#
Vertex Cover Reconfiguration and Beyond^{ †}

^{1}

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^{‡}

## Abstract

**:**

## 1. Introduction

## 2. Preliminaries

**Proposition**

**1.**

**Proof.**

**Proposition**

**2.**

## 3. Representing Reconfiguration Sequences

**Proposition**

**3.**

## 4. Hardness Results

$(t,d)$-bipartite constrained crown | |

Input: | A bipartite graph $G=(A\cup B,E)$ and two positive integers t and d |

Parameters: | t and d |

Question: | Does G have a $(t,d)$-constrained crown $(W,H)$ such that $W\subseteq A$ and $H\subseteq B$? |

**Lemma**

**1.**

**Proof.**

**Theorem**

**1.**

**Proof.**

- (1)
- Add each vertex ${y}_{i}$ such that ${b}_{i}\in H$. The resulting vertex cover size is $|A|+d+t+|H|$.
- (2)
- Remove $d+|H|$ vertices ${x}_{i}$ such that ${a}_{i}\in W$. The resulting vertex cover size is $|A|+t$.
- (3)
- Add each vertex from V. The resulting vertex cover size is $|A|+d+2t$.
- (4)
- Remove each vertex from U. The resulting vertex cover size is $|A|+t$.
- (5)
- Add each vertex removed in Phase 2. The resulting vertex cover size is $|A|+d+t+|H|$.
- (6)
- Remove each vertex added in Phase 1. The resulting vertex cover size is $|A|+d+t$.

## 5. Polynomial-Time Algorithms

**Definition**

**1.**

- (1)
- $|{A}^{\prime}|\le |A|$;
- (2)
- For every node ${A}^{\u2033}$ in β, $|{A}^{\u2033}|\le |A|+c$;
- (3)
- For every node ${A}^{\u2033}$ in β, ${A}^{\u2033}$ is obtained from its predecessor by either the removal or the addition of a single vertex in the symmetric difference of the predecessor and B;
- (4)
- No vertex is touched more than once in the course of β.

**Proposition**

**4.**

**Lemma**

**2.**

- (1)
- $|S|\le k-c$;
- (2)
- $|T|\le k-c$; and
- (3)
- For any two vertex covers A and B of G such that $|A|\le k-c$ and $|B|\le k-c$, either $A\stackrel{c,B}{\leftrightarrow}{A}^{\prime}$ or $B\stackrel{c,A}{\leftrightarrow}{B}^{\prime}$, where ${A}^{\prime}$ and ${B}^{\prime}$ are vertex covers of G.

**Proof.**

#### 5.1. Trees

**Theorem**

**2.**

**Proof.**

**Case (1):**$|S|=k$, $|T|=k$, S is non-minimal and T is non-minimal. When both S and T are of size k and are non-minimal, then each must contain at least one removable vertex. Hence, by removing such vertices, we can transform S and T into vertex covers ${S}^{\prime}$ and ${T}^{\prime}$, respectively, of size $k-1$. We let u and v be removable vertices in S and T, respectively, and we set ${S}^{\prime}=S\backslash \left\{u\right\}$ and ${T}^{\prime}=T\backslash \left\{v\right\}$.

- If $u\in {S}_{R}$ and $v\in {T}_{A}$, then the length of a shortest reconfiguration sequence from ${S}^{\prime}$ to ${T}^{\prime}$ will be $|{S}^{\prime}\Delta {T}^{\prime}|=|S\Delta T|-2$. Therefore, accounting for the two additional removals, the length of a shortest path from S to T will be equal to $|S\Delta T|$.
- If $u\in {S}_{R}$ and $v\in {C}_{ST}$, then the length of a shortest reconfiguration sequence from ${S}^{\prime}$ to ${T}^{\prime}$ will be $|{S}^{\prime}\Delta {T}^{\prime}|=|S\Delta T|-1$. Since v is in ${C}_{ST}$, it must be removed and added back. Therefore, the length of a shortest path from S to T will be equal to $|S\Delta T|+2$. The same is true when $u\in {C}_{ST}$ and $v\in {T}_{A}$ or when $u=v$ and $u\in {C}_{ST}$.
- Otherwise, when $u\in {C}_{ST}$, $v\in {C}_{ST}$ and $u\ne v$, the length of a shortest path from S to T will be $|S\Delta T|+4$, since we have to touch two vertices in ${C}_{ST}$ (i.e., two extra additions and two extra removals).

**Case (2):**$|S|=k$, $|T|=k-1$ and S is non-minimal (similar arguments hold for the symmetric case where $|S|=k-1$, $|T|=k$, and T is non-minimal). Since $|T|=k-1$, we only need to reduce the size of S to $k-1$ in order to apply Lemma 2. Since S is non-minimal, it must contain at least one removable vertex. We let u be a removable vertex in S, and we set ${S}^{\prime}=S\backslash \left\{u\right\}$.

- If $u\in {S}_{R}$, then the length of a shortest reconfiguration sequence from ${S}^{\prime}$ to T will be $|{S}^{\prime}\Delta T|=|S\Delta T|-1$. Therefore, accounting for the additional removal, the length of a shortest path from S to T will be equal to $|S\Delta T|$.
- If $u\in {C}_{ST}$, then the length of a shortest reconfiguration sequence from ${S}^{\prime}$ to T will be $|{S}^{\prime}\Delta T|=|S\Delta T|$. Since v is in ${C}_{ST}$, it must be removed and added back. Therefore, the length of a shortest path from S to T will be equal to $|S\Delta T|+2$.

#### 5.2. Cactus Graphs

**Proposition**

**5.**

**Proposition**

**6.**

**Proposition**

**7.**

**Lemma**

**3.**

- (1)
- $G[{S}_{R}\cup {T}_{A}]$ has a vertex $v\in {S}_{R}$ ($v\in {T}_{A}$) such that $|{N}_{G[{S}_{R}\cup {T}_{A}]}\left(v\right)|\le 1$; or
- (2)
- there exists a cycle Y in $G[{S}_{R}\cup {T}_{A}]$ such that all vertices in $Y\cap {S}_{R}$ ($Y\cap {T}_{A}$) have degree exactly two in $G[{S}_{R}\cup {T}_{A}]$.

**Proof.**

**Lemma**

**4.**

**Proof.**

**Lemma**

**5.**

**Proof.**

**Theorem**

**3.**

**Proof.**

## 6. $\mathsf{FPT}$ Algorithms

Vertex cover compression | |

Input: | A graph $G=(V,E)$ and a vertex cover C of G such that $|C|=k\ge 1$ |

Parameters: | k |

Question: | Does G have a vertex cover ${C}^{\prime}$ of size $k-1$? |

#### 6.1. Compression via Reconfiguration

**Theorem**

**4.**

**Proof.**

#### 6.2. $\mathsf{NP}$-Hardness on Four-Regular Graphs

**Proposition**

**8.**

**Proof.**

**Proposition**

**9.**

**Proof.**

**Lemma**

**6.**

**Proof.**

- (1)
- Add all k beads in ${L}_{1}$. Since S is a vertex cover of ${W}_{k}$, we know that the additional k beads will result in a vertex cover of size $3{k}^{2}+k$.
- (2)
- Add both vertices in ${p}_{1,1}^{u}$, and remove both vertices in ${p}_{1,1}^{l}$. The removal of both vertices in ${p}_{1,1}^{l}$ is possible since we added all their neighbours in ${L}_{1}$ (Step (1)) and ${U}_{1}$. The size of a vertex cover reaches $3{k}^{2}+k+2$ after the additions and then drops back to $3{k}^{2}+k$.
- (3)
- Repeat Step (2) for all sequin pairs ${p}_{i,1}^{u}$ and ${p}_{1,i}^{l}$ for $2\le i\le k$. The size of a vertex cover is again $3{k}^{2}+k$ once Step (3) is completed. Step (2) is repeated a total of k times. After every repetition, we have a vertex cover of ${W}_{k}$ since all beads in ${L}_{1}$ were added in Step (1), and the remaining neighbours of each sequin pair in ${U}_{i}$ are added prior to the removals.
- (4)
- Add both vertices in ${p}_{1,2}^{u}$, and remove vertex ${b}_{1,2}^{u}$.
- (5)
- Add ${b}_{2,1}^{l}$ and ${b}_{2,2}^{l}$. At this point, the size of a vertex cover is $3{k}^{2}+k+3$.
- (6)
- Remove both vertices in ${p}_{2,1}^{l}$.
- (7)
- Repeat Steps (4), (5) and (6) until all beads in ${L}_{2}$ have been added and the sequin pairs removed. When we reach the last sequin pair in ${L}_{2}$, ${b}_{2,1}^{l}$ was already added, and hence, we gain a surplus of one, which brings the vertex cover size back to $3{k}^{2}+k$.
- (8)
- Repeat Steps (4) to (7) for every remaining necklace in L.

**Theorem**

**5.**

**Proof.**

#### 6.3. $\mathsf{FPT}$ Algorithm for Graphs of Bounded Degree

- Every vertex in $V\left(\sigma \right)$ is coloured red; and
- Every vertex in ${N}_{G}\left(V\left(\sigma \right)\right)$ is coloured blue.

**Proposition**

**10.**

**Proof.**

**Proposition**

**11.**

**Theorem**

**6.**

#### 6.4. $\mathsf{FPT}$ Algorithm for Nowhere Dense Graphs

**Theorem**

**7.**

**Proof.**

## 7. Conclusions

## Acknowledgments

## Author Contributions

## Conflicts of Interest

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Mouawad, A.E.; Nishimura, N.; Raman, V.; Siebertz, S.
Vertex Cover Reconfiguration and Beyond. *Algorithms* **2018**, *11*, 20.
https://doi.org/10.3390/a11020020

**AMA Style**

Mouawad AE, Nishimura N, Raman V, Siebertz S.
Vertex Cover Reconfiguration and Beyond. *Algorithms*. 2018; 11(2):20.
https://doi.org/10.3390/a11020020

**Chicago/Turabian Style**

Mouawad, Amer E., Naomi Nishimura, Venkatesh Raman, and Sebastian Siebertz.
2018. "Vertex Cover Reconfiguration and Beyond" *Algorithms* 11, no. 2: 20.
https://doi.org/10.3390/a11020020