Completing colored graphs to meet a target property

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Completing colored graphs to meet a target property✩ Kathryn Cook a , Elaine M. Eschen a,∗ , R. Sritharan b , Xiaoqiang Wang a a

Lane Department of Computer Science and Electrical Engineering, P.O. Box 6109, West Virginia University, Morgantown, WV 26506, United States b Computer Science Department, The University of Dayton, Dayton, OH 45469, United States

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Article history: Received 5 December 2015 Received in revised form 23 December 2016 Accepted 20 January 2017 Available online xxxx Keywords: Completion problem Sandwich problem Chordal graph Strongly chordal graph Circular-arc graph

abstract We consider the problem of deciding whether a k-colored graph can be completed to have a given property. We establish that, when k is not fixed, the completion problems for (Helly) circular-arc graphs, even (Helly) proper or (Helly) unit circular-arc graphs, are NP-complete. When k is fixed, in the case of completion to a (Helly) circular-arc graph, (Helly) proper circular-arc graph, or (Helly) unit circular-arc graph we fully classify the complexities of the problems. We also show that deciding whether a 3-colored graph can be completed to be strongly chordal can be done in O(n2 ) time. As a corollary of our results, the sandwich problems for Helly circular-arc graphs, Helly proper circular-arc graphs, and Helly unit circular-arc graphs are NP-complete. © 2017 Elsevier B.V. All rights reserved.

1. Introduction The graphs that we consider are simple and any vertex coloring considered is proper. A colored graph is a graph properly colored with an arbitrary number of colors. A k-colored graph is a graph properly colored with k colors. In the Π sandwich problem, given graphs G1 = (V , E1 ) and G2 = (V , E2 ) on the same vertex-set with E1 ⊆ E2 and a property Π , the question is whether there is a graph G = (V , E ) that has property Π and also E1 ⊆ E ⊆ E2 holds. The set E1 contains the required edges while the set E2 \ E1 contains the optional edges. It is seen that when E1 = E2 the sandwich problem is the same as the recognition problem for property Π . The sandwich problem was introduced in 1995 by Golumbic et al. in [14] where they studied the problem for the property of membership in several classes of perfect graphs. Since then, the sandwich problem has been studied for a variety of NP properties, and a number of published papers can be found on the topic. In the Π completion of a colored graph problem, given a property Π and a graph G = (V , E ) with a proper vertex coloring c : V → Z , the question is whether there exists a supergraph G′ = (V , E ′ ) of G that has property Π and also is properly colored by c. When such a G′ exists, we say G′ is a Π completion of G and G admits a Π completion. By taking the edges in E to be the required edges and {xy | x ∈ V , y ∈ V , xy ̸∈ E , and c (x) ̸= c (y)} to be the set of optional edges, it is seen that the Π completion of a colored graph is a restriction of the sandwich problem for property Π . There are sandwich problems that are NP-complete (such as for comparability graphs), whose corresponding colored graph completion problem is trivial (as every complete k-partite graph is a comparability graph). However, there are also NP-complete sandwich problems whose colored graph completion version remains hard. Π completion of colored graphs has been studied inside and outside the context of the sandwich problem.

✩ A preliminary version of this paper appeared in the Proceedings of the 39th International Workshop on Graph-Theoretic Concepts in Computer Science (WG), 2013. ∗ Corresponding author. E-mail addresses: [email protected] (E.M. Eschen), [email protected] (R. Sritharan), [email protected] (X. Wang).

http://dx.doi.org/10.1016/j.dam.2017.01.021 0166-218X/© 2017 Elsevier B.V. All rights reserved.

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It was established in [28] that the perfect phylogeny problem is equivalent to the problem of chordal completion of colored graphs. It was subsequently shown that this problem is NP-complete [4,5], which in turn implies that the sandwich problem for chordal graphs is also NP-complete. The sandwich problems for strongly chordal graphs and chordal bipartite graphs were two of the problems whose complexities were left as an open question in [14]. It was shown in [7,24] that these problems are NP-complete; these proofs established that the problems of strongly chordal completion of colored graphs and chordal bipartite completion of colored graphs are NP-complete. The complexity of chordal completion of a colored graph has been studied when the input graph is colored with k colors, where k is constant. For the case that k = 3, several linear-time algorithms are known [6,16,17]. It is also known that for every fixed k, the problem can be solved in polynomial time [22]. Due to potential applications to the problem of DNA physical mapping [3,13], the problem of interval completion of a colored graph, and its variations, have been studied extensively in the literature. It is known that the problems of interval completion of a colored graph [3] and unit interval completion of a colored graph [13] are NP-complete. Several results are known for interval completion and unit interval completion of a colored graph when the input is restricted. Let k be the number of colors used on the input graph. In contrast to the case for chordal graphs, the best known algorithm for the interval completion problem when k = 3 runs in O(n2 ) time [3]. In further contrast, the interval completion of a k-colored graph problem has been shown to be NP-complete for every fixed k ≥ 4 [3]. In fact, the interval completion of a colored graph problem is NP-complete even for four-colored caterpillars [1]. Similarly, it is known that the unit interval completion of a colored graph is solvable in polynomial time for caterpillars with hair length less than 2 [2], while it is NP-complete for caterpillars with hair length at least 2 [2]. The main contributions of this paper are as follows: First, we provide simple proofs that, given a graph colored with an arbitrary number of colors, each of the problems (Helly) circular-arc completion, (Helly) proper circular-arc completion, and (Helly) unit circular-arc completion is NP-complete. Later we prove these problems remain NP-complete when the number of colors is fixed and at least 3. Then, we provide a full classification of the complexity of the problem of completing a colored graph to be a circular-arc graph when the number of colors used is fixed. Specifically, given a k-colored graph, we show that when k = 2, there is an O(n)-time algorithm for the problem, but when k = 3, the problem is NP-complete; in turn, the problem remains NP-complete for every fixed k, k ≥ 3. We provide an identical classification for the problem of completing a colored graph to be a Helly circular-arc graph, proper circular-arc graph, unit circular-arc graph, Helly proper circular-arc graph, or Helly unit circular-arc graph. To the best of our knowledge these are the first instances of a colored graph completion problem on 3-colored graphs that are hard. We also show that deciding whether a 3-colored graph admits a strongly chordal completion can be decided in O(n2 ) time; our algorithm is based on a characterization of bi-connected 3-colored graphs that admit a strongly chordal completion. We conjecture that the corresponding problem for 4-colored graphs is NP-complete. It is known that the sandwich problems for interval graphs, circular-arc graphs, and proper (resp., unit) circular-arc graphs are NP-complete [14]. It follows from our results that the sandwich problem for Helly circular-arc graphs and the sandwich problem for Helly proper (resp., unit) circular-arc graphs are NP-complete. 2. Definitions All graphs considered in this paper are finite, undirected, and simple (i.e., without loops and multiple edges). For notations and terminology not defined here, we refer to [29]. Let G = (V , E ) be a graph. We use n to refer to |V | and m to refer to |E |. For S ⊆ V , G[S ] refers to the subgraph of G induced by S. For a vertex v ∈ V the (open) neighborhood of v is the set N (v) = {u ∈ V | uv ∈ E }, while the closed neighborhood of v is N [v] = N (v) ∪ {v}. We use NG (v) and NG [v] to refer to an open (resp., closed) neighborhood in G if necessary for clarity. Vertex x is simplicial if N (x) is a clique. The complete graph with k vertices is denoted by Kk . The complete bipartite graph B(X , Y , E ), where |X | = 1, |Y | = 3, and E = {xy | x ∈ X andy ∈ Y }, is denoted by K1,3 (also known as the claw). A path with k vertices is denoted Pk . G is chordal if every cycle with at least 4 vertices in G has a chord. For k ≥ 1, a k-tree is defined recursively as follows: Kk+1 is a k-tree. Given a k-tree Tn on n vertices, a k-tree Tn+1 on n + 1 vertices can be constructed by adding a new vertex whose neighborhood is a clique of size k in Tn . A partial k-tree is a subgraph of a k-tree. An n-sun is the graph on 2n vertices (n ≥ 3) whose vertex set can be partitioned into W = {w0 , . . . , wn−1 } and U = {u0 , . . . , un−1 } such that U is a clique, W is an independent set, and ui is adjacent to wj if and only if i = j or i = j + 1 (mod n). A graph is strongly chordal if it is chordal and does not contain an n-sun, n ≥ 3. G is an interval graph if every x ∈ V can be mapped to an interval Ix on the real line such that xy ∈ E if and only if Ix ∩ Iy ̸= ∅; {Ix | x ∈ V } is an interval model for G. G is a circular-arc graph if every x ∈ V can be mapped to an arc Ax on a circle such that xy ∈ E if and only if Ax ∩ Ay ̸= ∅; {Ax | x ∈ V } is a circular-arc model for G. A circular-arc graph G is a proper circular-arc graph if G has a circular-arc model in which no arc properly contains another arc. A circular-arc graph G is a unit circular-arc graph if G has a circular-arc model in which all arcs have unit length. Proper interval graphs and unit interval graphs are similarly defined. By definition, unit circular-arc graphs (resp., unit interval graphs) are a subclass of proper circular-arc graphs (resp., proper interval graphs). In circular-arc models and proper circular-arc models the arcs can be adjusted so that no two arcs share an endpoint. Thus, these classes remain the same whether arcs are open or closed. In the case of unit circular-arc graphs (resp., unit interval graphs), we assume the original definition for circular-arc models (resp., interval models) in which arcs (resp., intervals) are restricted to be all closed (or all open). Under this assumption, we have the following relationships between classes. Roberts [23] proved a graph is a unit interval graph if and only if it is a

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proper interval graph. Tucker [26] proved that the class of unit circular-arc graphs is properly included in the class of proper circular-arc graphs. If a mix of open and closed arcs is permitted, then every proper circular-arc graph is a unit circular-arc graph [26]. A set of pairwise intersecting arcs in a model for a circular-arc graph has the Helly property if they all have a common point of intersection. A circular-arc graph is a Helly circular-arc graph if it admits a circular-arc model in which any set of arcs corresponding to a clique has the Helly property. It is well known [12] that G is a Helly circular-arc graph if and only if there is a circular ordering of the maximal cliques of G that has the circular ones property (i.e., for any vertex x of G, the maximal cliques containing x are circularly consecutive in the ordering). A circular-arc graph G is a Helly proper circular-arc graph if G has a circular-arc model that is simultaneously Helly and proper. A circular-arc graph G is a Helly unit circular-arc graph if G has a circular-arc model that is simultaneously Helly and unit. Tucker [26] gives examples of proper circular-arc graphs that are Helly proper circular-arc graphs and not unit circular-arc graphs. A set of pairwise intersecting intervals on the real number line always has the Helly property; thus, interval graphs are always Helly. 3. Circular-arc completions 3.1. Completing arbitrarily colored graphs Using the fact that interval colored graph completion [3] and unit interval colored graph completion [13] are NP-complete problems and the following theorem due to Tucker, we provide simple proofs that, given a graph colored with an arbitrary number of colors, each of the problems (Helly) circular-arc completion, (Helly) proper circular-arc completion, and (Helly) unit circular-arc completion is NP-complete. Later we prove these problems remain NP-complete when the number of colors is fixed and at least 3. Theorem 3.1 ([27]). Suppose the vertex set of a circular-arc graph G can be partitioned into two cliques. Then in any circular-arc model there exist two points p1 and p2 such that each circular arc contains at least one of them. Theorem 3.2. Each of the following problems is NP-complete: (Helly) circular-arc completion of a colored graph, (Helly) proper circular-arc completion of a colored graph, and (Helly) unit circular-arc completion of a colored graph. Proof. Since circular-arc graphs [27] and Helly circular-arc graphs [12] can be recognized in polynomial time, the colored graph completion problems for these classes are in NP. We show that the problem of deciding whether a colored graph G admits a circular-arc completion is NP-complete by reducing the interval completion of a colored graph problem to it. Let colored graph G be an instance of the interval completion problem. Suppose G is colored with k colors. We construct an instance G′ of the circular-arc completion problem by forming the disjoint union of G and two complete graphs D1 and D2 of size k; each complete graph is colored with the k colors used on G. Clearly, if G admits an interval completion, then G′ has a circular-arc completion. Now suppose G′ admits a circular-arc completion H. Consider a circular-arc model for H and the induced model for the subgraph H ′ of H induced by the vertices of D1 ∪ D2 . Theorem 3.1 applies to H ′ and p1 and p2 must each be contained in exactly k arcs with distinct colors. Hence, in the circular-arc model for H no arc associated with a vertex of G contains p1 or p2 . It follows that G has an interval completion. A similar proof shows that the Helly circular-arc colored graph completion problem is NP-complete. We use the same reduction from the interval completion of a colored graph problem and the fact that all interval graphs are Helly. Clearly, if G admits an interval completion, then G′ has a Helly circular-arc completion. For the other direction, suppose G′ has a Helly circular-arc completion; then the rest of the proof is the same. Since proper circular-arc graphs [8,15,25], unit circular-arc graphs [10], and Helly proper (resp., Helly unit) circular-arc graphs [18–20] can be recognized in polynomial time the remaining problems are in NP. A graph is a unit interval graph if and only if it is a proper interval graph [23]. The unit interval completion of a colored graph problem is known to be NPcomplete [13]. Reductions from unit interval completion of a colored graph similar to the reductions used above establish the theorem.  3.2. Completing 2-colored graphs A short caterpillar is a tree obtained by starting with a path on k vertices, k ≥ 1, and then optionally attaching one or more vertices of degree one to the vertices on the path. A bug is a graph obtained by starting with a cycle on 2k vertices, k ≥ 2, and then optionally attaching one or more vertices of degree one to the vertices on the cycle. Short caterpillars and bugs have unique 2-colorings (up to the switching of the color classes). The tree T2 is obtained from the graph claw by subdividing each edge once (see Fig. 1). Observation 3.1. The tree T2 is not a circular-arc graph. Any tree that is not an interval graph must contain T2 as an induced subgraph. Observation 3.2. A 2-colored circular-arc graph G that contains an induced cycle C on 2k vertices, k ≥ 2, is connected. Further, a vertex of G that is not on C must be adjacent to exactly one vertex on C and there is no edge in G with both endpoints not on C .

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Fig. 1. Tree T2 and a 3-colored kite (vertex subscript i denotes color i).

Lemma 3.1. Let G be a 2-colored graph. G is a circular-arc graph if and only if G is either the disjoint union of short caterpillars or G is a bug. Proof. It is easily seen that the disjoint union of short caterpillars is a circular-arc graph and a bug is a circular-arc graph. For the other direction, suppose G is a 2-colored circular-arc graph. If G is a forest, then by Observation 3.1, G cannot have a T2 as an induced subgraph. Therefore, G is the disjoint union of short caterpillars. If G is not a forest, then it contains an induced cycle C on 2k vertices, k ≥ 2. Then by Observation 3.2, G is a bug.  Observation 3.3. The tree K1,3 is not a proper circular-arc graph. Lemma 3.2. Let G be a 2-colored graph. G is a proper circular-arc graph if and only if G is either the disjoint union of paths or G is a chordless cycle on 2k vertices, k ≥ 2. Proof. The proof is similar to that of Lemma 3.1.



Corollary 3.1. Every 2-colored proper circular-arc graph is a unit circular-arc graph. Theorem 3.3. Let G be a 2-colored graph. Then, the following are equivalent: (i) G admits a circular-arc completion. (ii) G admits a Helly circular-arc completion. (iii) G is either the disjoint union of short caterpillars or G is the disjoint union of a bug with isolated vertices of either color. Proof. (i) ⇒ (iii) Let G be a 2-colored graph that admits a circular-arc completion. Suppose G is a forest. We claim that G cannot contain a T2 . Suppose it does. Then, as T2 is not a circular-arc graph (see Observation 3.1), edges were added among vertices of the T2 during the completion. However, adding any edge between two vertices of different colors in a T2 creates an induced cycle on 4 vertices and the T2 has an edge with both endpoints not on this cycle, a contradiction of Observation 3.2. Therefore, G is a forest that does not contain a T2 , and hence, G is the disjoint union of short caterpillars. Now suppose G is not a forest. Let H be a component of G that contains a cycle. We claim that H itself must be a circular-arc graph. Suppose not. Then, edges were added to H during the completion to produce a circular-arc graph H ′ . By Lemma 3.1, H ′ is bug. Since H itself is connected, deletion of any of the added edges from H ′ results in a connected graph. Therefore, any edge of H ′ with one endpoint on the cycle of H ′ and the other endpoint not on the cycle was already present in H. Since at least one edge was added to H, this implies that not all the edges of the unique cycle in H ′ , and hence the unique cycle in H, were present in H, a contradiction. Therefore, H is a circular-arc graph, and by Lemma 3.1, H itself is a bug. It then follows, that the remaining components of G must be isolated vertices of either color. (iii) ⇒ (ii) The disjoint union of a bug and isolated vertices of either color can be completed to be a bug. By Lemma 3.1, each of a bug and the disjoint union of short caterpillars admits a circular-arc model. As the graph is 2-colored, the model is indeed a Helly circular-arc model. (ii) ⇒ (i) is trivial.  Corollary 3.2. Let G be a 2-colored graph. Then, the following are equivalent: (i) G admits a proper circular-arc completion. (ii) G admits a Helly proper circular-arc completion. (iii) G is either the disjoint union of paths or G is a chordless cycle on 2k vertices, k ≥ 2. It follows from Lemma 3.2, Corollary 3.1, and Corollary 3.2 that a 2-colored graph G admits a proper circular-arc completion if and only if G itself is a unit circular-arc graph. Corollary 3.3. Whether a 2-colored graph admits a circular-arc completion, Helly circular-arc completion, proper circular-arc completion, or Helly proper circular-arc completion can be tested in O(n) time.

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Fig. 2. A schematic of the transformation: the graph G(I ).

3.3. Completing 3-colored graphs We show that the problem of deciding whether a 3-colored graph G admits a circular-arc completion (or Helly circulararc completion) is NP-complete by a transformation from the strongly NP-complete problem 3-partition [11]. We also give NP-completeness results for the completion to several other subclasses of circular-arc graphs. Our transformation uses ideas from [3]. However, the proof in [3] is for interval completion of a 4-colored graph; we require only three colors. Problem 3-partition 3m Instance: Integers m ∈ N and Q ∈ N, a sequence s1 , . . . , s3m ∈ N such that i=1 si = mQ and 14 Q < si < 12 Q , 1 ≤ i ≤ 3m.  Question: Can the set {1, . . . , 3m} be partitioned into m disjoint sets S1 , . . . , Sm such that for 1 ≤ j ≤ m, i∈Sj si = Q ? Note, if the answer is yes, then each set Si is a 3-set. Given an instance I of the 3-partition problem, we describe the construction of the 3-colored graph G(I ). Then, we prove some properties of the construction. Finally, we establish that I admits a 3-partition if and only if G(I ) admits a (Helly) circular-arc completion. Suppose we are given m, Q , s1 , s2 , . . . , s3m ∈ N as the instance I of 3-partition. The graph G(I ) is constructed as follows: C-cliques: Clique Ci consists of the vertices ci,1 , ci,2 , and ci,3 where vertex ci,j is colored j. Add cliques C0 , C1 , . . . , Cm−1 to G(I ). Note that when clique Ci+1 is referred to, and wherever appropriate, the arithmetic is done modulo m. Tracks: Track-i consists of the path di,1 di,2 · · · di,24Q where vertex di,j is colored 1 if j mod 3 = 1, colored 2 if j mod 3 = 2, and colored 3 if j mod 3 = 0. Thus, the color pattern 123123 . . . repeats on the path. Construct track-0, track-1, . . . , track-m − 1. Identify vertex ci,1 of Ci with vertex di,1 of track-i, for 0 ≤ i ≤ m − 1. Also, identify vertex di,24Q of track-i with vertex c(i+1),3 of Ci+1 , for 0 ≤ i ≤ m − 1. Thus, the part of G(I ) formed so far consists of cyclically ordered cliques C0 , C1 , . . . , Cm−1 where Ci is connected to Ci+1 by track-i. W-paths: Corresponding to si , construct the path W i = ei,1 ei,2 · · · ei,24si −2 . Vertex ei,j is colored 3 if j mod 3 = 1, colored 1 if j mod 3 = 2, and colored 2 if j mod 3 = 0. Thus, the color pattern 312312 . . . repeats on the path with each of the vertices ei,1 and ei,24si −2 colored 3. Add the paths W 1 , W 2 , . . . , W 3m to G(I ). In summary, the 3-colored graph G(I ) consists of the m C-cliques cyclically connected by the m tracks in disjoint union with the 3m W-paths. We refer to the part of G(I ) consisting of the C-cliques connected by the tracks as the circle-gadget. The reader is referred to Fig. 2 for a schematic. In the transformation in [3] from 3-partition to the interval completion of 4-colored graphs the weight gadgets are paths. The graph G in the constructed instance includes two cliques of size 4, and a vertex that is adjacent to exactly one vertex in

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each of these cliques and to one endpoint of each weight path. This forces the weight paths to lie between the cliques of size 4 in any interval model of an interval completion of G. In the case of completion to a circular-arc graph, this type of structure in the constructed graph would over-constrain the completion; hence, we are able to construct a 3-colored graph. Let 3-colored-kite be the 3-colored graph given in Fig. 1, where each vertex with subscript i has color i. Claim 3.1. Let A be a model of any circular-arc completion of a 3-colored-kite. Then, the arcs corresponding to the vertices of the clique {x1 , c2 , c3 } must have the Helly property in A. Proof. By way of contradiction, let A be a model for a circular-arc completion of the 3-colored-kite in which the arcs corresponding to the vertices x1 , c2 , and c3 do not have the Helly property and hence cover the circle. Without loss of generality, assume that the vertices are ordered c2 , x1 , c3 in the clockwise direction with respect to the segment of each arc that does not contain a point on the other two arcs. Then, there is a point p on Ax1 that is also on Ax2 and Ax3 . Similarly, there exists a point q on Ac2 that is also on Ax3 and Ay1 . Likewise, there exists a point r on the Ac3 that is also on Ay1 and Ay2 . Finally, there exists a point s on Ax1 that is also on Ay2 and Ay3 . It then follows that Ax3 must extend in the clockwise direction from q to p, and Ay2 must extend in the counterclockwise direction from r to s. Now, Ax3 and Ac3 constrain Ay3 to be contained in Ax1 . Thus, Ay3 cannot intersect Az1 , a contradiction.  Claim 3.2. Let A be a model of any circular-arc completion of the circle-gadget. Then, A is a Helly circular-arc model that covers the circle. Proof. As the 3-colored-kite is an induced subgraph of the circle-gadget, we can deduce from Claim 3.1 that the arcs corresponding to the vertices of each Ci must have the Helly property; let pi be a point that is on the arcs corresponding to ci,1 = di,1 , ci,2 , and di−1,24Q = ci,3 . Observe that vertices on track-i involve all three colors and each point pi is on arcs of vertices colored 1, 2, and 3. Therefore, the cyclic order of the points {pi | 0 ≤ i ≤ m − 1} must be p0 , p1 , . . . , pm−1 ; without loss of generality, let this be the cyclic order of the points in the clockwise direction. Then, for the same reasons, the arcs corresponding to the vertices di,2 , . . . , di,24Q −1 of track-i are constrained to be in the segment (pi , pi+1 ) of the circle in the clockwise direction. It then follows that no three arcs of A cover the circle and A is a Helly circular-arc model that covers the circle.  Corollary 3.4. Let A be a model of any circular-arc completion of G(I ). Then, A is a Helly circular-arc model. Proof. Let pi be a point on each of the arcs corresponding to vertices of Ci . As the path W j involves vertices colored 1, 2, and 3, the arcs corresponding to the vertices of W j are constrained to be in the segment (pi , pi+1 ), for some i, of the circle in the clockwise direction. This, in conjunction with Claim 3.2, yields the conclusion.  In the next two lemmata, we employ ideas from [3] where it is shown that the interval completion problem for 4-colored graphs is NP-complete. Lemma 3.3. If I admits a 3-partition, then G(I ) admits a circular-arc completion. Further, any model for a circular-arc completion of G(I ) is a Helly circular-arc model. Proof. Suppose the set {1, . . . , 3m} can be partitioned into m disjoint sets S1 , . . . , Sm such that for 1 ≤ j ≤ m,



i∈Sj

si = Q .

We will demonstrate a Helly circular-arc completion H of G(I ) by providing a circular ordering R of the maximal cliques of H with the circular ones property. That any model for a circular-arc completion of G(I ) is a Helly circular-arc model follows from Corollary 3.4. We begin with C0 , C1 , . . . , Cm−1 in cyclic order as R. Without loss of generality, let S1 = {1, 2, 3}. We show how to construct the part R1 of R so that it starts with C1 , ends with C2 , and incorporates the paths W 1 , W 2 , and W 3 , the cliques C1 , C2 , and track-1. It then follows that any Si can similarly be incorporated between Ci and Ci+1 to obtain the full ordering R. Intuitively, the cliques in R1 are obtained by interleaving each of the paths W 1 , W 2 , and W 3 one by one with track-1 to create an interval graph so that the cliques involving vertices of a W i are disjoint from those involving vertices of a W j , where {i, j} ⊆ {1, 2, 3}. The resulting cliques are of the types: triangle {c1,1 , c1,2 , e1,1 }; triangle with an edge of a W r (indicated in bold) and a vertex of track-1; triangle with an edge of track-1 (indicated in bold) and a vertex of a W r ; clique of size 2 that consists of a track-1 edge (indicated in bold) that precedes or follows the interleaving of a W i . Specifically, the order of the cliques in R1 is as follows (see also Fig. 3): C1 , {c1,2 , d1,1 , e1,1 }, {d1,1 , d1,2 , e1,1 }, {e1,1 , e1,2 , d1,2 }, {d1,2 , d1,3 , e1,2 }, {e1,2 , e1,3 , d1,3 }, {d1,3 , d1,4 , e1,3 }, {e1,3 , e1,4 , d1,4 }, . . . ,

{d1,24s1 −3 , d1,24s1 −2 , e1,24s1 −3 }, {e1,24s1 −3 , e1,24s1 −2 , d1,24s1 −2 }, {d1,24s1 −2 , d1,24s1 −1 , e1,24s1 −2 }, {d1,24s1 −1 , d1,24s1 }, {d1,24s1 , d1,24s1 +1 }, {d1,24s1 +1 , d1,24s1 +2 , e2,1 }, {e2,1 , e2,2 , d1,24s1 +2 }, . . . , {e2,24s2 −3 , e2,24s2 −2 , d1,24(s1 +s2 )−2 }, {d1,24(s1 +s2 )−2 , d1,24(s1 +s2 )−1 , e2,24s2 −2 }, {d1,24(s1 +s2 )−1 , d1,24(s1 +s2 ) }, {d1,24(s1 +s2 ) , d1,24(s1 +s2 )+1 }, {d1,24(s1 +s2 )+1 , d1,24(s1 +s2 )+2 , e3,1 }, {e3,1 , e3,2 , d1,24(s1 +s2 )+2 }, . . . , {e3,24s3 −3 , e3,24s3 −2 , d1,24(s1 +s2 +s3 )−2 }, {d1,24(s1 +s2 +s3 )−2 , d1,24(s1 +s2 +s3 )−1 , e3,24s3 −2 }, {d1,24(s1 +s2 +s3 )−1 , d1,24(s1 +s2 +s3 ) }, C2 . As 24(s1 + s2 + s3 ) equals 24Q , the above mentioned cliques are indeed feasible and it is easily verified that the vertices in every clique are of distinct colors.



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Fig. 3. A schematic of the circular-arc completion of G(I ).

Corollary 3.5. If I admits a 3-partition, then G(I ) admits a (Helly) unit circular-arc completion (and thus, admits a (Helly) proper circular-arc completion). Proof. The Helly circular-arc model for the circular-arc completion of G(I ) of Lemma 3.3 can be constructed using only unit length arcs. The vertex sets of the interval graphs created when a W -path is interleaved with a track are disjoint. Note that the interval graph involving the first W -path interleaved on track i includes Ci . The maximal cliques of these interval graphs form a sequence of triangles. It is easy to see that each of these interval graphs can be realized by a unit interval model. It remains to place these interval models on the circle, linking them with edges or length 2 paths whose middle vertices have color 3. A diagram of a unit circular-arc model is given in Fig. 4. In the diagram, the arcs in each group (colors 1, 2, and 3) share a segment of the circle and the groups from top to bottom follow one another in the clockwise direction on the circle. In the model each arc is comprised of five equal length subunits. The sequence of triangles for each interleaved W -path W i ends (in the clockwise direction) in a track vertex of color 2. The arc for this vertex extends in the clockwise direction two subunits beyond the other arcs of the sequence of triangles. Likewise, the sequence of triangles for each interleaved W -path W i begins (in the counterclockwise direction) in either a track vertex of color 1 or a C-clique vertex of color 3, and these arcs extend in the counterclockwise direction two subunits beyond the other arcs of the sequence. Thus, the connecting edges and paths between interval graphs can be modeled by unit length arcs.  Lemma 3.4. If G(I ) admits a (Helly) circular-arc completion, then I admits a 3-partition. Proof. Let H be a circular-arc completion of G(I ) and let A be a circular-arc model for H. It follows from Corollary 3.4 that A is a Helly circular-arc model. Consider a circular ordering R of the cliques of H such that for any vertex x of H, the cliques containing x are consecutive in R. Since the paths {W j } are disjoint from the circle-gadget in G(I ) and the vertices of a Ci are colored 1, 2, and 3, the endpoints of each edge of W j are part of a clique that is between Ci and Ci+1 in the clockwise direction, for some i, in R. Since the arcs corresponding to the vertices of the circle-gadget cover the circle (see Claim 3.2), every clique in R must contain a ci,s or a di,r vertex colored 1, 2, or 3. It then follows that the endpoints of an edge ej,k ej,k+1 of W j that are colored 1 and 2 must be part of a clique of R that contains a track vertex di,r colored 3. Further, since the last

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Fig. 4. Schematic of the unit circular-arc model for the completion of G(I ).

edge ej,24sj −3 ej,24sj −2 of W j has endpoints colored 2 and 3, r ≤ 24Q − 3. Corresponding to each such vertex di,r of track-i that is colored 3, there exists a segment S of R where each clique in S contains vertex di,r . Clearly, there are 8Q − 1 such segments in track-i, all the segments are between Ci and Ci+1 , and the segments are pairwise disjoint. Now, we claim that there is at most one clique in S corresponding to vertex di,r that contains the endpoints of an edge ej,k ej,k+1 of W j where ej,k is colored 1 and ej,k+1 is colored 2. Let Da be the last clique in the counterclockwise direction that contains the vertex di,r and let Db be the last clique in the clockwise direction that contains the vertex di,r . By way of contradiction, suppose there is another edge ep,q ep,q+1 of W p with ep,q colored 1 and ep,q+1 colored 2 such that clique Dc contains di,r , ej,k , and ej,k+1 and clique Dd contains di,r , ep,q , and ep,q+1 . Suppose, without loss of generality, that the ordering of the cliques in the clockwise direction from Da is: Da , Dc , Dd , Db . Note that it may be the case that Da = Dc or Db = Dd . We will show that this implies that either the arc for ej,k is contained in the arc for di,r or the arc for ej,k+1 is contained in the arc for di,r . This yields a contradiction as each of ej,k and ej,k+1 is adjacent to some vertex on W j colored 3. Let Dap be the first clique distinct from Da encountered when moving in the counterclockwise direction from Da . Observe that Dap must contain a vertex of color 1 or a vertex of color 2 from the circle-gadget (see Claim 3.2). If Dap contains a vertex of color 1, then the arc for ej,k is constrained to be contained in the arc for di,r ; otherwise, the arc for ej,k+1 is constrained to be contained in the arc for di,r . Let Bi be the indices of paths from {W j } whose edges are part of a clique between Ci and Ci+1 in the clockwise direction. Let Bi = {i1 , i2 , . . . , ir }; we may assume r ≥ 1. As each W j contains 8sj − 1 edges whose endpoints are colored 1 and 2, the total number of such edges corresponding to the W j paths whose indices are in Bi is (8si1 − 1) + (8si2 − 1) + · · · + (8sir − 1). It follows that 8(si1 + si2 + · · · + sir ) − r ≤ 8Q − 1. For each si , as si ∈ N and si > Q4 , we have si ≥ Q4 + 14 . Thus, 8Q − 1 ≥ 8(si1 + si2 + · · · + sir ) − r ≥ 8( Q4 + 14 )r − r. Since 8Q − 1 ≥ 8( Q4 + 41 )r − r, we have r ≤ 3. And since 8Q − 1 ≥ 8(si1 + si2 + · · · + sir ) − r, it follows from integrality that (si1 + si2 + · · · + sir ) ≤ Q . Thus, there is a partition of {1, . . . , 3m} into sets S1 , S2 , . . . , Sm such that the sum of the members of {sj | j ∈ Si }, for 1 ≤ i ≤ m, is at most Q . However, as the sum of s1 through s3m is mQ , members of {sj | j ∈ Si }, for 1 ≤ i ≤ m, indeed sum to exactly Q and I admits a 3-partition.  Corollary 3.6. If G(I ) admits a (Helly) proper (resp., unit) circular-arc completion, then I admits a 3-partition.

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Theorem 3.4. Given a 3-colored graph G, deciding each of the following is NP-complete:

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Does G admit a circular-arc completion? Does G admit a Helly circular-arc completion? Does G admit a proper circular-arc completion? Does G admit a Helly proper circular-arc completion? Does G admit a unit circular-arc completion? Does G admit a Helly unit circular-arc completion?

Proof. Since circular-arc graphs [27], Helly circular-arc graphs [12], proper circular-arc graphs [8,15,25], unit circulararc graphs [10], and Helly proper (resp., unit) circular-arc graphs [18–20] can be recognized in polynomial time, these completion problems are in NP. Since 3-partition is NP-complete in the strong sense, the transformation provided above, whose correctness follows from Lemma 3.3, Corollary 3.5 and Lemma 3.4, is a polynomial-time transformation.  Corollary 3.7. The sandwich problems for Helly circular-arc graphs, Helly proper circular-arc graphs, and Helly unit circular-arc graphs are NP-complete. Corollary 3.8. For every fixed k ≥ 3, deciding whether a k-colored graph admits a circular-arc completion, Helly circular-arc completion, proper (resp., unit) circular-arc completion, or Helly proper (resp., unit) circular-arc completion is NP-complete. Proof. Let G be a k-colored graph. Construct graph H from G by adding an isolated vertex v colored k + 1. If H admits a circular-arc completion of a particular type, then G admits a circular-arc completion of the same type: we just have to delete the arc Av corresponding to v . Suppose G admits a circular-arc completion G′ of a particular type. We can obtain a circular-arc completion H ′ of the same type for H by starting with a model for G′ . We choose an arbitrary arc Ax in the model for G′ and take Av identical to Ax .  4. Strongly chordal completions We consider the problem of strongly chordal completion of a colored graph, when the input graph is k-colored. The problem is NP-complete when k is not a constant [7]. When k is fixed and k = 2, the problem is trivial. We present an O(n2 )time algorithm for the problem when k = 3. The chordal completion of a 3-colored graph is well studied and multiple linear-time algorithms are known for it [6,16,17]. Observe that if a 3-colored chordal graph contains a sun, then it must be a 3-sun. Thus, given a 3-colored graph G, the problem of determining whether G admits a strongly chordal completion is equivalent to the problem of determining whether G admits a chordal completion that does not contain a 3-sun. The following is known about chordal completion of colored graphs. Proposition 4.1 ([6,22]). Let G be a (k + 1)-colored graph. G admits a chordal completion if and only if G admits a chordal completion that is a k-tree. We note that the corresponding statement for strongly chordal completion is not true. Construct G as follows: start with the P5 abcde colored with colors 1 and 2 and add vertex f colored 3 so that f is adjacent to each of a, b, c , d, e. Finally, add vertex g colored 3 adjacent only to c. We claim that every way G can be completed to be a 2-tree contains a 3-sun, and hence, G does not admit any strongly chordal completion that is a 2-tree. However, G clearly admits a strongly chordal completion, namely itself. However, when the input graph is bi-connected, the following shows that it is enough to consider completions to 2-trees. Proposition 4.2. If G is a k-connected, (k + 1)-colored, chordal graph with at least k + 1 vertices, then G is a k-tree. Proof. The proof is by induction on the number of vertices. Let G be as in the hypothesis of the proposition. It is well known [9] that a chordal graph always has a simplicial vertex. Let v be a simplicial vertex of G; by definition, N (v) is a clique. If G has k + 1 vertices, then G is a complete graph, and hence, is a k-tree. Now suppose G has more than k + 1 vertices. Observe that, as G is k-connected, |N (v)| ≥ k; on the other hand, as v is simplicial and G is (k + 1)-colored, |N (v)| ≤ k. Therefore, |N (v)| = k. Since a simplicial vertex is not in any minimal cutset of G, H = G \ v is also k-connected. H is also chordal and has at least k + 1 vertices. Now consider a simplicial vertex v ′ in H. Since H is k-connected and has at least k + 1 vertices, |NH (v ′ )| ≥ k; thus, NH [v ′ ] induces a (k + 1)-colored subgraph in G, which implies H is (k + 1)-colored. Applying the inductive hypothesis to H and adding vertex v to H so that it is adjacent to members of NG (v), we see that G is a k-tree.  Let G be a 3-colored graph given as input. It is clear that G can be completed to be strongly chordal if and only if each bi-connected component of G can be completed to be strongly chordal. So, we assume that the input graph is bi-connected. Further, given Proposition 4.2, the problem reduces to testing whether G admits a 2-tree completion that avoids a 3-sun. Therefore, it is necessary that G is a partial 2-tree.

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A tree of cycles [6] is a graph defined recursively as follows: a chordless cycle is a tree of cycles. In this context, a triangle is considered to be a chordless cycle. Given a tree of cycles Tk with k chordless cycles, a tree of cycles Tk+1 with k + 1 chordless cycles can be constructed by adding a new chordless cycle and identifying the end vertices of an edge of the new cycle with the end vertices of an edge in Tk . In a tree of cycles an attach edge is an edge shared by two or more chordless cycles. The cell-completion G′ of a graph G [6] is obtained from G by adding edges between all pairs of vertices {x, y} for which G[V − {x, y}] has at least three components. Given a bi-connected partial 2-tree G, the cell-completion G′ of G is a subgraph of any completion of G to a 2-tree [6]. When the cell-completion of a 3-colored graph G is properly colored, it is called a 3-colored cell-completion of G. Theorem 4.1 ([6]). Bi-connected graph G is a partial 2-tree if and only if its cell-completion G′ is a tree of cycles. Further, the cell-completion of G can be computed in linear time. Observation 4.1. Let G be a bi-connected 3-colored graph. G admits a strongly chordal completion if and only if G has a 3-colored cell-completion that admits a strongly chordal completion. Let G be a bi-connected 3-colored graph. Bodlaender and Kloks [6] give a characterization of when the cell-completion G′ of G admits a chordal completion: (i) G must be a partial 2-tree, and hence, G′ is a tree of cycles, (ii) G′ must be properly colored, and (iii) every chordless cycle in the cell-completion must use all three colors. They also give a linear-time algorithm for the 3-colored graph chordal completion problem [6]. Bodlaender and de Fluiter [3] give a characterization of when the cell-completion G′ of G admits an interval completion: (i) G must be a partial 2-path (i.e., G has pathwidth at most 2), and hence, G′ is a path of cycles (a special linear structure in which each cycle has at most two attach edges), (ii) G′ must be properly colored, and (iii) each cycle in G′ admits a ‘‘compatible’’ interval completion (cf. Corollary 4.2). An O(n2 )-time algorithm is also given for the 3-colored graph interval completion problem [3]. We provide a similar characterization for when the cell-completion G′ of G admits a strongly chordal completion: (i) G must be a partial 2-tree, and hence, G′ is a tree of cycles, (ii) G′ must be properly colored, (iii) each cycle in G′ has at most two attach edges, and (iv) each cycle in G′ admits a ‘‘compatible’’ interval completion (cf. Corollary 4.2). In the case of completion to strongly chordal graphs, the structure of the tree of cycles becomes restricted as well as the completions of the individual chordless cycles. The structure of the tree of cycles lies between a path of cycles and the unrestricted tree of cycles structure for the chordal case. This characterization provides the basis for our O(n2 )-time algorithm to decide whether a 3-colored graph can be completed to be strongly chordal. Observation 4.2. A 3-coloring of a 3-sun is unique (up to naming the colors) and it is not possible to add edges to a 3-colored 3-sun while maintaining chordality and a proper coloring. Let G′ be a 3-colored cell-completion of a bi-connected 3-colored graph G. The following lemma establishes that it is sufficient to complete each chordless cycle of G′ to a strongly chordal graph ensuring that when the completions of chordless cycles are ‘‘glued’’ together at the attach edges of G′ , no 3-sun is created. Lemma 4.1. Let G′ be a 3-colored cell-completion of a bi-connected 3-colored graph G. G′ can be completed to be strongly chordal if and only if G′ can be completed to a strongly chordal graph H by completing each chordless cycle C of G′ to a strongly chordal induced subgraph C ′ of H without creating a 3-sun S in H in which each edge of S is either an attach edge of G′ or has both its endpoints on the same chordless cycle of G′ . Proof. Suppose each chordless cycle C of G′ can be completed to a strongly chordal graph C ′ without creating a 3-sun S in H each of whose edges is either an attach edge of G′ or has both its endpoints on the same chordless cycle of G′ . Then, the endpoints of an attach edge of G′ is a cutset in H. Since an induced cycle cannot pass through a cutset that is a clique, H is chordal. Further, as H cannot have a 3-sun, H is strongly chordal. Now, suppose H is a strongly chordal completion of G′ . Let H − be the graph obtained from H by deleting every edge e such that e is not an attach edge of G′ and the endpoints of e are in different chordless cycles of G′ . For a chordless cycle C in G′ , since H [V (C )] = H − [V (C )], the completion of C in H − is strongly chordal. Suppose a 3-sun S each of whose edges is either an attach edge of G′ or has both its endpoints on the same chordless cycle of G′ has been created as an induced subgraph of H − . Then, as S did not exist in H, some of the deleted edges can be added between the vertices of S so that S is no longer an induced subgraph of H − ; but, by Observation 4.2, this is impossible.  Observation 4.3. Let C be 3-colored chordless cycle. Then, any chordal completion C ′ of C is a triangulation, and hence, is a 2-tree. Further, every edge of C is in a unique triangle in C ′ . In the following Lemmata 4.2, 4.3, and 4.4, and Corollaries 4.1 and 4.2, we assume a yes instance G of the strongly chordal completion of a 3-colored biconnected graph problem. Thus, G is a 3-colored bi-connected graph that is a partial 2-tree (by Proposition 4.2). Let G′ be a 3-colored cell-completion of G. By Theorem 4.1, G′ is a tree of cycles. Let H be a strongly chordal completion of G′ obtained by completing each chordless cycle C of G′ to a strongly chordal induced subgraph of H (see Lemma 4.1). Given a chordless cycle C of G′ , let C ′ = H [V (C )]. By Observation 4.3, C ′ is a triangulation of C . Let B be a cycle (not necessarily induced) in C ′ (B can be C ); let B′ = C ′ [V (B)]. Let A be a triangle in B′ . A is a border triangle of B′ if it includes two edges of B. A is an interior triangle of B′ if it includes exactly one edge of B. An exterior triangle of B′ consists of an edge uv of B, and edges uw and vw in H where w is not on B.

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Lemma 4.2. For a chordless cycle C in G′ , let C ′ be the triangulation of C in H. Let B be a cycle in C ′ and B′ = C ′ [V (B)]. If an edge e = uv of B is in an exterior triangle of B′ , then it is in a border triangle of B′ . Proof. If B has four vertices, then the statement is clearly true; so, assume B has at least five vertices. By way of contradiction, suppose edge e = uv is not in a border triangle of B′ , and hence, is in the interior triangle {u, v, w}. Without loss of generality, let v be on the clockwise side of u. Let B1 be the cycle formed by the clockwise segment of B from v to w and edge vw ; let B2 be the cycle formed by the counterclockwise segment of B from u to w and edge uw . Note that each of B1 and B2 has at least three vertices. Let T1 be the triangle that edge vw is part of in the induced triangulation of B1 and let T2 be the triangle that edge uw is part of in the induced triangulation of B2 . Finally, let T3 be the exterior triangle of B′ that includes uv . Then, the triangles {u, v, w}, T1 , T2 , and T3 induce a 3-sun, a contradiction.  Lemma 4.3. For a chordless cycle C in G′ , let C ′ be the triangulation of C in H. Let B be a cycle in C ′ and B′ = C ′ [V (B)]. If each of the edges uv , vw on B is in an exterior triangle of B′ , then v is adjacent to all the other vertices of B. Further, there cannot be another edge on B in an exterior triangle of B′ . Proof. We prove by induction on number of vertices on B that v must be adjacent to all the other vertices of B. The statement is clearly true when B has three vertices. Consider a cycle B with k ≥ 4 vertices and assume that the statement is true for all smaller cycles. Let u, v , w be the clockwise ordering of those vertices on B. We claim edge uw is not part of B′ ; suppose it is. Let B1 be the cycle formed by the clockwise segment of B from w to u and the edge uw . As B1 has at least three vertices, edge uw is part of a triangle T1 in the induced triangulation of B1 . Let the exterior triangles associated with edges uv and vw be {u, v, y} and {v, w, z }; y and z are distinct since G′ is a tree of cycles. Then, the triangles T1 , {u, v, w}, {u, v, y}, and {v, w, z } induce a 3-sun. Let x be the vertex immediately following u on B in the counterclockwise direction. As uv is in an exterior triangle of B′ , by Lemma 4.2, uv is in a border triangle of B′ . Therefore, as u is not adjacent to w in B′ , v must be adjacent to x in B′ . Let B2 be the cycle formed by the clockwise segment of B from v to x and the edge xv ; clearly, B2 has fewer vertices than B and edges xv and vw are in triangles exterior to B′2 = B′ [V (B2 )]. Applying the inductive hypothesis to B2 , we have that v is adjacent to all the other vertices of B2 . It is now seen that another edge of B that is in an exterior triangle of B′ creates a 3-sun.  Lemma 4.4. For a chordless cycle C in G′ , let C ′ be the triangulation of C in H. Let B be a cycle in C ′ and B′ = C ′ [V (B)]. If B has edge e = uv that is in an exterior triangle of B′ , then there exists a linear ordering of the triangles of B′ such that the ordering starts with a border triangle containing uv , ends with a border triangle, each of the other triangles is an interior triangle, and for each vertex x on B, the triangles that contain x are consecutive in the ordering. Further, if B has edge f ̸= e that is also in an exterior triangle of B′ , then f must be in the terminal border triangle in the ordering. Proof. If B has edges e and f that have a common endpoint and each of e and f is in an exterior triangle of B′ , then the statement follows from Lemma 4.3. We can now assume that no edge of B that shares an endpoint with e is in an exterior triangle of B′ . We prove the statement by induction on the number of vertices on B. The statement is clearly true when B has three or four vertices. Consider a cycle B with k > 4 vertices and assume that the statement is true for all smaller cycles. Let uv be an edge of B that is in an exterior triangle of B′ . By Lemma 4.2, uv is in a border triangle of B′ . Without loss of generality, assume vertex u is adjacent to the vertex w that is adjacent to v in clockwise order on B. Now, consider cycle B1 formed by the clockwise segment of B from w to u and the edge uw . B1 has fewer vertices than B and edge uw of B1 is in an exterior triangle of B′1 = B′ [V (B1 )]. We now apply the inductive hypothesis to B1 to get an ordering of triangles of the induced triangulation of B1 and prepend the triangle {u, v, w} to it (which makes the border triangle of the induced triangulation of B1 that contains edge uw an interior triangle of B′ ) to get the desired ordering for B′ . By Lemma 4.2, an edge f on B that is also in an exterior triangle of B′ must be in the terminal border triangle in the ordering.  Corollary 4.1. For a chordless cycle C in G′ , let C ′ be the triangulation of C in H. Then, C has at most two edges each of which is in an exterior triangle of C ′ . Specifically, a chordless cycle in G′ has at most two attach edges. Proof. The proof follows from Lemmas 4.3, 4.4, and the observation that every attach edge of a chordless cycle C of G′ is in some exterior triangle of C ′ .  Let G′ be a 3-colored cell-completion of a bi-connected 3-colored graph G. A chordless cycle C in G′ is of type-i, i = 0, 1, 2, if C has i attach edges. Corollary 4.2. For a chordless cycle C in G′ , let C ′ be the triangulation of C in H. Then, C ′ is an interval graph that admits a linear ordering of its triangles such that the first and last are border triangles of C ′ , the rest are interior triangles of C ′ , and for every vertex x in C , the triangles containing x are consecutive in the ordering. Further, (i) If C is of type-2 with attach edges e and f , then e is in the first triangle and f is in the last triangle. (ii) If C is of type-1 with the attach edge e, then e is in the first triangle and some edge in E (C ) is in the last triangle. (iii) If C is of type-0 with no attach edges, then some edge in E (C ) is in the first triangle and some edge in E (C ) is in the last triangle.

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Proof. In the case of (i) and (ii), that C ′ is an interval graph with the required properties follows from Lemma 4.4. Now, we prove the same for (iii). If C had three vertices, the statement is clearly true. So, assume C has at least four vertices. Consider strongly chordal completion C ′ of C ; it is a triangulation. Therefore, by the pigeon hole principle, some triangle {u, v, w} of C ′ is a border triangle; let u, v , w be the clockwise ordering of these vertices. Let C1 be the cycle formed by the clockwise segment of C from w to u and edge uw ; observe that edge uw is in an exterior triangle of C1 . Applying Lemma 4.4 to C1 , we get the desired conclusion.  Let C be a chordless cycle of type-i, i = 0, 1, 2, in a 3-colored cell-completion G′ of a bi-connected graph G. A triangulation C of C is a compatible interval completion if C ′ satisfies the part of Corollary 4.2 that applies to a chordless cycle of type-i. ′

Theorem 4.2. Let G′ be a 3-colored cell-completion of a bi-connected 3-colored graph G and H be an edge-supergraph of G′ that is properly colored. Then, H is strongly chordal if and only if, for each chordless cycle C in G′ , C has at most two attach edges and the completion C ′ of C in H is a compatible interval completion. Proof. Suppose G′ admits a strongly chordal completion H and let C be a chordless cycle in G′ . Then, by Corollary 4.1, C has at most two attach edges. By Corollary 4.2, the completion C ′ of C in H is a compatible interval completion. Now, suppose H is an edge-supergraph of G′ that is properly colored such that for every chordless cycle C in G′ , C has at most two attach edges and the completion C ′ of C in H is a compatible interval completion. Note that the endpoints of an attach edge form a clique cutset of H. Therefore, no induced cycle can cross such a cutset and H is chordal. Thus, for any chordless cycle C in G′ , the completion C ′ of C in H is a triangulation. Suppose a 3-sun consisting of the triangles xyz, pxy, qyz, and rxz existed in H. Then, the 3-sun must involve attach edges and vertices from at least two chordless cycles of G′ . Further, the only candidates for attach edges are xy, yz, and xz and at most two of these can be attach edges. Suppose xy is the only attach edge. Since the subgraph induced in H by {x, y, q, r , z } is bi-connected and there is no other attach edge in this subgraph, x, y, q, r , z lie on a single chordless cycle C of G′ . Let C ′ be the triangulation of C in H. It is impossible for the triangle xyz to be a border triangle in C ′ (as either the edges xr and yz will cross or the edges xz and yq will cross), contradicting C ′ being a compatible interval completion. Now, suppose each of xy and yz is an attach edge. Since the subgraph induced in H by {x, y, r , z } is bi-connected and there is no other attach edge in this subgraph, x, y, r , z lie on a single chordless cycle C of G′ . Let C ′ be the triangulation of C in H. In this triangulation with at least two triangles, the attach edges xy and yz are in the same triangle, contradicting C ′ being a compatible interval completion.  The results of this section establish the following theorem, which is the basis for Algorithm stc-completion. Theorem 4.3. Let G be a bi-connected 3-colored graph. The cell-completion G′ of G admits a strongly chordal completion if and only if (i) G is a partial 2-tree (and hence, G′ is a tree of cycles), (ii) G′ is properly colored, (iii) each cycle in G′ has at most two attach edges, and (iv) each cycle in G′ admits a compatible interval completion. Theorem 4.4. Algorithm stc-completion is correct and it can be implemented to run in O(n2 ) time. Thus, deciding whether a 3-colored graph can be completed to be strongly chordal can be done in O(n2 ) time. Proof. Since the cell-completion adds edges that must be in every 2-tree completion of a bi-connected partial 2-tree, if the cell-completion is not properly colored then the answer is no. The rest of the correctness of the algorithm follows from Theorem 4.3. Testing whether G is a partial 2-tree, and if so completing it to a 2-tree H, can be done in linear time [21]. It is shown in [6] that from G and H, the cell-completion G′ of G can be computed in linear time and also a list of the chordless cycles in G′ can be found in linear time. An algorithm to test whether a 3-colored chordless cycle admits a compatible interval completion in O(n2 ) time appears in [3]. The bi-connected components of a graph can be found in linear time using depth-first-search.  We note that from the details in the paper, when a 3-colored graph can be completed to be strongly chordal graph H, the graph H itself can be constructed in O(n2 ) time. 5. Conclusions We establish that when the number of colors is arbitrary the colored graph completion problems for (Helly) circulararc graphs, even (Helly) proper or (Helly) unit circular-arc graphs, are NP-complete. We provide a full classification of the complexities of the problems of completing a colored graph to be a (Helly) circular-arc graph, (Helly) proper circular-arc graph, or (Helly) unit circular-arc graph when the number of colors used is fixed. In the proof that deciding whether a 3-colored graph has a circular-arc completion is NP-complete, the graph that we construct is disconnected. We remark that the problem is unlikely to be easy for connected graphs. We also show that deciding whether a 3-colored graph admits a strongly chordal completion can be decided in O(n2 ) time; our algorithm is based on a characterization of bi-connected 3-colored graphs that admit a strongly chordal completion. We conjecture that the corresponding problem for 4-colored graphs is NP-complete. It follows from our results that the sandwich problems for Helly circular-arc graphs, Helly proper circular-arc graphs, and Helly unit circular-arc graphs are NP-complete.

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Algorithm 1 stc-completion input: Bi-connected 3-colored graph G output: yes if G has a strongly chordal completion, no otherwise if G is not a partial 2-tree then output no; stop else Compute any 2-tree completion H of G end if Use G and H to compute the cell-completion G′ of G if G′ is not properly colored then output no; stop else Construct list L of chordless cycles in G′ end if if some chordless cycle in L has more than two attach edges then output no; stop end if for each chordless cycle C in L do if C does not admit a compatible interval completion then output no; stop end if end for output yes References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29]

C. Àlvarez, J. Díaz, M. Serna, The hardness of intervalizing four colored caterpillars, Discrete Math. 235 (2001) 245–253. C. Àlvarez, M. Serna, On the proper intervalization of colored caterpillar trees, RAIRO-Theor. Inform. Appl. 43 (2009) 667–686. H. Bodlaender, B. de Fluiter, On intervalizing k-colored graphs for DNA physical mapping, Discrete Appl. Math. 71 (1996) 55–77. H.L. Bodlaender, M.R. Fellows, M.T. Hallett, H.T. Wareham, T.J. Warnow, The hardness of perfect phylogeny, feasible register assignment and other problems on thin colored graphs, Theoret. Comput. Sci. 244 (2000) 167–188. H.L. Bodlaender, M.R. Fellows, T.J. Warnow, Two strikes against perfect phylogeny, in: W. Kuich (Ed.), Proc. 19th ICALP, in: Lect. Notes Comput. Sci., vol. 623, 1992, pp. 273–283. H. Bodlaender, T. Kloks, A simple linear time algorithm for triangulating three-colored graphs, J. Algorithms 15 (1993) 160–172. C.M.H. de Figueiredo, L. Faria, S. Klein, R. Sritharan, On the complexity of the sandwich problems for strongly chordal graphs and chordal bipartite graphs, Theoret. Comput. Sci. 381 (2007) 57–67. X. Deng, P. Hell, J. Huang, Linear-time representation algorithms for proper circular arc graphs and proper interval graphs, SIAM J. Comput. 25 (1996) 390–403. G.A. Dirac, On rigid circuit graphs, Abh. Math. Sem. Univ. Hamburg 25 (1961) 71–76. G. Durán, A. Gravano, R.M. McConnell, J. Spinrad, A. Tucker, Polynomial time recognition of unit circular-arc graphs, J. Algorithms 58 (2006) 67–78. M.R. Garey, D.S. Johnson, Computers and Intractability. A Guide to NP-Completness, Freeman, New York, 1979. F. Gavril, Algorithms on circular-arc graphs, Networks 4 (1974) 357–369. P.W. Goldberg, M.C. Golumbic, H. Kaplan, R. Shamir, Four strikes against physical mapping, J. Comput. Biol. 2 (1995) 139–152. M.C. Golumbic, H. Kaplan, R. Shamir, Graph sandwich problems, J. Algorithms 19 (1995) 449–473. P. Hell, J. Bang-Jensen, J. Huang, Local tournaments and proper circular arc graphs, Lecture Notes in Comput. Sci. 450 (1990) 101–108. R. Idury, A. Schaffer, Triangulating three-colored graphs in linear time and linear space, SIAM J. Discrete Math. 2 (1993) 289–293. S.K. Kannan, T.J. Warnow, Triangulating 3-colored graphs, SIAM J. Discrete Math. 5 (1992) 249–258. M.C. Lin, F. Soulignac, J.L. Szwarcfiter, Proper Helly circular arc graphs, in: Proceedings 33rd International Workshop on Graph Theoretic Concepts in Computer Science (WG2007), Jena, Germany, in: Lect. Notes Comput. Sci., vol. 4769, 2007, pp. 248–257. M.C. Lin, F.J. Soulignac, J.L. Szwarcfiter, The clique operator on circular-arc graphs, Discrete Appl. Math. 158 (2010) 1259–1267. M.C. Lin, F.J. Soulignac, J.L. Szwarcfiter, Normal Helly circular-arc graphs and its subclasses, Discrete Appl. Math. 161 (2013) 1037–1059. J. Matousek, R. Thomas, Algorithms finding tree-decompositions of graphs, J. Algorithms 12 (1991) 1–22. F.R. McMorris, T.J. Warnow, T. Wimer, Triangulaing vertex-colored graphs, SIAM J. Discrete Math. 7 (1994) 296–306. F.S. Roberts, Indifference graphs, in: F. Harary (Ed.), Proof Techniques in Graph Theory, Academic Press, New York, 1969, pp. 139–146. R. Sritharan, Chordal bipartite completion of colored graphs, Discrete Math. 308 (2008) 2581–2588. A.C. Tucker, Matrix characterizations of circular-arc graphs, Pacific J. Math. 39 (1971) 535–545. A. Tucker, Structure theorems for some circular-arc graphs, Discrete Math. 7 (1974) 167–195. A.C. Tucker, An efficent test for circular arc graphs, SIAM J. Comput. 9 (1980) 1–24. T.J. Warnow, Combinatorial algorithms for constructing phylogenetic trees (Ph.D. thesis), University of California, Berkeley, 1991. D.B. West, Introduction to Graph Theory, second ed., Prentice Hall, New Jersey, 2001.