Cluster deletion revisited

TL;DR

This paper presents a new fixed-parameter algorithm for the Cluster Deletion problem, improving the computational efficiency for determining minimal edge removals to form a cluster graph.

Contribution

The authors develop an algorithm with a running time of O*(1.404^k) for the Cluster Deletion problem, enhancing previous methods.

Findings

Algorithm runs in O*(1.404^k) time

Improves efficiency over previous algorithms

Advances fixed-parameter tractability for Cluster Deletion

Abstract

In the Cluster Deletion problem the input is a graph $G$ and an integer $k$, and the goal is to decide whether there is a set of at most $k$ edges whose removal from $G$ results a graph in which every connected component is a clique. In this paper we give an algorithm for Cluster Deletion whose running time is $O^*(1.404^k)$.

Full text

Keywords

graph algorithms, parameterized complexity, branching algorithms.

1 Introduction

A graph $G$ is called a cluster graph if every connected component of $G$ is a clique. In the Cluster Deletion problem the input is a graph $G$ and an integer $k$, and the goal is to decide whether there is a set of at most $k$ edges whose removal from $G$ results a clique graph.

A graph $G$ is a cluster graph if and only if there is no induced $P_{3}$ in $G$, where $P_{3}$ is a path on 3 vertices. Therefore, there is a simple $O^{*}(2^{k})$-time branching algorithm for Cluster Deletion: If the current graph is not a cluster graph, find an induced $P_{3}$ in the graph and create two new instances by removing each edge of the path. Faster algorithms for Cluster Deletion were given in [4, 3, 2, 1]. First, an $O^{*}(1.77^{k})$-time algorithm was given in [4]. An $O^{*}(1.53^{k})$-time algorithm was given in [4], and an $O^{*}(1.47^{k})$-time algorithm was given in [2]. Finally, Böcker and Damaschke [1] gave an algorithm with a claimed running time of $O^{*}(1.415^{k})$.

In this paper we show that there is an error in the analysis of the algorithm of Böcker and Damaschke. We give a corrected analysis that shows that the running time of the algorithm is $O^{*}(1.415^{k})$ as claimed in [1]. Additionally, we give an algorithm for Cluster Deletion whose running time is $O^{*}(1.404^{k})$.

2 Preliminaries

For set of vertices $S$ in a graph $G$, $G[S]$ is the subgraph of $G$ induced by $S$ (namely, $G[S]=(S,E\cap(S\times S))$). For a set of edges $F$, $G-F$ is the graph obtained from $G$ by deleting the edges of $F$. For a vertex $v$, $N(v)$ is the set of vertices that are adjacent to $v$.

Let $P_{3}$ denote a path on 3 vertices, and $C_{4}$ denote a chordless cycle on 4 vertices.

A graph $G$ is an $\alpha$-almost clique if there is a set $X\subseteq V(G)$ of size at most $\alpha$ such that $G[V(G)\setminus X]$ is a clique.

Lemma 1** (Damaschke [2]).**

Let $\alpha$ be a constant. There is a polynomial time algorithm that given an $\alpha$-almost clique $G$, finds a set of edges $S$ of minimum size such that $G-S$ is a cluster graph.

3 The algorithm of Böcker and Damaschke

In this section we describe the algorithm of Böcker and Damaschke [1] and give a corrected analysis of the algorithm. The algorithm is a branching algorithm. When we say that the algorithm branches on sets $S_{1},\ldots,S_{p}$, we mean that the algorithm is called recursively on the instances $(G-S_{1},k-|S_{1}|),\ldots,(G-S_{p},k-|S_{p}|)$.

The algorithm uses the following branching rules.

(B1)  Suppose that $P$ is an induced path in $G$ with at least 7 vertices, and let $e_{1},e_{2},\ldots,e_{p}$ be the edges of $P$ (in order). Branch on $\{e_{1},e_{3},\ldots\}$ and $\{e_{2},e_{4},\ldots\}$.

We note that Rule (B1) is a simplified version of the corresponding rule given in [1]. To show the safeness of this rule, suppose that $(G,k)$ is a yes instance and let $S$ be a solution of $(G,k)$ (namely, $G-S$ is a cluster graph and $|S|\leq k$). Since $V(e_{1})\cup V(e_{2})$ induces a $P_{3}$ in $G$, $S$ contains at least one edge from $e_{1},e_{2}$. Suppose that $e_{1}\in S$. The set $V(e_{2})\cup V(e_{3})$ induces a $P_{3}$ in $G$, so either $e_{2}\in S$ or $e_{3}\in S$. Therefore, the set $S_{2}=(S\setminus\{e_{2}\})\cup\{e_{3}\}$ is also a solution of $(G,k)$ (since the deletion of $e_{3}$ from $G$ does not generate new induced $P_{3}$’s). By repeating this process we obtain a solution $S^{\prime}$ such that $\{e_{1},e_{3},\ldots\}\subseteq S^{\prime}$. Similarly, if $e_{2}\in S$ we can obtain a solution $S^{\prime}$ such that $\{e_{2},e_{4},\ldots\}\subseteq S^{\prime}$. Therefore, Rule (B1) is safe.

The branching vector of Rule (B1) is at least $(3,3)$ and the branching number is less than 1.26.

For an edge $e$, let $F_{e}$ be a set containing all edges $e^{\prime}$ such that $V(e)\cup V(e^{\prime})$ induces a $P_{3}$.

(B2)  If there is an edge $e$ for which $|F_{e}|\geq 4$, branch on $\{e\}$ and $F_{e}$.

The branching vector of Rule (B2) is at least $(1,4)$ and the branching number is less than 1.381.

(B3)  Let $v_{1},v_{2},v_{3},v_{4},v_{1}$ be an induced $C_{4}$ in $G$. Branch on $\{(v_{1},v_{2}),(v_{3},v_{4})\}$ and $\{(v_{2},v_{3}),(v_{4},v_{1})\}$.

The branching vector of Rule (B3) is $(2,2)$ and the branching number is less than 1.415.

The algorithm applies Rules (B1)–(B3) until none of these rule is applicable. If $G$ is a graph on which Rules (B1)–(B3) are not applicable and $G$ is not a cluster graph, the algorithm applies a new branching rule, denoted (B3), which is described below.

In Rule (B3), the algorithm finds an induced $P_{3}$ $u,v,w$. To describe the rule, we define the following sets. Let $A=\{u,v,w\}$. Let $B$ the set of all vertices not in $A$ with one or two neighbors in $A$. Let $B_{u}$ (resp., $B_{w}$) be the set of all vertices $b\in B$ such that $b$ is adjacent to exactly one vertex from $u,v$ (resp., from $w,v$). Clearly, $B=B_{u}\cup B_{w}$. Let $C$ be the set of all vertices with three neighbors in $A$. Let $D$ be the set of all vertices not in $A\cup B\cup C$ that have at least one neighbor in $C$.

For $i\geq 1$, let $B_{i}$ be a set containing every vertex $x$ not in $A\cup B\cup C\cup D$ such that the minimum distance between $x$ and a vertex in $B$ is exactly $i$. Additionally, denote $B_{0}=B$ and $B_{-1}=A$. For a vertex $x\in B_{i}$, let $N_{\mathrm{next}}(x)$, $N_{\mathrm{same}}(x)$, and $N_{\mathrm{prev}}(x)$ denote the sets of neighbors of $x$ in $B_{i+1}$, $B_{i}$, and $B_{i-1}$, respectively. Note that by definition, $N_{\mathrm{prev}}(x)\neq\emptyset$ for every $x\in\bigcup_{i\geq 0}B_{i}$. For $i\geq 0$, let $E_{i}$ be the set of all edges with one endpoint in $B_{i}$ and the other endpoint in $B_{i+1}$.

The following lemmas are proved in [1].

Lemma 2**.**

If $G$ is a graph in which Rule (B3) is not applicable then $|B_{u}|\leq 2$ and $|B_{w}|\leq 2$.

Lemma 3**.**

If $G$ is a graph in which Rule (B3) is not applicable then $C$ is a clique.

Lemma 4**.**

Let $G$ be a graph in which Rule (B3) is not applicable and $C$ is clique. Then, $|D|\leq 3$. Additionally, for every $x\in D$, $C\subseteq N(x)\subseteq B\cup C\cup D$.

We now give several lemmas which will be used in the analysis of Rule (B3).

Lemma 5**.**

If $G$ is a graph in which Rule (B3) cannot be applied, $|N_{\mathrm{next}}(x)|\leq 2$ for every $x\in\bigcup_{i\geq 0}B_{i}$.

Proof.

We first consider the case $x\notin B_{0}$. Let $y\in N_{\mathrm{prev}}(x)$ and $z\in N_{\mathrm{prev}}(y)$. The set $F_{(x,y)}$ contains the edge $(y,z)$ and the edge $(x,x^{\prime})$ for every $x^{\prime}\in N_{\mathrm{next}}(x)$. Therefore, $|N_{\mathrm{next}}(x)|\leq|F_{(x,y)}|-1\leq 2$.

We now consider the case $x\in B_{0}=B$. By the definition of $B$, there are vertices $y,z\in A$ such that $y$ is adjacent to $x$ and $z$, and $z$ is not adjacent to $x$. Therefore, the set $F_{(x,y)}$ contains the edge $(y,z)$ and the edge $(x,x^{\prime})$ for every $x^{\prime}\in N_{\mathrm{next}}(x)$. We obtain again that $|N_{\mathrm{next}}(x)|\leq 2$. ∎

By Lemma 5 we have that $|B_{1}|\leq 2|B|\leq 8$ and $|E_{0}|\leq 8$.

Let $j$ be the minimum index such that there is an edge $e\in E_{j}$ for which $|F_{e}|\geq 3$ (we note that while $G$ does not have an edge $e$ with $|F_{e}|>3$, we will later use this definition on graphs which can have such edges). If no such index exists, $j=\infty$.

Lemma 6**.**

If $j\geq 1$, $|N_{\mathrm{next}}(x)|\leq 1$ for every $x\in B_{1}\cup\cdots\cup B_{j}$.

Proof.

Let $y\in N_{\mathrm{prev}}(x)$ and $z\in N_{\mathrm{prev}}(v)$. As in the proof of Lemma 5, $|N_{\mathrm{next}}(x)|\leq|F_{(x,y)}|-1$. Since the edge $(x,y)$ is in $E_{0}\cup\cdots\cup E_{j-1}$, by the definition of $j$, $|F_{(x,y)}|\leq 2$ and the lemma follows. ∎

Lemma 7**.**

If $j\geq 2$, $|N_{\mathrm{same}}(x)|\leq 1$ for every $x\in B_{2}\cup\cdots\cup B_{j}$. Additionally, if $|N_{\mathrm{same}}(x)|=1$ then $N_{\mathrm{next}}(x)=\emptyset$.

Proof.

Let $y\in N_{\mathrm{prev}}(x)$ and $z\in N_{\mathrm{prev}}(v)$. Since $y\in B_{1}\cup\cdots\cup B_{j-1}$, by Lemma 6 we have that $N_{\mathrm{next}}(y)=\{x\}$. Therefore, for every $x^{\prime}\in N_{\mathrm{same}}(x)$ we have that $(x,x^{\prime})\in F_{(x,y)}$. Since $F_{(x,y)}$ also contains every $x^{\prime}\in N_{\mathrm{next}}(x)$, we conclude that ‘$|N_{\mathrm{same}}(x)|+|N_{\mathrm{next}}(x)|\leq|F_{(x,y)}|-1\leq 1$, and the lemma follows. ∎

Lemma 8**.**

If $j\geq 3$, $|N_{\mathrm{prev}}(x)|\leq 2$ for every $x\in B_{3}\cup\cdots\cup B_{j}$. Additionally, if $|N_{\mathrm{prev}}(x)|=2$ then $N_{\mathrm{next}}(x)=N_{\mathrm{same}}(x)=\emptyset$.

Lemma 9**.**

If $j\geq 2$, if $x\in B_{2}$ and $N_{\mathrm{next}}(x)\neq\emptyset$ then $|N_{\mathrm{prev}}(x)|\leq 2$. Additionally, let $x^{\prime}$ be the unique vertex in $N_{\mathrm{next}}(x)$. Then, $N(x^{\prime})=\{x\}$.

The proofs of Lemma 8 and Lemma 9 are similar to the proof of Lemma 7 and were thus omitted.

From the lemmas above we obtain the following corollaries.

Corollary 10**.**

$|E_{j}|\leq|E_{j-1}|\leq\cdots\leq|E_{0}|$.

Corollary 11**.**

If $j\geq 2$, $G[B_{2}\cup\cdots\cup B_{j}]$ is a collection of at most $|E_{1}|$ disjoint paths.

When we consider a subgraph $G^{\prime}$ of $G$, we use $G^{\prime}$ in superscript to refer to a set (or an integer) defined for $G^{\prime}$. For example, the set of all vertices $b$ such that $b\notin A$ and $b$ is adjacent in $G^{\prime}$ to one or two vertices in $A$ is denoted $B^{G^{\prime}}$.

We now describe Rule (B3). Recall that the algorithm first finds an induced $P_{3}$ $u,v,w$. The main idea is to disconnect $u,v,w$ from the rest of the graph. If $j\neq\infty$, the algorithm picks an arbitrary edge $e\in E_{j}$ such that $|F_{e}|\geq 3$. Then, the algorithm branches on $\{e\}$ and $F_{e}$. Namely, the algorithm builds two instances $(G_{1},k_{1})=(G-\{e\},k-1)$ and $(G_{2},k_{2})=(G-F_{e},k-|F_{e}|)$. This process is repeated on each of these two instances: Let $(G_{i},k_{i})$ be one of the two instances. If $j^{G_{i}}\neq\emptyset$, the algorithm picks an arbitrary edge $e_{i}\in E_{j^{G_{i}}}^{G_{i}}$ such that $|F_{e_{i}}^{G_{i}}|\geq 3$ and creates two new instances: $(G_{i1},k_{i1})=(G_{i}-\{e_{i}\},k_{i}-1)$ and $(G_{i2},k_{i2})=(G_{i}-F_{e_{i}}^{G_{i}},k_{i}-|F_{e_{i}}^{G_{i}}|)$. This is repeated until every instance $(G_{i_{1}\cdots i_{p}},k_{i_{1}\cdots i_{p}})$ satisfies $j^{G_{i_{1}\cdots i_{p}}}=\infty$. An instance $(G_{i_{1}\cdots i_{p}},k_{i_{1}\cdots i_{p}})$ with $j^{G_{i_{1}\cdots i_{p}}}=\infty$ generated by Rule (B3) will be called a first stage instance.

Next, on each first stage instance $(G^{\prime},k^{\prime})=(G_{i_{1}\cdots i_{p}},k_{i_{1}\cdots i_{p}})$, the algorithm repeatedly applies Rule (B3) as follows. Since $j^{G^{\prime}}=\infty$, we have by Corollary 11 that $G^{\prime}[\bigcup_{i\geq 2}B_{i}^{G^{\prime}}]$ is a collection of at most $|E_{1}^{G^{\prime}}|$ disjoint paths. On each path of these paths that has at least 7 vertices, the algorithm applies Rule (B3). If there are $q$ such paths, this generates $2^{q}$ instances from $(G^{\prime},k^{\prime})$. These instances will be called second stage instances.

Let $(G^{\prime\prime},k^{\prime\prime})$ be a second stage instance and let $(G^{\prime},k^{\prime})$ be the first stage instance from which $(G^{\prime\prime},k^{\prime\prime})$ was generated. Note that $G^{\prime\prime}[\bigcup_{i\geq 2}B_{i}^{G^{\prime\prime}}]$ is a collection of at most $|E_{1}^{G^{\prime\prime}}|=|E_{1}^{G^{\prime}}|$ disjoint paths, where each path contains at most 6 vertices. Let $H$ be the connected component of $v$ in $G^{\prime\prime}$. We will later show that $G^{\prime\prime}$ is an $O(1)$-almost clique. The algorithm uses the algorithm of Lemma 1 to find a set $S^{G^{\prime\prime}}\subseteq E(H)$ of minimum size such that $H-S^{G^{\prime\prime}}$ is a cluster graph. Then, the algorithm makes a recursive call on $(G^{\prime\prime}-S^{G^{\prime\prime}},k^{\prime\prime}-|S^{G^{\prime\prime}}|)$. Note that the size of $S$ is at least 1 since $u,v,w$ is an induced $P_{3}$ in $H$. This is repeated for every second stage instance. The instances $(G^{\prime\prime}-S^{G^{\prime\prime}},k^{\prime\prime}-|S^{G^{\prime\prime}}|)$ will be called third stage instances.

We now show that $H$ is an $O(1)$-almost clique. By Lemma 4, $V(H)=A\cup C\cup D\cup\bigcup_{i\geq 0}B_{i}^{G^{\prime\prime}}$. We have that $|A|=3$ and $|D|\leq 3$ (Lemma 4). Additionally, $B_{0}^{G^{\prime\prime}}\cup B_{1}^{G^{\prime\prime}}\subseteq B_{0}\cup B_{1}$ and therefore $|B_{0}^{G^{\prime\prime}}|+|B_{1}^{G^{\prime\prime}}|\leq|B_{0}|+|B_{1}|\leq 12$. Moreover, $|\bigcup_{i\geq 2}B_{i}^{G^{\prime\prime}}|\leq 6|E_{1}^{G^{\prime\prime}}|=6|E_{1}^{G^{\prime}}|$ and $|E_{1}^{G^{\prime}}|=O(1)$ (since every edge in $E_{1}^{G^{\prime}}$ is an edge in $G[B\cup B_{1}]$). Therefore, by Lemma 3, $H$ is an $O(1)$-almost clique.

We note that in the analysis of Rule (B3) in [1], it is claimed that if $(G_{i_{1}\cdots i_{p}},k_{i_{1}\cdots i_{p}})$ is a first stage instance then $p\leq 4$. However, this is not true. Suppose that $B=\{b_{1},b_{2},b_{3},b_{4}\}$, $B_{1}=\{c_{1},c_{2},c_{3},c_{4}\}$, $B_{2}=\{d_{1},\ldots,d_{6}\}$, $N(b_{1})=\{u,b_{2},c_{1},c_{2}\}$, $N(b_{2})=\{u,b_{1},c_{1},c_{2}\}$, $N(b_{3})=\{w,b_{4},c_{3},c_{4}\}$, $N(b_{4})=\{w,b_{3},c_{4},c_{4}\}$, $N(c_{1})=\{b_{1},b_{2},c_{2},d_{1},d_{2}\}$, $N(c_{2})=\{b_{1},b_{2},c_{1},d_{3}\}$, $N(c_{3})=\{b_{3},b_{4},c_{4},d_{4},d_{5}\}$, and $N(c_{4})=\{b_{3},b_{4},c_{3},d_{6}\}$. In this graph, Rule (B3) generates an instance $(G_{11111111},k_{11111111})$ by deleting one by one all 8 edges between $B$ and $B_{1}$.

We now bound the branching number of Rule (B3). We consider several cases. In the first case, suppose that $j\geq 2$ and $j\neq\infty$. Recall that the algorithm picks an edge $e=(x,y)\in E_{j}$ and generates the instances $(G_{1},k_{1})=(G-\{e\},k-1)$ and $(G_{2},k_{2})=(G-F_{e},k-|F_{e}|)$. Clearly, $|E_{j}^{G_{1}}|=|E_{j}|-1$. We now show that $|E_{j}^{G_{2}}|\leq|E_{j}|-1$. Since $G_{2}$ is obtained from $G$ by erasing the edges in $F_{e}$, and all the edges in $E_{j-1}$ that are incident on $x$ are in $F_{e}$, we have that $x\notin B_{j}^{G_{2}}$. Therefore, $(x,y)\notin E_{j}^{G_{2}}$. We now claim that $E_{j}^{G_{2}}$ does not contain an edge that is not in $E_{j}$. Suppose conversely that $e^{\prime}$ is such an edge. Then $x$ must be in $B_{j+1}^{G_{2}}$ and $e^{\prime}=(x^{\prime},x)$ for some $x^{\prime}\in B_{j}^{G_{2}}$. Therefore, in $G$ we have $x^{\prime}\in N_{\mathrm{same}}(x)$. This contradicts Lemma 7. Thus, $E_{j}^{G_{2}}$ does not contain edges that are not in $E_{j}$. It follows that $|E_{j}^{G_{2}}|\leq|E_{j}|-1$. By Corollary 10, $|E_{j^{G_{i}}}^{G_{i}}|\leq|E_{j}^{G_{i}}|\leq|E_{j}|-1$ for $i=1,2$. Using the same arguments, for every first stage instance $(G^{\prime},k^{\prime})=(G_{i_{1}\cdots i_{p}},k_{i_{1}\cdots i_{p}})$ we have that $|E_{j^{G^{\prime}}}^{G^{\prime}}|\leq|E_{j}|-p$. This implies that $p\leq|E_{j}|$. Therefore, the number of first stage instances generated by Rule (B3) is at most $2^{|E_{j}|}$.

Suppose for example that $|E_{j}|=2$. In the worst case, Rule (B3) generates four first stage instances: $(G_{11},k_{11})$, $(G_{12},k_{12})$, $(G_{21},k_{21})$, and $(G_{22},k_{22})$. Additionally, $k_{11}=k-1$, $k_{12}=k_{21}=k-4$, and $k_{22}=k-6$. Suppose that Rule (B3) is not applied on any of the four instances above. Then, the algorithm generates a third stage instance $(G_{i_{1}i_{2}}-S^{G_{i_{1}i_{2}}},k_{i_{1}i_{2}}-|S^{G_{i_{1}i_{2}}}|)$ from each first stage instance $(G_{i_{1}i_{2}},k_{i_{1}i_{2}})$. Since $|S^{G_{i_{1}i_{2}}}|\geq 1$ for every $i_{1},i_{2}$, we obtain that $k_{11}-S^{G_{11}}\leq k-3$, $k_{12}-S^{G_{12}}\leq k-5$, $k_{21}-S^{G_{21}}\leq k-5$, $k_{22}-S^{G_{22}}\leq k-7$. In other words, the branching vector of Rule (B3) in this case is at least $(3,5,5,7)$.

To analyze the case in which Rule (B3) is applied (at least once) on at least one of the four first stage instances note that for the sake of the analysis, we can assume for the sake of the analysis that the applications of Rule (B3) are done after the application of Rule (B3). For example, suppose that Rule (B3) is applied only on the instance $(G_{11},k_{11})$ and it is applied once on this instance, generating instances $(G^{\prime}_{11},k^{\prime}_{11})$ and $(G^{\prime\prime}_{11},k^{\prime\prime}_{11})$, where $k^{\prime}_{11}=k^{\prime\prime}_{11}=k_{11}-3$. Therefore, the branching vector of Rule (B3) in this case is at least $(6,6,5,5,7)$. However, we can assume for the analysis that Rule (B3) generates only four third stage instances, namely the instances $(G_{i_{1}i_{2}}-S^{G_{i_{1}i_{2}}},k_{i_{1}i_{2}}-|S^{G_{i_{1}i_{2}}}|)$, and then the algorithm applies Rule (B3) on $(G_{11}-S^{G_{11}},k_{11}-|S^{G_{11}}|)$. The branching vectors for these two rules are $(3,5,5,7)$ and $(3,3)$, respectively.

For a general value of $|E_{j}|$, we have that in the worst case Rule (B3) generates $2^{|E_{j}|}$ third stage instances (as discussed above, we can assume that Rule (B3) was not applied on the first stage instances). The branching vector is at least $R(|E_{j}|)$, where $R(p)$ is a vector of length $2^{p}$ in which the value $p+1+2i$ appears $\binom{p}{i}$ times, for $i=0,\ldots,p$. We note that this bound is not good enough for our purpose. Even for $|E_{j}|=5$, we have that the branching vector $R(5)$ has branching number of approximately 1.406 and we need a branching number less than 1.404. The solution is to give better bounds on the sizes of the sets $S_{G^{\prime\prime}}$. This will be discussed below.

Now consider the case $j=1$. In this case $|E_{1}^{G_{1}}|=|E_{1}|-1$ as in the first case. However, we now can have $|E_{1}^{G_{2}}|\geq|E_{1}|$. This occurs if $N_{\mathrm{same}}(x)\cap N_{\mathrm{prev}}(y)\neq\emptyset$. In this case, $x$ belongs to $B^{G_{2}}_{2}$, and for every $x^{\prime}\in N_{\mathrm{same}}(x)\cap N_{\mathrm{prev}}(y)$, the edge $(x,x^{\prime})$ is in $E_{1}^{G_{2}}$ and not in $E_{j}$. Note that in this case we have that $x$ and $y$ are not adjacent in $G_{2}$ to vertices in $B_{2}^{G_{2}}\cup B_{3}^{G_{2}}$ (If $z$ is adjacent in $G_{2}$ to $x$ or $y$ then $z$ must be adjacent to both $x$ and $y$, otherwise the edge between $z$ and $x$ or $y$ is in $F_{(x,y)}$. $z$ is also adjacent to $x$ and $y$ in $G$. Therefore, $z$ is in $B_{1}$ or $B_{2}$. By Lemma 6, $z\notin B_{2}$, so $z\in B_{1}$). Therefore, we can ignore the edges $(x^{\prime},x)$ and $(x^{\prime},y)$. Formally, if the case above occurs, we mark the edges $(x^{\prime},x)$ and $(x^{\prime},y)$ for every $x^{\prime}\in N_{\mathrm{same}}(x)\cap N_{\mathrm{prev}}(y)$. We now change the definition of $E_{1}$ to include all the unmarked edges with one endpoint in $B_{1}$ and the other endpoint in $B_{2}$. For this new definition we have $|E_{1}^{G_{2}}|\leq|E_{1}|-1$. Additionally, Corollary 10 remains true. Therefore, the analysis of the case $j=1$ is the same as the analysis for the case $j\geq 2$. That is, for a specific value of $|E_{1}|$, the worst branching vector is at least $R(|E_{1}|)$.

We now consider the case $j=0$. To handle this case (and to get a better bound on the branching number for the case $j\geq 1$), we use a Python script. The script goes over possible cases for the graph $G[A\cup B\cup B_{1}\cup B_{2}]$ and for each case it computes a branching vector that gives an upper bound on the branching number for this case. Formally, a configuration graph is a graph $J$ whose vertices are partitioned into 4 sets: (1) A set $A^{J}$ that contains 3 vertices $u^{\prime},v^{\prime},w^{\prime}$ which form a $P_{3}$. (2) A set $B^{J}$ that contains vertices that are adjacent to 1 or 2 vertices of $A^{J}$. (3) A set $B_{1}^{J}$ that contains vertices that are adjacent to at least one vertex in $B^{J}$ and are not adjacent to vertices in $A^{J}$. (4) A set $B_{2}^{J}$ such that every vertex in $B_{2}^{J}$ has exactly one neighbor and this neighbor is in $B_{1}^{J}$. Additionally, $|F_{e}^{J}|\leq 3$ for every edge $e$ with at least one endpoint in $A^{J}\cup B^{J}$. We note that the restriction on the degree of the vertices in $B_{2}^{J}$ is required in order to restrict the number of configuration graphs.

Let $G$ be a graph with an induced $P_{3}$ $u,v,w$. We say that $G$ matches a configuration graph $J$ if there is a bijection $\phi\colon A\cup B\cup B_{1}\to A^{J}\cup B^{J}\cup B_{1}^{J}$ such that (1) $\phi$ is an isomorphism between $G[A\cup B\cup B_{1}]$ and $J[A^{J}\cup B^{J}\cup B_{1}^{J}]$. (2) $\phi$ maps $u,v,w$ to $u^{\prime},v^{\prime},w^{\prime}$, respectively. (3) For a vertex $c\in B_{1}$, the number of neighbors of $c$ in $B_{2}$ is equal to the number of neighbors of $\phi(c)$ in $B_{2}^{J}$.

The script goes over all possible configuration graphs. For each configuration graph $J$, the script builds a vector whose branching number is an upper bound on the branching number of Rule (B3) when it is applied on a graph $G$ which matches $J$.

For each configuration graph $J$ the script generates a branching vector $R$ as follows. Suppose that $G$ is a graph that matches $J$. The script generates graphs of the form $J_{i_{1}\cdots i_{p}}$ like the generation of first stage instances in Rule (B3), except that now the process ends when $j^{J_{i_{1}\cdots i_{p}}}>0$. A graph $J_{i_{1}\cdots i_{p}}$ with $j^{J_{i_{1}\cdots i_{p}}}>0$ generated by the script will be called a first stage configuration graph. Consider some first stage configuration graph $J^{\prime}=J_{i_{1}\cdots i_{p}}$. This graph corresponds to the instance $(G^{\prime},k^{\prime})=(G_{i_{1}\cdots i_{p}},k_{i_{1}\cdots i_{p}})$ that Rule (B3) generates when it is applied on the graph $G$ and the induced path $u,v,w$. Note that if $j^{J^{\prime}}\neq\infty$ then $(G^{\prime},k^{\prime})$ is not a first stage instance and Rule (B3) will continue generating instances from $(G^{\prime},k^{\prime})$. By the analysis of the case $j\geq 1$ above, at the worst case, the number of first stage instances generated from $(G^{\prime},k^{\prime})$ is $2^{|E_{1}^{G^{\prime}}|}=2^{|E_{1}^{J^{\prime}}|}$. The vector $R(|E_{1}^{J^{\prime}}|)$ gives a lower bound on the differences between the parameter $k_{i_{1}\cdots i_{p}}$ and the parameters of these $2^{|E_{1}^{G^{\prime}}|}$ instances. Therefore, the vector $R$ can be the concatenation of $R(|E_{1}^{J_{i_{1}\cdots i_{p}}}|)$ for every first stage graph $J_{i_{1}\cdots i_{p}}$. We can get a better branching vector by giving a better bound on $|S^{G^{\prime\prime}}|$ for the graphs $G^{\prime\prime}$ of the second stage instances $(G^{\prime\prime},k^{\prime\prime})$ that are generated by Rule (B3) from an instance $(G^{\prime},k^{\prime})=(G_{i_{1}\cdots i_{p}},k_{i_{1}\cdots i_{p}})$, where $G^{\prime}$ matches the first stage configuration graph $J^{\prime}=J_{i_{1}\cdots i_{p}}$. For our purpose, we it was suffices to give a simple bound based on the edges between $A^{J^{\prime}}$ and $B^{J^{\prime}}$. For example, suppose that there are vertices $b^{\prime}_{1},b^{\prime}_{2}\in B^{J^{\prime}}$ such that $N(b^{\prime}_{1})\cap A^{J^{\prime}}=\{u^{\prime}\}$ and $N(b^{\prime}_{2})\cap A^{J^{\prime}}=\{w^{\prime}\}$. Then, in $G$ there are vertices $b_{1},b_{2}\in B$ such that $N(b_{1})\cap A=\{u\}$ and $N(b_{2})\cap A=\{w\}$. The edges $(b_{1},u)$ and $(b_{2},w)$ remain in every graph $G^{\prime\prime}$ of a second stage instance $(G^{\prime\prime},k^{\prime\prime})$ generated from $(G^{\prime},k^{\prime})$. Therefore, each such graph contains two edge disjoint induced $P_{3}$s ($b_{1},u,v$ and $b_{2},w,v$). Therefore, $|S^{G^{\prime\prime}}|\geq 2$. We use similar lower bounds in case of other edges between $A^{J^{\prime}}$ and $B^{J^{\prime}}$.

All the branching vectors generated by the script have branching numbers less than 1.393. Therefore, the branching number of Rule (B3) is less then 1.393.

Since all the branching rules of the algorithm have branching numbers less than 1.415, it follows that the running time of the algorithm is $O^{*}(1.415^{k})$.

4 New algorithm

Our algorithm first applies Rule (B3) and Rule (B3) until these rule cannot be applied (note that Rule (B3) is not applied). Let $G$ be a graph in which these rules cannot be applied. If $G$ does not contain induced $C_{4}$ then the algorithm applies Rule (B3). Otherwise, the algorithm applies a new branching rule, denoted (B3), which is as follows. The algorithm picks an induced $C_{4}$ $u,v,w,u^{\prime},u$. Note that $u,v,w$ is an induced $P_{3}$. Let $B_{u}$ and $C$ be the sets defined in the previous section. We have that $|B_{u}|\geq 1$ since $u^{\prime}\in B_{u}$. If $C$ is a clique then the algorithm applies Rule (B3) on $u,v,w$. Otherwise, let $x,y\in C$ be two non-adjacent vertices. Note that $u,x,w,y,u$ is an induced $C_{4}$. The algorithm applies Rule (B3) on the cycle $u,x,w,y,u$. This generates two instances: $(G_{1},k_{1})=(G-\{(u,x),(w,y)\},k-2)$ and $(G_{2},k_{2})=(G-\{(x,w),(y,u)\},k-2)$.

Consider an instance $(G_{i},k_{i})$ and note that $u,v,w$ is an induced $P_{3}$ in $G_{i}$. If Rule (B3) is applicable on $(G_{i},k_{i})$, the algorithm applies Rule (B3). Now suppose that Rule (B3) is not applicable. In $G_{i}$, the vertex $x$ is adjacent to $v$ and not adjacent to $u$. Therefore, $x\in B_{u}^{G_{i}}$. We also have $u^{\prime}\in B_{u}^{G_{i}}$, so by Lemma 2 we have that $B_{u}^{G_{i}}=\{u^{\prime},x\}$. Additionally, $x,y\notin C^{G_{i}}$. If $C^{G_{i}}$ is a clique then the algorithm applies Rule (B3) on the graph $G_{i}$ and the path $u,v,w$. Now suppose that $C^{G_{i}}$ is not a clique. let $x_{i},y_{i}\in C^{G_{i}}$ be two non-adjacent vertices. We have that $u,x_{i},w,y_{i},u$ is an induced $C_{4}$ in $G_{i}$. The algorithm applies Rule (B3) on the cycle $u,x_{i},w,y_{i},u$. This generates two instances: $(G_{i1},k_{i1})=(G_{i}-\{(u,x_{i}),(w,y_{i})\},k_{i}-2)$ and $(G_{i2},k_{i2})=(G_{i}-\{(x_{i},w),(y_{i},u)\},k_{i}-2)$. Consider the instance $G_{i1}$. Again, we have that $u,v,w$ is an induced $P_{3}$ in $G_{i1}$. We now have that $B_{u}^{G_{i1}}=\{u^{\prime},x,x_{i}\}$. By Lemma 2, Rule (B3) is applicable on $(G_{i1},k_{i1})$. Using the same arguments, Rule (B3) is applicable on $(G_{i2},k_{i2})$. Thus, the algorithm applies Rule (B3) on $(G_{i1},k_{i1})$ and on $(G_{i2},k_{i2})$.

We now analyze the branching number of Rule (B3). There are three cases that we need to consider. In the first case, the algorithm generates four instances $G_{11}$, $G_{12}$, $G_{21}$, and $G_{22}$, and then applies Rule (B3) on each of these instances. Therefore, the branching vector in this case is $(5,8,5,8,5,8,5,8)$ and the branching number is less than 1.404. In the second case, the algorithm generates, without loss of generality, the instances $G_{11}$, $G_{12}$, and $G_{2}$. The algorithm then applies Rule (B3) on $G_{11}$ and $G_{12}$, and applies Rule (B3) or Rule (B3) on $G_{2}$. The worst case is when Rule (B3) is applied on $G_{2}$. The branching vector in this case is $(5,8,5,8,3,6)$, and the branching number is less than 1.402. In the third case, the algorithm generates the instances $G_{1}$ and $G_{2}$ and applies Rule (B3) or Rule (B3) on each of these instances. The worst branching vector in this case is $(3,6,3,6)$ and the branching number is less than 1.398. Therefore, the branching number of Rule (B3) is less than 1.404.

Since all the branching rules of the algorithm have branching numbers less than 1.404, it follows that the running time of the algorithm is $O^{*}(1.404^{k})$.