppled Ryan onto the grass again. Tybalt stood over him. "You wasted your energy on it. If there were another attack, you'd be useless. You need to learn moderation, if not for your own sake then for hers."
Ryan glared daggers at him, no small feat considering that his vision was so blurry. "I don't even know what you're talking about."
"You'll learn soon enough. This was the first attack in daylight. There will be more to follow, I can tell you that already. The time is coming when you might need to give your life to protect her. Can you do that?"
He was so tired…
"Go away," Ryan closed his eyes and curled into a fetal position. No longer able to fight it, he gave in to the overwhelming urge to sleep.
"Well, that didn't go as I had planned…" Tybalt pulled out the necklace and stared at it as though he expected it to give him all the answers to his problems. It remained stubbornly silent, however, leaving him to deal with things himself. He gave a long-suffering sigh and put it back away. He then turned his gaze toward the sky and closed his eyes. The sun warmed his face and a gentle breeze ruffled his hair in weak encouragement. Give it time, it whispered. Give them time…
This was enough for Tybalt.

redpajama_set_name: RedPajamaCommonCrawl


ITMO University is one of Russia's top technological universities. As a sign of its commitment to international cooperation, on the 26th November 2014, ITMO University launched the first European Representative Office ever created by a Russian university in Brussels - in the heart of EU, close to the European Parliament and the Directorate-General for Research and Innovation, on the famous rue du Luxembourg.
ITMO University Representative Office is the point of contact for European partners interested in educational, scientific, and technological cooperation with the University.
Development of partnerships between ITMO University spin-off companies and members of the European innovation ecosystem.


The majority of the people do not read. They don't read books. Reading books is a way to grew, to know new things, and even though it's not the ultimate solution, it's something which can't hurt. It's not demonstrated but if people will read more there should be less crap around. Or at least is what we want to believe.
Most of the commuters share the same space, but even if they are only few cm of space dividing them, they will not talk to each other. Never. Only if something hilarious or really funny will happen able to "break the ice".
The proposed initiative consists in one action: give a book as a gift to a complete stranger. You can either buy the book (or took one of your old one). You can write a personal message inside or not. There is an official facebook group here.
This will happen the 26th of March. I'll do it. It looks fun.


text:
\section{Introduction}\label{s-intro}

In order to increase our understanding of graph classes, it is natural to consider some notion of ``width'' and to research what properties graph classes of bounded width may have. We say that a graph class has {\it bounded} width (for some specific width parameter) if there exists a constant~$c$ such that the width of every graph in the class is at most~$c$.
In particular, this type of structural research has been done in the context of graph containment problems, where the aim is to determine whether one graph~$H$ appears as a ``pattern'' inside some other graph~$G$.
Here, a pattern is defined by specifying a set of graph operations that may be used to obtain $H$ from $G$.
For instance, a graph~$G$ contains a graph~$H$ as a {\em minor} if~$H$ can be obtained from~$G$ via a sequence of vertex deletions, edge deletions and edge contractions.

Tree-width and its {\it linear} variant, path-width, are the best-known graph width parameters due to their relevance for graph minor theory~\cite{GMXX}.
Rank-width is another well-known parameter, introduced by Oum and Seymour~\cite{OS06}.
The rank-width of a graph~$G$ expresses the minimum width~$k$ of a tree-like structure obtained by recursively splitting the vertex set of $G$ in such a way that each cut induces a matrix of rank at most~$k$ (see Section~\ref{s-pre} for a formal definition).
Rank-width is more general than tree-width in the sense that every graph class of bounded tree-width has bounded rank-width, but there are classes for which the reverse does not hold, for example, the class of all complete graphs~\cite{CO00}.

The notion of rank-width has important algorithmic implications, as many {\sf NP}-complete decision problems are known to be polynomial-time solvable not only for graph classes of bounded tree-width, but also for graph classes of bounded rank-width; see~\cite{CourcelleMR00,EGW01,GK03,KR03,Ra07} for a number of meta-theorems capturing such decision problems.
Rank-width is equivalent to clique-width~\cite{OS06}, another important and well-studied width parameter.
Linear rank-width is a linearized variant of rank-width, known to be equivalent to linear clique-width (see, for example,~\cite{Ou17}) and to be closely related to the trellis-width of linear codes~\cite{Ka08}. We formally define the notions of rank-width and linear rank-width in Section~\ref{s-pre}.

The problem of determining whether a given graph has linear rank-width at most~$k$ for some given integer~$k$ is {\sf NP}-complete (this follows from a result of Kashyap~\cite{Ka08}).
On the positive side, Jeong, Kim, and Oum~\cite{JKO16} gave an FPT algorithm for deciding whether a graph has linear rank-width at most~$k$.
Ganian~\cite{Ga11} and Adler, Farley, and Proskurowski~\cite{AFP13} characterized the graphs of linear rank-width at most~$1$.
Recently, Ne{\v{s}}et{\v{r}}il et al.~\cite{Nesetril19} showed that every class of bounded linear rank-width is linearly $\chi$-bounded.
However, our knowledge on linear rank-width, the topic of this paper, is still limited.

\subsection*{Motivation}

To increase our understanding of rank-width and linear rank-width, we may want to verify if classical results for tree-width and path-width stay valid when we replace tree-width with rank-width and path-width with linear rank-width. The following two structural results, related to path-width and tree-width, form the core of the Graph Minor Structure Theorem. Here, a graph~$G$ is {\it $H$-minor-free} for some graph~$H$ if $G$ does not contain $H$ as a minor.

\begin{theorem}[Robertson and Seymour~\cite{GMI}]\label{t-rs83}
For every tree~$T$, the class of $T$-minor-free graphs has bounded path-width.
\end{theorem}

\begin{theorem}[Roberson and Seymour~\cite{GMV}]\label{t-rs94}
For every planar graph~$H$, the class of $H$-minor-free graphs has bounded tree-width.
\end{theorem}

\noindent
It is known that edge deletions and contractions may increase the rank-width and linear rank-width~\cite{Co14}.
Hence, working with minors is not a suitable approach for understanding rank-width and linear rank-width. Therefore, Oum~\cite{Ou05} proposed the notions of vertex-minors and pivot-minors, two closely related notions, which were called $\ell$-reductions and $p$-reductions, respectively, in~\cite{Bouchet1994}. Taking vertex-minors or pivot-minors does {\it not} increase the rank-width or linear rank-width of a graph~\cite{Ou05}.

\begin{figure}
  \begin{center}
    \tikzstyle{v}=[circle,draw,fill=black!50,inner sep=0pt,minimum width=4pt]
  \begin{tikzpicture}
    \draw (0,0)node[v](v){};
    \foreach \i in {1,2,3,4,6,7} {
      \draw (90-360/14+360*\i/7:1) node [v] (v\i){};
      }
    \draw (0,-0.5) node [v] (v5){};
      \node at (v7) [label=above:$v$]{};
      \node at (v1) [label=above:$u$]{};
      \draw (v3)--(v2)--(v1)--(v7)--(v6)--(v4);
      \foreach \i in {7,1,2} {
      \draw (v5)--(v\i);
    }
    \draw (v7)--(v)--(v1);
    \draw (v1)--(v3)--(v6);
    \node at (v4) [label=below:$G$]{};
  \end{tikzpicture}
  $\quad\quad$
  \begin{tikzpicture}
    \draw (0,0)node[v](v){};
    \foreach \i in {1,2,3,4,6,7} {
      \draw (90-360/14+360*\i/7:1) node [v] (v\i){};
      }
    \draw (0,-0.5) node [v] (v5){};
      \node at (v7) [label=above:$u$]{};
      \node at (v1) [label=above:$v$]{};
      \draw (v3)--(v2)--(v1)--(v7)--(v6)--(v4);
      \foreach \i in {7,6,1,3} {
      \draw (v5)--(v\i);
    }
    \draw (v1)--(v3);
    \draw (v6)--(v2);
    \draw (v7)--(v)--(v1);
    \foreach \i in {2,3,6}{
      \draw (v\i)--(v);
    }
    \node at (v4) [label=below:$G\pivot uv$]{};
  \end{tikzpicture}
  \end{center}
\caption{A graph before and after pivoting an edge (the example is taken from~\cite{DDJKKOP18}).}
\label{f-pivotexample}
\end{figure}

To define the notions of a vertex-minor and pivot-minor, we need some terminology. The \emph{local complementation} at a vertex~$u$ in a graph~$G$ replaces every edge of the subgraph induced by the neighbours of~$u$ with a non-edge, and vice versa.
The resulting graph is denoted by $G*u$.
An \emph{edge pivot} is the operation that takes an edge~$uv$, first applies a local complementation at~$u$, then at~$v$, and then at~$u$ again.
We denote the resulting graph $G\wedge uv=G*u*v*u$.
It is known that $G*u*v*u=G*v*u*v$~\cite{Ou05}, and thus $G\pivot uv=G \pivot vu$.
An alternative definition of the edge pivot operation is as follows.
Let~$S_u$ be the set of neighbours of~$u$ that are non-adjacent to~$v$ and let $S_v$ be the set of neighbours of~$v$ that are non-adjacent to~$u$, whereas we denote the set of common neighbours of~$u$ and~$v$ by $S_{uv}$.
We replace every edge between any two vertices in distinct sets from $\{S_u\setminus \{v\},S_v\setminus \{u\},S_{uv}\}$ by a non-edge and vice versa.
Afterwards, we delete every edge between~$u$ and~$S_u$ and add every edge between~$u$ and~$S_v$.
We also delete every edge between~$v$ and~$S_v$ and add every edge between~$v$ and~$S_u$.
We refer to \figurename~\ref{f-pivotexample} for an example.

A graph $H$ is a \emph{vertex-minor} of a graph $G$ if $H$ can be obtained from $G$ by a sequence of local complementations and vertex deletions.
A graph $H$ is a \emph{pivot-minor} of a graph $G$ if $H$ can be obtained from $G$ by a sequence of edge pivots and vertex deletions.
Hence $H$ is a vertex-minor of $G$ if $H$ is a pivot-minor of $G$, but the reverse is not necessarily true.
A graph is {\it $H$-vertex-minor-free} if it contains no vertex-minor isomorphic to $H$, and similarly,
a graph is {\it $H$-pivot-minor-free} if it contains no pivot-minor isomorphic to $H$.

It is natural to ask whether parallel statements of Theorems~\ref{t-rs83} and \ref{t-rs94} exist for rank-width or linear rank-width in terms of vertex-minors or pivot-minors.
Below we discuss the state-of-the-art for this research direction.

\subsection*{Related Work}

A \emph{circle graph} is the intersection graph of chords on a circle, and it is known that the class of circle graphs is closed under taking vertex-minors.
Bouchet~\cite{Bouchet1994} characterized circle graphs in terms of three forbidden vertex-minors.
Oum~\cite{Ou05} showed that the class of circle graphs has unbounded rank-width, and
asked, as an analog to Theorem~\ref{t-rs94} for the vertex-minor relation,
whether for every circle graph $H$, the class of $H$-vertex-minor-free graphs has bounded rank-width. Recently, Geelen et al.~\cite{GKMW2019} gave an affirmative answer to this question.

\begin{theorem}[Geelen, Kwon, McCarty, and Wollan~\cite{GKMW2019}]\label{thm:GKMW2019}
For every circle graph $H$, the class of $H$-vertex-minor-free graphs has bounded rank-width.
\end{theorem}

\noindent
Every pivot-minor of a graph is also a vertex-minor. Hence, for every graph~$H$, the class of $H$-vertex-minor-free graphs is contained in the class of $H$-pivot-minor-free graphs. This leads to the question whether we can strengthen Theorem~\ref{thm:GKMW2019} by replacing the vertex-minor relation with the pivot-minor relation. However, this is not true.
In order to see this, we first observe that
bipartite graphs are closed under taking pivot-minors~\cite{Ou05}.
Hence, any bipartite graph does not contain a non-bipartite circle graph as a pivot-minor.
Now consider the class~${\cal G}$ of $n\times n$ grids, which has unbounded rank-width.
As ${\cal G}$ is a subclass of bipartite graphs,
${\cal G}$ is $H$-pivot-minor-free for every non-bipartite circle graph~$H$ (such as, for example, $H=K_3$). Hence, for every non-bipartite graph $H$, the class of $H$-pivot-minor-free graphs has unbounded rank-width. This means we can only hope for strengthening Theorem~\ref{thm:GKMW2019} by considering bipartite circle graphs~$H$, and Oum~\cite{Oum2009} conjectured the following analog to Theorem~\ref{t-rs94} for the pivot-minor relation:

\begin{conjecture}[Oum~\cite{Oum2009}]\label{conj1}
For every bipartite circle graph $H$, the class of $H$-pivot-minor-free graphs has bounded rank-width.
\end{conjecture}

\noindent
So far, Conjecture~\ref{conj1} has been confirmed for bipartite graphs~\cite{Ou05}, circle graphs~\cite{Oum2009}, and line graphs~\cite{Oum2009}.
Should Conjecture~\ref{conj1} hold for all graphs, then it implies both Theorem~\ref{t-rs94}
and~\ref{thm:GKMW2019}~\cite{Ou05}.

We now turn to linear rank-width, for which Kant\'e and Kwon~\cite{KanteK2018} conjectured the following analog to Theorem~\ref{t-rs83} for the vertex-minor relation:

\begin{conjecture}[Kant\'e and Kwon~\cite{KanteK2018}]\label{conj2}
For every tree $T$, the class of $T$-vertex-minor-free graphs has bounded linear rank-width.
\end{conjecture}

\noindent
So far, Conjecture~\ref{conj2} has been confirmed for every class of graphs whose prime graphs, with respect to split decompositions, have bounded linear rank-width~\cite{KanteK2018}, such as, for example, distance-hereditary graphs (prime distance-hereditary graphs have size at most $3$).
Moreover, Conjecture~\ref{conj2} holds for every tree $T$ that is a path~\cite{KMOW19}.

\subsection*{Our Focus and Results}

We focus on the remaining analog, namely the analog to Theorem~\ref{t-rs83} for the pivot-minor relation. We first prove that we cannot hope for a result that holds for every tree~$T$.
 A {\em caterpillar} is a tree that contains a path~$P$, such that every vertex not on~$P$ has a neighbour in~$P$.

\begin{theorem}\label{t-main1}
If~$T$ is a tree that is not a caterpillar, then the class of $T$-pivot-minor-free distance-hereditary graphs has unbounded linear rank-width.
\end{theorem}

\noindent
Due to Theorem~\ref{t-main1}, we conjecture the following:

\begin{conjecture}\label{q-1}
For every \emph{caterpillar} $T$, the class of $T$-pivot-minor-free graphs has bounded linear rank-width.
\end{conjecture}

\noindent
In contrast, the aforementioned result of Kwon et al.~\cite{KMOW19} confirming Conjecture~\ref{conj2} if $T$ is a path implies that Conjecture~\ref{conj2} {\it has} been confirmed if $T$ is a caterpillar: every caterpillar~$T$ is a pivot-minor of some path $P$~\cite[Theorem 4.6]{ko14} and consequently, if $T$ is a caterpillar, then the class of $T$-vertex-minor-free graphs is contained in the class of $P$-vertex-minor-free graphs.

By the fact that every caterpillar~$T$ is a pivot-minor of some path $P$ and the fact that every path $P$ is a caterpillar by definition, we can also formulate Conjecture~\ref{q-1} as follows:

\medskip
\noindent
{\bf Conjecture~\ref{q-1} (alternative formulation).}
{\it For every \emph{path} $P$, the class of $P$-pivot-minor-free graphs has bounded linear rank-width.}

\medskip
\noindent
We make two contributions to Conjecture~\ref{q-1}. We first show, in Section~\ref{pndh}, that Conjecture~\ref{q-1} holds for distance-hereditary graphs.

\begin{theorem}\label{thm:pmdh}
Let $n\ge 2$ be an integer. Every $P_n$-pivot-minor-free distance-hereditary graph has linear rank-width at most $2n-3$.
\end{theorem}

\noindent
Theorems~\ref{t-main1} and~\ref{thm:pmdh}, together with the fact that every caterpillar  is a pivot-minor of some path, yields the following dichotomy.

\begin{corollary}\label{c-dicho}
For every tree $T$, the class of $T$-pivot-minor-free distance-hereditary graphs has bounded linear rank-width if and only if $T$ is a caterpillar.
\end{corollary}

\noindent
If a graph $G$ is $P_4$-pivot-minor-free, then $G$ has no induced subgraph isomorphic to $P_4$. This means that $G$ is distance-hereditary.
Hence, Theorem~\ref{thm:pmdh} has the following consequence:

\begin{corollary}\label{cor:p4pm}
Every $P_4$-pivot-minor-free graph has linear rank-width at most $3$.
\end{corollary}
\noindent
Below we give a short, alternative proof of Corollary~\ref{cor:p4pm} after introducing a more general strategy.
A graph~$G$ is {\it $H$-free} if $G$ does not contain the graph $H$ as an induced subgraph, and  $G$ is {\it $(H_1,\ldots,H_p)$-free} for some set of graphs $\{H_1,\ldots,H_p\}$ if $G$ is $H_i$-free for every $i\in \{1,\ldots,p\}$.
We can now try to obtain for a caterpillar~$T$, a constant bound on the linear rank-width of a $T$-pivot-minor-free graph by adapting the following {\it general} strategy:

\bigskip
\noindent
\begin{boxedminipage}{.92\textwidth}
{\bf Step 1.} Show that the class of $T$-pivot-minor-free graphs is a subclass of a class of\\
\hspace*{1.3cm} $(H_1,H_2)$-free graphs for some graphs $H_1$ and $H_2$.\\[3pt]
{\bf Step 2.} Show that this class of $(H_1,H_2)$-free graphs has bounded linear rank-width.
\end{boxedminipage}

\bigskip
\noindent
An advantage of this strategy is that it will lead to a stronger result that forms the start of a {\it systematic} study into boundedness of linear rank-width of $(H_1,H_2)$-free graphs.
This would address Open Problem~7.5 in~\cite{DJP19}, which asks for such a result.
We refer to Section~\ref{sec:conclusion} for a further discussion on this.

To illustrate our general strategy for the case where $H=P_4$, we can do as follows.
In Step~1, we observe that every $P_4$-pivot-minor graphs is $(P_4, \text{\textnormal{dart}})$-free
(see Figure~\ref{f-three} for an illustration of the dart).
In Step~2, we use a result of Brignall, Korpelainen, and Vatter~\cite{BKV17} who showed that
a class of $(P_4,H)$-free graphs has bounded linear rank-width if and only if $H$ is a threshold graph.
Hence, as the dart is a threshold graph,
 %
the class of $(P_4, \text{\textnormal{dart}})$-free graphs, and thus the class of
$P_4$-pivot-minor-free graphs, has bounded linear rank-width.

Whether $P_n$-pivot-minor-free graphs have bounded linear rank-width for $n\geq 5$ remains a challenging open question. In the remainder, we focus on the other tree $T$ on four vertices besides the $P_4$, which is the {\em claw} $K_{1,3}$ (the $4$-vertex star).
We will prove the following result.

\begin{theorem}\label{t-claw}
Every claw-pivot-minor-free graph has linear rank-width at most~$59$.
\end{theorem}

\noindent
As a consequence, we have verified (the original formulation of) Conjecture~\ref{q-1} for every caterpillar~$T$ on at most four vertices. Since every tree on at most four vertices is a caterpillar we have in fact showed that if $T$ is a tree on at most $4$ vertices, then the class of $T$-pivot-minor-free graphs has bounded linear rank-width.

We first explain how we perform Step~1.
In our previous paper~\cite{DDJKKOP18},
we proved that a graph is claw-pivot-minor-free if and only if it is $(\mbox{bull},\mbox{claw}, P_5, W_4, \bw_3)$-free; see  Figure~\ref{f-three} for pictures of these graphs.
It is readily seen that for proving boundedness of linear rank-width of some graph class~${\cal G}$ one may restrict to connected graphs from ${\cal G}$.
 We showed in~\cite{DDJKKOP18} that a graph~$G$ is $(\text{\textnormal{bull}},\text{\textnormal{claw}},P_5)$-free if and only if every component of~$G$ is $3P_1$-free (the graph $3P_1$ consists of three isolated vertices).
Hence, we derive the following result, in which we specify the graphs $H_1$ and $H_2$ of Step~1 as
$H_1=3P_1$ and $H_2=W_4$.

\begin{figure}
\begin{center}
\tikzstyle{v}=[circle,draw,fill=black!50,inner sep=0pt,minimum width=4pt]
$\quad\quad$
\begin{tikzpicture}
\draw (0,0) node [v] (v1){};
\draw (1,0) node [v] (v2){};
\draw (1,1) node [v] (v3){};
\draw (0,1) node [v] (v4){};
\draw (.5,.5) node[v] (v) {};
\draw (v4)--(v1)--(v)--(v2)--(v3);
\draw (0.5,0) node[label=below:$P_5$]{};
\end{tikzpicture}
$\quad\quad$
\begin{tikzpicture}
\draw (0,0) node [v] (v1){};
\draw (1,0) node [v] (v2){};
\draw (1,1) node [v] (v3){};
\draw (0,1) node [v] (v4){};
\draw (.5,.5) node[v] (v) {};
\draw (v4)--(v1)--(v)--(v2)--(v3);
\draw (v1)--(v2);
\draw (0.5,0) node[label=below:bull]{};
\end{tikzpicture}
$\quad\quad$
\begin{tikzpicture}
\draw (0,0) node [v] (v1){};
\draw (1,0) node [v] (v2){};
\draw (1,1) node [v] (v3){};
\draw (0,1) node [v] (v4){};
\draw (.5,.5) node[v] (v) {};
\foreach \i in {1,2,3,4} {
\draw (v)--(v\i);
}
\draw (v1)--(v2)--(v3)--(v4)--(v1);
\draw (0.5,0) node[label=below:$W_4$]{};
\end{tikzpicture}
$\quad\quad$
\begin{tikzpicture}
\draw (0,0.25) node [v] (v1){};
\draw (1,0.25) node [v] (v2){};
\draw (0.5,0.5) node [v] (v3){};
\draw (0.5,0) node [v] (v4){};
\draw (0.5,1) node[v] (v) {};
\draw (v1)--(v4)--(v2)--(v3)--(v1);
\draw (v)--(v3)--(v4);
\draw (0.5,0) node[label=below:dart]{};
\end{tikzpicture}
$\quad\quad$
\begin{tikzpicture}
\begin{scope}[xshift=-1.5cm]
\foreach \i in {1,2} {
\draw (120*\i:.7) node[v](a\i){};
}
\foreach \i in {3} {
\draw (-10:.7) node[v](a\i){};
}
\draw (a1)--(a2)--(a3)--(a1);
\end{scope}
\begin{scope}[xshift=1.5cm]
\draw (10:1) node[v] (b4){};
\foreach \i in {1,2} {
\draw (180-120*\i:.7) node[v](b\i){};
\draw (a\i)--(b\i)--(b4);
}
\foreach \i in {3} {
\draw (190:.7) node[v](b\i){};
\draw (a\i)--(b\i)--(b4);
}
\draw (b1)--(b2)--(b3)--(b1);
\end{scope}
\draw (-90:.4) node[label=below:$\bw_3$]{};
\end{tikzpicture}
\end{center}
\caption{The graphs $P_5$, bull, $W_4$, dart, and $\bw_3$.}\label{f-three}
\end{figure}

\begin{theorem}[Dabrowski et al.~\cite{DDJKKOP18}]\label{t-known}
Let $G$ be a connected graph. Then $G$ is claw-pivot-minor-free if and only if $G$ is $(3P_1,W_4,\bw_3)$-free. In particular,
the class of connected claw-pivot-minor-free graphs belongs to the class of $(3P_1,W_4)$-free graphs.
\end{theorem}

\begin{figure}[b]
   \tikzstyle{v}=[circle,draw,fill=black!50,inner sep=0pt,minimum width=4pt]
  \centering
   \begin{tikzpicture}[scale=0.6]
        \node [v]  (b1) at (1-4, 0-10){};
        \node [v]  (b3) at (2-4, 2-10){};
        \node [v]  (b4) at (0-4, 2-10){};
        \node [v]  (b5) at (1-4, 1-10){};
        \node [v]  (b6) at (0.5-4, 1.5-10){};
        \node [v]  (b7) at (1.5-4, 1.5-10){};

        \draw(b5)--(b6)--(b4);
        \draw(b5)--(b7)--(b3);
        \draw(b5)--(b1);
         \draw (1-4,-1-10) node [below] {$S_{1,2,2}$};

        \node [v]  (d1) at (1, 0-10){};
        \node [v]  (d3) at (2, 2-10){};
        \node [v]  (d4) at (0, 2-10){};
        \node [v]  (d6) at (0.5, 1.5-10){};
        \node [v]  (d7) at (1.5, 1.5-10){};

        \draw(d6)--(d4);
        \draw(d7)--(d3);
           \draw (1,-1-10) node [below] {$P_1+2P_2$};
\end{tikzpicture}
   \caption{The graphs $S_{1,2,2}$ and $P_1+2P_2$; note that $P_1+2P_2$ is an induced subgraph of $S_{1,2,2}$.}
  \label{fig:specialgraph}
\end{figure}

\noindent
As Step~2, we must prove that $(3P_1,W_4)$-free graphs have bounded linear rank-width.
With an eye on a future classification of boundedness of linear rank-width for $(H_1,H_2)$-free graphs, we aim to prove boundedness for classes of $(H_1,H_2)$-free graphs as large as possible.
For integers $1\leq i\leq j\leq k$, let~$S_{i,j,k}$ denote the {\it subdivided claw}, which is the graph obtained from the claw by subdividing its three edges
$i-1$ times, $j-1$ times and $k-1$ times, respectively; see also Figure~\ref{fig:specialgraph} and note that $S_{1,1,1}=K_{1,3}$.
The {\it complement} of a graph $G=(V,E)$ is the graph $\overline{G}=(V, \{uv\; |\; u,v\in V\; \mbox{with}\; u\neq v \; \mbox{and}\; uv\notin E\} )$. The graph $K_3$ denotes the triangle.
As $3P_1=\overline{K_3}$ and $W_4=\overline{P_1+2P_2}$ is an induced subgraph of
the complement of $\overline{S_{1,2,2}}$, the class of $(3P_1,W_4)$-free graphs is contained in the class of $(\overline{K_3},\overline{S_{1,2,2}})$-free graphs.
We prove the following result in Section~\ref{claw}:

\begin{theorem}\label{t-maintwosubgraphs}
Every $(K_3, S_{1,2,2})$-free graph has linear rank-width at most~$58$.
\end{theorem}

\noindent
As observed in Lemma~\ref{lem:complement} in Section~\ref{s-pre}, complementing a graph may increase its linear rank-width by at most~$1$. Hence, Theorems~\ref{t-known} and~\ref{t-maintwosubgraphs} and this observation imply Theorem~\ref{claw}.

Dabrowski et al.~\cite{DabrowskiDP2016} proved that the class of $(K_3, S_{1,2,2})$-free graphs has bounded rank-width. As every class of bounded linear rank-width has bounded rank-width, but the reverse is not necessarily true, Theorem~\ref{t-maintwosubgraphs} is a strengthening of their result.
Moreover, Theorem~\ref{t-maintwosubgraphs} is tight in the sense that even
the class of $S_{1,2,3}$-free bipartite graphs is known to have unbounded linear rank-width~\cite{BogdanLZ18}.

It remains to prove Theorems~\ref{t-main1},~\ref{thm:pmdh} and~\ref{t-maintwosubgraphs}, which we do in Sections~\ref{sec:caterpillar},~\ref{pndh} and~\ref{claw}, respectively.
We discuss future research in Section~\ref{sec:conclusion}.

\section{Preliminaries}\label{s-pre}

	In this paper, all graphs have no loops and no multiple edges.
	For a graph~$G$, let~$V(G)$ and~$E(G)$ denote the vertex set and edge set of~$G$, respectively.
	For $S\subseteq V(G)$, let~$G[S]=(S,\{uv\; |\; uv\in E, u,v\in S\})$ denote the subgraph of~$G$ induced by~$S$.
	For convenience, we write $G[v_1, v_2, \ldots, v_m]$ for $G[\{v_1, v_2, \ldots, v_m\}]$.
	A graph~$H$ is an \emph{induced subgraph} of $G$ if $H=G[S]$ for some $S\subseteq V(G)$.
	For a vertex $v\in V(G)$, we let $G- v$ be the graph obtained from~$G$ by removing~$v$.
	For a set $S\subseteq V(G)$, we let $G-S$ be the graph obtained by removing all vertices in~$S$.
	For an edge $e\in E(G)$, we let $G-e$ be the graph obtained by removing~$e$. For
	a set $F\subseteq E(G)$, we let $G-F$ be the graph obtained by removing all edges in $F$.

The set of neighbours of a vertex~$v$ in a graph~$G$ is denoted by~$N_G(v)$.
The size of~$N_G(v)$ is the \emph{degree} of~$v$.
	For a set $A\subseteq V(G)$, we let~$N_G(A)$ denote the set of all vertices in $G-A$ that have a neighbour in~$A$.
	Two vertices~$v$ and~$w$ in~$G$ are \emph{twins} if $N_G(v)\setminus \{w\}=N_G(w)\setminus \{v\}$.
	We say that twins~$v$ and~$w$ are \emph{false twins} if~$v$ is not adjacent to~$w$.
An edge~$e$ of a connected graph~$G$ is a \emph{cut edge} if $G-e$ is disconnected.

Let $A$ and $B$ be two disjoint vertex subsets of a graph $G$.
We let $G\times (A,B)$ be the graph  obtained from $G$ by taking a \emph{bipartite complementation} between $A$ and $B$, that is, by replacing each edge between a vertex of $A$ and a vertex of $B$ by a non-edge, and vice versa.
We say that~$A$ is \emph{complete} to~$B$ if for every $a\in A$ and every $b\in B$, $a$ is adjacent to~$b$, whereas~$A$ is \emph{anti-complete} to~$B$ if for every $a\in A$ and every $b\in B$, $a$ is not adjacent to~$b$.
If~$A$ is complete or anti-complete to~$B$, then $A$ is \emph{trivial} to~$B$.
If $A$ consists of one vertex~$v$, then we say that~$v$ is complete or anti-complete to~$B$.

A set~$F$ of edges is a \emph{matching} in a graph $G$ if no two edges in~$F$ have a common end-vertex.  A set $S$ of vertices in a graph $G$ is an \emph{independent set} if no two vertices in $S$ are adjacent, whereas $S$ is a \emph{clique} if every pair of vertices in $S$ is adjacent.
The complete graph $K_n$ is the graph on $n$ vertices that form a clique.
The complete bipartite graph~$K_{n,m}$ is the bipartite graph
with a bipartition $(A,B)$ such that $\abs{A}=n$, $\abs{B}=m$, and~$A$ is complete to~$B$.  The graph~$W_n$ is the graph on $n+1$ vertices that is obtained from a cycle on $n$ vertices by
adding one vertex that is made adjacent to all vertices in the cycle.
The \emph{length} of a path is the number of edges in the path.

For two graphs~$G$ and~$H$, we let $G+\nobreak H=(V(G)\cup V(H),E(G)\cup E(H))$ be the disjoint union of~$G$ and~$H$. We let $pG$ denote the disjoint union of~$p$ copies of~$G$.
The {\it subdivision} of an edge $uv$ in a graph removes the edge $uv$ and introduces a new vertex $w$ that is made adjacent (only) to $u$ and $v$.

A graph~$H$ is the \emph{$1$-subdivision} of a graph~$G$ if~$H$ is obtained from~$G$ by subdividing each edge of $G$ exactly once.
 A graph~$G$ is \emph{distance-hereditary} if for every connected induced subgraph~$H$ of~$G$ and every two vertices $v,w$ in~$H$, the distance between~$v$ and~$w$ in~$H$ is the same as the distance in~$G$.

Let~$G=(V,E)$ be a graph with vertices $x_1,\ldots,x_n$.
Let~$A=A_G$ denote the \emph{adjacency matrix} of~$G$, that is, entry $A_G(i,j)=1$ if $x_i$ is adjacent to $x_j$ and $A_G(i,j)=0$ if $x_i$ is not adjacent to $x_j$.
For a subset $X\subseteq V(G)$, the matrix $A[X,V\setminus X]$ is the $|X|\times |V\setminus X|$ submatrix of $A$ restricted to the rows of $X$ and the columns of $V\setminus X$.
The \emph{cut-rank function} of~$G$ is the function $\cutrk_G:2^{V}\rightarrow \mathbb{N}$ such that for each $X\subseteq V$,  \[\cutrk_G(X):=\rank(A_G[X,V\setminus X]),\] where we compute the rank {\it over the binary field}.
A {\it linear ordering} of $G$ is a permutation of the vertices of $G$.
The \emph{width} of a linear ordering $(x_1,\ldots, x_n)$ of~$G$ is defined as $\max_{1\le i\le n}\{\cutrk_G(\{x_1,\ldots,x_i\})\}$.
The \emph{linear rank-width} $\lrw(G)$ of $G$ is the minimum width over all linear orderings of~$G$.

Let $X$ and $Y$ be two disjoint subsets of vertices of a graph.
For an ordering $(x_1, \ldots, x_n)$ of the vertices of $X$ and an ordering $(y_1, \ldots, y_m)$ of the vertices of $Y$, we define the ordering
	\[(x_1, \ldots, x_n)\oplus (y_1, \ldots, y_m):=(x_1, \ldots, x_n, y_1, \ldots, y_m).\]

The cut-rank function is invariant under taking local complementation. This implies that the linear rank-width of a graph does not increase when taking its vertex-minor.

\begin{lemma}[Bouchet~\cite{Bouchet1989}; See Oum~\cite{Ou05}]\label{lem:bouchet}
If $G$ is obtained from $H$ by a sequence of local complementations, then
$\cutrk_G(X)=\cutrk_H(X)$ for all $X\subseteq V(G)$.
So, if $G$ is a vertex-minor of $H$, then $\lrw(G)\le \lrw(H)$.
\end{lemma}

We need three structural lemmas on linear rank-width.
Recall that the complement of a graph $G=(V,E)$ is the graph $\overline{G}=(V,\{uv\; |\; u,v\in V \; \mbox{with}\; u\neq v \; \mbox{and}\; uv\notin E\} )$.

\begin{lemma}\label{lem:complement}
	If~$G$ has linear rank-width~$k$, then~$\overline{G}$ has linear rank-width at most~$k+\nobreak 1$.
	\end{lemma}

	\begin{proof}
	Let~$H$ be the graph obtained from~$G$ by adding a vertex $a$ complete to $V(G)$.
	Observe that $\overline{G}=(H*a)[V(G)]$. Then, $\lrw(\overline{G})\leq \lrw(H*a)=\lrw(H)\le \lrw(G)+1$.
	Here, the second step follows from Lemma~\ref{lem:bouchet}, and the third step follows from the fact that adding one vertex to a graph may increase the linear rank-width by at most
	one.	\end{proof}

\begin{lemma}\label{lem:bipartitecomplement}
	Let~$G$ be a graph and~$A$ and~$B$ be two disjoint vertex subsets of~$G$.
	If~$G$ has linear rank-width~$k$, then $G\times (A,B)$ has linear rank-width at most~$k+\nobreak 2$.
	\end{lemma}
	\begin{proof}
	Let~$H$ be the graph obtained from~$G$ by adding two adjacent vertices~$a$ and~$b$ such that
	\begin{itemize}
	\item $N_H(a)\cap V(G)=A$ and $N_H(b)\cap V(G)=B$.
	\end{itemize}
	Observe that $G\times (A,B)=(H\pivot ab)[V(G)]$.
	Since adding two vertices may increase the linear rank-width by at most two,
	we have $\lrw(H\pivot ab)=\lrw(H)\le \lrw(G)+2$.
	Therefore, we have $\lrw(G\times (A,B))\le \lrw(G)+2$.
	\end{proof}

Let~$I$ be a set of pairwise twins in a graph~$G$.
We define~$G//I$ as the graph obtained from~$G$ by replacing~$I$ with one vertex having~$N_G(I)$ as its neighbourhood.

\begin{lemma}\label{lem:twinreduced}
Let~$G$ be a graph and $I_1, I_2, \ldots, I_m$ be pairwise vertex-disjoint subsets of~$V(G)$ such that each~$I_i$ is a set of pairwise twins in~$G$.
Then $\lrw(G)\le \lrw(G//I_1 //I_2 // \cdots //I_m)+1$.
\end{lemma}
\begin{proof}
Let $H:=G//I_1 //I_2 //\cdots //I_m$.
For each $j\in \{1, \ldots, m\}$, let~$w_j$ be the vertex kept from~$I_j$ in~$H$.
Suppose that~$L_H$ is a linear ordering of~$H$ with the optimal width.
We obtain a linear ordering $L_G$ of~$G$ from~$L_H$ by replacing each vertex $w_j\in \{w_1, \ldots, w_m\}$ with any linear ordering of~$I_j$.
Let $L_G:=(v_1, v_2, \ldots, v_n)$.

We observe that for each vertex partition $\mathcal{P}_i=(\{v_1, \ldots, v_i\}, \{v_{i+1}, \ldots, v_n\})$ of~$G$,
at most one set of $I_1, I_2, \ldots, I_m$ may have a vertex in both parts.
Suppose that a set $I_j$ has a vertex in both parts.
From $\mathcal{P}_i$, by replacing all other sets $I_{j'}$ with $w_{j'}$ and adding $w_j$ to one of the parts properly,
we can obtain a vertex partition of $H$ that is induced from the ordering $L_H$.
But since $I_j$ has one vertex in each part, $\rank (A(G)[\{v_1, \ldots, v_i\}, \{v_{i+1}, \ldots, v_n\}])$ could
be at most one larger than the rank of the matrix from the corresponding vertex partition of $H$.
It shows that $\rank (A(G)[\{v_1, \ldots, v_i\}, \{v_{i+1}, \ldots, v_n\}])\le \lrw(H)+1$.
We conclude that $\lrw(G)\le \lrw(H)+1$.
\end{proof}

A \emph{path decomposition} of a graph $G$ is a pair $(P,\mathcal{B})$, where $P$ is a path and $\mathcal{B}=(B_t)_{t\in V(P)}$ is a family of subsets $B_t\subseteq V(G)$, satisfying the following.
\begin{enumerate}
	\item For every $v\in V(G)$ there exists a $t\in V(P)$ such that $v\in B_t$.
	\item For every $uv\in E(G)$ there exists a $t\in V(P)$ such that $\{u,v\}\subseteq B_t$.
	\item For every $v\in V(G)$ the set $\{t\in V(P)\mid v\in B_t\}$ induces a subpath of $P$.
\end{enumerate}
The \emph{width} of a path decomposition $(P,\mathcal{B})$ is defined as $\max\{|B_t|\mid t\in V(P)\}-1\}$.
The \emph{path-width} of $G$ is the minimum width among all path decompositions of $G$.

We finish this section by proving that every tree with linear rank-width~$1$ is a caterpillar.

\begin{theorem}[Adler and Kant\'e~\cite{AdlerK2015}]\label{thm:lrwpwtree}
	For every tree $T$, the linear rank-width of a tree~$T$ is equal to the path-width of $T$.
\end{theorem}

\begin{theorem}[Takahashi, Ueno, and Kajitani \cite{Takahashi1994}]\label{thm:charpw}
 Let $G$ be a tree, $k$ be a positive integer, and $v$ be a vertex of $G$.
 Then $G$ has path-width at most $k$ if and only if for every vertex $v$, $G-v$ has at most two connected components with path-width exactly $k$ and all other connected components of $G$ have path-width less than $k$.
\end{theorem}

	\begin{lemma}\label{lem:charlrw1tree}
	A tree has linear rank-width at most~$1$ if and only if it is a caterpillar.
      \end{lemma}

      \begin{proof}
      Let $T$ be a tree.
      First suppose that $T$ is not a caterpillar. Then $T$ contains $S_{2,2,2}$ as an induced subgraph.
      Thus, $T$ contains a vertex $v$ such that $T-v$ contains at least three connected components each containing an edge.
      So, $T-v$ contains three connected components having path-width at least $1$.
      By Theorem~\ref{thm:charpw}, $T$ has path-width at least $2$, and by Theorem~\ref{thm:lrwpwtree},
      $T$ has linear rank-width at least $2$.

      Now suppose that $T$ is a caterpillar.
      We prove by induction on $\abs{V(T)}$ that $T$ has path-width at most~$1$.
      We may assume that $T$ has at least two vertices.
      Note that for every vertex $v$, $T-v$ has at most two connected components having an edge, which are still caterpillars,
      and all the other connected components are isolated vertices.
      So, $T-v$ has at most two connected components having path-width $1$ by induction, and all the other connected components have path-width $0$.
      Thus, by Theorem~\ref{thm:charpw}, $T$ has path-width at most $1$. This proves the claim.
       We now apply Theorem~\ref{thm:lrwpwtree} to conclude that every caterpillar has linear rank-width at most $1$.
      \end{proof}

\section{The Proof of Theorem~\ref{t-main1}}\label{sec:caterpillar}

In this section, we prove Theorem~\ref{t-main1}, which states that for every tree $T$ that is not a caterpillar, the class of $T$-pivot-minor-free graphs has unbounded linear rank-width.

		\begin{figure}[t]
	\centerline{
	\includegraphics[scale=0.6]{classt}}
	\caption{The graph
	$H\in {\cal C}$ is obtained from a tree $F$ by subdividing each edge once, and then applying a local complementation at every vertex of degree at least $3$ in $F$.}
	\label{fig:classt}
	\end{figure}

	Let~$\mathcal{C}$ be the class of graphs that can be obtained from the~$1$-subdivision of a tree by applying a local complementation at every vertex of degree at least~$3$.
	We give an example of a graph in ${\cal C}$ in Figure~\ref{fig:classt}.
Our proof of Theorem~\ref{t-main1} consists of the following parts:

\begin{enumerate}
\item we show that ${\cal C}$ has unbounded linear rank-width;
\item we show that every graph in ${\cal C}$ is a distance-hereditary graph with some additional properties needed to prove the third step; and
\item we show that  every graph in ${\cal C}$ is $T$-pivot-minor-free whenever $T$ is a tree that is not a caterpillar.
\end{enumerate}

We start with the following lemma that proves the first part.

\begin{lemma}\label{l-c}
The class~${\cal C}$ has unbounded linear rank-width.
\end{lemma}
\begin{proof}
Adler and Kant\'e~\cite{AdlerK2015} proved that trees have unbounded linear rank-width.
Note that the 1-subdivision $H$ of a graph $G$ contains $G$ as a vertex-minor:  for every subdivided vertex in $H$, perform a local complementation and remove it; this yields $G$.
So, by Lemma~\ref{lem:bouchet},
the class of $1$-subdivisions of trees also has unbounded linear rank-width.
	As local complementations do not change the linear rank-width of a graph by Lemma~\ref{lem:bouchet},  this means that
	$\mathcal{C}$ has unbounded linear rank-width.
\end{proof}

We will now prove that ${\cal C}$ is a subclass of the class of distance-hereditary graphs with some additional useful properties.
	In order to do this, we need the notion of a canonical split decomposition of a graph~\cite{CunninghamE80}, which we define below.

\begin{figure}
\centerline{\includegraphics[scale=0.5]{example} \quad\quad
\includegraphics[scale=0.5]{example2} }
\caption{For the graphs $F$ and $H$ in Figure~\ref{fig:classt}, the decompositions $D_{F'}$ and $D_H$ are the canonical split decompositions of $F'$ and $H$, respectively, where $F'$ is the $1$-subdivision of $F$.
Dashed edges denote marked edges and each circle denotes a bag.}
\label{fig:decomposition}
\end{figure}

	A vertex partition~$(X,Y)$ of a connected graph $G$ is a \emph{split} of~$G$ if $\abs{X}\ge 2, \abs{Y}\ge 2$, and~$N_G(Y)$ is complete to~$N_G(X)$.
	A connected graph $G$ on at least five vertices is \emph{prime} if it has no split.
		A connected graph~$D$ with a distinguished set of edges~$M(D)$ is a \emph{marked graph} if~$M(D)$ is a matching and each edge in~$M(D)$ is a cut edge.
	An edge in~$M(D)$ is a \emph{marked edge}, and every other edge of $D$ is an \emph{unmarked edge}.
	A vertex incident with a marked edge is a \emph{marked vertex}, and every other vertex of $D$ is an \emph{unmarked vertex}.
	Each connected component of $D-M(D)$ is a \emph{bag} of~$D$. If a marked edge $e$ is incident with a vertex of a bag~$B$, we say that $B$ is
        \emph{incident} with $e$.  A bag $B_1$ of $D$ is a neighbour bag of a bag $B_2$ of $D$ if there is a marked edge incident with both $B_1$ and $B_2$.
        The \emph{decomposition tree} of $D$ is the graph obtained from $D$ by contracting each bag into a vertex.

	If~$G$ has a split~$(X,Y)$, we construct a marked graph~$D$ on the vertex set $V(G) \cup \{x_1,y_1\}$ for some new vertices~$x_1$ and~$y_1$ such that
	\begin{itemize}
	\item for every two
	distinct vertices $x,y$ with $\{x,y\}\subseteq X$ or $\{x,y\}\subseteq Y$, it holds that $xy\in E(G)$ if and only if $xy\in E(D)$,
	\item $x_1y_1$ is a new marked edge,
	\item $X$ is anti-complete to~$Y$,
	\item $x_1$ is complete to $N_G(Y)$ (with only unmarked edges) and has no neighbours in $V(G)\setminus N_G(Y)$,
	\item $y_1$ is complete to $N_G(X)$ (with only unmarked edges) and has no neighbours in $V(G)\setminus N_G(X)$.
	\end{itemize}
	The graph~$D$ is a \emph{simple decomposition} of~$G$.
	A \emph{split decomposition} of a connected graph~$G$ is a marked graph~$D$ defined inductively to be either~$G$ or a marked graph obtained from a split decomposition~$D'$ of~$G$ by replacing a bag of~$D'$ with its simple decomposition.
We give an example of a split decomposition in Figure~\ref{fig:decomposition}.

	For a marked edge $xy$ of a marked graph $D$, the \emph{recomposition
  of $D$ along $xy$} is the marked graph $(D\wedge xy) -\{x,y\}$, where when we pivot $xy$, we add unmarked edges between $N_D(x)\setminus \{y\}$ and $N_D(y)\setminus \{x\}$.  This operation can be seen as merging two adjacent bags $B_1$ and $B_2$ into one bag $B$
  where the union of $B_1$ and $B_2$ was a simple decomposition of $B$.
  It is not hard to see that  if $D$ is a split decomposition of $G$, then $G$ can be obtained from $D$ by
recomposing along all the marked edges.

	A split decomposition~$D$ is \emph{canonical} if each bag of~$D$ is either a prime graph, a star, or a complete graph, and
recomposing every
marked edge in~$D$ results in a split decomposition having a bag that is neither a prime graph, a star, nor a complete graph.
We note that the split decompositions in Figure~\ref{fig:decomposition} are canonical.
	We say that a bag is a {\it star bag} if it is a star and a {\it complete} bag if it is a complete graph.

	\begin{theorem}[Cunningham and Edmonds~\cite{CunninghamE80}] \label{thm:CED}
	Every connected graph has a unique canonical split decomposition, up to isomorphism.
	\end{theorem}

Bouchet~\cite{Bouchet1988} described how split decompositions change under applying local complementations.
A vertex~$v$ in a split decomposition~$D$ \emph{represents} an unmarked vertex~$x$ (or is a \emph{representative} of~$x$) if either $v=x$ or there is a path of even length from~$v$ to~$x$ in~$D$ starting with a marked edge such that marked edges and unmarked edges appear alternately in the path.
Two unmarked vertices~$x$ and~$y$ are \emph{linked} in~$D$ if there is a path from~$x$ to~$y$ in~$D$ such that unmarked edges and marked edges appear alternately in the path.

A \emph{local complementation} at an unmarked vertex~$x$ in a split decomposition~$D$, denoted by~$D*x$, is the operation that replaces each bag~$B$ containing a representative~$w$ of~$x$ with~$B*w$.

\begin{lemma}[Bouchet~\cite{Bouchet1988}]\label{lem:localdecom}
Let~$D$ be the canonical split decomposition of a connected graph~$G$.
If~$x$ is an unmarked vertex of~$D$, then~$D*x$ is the canonical split decomposition of~$G*x$.
\end{lemma}

Let~$x$ and~$y$ be linked unmarked vertices in a split decomposition~$D$, and let~$P$ be the path in~$D$ linking~$x$ and~$y$ where unmarked edges and marked edges appear alternatively.
Observe that such a path is unique.
The \emph{pivoting on~$xy$ of~$D$}, denoted by $D\wedge xy$, is the split decomposition obtained as follows: for each bag~$B$ containing an unmarked edge~$vw$ of~$P$, we replace~$B$ with $B\wedge vw$.

\begin{lemma}[Adler, Kant\'e, and Kwon~\cite{AdlerKK2016}]\label{lem:pivotdecom}
	Let~$D$ be the canonical split decomposition of a connected graph~$G$.
	If $xy\in E(G)$, then $D\wedge xy$ is the canonical split decomposition of $G\wedge xy$.
\end{lemma}

An \emph{internal} edge of a tree is an edge that is not incident with a vertex of degree~$1$.
We also need the following result due to Bouchet~\cite{Bouchet1988}.

\begin{lemma}[Bouchet~\cite{Bouchet1988}]\label{thm:bouchettree}
  A canonical split decomposition of a tree $T$ can be constructed by replacing each internal edge of $T$ by a path of length $3$, the middle edge of the path  being the
  marked edge.
\end{lemma}

A bag of a split decomposition is a \emph{branching bag} if it is incident with at least three marked edges.
Let $\mathcal{M}$ be the set of all distance-hereditary graphs, in which each connected component admits a canonical split decomposition with the property that every branching bag is a complete bag.
We will prove that ${\cal C}$ is a subclass of ${\cal M}$, so every graph of ${\cal C}$ is a distance-hereditary graph with the additional property that  every branching bag in its canonical split decomposition is a complete bag.
In order to do this we need one more lemma.

\begin{lemma}[Bouchet~\cite{Bouchet1988}]\label{l-bouch}
A connected graph is distance-hereditary if and only if every bag of its canonical split decomposition is either a star or a complete graph.
\end{lemma}

\begin{lemma}\label{l-subset}
It holds that  $\mathcal{C}\subseteq \mathcal{M}$.
\end{lemma}

\begin{proof}
Let $G\in {\cal C}$.
By the definition of~$\mathcal{C}$, we find that $G$ is obtained from the $1$-subdivision~$T$ of some tree by performing local complementations at vertices of degree at least~$3$.
By Lemma~\ref{thm:bouchettree}, we can first construct the canonical split decomposition of~$T$.
By Lemma~\ref{lem:localdecom} we can then perform local complementations at corresponding unmarked vertices in the canonical split decomposition of $T$ to obtain the canonical split decomposition of~$G$ (we refer again to Figure~\ref{fig:decomposition} for an example).
By construction, every bag of the canonical split decomposition of $G$ is a star or a complete bag.
Hence, by Lemma~\ref{l-bouch}, we find that $G$ is a distance-hereditary graph.
From our construction we also note that every branching bag is complete.
We conclude that $G\in {\cal M}$.
\end{proof}

We will now prove that for every tree $T$ that is not a caterpillar, ${\cal M}$ is $T$-pivot-minor-free.
In order to do this we first show that $\mathcal{M}$ is closed under taking pivot-minors.

\begin{lemma}\label{lem:closed}
The class $\mathcal{M}$ is closed under taking pivot-minors.
\end{lemma}
\begin{proof}
	Let $G\in \mathcal{M}$.
	It is sufficient to show the following:
	\begin{enumerate}[(1)]
	\item for $v\in V(G)$, $G-v$ is in $\mathcal{M}$,
	\item for an edge $wz\in E(G)$, $G\pivot wz$ is in $\mathcal{M}$.
	\end{enumerate}
	We may assume that $G$ is connected. Let $D$ be the canonical split decomposition of $G$.

	We first show (2). Let $wz$ be an edge of $G$.
	By Lemma~\ref{lem:pivotdecom},
	$D\wedge wz$ is the canonical split decomposition of $G\wedge wz$.
	Let~$P$ be the path in~$D$ linking~$w$ and~$z$ where unmarked edges and marked edges appear alternatively.
	By the definition of pivoting in a canonical split decomposition,
	 we obtain $D\wedge wz$ from $D$ as follows: for each bag~$B$ containing an unmarked edge~$xy$ of~$P$, we replace~$B$ with $B\wedge xy$.
	 It is easy to observe that if $B$ is a complete bag, then $B\wedge xy$ is again a complete bag,
	 and if $B$ is a star bag, then $B\wedge xy$ is again a star bag.
	Therefore, $G\wedge wz$ is again contained in $\mathcal{M}$.

	It remains to prove (1). Suppose that $v\in V(G)$ and let $B$ be the bag containing $v$.
	We may assume that $D$ has at least two bags; otherwise, it is trivial.

\medskip
\noindent
{\bf Case 1.} $B$ is a complete bag.\\
	If $\abs{V(B)}\ge 4$, then after removing $v$ in $G$,
we find that $B$
	 is still a complete bag of size at least $3$.
	So, $D-v$ is a canonical split decomposition of $G-v$.
	If $\abs{V(B)}=3$, then $B-v$ is merged with one of the neighbour bags of $B$.
	This process does not change the type of the neighbour bag.
	It is possible that the two neighbour bags $B_1$ and $B_2$ of $B$ in $D$ are star bags,
	and they can be merged after $B-v$ is merged with a neighbour bag.
	In this case, each of $B_1$ and $B_2$ has at most two neighbour bags,
	and after merging $B_1$ and $B_2$, it is again a star bag that has at most two neighbour bags.
	Thus, $G-v$ is in $\mathcal{M}$ again.

\medskip
\noindent
{\bf Case 2.} $B$ is a star bag and $v$ is leaf of $B$.\\
	If $\abs{V(B)}\geq 4$, then $D-v$ is the canonical split decomposition of $G-v$.
	If $\abs{V(B)}=3$, then $B-v$ is merged with one of the neighbour bags of $B$.
	This process does not change the type of the neighbour bag.
	There might be two cases where the two neighbour bags $B_1$ and $B_2$ of $B$ in $D$
	are merged after $B-v$ is merged with a neighbour bag.
	If $B_1$ and $B_2$ are complete bags, then the merged bag becomes a complete bag.
	If $B_1$ and $B_2$ are star bags, then each of them has at most two neighbour bags in $D$,
	and after merging, the new bag is again a star bag that has at most two neighbour bags.
	So, $G-v$ is in $\mathcal{M}$.

\medskip
\noindent
{\bf Case 3.} $B$ is star bag and $v$ is the center of $B$.\\
	Then $v$ is a cut vertex of $G$; that is, $G-v$ is disconnected.
	Furthermore, each component of $G-v$ either consists of a single vertex, or it admits a split decomposition obtained from a connected component of
	$D-V(B)$ by removing a leaf of a star bag or a vertex in a complete bag. Thus, each component of $G-v$ is in $\mathcal{M}$,
	and thus, $G-v$ is also in $\mathcal{M}$.
	This conclude the proof of the lemma.
\end{proof}

We also need a known characterization of graphs of linear rank-width at most~$1$ in terms of their canonical split decompositions.

\begin{lemma}[Kant\'e and Kwon~{\cite{KanteK2018}}]\label{thm:charlrw1split}
Let $G$ be a connected graph with canonical split decomposition~$D$.
Then $G$ has linear rank-width at most~$1$  if and only if~$G$ is distance-hereditary and
the decomposition tree of  $D$ is a path.
\end{lemma}

We are now ready to prove Theorem~\ref{t-main1}.

\medskip
\noindent
{\bf Theorem~\ref{t-main1} (restated).}
{\it If~$T$ is a tree that is not a caterpillar, then the class of $T$-pivot-minor-free distance-hereditary graphs has unbounded linear rank-width.}

\begin{proof}
Let~$T$ be a tree that is not a caterpillar. As ${\cal C}\subseteq {\cal M}$ by Lemma~\ref{l-subset} and ${\cal C}$ has unbounded linear rank-width by Lemma~\ref{l-c}, it follows that ${\cal M}$ has unbounded linear rank-width. Moreover, ${\cal M}$ is a subclass of the class of distance-hereditary graphs.
Hence, to prove the theorem it remains to show that every graph in ${\cal M}$ is $T$-pivot-minor-free.

For contradiction, assume that~$T$ is a pivot-minor of some graph~$H\in \mathcal{M}$.
As $\mathcal{M}$ is closed under pivot-minors by Lemma~\ref{lem:closed}, we find that
$T\in \mathcal{M}$.
Let $L_T$ be a canonical split decomposition of $T$.
Since $T$ is a tree, $L_T$ has no complete bags.
So, by the definition of $\mathcal{M}$, $L_T$ has no branching bags, and thus, the decomposition tree of $L_T$ is a path.
Since $T$ is distance-hereditary
and the decomposition tree of $L_T$ is a path,
$T$ has linear rank-width at most $1$ by Lemma~\ref{thm:charlrw1split}.
Thus, by Lemma~\ref{lem:charlrw1tree}, $T$ is a caterpillar, a contradiction.
\end{proof}

\section{The Proof of Theorem~\ref{thm:pmdh}}\label{pndh}

In this section, we prove Theorem~\ref{thm:pmdh}, which states that $P_n$-pivot-minor-free distance-hereditary graphs have bounded linear rank-width.

To prove Theorem~\ref{thm:pmdh}, we use the canonical split decomposition of a distance-hereditary graph, discussed in Section~\ref{sec:caterpillar}.
A sequence $B_1, B_2, \ldots, B_n$ of distinct bags in a canonical split decomposition is a \emph{path of bags} if for each $i\in \{1, \ldots, n-1\}$, $B_{i+1}$ is a neighbour bag of $B_i$.
By the definition of a canonical split decomposition,
we may observe that if it contains a path of bags, then at least half of the bags are star bags. By applying some pivot operations,
we can extract a long path as a pivot-minor in this case.
So, we may assume that the decomposition tree of the canonical split decomposition of a given graph has no long path.
We use the following result which relates the path-width of a decomposition tree and the linear rank-width of the graph to conclude the theorem.

\begin{proposition}[Kant\'e and Kwon~{\cite{KanteK2018}}]\label{Kantek2018}
Let $D$ be the canonical split decomposition of a connected distance-hereditary graph $G$, and $T_D$ be the decomposition tree of $D$. Then $\frac{1}{2}\pw(T_D)\le \lrw(G) \le \pw(T_D) + 1$.
\end{proposition}

We also use a tight version of Theorem~\ref{t-rs83}. We will use the fact that if a graph contains a minor isomorphic to $P_n$,
then it contains a subgraph isomorphic to $P_n$ as well.

\begin{theorem}[Bienstock, Robertson, Seymour, and Thomas~\cite{BienstockRST1991}]\label{t-brst91}
For every tree~$T$ on $n$ vertices, the class of $T$-minor-free graphs has path-width at most $n-2$.
\end{theorem}

We are now ready to prove Theorem~\ref{thm:pmdh}.

\medskip
\noindent
{\bf Theorem~\ref{thm:pmdh} (restated).}
{\it Let $n\ge 3$ be an integer. Every $P_n$-pivot-minor-free distance-hereditary graph has linear rank-width at most $2n-5$.}

\begin{proof}
Let $G$ be a distance-hereditary graph having no pivot-minor isomorphic to $P_n$.
We will show that $G$ has linear rank-width at most $2n-3$.
We may assume that $G$ is connected.
Let $D$ be the canonical split decomposition of $G$ and $T_D$ be its decomposition tree.

We claim that $D$ has no path with $2n-4$ bags.
Assume that such a path of bags $B_1, B_2, \ldots, B_{2n-4}$ exists. As no two complete bags are neighbour bags in a canonical split decomposition,
at most $n-2$ bags in the sequence are complete bags.
Thus, there are at least $n-2$ bags that are star bags.
Let $B_{i_1}, B_{i_2}, \ldots, B_{i_t}$ be the sequence of all star bags in $B_1, B_2, \ldots, B_{2n-4}$ where
$1\le i_1<i_2< \cdots <i_t\le 2n-4$.
For convenience, we assign $B_0=B_{2n-3}=\emptyset$.

We claim that for $1\le k\le t$, there is a graph $G_k$ pivot-equivalent to $G$ with a canonical split decomposition $D_k$
such that
\begin{itemize}
	\item[($\ast$)] for every $1\le k'\le k$, the bag $D_k[V(B_{i_{k'}})]$ is a star bag whose center has no neighbour in $V(B_{i_{k'}-1})\cup V(B_{i_{k'}+1})$.
\end{itemize}
For $k=1$, assume that $D$ does not satisfy the property $(\ast)$.
We choose a vertex $v$ of $G$ represented by a vertex $v'$ in $B_{i_1}$ that has no neighbour in $V(B_{i_{1}-1})\cup V(B_{i_{1}+1})$,
and we choose a vertex $w$ of $G$ represented by the center $w'$ of $B_{i_1}$.
Then $v$ is adjacent to $w$ in $G$. In $G\pivot vw$, the center of $B_{i_1}\pivot v'w'$ has no neighbour in $V(B_{i_{1}-1})\cup V(B_{i_{1}+1})$.
Thus, $G_1=G\pivot vw$ and $D_1=D\pivot vw$ satisfy $(\ast)$.

Now, assume that $k>1$ and the property ($\ast$) is satisfied for $k-1$.
If the center of $D_{k-1}[V(B_{i_k})]$ has no neighbour in $V(B_{i_{k}-1})\cup V(B_{i_{k}+1})$,
then $G_k=G_{k-1}$ and $D_k=D_{k-1}$ satisfy $(\ast)$.
So, we may assume that
the center of $D_{k-1}[V(B_{i_k})]$ has a neighbour in $V(B_{i_{k}-1})\cup V(B_{i_{k}+1})$.
We distinguish between two cases.

\medskip
\noindent
\textbf{Case 1.} The center of $D_{k-1}[V(B_{i_k})]$ has a neighbour in $V(B_{i_{k}-1})$.

Note that if $D_{k-1}[V(B_{i_{k}-1})]$ is a star bag, then by the inductive hypothesis,
its center has no neighbour in $V(B_{i_k})$. But it is not possible by the definition of a canonical split decomposition.
Thus, $D_{k-1}[V(B_{i_{k}-1})]$ is a complete bag. We choose a vertex $v$ in $G_{k-1}$ represented by a vertex in $D_{k-1}[V(B_{i_{k}-1})]$ having no neighbour in $V(B_{i_{k}-2})\cup V(B_{i_{k}})$
and choose a vertex $w$ in $G_{k-1}$ represented by a vertex in $D_{k-1}[V(B_{i_k})]$ having no neighbour in $V(B_{i_{k}-1})\cup V(B_{i_{k}+1})$.
Observe that $v$ is adjacent to $w$ in $G_{k-1}$, because the center of $D_{k-1}[V(B_{i_k})]$ has a neighbour in $V(B_{i_k-1})$.
Thus, in $D_{k-1}\pivot vw$, the bag induced by $V(B_{i_k})$ is a star bag whose center has no neighbour in $V(B_{i_{k}-1})\cup V(B_{i_{k}+1})$.
As the bags on $V(B_{i_1}), \ldots, V(B_{i_{k-1}})$ are not changed by this pivot operation, $G_k=G_{k-1}\pivot vw$ and $D_k=D_{k-1}\pivot vw$ satisfy $(\ast)$.

\medskip
\noindent
\textbf{Case 2.} The center of $D_{k-1}[V(B_{i_k})]$ has a neighbour in $V(B_{i_{k}+1})$.

We choose a vertex $v$ in $G_{k-1}$ represented by a vertex in $D_{k-1}[V(B_{i_{k}+1})]$ having no neighbour in $V(B_{i_{k}})\cup V(B_{i_{k}+2})$
and choose a vertex $w$ in $G_{k-1}$ represented by a vertex in $D_{k-1}[V(B_{i_k})]$ having no neighbour in $V(B_{i_{k}-1})\cup V(B_{i_{k}+1})$ and linked to $v$ in $D_{k-1}$.
Such a vertex $w$ exists, because  the center of $D_{k-1}[V(B_{i_k})]$ has a neighbour in $V(B_{i_{k}+1})$,
and thus, if $D_{k-1}[V(B_{i_k})]$ is a star, then its center has a neighbour in $V(B_{i_k})$.
By a similar reason, $G_k=G_{k-1}\pivot vw$ and $D_k=D_{k-1}\pivot vw$ satisfy $(\ast)$.

\medskip
\noindent
Hence, we have found that the claim holds.

\medskip
\noindent
Now, in $D_t$, let $v_j$ be a vertex of $G_t$ represented by the center of $D_t[V(B_{i_j})]$ for each $1\le j\le t$,
and let $v_0$ be a vertex of $G_t$ represented by a leaf of $D_t[V(B_{i_1})]$ which has no neighbour in $V(B_{i_1+1})$,
and let $v_{t+1}$ be a vertex of $G_t$ represented by a leaf of $D_t[V(B_{i_t})]$ which has no neighbour in $V(B_{i_t-1})$.
It is not difficult to check that $v_0v_1v_2 \cdots v_tv_{t+1}$ is an induced path of $G_t$ on $t+2\ge n$ vertices.
This contradicts the assumption that $G$ has no pivot-minor isomorphic to $P_n$.
We conclude that $D$ has no path with $2n-4$ bags $B_1, B_2, \ldots, B_{2n-4}$.

The above means that the decomposition tree $T_D$ has no path on $2n-4$ vertices. By Theorem~\ref{t-brst91},
$T_D$ has path-width at most $2n-6$. By Proposition~\ref{Kantek2018}, we find that $G$ has linear rank-width at most $2n-5$.
\end{proof}

\section{The Proof of Theorem~\ref{t-maintwosubgraphs}}\label{claw}

In this section, we prove Theorem~\ref{t-maintwosubgraphs}, which states that the class of $(K_3,S_{1,2,2})$-free graphs has linear rank-width at most~$58$. We prove the following statements in this order:

\begin{enumerate}
\item bipartite $2P_2$-free graphs, which form a subclass of bipartite $(P_1+2P_2)$-free graphs, have linear rank-width at most~$1$;
\item bipartite $(P_1+2P_2)$-free graphs, which form a subclass of bipartite $S_{1,2,2}$-free graphs, have linear rank-width at most~$3$;
\item bipartite $S_{1,2,2}$-free graphs have linear rank-width at most~$3$;
\item non-bipartite $(K_3,C_5,S_{1,2,2})$-free graphs have linear rank-width at most~$3$; and
\item  $(K_3,S_{1,2,2})$-free graphs with an induced $C_5$ have linear rank-width at most~$58$.
\end{enumerate}

Note that statements 3--5 cover all cases for proving Theorem~\ref{t-maintwosubgraphs}.
So, we first consider bipartite $2P_2$-free graphs.

	\begin{lemma}\label{lem:bipartitechain}
	Every bipartite $2P_2$-free graph has linear rank-width at most~$1$.
	\end{lemma}
	\begin{proof}
	Let~$G$ be a bipartite $2P_2$-free graph with bipartition~$(A,B)$.
It is well known~\cite{Yannakakis1982} that a bipartite graph with bipartition $(X_1,X_2)$ is $2P_2$-free if and only if it is a bipartite chain, that is, for each $i\in \{1,2\}$, the neighbourhoods of the vertices in~$X_i$ can be ordered linearly with respect to the inclusion relation.
We may assume that $G$ is connected.
Hence, as $G$ is $2P_2$-free, we can define a sequence
	$A_1, A_2, \ldots, A_m$ of pairwise vertex-disjoint subsets of~$A$ such that
	\begin{itemize}
	\item $A_1\cup A_2\cup \cdots \cup A_m=A$,
	\item each~$A_i$ is a maximal set of pairwise twins in~$G$,
	\item for integers $i,j\in \{1, \ldots, m\}$ with $i<j$, $N_G(A_i)\subsetneq N_G(A_j)$.
	\end{itemize}
	If $m=1$, then $G$ is complete bipartite. In this case, we take a linear ordering $L_1$ of $A_1$ and a linear ordering $L_2$ of $V(G)\setminus A_1$ arbitrarily.
	It is not hard to see that $L_1\oplus L_2$ is a linear ordering of width at most $1$. Hence $G$ has linear rank-width at most~$1$.

	Now suppose that $m\geq 2$. In this case, $G$ is not complete bipartite.
	Notice that for each $i\in \{2, \ldots, m\}$, there is a vertex~$v\in B$ that has a neighbour in~$A_i$ but does not have a neighbour in $A_1\cup \cdots \cup A_{i-1}$; otherwise, vertices in $A_{i-1}\cup A_i$ have the same neighbourhood in~$B$, which contradicts the maximality of~$A_i$.
	For each $i\in \{2, \ldots, m\}$, let $B_i:=N_G(A_i)\setminus N_G(A_{i-1})$, and let $B_1:=N_G(A_1)$.
	Since~$G$ is connected, we have $B=B_1\cup B_2\cup \cdots \cup B_m$.

	For each $i\in \{1, \ldots, m\}$, let~$L^A_i$ be
	an ordering of~$A_i$ and~$L^B_i$ be an ordering of~$B_i$.
	It is not difficult to check that the linear ordering
	$L^B_1\oplus L^A_1\oplus L^B_2 \oplus L^A_2\oplus \cdots \oplus L^B_m\oplus L^A_m$ has width at most~$1$.
	\end{proof}

We now consider bipartite $(P_1+2P_2)$-free graphs and show the following lemma.

\begin{lemma}\label{lem:bipartitep5free}
Every bipartite $(P_1+\nobreak 2P_2)$-free graph has linear rank-width at most~$3$.
\end{lemma}
\begin{proof}
	Let~$G$ be a bipartite $(P_1+\nobreak 2P_2)$-free graph with bipartition~$(A,B)$.
	Then $G\times (A,B)$ is $P_5$-free. We may assume without loss of generality that $G\times (A,B)$ is connected. Then, as $G\times (A,B)$ is also bipartite, $G\times (A,B)$ is readily seen to be $2P_2$-free.
	By Lemma~\ref{lem:bipartitechain}, $G\times (A,B)$ has linear rank-width at most~$1$.
	By Lemma~\ref{lem:bipartitecomplement}, $G=G\times (A,B)\times (A,B)$ has linear rank-width at most~$3$.
\end{proof}

We now consider $S_{1,2,2}$-free bipartite graphs and need two results by Lozin~\cite{Lozin2000}.

\begin{lemma}[Lozin~\cite{Lozin2000}]\label{lem:lozin1}
Every connected bipartite $(S_{1,2,2},P_7)$-free graph is $(P_1+\nobreak 2P_2)$-free.
\end{lemma}

\begin{lemma}[Lozin~\cite{Lozin2000}]\label{lem:lozin2}
Let~$G$ be a bipartite $S_{1,2,2}$-free graph with no twins. If $G$ contains an induced~$P_7$,
then~$G$ is $K_{1,3}$-free
(and thus $G$ has maximum degree at most~$2$).
\end{lemma}

\begin{proposition}\label{t-clawpmfree1}
Every bipartite $S_{1,2,2}$-free graph has linear rank-width at most~$3$.
\end{proposition}
\begin{proof}
Let~$G$ be a bipartite $S_{1,2,2}$-free graph with bipartition~$(A,B)$.
We may assume that~$G$ is connected.
If~$G$ is $P_7$-free, then by Lemma~\ref{lem:lozin1}, $G$ is $(P_1+\nobreak 2P_2)$-free, and by Lemma~\ref{lem:bipartitep5free}, $G$ has linear rank-width at most~$3$.
Thus, we may assume that~$G$ contains an induced subgraph isomorphic to~$P_7$. Note that $G$ is not a complete bipartite graph.

Let $I_1, I_2, \ldots, I_m$ be the vertex partition of~$G$ such that each~$I_i$ is a maximal set of pairwise twins in~$G$.
Since~$G$ is connected and bipartite, each~$I_i$ is contained in one of $A$ or~$B$.
Let $G_1:=G//I_1 //I_2 // \cdots //I_m$.
Note that~$G_1$ is also connected.
We claim that~$G_1$ has no twins. Note that $G_1$ is not an edge, because $G$ is not a complete bipartite graph. So, $G_1$ has at least three vertices.
Moreover, $G_1$ still has an induced subgraph isomorphic to $P_7$, as $P_7$ has no twins.

Suppose for contradiction that~$G_1$ has two twins~$v_1$ and~$v_2$, and assume that~$v_1$ and~$v_2$ were identified from~$I_{i_1}$ and~$I_{i_2}$ for some~$i_1$ and~$i_2$, respectively.
Since each~$v_i$ has a neighbour and $G_1$ has at least three vertices, $v_1$ and~$v_2$ are in the same part of the bipartition, and thus~$I_{i_1}$ and~$I_{i_2}$ are contained in the same part of the bipartition of~$G$.
Thus~$I_{i_1}$ and~$I_{i_2}$ have the same neighbourhoods in~$G$, contradicting the fact that they are maximal sets of pairwise twins in~$G$.
So, $G_1$ has no twins.
Then $G_1$ has linear rank-width at most~$2$ because by Lemma~\ref{lem:lozin2} every vertex has degree at most $2$.
By Lemma~\ref{lem:twinreduced}, we conclude that $G$ has linear rank-width at most~$3$.
\end{proof}

We now consider $(K_3,C_5,S_{1,2,3})$-free graphs and need the following result as a lemma.

\begin{lemma}[Dabrowski, Dross, and Paulusma~\cite{DabrowskiDP2016}]\label{lem:c5free}
Let $G$ be a connected $(K_3,C_5,S_{1,2,3})$-free graph that does not contain a pair of false twins.
Then $G$ is either bipartite or an induced cycle.
\end{lemma}

	\begin{proposition}\label{prop:clawpmfree2}
	Every non-bipartite $(K_3, C_5, S_{1,2,2})$-free graph has linear rank-width at most~$3$.
	\end{proposition}
	\begin{proof}
	Let~$G$ be a connected non-bipartite $(K_3, C_5, S_{1,2,2})$-free graph.
	By Lemma~\ref{lem:c5free},
	$G$ is a graph obtained from an induced cycle $C=c_1c_2 \cdots c_kc_1$ by adding false twins.
	 For each $i\in \{1, \ldots, k\}$, let~$U_i$ be the maximal set of false twins containing~$c_i$ in~$G$.
	 As $G//U_1 //U_2// \cdots //U_k$ is isomorphic to $C$ and $C$ has linear rank-width at most $2$,
	 by Lemma~\ref{lem:twinreduced}, we find that $G$ has linear rank-width at most $3$.
	\end{proof}

We now consider  $(K_3, S_{1,2,2})$-free graphs that contain an induced~$C_5$. We first introduce some additional terminology and lemmas.
A graph is \emph{$3$-partite} if its vertex set can be partitioned into three independent sets.
The following lemma follows from a theorem of Bogdan et al~\cite{BogdanLZW18}.

\begin{lemma}\label{lem:3partitegraph}
Let~$G$ be a $3$-partite graph with vertex partition $(V_1, V_2, V_3)$ such that
\begin{itemize}
\item[(a)] for every $a\in V_1$, $b\in V_2$, $c\in V_3$, $G[a,b,c]$ is isomorphic neither to~$K_3$ nor to~$3P_1$,
\item[(b)] $G[V_1\cup V_2]$, $G[V_1\cup V_3]$, and $G[V_2\cup V_3]$ are $2P_2$-free.
\end{itemize}
Then~$G$ has linear rank-width at most~$3$.
\end{lemma}

\begin{proof} Bogdan et al~\cite{BogdanLZW18} proved that every $n$-vertex graph satisfying conditions (a) and (b) admits a linear ordering $x_1,\ldots,x_n$
  and a labeling $\ell:V(G)\to \{a,b,c\}$ such that $x_ix_j\in E(G)$ if and only if $i<j$ and $(\ell(x_i),\ell(x_j))\in \{(a,b),(b,c),(c,a)\}$. Each cut of
  this linear ordering has cut-rank at most $3$, because it has at most three different rows.
\end{proof}

Let $G$ be a graph and $V_1, V_2, V_3$ be
three pairwise
disjoint independent sets of $G$. We denote the subgraph of $G$ induced by $V_1\cup V_2\cup V_3$ as $G[V_1, V_2, V_3]$.
Moreover, if $G[V_1, V_2, V_3]$ satisfies conditions~(a) and~(b) in Lemma~\ref{lem:3partitegraph}, then we call $G[V_1, V_2, V_3]$  \emph{nice}.

We are now ready to prove the following result. We note that Brandst\"adt, Mahfud and Mosca~\cite{BrandstadtMM16} gave an alternative proof of the result  from~\cite{DabrowskiDP2016} that shows that $(K_3,S_{1,2,2})$-free graphs have bounded rank-width. Some parts of the proof of our result below are similar to parts of the proof of~\cite{BrandstadtMM16}. As we need to use slightly different arguments, we have chosen to keep our proof self-contained. However, we explicitly indicate whenever there is overlap between our arguments and the ones used in~\cite{BrandstadtMM16}.

	\begin{proposition}\label{p-c5}
	Every $(K_3, S_{1,2,2})$-free graph that contains an induced~$C_5$ has linear rank-width at most~$58$.
	\end{proposition}

        \begin{proof}
	Let~$G$ be a $(K_3, S_{1,2,2})$-free graph that contains an induced subgraph $C$ isomorphic to~$C_5$.
	We may assume without loss of generality that~$G$ is connected.
We write $C=c_1c_2c_3c_4c_5c_1$ and interpret subscripts modulo~$5$.
	Let $U:=V(G)\setminus V(C)$.

	Since~$G$ is $K_3$-free, every vertex in~$U$ has either no neighbours in~$C$ or exactly one neighbour or two neighbours that are not consecutive in~$C$.

We claim that every vertex of $U$ has a neighbour in $C$.
This can be seen as follows. For contradiction, assume that $U$ contains a vertex that has no neighbour in $C$. As $G$ is connected, this means that $U$ contains two vertices
 $w_1$ and $w_2$, such that $w_1$ has a neighbour in $C$ and $w_2$ is adjacent to $w_1$ but $w_2$ has no neighbour in $C$.
	If $w_1$ has exactly one neighbour $c_i$ in $C$, then $G[c_{i-2}, c_{i-1}, c_i, c_{i+1}, w_1, w_2]$ is isomorphic to $S_{1,2,2}$.
	If $w_1$ has two neighbours $c_{i-1}, c_{i+1}$ in $C$, then $G[c_{i-3}, c_{i-2}, c_{i-1}, c_i, w_1, w_2]$ is isomorphic to $S_{1,2,2}$.
However, both cases are not possible, as	 $G$ is $S_{1,2,2}$-free. Hence, we conclude that every vertex in $U$ has a neighbour in $C$.

	Consequently, we can partition~$U$ into ten parts $\{V_1, \ldots, V_5, W_1, \ldots, W_5\}$ such that for each $i\in \{1, \ldots ,5\}$,
	\begin{itemize}
	\item $V_i$ is the set of vertices whose unique neighbour in~$C$ is~$c_i$, and
	\item $W_i$ is the set of vertices that are adjacent to~$c_{i-1}$ and~$c_{i+1}$.
	\end{itemize}
	Each set in $\{V_1, \ldots, V_5, W_1, \ldots, W_5\}$ is an independent set as~$G$ is $K_3$-free.
	We verify basic relations between these parts.
	Let $i\in \{1,\ldots, 5\}$.
	\begin{enumerate}
	\item[(1)] $V_i$ is complete to $V_{i-1}\cup V_{i+1}$, and anti-complete to $V_{i-2}\cup V_{i+2}$.

	For contradiction, suppose that there are $a\in V_i$ and $b\in V_{i+1}$ that are not adjacent.
	Then $G[c_{i-2}, c_{i-1}, c_i, c_{i+1}, a, b]$ is isomorphic to $S_{1,2,2}$.
	So, there are no such vertices. This implies that $V_i$ is complete to $V_{i+1}$, and also complete to $V_{i-1}$.
	Suppose that there are $a\in V_i$ and $b\in V_{i+2}$ that are adjacent.
	Then $G[c_{i-2}, c_{i-1}, c_i, c_{i+1}, a, b]$ is isomorphic to $S_{1,2,2}$.
	Hence, $V_i$ is anti-complete to $V_{i+2}$ and also anti-complete to $V_{i-2}$.\dia

	\item[(2)] $W_i$ is anti-complete to $W_{i-2}\cup W_{i+2}$.

	This is because $G$ is $K_3$-free.\dia
	\item[(3)] $W_i$ is complete to~$V_i$, and anti-complete to $V_{i-1}\cup V_{i+1}$.

	Assume that there are vertices $a\in W_i$ and $b\in V_i$ that are not adjacent to each other.
	Then $G[c_{i-3}, c_{i-2}, c_{i-1}, c_i, a, b]$ is isomorphic to $S_{1,2,2}$.
	This implies that $W_i$ is complete to $V_i$.
	As $G$ is $K_3$-free, $W_i$ is anti-complete to $V_{i-1}\cup V_{i+1}$.	\dia

	\item[(4)] $G[V_i\cup W_{i+2}]$, $G[V_i\cup W_{i-2}]$, $G[W_i\cup W_{i+1}]$ are $2P_2$-free.

	Suppose that there are $a_1, a_2\in V_i$ and $b_1, b_2\in W_{i+2}$ such that $a_1b_1, a_2b_2\in E(G)$ and $a_1b_2, a_2b_1\notin E(G)$.
	Then $G[c_{i-1}, c_i, a_1, a_2, b_1, b_2]$ is isomorphic to $S_{1,2,2}$. So, $G[V_i\cup W_{i+2}]$ is $2P_2$-free, and similarly,
	$G[V_i\cup W_{i-2}]$ is $2P_2$-free.
Assume that there are $a_1, a_2\in W_i$ and $b_1, b_2\in W_{i+1}$ such that $a_1b_1, a_2b_2\in E(G)$ and $a_1b_2, a_2b_1\notin E(G)$.
	Then $G[c_{i-2}, c_{i-1}, a_1, a_2, b_1, b_2]$ is isomorphic to $S_{1,2,2}$, a contradiction. \dia

	\item[(5)] For $v\in V_i, w\in W_{i+2}, z\in W_{i-2}$, $\{v,w,z\}$ is not an independent set.

	Consider that such $v,w,z$ forming an independent set exist.
	Then $G[c_{i-1}, c_i, c_{i+1}, a, b, c]$ is isomorphic to $S_{1,2,2}$, a contradiction. \dia
      \end{enumerate}

      By Claims (4) and (5), we deduce that for each $1\leq i \leq 5$,
      $G[V_i,W_{i-2},W_{i+2}]$ is a nice $3$-partite graph.
      Moreover, $V_i$ is complete to
      $V_{i-1}, V_{i+1}, W_i$ and anti-complete to $V_{i+2}, V_{i+3},W_{i+1}$.
      By doing bipartite complementations between $V_i$ and $V_{i+1}\cup W_i$ for each $1\le i\le 5$,
      we can assume that each edge belongs to $G[V_i,W_{i+2},W_{i-2}]$ for some $1\leq i \leq 5$.

  Our goal is now to compute a refined set of
      pairwise non-intersecting $3$-partite graphs by doing a small number of  bipartite complementations such that each new $3$-partition graph satisfies conditions~(a) and~(b) of Lemma \ref{lem:3partitegraph}.

      We first observe that each $G[V_i,W_{i-2},W_{i+2}]$ intersects only $G[V_{i+1},W_{i-2},W_{i-1}]$ and $G[V_{i-1}, W_{i+2},W_{i+1}]$.  We now aim for each $1\leq i
      \leq 5$ to split $V_i$, $W_{i-2}$ and $W_{i+2}$ in such a way that we can construct the non-intersecting desired $3$-partite graphs after some bipartite
      complementations. We will use the same construction as in \cite{BrandstadtMM16}, but the way we use the different sets differs from \cite{BrandstadtMM16}.

      For each $1\leq i \leq 5$, let us define the following partition of $W_{i}$:
      \begin{align*}
        W_{i}^- &= \{x\in W_{i}\mid x\ \textrm{has a non-neighbour in}\  W_{i+1}\},\\
        W_{i}^+ &=\{x\in W_{i}\setminus W_{i}^- \mid x\ \textrm{has a non-neighbour in}\ W_{i-1}\},\\
        W_{i}^* &= W_{i}\setminus (W_{i}^-\cup W_{i}^+).
      \end{align*}
      By definition, $W_i^+\cup W_i^*$ is complete to $W_{i+1}$ and $W_i^-$ is complete to $W_{i+1}^*$.
      We claim that $W_i^-$ is also complete to $W_{i+1}^-$.
      Assume that a vertex $x\in W_i^-$ has a non-neighbour $y$
          in $W_{i+1}^-$. Then, by definition $y$ has a non-neighbour $z$ in $W_{i+2}$. Therefore, $G[c_{i+1},x,c_{i-1},z,c_{i+2},y]$ is isomorphic to $S_{1,2,2}$ because $x$
          is not adjacent to $z$ by (2), a contradiction.
	Thus, $W_i^-$ is complete to $W_{i+1}^-$.

      We now show some relationships between the $V_i$'s and $W_i$'s, which were also proven in \cite{BrandstadtMM16}, but we add the proofs for completeness.  Let $1\leq i \leq 5$.
      \begin{enumerate}[(a)]
        \item $V_i$ is anti-complete to $W_{i+2}^+$ and $W_{i-2}^-$.

 For contradiction, assume that a vertex $x\in V_i$ has a neighbour $z\in W_{i+2}^+\cup W_{i-2}^-$.
First suppose that $z\in W_{i+2}^+$.
Then, by definition, $z$ has a non-neighbour $y\in W_{i+1}$. However, now $G[c_i,c_{i-1},
          x,z,y,c_{i+2}]$ is isomorphic to $S_{1,2,2}$, a contradiction.  Now suppose that $z\in W_{i-2} ^-$. Then, by definition, $z$ has a non-neighbour $y\in W_{i-1}$. However, now
          $G[c_i,y,c_{i-2},x,z,c_{i+1}]$ is
          isomorphic to $S_{1,2,2}$, another contradiction. \dia

        \item If a vertex $x\in V_i$ has a neighbour in $W_{i+2}^*$ (resp. $W_{i-2}^*$), then $x$ is complete to $W_{i+2}^-$ and anti-complete to $W_{i-2}$ (resp. is
          complete to $W_{i-2}^+$ and anti-complete to $W_{i+2}$).

          Assume $x\in V_i$ has a neighbour $y$ in $W_{i+2}^*$. Because $W_{i+2}^*$ is complete to $W_{i+3}=W_{i-2}$, we find that
          $x$ is anti-complete to $W_{i-2}$. Assume that $x$ has
          a non-neighbour $z$ in $W_{i+2}^-$. By definition, there is a vertex $w\in W_{i-2}$ not adjacent to $z$, but adjacent to $y$. Then, $G[y,w,c_{i-1},x,c_{i+1},z]$
          is isomorphic to $S_{1,2,2}$, a contradiction.

          Assume now that $x\in V_i$ has a neighbour $y$ in $W_{i-2}^*$. Because by definition $W_{i-2}^*$ is complete to $W_{i-3}=W_{i+2}$, we find that $x$ is anti-complete to
          $W_{i+2}$. Suppose that $x$ has a non-neighbour $z$ in $W_{i-2}^+$.  By definition, $z$ has a non-neighbour $w$ in $W_{i+2}$, which is adjacent to $y$. Again,
          $G[y,x,w,c_{i+1},c_{i-1},  z]$ is isomorphic to $S_{1,2,2}$, a contradiction. \dia
        \end{enumerate}

\noindent
  	For each $1\leq i \leq 5$, let
        \begin{align*}
          V_i^1 & =  \{x\in V_i\mid x\ \textrm{has a neighbour in $W_{i+2}^*$}\},\\
          V_i^2 &= \{x\in V_i\mid x\ \textrm{has a neighbour in $W_{i-2}^*$}\},\\
          V_i^3 &= \{x\in V_i\mid x\ \textrm{has no neighbour in}\ W_{i+2}^*\ \textrm{and}\ W_{i-2}^*\}.
        \end{align*}
          From (a), we know that for each $x\in V_i$, $x$ can have neighbours in only $W_{i+2}^- \cup W_{i+2}^*$ and in $W_{i-2}^+\cup W_{i-2}^*$.
          From (b), every vertex in $V_i^1$ is complete to $W_{i+2}^-$ and anti-complete to $W_{i-2}$,
          and every vertex in $V_i^2$ is complete to $W_{i-2}^+$ and anti-complete to $W_{i+2}$.
          Thus, $V_i^1, V_i^2, V_i^3$ are disjoint sets.

               The common neighbours between a vertex $x\in V_i$ and $y\in V_{i+1}$ are in $W_{i-2}^*$ and the common neighbours between $x\in
        V_i$ and $y\in V_{i-1}$ are in $W_{i+2}^*$.
For each $1\leq i \leq 5$, we define the following $3$-partite graphs (by ignoring duplicates):
        \begin{align*}
          G_i^1 & = G[V_i^1, V_{i-1}^2, W_{i+2}^*],\\
          G_i^2 & = G[V_i^2,V_{i+1}^1, W_{i-2}^*],\\
          G_i^3 &= G[V_i^3,W_{i+2}^-,W_{i-2}^+].
        \end{align*}

	Now, we do some bipartite complementations.
       First we do bipartite complementations between
        $W_i^+\cup W_i^*$ and $W_{i+1}$, and between $W_i^-$ and $W_{i+1}^-\cup W_{i+1}^*$ for each $i$.
        The resulting graph has the property that the edges between $W_i$ and $W_{i+1}$ are always between $W_i^-$ and $W_{i+1}^+$.
	Secondly, we do bipartite complementations between $V_i^1$ and $W_{i+2}^-$, and $V_i^2$ and $W_{i-2}^+$ for each $i$.
	Lastly, we remove $C$.
	Observe that the edges of the remaining graph are all contained in one $G_i^j$ for some $1\leq i \leq 5$
        and $1\leq j\leq 3$, and by definition the $G_i^j$'s are pairwise disjoint (after removing the duplicates).

        Because all of the graphs $G_i^j$ are disjoint, the linear rank-width of the resulting graph is the maximum linear rank-width of its connected components. Since each connected component is a nice
        $3$-partite graph,
        by Lemma \ref{lem:3partitegraph}, we can conclude that the linear rank-width of the resulting graph is at most $3$. We will now count the number of
        times we applied bipartite complementations. For each $1\leq i\leq 5$, we did one bipartite complementation to keep only the edges between $V_i$ and $W_{i-2}\cup W_{i+2}$,
        and then two bipartite
        complementations to keep the edges between $W_i^-$ and $W_{i+1}^+$, and finally two bipartite complementations to remove the edges between $V_i^1$ and $W_{i+2}^-$, and $V_i^2$ and $W_{i-2}^+$, that is, in total five bipartite complementations, resulting to a total of $25$ bipartite complementations. So, by Lemma \ref{lem:bipartitecomplement}, the graph $G-V(C)$ has linear rank-width at most
        $3+2*25=53$. Because $|V(C)|=5$, we can conclude that the linear rank-width of $G$ is at most $58$.
	\end{proof}

We are now ready to prove Theorem~\ref{t-maintwosubgraphs}.

\medskip
\noindent
{\bf Theorem~\ref{t-maintwosubgraphs} (restated).}
{\it Every $(K_3, S_{1,2,2})$-free graph has linear rank-width at most~$58$.}

\begin{proof}
Let $G$ be a $(K_3, S_{1,2,2})$-free graph, that can be assumed connected. If $G$ is bipartite, we use Proposition~\ref{t-clawpmfree1}.
If $G$ is non-bipartite but $C_5$-free, we use Proposition~\ref{prop:clawpmfree2}. In the remaining case, we use Proposition~\ref{p-c5}.
\end{proof}

\section{Concluding Remarks}\label{sec:conclusion}

In this paper we researched the relationship between pivot-minors and boundedness of linear rank-width. We first proved that for every tree~$T$ that is not a caterpillar, the class of $T$-pivot-minor-free graphs has unbounded linear rank-width.
We then posed Conjecture~\ref{q-1}, which states that an affirmative answer can be found whenever $T$ {\it is} a caterpillar.
We were only able to give an affirmative answer to this conjecture that holds for every caterpillar $T$, if the class of $T$-pivot-minor-free graphs is, in addition, distance-hereditary as well.
We also proved that the class of $K_{1,3}$-pivot-minor-free graphs has bounded linear rank-width.
As a next step for proving Conjecture~\ref{q-1}, it seems natural to consider the case where $T=K_{1,r}$ for $r\geq 4$.
We could also prove Conjecture~\ref{q-1} for $P=P_4$.
Since Conjecture~\ref{q-1} is equivalent to the alternative conjecture that for every path $P$, the class of $P$-pivot-minor-free graphs has bounded linear rank-width, the case where $P=P_5$ is another interesting open case.

For obtaining our results (in particular, the case where $T=K_{1,3}$) we followed a general strategy consisting of two steps.  We believe this strategy is also useful for making further progress towards Conjecture~\ref{q-1}. However, Step~1 of the strategy requires us to find a hereditary graph class (class of graphs that can be characterized by a set of forbidden induced subgraphs) that contains the class of $T$-pivot-minor-free graphs under consideration. In general, finding an appropriate hereditary graph class is a challenging task.

The fact that $K_{1,3}$-pivot-minor-free graphs have bounded linear rank-width follows from a stronger result that we showed, namely that $(K_3,S_{1,2,2})$-free graphs have bounded linear rank-width. Showing this stronger result will be useful for a systematic study on the boundedness of linear rank-width of  $(H_1,H_2)$-free graphs.
Such a classification  already exists for $H$-free graphs, as observed in~\cite{DJP19}: for a graph $H$, the class of $H$-free graphs has bounded linear rank-width if and only if~$H$ is a subgraph of $P_3$ not isomorphic to $3P_1$.
We note that similar classifications also exist for other width parameters: for the tree-width of $(H_1,H_2)$-free graphs~\cite{BBJPPV20}, the rank-width of $H$-free graphs
(see~\cite{DJP19}), rank-width of $H$-free bipartite graphs~\cite{DP14,Lo02,LV08}, and up to five non-equivalent open cases, rank-width of $(H_1,H_2)$-free graphs (see~\cite{BDJP19} or~\cite{DJP19}), mim-width of $H$-free graphs, whereas there is still an infinite number of open cases left for the mim-width of $(H_1,H_2)$-free graphs~\cite{BHMPP20}.

We leave a systematic study into boundedness of linear rank-width of $(H_1,H_2)$-free graphs for future research. Here, we only collect known results. The class of $(H_1, H_2)$-free graphs has bounded linear rank-width if
\begin{itemize}
	\item one of $H_1$ and $H_2$ is a subgraph of $P_3$ that is not isomorphic to $3P_1$~\cite{DJP19},
	\item $(H_1, H_2)=(K_3, S_{1,2,2})$ and $(3P_1, \overline{S_{1,2,2}})$ (Theorem~~\ref{t-maintwosubgraphs})
	\item $(H_1, H_2)=(P_4, F)$ where $F$ is a threshold graph~\cite{BKV17}.
\end{itemize}
The class of $(H_1, H_2)$-free graphs has unbounded linear rank-width if
\begin{itemize}
	\item $(H_1, H_2)=(K_3, S_{1,2,3})$ or $(3P_1, \overline{S_{1,2,3}})$~\cite{BogdanLZ18},
	\item $(H_1, H_2)=(P_4, F)$ where $F$ is not a threshold graph~\cite{BKV17},
	\item all known cases where $(H_1, H_2)$-free graphs have unbounded rank-width.
\end{itemize}

redpajama_set_name: RedPajamaArXiv


A despatch from our correspondent in Sri Lanka.
I would very much appreciate it if my Our Salon friends could read and comment on this.
MICK HUCKNALL WAS THE FIRST TO COMMENT!!!
Been a while. How's life in Sri Lanka?
Here's looking at you kid. Glad you're here.
Hi Padraig. Glad to see you here. I don't know when I'll fully move but Open is making the decision easier day by day.
Padraig, I think I can safely say that Our Salon won't have the spam and errors that Open experiences. As far a respect - as the number of members increases, I have no doubt there will be negative interactions between members (it's already happened) - I don't see lorianne ever responding that way unless she's joking.
Welcome to Our Salon, Padraig! I hope you enjoy your stay.