Towards a Complexity Dichotomy for Colourful Components Problems on k-caterpillars and Small-Degree Planar Graphs

Abstract

A connected component of a vertex-coloured graph is said to be colourful if all its vertices have different colours, and a graph is colourful if all its connected components are colourful. Given a vertex-coloured graph, the Colourful Components problem asks whether there exist at most p edges whose removal makes the graph colourful, and the Colourful Partition problem asks whether there exists a partition of the vertex set with at most p parts such that each part induces a colourful component. We study the problems on k-caterpillars (caterpillars with hairs of length at most k) and explore the boundary between polynomial and NP-complete cases. It is known that the problems are NP-complete on 2-caterpillars with unbounded maximum degree. We prove that both problems remain NP-complete on binary 4-caterpillars and on ternary 3-caterpillars. This answers an open question regarding the complexity of the problems on trees with maximum degree at most 5. On the positive side, we give a linear time algorithm for 1-caterpillars with unbounded degree, even if the backbone is a cycle, which outperforms the previous best complexity for paths and widens the class of graphs. Finally, we answer an open question regarding the complexity of Colourful Components on graphs with maximum degree at most 5. We show that the problem is NP-complete on 5-coloured planar graphs with maximum degree 4, and on 12-coloured planar graphs with maximum degree 3. Since the problem can be solved in polynomial-time on graphs with maximum degree 2, the results are the best possible with regard to the maximum degree.

1 Introduction

A coloured graph is a graph whose vertices are (not necessarily properly) coloured. A connected component of a coloured graph is a colourful component if all its vertices have different colours. A graph is said to be colourful if all its connected components are colourful.

In this paper we focus on two closely related problems where a coloured graph and a positive integer p are given as inputs: the Colourful Components problem asks if there exist at most p edges whose removal makes the graph colourful; the Colourful Partition problem is to decide if there exists a partition of the vertex set with at most p parts such that each part induces a colourful component in the graph.

One key problem in comparative genomics is to partition a set of genes into orthologous genes, which are sets of genes in different species that have evolved through speciation events only, i.e. originated by vertical descent from a single gene in the last common ancestor. The problem has been modelled as a graph problem where orthologous genes translate into colourful components in the graph [1, 15]. The vertices of the graph represent the genes, and a colour is given to each vertex to symbolise the specie the corresponding gene belongs to. An edge between two vertices is present in the graph if the two corresponding genes are (sufficiently) similar. The quality of a partition of a set of genes into orthologous genes can be expressed in different ways. Minimising the number of similar genes in different subsets of the partition is a well studied variant [4, 5, 8, 13, 15], and it corresponds to minimising the number of edges between the colourful components (as in Colourful Components). Alternatively, one can ask for a partition of minimum size, i.e which contains the minimum number of orthologous genes, or equivalently the minimum number of colourful components [1, 5, 6] (as in Colourful Partition). Another variant, not studied in this paper, considers the objective function that maximises the number of edges in the transitive closure [1, 6, 13].

Now, we give the formal definitions of the problems considered herein.

Colourful Components  

Input: :

A vertex-coloured graph $G=(V,E)$, a positive integer p.

Question: :

Are there at most p edges in E whose removal makes G colourful?

Colourful Partition 

Input: :

A vertex-coloured graph $G=(V,E)$, a positive integer p.

Question: :

Is there a partition of V with at most p parts s.t. each part induces a colourful component in G?

It is interesting to notice the similarities between Colourful Components and the Multicut [3, 12] and Multi-Multiway Cut [2] problems. In the Multicut problem, a graph and a set of pairs of vertices are given and the goal is to minimise the number of edges to remove in order to disconnect each pair of vertices. In the Multi-Multiway Cut problem, a graph and sets of vertices are given and the goal is to minimise the number of edges to remove in order to disconnect all paths between vertices from the same vertex set. Thus, Colourful Components is a special case where the sets of vertices form a partition.

Both Colourful Components and Colourful Partition problems can be compared to the Graph Motif problem [7]. This problem takes a coloured graph and a multiset of colours M (the motif) as input, and the goal is to determine whether there exists a connected subgraph S such that the multiset of colours used by the vertices in S corresponds exactly to M. If M is a set (where each colour appears at most once), then M is said to be colourful.

In this paper, all graphs are simple. We assume that a coloured graph $G=(V,E)$ is always associated with a colouring function c from V to a set of colours, hence for each vertex $u\in V$, c(u) is the colour of the vertex u. The colour multiplicity of G corresponds to the maximum number of occurrence of any colour in the graph. If G contains at most $\ell$ colours we say that G is an $\ell$-coloured graph. To simplify the notations, we may say that a vertex u belongs to a path P in G if there exists an edge $e \in P$ with $u \in e$. A path P in G between two vertices u and v is called a bad path if $c(u) = c(v) = \gamma$ and u, v are the only two vertices of colour $\gamma$ in P. Hence, a connected component is colourful if and only if it does not contain a bad path. Lastly, given a set of edges $S \subseteq E$, we denote by $G-S$ the graph $(V,E \setminus S)$, and for a vertex $u \in V$, N[u] is the closed neighbourhood of u.

A k-caterpillar, also commonly called a caterpillar with hairs of length at most k [10], is a tree in which all the vertices are within distance k of a central path, called the backbone. Similarly, we define a cyclic k-caterpillar as a k-caterpillar whose backbone is a chordless cycle. Note that 2-caterpillars are also known as lobster graphs.

Observe that, on a tree, there is a solution to Colourful Components with p edges if and only if there is a solution to Colourful Partition with $p+1$ parts. However, this is not the case on general graphs [5]. Both problems are known to be NP-complete on subdivided stars [6], trees of diameter at most 4 [4], and trees with maximum degree 6 [5]. Trees of diameter at most 4 are in fact a subclass of 2-caterpillars, so both problems are NP-complete on 2-caterpillars when the maximum degree is unbounded. In Sect. 2.1, we prove that Colourful Components and Colourful Partition are NP-complete on binary 4-caterpillars and on ternary 3-caterpillar, hence with the maximum degree at most 3 or 4. This answers an open question, proposed in [5], regarding the complexity of the problems on trees with maximum degree at most 5. Nonetheless, we propose a linear time algorithm for Colourful Components and Colourful Partition on 1-caterpillars and cyclic 1-caterpillars with unbounded degree in Sect. 2.2. This result improves the complexity of the known quadratic-time algorithm for paths [6] and widens the class of graphs. We, therefore, obtain a complete complexity dichotomy for the problems on k-caterpillars with regard to k and the maximum degree in the graph.

We also consider the complexity of Colourful Components in planar graphs with small degree. It is known that the problem is NP-complete on 3-coloured graphs with maximum degree 6 [4], while Colourful Partition is NP-complete on 3-coloured 2-connected planar graphs with maximum degree 3 [5]. However, it was an open question whether Colourful Components is NP-complete on $\ell$-coloured graphs with maximum degree at most 5. In Sect. 3, we answer that question and show that Colourful Components is NP-complete on 5-coloured planar graphs with maximum degree 4 and on 12-coloured planar graphs with maximum degree 3. As Colourful Components is polynomial-time solvable on graphs with maximum degree 2, our result is the best possible with regard to the maximum degree.

2 Complexity on k-caterpillars

In this section, we focus on the complexity of Colourful Components and Colourful Partition on k-caterpillars, depending on the value of k and the maximum degree of the graphs.

2.1 NP-completeness

First, we show that Colourful Components and Colourful Partition are NP-complete on binary 4-caterpillars and ternary 3-caterpillars. We recall that a binary tree (resp. ternary) is a rooted tree in which each vertex has no more than two children (resp. three children). We propose a reduction from 3-SAT with at most four occurrence of each variable, known as 3, 4-SAT, which is proved NP-complete in [14].

Construction 1

Consider an instance $\phi$ of 3, 4-SAT, that is, a set of m clauses $C_1, C_2, \dots , C_m$ on n variables $x_1, x_2, \dots , x_n$, where each clause contains exactly three literals and where each variable appears at most four times.

For each variable $x_i$, we define a variable gadget. Firstly, we create two vertices labelled $x_i$ and $\bar{x}_i$, which are the roots of two binary trees $T_{x_i}$ and $T_{\bar{x}_i}$, respectively. If a clause $C_j$ contains the literal $x_i$, then we create a vertex labelled $x_{i,j}$ in $T_{x_i}$. Similarly, if a clause $C_j$ contains the literal $\bar{x}_i$, then there we create a vertex labelled $\bar{x}_{i,j}$ in $T_{\bar{x}_i}$. All created vertices are connected in such a way that $T_{x_i}$ and $T_{\bar{x}_i}$ are binary trees of depth at most 2. We assume that all the vertices in the trees, except for $x_i$ and $\bar{x}_i$, correspond to one literal in a clause. Finally, we connect $x_i$ and $\bar{x}_i$ to a same new vertex $r_{x_i}$. Notice that the gadget is a binary tree of depth at most 3 (see Fig. 1).

For each clause $C_j$, we define a clause gadget. Let $\ell _{1}$, $\ell _{2}$ and $\ell _{3}$ be the literals in $C_j$. We create four vertices $y_j$, $y'_j$, $z_j$ and $z'_j$, three vertices labelled $\ell _{1,j}$, $\ell _{2,j}$ and $\ell _{3,j}$ representing the literals in $C_j$, and one extra vertex $r_{C_j}$. Then, we add the edges $\{\ell _{1,j}, y_j\}$, $\{\ell _{2,j}, y'_j\}$, $\{\ell _{3,j}, z'_j\}$, $\{y_j, z_j\}$, $\{y'_j, z_j\}$, and the edges $\{z_j, r_{C_j}\}$ and $\{z'_j, r_{C_j}\}$. Notice that the gadget is a binary tree of depth 3 (see Fig. 1).

Now, we describe two slightly different ways to obtain a tree T containing all the variable and clause gadgets.

• To get T as a binary 4-caterpillar, create a central path with $n+m$ new vertices and connect each of the $n+m$ vertices $r_{x_i}$ and $r_{C_j}$ to a different vertex of the central path.

• To get T as a ternary 3-caterpillar, connect the $n+m$ vertices $r_{x_i}$ and $r_{C_j}$ together (in any order) to create a central path.

In both cases, the central path corresponds to the backbone of T. We set the root r of T such that it belongs to the backbone of T and has minimum degree, hence two or three children, if T is a binary or ternary caterpillar, respectively.

Finally, we assign a colour to each vertex in T. For each gadget representing a variable $x_i$, let $c(x_i) = c(\bar{x}_i)$ be a new colour. Also, for each vertex $\widetilde{x}_{i,j} \in \{ x_{i,j}, \bar{x}_{i,j} \}$, let $c(\widetilde{x}_{i,j})$ be a new colour. Then, for each gadget representing a clause $C_j$, set the colour of each leaf $\ell _{k,j}$ such that if $\ell _{k,j} = x_i$, then $c(\ell _{k,j}) := c(x_{i,j})$, but if $\ell _{k,j} = \bar{x}_i$, then $c(\ell _{k,j}) := c(\bar{x}_{i,j})$. Furthermore, let $c(y_j) = c(y'_j)$ and $c(z_j) = c(z'_j)$ be two new colours. Lastly, all the vertices in T which do not belong to any gadget are given new colours. Notice that there are no such vertices if T is a 3-caterpillar. Obviously, in both cases, T is a coloured caterpillar with colour-multiplicity 2.

Note that Construction 1 can be done in polynomial time.

Fig. 1.

Examples of gadgets used in Construction 1. On the left, the variable gadget of $x_1$, appearing as a positive literal in $C_{2}$, $C_{4}$ and $C_{5}$, and as a negative literal in $C_{3}$. On the right, the clause gadget of $C_2$.

Theorem 1

Colourful Components and Colourful Partition are NP-complete on coloured ternary 3-caterpillars with colour-multiplicity 2 and on coloured binary 4-caterpillars with colour-multiplicity 2.

Proof

Obviously, the problem is in NP. Let $\phi$ be an instance of 3, 4-SAT with m clauses and n variables. We transform $\phi$ into a coloured tree T as described in Construction 1 such that T has colour multiplicity 2 and is a coloured binary 4-caterpillar or a coloured ternary 3-caterpillar. We claim that there is a solution to 3, 4-SAT on $\phi$ if an only if there is a set of exactly $n+2m$ edges in T whose removal makes T colourful.

Let $\beta$ be a solution to 3, 4-SAT on $\phi$. We define the set of edges S as follows:

• For each variable $x_i$, the set S contains the edge $\{r_{x_i},x_i\}$ if $x_i = True$ in $\beta$, or $\{r_{x_i},\bar{x}_i\}$ if $x_i = False$ in $\beta$.

• For each clause $C_j$, the set S contains two edges: one from the path between $y_j$ and $y'_j$, and one from the path between $z_j$ and $z'_j$. Moreover, in $G-S$, the leaf $\ell _{k,j}$ which belongs to the same connected component as the vertex in $r_{C_j}$ must correspond to (one of) the literal(s) satisfying the clause $C_j$ in $\beta$.

Clearly, the set S contains $n+2m$ edges. We denote by $\mathcal {F}$ the forest $T-S$, and by $T'$ the connected component in $\mathcal {F}$ containing the root r. Obviously, two vertices of the same colour from a same variable gadget do not both belong to a same connected component of $\mathcal {F}$, and the same is true for a clause gadget. Also, note that two vertices of different variable gadgets do not have the same colour, and similarly for vertices of different clause gadgets. Lastly, observe that two vertices of two different gadgets belong to the same connected component if and only if they are connected through the backbone, which is in $T'$. Thus, if there exist two vertices of the same colour in a same connected component of $\mathcal {F}$, one must be from a variable gadget and the other one from a clause gadget, and they both necessarily belong to $T'$. Without loss of generality, consider $x_{i,j}$ from the variable gadget of $x_i$ and $\ell _{k,j}$ from the clause gadget of $C_j$, such that $x_{i,j}, \ell _{k,j} \in T'$. To prove a contradiction, assume that $c(x_{i,j}) = c(\ell _{k,j})$. Note that the literal represented by $\ell _{k,j}$ is $x_{i,j}$, otherwise the two vertices would not have the same colour. Since $\ell _{k,j}$ is in $T'$, it is connected to the vertex $r_{C_j}$ of the clause gadget, hence $\ell _{k,j}$ satisfies the clause $C_j$. Therefore, $x_{i}$ satisfies the clause $C_j$, and the variable $x_i = True$ in $\beta$. By construction, this implies that the edge $\{r_{x_i},x_i\}$ belongs to S, and therefore that the subtree $T_{x_i}$, containing $x_{i,j}$, is not part of $T'$, which is a contradiction.

Let S be a solution to Colourful Components on T such that $|S| = n+2m$. Observe that one needs to remove at least one edge per variable gadget to put the vertices $x_i$ and $\bar{x}_i$ into different connected components, and at least two edges per clause gadget, the first one to disconnect $y_j$ and $y'_j$ and the second one to disconnect $z_j$ and $z'_j$. Since $|S| = n+2m$, S must only contain n edges from variable gadgets and 2m edges from clause gadgets. We denote by $T'$ the connected component of $T- S$ containing the root r. Notice that, for each variable gadget, either $x_i$ or $\bar{x}_i$ belongs to $T'$, but not both. Also, for each clause gadget, exactly one leaf $\ell _{k,j}$ belongs to $T'$. We construct the solution $\beta$ to 3, 4-SAT on $\phi$ such that, for each variable gadget, if $\{r_{x_i},x_i\} \in S$, then we set $x_i := True$ in $\beta$, and if $\{r_{x_i},\bar{x}_i\} \in S$, then we set $x_i := False$ in $\beta$. To prove a contradiction, assume that there is a clause $C_j$ which is not satisfied in $\phi$ with regard to $\beta$. Consider the leaf $\ell _{k,j} \in T'$ from the gadget clause of $C_j$, and assume without loss of generality that $\ell _{k,j} = x_{i}$. If $C_j$ is not satisfied, then the variable $x_{i} := False$ in $\beta$. This means that S contains the edge $\{r_{x_i},\bar{x}_{i}\}$, but not the edge $\{r_{x_i},x_{i}\}$, and thus $x_{i,j} \in T'$. However, since $c(x_{i,j}) = c(\ell _{k,j})$, then S is not a solution for T, a contradiction.   $\square$

2.2 Polynomiality

Now, we show that Colourful Components and Colourful Partition can be solved in linear time on 1-caterpillars with unbounded maximum degree, even if the backbone is a chordless cycle. To simplify the notations, we use the term general caterpillars to denote both 1-caterpillars and cyclic 1-caterpillars.

We consider the vertices in the backbone as internal vertices of stars, hence vertices of degree 1 are the leaves of a star whose internal vertex belongs to the backbone. We assume that the edges and the vertices in the backbone are either linearly of cyclically ordered, if the backbone is a path or a cycle, respectively.

Remark 1

Consider a coloured general caterpillar. If two vertices of a star have the same colour, then at least one of these vertices is a leaf and it must belong to a different colourful component than the internal vertex of the star. Hence, a general caterpillar can be preprocessed in such a way that, for each such leaf, we add its adjacent edge to a set $S_p$. This procedure is repeated until there is no such leaf in $G-S_p$. At the end of the preprocessing, each star in $G-S_p$ is a colourful star, that is, only contains vertices with different colours.

If a coloured general caterpillar is not colourful, then it contains either one or at least two colours that appear more than once. We deal with these two cases independently in the following lemmas.

Lemma 1

Colourful Components and Colourful Partition can be solved in linear time in coloured general caterpillars where exactly one colour appears at least twice.

Lemma 2

Let G be a coloured general caterpillar with only colourful stars such that at least two colours appear at least twice in G. Then there exists an optimal solution S of Colourful Components in G such that $S \subseteq B$, where B is the backbone of G.

Let G be a coloured general caterpillar with backbone B and only colourful stars. We say that a bad path P between two vertices of colour $\gamma$ in G is a colour-critical bad path if and only if there is no other bad path $P'$ between two vertices of colour $\gamma$ such that $P' \cap B \subseteq P$. Hence, two colour-critical bad paths with endpoints of colour $\gamma$ do not have any common edge in the backbone B.

Remark 2

Let G be a coloured general caterpillar with only colourful stars such that at least two colours appear twice. We denote by B the backbone of G. Lemma 2 guarantees that there exists an optimal solution S to Colourful Components on G such that $S \subseteq B$. It is clear that if each colour-critical bad path contains an edge in S and $S \subseteq B$, then S is a solution to Colourful Components on G. Hence, there is an optimal solution to Colourful Components that contains only edges in B that also belong to some colour-critical bad path.

Now, the idea is to define a circular-arc graph H (an intersection graph of a collection of arcs on the circle) based on the colour-critical bad paths of G. A minimum clique cover $\mathcal {Q}$ of H, which is a partition of the vertex set into a minimum number of cliques, can be obtained in linear time [9]. We show that $\mathcal {Q}$ can be translated back into an optimal solution to Colourful Components and Colourful Partition on G in linear time.

Lemma 3

Let G be a coloured general caterpillar with only colourful stars, and A be the multiset of pairs returned by Algorithm 1. Then there is a bijection between the set of colour-critical bad paths in G and the multiset A.

Proof

Let B be the backbone of $G=(V,E)$. A colour-critical bad path P from a to b is detected in Algorithm 1 at Line 11, when b is found to have the same colour $\gamma$ as a (the last recorded vertex of colour $\gamma$). Let x be the internal vertex of the star to which a belongs, and y for b, respectively. When b is considered in the algorithm, the pair (L[c(b)], y) is added to A at Line 12, and since $L[c(b)] = x$ then $(x,y) \in A$. Thus, the arc with endpoints $(x,y) \in A$ corresponds to the colour-critical bad path P from a to b in G.

An ordered pair (x, y) in A refers to two vertices x and y in V that are internal vertices of two stars. If such a pair exists, then there are two vertices a and b with the same colour $\gamma$, such that a belongs to the same star as x and b to the same star as y, and in the path P from a to b, with regard to the order on B, there is no other vertex w with colour $\gamma$ in a star whose internal vertex is in P (since the last seen vertex of colour $\gamma$, before b, is $L[c(b)] = a$). Thus, the path P is a colour-critical bad path and it corresponds to the pair (x, y) in A.   $\square$

Lemma 4

Algorithm 1 runs in linear time.

Proof

Let G be a coloured general caterpillar with only colourful stars and backbone B, and let A be the multiset of ordered pairs obtained by Algorithm 1 with input G.

In Algorithm 1, when a colour is detected for the first time at Line 9, the internal vertex u of the star is stored in the variable end. If the backbone is a chordless cycle, i.e. if G is a cyclic caterpillar, the second time that the vertex end is considered in the main loop the algorithm sets the variable proceed to false at Line 14. If the backbone is a path, i.e. if G is a caterpillar, the algorithm considers each vertex exactly once and sets the variable proceed to false at Line 17. Thus, Algorithm 1 runs in linear time.   $\square$

Theorem 2

Colourful Components and Colourful Partition can be solved in linear time on coloured general caterpillars.

Proof

Let $G=(V,E)$ be a coloured general caterpillar with backbone B. First, we prove that a solution to Colourful Components on G can be found in linear time. We apply the preprocessing to G, as defined in Remark 1, and denote by $S_p$ the set of edges that have been removed. Hence, $G-S_p$ contains only colourful stars. If $G-S_p$ is colourful, then $S_p$ is an optimal solution to Colourful Components. Otherwise, denote by $G'=(V',E')$ the connected component of $G-S_p$ which contains the backbone B. If $G'$ contains exactly one colour that appears more than once, then according to Lemma 1 Colourful Components and Colourful Partition can be solved in linear time. Therefore, we assume that $G'$ contains at least two colours that appear at least twice. Let A be the multiset of ordered pairs obtained by Algorithm 1 with input $G'$.

According to Lemma 4, A can be obtained in linear time. According to Lemma 3, each ordered pair (x, y) in A corresponds to a colour-critical bad path P from x to y in G. These paths can be seen as arcs on the circle, which represent a circular-arc graph $H=(X,F)$. Let $\mathcal {Q}$ be a minimum clique cover of H obtained in linear time [9], and $S'$ be an empty set of edges. Choose a clique $Q_i \in \mathcal {Q}$. From our construction of H, each vertex $u \in Q_i$ corresponds to a colour-critical bad path $P_u$ in G. Let $D_i := \bigcap _{u \in Q_i} P_u$, and notice that $|D_i \cap B| >0$. Then, choose an edge $e \in D_i \cap B$, and add e to $S'$. We claim that, once each clique in $\mathcal {Q}$ has been processed, thus once $|S'| = |\mathcal {Q}|$, the set $S'$ is an optimal solution to Colourful Components on G. Notice that $S'$ can be computed in linear time.

As stated before, each colour-critical bad path in $G'$ maps to an ordered pair in A, which corresponds to a vertex of H. Hence, a clique $Q_i$ in H corresponds to a set of colour-critical bad paths sharing a common subpath $D_i$. The set $S'$ contains an edge in $D_i \cap B$ for each $Q_i \in \mathcal {Q}$, hence there is no colour-critical bad path in $G'-S'$. Since $S' \subset B$ and each coloured-minimal bad paths have an edge in $S'$, then following Remark 2 $S'$ is a solution to Colourful Components on $G'$. Moreover, since Q is optimal, $S'$ is an optimal solution on $G'$. Let $S := S_p \cup S'$, and note that S is an optimal solution to Colourful Components on G.

Since $|E| \in \mathcal {O}(|V|)$, we can detect each connected component of $G-S$ in linear time (for instance, with a breadth-first search). Thus, we can construct the partition $\pi$ of V such that each part corresponds to a connected component of $G-S$ in linear time. Obviously, $\pi$ is a solution to Colourful Partition on G. Since S is optimal, due to the structure of G, the partition $\pi$ is optimal.    $\square$

3 Colourful Components on Small-Degree Planar Graphs

In [4], the authors prove that Colourful Components is NP-complete even when restricted to 3-coloured graphs with maximum degree 6. Using a similar reduction from Planar 3-SAT, we show that the vertices of degree 6 can be replaced with gadgets only containing vertices of degree 4, or 3, if we relax the number of colours from 3 to 5, or 12, respectively.

An instance of Planar 3-SAT is a 3-CNF formula in which the bipartite graph of variables and clauses is planar. Planar 3-SAT has been proved NP-complete in [11].

Construction 2

Given an instance of Planar 3-SAT $\phi$, that is a set of m clauses $C_1,C_2,\dots ,C_m$ on n variables, we construct the graph $G=(V,E)$ such that:

• For each variable x in $\phi$, let $m_x$ denotes the number of clauses in which x appears. We construct a cycle of length $4m_x$ in G with vertices $V_{x} := \{x_j^1, x_j^2, x_j^3, x_j^4 ~|~ x \in C_j\}$ with an arbitrary fixed cyclic ordering of the clauses containing x. The vertices are coloured alternatively with two colours $c_o$ and $c_e$, that is, $c(x_j^1) = c(x_j^3)= c_o$ and $c(x_j^2) = c(x_j^4)= c_e$, for all j such that $x\in C_j$.

• For each clause $C_j$ containing three variables p, q and r, we construct a clause gadget. We propose two types of gadgets:

  • The gadget $\mathcal {A}_j^4$ is made of a cycle of length 3, with vertices $a_j^1$, $a_j^2$ and $a_j^3$ such that each $a_j^i$ is given colour i, different from $c_o$ and $c_e$. We define how the vertices from $V_p$ are connected to $\mathcal {A}_j^4$. If the variable p appears as a positive literal in $C_j$, connect the vertices $p_j^1$ to $a_j^1$ and $p_j^2$ to $a_j^2$. Otherwise, if p occurs as a negative literal, connect the vertices $p_j^2$ to $a_j^1$ and $p_j^3$ to $a_j^2$. Do the same for the variables q and r by connecting the corresponding vertices in $V_{q}$ to $a_j^2$ and $a_j^3$, and the corresponding vertices in $V_{r}$ to $a_j^3$ and $a_j^1$. Notice that the vertices in $\mathcal {A}_j^4$ have degree 4.

  • The gadget $\mathcal {A}_j^3$ is made of a cycle of length 9 with vertices labelled $a_j^1,\dots ,a_j^9$ and an additional vertex $a_j^{10}$ connected to $a_j^2$, $a_j^5$ and $a_j^8$. We set the colour i to each vertex $a_j^i$, different from $c_o$ and $c_e$. We define how the vertices from $V_p$ are connected to $\mathcal {A}_j^3$. If the variable p appears as a positive literal in $C_j$, connect the vertices $p_j^1$ to $a_j^1$ and $p_j^2$ to $a_j^3$. Otherwise, the variable p occurs as a negative literal, connect the vertices $p_j^2$ to $a_j^1$ and $p_j^3$ to $a_j^3$. Do the same for the variables q and r by connecting the corresponding vertices in $V_q$ to $a_j^4$ and $a_j^6$, and the corresponding vertices in $V_r$ $a_j^7$ and $a_j^9$. Notice that the vertices in $\mathcal {A}_j^3$ have degree 3.

  See Fig. 2 for an example of the gadgets.

Since the bipartite graph of variables and clauses of $\phi$ is planar and each vertex can be replaced by a clause or vertex gadget, with a correct cyclic ordering of the clauses for each variable, the resulting graph G is planar.

Note that Construction 2 can be done in polynomial time.

Fig. 2.

Two possible clause gadgets of a clause $C_j := (p \vee \bar{q} \vee r)$. White vertices have colour $c_o$, grey vertices have colour $c_e$, and each $a_j^i$ is given colour i, different from $c_o$ and $c_e$.

Theorem 3

Colourful Components is NP-complete on 5-coloured planar graphs with maximum degree 4 and on 12-coloured planar graphs with maximum degree 3.

4 Conclusion

This paper proposes of complete dichotomy of the computational complexity of Colourful Components and Colourful Partition on k-caterpillars. The NP-completeness of the problems on 2-caterpillars with unbounded degree demonstrates the inherent complexity of the problems. We prove that both problems remain NP-complete on ternary 3-caterpillars and on binary 4-caterpillars, where both the maximum degree and the hair length are bounded by small constants. Nevertheless, our linear-time algorithm for both problems on general 1-caterpillars, with unbounded degree, generalises the class of paths and cycles, and beats the complexity of the previous best known algorithm for paths. An interesting question is to answer whether the problems remain NP-complete on binary 3-caterpillars and on 2-caterpillars with bounded degree.

We also prove that Colourful Components is NP-complete on 5-coloured planar graphs with maximum degree 4 and on 12-coloured planar graphs with maximum degree 3. A natural question is to ask whether the problem remains NP-complete when the number of colours is decreased but the maximum degree is 3 or 4.