On the chromatic number of disjointness graphs of curves

Let $\omega(G)$ and $\chi(G)$ denote the clique number and chromatic number of a graph $G$, respectively. The {\em disjointness graph} of a family of curves (continuous arcs in the plane) is the graph whose vertices correspond to the curves and in which two vertices are joined by an edge if and only if the corresponding curves are disjoint. A curve is called {\em $x$-monotone} if every vertical line intersects it in at most one point. An $x$-monotone curve is {\em grounded} if its left endpoint lies on the $y$-axis. We prove that if $G$ is the disjointness graph of a family of grounded $x$-monotone curves such that $\omega(G)=k$, then $\chi(G)\leq \binom{k+1}{2}$. If we only require that every curve is $x$-monotone and intersects the $y$-axis, then we have $\chi(G)\leq \frac{k+1}{2}\binom{k+2}{3}$. Both of these bounds are best possible. The construction showing the tightness of the last result settles a 25 years old problem: it yields that there exist $K_k$-free disjointness graphs of $x$-monotone curves such that any proper coloring of them uses at least $\Omega(k^{4})$ colors. This matches the upper bound up to a constant factor.


Introduction
Given a family of sets, C, the intersection graph of C is the graph, whose vertices correspond to the elements of C, and two vertices are joined by an edge if the corresponding sets have a nonempty intersection. Also, the disjointness graph of C is the complement of the intersection graph of C, that is, two vertices are joined by an edge if the corresponding sets are disjoint.
As usual, we denote the clique number, the independence number, and the chromatic number of a graph G by ω(G), α(G) and χ(G), respectively.

Families of curves
A curve or string in R 2 is the image of a continuous function φ : [0, 1] → R d . A curve C ⊂ R 2 is called x-monotone if every vertical line intersects C in at most one point. Note that any convex set can be approximated arbitrarily closely by x-monotone curves, so the notion of x-monotone curve extends the notion of convex sets. We say that C is grounded at the curve L if one of the endpoints of C is in L, and this is the only intersection point of C and L.
A grounded x-monotone curve is an x-monotone curve that is contained in the half-plane {x ≥ 0}, and whose left endpoint lies on the vertical line {x = 0}. It was first suggested by Erdős in the 1970s, and remained the prevailing conjecture for 40 years, that the family of intersection graphs of curves (the family of so-called "string graphs") is χ-bounded [3,21]. There were many promising facts pointing in this direction. Extending earlier results of McGuinness [28], Suk [37], and Lasoń et al. [27], Rok and Walczak [35,36] proved the conjecture for grounded families of curves. Nevertheless, in 2014, Pawlik et al. [34] disproved Erdős's conjecture. They managed to modify Burling's above-mentioned construction to obtain a sequence of finite families of segments in the plane whose intersection graphs, G n , are triangle-free (that is, ω(G n ) = 2), but their chromatic numbers tend to infinity, as n → ∞.
Recently, Pach, Tardos and Tóth [32] proved that the family of disjointness graphs of curves in the plane is not χ-bounded either; see also [30]. However, the situation is different if we restrict our attention to x-monotone curves. It was shown in [33,26] that the family of disjointness graphs of x-monotone curves in the plane is χ-bounded with a bounding function f (k) = k 4 . For grounded x-monotone curves, the same proof provides a better bounding function: f (k) = k 2 . These results proved 25 years ago were not likely to be tight. However, in spite of many efforts, no-one has managed to improve them or to show that they are optimal.

Our results
The aim of the present paper is to fill this gap. We proved, much to our surprise, that the order of magnitude of the last two bounds cannot be improved. In fact, in the case of grounded x-monotone curves, we determined the exact value of the best bounding function for every k ≥ 2. To the best of our knowledge, this is the first large family of non-perfect geometric disjointness graphs, for which one can precisely determine the best bounding function.
Theorem 1. Let G be the disjointness graph of a family of grounded x-monotone curves.
It turns out that disjointness graphs of grounded x-monotone curves can be characterized by two total orders defined on their vertex sets that satisfy some special properties. This observation is the key idea behind the proof of the above two theorems.
The disjointness graph of any collection of x-monotone curves, each of which intersects a given vertical line (the y-axis, say), is the intersection of two disjointness graphs of grounded x-monotone curves. The methods used for proving Theorems 1 and 2 can be extended to such disjointness graphs and yield sharp bounds.
Theorem 3. Let G be the disjointness graph of a family C of x-monotone curves such that all elements of C have nonempty intersection with a vertical line l. If ω(G) = k, then 3 . Theorem 4. For every positive integer k ≥ 2, there exists a family C of x-monotone curves such that all elements of C have nonempty intersection with a vertical line l, the disjointness graph G of C satisfies ω(G) = k, and χ(G) = k+1 2 k+2 3 . As we have mentioned before, according to [33,26], k 4 is a bounding function for disjointness graphs of any family of x-monotone curves. Theorem 4 implies that the order of magnitude of this bounding function is best possible. Actually, we can obtain a little more.
Theorem 5. For any positive integer k, let f (k) denote the smallest m such that any K k+1 -free disjointness graph of x-monotone curves can be properly colored with m colors. Then we have Here the lower and upper bounds differ by a factor of less than 6, and there is some hope that one can determine the exact value of f (k). The lower bound follows directly from Theorem 4.
Our paper is organized as follows. In Section 2, we prove Theorem 1 and the upper bound in Theorem 5. The existence of the graphs satisfying Theorem 2 is proved in Section 3, using probabilistic techniques. The proofs of Theorems 3 and 4 are presented in Sections 4 and 5, respectively. The last section contains open problems and concluding remarks.
Proof. Fix an ordering < of V (G) such that G < is a semi-comparability graph. For every v ∈ V (G), let f (v) denote the size of the largest clique with minimal element v. Then The main observation is that G[V i ] is a partial order. Indeed, suppose to the contrary that there exist 3 vertices a, b, c ∈ V i such that a < b < c and ab, bc ∈ E(G), but ac ∈ E(G). Let C ⊂ V (G) be a clique of size i with minimal element c. If d ∈ C \ {c}, then b and d must be joined by an edge, otherwise the quadruple a, b, c, d satisfies the conditions ab, bc, cd ∈ E(G) and ac, bd ∈ E(G). Thus, b is joined to every vertex in C by an edge, which means that C ∪ {b} is a clique of size i + 1 with minimal element b, contradicting our assumption that b ∈ V i .
Hence, every G[V i ] is a partial order. Using the fact that G[V i ] does not contain a clique of size i + 1, by Mirsky's theorem [7] we obtain that 2 , as required.
The combination of Lemmas 7 and 8 immediately implies Theorem 1. Next, we prove the upper bound in Theorem 5.

54:5
Theorem 9. Let G be the disjointness graph of a collection of x-monotone curves with ω(G) = k. Then we have χ(G) ≤ k 2 k+1 2 . Proof. Let C be a collection of x-monotone curves satisfying the conditions in the theorem. For any γ ∈ C, let x(γ) denote the projection of γ to the x-axis. For α, β ∈ C, let α ≺ β if min x(α) < min x(β) and max x(α) < max x(β).
Suppose that α and β are disjoint. Let α < 1 β if α ≺ β and α is below β, that is, if on every vertical line that intersects both α and β, the intersection point of α lies below the intersection point of β. Let α < 2 β if α ≺ β and β is below α. Clearly, < 1 and < 2 are partial orders.
As ω(G) ≤ k, the size of the longest chains with respect to < 1 and < 2 is at most k. Therefore, the vertices of G can be colored with k 2 colors such that each color class is an antichain in both < 1 and < 2 .
It remains to show that each of these color classes can be properly colored with k+1 2 colors. Let C ⊂ C such that no two elements of C are comparable by < 1 or < 2 . Then, if α, β ∈ C , then either α and β intersect, or one of the intervals x(α) or x(β) contains the other. In either case, x(α) and x(β) have a nonempty intersection, so any two elements of {x(γ) : γ ∈ C } intersect. Hence, γ∈C x(γ) is nonempty, and there exists a vertical line l that intersects every element of C . Let G denote the disjointness graph of C . Order the elements of C with respect to their intersections with l, from bottom to top. We claim that the resulting ordered graph G < is a semi-comparability graph. Indeed, suppose to the contrary that there are four vertices a, b, c, d ∈ V (G ) such that a < b < c < d and ab, bc, cd ∈ E(G ), but ac, bd ∈ E(G ). Without loss of generality, suppose that the length of x(b) is larger than the length of x(c); the other case can be handled similarly. As bc ∈ E(G ), we have x(c) ⊂ x(b) and b is below c, so every vertical line intersecting c intersects b as well, and its intersection with b lies below its intersection with c. Also, as ab ∈ E(G ), we have that a is below b. But then a and c must be disjoint, contradicting the condition ac ∈ E(G ).
Thus, we can apply Lemma 8 to conclude that G can be properly colored with k+1 2 colors. This completes the proof.
Let g(n) denote the maximal number m such that every collection of n convex sets in the plane contains m elements that are either pairwise disjoint, or pairwise intersecting. Larman et al. [26] proved that g(n) ≥ n 1/5 , while the best known upper bound, due to Kynčl [25] is g(n) < n log 8/ log 169 ≈ n 0.405 . Theorem 9 implies the following modest improvement on the lower bound. Corollary 10. Every collection of n x-monotone curves (or convex sets) in the plane contains ((2 + o(1))n) 1/5 ≈ 1.15n 1/5 elements that are either pairwise disjoint or pairwise intersecting.
Proof. In every graph G on n vertices, we have α(G)χ(G) ≥ n. In view of Theorem 9, this implies that if C is a collection of n x-monotone curves and G is the disjointness graph of C, then we have Therefore, max{α(G), ω(G)} ≥ ((2 + o(1))n) 1/5 , as claimed.
Many attempts were made to improve the order of magnitude of the lower bound on g(n). It appeared to be conceivable to cover the disjointness graph G of any collection of x-monotone curves with fewer than 4 comparability graphs, which would have yielded χ(G) ≤ (ω(G)) 3 and g(n) ≥ n 1/4 . These hopes are shattered by Theorem 4. Magical graphs -Proof of Theorem 2 The converse of Lemma 7 is not true: not every semi-comparability graph can be realized as the disjointness graph of a collection of grounded x-monotone curves. See Section 6, for further discussion. To characterize such disjointness graphs, we need to introduce a new family of graphs. A graph G <1,<2 with two total orderings, < 1 and < 2 , on its vertex set is called doubleordered. If the orderings < 1 , < 2 are clear from the context, we shall write G instead of G <1,<2 .
A graph G is said to be magical, if there exist two total orders < 1 , < 2 on V (G) such that G <1,<2 is magical. In this case, we say that the pair (< 1 , < 2 ) witnesses G.
It easily follows from the above definition that if G <1,<2 is magical, then G <1 is a semi-comparability graph.

Lemma 12. If C is a collection of grounded x-monotone curves, then the disjointness graph of C is magical.
Proof. Let G be the disjointness graph of C, and identify the vertices of G with the elements of C. For any γ ∈ C, let (0, y γ ) be the endpoint of γ lying on the vertical axis {x = 0}, and let (x γ , y γ ) be the other endpoint of γ.
Define the total orderings < 1 and < 2 on V (G), as follows. Let γ < 1 γ if and only if y γ < y γ , and let γ < 2 γ if and only if x γ < x γ .
Suppose that for a triple a, b, c ∈ C we have that a < 1 b < 1 c and ab, bc ∈ E(G), but ac ∈ E(G). Then a and c intersect. Hence, a, c, and the ground curve {x = 0} enclose a region A, and b ⊂ A. This implies that the x-coordinate of the right endpoint of b is smaller than the x-coordinates of the right endpoints of a and c. Therefore, we have b < 2 a and b < 2 c, showing that G is magical. Lemma 13. Let G be a magical graph. Then there exists a family C of grounded x-monotone curves such that the disjointness graph of C is isomorphic to G.
Proof. Let n be the number of vertices of G. Let < 1 and < 2 be total orderings on V (G) witnessing that G is magical. For any vertex v ∈ V (G), let y(v) ∈ [n] denote the position of v in the ordering < 1 , and let x(v) denote the position of v in the ordering < 2 .
For any v ∈ V (G), we define an x-monotone curve C v , which will be composed of , and ends at the point (i, y(v)). The pieces C v (i) are defined as follows. Let there is an edge between u i and v, then let C v (i) be the horizontal line segment connecting (i − 1, y(v)) and (i, y(v)). Otherwise, let C v (i) be the polygonal curve consisting of two segments whose three vertices are See Figure 1 for an illustration. One can easily check the following property of the curves (ii) Exactly one of v and w is joined to u i in G. Without loss of generality, assume that it is w.
If v and w are not joined by an edge in G, then C v (min{x(v), x(w)}) and C w (min{x(v), x(w)}) intersect by definition, so C v and C w have a nonempty intersection.
Our task is reduced to showing that if v and w are joined by an edge, then C v and C w do not intersect. Suppose to the contrary that C v and C w intersect. Then there exists y(w). Without loss of generality, let y(u i ) ≤ y(v), y(w), the other case can be handled in a similar manner. Again, without loss of generality, we can assume that y(w) < y(v). Then C v (i) intersects C ui (i), and C w (i) is disjoint from C ui (i), or equivalently, u i w ∈ E(G), but u i v ∈ E(G). However, this is impossible, because wv ∈ E(G), so the triple u i , w, v would contradict the assumption that G is magical.
By Lemma 13, in order to prove Theorem 2, it is enough to verify the corresponding statement for magical graphs. In other words, we have to prove the following. The rest of this section is devoted to the proof of this theorem. The proof is probabilistic and is inspired by a construction of Korándi and Tomon [18]. We shall consider a random double-ordered graph with certain parameters, and show that the smallest magical graph covering its edges meets the requirements in Theorem 14. To accomplish this plan, we first examine how the smallest magical graph covering the edges of a given double-ordered graph looks.
Let G <1,<2 be a double-ordered graph. A sequence of vertices x 1 , . . . , and either x 1 < 2 x 2 , . . . , x r−1 or x r < 2 x 2 , . . . , x r−1 . See Figure 2.  Let x 1 , . . . , x r be a mountain-path in G with x 1 < 2 x r . Using the definition of magical graphs, it is easy to prove by induction on i that x 1 and x i are joined by an edge in E(H), for every i > 1. Therefore, we have x 1 x r ∈ E(H). (We can proceed similarly if x r < 2 x 1 .) With a slight abuse of notation, from now on let H = H <1,<2 denote the doubleordered graph on V (G), in which u and v are joined by an edge if and only if there exists a mountain-path connecting u to v. We will show that H is magical, that is, for every triple u, v, w ∈ V (G), the following holds: if u < 1 v < 1 w such that uv, vw ∈ E(H), and u < 2 v or w < 2 v, then uw ∈ E(H). As uv, vw ∈ E(H), there exist two mountain-paths u = x 1 , x 2 , . . . , x r = v and v = x r , x r+1 , . . . , x s = w. By the assumption that u < 2 v or w < 2 v, the path u = x 1 , . . . , x s = w is a mountain-path as well, so uw ∈ E(H).
For the rest of the discussion, we need to introduce a few parameters that depend on k. Set λ = 1/k 2 , t = 100k 2 log k, h = 3t k 2 k 2k 2 +8 , n = 9h and p = t/n.
be a set of n arbitrary points in the interior of the unit square [ak + b, ak + b + 1] × [bk + a, bk + a + 1] with distinct x and y coordinates, see Figure 3. Let V = (a,b)∈S A a,b , and let < 1 and < 2 be the total orderings on V induced by the x and y coordinates of the elements of V , respectively. A pair of vertices {u, v} in V is called available if u ∈ A a,b , v ∈ A a ,b with (a, b) = (a , b ).
Let G 0 denote the random graph on V in which every available pair of vertices is connected by an edge with probability p, independently from each other. G 0 does not have any edge whose endpoints belong to the same set A a,b . Let G <1,<2 be the minimal magical graph on V containing all edges of G 0 . Claim 16. With probability at least 2/3, G has no independent set larger than (1 + λ)n.
Proof. As G 0 is a subgraph of G , it is enough to show that G 0 has no independent set of size greater than (1 + λ)n, with probability at least 2/3.
Let I ⊂ V such that |I| > (1 + λ)n. Then there are at least λn 2 /2 available pairs of vertices, whose both endpoints belong to I. Indeed, if u ∈ A a,b , then {u, v} is available for every v ∈ (I \ A a,b ), so there are at least |I \ A a,b | ≥ λn available pairs containing u. Hence, the total number of available pairs in I is at least |I|λn/2 > λn 2 /2.
If uv, vw, uw ∈ E(G ), then there exist three mountain-paths, P u,v , P v,w and P u,w , with endpoints {u, v}, {v, w} and {u, w}, respectively. See Figure 3. Note that each of these paths intersects every A a,b in at most one vertex. As u < 1 v < 1 w, the only vertex in the intersection of P uv and P vw is v. Moreover, P uw cannot contain v as v < 2 u and v < 2 w.

54:10 On the Chromatic Number of Disjointness Graphs of Curves
Consider the graph P = P uv ∪ P vw ∪ P uw . It is a connected graph, but not a tree, because there are two distinct paths between u and w: P uv ∪ P vw and P uw . Hence, we have |E(P )| ≥ |V (P )|. Let P denote the set of all such graphs P that appear in G 0 with positive probability. Then P({u, v, w} induces a triangle in G ) = P(P is a subgraph of G 0 for some P ∈ P) ≤ P ∈P P(P is a subgraph of G 0 ).
For a fixed P ∈ P, every edge of P is present in G 0 independently with probability p. Hence, the probability that P is a subgraph of G 0 is p |E(P )| , which is at most p |V (P )| . The number of graphs in P with exactly m vertices is at most |V | m−3 < (k 2 n) m−3 , as each member of P contains the vertices u, v, w. Finally, every member of P has at most 3|S| ≤ 3k 2 vertices, so we can write Since the number of holes in V is at most |V | 3 < |V | 3 < k 6 n 3 , we obtain Applying Markov's inequality, the probability that V contains more than 3h holes that induce a triangle in G is at most 1/3. Hence, there exists a magical graph G on V such that G has no independent set of size (1 + λ)n, and G contains at most 3h triangles whose vertices form a hole. By deleting a vertex of each such hole in G , we obtain a magical graph G with at least |S|n − 3h vertices, which has no triangle whose vertices form a hole, and no independent set of size (1 + λ)n.
First, we show that χ(G) ≥ |S| = k+1 2 . Indeed, if χ(G) ≤ |S| − 1, then G contains an independent set of size contradiction. It remains to prove that ω(G) = k. Clearly, ω(G) ≥ k, otherwise, by Lemma 8, we would have χ(G) ≤ k 2 , contradicting the last paragraph. Thus, we have to show that G has no clique of size k + 1. For this, we need the following observation.
Claim 18. Let K be a subset of S that does not contain three points (a 1 , b 1 ), (a 2 , b 2 ), (a 3 , b 3 )  such that a 1 < a 2 ≤ a 3 and b 2 ≤ b 1 and b 2 < b 3 . Then we have |K| ≤ k.
Proof. We call (a 1 , b 1 ), (a 2 , b 2 ), (a 3 , b 3 ) a bad triple, if a 1 < a 2 ≤ a 3 and b 2 ≤ b 1 and b 2 Let S = S k . We prove the claim by induction on k. For k = 1, the claim is trivial. Suppose that k ≥ 2 and that the statement has already been verified for k − 1. We distinguish two cases.
Then |K | ≥ |K| − 1 and K ⊂ S k−1 does not contain a bad triple. Thus, by the induction hypothesis, we have |K | ≤ k − 1, which implies that |K| ≤ k.

54:11
Case 2: K contains 2 distinct elements of the form (k, b) and (k, b ), where b < b . Then K cannot contain (a, k) for any a ∈ [k − 1], otherwise (a, k), (k, b), (k, b ) would be a bad triple. Thus, K contains at most one element from the row {(a, k) : a ∈ [k]} (it might contain (k, k)). Let Again, |K | ≥ |K| − 1 and K ⊂ S k−1 does not contain a bad triple. By the induction hypothesis, we have |K | ≤ k − 1 and, hence, |K| ≤ k.
Now we are in a position to finish the proof of Theorem 14. Let G denote the magical graph obtained from G by deleting a vertex from each of its holes that form a triangle (see right before Claim 18). Suppose that C ⊂ V is a clique in G. Then C does not contain a hole and it intersects each A a,b in at most one vertex.
The condition that C does not contain a hole implies that K does not contain three points (a 1 , b 1 ), (a 2 , b 2 ), (a 3 , b 3 ) such that a 1 < a 2 ≤ a 3 and b 2 ≤ b 1 and b 2 < b 3 . Hence, by Claim 18, we have |C| = |K| ≤ k. This completes the proof of Theorem 14 and, hence, the proof of Theorem 2.

Bounding function for curves that intersect a vertical line-Proof of Theorem 3
A triple-ordered graph is a graph G <1,<2,<3 with three total orders < 1 , < 2 , < 3 on its vertex set.
By Lemmas 12 and 13, it is not hard to characterize disjointness graphs of x-monotone curves intersected by a vertical line. Lemma 20. Let C be a collection of x-monotone curves such that each member of C intersects the vertical line l. Then the disjointness graph of C is double-magical.
We can just as easily prove the converse of Lemma 20, using Lemma 13.

Lemma 21.
Let G be a double-magical graph. Then there exists a collection of curves C such that each member of C has a nonempty intersection with the vertical line {x = 0}, and the disjointness graph of C is isomorphic to G.
(1) If ab ∈ E(G), then a and b are comparable by precisely one of these 4 partial orders. Proof. Let < 1 , < 2 , < 3 be total orders on V (G) witnessing G, and let ≺ 1 , ≺ 2 , ≺ 3 , ≺ 4 denote the partial orders defined above. Clearly, there is no chain of length k + 1 with respect to any of the partial orders ≺ i , because that would contradict the assumption ω(G) = k.
For h = 1, . . . , k, let S h denote the set of vertices v ∈ V (G) for which the size of a longest ≺ 1 -chain with maximal element v is k − h + 1. Then the sets S 1 , . . . , S k form a partition of V (G), where each S h is a ≺ 1 -antichain that contains no clique of size h + 1. Indeed, suppose that C ⊂ S h induces a clique of size h + 1 in G, and consider the smallest vertex v ∈ C with respect to the order < 1 . There exists a ≺ 1 -chain D of size k − h + 1 ending at v. This implies that for every a ∈ D and b ∈ C, we have a ≺ 1 v and v ≺ i b for some i ∈ {2, 3, 4}. Then, by (2), we would have ab ∈ E(G). Hence, D ∪ C would induce a clique of size k + 1, contradiction.
For h = 1, . . . , k and m = 1, . . . , h, let S h,m denote the set of vertices in S h for which the largest ≺ 2 -chain in S h with smallest element v has size h − m + 1. As ω(G[S h ]) ≤ h, the sets S h,1 , . . . , S h,h are ≺ 1 -and ≺ 2 -antichains partitioning S l . Further, S h,m contains no clique of size m + 1. Otherwise, if C ⊂ S h,m forms a clique of size m + 1 in G, then consider the largest vertex v ∈ C with respect to the order < 1 . There exists a ≺ 2 -chain D of size h − m + 1 whose smallest element is v. Hence, for every a ∈ C and b ∈ D, we have a ≺ i v and v ≺ 2 b for some i ∈ {3, 4}, which implies, by (3), that ab ∈ E(G). Hence, C ∪ D would induce a clique of size h + 1 in S h , contradiction.
Thus, we obtained that S h,m is a ≺ 1 -and ≺ 2 -antichain, which does not contain a clique of size m + 1. In particular, the size of the longest ≺ 3 -and ≺ 4 -chains in S l,m is at most m. This means that G[S h,m ] can be properly colored with m 2 colors. Indeed, set the color of v ∈ S h,m to be φ(v) = (r, q), where r is the size of the largest ≺ 3 -chain with smallest element v, and q is the size of the largest ≺ 4 -chain with smallest element v.