Optimality Program in Segment and String Graphs

Planar graphs are known to allow subexponential algorithms running in time $$2^{O(\sqrt{n})}$$ 2 O ( n ) or $$2^{O(\sqrt{n} \log n)}$$ 2 O ( n log n ) for most of the paradigmatic problems, while the brute-force time $$2^{\varTheta (n)}$$ 2 Θ ( n ) is very likely to be asymptotically best on general graphs. Intrigued by an algorithm packing curves in $$2^{O(n^{2/3}\log n)}$$ 2 O ( n 2 / 3 log n ) by Fox and Pach (SODA’11), we investigate which problems have subexponential algorithms on the intersection graphs of curves (string graphs) or segments (segment intersection graphs) and which problems have no such algorithms under the Exponential Time Hypothesis (ETH). Among our results, we show that, quite surprisingly, 3-Coloring can also be solved in time $$2^{O(n^{2/3}\log ^{O(1)}n)}$$ 2 O ( n 2 / 3 log O ( 1 ) n ) on string graphs while an algorithm running in time $$2^{o(n)}$$ 2 o ( n ) for 4-Coloring even on axis-parallel segments (of unbounded length) would disprove the ETH. For 4-Coloring of unit segments, we show a weaker lower bound, excluding a $$2^{o(n^{2/3})}$$ 2 o ( n 2 / 3 ) algorithm (under the ETH). The construction exploits the celebrated Erdős–Szekeres theorem. The subexponential running time also carries over to Min Feedback Vertex Set, but not to Min Dominating Set and Min Independent Dominating Set.


Introduction
Most combinatorial optimization and decision problems admit subexponential algorithms when restricted to planar graphs. More precisely, they can be solved in time 2 O( √ n) , or 2Õ ( √ n) on planar graphs with n vertices, while under the ETH (Exponential Time Hypothesis, which asserts that 3-Sat cannot be solved in subexponential time [22,24]) they do not admit an algorithm running in time 2 o(n) on general graphs. The former is due to the facts that planar graphs have treewidth O( √ n) and that we have efficient algorithms parameterized by the treewidth tw of the graph, namely running in 2 O(tw) n O (1) , or 2Õ (tw) n O (1) .
The so-called bidimensionality theory [10,[12][13][14] pushes this speed-up further by yielding (1) algorithms where k is the targeted size of a solution (think for example of the problems of finding a maximum independent set or a minimum dominating set of size k). In a nutshell, it exploits a deep structural result by Robertson, Seymour, and Thomas [36]: planar graphs with treewidth tw have a Θ(tw)-by-Θ(tw) grid as a minor (i.e., any graph obtained by deleting vertices and edges, and contracting edges). Thus, if the presence of a large grid minor makes the problem trivial (as in, one can always answer yes or always answer no), then one only has to solve efficiently instances with low treewidth; which, as we noted, can often be done. The claimed running time is obtained by defining large grids as Θ( √ k)-by-Θ( √ k), since their absence as minors imply that the treewidth is in O( √ k). The bidimensionality theory is extremely versatile. It also gives approximation schemes and linear vertex kernels and could be generalized to graphs with bounded genus and graphs excluding a fixed minor [11].
A natural line of research is to generalize or extend the subexponential (parameterized) algorithms to classes of graphs which do not fall into those categories. For geometric intersection graphs, the situation is much richer than for planar graphs. For instance, Marx and Pilipczuk already observed that packing problems (of the kind of Max Independent Set) are more broadly subject to subexponential algorithms -running typically in n O( √ k) -than covering problems (of the kind of Min Dominating Set) -for which n O(k) is essentially optimal under the ETH [31,32].
We briefly survey the existing results in the design of subexponential algorithms on geometric intersection graphs. A prominent role is played by intersection graph of families of fat objects, i.e., objects for which the aspect ratio (their length divided by their width) is bounded. We highlight that fat objects, and in particular disks and squares, often allow faster algorithms and the so-called square-root phenomenon. As we will see, subexponential algorithms are less frequent on intersection graphs of curves and segments but nevertheless present such as exemplified by Max Independent Set, 3-Coloring, and Min Feedback Vertex Set.

Subexponential algorithms on geometric intersection graphs.
By a ply of a family of geometric objects we denote the maximum number of objects covering a single point. Smith and Wormald show that for any collection of n convex fat objects with ply p there is a balanced separator of size O( √ np) [40]. This leads to subexponential algorithms when the ply is constant, or in general for problems becoming trivial when the ply is too large, such as k-Coloring. The 2Õ ( √ nk) -time algorithm that this win-win provides for coloring n fat objects, say disks, with k colors is shown essentially optimal under the ETH by Biró et al. [5].
A next step may consist of designing FPT 1 or XP 2 algorithms where the dependency in the parameter is subexponential (for problems of the form "find k vertices such that..."). Using a shifting argument à la Baker [4], Alber and Fiala obtain a n O( √ k) -time to decide if one can find k disjoint unit disks or squares among n [3]. Marx and Pilipczuk generalize this result to packing k disjoint polygons among n in the same time [31,32]. Their approach is based on guessing a small separator in the medial axis (i.e., the Voronoi diagram of polygons) of a supposed solution, as suggested by Adamaszek and Wiese and Har-Peled to obtain QPTAS for geometric packing problems [1,2,21].
Marx showed that Max Independent Set and Min Dominating Set in the intersection graphs of disks or squares are W[1]-complete, and therefore unlikely to be FPT [30]. Those reductions also show that the n O( √ k) algorithms [31,32] are essentially optimal under the ETH. Fomin et al. [17] observed that unit disks of bounded degree have treewidth O( √ n) and used this fact to extend bidimensionality to unit disk graphs for a handful of problems. Recently, a superset of the previous authors gave 2 O( √ k) n O(1) -time algorithms for k-Feedback Vertex Set, k-Path, k-Cycle, Exact k-Cycle [16]. 1 with running time f (k)n O(1) 2 with running time n f (k)

XX:3
Non-fat objects: segments and strings. Segment intersection graphs (or segment graphs in short) are the intersection graphs of straight-line segments in the plane. They are called unit segments if all the segments of a representation share the same length. For a fixed integer k, k-Dir is defined as the set of intersection graphs of segments, each parallel to one of fixed k directions. Strings graphs are the intersection graphs of simple curves in the plane. Those curves can be assumed polygonal without loss of generality. The vertices of the polygonal curves in a geometric representation are called geometric vertices not to confuse them with the actual vertices of the graph. As shown by Kratochvíl and Matoušek, there are string graphs with n vertices, which require 2 Ω(n) geometric vertices in any string representation with polygonal curves [26].
A systematic study of segment graphs and their subclasses was initiated by Kratochíl and Matoušek [27]. It is interesting to point out that every planar graph is a segment graph, as shown by Chalopin and Gonçalves [9] (this was a long-standing conjecture by Scheinerman [39]).
The class of string graphs is very general, as it includes split graphs (i.e., graphs whose vertices can be partitioned into two sets inducing a clique and an independent set), intersection graphs of bodies (i.e., compact shapes with non-empty interior), or incomparability graphs (i.e., graphs whose vertex set is given by the set of elements of a poset, and edges join elements that are incomparable).
Biró et al. showed that even though coloring disks or more generally fat objects with a constant number of colors can be solved in 2Õ ( √ n) [5], 6-coloring axis-parallel segments (2-Dir) in time 2 o(n) would refute the ETH. This suggests that subexponential algorithms are less frequent on the intersection graphs of non-fat objects such as segments and strings. On the other hand, Fox and Pach presented a subexponential algorithm for Max Independent Set on string graphs [18]. Their approach uses a win-win strategy and is based on the existence of balanced separators in string graphs. Fox, Pach, and Tóth showed that string graphs with m edges have balanced separators of size O(m 3/4 log m), and conjectured that there is always a separator of size O( √ m) [20]. Matoušek showed that string graphs admit a balanced separator of size O( √ m log m) [34]. Finally, very recently Lee improved the result of Matoušek, proving the conjecture.
Theorem 1 (Lee [28]). Every string graph with m edges has a balanced separator of size O( √ m). Moreover, it can be found in polynomial time, provided that the geometric representation is given.
Let us point out that this result generalizes the famous planar separator theorem by Lipton and Tarjan [29], as planar graphs are string graphs and the number of edges in a planar graph is linear in the number of vertices. This also shows that Theorem 1 is best possible (up to the constants), as the planar separator theorem is asymptotically tight.

Our contributions.
We show that the subexponential algorithm for Max Independent Set in string graphs by Fox and Pach [18], running in time 2Õ (n 2/3 ) , can be extended to 3-Coloring and Min Feedback Vertex Set. As in the algorithm of Fox and Pach, the central idea is a winwin: either the graph is rather sparse and the separator of Theorem 1 gives a speed-up, or the graph has a high-degree vertex (used for 3-Coloring) or a large biclique (used for Min Feedback Vertex Set) and an efficient branching can be performed. Refining a lower bound of Biro et al. [5], we complement this former result by showing that for any k 4, k-Coloring cannot be solved in 2 o(n) even on axis-parallel segments, unless the ETH fails. The reduction relies on having segment lengths with two different orders of magnitude. We therefore ask if unit segments could allow a faster algorithm for k-Coloring for k 4. Under the ETH, we provide a stronger lower bound than the one for planar graphs (which refutes a running time 2 o( √ n) ) and show that unit segments cannot be k-colored in 2 o(n 2/3 ) for any k 4. Our construction uses the fact, closely related to the famous Erdős-Szekeres [15] theorem, that any permutation on n totally ordered elements can be partitioned into O( √ n) monotone subsequences (see Brandstädt and Kratsch [7]).
We then give tight ETH lower bounds for Min (Connected) Dominating Set and Min Independent Dominating Set on segments and Max Clique on strings. For that, we design reductions whose number n of produced segments is linear in N + M from satisfiability problems with N variables and M clauses. Indeed, the sparsification lemma of Impagliazzo et al. [23] implies that those satisfiability problems are not solvable in 2 o(N +M ) unless the ETH fails; which enables us to conclude that the problems are not solvable in 2 o(n) under the ETH, on graphs with n vertices.
Although the NP-hardness of the aforementioned problems is known for segment intersection graphs [8,42], getting such linear reductions might be difficult.
For instance, while it is known that planar graphs are a subclass of segment intersection graphs [9], implying the NP-hardness of all the problems of Table 1 except k-Coloring for k 4 and Max Clique, this fact does not serve our purpose since they can be solved in time 2 O( √ n) on planar graphs. The situation is an interesting intermediate between planar and general graphs. Our objects can intersect but we cannot afford crossover gadgets (at least not quadratically many). Certain intersections create unwanted edges of which we have to tame the importance. It is also noteworthy that segment/string graphs cannot be expanders since if they have constant degree, by Theorem 1, they have treewidthÕ( √ n). Hence, we are deprived of the usual hardest instances.

Problem
Upper bound Lower bound  Table 1 Upper and lower bounds for classical problems on string and segment graphs. The upper bounds work on string graphs. The lower bounds are designed on segment graphs, unless precised otherwise. Our results are written in bold. By p we denote the number of geometric vertices if a geometric representation is given.

Geometric representation and robust algorithms.
In case of graphs with geometric representations, it is important to distinguish between a geometric intersection graph (i.e., a pure abstract structure, for which we know that some geometric representation exists), and the representation itself. Note that this is not the case with planar graphs, as finding a plane embedding can be done in linear time [6].
Finding a segment or string representation of a graph was shown to be NP-hard by Kratochvíl [25], and Kratochvíl and Matoušek [27], respectively. However, it was very unclear if the problems are in NP (which is usually the trivial part of an NP-completeness proof). As mentioned above, Kratochvíl and Matoušek [26] showed that some string graphs require a representation of exponential size, which proved that the simple idea of exhaustively guessing the representation cannot work for this problem. Finally, the NP-membership of recognizing string graphs was proven by Schaefer, Sedgwick, and Štefankovič [37].
The story of recognizing segment graphs is even more interesting. On the first sight, the situation seems simpler than for strings, as the number of geometric points in a segment representation is clearly polynomial in n. However, it appears that there are segment graphs, whose every segment representation requires points with coordinates doubly exponential in n, i.e., using 2 Ω(n) digits (see Kratochvíl and Matoušek [27], and McDiarmid and Müller [35]). Finally, the problem was shown to be complete for the class ∃R (see Schaefer and Štefankovič [38]), i.e., the class of problems reducible in polynomial time to deciding if a given existential formula over the reals is true. This is a strong evidence that the problem is not in NP. For a very nice exposition of the ∃R-completeness proof, see Matoušek [33].
All this shows that a requirement of an explicit geometric representation of an input graph may be a serious drawback of an algorithm. We call an algorithm robust if it takes only an abstract structure as an input, and either computes the solution, or concludes (correctly) that the input graph does not belong to the desired class. On the one hand, our algorithms (see Section 2) are robust, but work slightly faster if the input is given along with the geometric representation. On the other hand, the lower bounds (see Section 3) hold even if the geometric representation is given explicitly.

Upper bounds
Fox and Pach showed that, on string graphs, a maximum independent set can be computed in subexponential time: [18]). Max Independent Set can be solved in time 2 O(n 2/3 log n) in string graphs with n vertices.
In their paper, they give a worse running time than the one claimed above. This is because they used the O(m 3/4 log m) separator theorem [20], which has been recently improved to O( √ m) [28]. The algorithm is a simple win-win argument. If there is a vertex with degree at least n 1/3 , then either removing it or selecting it and removing its neighbors gives a branching F (n) . Otherwise, if all the vertices have degree smaller than n 1/3 , the graph is rather sparse and the balanced separator of size O( √ m) = O(n 2/3 ) provides an efficient divide-and-conquer. The threshold n 1/3 is computed so that it balances the running time of those two subroutines and gives the claimed overall asymptotic time.
This result was somewhat improved by Marx and Pilipczuk [31,32] based on an approach introduced by Adamaszek, Har-Peled, and Wiese [1] to get QPTAS for geometric problems. However, their algorithm necessitates that the string graph is given with a representation by polygonal curves on a polynomial number of geometric vertices. (1) in string graphs with n vertices, where the strings are given as polygonal curves on a total of p geometric vertices.

Theorem 3 (Marx & Pilipczuk [31]). Max Independent Set can be solved in time
In a nutshell, the idea is to exhaustively guess a small balanced face-separator in the Voronoi diagram of a supposed (although not known) fixed solution, and solve recursively the two subinstances in the inside and outside of this separator.
If this approach does not seem to generalize easily to coloring problems, the win-win of Fox and Pach can be transported to 3-Coloring with a bit more arguments. 2 O(n 2/3 log n) , if the geometric representation is given, 2 O(n 2/3 log 2 n) , even without geometric representation.
and we want to find a proper coloring, in which every vertex receives a color from its list). Observe that without loss of generality we can assume that each list has two or three elements. Indeed, if there is a vertex with just one allowed color, we can fix this color and remove it from the list of each of its neighbors. Let N be the sum of the lengths of the lists; clearly 2n N 3n. First, assume that G has no vertex with degree larger than n 1/3 , then the number m of edges is O(n 4/3 ). By Theorem 1, G has a balanced separator of size O( √ m) = O(n 2/3 ). We can find this separator in polynomial time, if the representation is given, or by exhaustive guessing in time n O(n 2/3 ) = 2 O(n 2/3 log n) , without using a representation. Then we list all possible colorings of the separator and proceed with a standard divide-and-conquer approach. The depth of the recursion is O(log n), so the total time complexity of this step is 2 O(n 2/3 log 2 n) , or 2 O(n 2/3 log n) if we use the geometric representation to find a separator.
If there is a vertex v of degree at least n 1/3 , then one among the lists: 2, 3} appears on at least n 1/3 /4 of its neighbors. Thus there are two colors (say, 1 and 2) that appear in lists of at least n 1/3 /4 of neighbors of n. Since the list of v has size at least two, one of these colors (say 1) appears on the list of v. We branch into two possibilities: choosing the color 1 for v (then we exclude 1 from the lists of all neighbor of v), and not choosing 1 for v (then we remove 1 from the list of v). The complexity F of this step is given by the recursion Combining these two cases gives the claimed time complexity. Finally, observe that if the input graph is not a string graph, then the exhaustive search for a separator might fail, and then we can report a wrong input instance.
For Min Feedback Vertex Set, there is no obvious subexponential branching on a high-degree vertex. Instead, we use the following theorem by Lee.
Theorem 5 (Lee [28]). There is a constant c such that for any t 1, K t,t -free string graphs on n vertices have fewer than c · t log t · n edges.
It is worth mentioning that Fox and Pach [19, Theorem 5] obtained a slightly weaker result with log O(1) t instead log t. Proof. The proof is similar to the proof of Theorem 4. Let G be the input graph. If G has fewer than c/3 · n 4/3 log n edges (where c is a constant from Theorem 5), then by Theorem 1 there is a balanced separator of size O( √ m) = O(n 2/3 log 1/2 n), and a divideand-conquer approach yields a running time 2 O(n 2/3 log 3/2 n) (with the representation, we use it to find the separator), or 2 O(n 2/3 log 5/2 n) (without the representation, we guess the separator exhaustively).
Otherwise, by Theorem 5, there is a subgraph of G isomorphic to the biclique K n 1/3 ,n 1/3 .
We can find it by exhaustive guessing in time n n 2/3 · poly(n) = 2 O(n 2/3 log n) . Observe that any feedback vertex set of G must contain all but one vertex of one bipartition class of the biclique. Guessing which vertex is not chosen into the solution gives us a branching algorithm, whose complexity is given by the recursion F (n) 2 O(n 2/3 log n) + 2n 1/3 F (n − n 1/3 + 1), which is solved by F (n) = 2 O(n 2/3 log n) . If the exhaustive search for a separator or a biclique fails, then we can correctly report that the input graph is not a string graph.

Lower bounds
Rather surprisingly, the win-win for 3-Coloring abruptly ceases to work for k-Coloring for every k 4. First, let us consider the List 4-Coloring. Following Kratochvíl and Matoušek [27], by Pure 2-Dir we denote graphs admitting a 2-Dir representation in which parallel segments do not intersect. Observe that such a graph is bipartite. Proof. Let Φ be a 3-Sat formula with n variables v 1 , v 2 , . . . , v n and m clauses C 1 , C 2 , . . . , C m . By repeating some literals in a clause, we may assume that each clause contains exactly three literals. For a clause C i , let v i 1 , v i 2 , v i 3 denote the variables of C i . We construct a 2-Dir graph G with lists L of colors from the set {1, 2, 3, 4}, such that Φ is satisfiable if and only if G is list-colorable with respect to the lists L.
For each variable v i , we introduce a horizontal segment called x i . For each clause C i we introduce three vertical segments y i 1 , y i 2 , y i 3 , corresponding to v i 1 , v i 2 , and v i 3 , respectively. We arrange them in a grid-like way (see Figure 1). One may observe that the intersection graph induced by those segments is a biclique. We set the lists of each x i to {1, 2} and the lists of each y i 1 , y i 2 , y i 3 to {3, 4}.
The colors 1 and 2 used for coloring x i will be interpreted, respectively, as true and false values given to v i , while the colors 3 and 4 given to y i j will be interpreted, respectively, as true and false values given to the literal corresponding to v i j . To ensure this, we need to introduce equality gadgets and inequality gadgets. If the variable v i appears positively in the clause C j as its -th literal, then at the crossing point of x i and y j we put the equality gadget ensuring that in any feasible coloring of G, the color of x i is 1 (2, respectively) if and only if the color of y i is 3 (4, respectively). On the other hand, if v i appears negatively in C j as its -th literal, then at the crossing point of x i and y j we put the inequality gadget ensuring that in any feasible coloring of G, the color of x i is 1 (2, respectively) if and only if the color of y j is 4 (3, respectively).
The equality gadget consists of 3 segments, arranged as depicted on Figure 2. Consider the equality gadget (left lists on Figure 2) and suppose x i gets the color 1. Then a receives color 3, and c gets the color 4. Thus the only choice for the color for y j is 3. The coloring can be extended by coloring b to 2. The other cases are symmetric. The inequality gadget is analogous and uses the right lists on Figure 2.
Lists in the equality gadget.
vertex list Lists in the inequality gadget.

Figure 2
Equality and inequality gadgets. The arrangement of segments is the same in both gadget, the only difference is the lists.
The only thing left is to ensure that the coloring of y j 1 , y j 2 , y j 3 exists if and only if C j is satisfied. This is ensured by the satisfiability gadget depicted in Figure 3, attached to the top ends of y 1 j , y j 2 , and y j 3 . Note that the gadget can be colored if and only if one of y j 1 , y j 2 , y j 3 gets color 3, which is equivalent to one literal of C j being set to true. The non-list version is obtained analogously to the hardness for 6-Coloring in [5]. We include the proof to make the paper self-contained. Proof. We modify the construction from the proof of Theorem 7. We first introduce k overlapping segments R 1 , R 2 , . . . , R k , whose coloring will serve as a reference coloring. Since these segments are pairwise intersecting, each of them receives a different color. We will denote by i ∈ [k] the color assigned to R i . Now, for each segment v of G, we want to simulate the list L(v) from the instance of List 4-Coloring constructed in the proof of Theorem 7. For every color i / ∈ L(v), we want to introduce a segment s i intersecting v, which will always receive color i.
To achieve this, we first need to transport the reference coloring to every gadget. We split it into two parts -we will separately transport colors 1 and 2, and colors greater than 2. The overall high-level idea is depicted in Figure 4. Observe that this already simulates the lists for every x i . Such a construction relies on a constant-size gadget, which allows us to turn or split the reference coloring. The construction of this gadget is depicted in Figure 5. Note that the number of segments in this gadget is constant if k is constant. Moreover, turning or splitting the reference coloring of fewer than k colors can be obtained by a simple adaptation of the turning gadget. Indeed, suppose we want to introduce a turning gadget for the set of colors C ⊆ [k], with |C| = k < k and we have k overlapping segments carrying these colors. We introduce k − k dummy segments, overlapping these segments. The dummy segments will clearly receive colors from [k] \ C, but we do not know which segment will get which color. Now we introduce a turning gadget for k colors. We know which segments leaving the turning gadget get colors in C (and we precisely know which segment gets which color). We do not need he remaining segments anymore, so we can finish them as soon as they leave the turning gadget. Now, the only thing left is to connect every segment in every gadget to appropriate segments carrying the reference coloring (note that each y j belongs to some (in)equality gadget). This can easily be done using a constant number of additional segments per gadget: we introduce a turning gadget for k colors, and finish the segments that are not needed anumore before they intersect the segments in gadgets (see Figure 6).
The total size of the construction increases by a constant factor, as we introduce O(n)  Observe that the construction in the proof of Theorem 7 cannot be performed with segments of bounded lengths, since segments x i and y j k need to have length O(n) (while the segments inside the gadgets can have unit length). For unit segments, we show the following weaker lower bound. Proof. Consider a 3-Sat instance with variables v 1 , v 2 , . . . , v n and clauses C 1 , C 2 , . . . , C m , where m = Θ(n). By duplicating some literals if necessary, we may assume that each clause contains exactly three literals.
Let us start with a reduction to List k-Coloring in unit 2-Dir. For each clause C j we introduce three vertical unit segments, each corresponding to one literal in C j . These segments will be called literal segments. We place them in such a way that the distance between the leftmost and the rightmost literal segment is slightly smaller than 1/2. The ordering of the segments is the following: first, the literal segments corresponding to the clause C 1 , then literal segments corresponding to the clause C 2 and so on. Moreover, they are slightly shifted vertically, so that the y-coordinates of their top endpoints form an increasing sequence. We set the list of possible colors for each literal segment to {3, 4} and for each three literal segments corresponding to a single clause, we introduce a satisfiability gadget already shown in Figure 3. We will interpret 3 as assigning the value true to the particular literal (and 4 will correspond to false). Figure 7 shows how the placement of literal segments and satisfiability gadgets. Analogously, for each variable v, we introduce a vertical segment for each occurrence of v (we call these segments occurrence segments) with list {3, 4}. The segments are placed in such a way that the distance between the leftmost and the rightmost occurrence segment is slightly smaller than 1/2. Moreover, leftmost occurence segments correspond to v 1 , then we put the segments for v 2 etc. For a variable v i for i ∈ [n], we introduce a horizontal segment, intersecting the occurrence segments of variables v i , v i+1 , . . . , v n . These segments will be called variable segments. The variable segments are pairwise disjoint and each has list {1, 2}. Again, we will interpret 1 as the value true given to a variable, and 2 will denote false. Now we need to ensure that the truth assignment defined by the coloring of occurrence segments is consistent. For this, we will use a slightly modified version of the (in)equality segment introduced in Figure 2. The modified gadget is shown in Figure 8. For a variable v and its positive occurrence, we introduce an equality Lists in the equality gadget.
vertex list Lists in the inequality gadget. gadget joining the occurrence segment and the variable segment. Analogously, we introduce an inequality gadget joining the variable segment and the occurrence segment corresponding to a negative occurrence. The placement of all these segments is shown in Figure 9.
Now we need to make sure that the truth assignment given by the coloring of literal segments is consistent with the truth assignment given by the coloring of occurrence segments. We will do it in a very similar way as we synchronized the colorings of occurrence We observe that this is quite easy to do if σ is either increasing or decreasing. As shown by Brandstädt and Kratsch [7], each permutation of [3m] can be partitioned into at most z := 6m + 1/4 − 1/2 = O( √ n) monotone sequences. Their proof is constructive and can be easily transformed into a polynomial algorithm finding such a partition. Let σ 1 , σ 2 , . . . , σ z be the partition of σ, where each σ i is monotone. We introduce z layers, each corresponding to one σ i . The ith layer is responsible to synchronize the colorings of literal segments and occurrence segments corresponding to elements of σ i (we call such segments important for layer i). Figure 10 shows a single layer. Note that each layer contains copies of all literal segments and occurrence segments. They appear in two groups -literal segments on the left and occurrence segments on the right. The distance between leftmost and rightmost segment in one group is slightly less than 1/2, and the distance between the rightmost literal segment and the leftmost occurrence segment is slightly less than 2. This allows us to fit two equality gadgets, whose horizontal segments are collinear but non-overlapping (see Fig. 10 and notice that we can adjust the distances within groups and between the groups, so that the distance between the horizontal segments is always positive and smaller than 1). For each such a pair we introduce k − 1 overlapping segments with lists {1, 2, . . . , k}, intersecting both of them and nothing else. literal segments occurrence segments The last thing to do is to connect the literal segments (with (in)equality gadgets), layers, and occurrence segments with satisfiability gadgets. We place the occurrence segments at the bottom, and then we introduce layers in such a way that the corresponding vertical segments are collinear but non-intersecting. Finally, the literal segments with satisfiability gadgets are put on top. Let s 1 and s 2 be collinear vertical segments and let s 1 be above s 2 . If s 1 is an important literal segment of the ith layer, then there are two segments a, b belonging to the appropriate equality gadget, which are covering the lower endpoint of s 1 . We introduce k − 3 vertical segments q 1 , q 2 , . . . , q k−3 , intersecting only s 1 ,a,b (so its upper endpoint is below the horizontal segment of the equality gadget, where a and b belong), and set their lists to {1, 2, . . . , k}. We adjust the distance between the layers, so that a, b, and q intersect s 2 , but no other vertex of its layer.
The situation is analogous if s 2 is an important occurrence segment of ith layer. Finally, if none of s 1 , s 2 is an important segment of a layer, we introduce k − 1 overlapping segments q 1 , q 2 , . . . , q k−1 with lists {1, 2, . . . , k}, intersecting s 1 , s 2 , and no other previously constructed segment. This way s 1 and s 2 are non-adjacent, but they are both intersected by three pairwise intersecting segments. This ensures that s 1 and s 2 will receive the same color. The If we want to obtain a reduction to the non-list k-Coloring for any k 4, we need to transport the reference coloring to each gadget. However, this cannot be done for segments a, b in our (in)equality gadgets (recall Theorem 8 and Figure 8), which have non-trivial lists and are fully covered by other segments -note that if a segment in color 4 intersects a, it will also intersect x (note that the segments with lists {1, 2, . . . , k}, even if they are fully covered, are not problematic since they do not need the reference coloring). But if we use a third direction, we can make a, b intersect x and y (again, we use notation in Figure 8) and no other vertex with non-trivial list in our grid-like structure -it is enough to choose their slope to be "almost vertical" (see Fig. 11). This shows the claimed lower bound, for every k 4, for k-Coloring of unit 3-Dir graphs and completes the proof.
We show that on segment graphs, Min Dominating Set, Min Connected Dominating Set, and Min Independent Dominating Set are unlikely to admit a subexponential algorithm. Put M pairwise-disjoint small segments on a circle as shown in Figure 12; M slightly perturbed points work as long as no two pairs define the same direction. Each small segment s(C j ) represents a distinct clause C j . For each literal σx i , where x i is one of the N variables appearing in φ and σ ∈ {∅, ¬}, we add a segment s(σx i ) crossing only the two small segments s(C j ) and s(C j ) corresponding to the clauses this literal satisfies. We prolong all the segments corresponding to literals to make them pairwise intersect.
For each pair of literals x i , ¬x i , we add a small segment s(i) near the intersection of s(x i ) and s(¬x i ) which intersects only s(x i ) and s(¬x i ). This finishes the construction (see Figure 12). Note that the total number of segments is n := 3N + M = Θ(N ). We claim that there is a dominating set of size N in the intersection graph if and only if φ is satisfiable.
Indeed, to dominate all the segments s(i) for i ∈ [N ], one has to take exactly one of s(x i ) and s(¬x i ). The choice of the literals will only dominate all the segments s(C j ) for j ∈ [M ] if the chosen dominating set corresponds to a satisfying assignment. The reverse direction is straightforward.
Moreover, as every pair of segments representing literals intersects, the dominating set encoding the satisfiable assignment is connected (it even induces a clique). Figure 12 An example with 9 clauses and 6 variables. The segments σ(xi) and σ(¬xi) can be inferred from σ(i) or from the colors (although we only specified which is which for variable x1). Observe that the light blue and purple pairs do not intersect inside the circle so those segments should be prolonged outside it until they meet (this part is not shown in the picture).
The hardness proof for Min Independent Dominating Set has to be quite different and is trickier. Proof. We reduce from 3-SAT on N variables and M clauses, where each literal appears at most four times, and each clause contains exactly three literals. This restriction was shown NP-complete by Tovey [41]. We can further assume that each literal appears at most three times. Indeed, if the same literal appears four times, then by definition its negation cannot appear in the whole instance. Hence, the variable can be set so as to satisfy the four corresponding clauses.
The variable gadget G(x i ) for variable x i consists of three parallel segments T i representing positive occurrences crossing three parallel segments F i representing negative occurrences. Those two sets of segments intersect three dummy pairs of parallel segments as shown on Figure 13. Note that even if a literal appears strictly fewer than three times, we keep exactly three parallel segments to encode it.
The clause gadget G(C j ) for the clause C j consists of three pairwise crossing segments, drawn in blue in Figure 13, each of which crosses one red segment. Each pair of blue and red segment corresponds to one of literals of C j . Additionally, all blue segments are intersected by four parallel dummy segments, and each pair of crossing blue and red segment has a private segment, crossing both of them. Now, for every literal (¬)x i belonging to the clause C j , we add a new segment denoted by (¬)x i ∈ C j . This segment crosses one non-dummy segment in G(x i ): a segment of T i if the literal is positive, and a segment of F i otherwise. Moreover, it crosses one literal (red) segment from G(C j ), and no other segments in variable and clause gadgets, see Figure 14. Those lastly introduced segments cross exactly once each literal segment and at most once each segment of i∈[n] T i ∪ F i .
We claim that such a constructed graph has an independent dominating set of size at most 3N + 3M if and only if the initial formula is satisfiable.
First, suppose that A is a satisfying assignment. If x i is set to true by A, we select the three segments of T i in the solution, otherwise we select the three segments of F i . Now consider a clause C j . Since A is satisfying, it contains at least one true literal. In G(C j ) we select the blue segment corresponding to a true literal and the red segments corresponding to the other two literals. Note that this way we select 3N + 3M segments and the selected set dominates all segments from all gadgets. Let us consider a segment (¬)x i ∈ C j . Either it is dominated by one of the selected red segments in G(C j ). Or it corresponds to a true literal of C j , so it is dominated by a selected segment in G(x i ).
On the other hand, assume that there is an independent dominating set S of size at most 3N + 3M . Notice that in order to dominate all dummy segments in G(x i ), we need to select at least three segments from G(x i ), and if we want to select exactly three, we need to choose either all segments in T i , or all segments in F i . Analogously, we need to select at least three

Figure 14
The overall picture. We only represented two clauses: C3 = ¬x2 ∨ x3 ∨ x5 and C4 = ¬x1 ∨ ¬x2 ∨ ¬x5. segments from each G(C j ), and if we want to select exactly three, we need do choose one of blue segments and thus we cannot choose its corresponding red segment. Note that since the total size of S is 3N + 3M , we need to select exactly three segments in each gadget, and no segment (¬)x i ∈ C j is selected.
We define the assignment A as follows: if segments T i are in S, then x i is set true, and otherwise x i is set false. Suppose that C j is not satisfied by A, which means that all its literals are false. This means that the three segments joining appropriate variable gadgets with G(C j ) are not dominated by segments in variable gadgets, so S must contain all three red segments from G(C j ). However, this way the horizontal segments from G(C j ) are not dominated, a contradiction.
The total number of segments is bounded by 12N + 13M + 3M = O(N + M ), so the claim holds. Proof. We reduce from 3-Sat with a linear number of clauses, where every clause contains exactly three literals. Let φ be an instance with N variables and M = Θ(N ) clauses. For any positive integers p and s, the co-cluster CC p,s = K s,s,...,s (p times) can be represented as in Figure 15. We encode the N variables by 2N curves representing true and false for each variable by a co-cluster CC N,2 and the M = Θ(N ) clauses by 3M curves each representing a distinct literal in a clause by a co-cluster CC M,3 . There are in total n := 2N + 3M = Θ(N ) curves. We make the 2N variable curves intersect the 3M literal curves in a grid-like way. They form an almost complete biclique K 2N,3M where 3M edges are removed.
More precisely, the literal curve c(l j i ) (j ∈ [M ], i ∈ [3]) intersects every variable curve but c(σx k ) (k ∈ [N ]) encoding the k-th variable with sign σ ∈ {∅, ¬} for which l j i and σw k are opposite literals (see Figure 16). It is easy to observe that there is a clique of size N + M if and only if φ has a satisfying assignment.   Figure 16 The representation of 3 clauses: x3 ∨ x1 ∨ ¬x6, x5 ∨ ¬x4 ∨ ¬x2, and ¬x3 ∨ x4 ∨ ¬x6.

Perspectives
We have started a generalized optimality program on segment and string graphs for the most principal graph problems. On the algorithmic side, we extended a subexponential algorithm for Max Independent Set on string graphs [18] to two other problems: 3-Coloring and Min Feedback Vertex Set. On the complexity side, we showed that no subexponential algorithm is likely for, among others, 4-Coloring, Min Dominating Set, and Min Independent Dominating Set. It is quite easy to obtain such lower bounds for string graphs. Extending those results to segments requires more ingenuity, and even more so when it comes to unit segments. A handful of questions remains unsettled. Can we improve the algorithm or give tight ETH lower bounds for the following problems: Max Independent Set without geometric representation, 3-Coloring, and Min Feedback Vertex Set on segments/strings? Can we show for Max Clique the same lower bound for segment graphs as we have for string graphs. The mere NP-hardness of Max Clique on segments [8] answered a long-standing open question. Hence, it is likely that getting a tight ETH hardness will be difficult. We would also find interesting to have, for a certain problem, an algorithm for segments (resp. unit segments) which beats the ETH lower bound on strings (resp. segments). So far, we only have candidate problems for such a "separation".
Finally, another natural continuation of this work is to determine which fixed-parameter tractable problems can be solved in time O * (2Õ (k 2/3 ) ) or O * (2Õ ( ) intersections) on string graphs due to the linear kernel yielding an equivalent instance on 2k vertices and the algorithm for Max Independent Set. The latter problem can be solved in n O( √ k) in segments or more generally in polygons of polynomial complexity [31], while Min Dominating Set on string graphs cannot be solved in time f (k)n o(k) , for any computable function f , unless the ETH fails (since this lower bound holds for split graphs).