# The exact chromatic number of the convex segment disjointness graph

Let P be a set of n points in strictly convex position in the plane. Let D_n be the graph whose vertex set is the set of all line segments with endpoints in P, where disjoint segments are adjacent. The chromatic number of this graph was first studied by Araujo, Dumitrescu, Hurtado, Noy, and Urrutia [2005] and then by Dujmović and Wood [2007]. Improving on their estimates, we prove the following exact formula: χ(D_n) = n - √(2n + 14) - 12.

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## 1 Introduction

Throughout this paper, is a set of points in strictly convex position in the plane. The convex segment disjointness graph, denoted by , is the graph whose vertex set is the set of all line segments with endpoints in , where two vertices are adjacent if the corresponding segments are disjoint. Obviously does not depend on the choice of . Now assume that consists of evenly spaced points on a unit circle in the plane. The graph was introduced by Araujo, Dumitrescu, Hurtado, Noy and Urrutia [1], who proved the following bounds on the chromatic number of :

 2⌊13(n+1)⌋−1⩽χ(Dn)

Both bounds were improved by Dujmović and Wood [10] to

 34(n−2)⩽χ(Dn)

In this paper we prove matching upper and lower bounds, thus concluding the following exact formula for .

###### Theorem 1.
 χ(Dn)=n−⌊√2n+14−12⌋.

Equivalently, , where is the unique integer satisfying .

Theorem 1 is trivial for , so we henceforth assume that . The proof of the lower bound in Theorem 1 is based on the observation that each colour class in a colouring of is a convex thrackle. We then prove that two maximal convex thrackles must share an edge in common. From this we prove a tight upper bound on the number of edges in the union of convex thrackles. Theorem 1 quickly follows. These results are presented in Section 2. The proof of the upper bound in Theorem 1 is given by an explicit colouring, which we describe in Section 3.

## 2 Proof of Lower Bound

A convex thrackle on is a geometric graph with vertex set such that every pair of edges intersect; that is, they have a common endpoint or they cross. Observe that a geometric graph on is a convex thrackle if and only if forms an independent set in . A convex thrackle is maximal if it is edge-maximal. As illustrated in Figure 1, it is well known and easily proved that every maximal convex thrackle

consists of an odd cycle

together with some degree vertices adjacent to vertices of . For each vertex in , let be the convex wedge with apex , such that the boundary rays of contain the neighbours of in . Then every degree-1 vertex of lies in a unique wedge and the apex of this wedge is the only neighbour of in ; see [8, Lemma 1] for a strengthening of these observations. See [6, 22, 2, 20, 9, 13, 18, 4, 12, 19, 14, 21, 7, 11, 15, 16, 23, 5] for more on thrackles in general. Note that it is immediate from the above observations that every convex thrackle satisfies . Conways’s famous thrackle conjecture says this property holds for all thrackles. Note that is an example of a musquash [3, 17].

The following lemma is the heart of the proof of the lower bound in Theorem 1. We therefore include two proofs.

###### Lemma 2.

Let and be maximal convex thrackles on . Let and . Assume that . Then there is an edge in , with one endpoint in and one endpoint in .

###### Combinatorial Proof of Lemma 2.

Define a directed bipartite multigraph with bipartition as follows. For each vertex , add a blue arc to , where is the unique vertex in for which . Similarly, for each vertex , add a red arc to , where is the unique vertex in for which . Since , every vertex of has outdegree . Thus contains a directed cycle . By construction, vertices in are not incident to an incoming and an outgoing edge of the same colour. Thus alternates between blue and red arcs. The red edges of form a matching as well as the blue edges, both of which are thrackles on the same set of points (namely, ). However, there is only one matching thrackle on a set of points in convex position. Therefore is a 2-cycle, which corresponds to an edge in , with one endpoint in and one endpoint in . ∎

Our second proof of Lemma 2 depends on the following topological notions. Let be the unit circle. For points , let be the clockwise arc from to in . A -action on is a homeomorphism such that for all . Say that is free if for all .

###### Lemma 3.

If and are free -actions of , then for some point .

###### Proof.

Let . If then we are done. Now assume that . Without loss of generality, appear in this clockwise order around . Parameterise with a continuous injective function , such that and . Assume that for all , otherwise we are done. Since is free, for all . Thus . Also , as otherwise for some . This implies that appear in this clockwise order around . In particular, with , we have . Thus . Hence for some . Since is a -action, . This is a contradiction since but . ∎

###### Topological Proof of Lemma 2.

Assume that lies on . Let be a maximal convex thrackle on . As illustrated in Figure 1, for each vertex in , let be a pair of closed intervals of defined as follows. Interval contains and bounded by the points of that are of the way towards the first points of in the clockwise and anticlockwise direction from . Let and be the neighbours of in , so that is before in the clockwise direction from . Let be the endpoint of in the clockwise direction from . Let be the endpoint of in the anticlockwise direction from . Then is the interval bounded by and and not containing . Define as follows. For each , map the anticlockwise endpoint of to the anticlockwise endpoint of , map the clockwise endpoint of to the clockwise endpoint of , and extend linearly for the interior points of and , such that and . Since the intervals and are disjoint, is a free -action of .

By Lemma 3, there exists such that . Let and so that and , where and are defined with respect to and respectively. Since , we have and . Thus . If then , implying . Thus . Hence . Without loss of generality, . Thus . If then is an endpoint of both and , implying , which is a contradiction. Thus contains points other than . It follows that and . Therefore the edge is in both and . Moreover one endpoint of is in and one endpoint is in . ∎

###### Theorem 4.

For every set of points in strictly convex position, the union of maximal convex thrackles on has at most edges.

###### Proof.

For a set of maximal convex thrackles on , define for , and let be the set of triples where and . The proof proceeds by induction on .

First suppose that . Thus for all distinct . By Lemma 2, and have an edge in common, with one endpoint in and one endpoint in . Hence distinct pairs of thrackles have distinct edges in common. Since every maximal convex thrackle has edges and we overcount at least one edge for every pair, the total number of edges is at most .

Now assume that . Thus there is a vertex and a pair of distinct thrackles and , such that . We now modify to create a new set of convex thrackles, as illustrated in Figure 2. First, replace by two consecutive vertices and on . Then, for each cycle with and (which includes ), replace by in , and add the edge to , where is the vertex in for which is inserted into . Now, replace by in , and add the edge to , where is the vertex in for which is inserted into . Finally, for each cycle with , if is the vertex in with , then replace the edge by and in . Let be the resulting set of thrackles. Then , and every element of arises from an element of (replacing by or , as appropriate). Thus . Since one edge is added to each thrackle, the number of edges in equals the number of edges in plus . By induction, has at most edges, implying has at most edges. ∎

In the language of Dujmović and Wood [10], Theorem 4 says that every -vertex graph with convex antithickness has at most edges.

We now show that Theorem 4 is best possible for all . Let be a set of vertices in with no two consecutive vertices in . If and are consecutive in this order in , then is a maximal convex thrackle, and has exactly edges in total.

###### Proof of Lower Bound in Theorem 1.

If then, there are convex thrackles whose union is the complete geometric graph on . Possibly add edges to obtain maximal convex thrackles with edges in total. By Theorem 4, . The quadratic formula implies the result. ∎

## 3 Proof of Upper Bound

Label the points of by in clockwise order. Denote by the line segment between points with , which is a vertex of . It will be convenient to adopt the matrix convention for indexing rows and columns in . That is, row is immediately below row , column is immediately to the right of column , and refers to the lattice point in row and column . Identify the vertex of with the lattice point where , which we represent as a unit square in our figures. Define . We may consider represented as a triangle-shaped polyomino as illustrated in Figure 3(a).

Now, two distinct vertices and in are adjacent if and only if or . In particular, for and to be non-adjacent, must lie in the nonshaded region in Figure 4. In particular, cannot be strictly southwest or strictly northeast of . Moreover, .

We conclude that every independent set of is a subset of some rectangle of the form (with the southwest corner removed). Namely, choose such that is maximal and is minimal. Then for each . In fact, it is straightforward to show that each maximal independent set forms a path from to for some , where each step in the path is of the form or . An example is given in Figure 3(a). Conversely, every such path is a maximal independent set. We refer to such a path as a maximal thrackle path; the corresponding set of line segments forms a maximal convex thrackle, as shown in Figure 3(b).

To summarize, the chromatic number of equals the minimum number of maximal thrackle paths that cover . For example, Figure 5 shows that it is possible to cover with ten thrackle paths. As a consequence, . Indeed, we have equality by the lower bound in Theorem 1.

For , define the following intervals:

 Nk:=[(k2)+1,(k+12)]andN′k:=[(k2)+1,(k+12)−1].

Thus, , , , etc. The sets form a partition of . Observe that and for each .

We now describe an infinite sequence of infinite paths covering the infinite polyomino . The final construction for is then obtained as a restriction of the covering to the set . For each and for each , let be the following path: start at , walk south to

 (((k+12)−i+12),i),

make one step east to

 (((k+12)−i+12),i+1),

then walk south to , and finally walk east through all the points in the -th row.

We now show that for each , the paths cover all the points in the -th column. Let . If then the path covers the -th column. If then the path covers the -th column. Now assume that and . Let . The path covers the topmost points in the -th column. The next points of the -th column lie in the rows . These rows are completely covered by the paths where . The remaining bottom part of the -th column from to is covered by .

Now consider the restriction of the paths to the triangular polyomino . Each intersection is a maximal thrackle path in . Let be the unique integer satisfying . Then the above construction gives a covering of the polyomino by thrackle paths, since a path exists for each , except for the values . The upper bound in Theorem 1 follows.

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