Parameterized Domination in Circle Graphs

A circle graph is the intersection graph of a set of chords in a circle. Keil [Discrete Appl. Math., 42(1):51–63, 1993] proved that Dominating Set, Connected Dominating Set, and Total Dominating Set are NP-complete in circle graphs. To the best of our knowledge, nothing was known about the parameterized complexity of these problems in circle graphs. In this paper we prove the following results, which contribute in this direction: Dominating Set, Independent Dominating Set, Connected Dominating Set, Total Dominating Set, and Acyclic Dominating Set are W[1]-hard in circle graphs, parameterized by the size of the solution. Whereas both Connected Dominating Set and Acyclic Dominating Set are W[1]-hard in circle graphs, it turns out that Connected Acyclic Dominating Set is polynomial-time solvable in circle graphs. If T is a given tree, deciding whether a circle graph G has a dominating set inducing a graph isomorphic to T is NP-complete when T is in the input, and FPT when parameterized by t=|V(T)|. We prove that the FPT algorithm runs in subexponential time, namely $2^{\mathcal{O}(t \cdot\frac{\log\log t}{\log t})} \cdot n^{\mathcal{O}(1)}$ , where n=|V(G)|. Dominating Set, Independent Dominating Set, Connected Dominating Set, Total Dominating Set, and Acyclic Dominating Set are W[1]-hard in circle graphs, parameterized by the size of the solution. Whereas both Connected Dominating Set and Acyclic Dominating Set are W[1]-hard in circle graphs, it turns out that Connected Acyclic Dominating Set is polynomial-time solvable in circle graphs. If T is a given tree, deciding whether a circle graph G has a dominating set inducing a graph isomorphic to T is NP-complete when T is in the input, and FPT when parameterized by t=|V(T)|. We prove that the FPT algorithm runs in subexponential time, namely $2^{\mathcal{O}(t \cdot\frac{\log\log t}{\log t})} \cdot n^{\mathcal{O}(1)}$ , where n=|V(G)|.


Introduction
A circle graph is the intersection graph of a set of chords in a circle (see Fig. 1 for an example of a circle graph G together with a circle representation of it).The class of circle graphs has been extensively studied in the literature, due in part to its applications to sorting [12] and VLSI design [30].Many problems which are NP-hard in general graphs turn out to be solvable in polynomial time when restricted to circle graphs.For instance, this is the case of Maximum Clique and Maximum Independent Set [18], Treewidth [25], Minimum Feedback Vertex Set [19], Recognition [20,31], Dominating Clique [23], or 3-Colorability [33].
But still a few problems remain NP-complete in circle graphs, like k-Colorability for k ≥ 4 [32], Hamiltonian Cycle [8], or Minimum Clique Cover [24].In this article we study a variety of domination problems in circle graphs, from a parameterized complexity perspective.A dominating set in a graph G = (V, E) is a subset S ⊆ V such that every vertex in V \ S has at least one neighbor in S. Some extra conditions ⋆ A preliminary conference version of this work appeared in the Proceedings of the 38th International Workshop on Graph-Theoretic Concepts in Computer Science (WG), Jerusalem, Israel, June 2012.The third author was partially supported by EPSRC Grant EP/G043434/1.The other authors were partially supported by AGAPE (ANR-09-BLAN-0159) and GRATOS (ANR-09-JCJC-0041) projects (France).can be imposed to a dominating set.For instance, if S ⊆ V is a dominating set and G[S] is connected (resp.acyclic, an independent set, a graph without isolated vertices, a tree, a path), then S is called a connected (resp.acyclic, independent, total, tree, path) dominating set.In the example of Fig. 1, vertices 1 and 5 (resp.3, 4, and 6) induce an independent (resp.connected) dominating set.The corresponding minimization problems are defined in the natural way.Given a set of graphs G, the Minimum G-Dominating Set problem consists in, given a graph G, finding a dominating set S ⊆ V (G) of G of minimum cardinality such that G[S] is isomorphic to some graph in G. Throughout the article, we may omit the word "Minimum" when referring to a specific problem.
For an introduction to parameterized complexity theory, see for instance [10,15,27].A decision problem with input size n and parameter k having an algorithm which solves it in time f (k) • n O (1) (for some computable function f depending only on k) is called fixed-parameter tractable, or FPT for short.The parameterized problems which are W [i]hard for some i ≥ 1 are not likely to be FPT [10,15,27].A parameterized problem is in XP if it can be solved in time f (k) • n g (k) , for some (unrestricted) functions f and g.The parameterized versions of the above domination problems when parameterized by the cardinality of a solution are also defined naturally.
Previous work.Dominating Set is one of the most prominent classical graphtheoretic NP-complete problems [17], and has been studied intensively in the literature.Keil [23] proved that Dominating Set, Connected Dominating Set, and Total Dominating Set are NP-complete when restricted to circle graphs, and Damian and Pemmaraju [9] proved that Independent Dominating Set is also NP-complete in circle graphs, answering an open question from Keil [23].
Hedetniemi, Hedetniemi, and Rall [21] introduced acyclic domination in graphs.In particular, they proved that Acyclic Dominating Set can be solved in polynomial time in interval graphs and proper circular-arc graphs.Xu, Kang, and Shan [34] proved that Acyclic Dominating Set is linear-time solvable in bipartite permutation graphs.The complexity status of Acyclic Dominating Set in circle graphs was unknown.
In the theory of parameterized complexity [10,15,27], Dominating Set also plays a fundamental role, being the paradigm of a W [2]-hard problem.For some graph classes, like planar graphs, Dominating Set remains NP-complete [17] but becomes FPT when parameterized by the size of the solution [2].Other more recent examples can be found in H-minor-free graphs [3] and claw-free graphs [7].
The parameterized complexity of domination problems has been also studied in geometric graphs, like k-polygon graphs [11], multiple-interval graphs and their complements [13,22], k-gap interval graphs [16], or graphs defined by the intersection of unit squares, unit disks, or line segments [26].But to the best of our knowledge, the pa-rameterized complexity of the aforementioned domination problems in circle graphs was open.
Our contribution.In this paper we prove the following results, which settle the parameterized complexity of a number of domination problems in circle graphs: • In Section 2, we prove that Dominating Set, Connected Dominating Set, Total Dominating Set, Independent Dominating Set, and Acyclic Dominating Set are W [1]-hard in circle graphs, parameterized by the size of the solution.
Note that Acyclic Dominating Set was not even known to be NP-hard in circle graphs.The reductions are from k-Colored Clique in general graphs.• Whereas both Connected Dominating Set and Acyclic Dominating Set are W [1]-hard in circle graphs, it turns out that Connected Acyclic Dominating Set is polynomial-time solvable in circle graphs.This is proved in Section 3.1.• Furthermore, if T is a given tree, we prove that the problem of deciding whether a circle graph has a dominating set isomorphic to T is NP-complete (Section 2.3) but FPT when parameterized by |V (T )| (Section 3.2).The NP-completeness reduction is from 3-Partition, and we prove that the running time of the FPT algorithm is subexponential.As a corollary of the algorithm presented in Section 3.2, we also deduce that, if T has bounded degree, then deciding whether a circle graph has a dominating set isomorphic to T can be solved in polynomial time.
Further research.Some interesting questions remain open.We proved that several domination problems are W [1]-hard in circle graphs.Are they W [1]-complete, or may they also be W [2]-hard? On the other hand, we proved that finding a dominating set isomorphic to a tree can be done in polynomial time.It could be interesting to generalize this result to dominating sets isomorphic to a connected graph of fixed treewidth.Finally, even if Dominating Set parameterized by treewidth is FPT in general graphs due to Courcelle's theorem [6], it is not plausible that it has a polynomial kernel in general graphs [5].It may be the case that the problem admits a polynomial kernel parameterized by treewidth (or by vertex cover) when restricted to circle graphs.

Hardness results
In this section we prove hardness results for a number of domination problems in circle graphs.In order to prove the W [1]-hardness of the domination problems, we provide two families of reductions.Namely, in Section 2.1 we prove the hardness of Dominating Set, Connected Dominating Set, and Total Dominating Set, and in Section 2.2 we prove the hardness of Independent Dominating Set and Acyclic Dominating Set.Finally, we prove the NP-completeness for trees in Section 2.3.For better visibility, some figures of this section have colors, but these colors are not indispensable for completely understanding the depicted constructions.Before stating the hardness results, we need to introduce the following parameterized problem, proved to be W [1]-hard in [13].

k-Colored Clique
Instance: A graph G = (V, E) and a coloring of V using k colors.Parameter: k.
Question: Does there exist a clique of size k in G containing exactly one vertex from each color?Note that in an instance of k-Colored Clique, we can assume that there is no edge between any pair of vertices colored with the same color.Also, we can assume that for each 1 ≤ i ≤ k, the number of vertices colored with color i is the same.Indeed, given an instance G, we can consider an equivalent instance G ′ obtained by putting together k! disjoint copies of G, one for each permutation of the color classes.
In a representation of a circle graph, we will always consider the circle oriented anticlockwise.Given three points a, b, c in the circle, by a < b < c we mean that starting from a and moving anticlockwise along the circle, b comes before c.In a circle representation, we say that two chords with endpoints (a, b) and (c, d) are parallel twins if a < c < d < b, and there is no other endpoint of a chord between a and c, nor between d and b.Note that for any pair of parallel twins (a, b) and (c, d), we can slide c (resp.d) arbitrarily close to a (resp.b) without modifying the circle representation.

Hardness of domination, and connected and total domination
We start with the main result of this section.Proof: We shall reduce the k-Colored Clique problem to the problem of finding a dominating set of size at most k(k + 1)/2 in circle graphs.Let k be an integer and let G be a k-colored graph on kn vertices such that n vertices are colored with color i for all 1 ≤ i ≤ k.For every 1 ≤ i ≤ k, we denote by x i j the vertices of color i, with 1 ≤ j ≤ n.Let us prove that G has a k-colored clique of size k if and only if the following circle graph C has a dominating set of size at most k(k + 1)/2.We choose an arbitrary point of the circle as the origin.The circle graph C is defined as follows: 2 for an illustration).Each cluster has n particular points denoted by 1, . . ., n following the order of the circle.These intervals are constructed in such a way that the origin is not in a section.
• Sections are numbered from 1 to k following the anticlockwise order from the origin.
Similarly, the clusters inside each section are numbered from 1 to k + 1.
we add a chord with endpoints c ij and c ′ ij , which we call the extremal chord of the j-cluster of the i-th section.
• For each 1 ≤ i ≤ k and 1 ≤ j ≤ k, we add chords between the j-th and the (j + 1)-th clusters of the i-th section as follows.For each 0 ≤ l ≤ n, we add two parallel twin chords, each having one endpoint in the interval ]l, l + 1[ of the j-th cluster, and the other endpoint in the interval ]l, l + 1[ of the (j + 1)-th cluster.These chords are called inner chords (see Fig. 3 for an illustration).We note that the endpoints of the inner chords inside each interval can be chosen arbitrarily.The interval ]0, 1[ is the interval between c ij and the point 1, and similarly ]n, n + 1[ is the interval between the point n and c ′ ij .• We also add chords between the first and the last clusters of each section.For each 1 ≤ i ≤ k and 1 ≤ l ≤ n, we add a chord joining the point l of the first cluster and the point l of the last cluster of the i-th section.For each 1 ≤ i ≤ k, these chords are called the i-th memory chords.• Extremal, inner, and memory chords will ensure some structure on the solution.On the other hand, the following chords will simulate the behavior of the original graph.
In fact, the n particular points in each cluster of the i-th section will simulate the behavior of the n vertices of color i in G. Let i < j.The chords from the i-th section to the j-th section are between the j-th cluster of the i-th section and the (i + 1)-th cluster of the j-th section.Between this pair of clusters, we add a chord joining the point h (in the i-th section) and the point l (in the j-th section) if and only if x i h x j l ∈ E(G).We say that such a chord is called associated with an edge of the graph G, and such chords are called outer chords.In other words, there is an outer chord in C if the corresponding vertices are connected in G.
Fig. 3. Representation of the chords between the j-th and the (j + 1)-th cluster of the i-th section.The higher chords are extremal chords.The others are inner chords and have to be replaced by two parallel twin chords.
Intuitively, the idea of the above construction is as follows.For each 1 ≤ i ≤ k, among the k + 1 clusters in the i-th section, the first and the last one do not contain endpoints of outer chords, and are only used for technical reasons (as discussed below).The remaining k − 1 clusters in the i-th section capture the edges of G between vertices of color i and vertices of the remaining k − 1 colors.Namely, for any two distinct colors i and j, there is a cluster in the i-th section and a cluster in the j-th section such that the outer chords between these two clusters correspond to the edges in G between colors i and j.The rest of the proof is structured along a series of claims.
Claim 1 If there exists a k-colored clique in G, then there exists a dominating set of size k(k + 1)/2 in C.
Proof: Assume that there is a k-colored clique K in G and let us denote by k i the integer such that x i ki is the vertex of color i in this clique.Let D be the following set of chords.For each section 1 ≤ i ≤ k, we add to D the memory chord joining the points k i of the first and the last clusters.We also add in D the outer chords associated with the edges of the k-colored clique.The set D contains k(k + 1)/2 chords: k memory chords and k(k − 1)/2 outer chords.Let us prove that D is a dominating set.
The extremal chords are dominated, since D has exactly one endpoint in each cluster.Indeed, there is an endpoint in the first and the last cluster of section because of the memory chords of D. There is an endpoint in the other clusters because of the outer chord associated with the edge of the k-colored clique.The inner chords are also dominated.Indeed, for each section i, the endpoint of the chord of D is k i , for all clusters j such that 1 ≤ j ≤ k + 1.Thus, for all 1 ≤ l ≤ n, the inner chords between the intervals ]l, l + 1[ of the j-th cluster and ]l, l + 1[ of the (j + 1)-th cluster are dominated by the chord of D with endpoint in the j-th section if l ≤ k i − 1, or by the chord of the (j + 1)-th section otherwise.
The outer chords are dominated by the memory chords of D, since the outer chords have their endpoints in two different sections.Finally, the memory chords are also dominated by the outer chords of D for the same reason.
In the following we will state some properties about the dominating sets in C of size k(k + 1)/2.Claim 2 A dominating set in C has size at least k(k + 1)/2, and a dominating set of this size has exactly one endpoint in each cluster.
Proof: The interval of a chord linking x and y, with x < y, is the interval [x, y].One can note that when ℓ chords have pairwise disjoint intervals, at least ⌈ℓ/2⌉ chords are necessary to dominate them.
The intervals [c i,j , c ′ i,j ] of the extremal chords are disjoint by assumption.Since there are k(k + 1) extremal chords, a dominating set has size at least k(k + 1)/2.And if there is a dominating set of such size, it must have exactly one endpoint in each interval, i.e., one endpoint in each cluster.Claim 3 A dominating set of size k(k+1)/2 in C contains no inner nor extremal chord.
Proof: Let D be a dominating set in C of size k(k + 1)/2.It contains no extremal chords, since both endpoints of an extremal chord are in the same cluster, which is impossible by Claim 2. If D contains an inner chord c, the parallel twin of c in C is dominated by some other chord c ′ .But then c ∪ c ′ intersect at most three clusters, which is again impossible by Claim 2.
By Claim 3, a dominating set in C of size k(k + 1)/2 contains only memory and outer chords.Thus, the unique (by Claim 2) endpoint of the dominating set in each cluster is one of the points {1, . . ., n}, and we call it the value of a cluster.Fig. 4 illustrates the general form of a solution.
Claim 4 Assume that C contains a dominating set of size k(k + 1)/2.Then, in a given section, the value of a cluster does not increase between consecutive clusters.
Proof: Assume that in a given arbitrary section, the value of the j-th cluster is l.The inner chords between the interval ]l, l + 1[ of the j-th cluster and the interval ]l, l + 1[ of the (j + 1)-th cluster have to be dominated.Since the value of the j-th cluster is l, they are not dominated in the j-th cluster.Therefore, in order to ensure the domination of these chords, the value of the (j +1)-th cluster is at most l.
Claim 5 Assume that C contains a dominating set of size k(k + 1)/2.Then, for each 1 ≤ i ≤ k, all the clusters of the i-th section have the same value.Proof: Let D be such a dominating set.In a given section, the endpoints of D in the first and the last clusters are endpoints of a memory chord, and for all l, they link the point l of the first cluster to the point l of the last one.Thus, the first and the last clusters have the same value.Since by Claim 4 the value of a cluster decreases between consecutive clusters, the value of the clusters of the same section is necessarily constant.
The value of a section is the value of the clusters in this section (note that it is well-defined by Claim 5).The vertex associated with the i-th section is the vertex x i k if the value of the i-th section is k.

Claim 6
If there is a dominating set in C of size k(k + 1)/2, then for each pair (i, j) with 1 ≤ i < j ≤ k, the vertex associated with the i-th section is adjacent in G to the vertex associated with the j-th section.Therefore, G has a k-colored clique.
Proof: Let i and j be two sections with i < j, and let x i k and x j l be the vertices associated with these two sections, respectively.By Claim 5, the chord of the dominating set in the j-th cluster of the i-th section has a well-defined endpoint k, and the chord of the dominating set in the (i − 1)-th cluster of the j-th section has a well-defined endpoint l.The vertex x i k associated with the i-th section is adjacent in G to the vertex x j l associated with the j-th section.Indeed, the chords having endpoints in these clusters are exactly the chords between these two clusters, and there is a chord if and only if there is an edge between the corresponding vertices in G.
Claims 1 and 6 together ensure that C has a dominating set of size k(k + 1)/2 if and only if G has a k-colored clique.The reduction can be easily done in polynomial time, and the parameters of the problems are polynomially equivalent.Thus, Dominating Set in circle graphs is W [1]-hard.This completes the proof of Theorem 1.
From Theorem 1 we can easily deduce the W [1]-hardness of two other domination problems in circle graphs.
Fig. 5. Gadget H i in interval I i used in the proof of Theorem 2, corresponding to a color class X i of the k-colored input graph G.The dashed chords correspond to non-edges of G.
Proof: In the construction of Theorem 1, if there is a dominating set of size k(k + 1)/2 in C, it is necessarily connected (see the form of the solution in Fig. 4).Indeed, the memory chords ensure the connectivity between all the chords with one endpoint in a section.Since there is a chord between each pair of sections, the dominating set is connected.Finally, note that a connected dominating set is also a total dominating set, as it contains no isolated vertices.

Hardness of independent and acyclic domination
We proceed to describe our second construction in order to prove parameterized reductions for domination problems in circle graphs.
Theorem 2. Independent Dominating Set is W [1]-hard in circle graphs.
Proof: We present a parameterized reduction from k-Colored Clique in a general graph to the problem of finding an independent dominating set of size at most 2k in a circle graph.Let G be the input k-colored graph with color classes X 1 , . . ., X k ⊆ V (G).Let x i 1 , . . ., x i n be the vertices belonging to the color class X i ⊆ V (G), in an arbitrary order.We proceed to build a circle graph H by defining its circle representation.Let I 1 , . . ., I k be a collection of k disjoint intervals in the circle, which will we associated with the k colors.For 1 ≤ i ≤ k, we proceed to construct an induced subgraph H i of H whose chords have all endpoints in the interval I i , which we visit from left to right.Throughout the construction, cf.Fig. 5 for an example with n = 4.
We start by adding two cliques on n vertices L i and R i , with chords l i 1 , . . ., l i n and r i 1 , . . ., r i n , respectively, in the following way.The endpoints of L i and R i are placed in three disjoint subintervals of I i , such that the first subinterval contains, in this order, the left endpoints of l i 1 , . . ., l i n .The second subinterval contains the right endpoints of L i and the left endpoints of R i , in the order Finally, the third subinterval contains, in this order, the right endpoints of r i 1 , . . ., r i n .The blue (resp.red) chords in Fig. 5 correspond to L i (resp.R i ).
For 1 ≤ j ≤ n, we define the interval v i j as the open interval between the right endpoint of l i j and the left endpoint of r i j ; cf. the thick intervals in Fig. 5.Such an interval v i j will correspond to vertex x i j of G.We also add two sets of 2k + 1 parallel twin chords whose left endpoints are placed exactly before the left (resp.right) endpoint of l i 1 (resp.r i 1 ) and whose right endpoints are placed exactly after the left (resp.right) endpoint of l i n (resp.r i n ); cf. the green chords in Fig. 5.We call these chords parallel chords.This completes the construction of H i .Finally, for each pair of vertices x i p , x j q of G such that i = j and {x i p , x j q } / ∈ E(G), we add to H a chord c i,j p,q between the interval v i p in H i and the interval v j q in H j ; cf. the dashed chords in Fig. 5.We call these chords outer chords.That is, the outer chords of H correspond to non-edges of G.This completes the construction of the circle graph H.
We now claim that G has a k-colored clique if and only if H has an independent dominating set of size at most 2k.
Indeed, let first K be a k-colored clique in G containing vertices x 1 j1 , x 2 j2 , . . ., x k j k , and let us obtain from K an independent dominating set S in H.For 1 ≤ i ≤ k, the set S contains the two chords l i ji and r i ji from H i .Note that S is indeed an independent set.For 1 ≤ i ≤ k, since both L i and R i are cliques, all the chords in L i and R i are dominated by S. Clearly, all parallel chords are also dominated by S. The only outer chords with one endpoint in H i which are neither dominated by l i ji nor by r i ji are those with its endpoint in the interval v i ji .Let c be such an outer chord, and suppose that the other endpoint of c is in H ℓ .As K is a clique in G, it follows that there is no outer chord in H with one endpoint in v i ji and the other in v ℓ j ℓ , and therefore necessarily the chord c is dominated either by l i ji or by r ℓ j ℓ .Conversely, assume that H has an independent dominating set S with |S| ≤ 2k.Note that for 1 ≤ i ≤ k, because of the two sets of 2k + 1 parallel chords in H i , at least one of the chords in L i and at least one of the chords in R i must belong to S, so |S| ≥ 2k.Therefore, it follows that |S| = 2k and that S contains in H i , for 1 ≤ i ≤ k, a pair of non-crossing chords in L i and R i .Note that in each H i , the two chords belonging to S must leave uncovered at least one of the intervals (corresponding to vertices) v i 1 , . . ., v i n .Let v i ji and v ℓ j ℓ be two uncovered vertices in two distinct intervals I i and I ℓ , respectively.By the construction of H, it holds that the vertices x i ji and x ℓ j ℓ must be adjacent in G, as otherwise the outer chord in H between the intervals v i ji and v ℓ j ℓ would not be dominated by S. Hence, a k-colored clique in G can be obtained by selecting in each H i any of the uncovered vertices.
The construction of Theorem 2 can be appropriately modified to deal with the case when the dominating set is required to induce an acyclic subgraph.
Proof: As in Theorem 2, the reduction is again from k-Colored Clique.From a kcolored G, we build a circle graph H that contains all the chords defined in the proof of Theorem 2, plus the following ones for each H i , 1 ≤ i ≤ k (cf.Fig. 6 for an illustration): we add another set of 2k + 1 parallel chords whose left (resp.right) endpoints are placed exactly before (resp.after) the right (resp.left) endpoint of l i 1 (resp.r i n ); cf. the middle green chords in Fig. 6.We call these three sets of 2k + 1 chords parallel chords.
Furthermore, we add a clique with n chords d i 1 , . . ., d i n such that for 1 ≤ j ≤ n the left (resp.right) endpoint of d i j is placed exactly after the left (resp.right) endpoint of l i j (resp.r i j ).Finally, for each such a chord d i j we add a parallel twin chord, denoted by di j .We call these 2t chords distance chords, and their union is denoted by D i ; cf. the brown edges in Fig. 6.This completes the construction of H.Note that a pair of chords l i j1 and r i j2 dominates all the distance chords in H i if and only if l i j1 and r i j2 do not cross, that is, if and only if j 1 ≤ j 2 .
We now claim that G has a k-colored clique if and only if H has an acyclic dominating set of size at most 2k.
Indeed, let first K be a k-colored clique in G.An independent (hence, acyclic) dominating set S in H of size 2k can be obtained from K exactly as explained in the proof Fig. 6.Gadget H i in interval I i used in the proof of Theorem 3, corresponding to a color class X i of the k-colored input graph G.The dashed chords correspond to non-edges of G.
of Theorem 2. Note that in each H i , the distance chords in D i are indeed dominated by S because the corresponding chords in L i and R i do not cross.Conversely, assume that H has an acyclic dominating set S with |S| ≤ 2k.First assume that S contains no outer chord.By the parallel chords in each H i (cf. the green chords in Fig. 6), it is easy to check that S must contain at least two chords in each H i , and therefore we have that |S| = 2k.We now distinguish several cases according to which two chords in a generic H i can belong to S. Let {u, v} = S ∩ V (H i ).Because of the parallel chords, it is clear that only one L i , R i , or D i cannot contain both u and v.It is also clear that no parallel chord can be in S.
Since v must dominate the twin chord di j1 , it follows by the construction of H i that j 2 ≤ j 1 , and therefore the chord r i j1 is dominated neither by u nor by v (cf.Fig. 6), a contradiction.The case u ∈ D i and v ∈ R i is similar.Therefore, we may assume w.l.o.g. that u = l i j1 and u = r i j2 .Note that such a pair of chords l i j1 and r i j2 dominates all the distance chords in H i if and only if l i j1 and r i j2 do not cross.Hence, as in the proof of Theorem 2, for each H i , 1 ≤ i ≤ k, the two chords belonging to S leave at least one uncovered interval v i ji (corresponding to vertex x i ji ), and in order for all the outer chords in H to be dominated, the union of the k uncovered vertices must induce a k-colored clique in G. Therefore, if H has an acyclic dominating set of size at most 2k with no outer chord, then G has a k-colored clique.Note that in this case S consists of an independent set.
Otherwise, the acyclic dominating set S contains some outer chord.Assume w.l.o.g. that S contains outer chords with at least one endpoint in each of H 1 , . . ., H p (with p ≥ 2, as each chord has two endpoints), and no outer chord with an endpoint in any of H p+1 , . . ., H k (only if p < k).By the arguments above, for p + 1 ≤ i ≤ k it follows that S contains exactly one chord in L i and exactly one chord in R i .For 1 ≤ i ≤ p, in order for all the parallel chords in H i to be covered, S must contain some chords in L i , R i , or D i .As by assumption |S| ≤ 2k, in at least one H i with 1 ≤ i ≤ p, S contains exactly one chord in L i , R i , or D i .By the construction of H, this chord must necessarily be a distance chord, as otherwise some parallel chords in H i would not be dominated by S (cf.Fig. 6).Assume w.l.o.g. that in H 1 only the distance chord d 1 j1 and one outer chord outgoing from interval v 1 j2 belong to S. But then if j 2 ≥ j 1 (resp.j 2 < j 1 ) the chord r 1 j2 (resp.l 1 j2 ) is not dominated by S, a contradiction.We conclude that S contains at least two outer chords in H 1 .
By a simple counting argument, as |S| ≤ 2k it follows that for 1 ≤ i ≤ p, S contains exactly one distance chord and two outer chords from each of H 1 , . . ., H p (and, in particular, |S| = 2k).But then the subgraph of H[S] induced by the chords belonging to V (H 1 ) ∪ . . .∪ V (H p ) has minimum degree at least two, and therefore it contains a cycle, a contradiction to the assumption that H[S] is acyclic.Thus, S cannot contain any outer chord, and the theorem follows.

NP-completeness for a given tree
The last result of this section is the NP-completeness when the dominating set is restricted to be isomorphic to a given tree.
Theorem 4. Let T be a given tree.Then {T }-Dominating Set is NP-complete in circle graphs when T is part of the input.
Proof: We present a reduction from the 3-Partition problem, which consists in deciding whether a given multiset of n = 3m integers I can be partitioned into m triples that all have the same sum B. The 3-Partition problem is strongly NP-complete, and in addition, it remains NP-complete even when every integer in I is strictly between B/4 and B/2 [17].Let I = {a 1 , . . ., a n } be an instance of 3-Partition, in which we can assume that the a i 's are between B/4 and B/2, and let B = n i=1 a i /m be the desired sum.Note that we can also assume that B is an integer, as otherwise I is obviously a No-instance.
We proceed to define a tree T and to build a circle graph G that has a {T }-dominating set S if and only if I is a Yes-instance of 3-Partition.Given I = {a 1 , . . ., a n }, let T be the rooted tree obtained from a root r to which we attach a path with a i vertices, for i = 1, . . ., n; see , where for better visibility these chords have been depicted outside the circle.Finally, for i ∈ {1, 2, . . ., mB} \ {B, 2B, . . ., mB}, we add a chord r i whose first endpoint is exactly after the first endpoint of b i (in the anticlockwise order, starting from any of the endpoints of the root r), and whose second endpoint is exactly before the second endpoint of b i+1 .These chords are called chain chords; cf. the red chords in Fig 7(b).Note that r i is adjacent to g i , g i+1 , b i , and b i+1 .This completes the construction of the circle graph G.Each one of the m connected components that remain in G after the removal of r and the parallel chords is called a block.
Conversely, let S be a {T }-dominating set S in G, and note that we can assume that the root of T has arbitrarily big degree.As the vertex of G corresponding to the chord r is the only vertex of G of degree more than 6, necessarily r belongs to S, and corresponds to the root of T .We claim that S contains no pendant chord.Indeed, by construction of G, exactly n of the branch chords are in S, which dominate exactly n pendant chords.As G[S \ {r}] consists of n disjoint paths, each attached to r through a branch chord, the total number of chords in these paths which are not branch chords is mB − n.These mB − n pendant or chain chords must dominate the pendant chords that are not dominated by branch chords, which are also mB − n many.Assume that a pendant chord b belongs to S. Since T is a tree, there must exist a path P in S between b and one of the branch chords, say g.Assume that P contains p chords, including b but not g.It is clear that b is the only pendant chord contained in P , as otherwise P would have a cycle.Therefore, P has p chords and dominates exactly p − 1 pendant chords that are not dominated by branch chords, which contradicts the fact that mB − n pendant or chain chords must dominate the mB − n pendant chords that are not dominated by branch chords.Hence, S contains no pendant chord, so S contains exactly mB − n chain chords.
Since T is a tree, each path in S made of consecutive chain chords intersects exactly one branch chord.As the a i 's are strictly between B/4 and B/2, each block has exactly 3 branch chords in S. The fact that chain chords are missing between consecutive blocks assures the existence of a 3-partition of I.More precisely, the restriction of S to each block defines the integers belonging to each triple of the 3-partition of I as follows.For a branch chord g i ∈ S, let P i be the path in S hanging from g i , which consists only of chain chords.Then, for each branch chord g i ∈ S, the corresponding integer is defined by the number of vertices in P i plus one.By the above discussion, these m triples define a 3-partition of I.The theorem follows.
To conclude this section, it is worth noting here that {T }-Dominating Set is W [2]hard in general graphs.This can be proved by an easy reduction from Set Cover parameterized by the number of sets, which is W [2]-hard [28].Indeed, let C be a collection of subsets of a set S, and the question is whether there exist at most k subsets in C whose union contains all elements of S. We construct a graph G as follows.First, we build a bipartite graph (A ∪ B, E), where there is a vertex in A (resp.B) for each subset in C (resp.element in S), and there is an edge in E between a vertex in A and a vertex in B if the corresponding subset contains the corresponding element.We add a new vertex v, which we join to all the vertices in A, and k + 1 new vertices joined only to v. It is then clear that G has a dominating set isomorphic to a star with exactly k leaves if and only if there is a collection of at most k subsets in C whose union contains all elements of S.

Polynomial and FPT algorithms
In this section we provide polynomial and FPT algorithms for finding dominating sets in a circle graph which are isomorphic to trees.Namely, in Section 3.1 we give a polynomialtime algorithm to find a dominating set isomorphic to some tree.This algorithm contains the main ideas from which the other algorithms in this section are inspired.In Section 3.2 we modify the algorithm to find a dominating set isomorphic to a given tree T in FPT time, the parameter being the size of T .By carefully analyzing its running time, we prove that this FPT algorithm runs in subexponential time.It follows from this analysis that if the given tree T has bounded degree (in particular, if it is a path), then the problem of find a dominating set isomorphic to T can be solved in polynomial time.

Polynomial algorithm for trees
Note that, in contrast with Theorem 5 below, Theorem 3 in Section 2.2 states that, if F is the set of all forests, then F -Dominating Set is W [1]-hard in circle graphs.This is one of the interesting examples where the fact of imposing connectivity constraints in a given problem makes it computationally easier, while it is usually not the case (see for instance [4,29]).In the following, by size of a set of chords, we mean the number of chords in it, i.e., the number of vertices of C in this set.We say that a forest F of C spans a region ab − cd if each of a, b, c, and d is an endpoint of some chord in F , and each endpoint of a chord of Note that an (ab − cd)-dominating forest with several connected components might not dominate some chord going from [a, c] to [d, b].This is not the case if F is connected, as stated in the following claim.We now state two properties that will be useful in the algorithm.Their correctness is proved below.
T1 Let F 1 and F 2 be two valid forests for two regions ab − cd and ef − gh, respectively, such that a If there is no chord with both endpoints either in [c, e] or in [f, d], then F 1 ∪ F 2 is valid for ab − gh (see Fig. 8).T2 Let F 1 and F 2 be two valid forests for two regions ab − cd and ef − gh, respectively (F 2 being possibly empty), and let uv be a chord such that u ≤ a ≤ c ≤ e ≤ g ≤ v ≤ h ≤ f ≤ d ≤ b, and such that there is no chord with both endpoints either in [u, a], or in [g, v], or in [v, h], or in [b, u].Then F 1 ∪ F 2 ∪ {uv} is a tree which is valid for df − ce.When F 2 is empty, we consider that e, f, g, h correspond to the point v.
Roughly speaking, the intuitive idea behind this two properties is to reduce the length of the circle in which we still have to do some computation (that is, outside the valid regions), which will be helpful in the dynamic programming routine.Again, the proof is structured along a series of claims.Before verifying the correctness of Properties T1 and T2, let us first state a useful general fact.
then the chord γ is dominated by F .Thus we can assume that u ≤ β < α.And note that by assumption, β ≤ c ≤ α.But in F , the chord with endpoint c is connected to the chord ω, thus there is a chord wz of F such that w ≤ β ≤ z ≤ α, and therefore the chord γ is dominated by F , which achieves the proof of the claim.
Note that the proof of Claim 8 is symmetric, and then the same result is still true for the intervals [a, b] and [b, a].Note also that the same result holds without the asumption that F is (ab − cd)-dominating.
Claim 9 Property T1 is correct.
Proof: Let us prove that F 1 ∪ F 2 is a forest and that it is valid for the region ab − gh.For an illustration refer to Fig. 8. Since F 1 and F 2 span ab − cd and ef − gh respectively, all the endpoints of the chords of still a forest and the connected components of the union are precisely the connected components of F 1 and the connected components of F 2 .
Since F 1 spans ab − cd, there is a chord of F 1 with endpoint a and one chord with endpoint b, and the same holds for F 2 and g, h.Then F 1 ∪ F 2 spans the region ab − gh. Since Therefore, F 1 ∪ F 2 is valid for the region ab − gh.
Claim 10 Property T2 is correct.
Proof: Let F 1 and F 2 be two valid forests for ab − cd and for ef − gh, respectively (F 2 being possibly empty), and let uv be a chord with endpoints u and v, such that For an illustration, refer also to Fig. 8. First note that T = F 1 ∪ F 2 ∪ {uv} is a tree.Indeed, as in Claim 9, one can prove that F 1 ∪ F 2 is a forest with exactly one chord with one endpoint in [a, g] and one endpoint in [h, b] per connected component.Thus, the addition of uv ensures that T is a tree.
Since F 1 and F 2 spans ab − cd and ef − gh respectively, T spans df − ce.Indeed, there are chords intersecting d and c in F 1 , chords intersecting e and f in F 2 , and all the chords are strictly inside [d, c] ∪ [e, f ].Note that when the forest F 2 is empty, there is a chord intersecting v, and thus the tree T spans df − ce.
The tree T spans df − ce, since there is exactly one chord with one endpoint in [d, c] and one endpoint in [e, f ], which is precisely the chord uv.
Let us prove that T is (df − ce)-dominating.First note that the chords with one endpoint in [u, v] and one endpoint in [v, u] are dominated by uv.By symmetry, we just have to prove that the chords with both endpoints in [u, v] are dominated by T .By symmetry again, we just have to prove that the chords with both endpoints in [u, c] are dominated.There is no chord with both endpoints in [u, a], the chords with both endpoints in [a, c] are dominated by F 1 , since F 1 is (ab − cd)-dominating.By Claim 8, the chords with one endpoint in [a, b] and one endpoint in [b, a] are dominated by F 1 , thus the chords with one endpoint in [u, a] and one in [a, c] are dominated.
Therefore, we conclude that T is a valid tree for df − ce.
For a region ab − cd, we denote by v f ab,cd (resp.v t ab,cd ) the least integer l for which there is a valid forest (resp.tree) of size l for ab − cd.If there is no valid forest (resp.tree) for ab − cd, we set v f ab,cd = +∞ (resp.v t ab,cd = +∞).Let us now describe our algorithm based on dynamic programming.With each region ab − cd, we associate two integers v 1 ab,cd and v 2 ab,cd .Algorithm 1 below calculates these two values for each region.We next prove that v 1 ab,cd = v f ab,cd and v 2 ab,cd = v t ab,cd , and that Algorithm 1 computes the result in polynomial time.
Algorithm 1 Dynamic programming for computing a dominating tree if there are two regions ab − cd and ef − gh such that v 1 ab,cd = j1 and v 1 ef,gh = j2 with j1 + j2 = j satisfying Property T1, with v 1 ab,gh = +∞ then v 1 ab,gh ← j if there is a region ab − cd and a chord uv such that v 1 ab,cd = j − 1 satisfying Property T2 with an empty second forest then dv,cv ← j if there are two regions ab − cd and ef − gh and a chord uv such that v 1 ab,cd = j1 and v 1 ef,gh = j2 with j1 Claim 11 For any region ab − cd, v 1 ab,cd = 1 (resp.v 2 ab,cd = 1) if and only if v f ab,cd = 1 (resp.v t ab,cd = 1).
Proof: Let ab − cd be a region such that v f ab,cd = 1.Therefore, there is a set of chords of size one which is valid for ab − cd.Let ω be this chord.Since ω spans a, b, c and b, ω has endpoints a, b, c, d.This implies that a = c and b = d, i.e., ω is precisely the chord ab.If v f ab,cd = 1, then a = c, b = d, and the chord ab exists, which corresponds exactly to the initialization of the algorithm.Conversely, it is clear that by definition the chord ab is valid for the region ab − ab.Thus, v 1 ab,cd = 1 if and only if v f ab,cd = 1.The same result holds for trees, since a forest of size one is a tree.
Claim 12 For any region ab − cd, v f ab,cd ≤ v 1 ab,cd (resp.v t ab,cd ≤ v 2 ab,cd ).
Proof: The claim is true for the initialization, since if v 1 ab,cd = 1 then v f ab,cd = 1.By induction it is still true for all integers k, since Properties T1 and T2 are correct, and when a value is affected in the dynamic programming of Algorithm 1, one of the two properties is applied.
Claim 13 For any region ab − cd, v t ab,cd ≥ v 2 ab,cd and v f ab,cd ≥ v 1 ab,cd .
Proof: Let us prove it by induction on j.Claim 11 ensures that the result is true for j = 1.Assume that for all j < k, if v f ab,cd = j then v 1 ac,bd ≤ j and that if v t ac,bd = j then v 2 ac,bd ≤ j.
Let us first prove that the induction step holds for trees.We now prove that if v t ab,cd = j, then v 2 ab,cd ≤ j.Let T be a valid tree of size j for the region ab − cd.Since T spans ab − cd, there is exactly one chord uv with one endpoint in [a, c] and one endpoint in [d, b].Let F 1 be the restriction of T to the chords with both endpoints in [a, c], and let F 2 be the restriction of T to the chords with both endpoints in [d, b].Note that T = F 1 ∪ F 2 ∪ {uv}.Let e, f (resp.g, h) be the points of the circle graph intersected by , and e, f (resp.g, h) are as near as possible from u (resp.v) (see Fig. 9 for an example).Let us denote by j 1 (resp.j 2 ) the size of F 1 (resp.F 2 ).Note that j 1 + j 2 = j − 1.Let us prove that F 1 is valid for ac − ef , that F 2 is valid for db − hg, and that Property T2 can be applied to F 1 , F 2 , and the chord uv.By symmetry, we just have to prove that F 1 is valid.There are chords intersecting e, f by definition of e, f , and chords intersecting a, c since T spans ab − cd.The endpoints of the chords are in [a, e] ∪ [f, c], since T spans ab − cd and e, f are the nearest points from u which are in T .Thus F 1 spans ac − ef .
If there is a connected component of F 1 with no chord from [a, e] to [f, c], then F 1 ∪F 2 ∪{uv} cannot be a tree, since it would not be connected.If a connected component of F 1 has two chords from [a, e] to [f, c], then F 1 ∪ {u, v} has a cycle, which contradicts the fact that T is a tree.Thus F 1 spans ac − ef .
Let us prove that F 1 is (ac − ef )-dominating.Indeed, if there is a chord with both endpoints in [a, e] which is not dominated by F 1 , it cannot be dominated by F 2 and uv, since none of their endpoints is in this interval.Thus T is not valid.Hence all the chords in the interval [a, e], and by symmetry also in the interval [f, c], are dominated by F 1 .Therefore F 1 is valid.By induction hypothesis, since the size of F 1 is at most j 1 , we have v 2 ac,ef ≤ j 1 , and the same holds for F 2 .Since T is valid, one can note that there is no chord with both endpoint either in [e, u], or [u, f ], or [h, v], or [v, g].Thus Property T2 can be safely applied and then v 2 ab,cd ≤ j, as we wanted to prove.Let us now prove that the induction step also holds for forests.Let us now consider a forest F for the region ab − cd.If the forest has exactly one connected component, that is, if it is a tree, then the inequality holds by the first part of the induction.
Assume now that F has at least two connected components.Let us now prove that if v f ab,cd = j then v 1 ab,cd ≤ j.Let ab − cd be a region with v f ac,bd = k.This means that there exists a forest F with one endpoint in a, b, c, d, since F spans ab − cd.Since v t ab,cd = j, F has at least two connected components.Note that the case when v t ab,cd = j is treated just above.
Since Let us denote by F 2 the set F \ F 1 (see Fig. 10 for an example).Let us prove that F 1 and F 2 are valid for ab − ef and for gh− cd, respectively.Since F is (ab − cd)-dominating, all the chords with both endpoints either in [a, e] or in [f, b] (resp.[g, c] or [h, d]) are dominated by F , thus by F 1 (resp.F 2 ).Therefore F 1 (resp.F 2 ) is (ab − ef )-dominating (resp.(gh − cd)-dominating).Thus, by induction hypothesis we have v 1 ac,ef = v f ac,ef and v 1 gh,cd = v f gh,cd .And since Property T1 can be applied for F 1 ∪ F 2 , by the safeness of Property T1, in Algorithm 1 we have v 1 ab,cd ≤ v f ab,cd .
Claims 12 and 13 together ensure that v f ab,cd = v 1 ab,cd and that v t ab,cd = v 2 ab,cd .Hence, by dynamic programming all the regions of a given size can be found in polynomial time.
Let us now explain how we can verify if there is a dominating set isomorphic to some tree of a given size k.This in particular will prove Theorem 5. Conversely, let T be a dominating tree of size k.Let uv be a chord of T which disconnects T .Thus T \{uv} has at least two connected components F 1 and F 2 .Let a, c be the two extremities of the first one, and let b, d be the extremities of the other ones (see Fig. 9 for an illustration).Let us prove that v t ab,cd ≤ k.Indeed, T is a tree by assumption, by definition it spans a, b, c, d, it spans ab − cd since the chord uv is the unique chord from [a, c] to [d, b], and it is (ab − cd)-dominating since T is a dominating tree.In addition, since T is a dominating tree, there is no chord with both endpoints either in the interval [b, a] or in [c, d], as otherwise such a chord would not be dominated by T .
By dynamic programming, Algorithm 1 computes in polynomial time the regions for which there is a valid tree of any size from 1 to n.Given a region ab − cd with v t ab,cd ≤ k, we just have to verify that there are no chords in the intervals [b, a] and [c, d], which can clearly be done in polynomial time.One can easily check that Algorithm 1 runs in time O(n 10 ), but we did not make any attempt to improve its time complexity.This completes the proof of Theorem 5.
As Algorithm 1 computes the regions for which there is a valid tree of any size from 1 to n, it can be slightly modified to obtain the following corollary.
Corollary 2. Let T k be the set of all trees of size exactly k.Then T k -Dominating Set can be solved in polynomial time in circle graphs.

FPT algorithm for a given tree
It turns out that when we seek a dominating set isomorphic to a given fixed tree T , the problem is FPT parameterized by |V (T )|.In order to express the running time of our algorithm, and to prove that it is subexponential in |V (T )|, we need some definitions.Let T be a tree, and let us root T at an arbitrary vertex r.Let v be a vertex of T .We denote by T [v] the subtree of T induced by v and the descendants of v in the rooted tree.Let v 1 , . . ., v l be the children of v in the tree T rooted at r.We define F (v) as the forest where we consider two isomorphic trees T [v i ] and T [v j ] as the same element.Suppose that F (v) contains exactly s non-isomorphic trees T 1 , . . ., T s , and that for 1 ≤ i ≤ s, there are exactly d i trees in F (v) which are isomorphic to T i (note that s i=1 d i = l).We define the following parameter, which corresponds to the number of non-isomorphic sub(multi)sets of the multiset F (v): Finally, we also define the following three parameters: Proof: The idea of the proof is basically the same as in the proof of Theorem 5.The main difference is that in the proof of Theorem 5, when Properties T1 or T2 are satisfied, we can directly apply them and still obtain a forest or a tree.In the current proof, when we make the union of two forests, we have to make sure that the union of the two forests is still a subforest of T , and that we can correctly complete it to obtain the desired tree T .
For obtaining that, we will apply the two properties stated below, whenever it is possible to create forests which are induced by the children of the same vertex of T .Let us first give some intuition on the algorithm.
In the following we consider the tree T rooted at an arbitrary vertex r.Let w 1 , . . ., w l be some vertices of T which are children of the same vertex y.The subforest of T induced by w 1 , . . ., w l , denoted by F (w 1 , . . ., w l ), is the forest Roughly speaking, the idea of the algorithm is to exhaustively seek, for each region ab − cd and any possible subforest F of F (v) for every vertex v in T , a valid forest for ab − cd isomorphic to F , and then try to grow it until hopefully obtaining the target tree T .Note that if a vertex v of T has k children, there are a priori 2 k possible subsets of children of v, which define 2 k possible types of subforests in F (v).But the key point is that if some of the trees in F (v) are isomorphic, some of the choices of subsets of subforests will give rise to the same tree.In order to avoid this redundancy, for each vertex v of T , we partition the trees in F (v) into isomorphism classes, and then the choices within each isomorphism class reduce to choosing the multiplicity of this tree, which corresponds to the parameter d i + 1 (as we may not choose any copy of it) defined before the statement of Proposition 1.Note that carrying out this partition into isomorphism classes can be done in polynomial time (in t) for each vertex of T , using the fact that one can test whether two rooted trees T 1 and T 2 with t vertices are isomorphic in O(t) time [1].
Therefore, if we proceed in this way, the number of such subforests for each vertex v ∈ V (T ) is at most α T r (v).As we repeat this procedure for every node of T , the cost of this routine per vertex is at most α T r = max {v∈V (T )} α T r (v).And as we chose the root arbitrarily, it follows that the function can be upper-bounded by α T = max {r root of T } α T r , which in turn can be upper-bounded by α t = max {T :|V (T )|=t} α T , which is a subexponential function by Proposition 1.We would like to note that this step is the unique non-polynomial part of the algorithm.
Let us now explain more precisely the outline of the algorithm.An induced subtree T 1 of the input circle graph is valid for a region ab − cd and a tree T [w], if it is valid for ab − cd, if, in addition, there is an isomorphism between T 1 and T [w] for which the unique chord between [a, c] and [b, d] corresponds to w.A forest F 1 is valid for a region ab − cd and F (w 1 , . . ., w l ), if it is valid for ab − cd, and if there is an isomorphism between F 1 and F (w 1 , . . ., w l ) for which the unique chord between [a, c] and [b, d] of each connected component T [w i ] corresponds to vertex w i .Let us now state the two properties that correspond to Properties T1 and T2 of Theorem 5. .When F 2 is empty, we consider that e, f, g, h correspond to the point v.
In the proof of Theorem 5, we have seen that the validity of the corresponding regions is satisfied.Thus, we just have to verify that the tree or the forest which is created is isomorphic to the target tree T , and that the chords with one endpoint in each side are children of the same vertex.The union of the two isomorphisms, and the fact that the chords with one endpoint in both sides are the children of y, ensures that both properties are true.Indeed, for example for Property F2, since the chords with one endpoint in each interval are exactly the children of y, it holds that the chord corresponding to the vertex y intersects exactly its children.
For each region ab − cd and each tree T [w], we define a boolean variable b t ab,cd,w , which is set to 'true' if and only if there is a valid tree for ab − cd and T [w].For each region ab − cd and each forest F (w 1 , . . ., w l ), we define a boolean variable b f ab,cd,w1,...,w l which is set to true if and only if there is a valid forest for ab − cd and F (w 1 , . . ., w l ).(For the sake of simplicity, we distinguish between trees and forests, but we would like to stress that it is not strictly necessary for the algorithm.) By a dynamic programming similar to Algorithm 1 in Theorem 5, we can compute all the regions ab − cd and all vertices v of T for which b t ab,cd,v = true (and the same for forests).If there is a region ab − cd for which b t ab,cd,r = true, and such that there is no chord with both endpoints either in [b, a] or in [c, d], then the tree T dominates all the chords in the input circle graph C. Indeed, the safeness of Properties F1 and F2 ensures that there is a valid tree isomorphic to T for the region ab − cd.And Claim 14 in the proof of Theorem 5 ensures that this tree is indeed a dominating tree.
Note that, indeed, the unique non-polynomial step of the algorithm consists in generating the collection of non-isomorphic subforests, which are at most α T many.Thus, the dynamic programming algorithm runs in time O(α T • n O (1) ).Again, we did not make any effort to optimize the degree of the polynomial in the running time.

4 GFig. 1 .
Fig. 1.A circle graph G on 8 vertices together with a circle representation of it.

Fig. 2 .
Fig. 2. Sections and clusters in the reduction of Theorem 1.

Theorem 1 .
Dominating Set is W [1]-hard in circle graphs, when parameterized by the size of the solution.

0 0 1 1 oFig. 4 .
Fig. 4. The general form of a solution in the reduction of Theorem 1.The thick chords are memory chords and the other ones are outer chords.The origin is depicted with a small "o".

Corollary 1 .
Connected Dominating Set and Total Dominating Set are W [1]hard in circle graphs, when parameterized by the size of the solution.
Fig 7(a) for an example with n = 9, m = 3, and B = 5.(In this figure, for simplicity not all the a i 's are between B/4 and B/2, but we assume that this fact is true in the proof.)Note that |V (T )| = mB + 1.The circle graph G is obtained as follows; see Fig 7(b) for the construction corresponding to the instance of Fig 7(a): We start with a chord r that will correspond to the root of T .Now we add mB parallel chords g 1 , . . ., g mB intersecting only with r.These chords are called branch chords; cf. the green chords in Fig 7(b).We can assume that the endpoints of the branch chords are ordered clockwise in the circle.For i = 1, . . ., mB, we add a chord b i incident only with g i .These chords are called pendant chords; cf. the blue chords in Fig 7(b)

Fig. 7 .
Fig. 7. Reduction in the proof of Theorem 4. (a) Instance I of 3-Partition, with n = 9, m = 3, and B = 5, together with the associated tree T .(b).Circle graph G built from I. (c) The thick chords define a {T }-dominating set S in G.

Theorem 5 .
Let T be the set of all trees.Then T -Dominating Set can be solved in polynomial time in circle graphs.In other words, Connected Acyclic Dominating Set can be solved in polynomial time in circle graphs.Proof: Let C be a circle graph on n vertices and let C be an arbitrary circle representation of C. We denote by P the set of intersections of the circle and the chords in this representation.The elements of P are called points.Without loss of generality, we can assume that only one chord intersects a given point.Given two points a, b ∈ P, the interval [a, b] is the interval from a to b in the anticlockwise order.Given four (nonnecessarily distinct) points a, b, c, d ∈ P, with a ≤ c ≤ d ≤ b, by the region ab − cd we mean the union of the two intervals [a, c] and [d, b].Note that these two intervals can be obtained by "subtracting" the interval [c, d] from the interval [a, b]; this is why we use the notation ab − cd.
split by a region ab − cd if for each connected component of F there is exactly one chord with one endpoint in [a, c] and one endpoint in [d, b].Given a region ab − cd, a forest F is (ab − cd)-dominating if all the chords of C with both endpoints either in the interval [a, c] or in the interval [d, b] are dominated by F .A forest is valid for a region ab − cd if it spans ab − cd, is split by ab − cd, and is (ab − cd)-dominating.

Claim 7 Fig. 8 .
Fig.8.On the left (resp.right), regions corresponding to Property T1 (resp.Property T2).Full lines correspond to real chords of C, dashed lines correspond to the limit of regions.Bold intervals correspond to intervals with no chord of C with both endpoints in the interval.

Claim 8
Let ab − cd be a region and let F be a valid forest for ab − cd.The chords with one endpoint in [c, d] and one endpoint in [d, c] are dominated by F .Proof: Let us now consider a chord γ with one endpoint α in [c, d] and one endpoint β in [d, c].First assume that β is in [b, a].Since F is split by ab − cd, there is a chord of F with one endpoint in [a, c] and one in [d, b], and such a chord dominates γ.Therefore, by symmetry, we can assume that β is in [a, c].Since F spans ab − cd, there is a chord in F with endpoint c.Since F is split by ab − cd, there is a chord ω = uv of F , in the same connected component as the chord with endpoint c, with one endpoint in [a, c] and one endpoint in [d, b]

F 1
is split by ab − cd, there is exactly one chord per connected component between [a, c] and [d, b], thus also between [a, g] and [h, b].The same holds for F 2 .Thus each connected component of F 1 ∪ F 2 has exactly one chord with one endpoint in [a, g] and the other one in [h, b].So F 1 ∪ F 2 is split by ab − gh.Let us now prove that F 1 ∪ F 2 is (ab − gh)-dominating.Let us verify that all the chords in the interval [a, g] are dominated by F 1 ∪ F 2 .By symmetry, the same will hold for the interval [h, b].All the chords with both endpoints in the interval [a, c] are dominated by F 1 , and those with both endpoints in the interval [e, g] are dominated by F 2 .By assumption, there is no chord in the interval [c, e].The chords with one endpoint in [c, d] and one endpoint in [d, c] are dominated by F 1 by Claim 8, and those with one endpoint in [e, f ] and one endpoint in [f, e] are dominated by F 2 .Thus all the chords with both endpoints in [a, c] are dominated by

Fig. 9 .
Fig.9.On the left the original tree, and on the right the partition of the tree into two forests and a chord uv.

Fig. 10 .
Fig. 10.On the left the original forest, and on the right the partition of the forest into the two forests.

F
spans ab − cd, all the endpoints of F are in [a, c] ∪ [d, c].Let F 1 be the connected component of the chord with endpoint a.The point e (resp.f ) is the point of [a, c] (resp.[d, b]) with an endpoint in F 1 , and such that there is no endpoint of F 1 after e (resp.before f ) in [a, c] (resp.[d, b]).Let g (resp.h) be the first endpoint of F after e in [a, c] (resp.before f in [d, b]).

Claim 14
Let k be a positive integer.There is a dominating tree of size at most k in C if and only if there is a region ab − cd such that v t ab,cd ≤ k and such that there is no chord strictly contained in [b, a] nor in [c, d].Proof: Assume that there is a region ab − cd with v t ab,cd ≤ k and no chord strictly contained in [b, a] nor in [c, d].Then by Claim 8, all the chords with one endpoint in [b, a] and one endpoint in [a, b] are dominated, and the same holds for the couple c, d.All the chords with both endpoints in [a, c] ∪ [d, b] are dominated by Claim 7, as T is valid for ab − cd.Since by assumption there is no chord strictly contained in [b, a] and in [c, d], all the chords of the circle graph C are dominated, as by assumption there is no chord in the other intervals.Thus, T is a dominating set.

F1 Let F 1
and F 2 be two valid forests for ab − cd and F (v 1 , . . ., v l ), and for ef − gh and F (w 1 , . . ., w m ), respectively.Assume in addition that a ≤ c ≤ e ≤ g ≤ h ≤ f ≤ d ≤ b.Assume also that, for all 1 ≤ i ≤ l, 1 ≤ j ≤ m, the vertices v i and w j are pairwise distinct and are children of the same vertex y of T .If there is no chord with both endpoints either in [c, e] or in [f, d], then F 1 ∪ F 2 is valid for ab − gh and F (v 1 , . . ., v l , w 1 , . . ., w m ).F2 Let F 1 and F 2 be two valid forests for ab − cd and F (v 1 , . . ., v l ), and for ef − gh and F (w 1 , . . ., w m ), respectively (F 2 being possibly empty), and let uv be a chord of the input graphC.Assume that u ≤ a ≤ c ≤ e ≤ g ≤ v ≤ h ≤ f ≤ d ≤ band that there is no chord with both endpoints either in [u, a], or in [g, v], or in [v, h], or in [b, u].Assume also that there exists a vertex y of T with exactly l + m children v 1 , . . ., v l , w 1 , . . ., w m Then F 1 ∪ F 2 ∪ {uv} is a tree which is valid for df − ce and T [y]