3-colorable planar graphs have an intersection segment representation using 3 slopes

Daniel Gonçalves

► To cite this version:

Daniel Gonçalves. 3-colorable planar graphs have an intersection segment representation using 3 slopes. Journal of Combinatorial Theory, Series B, 2025, 177, pp.234-256. 10.1016/j.jctb.2025.11.005. hal-05409566

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Submitted on 10 Dec 2025

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<figure><img src="image_1.png" /></figure>
3-colorable planar graphs have an intersection
segment representation using 3 slopes*

Daniel Gonçalves<sup>a</sup>

<sup>a</sup>LIRMM, Université de Montpellier, CNRS, Montpellier, France.

Abstract

In his PhD Thesis E.R. Scheinerman conjectured that planar graphs are inter-section graphs of line segments in the plane. This conjecture was proved with two different approaches by J. Chalopin and the author, and by the author, L. Isenmann, and C. Pennarun. In the case of 3-colorable planar graphs E.R. Scheinerman conjectured that it is possible to restrict the set of slopes used by the segments to only 3 slopes. Here we prove this conjecture by using an approach introduced by S. Felsner to deal with contact representations of planar graphs with homothetic triangles.

# 1 Introduction

In this paper, we consider intersection representations for planar graphs. A segment representation of a graph $G$ maps every vertex $v \in V(G)$ to a straight line segment $v$ of the plane so that two segments $u$ and $v$ intersect if and only if $uv \in E(G)$. In his PhD Thesis, E.R. Scheinerman [15] conjectured that every planar graph has a segment representation. This conjecture attracted a lot of attention. H. de Fraysseix and P. Ossona de Mendez [6] proved it for a large family of planar graphs, the planar graphs having a 4-coloring in which every induced cycle of length 4 uses at most 3 colors. In particular, this implies the conjecture for 3-colorable planar graphs. Then J. Chalopin and the author finally proved this conjecture [2]. More recently, a much simpler proof was provided by the author, L. Isenmann, and C. Pennarun [10]. Here we focus on segment representations of planar graphs with further restrictions.

In his PhD Thesis, E.R. Scheinerman [15] proved that every outerplanar graph has a segment representation where only 3 slopes are used, and where parallel segments do not intersect. Let us call such a representation a *3-slopes segment representation*. This result led E.R. Scheinerman conjecture [16] (see also [6]) that such representation exists for every 3-colorable planar graph. Later, several groups proved a related result on bipartite planar graphs [3, 7, 11]. They proved that every bipartite planar graph has

<footnote>*This is the extended version of a paper presented at WG '19 [8]. This research is partially supported by the ANR GATO, under contract ANR-16-CE40-0009.</footnote>
<figure><img src="image_2.png" /><figcaption>Figure 1: The octahedron and a 3-slopes contact representation. It is unique, up to vertex automorphism, up to scaling, and once the slopes are set.</figcaption></figure>

a 2-slopes segment representation, with the extra property that segments do not cross each other. Let us call such a representation a *2-slopes contact segment representation*. Later, de Castro *et al.* [1] considered a particular class of 3-colorable planar graphs. They proved that every triangle-free planar graph has a 3-slopes contact segment representation. Such a contact segment representation cannot be asked for any 3-colorable planar graph. Indeed, up to isomorphism, the octahedron has only one 3-slopes contact segment representation depicted in Figure 1, and one can easily check that this representation does not extend to the (3-colorable) graph obtained after gluing a copy of an octahedron in each of its faces. However, we will use 3-slopes contact segment representations in the proof of our main result.

**Theorem 1** Every 3-colored planar graph has a 3-slopes segment representation such that parallel segments correspond to the color classes.

As such a representation implies 3-colorability, and as 3-colorability is NP-complete already for planar graphs [17] we have that it is NP-hard to decide if a plane graphs has a 3-slopes segment representation in which no parallel segments intersect. Here, the NP-membership clearly follows from Theorem 1. The NP-membership is often non-trivial for segment representations. For example, we know that it is NP-hard to decide whether a graph $G$ admits a segment representation, actually it is even $\exists\mathbb{R}$-complete [13], but it is still open whether this problem belongs to NP or not.

It is easy to see that every 3-colored planar graph is the induced subgraph of some 3-colored triangulation. Indeed, adding a new vertex in a non-triangular face, one can reduce the maximum length of such a face, or reduce the number of colors around such a face. As a 3-slopes segment representation of a triangulation naturally induces such a representation for its induced subgraphs, we can restrict ourselves to the case of triangulations in the following. In Section 2 we provide some basic definitions. Section 3 is devoted to the so-called *triangular contact schemes*, which are particular 3-slopes contact segment representations of an auxiliary graph (to be formally defined there). It is shown that every 3-colorable triangulation admits such a scheme. Then, those schemes are used in Section 4 to build 3-slopes segment representations. Finally, we conclude with some remarks on 4-slopes segment representations.
## 2 Terminology

All graphs considered here are simple, that is without loops nor multiple edges. A triangulation is a plane graph where every face has size three. A triangulation $T$ is Eulerian if every vertex has even degree. It is well known that these triangulations are the 3-colorable triangulations [14]. Actually, these triangulations are uniquely 3-colorable (up to color permutation). Hence, their vertex set $V(T)$ is canonically partitioned into three independent sets $A, B$ and $C$. In the following we will denote the vertices of these sets respectively $a_i$ with $0 \le i < |A|$, $b_j$ with $0 \le j < |B|$, and $c_k$ with $0 \le k < |C|$. In such a triangulation $T$ any face is incident to one vertex $a_i$, one vertex $b_j$, and one vertex $c_k$, and these vertices appear in this order either clockwise or counterclockwise. In the following, the vertices of the outerface are always denoted by $a_0, b_0$ and $c_0$, and they appear clockwise in this order around $T$. As the orders of two adjacent faces are opposite, the dual graph of $T$ is bipartite. Given a Eulerian triangulation $T$ with face set $F(T)$, let us denote by $F_1(T)$ and $F_2(T)$ (or simply $F_1$ and $F_2$ if it is clear from the context) the face sets partitioning $F(T)$, such that no two adjacent faces belong to the same set, and such that $F_2(T)$ contains the outer face. Note that by construction for any face $f \in F_1(T)$ (resp. $f \in F_2(T)$) its vertices $a_i, b_j$ and $c_k$ appear in clockwise (resp. counterclockwise) order around $f$. Note that the vertices $a_0, b_0$ and $c_0$ appear in clockwise order around $T$, but in counterclockwise order with respect to the outer face. Let $n = |V(T)|$. As $T$ is a triangulation, by Euler's formula it has $2n - 4$ faces. Hence, as $T$'s dual is bipartite and 3-regular, $|F_1(T)| = |F_2(T)| = n - 2$.

In the following, we build 3-slopes segment representations. The 3 slopes used are expected to be distinct, but apart from that the exact 3 slopes considered do not matter. Indeed, for any two triples of slopes, $(s_1, s_2, s_3)$ and $(s'_1, s'_2, s'_3)$, there exists an affine map of the plane turning any 3-slopes segment representation using slopes $(s_1, s_2, s_3)$ into a 3-slopes segment representation using slopes $(s'_1, s'_2, s'_3)$. We denote by $\vec{a}$, $\vec{b}$, and $\vec{c}$ the vectors corresponding to slopes of the sets $A, B$, and $C$ respectively. The magnitude of these vectors is chosen such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.

## 3 TC-representations and TC-schemes

We begin with the definition of the particular type of 3-slopes contact representations illustrated in Figure 2.

**Definition 2** For the fixed vectors $\vec{a}, \vec{b}, \vec{c}$ as defined previously, a Triangular 3-slopes Contact segment representation (TC-representation for short) is a 3-slopes contact segment representation using the same slopes as $\vec{a}, \vec{b}$, and $\vec{c}$, and where:

* Three segments $a_0, b_0$, and $c_0$, form a triangle which contains all the other segments. Going clockwise around this outer-triangle, one successively follows $\alpha \vec{a}, \alpha \vec{b}$, and then $\alpha \vec{c}$, with $\alpha = 1$.

* Every inner region is a triangle, where each side is contained in a segment of the representation.
<figure><img src="image_3.png" /><figcaption>Figure 2: (left) Vectors $\vec{a}, \vec{b}$, and $\vec{c}$ (middle) A TC-representation $\mathcal{R}$ with various types of intersection points. The circled ones correspond to particular intersection points with more than three segments. (right) Its corner graph $C(\mathcal{R})$, where gray faces correspond to the degenerate faces (i.e. they correspond to intersection points in $\mathcal{R}$). The dark gray faces are particular degenerate faces. One has size six, and there are two faces of size three that correspond to the same intersection point.</figcaption></figure>

* Two parallel segments intersect in at most one point, and this point is an endpoint of both segments.

**Remark 3** In such a representation, an intersection point $\mathbf{p}$ is of one of the following four types (see Figure 2).

* The intersection point of 2 outer segments;

* the intersection point of 3 segments (with 3 distinct slopes) such that exactly two of them end at $\mathbf{p}$;

* the intersection point of 5 segments such that exactly four of them end at $\mathbf{p}$ (such a point will be generally considered as the merge of two intersection points of 3 segments); or

* the intersection point of 6 segments that have an end at $\mathbf{p}$.

**Definition 4** Given a TC-representation $\mathcal{R}$, let the corner graph $C(\mathcal{R})$ be the plane graph whose vertices correspond to the segments of the representation, and where two vertices are adjacent if and only if the corresponding segments form a corner of one of the inner triangles. The orders of the neighbors around a vertex $v$ correspond to the order of the segments around the segment $v$.

Note that the corner graph of a TC-representation has several properties. For example, since two parallel segments cannot form a corner of a triangle, they correspond to non-adjacent vertices. The slopes hence define a 3-coloring of the graph. Note also that the dual graph of $C(\mathcal{R})$ is bipartite. Indeed, such a plane graph has two types of
faces, one set contains the (triangular) faces corresponding to the inner regions of the
TC-representation, and the other set contains the outerface and the faces corresponding
to intersection points. Let us denote the latter faces degenerate faces, and note that
those faces have size three or six. A size six face $(a_i, b_j, c_k, a_{i'}, b_{j'}, c_{k'})$ comes from
the intersection point of six segments, and as those six segments go in distinct direc-
tions they do not intersect elsewhere, so this cycle has no chord in $C(\mathcal{R})$. Note that
going clockwise on the border of any inner region of $\mathcal{R}$, since $\vec{d} + \vec{b} + \vec{c} = \vec{0}$, one
successively follows $\alpha \vec{d}$, $\alpha \vec{b}$, and then $\alpha \vec{c}$, for some not necessarily positive value
$\alpha$. This value $\alpha$ is the size of this inner triangle. Note that the outer-triangle formed
by $a_0, b_0$ and $c_0$ has size 1. In our illustrations, the triangles of positive size have a
corner on the left. Let us now prove that any corner graph is 3-connected.

**Lemma 5** For any TC-representation $\mathcal{R}$, its corner graph $C(\mathcal{R})$ is 3-connected, and hence it has only one planar embedding with $a_0, b_0, c_0$ on the outerboundary in clockwise order.

**Proof.** Towards a contradiction, suppose $C(\mathcal{R})$ is not 3-connected for some TC-representation $\mathcal{R}$, and let $\{x, y\}$ be a separating set. The case where there is a vertex cut but no separating set of size two is trivial. We assume without loss of generality that $x$ is a $A$-vertex, then by symmetry we just have to consider the case where $y$ is a $A$-vertex, and the case where $y$ is a $C$-vertex.

If both $x$ and $y$ are $A$-vertices, let us prove that any connected component $H$ of $C(\mathcal{R}) \setminus \{x, y\}$ contains $b_0$. Indeed, let $b_j$ be the rightmost vertical segment of $H$. If $b_j \neq b_0$ there is a $C$-vertex $c_k$ such that $c_k$ intersects $b_j$ on its topmost point and such that $c_k$ continues further on the right. Indeed, the topmost inner triangle incident to the right of $b_j$ is incident to such segment $c_k$. Then the rightmost point of $c_k$ has to belong to some vertical segment on the right of $b_j$, a contradiction.

If $x$ is a $A$-vertex and $y$ is a $C$-vertex, we have to distinguish two subcases according
to how they intersect (or not).

Suppose first that $x$ and $y$ are not on the boundary of an inner triangle, of negative
size (i.e. pointing to the right). In this case, we can proceed as in the previous case by
considering the rightmost vertical segment of any connected component $H$ of $C(\mathcal{R}) \setminus \{x, y\}$. If $b_j \neq b_0$, the segment $b_j$ has (at least) 2$t$ incident segments going further
to the right, where $t \ge 1$ is the number of inner faces incident to $b_j$, on its right. If
some of these 2$t$ neighbors in $C(\mathcal{R})$ is distinct from $x$ and $y$, this segment brings us to
another $B$-vertex on the right, as in the previous case. If these 2$t$ neighbors correspond
exactly to $x$ and $y$, we have $t=1$ and we are in the case where $x$ and $y$ are on the
boundary of an inner triangle, of negative size, a contradiction.

Suppose now that $x$ and $y$ are on the boundary of an inner triangle, of negative
size (i.e. pointing to the right). In that case, we are going to prove that any connected
component $H$ of $C(\mathcal{R}) \setminus \{x, y\}$ contains the vertex $b_t$, corresponding to the leftmost
vertical segment; the one forming an inner triangle with $a_0$ and $c_0$. We proceed simi-
larly as above by letting $b_j$ be the leftmost vertical segment of $H$, and by considering
its (at least) 2$t$ incident segments going further to the left, where $t \ge 1$ is the number
of inner faces incident to $b_j$, on its left. If some of these neighbors in $C(\mathcal{R})$ is distinct
from $x$ and $y$, this segment would bring us to another $B$-vertex on the left, a contra-
diction. We are thus left with the case where $t=1$ and where there are only two such
<figure><img src="image_4.png" /><figcaption>Figure 3: From left to right. A TC-representation $\mathcal{R}$, with the size of its inner triangles; its corner graph $C(\mathcal{R})$, where gray faces are the degenerate faces; and two triangulations where $\mathcal{R}$ is a TC-scheme of both.</figcaption></figure>

neighbors, which would be $x$ and $y$. This is impossible because in that case $x$ and $y$ would be on the boundary of an inner triangle, of positive size, and it is impossible in a TC-representation for two segments to have more than one inner triangle in common. $\square$

**Lemma 6** Consider a Eulerian triangulation $T$ and a TC-representation $\mathcal{R}$ such that each segment of $\mathcal{R}$ corresponds to a distinct vertex of $T$. The following two statements are equivalent:

(1) $C(\mathcal{R})$ is a submap of $T$ with the same outer face as $T$, such that the vertices and edges of $T$ not in $C(\mathcal{R})$ lie inside degenerate faces of $C(\mathcal{R})$.

(2) The outerface and the non-degenerate faces of $C(\mathcal{R})$ are, respectively, the outerface and a subset of the inner faces of $T$.

If (1) (or equivalently (2)) is fulfilled, we call $\mathcal{R}$ a *TC-scheme of* $T$ (see Figure 3).

**Proof.** In both cases, the outerfaces of $C(\mathcal{R})$ and $T$ are asked to be the same, so we can focus on the inner faces.

(1) $\Rightarrow$ (2). If $C(\mathcal{R})$ and $T$ fulfill (1), then any non-degenerate face of $C(\mathcal{R})$ induces a cycle in $T$ (as $C(\mathcal{R})$ is a subgraph of $T$). Furthermore, this cycle does not contain any edge or vertex of $T$. It thus bounds an inner face of $T$.

(2) $\Rightarrow$ (1). In a corner graph $C(\mathcal{R})$, every edge belongs to a non-degenerate innerface (by definition). Hence, if $C(\mathcal{R})$ and $T$ fulfill (2), any edge of $C(\mathcal{R})$ is an edge of $T$. $C(\mathcal{R})$ is thus a subgraph of $T$. Furthermore, as they have same outerface, and as $C(\mathcal{R})$ has only one possible embedding (by Lemma 5) we have that $C(\mathcal{R})$ is a submap of $T$. The non-degenerate faces of $C(\mathcal{R})$ being inner faces of $T$ (by (2)) the vertices and edges of $T$ not in $C(\mathcal{R})$ have to lie inside the degenerate faces of $C(\mathcal{R})$. $\square$

The main ingredient in the proof of Theorem 1 is the following.
**Theorem 7** Every Eulerian triangulation T has a TC-scheme, and this scheme is unique.

To prove this theorem we use TC-schemes as defined in Lemma 6.(2). The definition in Lemma 6.(1) will be used in Section 4 to complete the proof of Theorem 1. To prove Theorem 7, we first model TC-schemes of T by means of a system of linear equations in Subsection 3.1. We then show in Subsection 3.2 that such a linear system always has a solution, and that this solution is unique (c.f. Lemma 8). Finally, we show in Subsection 3.3 that the solution of this linear system defines a TC-scheme of T (c.f. Lemma 11).

## 3.1 The linear system model

In a TC-representation all the triangles are homothetic, and they are included in the outer triangle which has size one. Note that we will have positive, negative, and degenerate (zero) homothets. Recall that the size of a triangle is its relative size with respect to the outer triangle. The variables of our linear system correspond to the sizes of the triangular regions. So for each face $f \in F_1$ we have a variable $x_f$. Informally, the value of $x_f$ will prescribe the size and shape of the corresponding triangle in a TC-representation. If $x_f < 0$, $x_f = 0$, or if $x_f > 0$ the corresponding triangle has a corner on the right, is missing, or has a corner on the left, respectively.

Let us denote by $F_1(v)$ the subset of faces of $F_1$ incident to $v$. As the outer triangle has size 1 and contains the other triangles, the faces in $F_1(a_0)$ should have non-negative sizes, and they should sum up to 1 (see Figure 4, left). We hence consider the following constraint.

$$ \sum_{f \in F_1(a_0)} x_f = 1 \qquad (a_0) $$

We add no constraint about the sign of these sizes. Note that similar constraints hold for $b_0$ and $c_0$.

$$ \sum_{f \in F_1(b_0)} x_f = 1 \qquad (b_0) $$

$$ \sum_{f \in F_1(c_0)} x_f = 1 \qquad (c_0) $$

Similarly, around an inner segment of a TC-representation all the triangles on one side have the same size sign, which is opposite to the other side (see Figure 4, right). Furthermore, by summing all of these sizes one should obtain 0. Hence, for any inner vertex u we consider the following constraint.

$$ \sum_{f \in F_1(u)} x_f = 0 \qquad (u) $$

In the following, Equation (a_j) will refer to Equation (u) where vertex u is replaced by a_j. Note that every face f ∈ F_1 is incident to exactly one vertex of A, one vertex of B, and one vertex of C. Hence by summing Equations (a_0), (a_1),..., (a_{|A|}), one obtains that $\sum_{f \in F_1} x_f = 1$. The same holds with Equations (b_0), (b_1),..., (b_{|B|}), or
<figure><img src="image_5.png" /><figcaption>Figure 4: (left) The size of the triangles around $a_0$. (right) The size of the triangles around some inner vertex $b_i$.</figcaption></figure>

with Equations $(c_0)$, $(c_1)$, ..., $(c_{|C|})$. Equations $(b_0)$ and $(c_0)$ are hence implied by the others, and thus we do not need to consider them anymore. Let us denote by $\mathcal{L}$ the obtained system of $n-2$ linear equations on $|F_1| = n-2$ variables.

## 3.2 $\mathcal{L}$ has a unique solution

Let us define the set $V' = V \setminus \{b_0, c_0\}$ of size $n-2$. Finding a solution to $\mathcal{L}$ is equivalent to finding a vector $S \in \mathbb{R}^{F_1}$ (that is a vector indexed by elements of $F_1$) such that $MS = I$, where $M \in \mathbb{R}^{V' \times F_1}$ (a square matrix indexed by elements of $V' \times F_1$) and $I \in \mathbb{R}^{V'}$ are defined by

$$M(x_i, f) = \begin{cases} 1 & \text{if } f \in F_1(x_i) \\ 0 & \text{otherwise.} \end{cases}$$

$$I(x_i) = \begin{cases} 1 & \text{if } x_i = a_0 \\ 0 & \text{otherwise.} \end{cases}$$

Given some bijective mappings $g_{V'} : [1, \dots, n-2] \to V'$ and $g_{F_1} : [1, \dots, n-2] \to F_1$, one can index the elements of $M$ by pairs $(i, j) \in [1, \dots, n-2] \times [1, \dots, n-2]$, and thus define the determinant of $M$. By the following lemma, $\mathcal{L}$ has a solution vector $S$, and this solution is unique.

**Lemma 8** The matrix $M$ is non-degenerate, i.e. $\det(M) \neq 0$.

The proof of this lemma is inspired by the work of S. Felsner [4] on contact representations with homothetic triangles. See also [5] for another proof using the same approach.

**Proof.** Let $T_M$ be the bipartite graph with independent sets $V'$ and $F_1$ such that $v \in V'$ and $f \in F_1$ are adjacent if and only if $v$ and $f$ are incident in $T$. Note that $M$ is the biadjacency matrix of $T_M$. From the embedding of $T$ one can easily embed $T_M$ in such a way that all the inner faces have size 6, and such that $a_0$ is on the outerboundary.

Note that every perfect matching of $T_M$ (if any) corresponds to a permutation $\sigma$ on $[1, \dots, n-2]$ defined by $\sigma(g_{V'}^{-1}(v)) = g_{F_1}^{-1}(f)$, for any edge $vf$ of the perfect matching. If the obtained permutation is even we call such perfect matching positive, otherwise it is negative. From the Leibniz formula for the determinant,

$$\det(M) = \sum_{\sigma \in S_{n-2}} \operatorname{sgn}(\sigma) \prod_{i \in [1, \dots, n-2]} M(g_{V'}(i), g_{F_1}(\sigma(i)))$$
one can see that $\det(M)$ counts the number of positive perfect matchings of $T_M$ minus its number of negative perfect matchings.

**Claim 9** *The graph $T_M$ admits at least one perfect matching.*

**Proof.** As $T_M$ is bipartite, and as $|V'| = |F_1|$, it suffices to show that $T_M$ has an $F_1$-saturating matching. This follows from Hall's marriage theorem, and the fact that for any set $X \subseteq F_1$ the set $Y \subset V'$ of vertices incident to a face in $X$ is such that $|Y| \ge |X|$. Let us show this below for any set $X \subseteq F_1$.

Consider the (planar) subgraph of $T$ with all the edges and all the vertices incident to a face of $X$. Then, triangulate this graph and denote by $T_X$ the obtained triangulation. Note that as any two faces of $X$ are not adjacent in $T_X$, this triangulation has at least $2|X|$ faces. Indeed, around each vertex there are at least twice as many faces as faces of $X$, and summing over every vertex one obtains the inequality. Together with the fact that $T_X$ has $2|V(T_X)| - 4$ faces,

$$2|V(T_X)| - 4 \ge 2|X|$$

and that $V(T_X) \subseteq Y \cup \{b_0, c_0\}$,

$$|Y| + 2 \ge |V(T_X)|$$

one obtains that

$$2|V(T_X)| - 4 + 2|Y| + 4 \ge 2|X| + 2|V(T_X)| \\ |Y| \ge |X|$$

□

Given a graph $G$ and a perfect matching $M$ of $G$, an alternating cycle $C$ is a cycle of $G$ with edges alternating between $M$ and $E(G) \setminus M$. Note that replacing in $M$ the edges of $M \cap C$ by the edges of $C \setminus M$ yields another perfect matching. We call such operation a *cycle exchange*. It is folklore that the perfect matchings of a graph are linked by cycle exchanges. Indeed, given any perfect matching $M_1$ of $G$ one can reach any perfect matching $M_2$, by a succession of cycle exchanges. Actually, for $T_M$ any such cycle has length congruent to $2 \pmod 4$.

**Claim 10** *For any perfect matching of $T_M$ and any of its alternating cycles $C$, we have that the length $\ell(C)$ of $C$ is congruent to $2 \pmod 4$.*

**Proof.** The subgraph $G$ of $T_M$ induced by the vertices and edges on or inside $C$ is such that all the inner faces have length 6, and it is routine from Euler's formula to show that $C$ has length congruent to $2 + 2|V_G| \pmod 4$, where $V_G$ is the vertex set of $G$. Indeed,

$$\ell(C) - 6 + 6|F_G| = \sum_{f \in F_G} \ell(f) = 2|E_G| = 2|V_G| + 2|F_G| - 4 \\ \ell(C) \equiv 2|V_G| + 2 \pmod 4$$
Finally, as the vertices of $G$ are paired by the perfect matching we have that $|V_G|$ is even. $\square$

The previous claim implies that all the perfect matchings of $T_M$ induce permutations of the same sign. Indeed, performing a $(4k+2)$-cycle exchange does not change the sign of the permutation, as it corresponds to performing $2k$ transpositions in the permutation. Hence, all the terms of $\det(M)$ have the same sign, and this sum has at least one non-zero term (by Claim 9). Thus $\det(M) \neq 0$. $\square$

### 3.3 A solution of $\mathcal{L}$ defines a TC-scheme

A TC-scheme $\mathcal{R}$ corresponds to a solution of $\mathcal{L}$, the linear system defined for a Eulerian triangulation $T$, if $\mathcal{R}$ is a TC-scheme of $T$ such that each face $f \in F_1$ corresponds to an inner triangle of $\mathcal{R}$ of size $x_f$, the solution of $\mathcal{L}$, unless $x_f = 0$. In other words, the inner regions of $\mathcal{R}$ correspond to non-zero valued faces of $F_1$.

Let us now proceed to the main result of this section.

**Lemma 11** Every Eulerian triangulation $T$ admits a TC-scheme $\mathcal{R}$ that corresponds to the solution of its linear system $\mathcal{L}$.

*Proof.* We proceed by induction on the number of faces $f \in F_1$ such that $x_f = 0$. We start with the case where every face $f \in F_1$ is such that $x_f \neq 0$.

**If every face $f \in F_1$ is such that $x_f \neq 0$,** we construct a TC-scheme $\mathcal{R}$ corresponding to the solution of $\mathcal{L}$ as follows. First let $\Delta$ be a triangle formed by three vectors $\vec{a}, \vec{b}$, and $\vec{c}$ in this order. The sides of $\Delta$ correspond to $a_0, b_0$, and $c_0$, respectively. For each face $f \in F_1$, let $\Delta_f$ be a homothetic copy of $\Delta$ with ratio $x_f$. The triangle $\Delta_f$ is thus obtained by following the vectors $x_f \vec{a}, x_f \vec{b}$, and $x_f \vec{c}$ in this order. We are going to show that all these triangles $\Delta_f$ can be arranged as a tiling of $\Delta$, forming a TC-representation of $T$ (i.e. such that $T = C(\mathcal{R})$).

Note that a necessary condition for this to work is that (1) every face of $f \in F_1$ around $a_0, b_0$, or $c_0$ is positive (i.e. $x_f > 0$), and that (2) for any inner vertex $v$ of $T$ its positive (resp. negative) incident faces in $F_1$ are consecutive around $v$. Otherwise, this would result in overlapping triangles $\Delta_f$ (see Figure 4). We first show that (1) and (2) are fulfilled, and then we show that this suffices to ensure the construction of $\mathcal{R}$.

Consider the incidence graph $I$, between vertices of $V(T)$ and faces of $F_2$. First note that this plane graph has only size six faces and that they are in bijection with the faces in $F_1$. Let us orient the edges of $I$ as follows. An edge $vf$ of $I$, with $v \in V(T)$ and $f \in F_2(T)$, is oriented from $v$ to $f$ if and only if the incident faces (which correspond to faces in $F_1$) have different signs. Note that for an inner vertex of $T$, $d^+(v) = 2k$ for some $k \ge 1$ (as $v$ is incident to positive and to negative faces in $T$), and that $d^+(f) = 1$ or $3$ for a face $f \in F_2$. The graph $I$ has $2n-2$ vertices (3 outer vertices of $T$, $n-3$ inner vertices of $T$, and $n-2$ faces of $F_2$) and $3n-6$ edges. The outerface $f^o$ of $T$ has outdegree 3 in $I$. Otherwise, among the three faces of $F_1$ incident to $a_0b_0, a_0c_0$, or $b_0c_0$ there would be positive ones and negative ones. This would imply that two of $a_0, b_0$, and $c_0$ have outdegree at least 2 in $I$. This would be
<figure><img src="image_6.png" /><figcaption>Figure 5: (a) Example of a Eulerian triangulation $T$ (dashed lines), with incidence graph $I$. The numbers correspond to the solution of $\mathcal{L}$. (b) The graph $I'$ obtained after (Step 1). (c) The graph $G^\Delta$.</figcaption></figure>

impossible as summing over the outdegrees of $a_0, b_0, c_0$, the inner vertices of $T$, and the faces of $F_2(T)$ would lead to $2 + 2 + 2(n-3) + (n-2) > 3n-6$. We thus have that $d^+(f^o) = 3$, and a counting argument gives us that the other faces $f$ of $F_2$ have outdegree one, that the outer vertices have outdegree zero, and that the inner vertices of $T$ have outdegree two. Thus, (1) and (2) are verified.

To construct the TC-representation of $T$, we define a plane graph $G^\Delta$ from $I$ by replacing $f^o$ with three vertices (Step 1), and for each vertex $v \in V(T)$, by turning its neighborhood in $I$ from a star into a path (Step 2).

(Step 1) The vertex $f^o$ is replaced by three new vertices $f_a^o, f_b^o$, and $f_c^o$ in such a way that $f_a^o$ is adjacent to $b_0$ and $c_0$ (see Figure 5). The six new edges are oriented towards the newly created vertices. Let us use $I'$ to denote this new oriented graph. Note that now every vertex $v \in V(T)$ has outdegree two, and that by assigning size $-1$ to the outerface, all faces incident to $v$ sum up to zero.

(Step 2) For each vertex $v \in V(T)$, its neighborhood in $I'$ is turned into a path $P_v$ whose ends are the out-neighbors of $v$. The in-neighbors are ordered as follows in $P_v$. We first denote by $f^+$ (resp. $f^-$) the out-neighbors of $v$ such that the face following $f^+$ (resp. $f^-$), around $v$ in clockwise order, has positive (resp. negative) size (i.e. solution in $\mathcal{L}$). Two in-neighbors $f, f'$ of $v$ are ordered along $P_v$ in such a way that $f$ is closer to $f^+$ than $f'$, if and only if the sum of the face sizes going around $v$ from $f^+$ to $f$ is lower than the sum from $f^+$ to $f'$. If the two sums are equal, then $f$ and $f'$ are merged into a single vertex (see Figure 6). As all the faces around $v$ have non-zero sizes, and as positive sizes are consecutive, a vertex $f$ is merged at most once while constructing $P_v$. Note that $f$ is merged with another face $f'$ on $P_v$ only if $vf$ is oriented towards $v$ and this occurs for only one neighbor of $f$, so $f$ is not merged to more faces while
<figure><img src="image_7.png" /><figcaption>Figure 6: An example of (Step 2) with a merge of $f$ and $f'$.</figcaption></figure>

constructing the other paths $P_u$.

The obtained plane graph is denoted by $G^\Delta$. Note that the inner faces of $G^\Delta$ correspond to faces of $F_1$, and we assign them the corresponding sizes. Note also that a face of $G^\Delta$ corresponding to the face $a_ib_jc_k \in F_1$, is bordered by three subpaths of paths $P_{a_i}$, $P_{b_j}$, and $P_{c_k}$. We now assign a positive length to each edge of $G^\Delta$ so that the length of each of these subpaths corresponds to the size of the face, forgetting the sign. For each edge $f f'$ of a path $P_v$, we assign the absolute value of the sum of the face sizes between $f$ and $f'$ around $v$ in $I'$.

By construction, $G^\Delta$ has three types of vertices:

• The vertices $f_a^o, f_b^o$, and $f_c^o$, which have degree two. Indeed, e.g. the vertex $f_a^o$ is at the end of $P_{b_0}$ and $P_{c_0}$.

• The vertices originating from a single vertex $f \in V(I) \setminus (V(T) \cup \{f^o\})$. As each such $f$ has in-degree two and out-degree one in $I$, it is at the end of two paths and in the middle of a third one.

• The vertices originating from two vertices $f, f' \in V(I) \setminus (V(T) \cup \{f^o\})$. By construction, such vertex is in the middle of a path, and has two paths ending on each side (corresponding to in-neighbors in $I$).

From the orientation of $I'$, note that the sign of the faces alternate around any of these vertices (see Figure 6). We now want to draw $G^\Delta$ planarly, in such a way that its inner faces are all homothetic to the triangle formed by following the three vectors $\vec{a}, \vec{b},$ and $\vec{c}$. More precisely, for a face $f$ of size $\alpha$ that is bordered by subpaths $P_a^f \subseteq P_{a_i}$, $P_b^f \subseteq P_{b_j}$, and $P_c^f \subseteq P_{c_k}$, the subpaths $P_a^f, P_b^f$ and $P_c^f$ should be mapped to vectors $\alpha \vec{a}, \alpha \vec{b}$, and $\alpha \vec{c}$, respectively, in such a way that the edge length along these paths are met. Note that there is no local obstruction to the existence of such embedding.

• Each edge $f f'$ of $G^\Delta$ is consistently embedded. Indeed, the length of $f f'$ is set in $G^\Delta$, and regardless of the considered incident face (as these faces have different signs), the vector $\overrightarrow{ff'}$ has the same direction.
* For the outer vertices $f_{\bar{a}}^o$, $f_{\bar{b}}^o$, and $f_{\bar{c}}^o$, their incident inner faces form an angle smaller than $\pi$ (e.g. for $f_{\bar{a}}^o$ the angle is the one from $\vec{c}$ to $-\vec{b}$). For any other outer vertex $f$, which necessarily corresponds to a single vertex of $I'$, its (three) incident inner faces form an angle of size exactly $\pi$. For example, if $f$ is in the middle of the path $P_{a_0}$ and at the end of paths $P_{b_j}$ and $P_{c_k}$, we know by (1) that the inner faces incident to $P_{a_0}$ are positive, while the third one is negative because the edges $fb_j$ and $fc_k$ are oriented towards $f$ in $I'$. Thus, the angles around $f$ go from $\vec{a}$, to $-\vec{c}$, to $\vec{b}$, and to $-\vec{a}$.

* For any inner vertex $f$ corresponding to a single vertex of $I'$, its (four) incident faces form an angle of size exactly $2\pi$. For example, if $f$ is in the middle of a path $P_{a_i}$ and at the end of paths $P_{b_j}$ and $P_{c_k}$, as the signs of the four faces alternate, the angles around $f$ go from $-\vec{a}$ to $\vec{a}$, to $-\vec{c}$, to $\vec{b}$, and back to $-\vec{a}$, for $x \in \{-1, +1\}$.

* Similarly, for an inner vertex $f$ originating from two vertices of $I'$, the sum of the 6 angles is again $2\pi$.

From these observations, a simple variant of Lemma 6 of [5] ensures the existence of such embedding. Alternatively, one could triangulate $G^\Delta$ to use this lemma directly.

Note that this embedding is such that for each vertex $a_i \in V(T)$ (resp. $b_j \in V(T)$ and $c_k \in V(T)$) the corresponding path $P_{a_i}$ (resp. $P_{b_j}$ and $P_{c_k}$) forms a segment parallel to $\vec{a}$ (resp. $\vec{b}$ and $\vec{c}$). As in $G^\Delta$ a vertex $f$ is in the middle of at most one path $P_v$, these segments do not cross. For any inner edge of $T$, say $a_ib_j$ incident to a face $f \in F_2(T)$, the paths $P_{a_i}$ and $P_{b_j}$ touch at the vertex $f$ of $G^\Delta$. For the outer edges the contact points are $f_{\bar{a}}^o$, $f_{\bar{b}}^o$, and $f_{\bar{c}}^o$. We thus have a TC-scheme of $T$.

**If some faces $f \in F_1$ are such that $x_f = 0$, consider a face $a_ib_jc_k \in F_1$ such that**
$x_{a_ib_jc_k} = 0$. Let $a'$, $b'$, $c'$ be the vertices belonging to a triangle adjacent to $a_ib_jc_k$ (see the left of Figure 7). Note that the vertices $a_i$ and $a'$ (resp. the vertices $b_j$ and $b'$, and the vertices $c_k$ and $c'$) have at least two common neighbors, $b_j$, $c_k$ (resp. $a_i$, $c_k$ and $a_i$, $b_j$). It may be the case that they have more common neighbors, but the planarity and the 3-coloring of $T$ forces one of these pairs to restrict to these two common neighbors only. Let us assume that $a_i$ and $a'$ have exactly two common neighbors, $b_j$, $c_k$. Let $T'$ be the Eulerian triangulation obtained from $T$ by deleting the edges $b_jc_k$, $a'b_j$ and $a'c_k$, and by merging $a_i$ and $a'$ (see Figure 7). The resulting vertex of $T'$ is denoted by $a'_i$. Note that by the choice of $a_i$, $a'$, $T'$ has no multiple edges. We can thus apply the induction on $T'$.

Let $\mathcal{L}'$ be the linear system defined for $T'$. Note that a solution of $\mathcal{L}$ clearly induces a solution of $\mathcal{L}'$. Indeed, every vertex $v \in V(T') \setminus \{a'_i, b_j, c_k\}$ is incident to the same faces as in $T'$, so they sum up to 0 (or to 1 for outer vertices). For $b_j$, or $c_k$ these vertices are incident to one less face of $F_1$, the face $a_ib_jc_k$, and as $x_{a_ib_jc_k} = 0$, their incident faces still sum up to 0 (or to 1) in $T'$. Similarly, as the faces of $F_1$ incident to $a'_i$ in $T'$ are the faces of $F_1$ incident to $a_i$ or to $a'$ in $T$, except $a_ib_jc_k$, they sum up to 0. As the solution of $\mathcal{L}'$ has one less 0 entry, we can apply the induction, and consider a
<figure><img src="image_8.png" /><figcaption>Figure 7: Reduction to a smaller triangulation $T'$ by identifying $a_i$ and $a'$.</figcaption></figure>

TC-scheme $\mathcal{R}'$ of $T'$ corresponding to this solution of $\mathcal{L}'$. We consider different cases according to whether $a_i$ and $a'$ have non-zero valued incident faces in $T$.

If $a'$ has only zero valued incident faces in $T$, then $\mathcal{R}'$ is also a TC-scheme for $T$ corresponding to $\mathcal{L}$. Note by Lemma 6.(2) that every non-degenerate inner face $f$ of $C(\mathcal{R}')$ is a face of $F_1(T')$. Renaming $a_i$ the vertex originated by the merging of $a_i$ and $a'$, instead of $a'_i$, all the faces of $F_1(T')$ with non-zero size, are also faces of $F_1(T)$. Hence, $\mathcal{R}'$ is also a TC-scheme for $T$. The fact that it corresponds to $\mathcal{L}$ is clear by the construction.

If $a_i$ has only zero valued incident faces in $T$, we proceed as above. The only difference is that we have to denote $a'$ the vertex originated by the merging of $a_i$ and $a'$.

If both $a_i$ and $a'$ have non-zero valued incident faces in $T$, let us divide the segment $a'_i$ of $\mathcal{R}'$ into two parts, one for each of $a_i$ and $a'$. Note that the faces of $F_1(T) \setminus \{a_i b_j c_k\}$ incident to $a_i$ (resp. $a'$) in $T$ correspond to consecutive triangles around $a'_i$. Furthermore, as their sizes sum up to 0 there is a point $p \in a'_i$ that divides $a'_i$ into two parts, $a_i$ and $a'$, such that the faces of $F_1 \setminus \{a_i b_j c_k\}$ incident to $a_i$ (resp. $a'$) in $T$ correspond to triangles with a side contained inside $a_i$ (resp. $a'$). Let us denote by $\mathcal{R}$ the obtained TC-representation. As every non-degenerate face $f$ of $C(\mathcal{R})$ corresponds to a face of $F_1(T)$ whose size is $x_f$, by Lemma 6.(2) we have that $\mathcal{R}$ is a TC-scheme of $T$ following the solution of $\mathcal{L}$. This concludes the induction step of the proof. $\square$

# 4 3-slopes segment representations

In this section, we use Theorem 7 to prove the main theorem of the article, Theorem 1. As already mentioned, it is sufficient to prove it for Eulerian triangulations. From now on, given a Eulerian triangulation with more than three vertices, let $a_1, b_1$ and $c_1$ be the vertices forming an inner face with vertices $b_0$ and $c_0$, with $a_0$ and $c_0$, and with $a_0$ and $b_0$, respectively. Theorem 1 follows from the following technical proposition.
<figure><img src="image_9.png" /><figcaption>Figure 8: (left) A 3-slopes segment representation inside a hexagon. Note that some intersection points lie inside the triangle $a_0b_0c_0$, while some are outside this triangle (here this is the case for the intersection of $b_1$ and $c_1$). (right) An illustration of its shape.</figcaption></figure>

**Proposition 12** For every sufficiently small $\epsilon > 0$, every simple Eulerian triangulation $T$ admits a 3-slopes segment representations $\mathcal{R}$ such that:

* The segments $a_0, b_0$, and $c_0$ form a triangle $\Delta$ of size 1 (its sides are obtained by following $\vec{a}, \vec{b}$, and $\vec{c}$).

* Every segment is contained in the hexagon centered on $\Delta$, obtained by successively following $(1-\epsilon)\vec{a}, -2\epsilon\vec{c}, (1-\epsilon)\vec{b}, -2\epsilon\vec{a}, (1-\epsilon)\vec{c}$, and $-2\epsilon\vec{b}$ (see Figure 8).

* The segments $a_1, b_1$, and $c_1$ have both endpoints on the border of the hexagon.

* No three segments intersect at the same point.

* For every inner edge $uv$ the segments $u$ and $v$ cross each other (i.e. their intersection point is not an endpoint).

Given such representation $\mathcal{R}$ of a triangulation having some inner vertices, we define the *shape* of $\mathcal{R}$ as the triplet $(s_a, s_b, s_c)$ of sizes of the triangles corresponding to $a_1b_0c_0, a_0b_1c_0, a_0b_0c_1$, respectively, in $\mathcal{R}$ (see the right part of Figure 8).

**Proof.** We proceed by induction on $|V(T)|$. The initial case of this induction, when $|V(T)| = 3$ clearly holds. We thus proceed to the induction step with a Eulerian triangulation $T$ such that $|V(T)| > 3$, and we assume that the proposition holds for any simple Eulerian triangulation with fewer vertices.

By Theorem 7, we consider a TC-scheme $\mathcal{R}$ of $T$, and we prolong the segments $a_1, b_1$, and $c_1$ to have their endpoints on the border of the hexagon. Then, to reach the sought representation, we resolve every degenerate point of $\mathcal{R}$. A $k$-degenerate point of $\mathcal{R}$, for $k = 3, 5$ or $6$, is a point $p$ belonging to $k$ segments. Here, resolving means that the segments of a 3-degenerate point (resp. a 5- or a 6-degenerate point) are moved to form a triangle (resp. two triangles, or a polygon) inside which we are going to draw
<figure><img src="image_10.png" /><figcaption>Figure 9: (left) A 3-degenerate point on $V$ (middle) Small perturbation of $R$ (right) The addition of a representation roughly inside the new triangle. The new segments are added in the hexagon with dashed gray border.</figcaption></figure>

a 3-slopes representation of the graph corresponding to this degenerate face of $C(\mathcal{R})$, this is possible by using the induction on this smaller graph. The degenerate points of $\mathcal{R}$ are resolved from left to right. This means that at a given stage of this process there is a vertical line (parallel with $\vec{b}$) $V$ such that on its left all the intersection points (except for the one of $a_0c_0$) are crossing points. This implies that on the left of $V$ the representation handles some small perturbations: Every segment, except $a_0$ and $c_0$, can slightly move in any direction, without creating or destroying any intersection point.

Let $V$ be the leftmost vertical line containing degenerate points. We resolve those degenerate points by slightly moving segments on the left of or on $V$, while maintaining the right side of the representation unchanged (except at the vicinity of $V$). We consider different cases according to the degenerate points on $V$.

**If $V$ contains a 3-degenerate point $p$ in the interior of a (vertical) segment $b_j$ and at the end of two segments $a_i$ and $c_k$ lying on the left of $V$, the situation is rather simple.** Move these segments $a_i$, $c_k$ a little to the left and slightly prolong them to intersect $b_j$ (see Figure 9). As there is no degenerate point on the left of $V$ these moves can be done while maintaining the existing intersections and avoiding new intersections. If $a_ib_jc_k$ is not a face of $T$, consider the triangulation $T'$ induced by the vertices in the cycle $a_ib_jc_k$ of $T$. By induction, $T'$ has a representation that can be drawn inside a hexagon centered on the newly formed triangle bordered by the segments $a_i$, $b_j$ and $c_k$.

**If $V$ contains a 5-degenerate point $p$ in the interior of a (vertical) segment $b_j$, the situation is similar to the previous one.** Move the segments on the left of $V$ as depicted in Figure 10. If the new triangle is not a face of $T$, we add a representation inside. We are now left with a single 3-degenerate point at $p$. This corresponds to the following case.
<figure><img src="image_11.png" /><figcaption>Figure 10: (left) A 5-degenerate point on $V$. (middle) A small perturbation of $R$. The small circle indicates the position of the original 5-degenerate point $p$, that is now a simple 3-degenerate point. (right) Addition of a 3-slope segment representation, if needed.</figcaption></figure>

<figure><img src="image_12.png" /><figcaption>Figure 11: (left) A 3-degenerate point $p$, in a small circle, with segments on the right of $V$. (middle) Slightly moving $b_j$ to the right, and addition of extra segments in the hexagon, if needed. (right) Slightly moving $b_j$ to the left.</figcaption></figure>
<figure><img src="image_13.png" /><figcaption>Figure 12: (left) A 3-degenerate point $\mathbf{p}$ at the end of $\mathbf{b}_t$. The cases where $\mathbf{a}_i$ end at $\mathbf{p}$ is similar to the case where $\mathbf{c}_k$ ends at this point, since the segment ending there is prolonged. (middle) Slightly moving $\mathbf{b}_t$ to the left, and addition of extra segments in the hexagon, if necessary. (right) Slightly moving $\mathbf{b}_t$ to the right.</figcaption></figure>

If $\mathcal{V}$ contains a 3-degenerate point $\mathbf{p}$ in the interior of a (vertical) segment $\mathbf{b}_j$ and at the end of two segments, $\mathbf{a}_i$ and $\mathbf{c}_k$, lying on the right of $\mathcal{V}$, one can move $\mathbf{b}_j$ slightly to the right or slightly to the left and resolve these points without changing the right part of the representation. The choice of moving $\mathbf{b}_j$ to the right or to the left is explained in the next paragraph, but we can assume this move to be arbitrarily small. Regardless of the direction $\mathbf{b}_j$ is moved, one has to prolong $\mathbf{a}_i$ and $\mathbf{c}_k$ to keep the intersections between these segments and $\mathbf{b}_j$ (see Figure 11). Note that in order to preserve the representation on the right of $\mathcal{V}$ the segments $\mathbf{a}_i$ and $\mathbf{c}_k$ are not moved, they are only prolonged across $\mathbf{p}$. Again, if $\mathbf{a}_i \mathbf{b}_j \mathbf{c}_k$ is not a face of $T$, we draw a representation inside the newly formed triangle. Note that if $\mathbf{b}_j$ moves to the right, the triangle bordered by $\mathbf{a}_i$, $\mathbf{b}_j$ and $\mathbf{c}_k$ has negative size, but it suffices to apply a homothety with negative ratio to obtain a representation that can be drawn inside.

**Consider now the degenerate points at the end of a (vertical) segment b_j of V.**
Let $b_1, b_2, \dots, b_t$ be a maximal sequence of segments on $\mathcal{V}$, from bottom to top, such that $b_j$ and $b_{j+1}$ intersect in a point. We are going to move these segments slightly to the right or slightly to the left of $\mathcal{V}$. The direction of these moves, right or left, and their exact magnitude will be set later, but remember that these moves can be made arbitrarily small. Recall that the 3-degenerate points in the interior of the segments $b_j$ with $1 \le j \le t$ can be dealt if the move of $b_j$ is sufficiently small (see the previous cases).

Consider first the upmost endpoint $\mathbf{p}$ of $\mathbf{b}_t$, and let $\mathbf{a}_i$ and $\mathbf{c}_k$ be the other segments meeting at this point. The point $\mathbf{p}$ is an endpoint for one of $\mathbf{a}_i$ or $\mathbf{c}_k$. We slightly prolong this segment across $\mathbf{p}$ (see Figure 12). Then, whatever the direction $\mathbf{b}_t$ is moved, by slightly prolonging $\mathbf{b}_t$, we maintain the desired intersection. Note that these prolongings ensure that the representation on the right of $\mathcal{V}$ is unchanged, such as segment $\mathbf{a}_0$ if $\mathbf{a}_i = \mathbf{a}_0$. Again, if $\mathbf{a}_i \mathbf{b}_t \mathbf{c}_k$ is not a face of $T$, we draw a representation inside the newly formed triangle. In this case, if $\mathbf{a}_i = \mathbf{a}_0$, note that as the move of $\mathbf{b}_t$ can be arbitrarily small, the representation of $T$ can remain inside the hexagon of Proposition 12. The case of the lowest endpoint of $\mathbf{b}_1$ is symmetrical.

Now we arbitrarily choose to move $\mathbf{b}_1$ towards the left of $\mathcal{V}$, and we assume that move to be arbitrarily small. In the following we will see how the direction chosen for this move, to the left, implies a direction for moving $\mathbf{b}_2$, which in turn implies a direction for moving $\mathbf{b}_3$, and so on until $\mathbf{b}_t$. Regarding the magnitude of these moves,
<figure><img src="image_14.png" /><figcaption>Figure 13: (left) A 5-degenerate point on $\mathcal{V}$ (middle) & (right) Small moves that resolve this point with respect to the directions taken by $b_j$ and $b_{j+1}$.</figcaption></figure>

they are all positively correlated, meaning that all these moves can be arbitrarily small, if the first one (of $b_1$) is chosen sufficiently small.

Consider now any intersection point $\mathbf{p}$ between $b_j$ and $b_{j+1}$, and let us distinguish various cases according to the other segments containing $\mathbf{p}$.

Suppose there is a segment $a_i$ going through $\mathbf{p}$. It is shown in Figure 13 how to resolve these two overlapped 3-degenerate points, in order to create two triangles, where one can add a small representation if needed. The case where there is a segment $c_k$ going through $\mathbf{p}$ is similar.

Assume now that six segments intersect at $\mathbf{p}$. Let $b_j$ be the one below $\mathbf{p}$, and let $a, c, b_{j+1}, a'$, and $c'$ be the other ones around $\mathbf{p}$ clockwise. The degenerate face corresponding to $\mathbf{p}$ in $C(\mathcal{R})$ is bounded by the cycle $(b_j, a, c, b_{j+1}, a', c')$, and there are several cases according to whether there are chords within this cycle in $T$. For all the cases, we assume that the move of $b_j$, its direction and its (arbitrarily small) magnitude is set, and we will see how it determines the direction and magnitude of the moves applied to $a, c, b_{j+1}, a'$, and $c'$.

Suppose that none of $ab_{j+1}, b_{j+1}c'$, or $ac'$ is a chord of $(b_j, a, c, b_{j+1}, a', c')$ in $T$ (see Figure 14), then we consider the subgraph of $T$ induced by the vertices on and inside this cycle. We add the edges $ab_{j+1}, b_{j+1}c'$, and $ac'$ outside the cycle, and we denote by $T'$ the obtained simple Eulerian triangulation. By the induction, we know that $T'$ admits a 3-slope segment representation $\mathcal{R}'$, and let $(s_a, s_b, s_c)$ be the shape of $\mathcal{R}'$. If $b_j$ is moved to the left of $\mathcal{V}$, we resolve the point $\mathbf{p}$ by moving the segments as depicted in Figure 14. Each of these moves are proportional to the move of $b_j$ towards the left, and the magnitude of each of these moves is prescribed by the shape $(s_a, s_b, s_c)$ in order to allow us to copy $\mathcal{R}'$ inside the triangle formed by $a, b_{j+1}$, and $c'$. Then we shorten $a, b_{j+1}$, and $c'$ to avoid intersections among them. The case where $b_j$ is moved to the right of $\mathcal{V}$ is depicted in the left of Figure 15. In that case, note that $\mathcal{R}'$ is not only rescaled to fit into the hexagon, but we first consider a homothety of $\mathcal{R}'$ with negative ratio.

If none of $b_jc, ca'$, or $a'b_j$ is a chord of $(b_j, a, c, b_{j+1}, a', c')$ we proceed similarly (see the middle and the right of Figure 15). The only difference is that we add the edge $b_jc, ca'$, or $a'b_j$ outside $(b_j, a, c, b_{j+1}, a', c')$ to obtain $T'$.

We now have to deal with the case where $(b_j, a, c, b_{j+1}, a', c')$ has two opposite
<figure><img src="image_15.png" /><figcaption>Figure 14: Resolution of a 6-degenerate point $\mathbf{p}$ when there is no chord $ab_{j+1}, b_{j+1}c'$, or $ac'$. (left) A 6-degenerate point on $\mathcal{V}$, and below, the triangulation $T'$ obtained from the vertices inside this degenerate face of the TC-scheme. (middle) The shape of $\mathcal{R}'$ and the moves made according to this shape and according to the move of $b_j$ towards the left. (right) The 3-slope segment representation $\mathcal{R}'$ is included, and the segments $a, b_{j+1}$, and $c'$ are shortened to avoid the artificially added edges $ab_{j+1}, b_{j+1}c'$, and $ac'$.</figcaption></figure>

<figure><img src="image_16.png" /><figcaption>Figure 15: (left) Resolution of a 6-degenerate point $\mathbf{p}$ when there is no chord $ab_{j+1}, b_{j+1}c'$, or $ac'$, and when $b_j$ is moved to the right. (middle) The case where none of $b_jc, ca'$, or $a'b_j$ is a chord and where $b_j$ is moved to the left. In this illustration there is a chord $ac'$, but this may not be the case, of course. (right) Same case with $b_j$ moved to the right of $\mathbf{p}$.</figcaption></figure>
<figure><img src="image_17.png" /><figcaption>Figure 16: (left) The case where $ac'$ and $a'c$ are chords, with $b_j$ moved to the left or to the right of $p$. (right) The case where $b_j c$ and $b_{j+1} c'$ are chords, with $b_j$ moved to the left or to the right of $p$. In the first case, note that $\mathcal{R}_2$ contains the intersection point of $a$ and $c'$ inside the triangle, but it could be the case that the representation $\mathcal{R}_2$ obtained by induction has this intersection point located outside the triangle. See for example the illustration on the right part, where the intersection point of $b_j$ and $c$ is outside the triangle.</figcaption></figure>

chords. If $ac'$ and $a'c$ are chords (see the left part of Figure 16), we consider two triangulations. Let $T_1$ be the one inside the cycle $(c, b_{j+1}, a')$ and let $T_2$ be the one obtained from the interior of the 5-cycle $(a', c', b_j, a, c)$ by adding the edges $a'b_j$ and $b_jc$ outside. By the induction, we know that $T_1$ and $T_2$ admit 3-slopes segment representations $\mathcal{R}_1$, and $\mathcal{R}_2$. We resolve the point $p$ by moving the segments as depicted in Figure 16, and the magnitude of each of these moves, except for $b_j$ and $b_{j+1}$, is prescribed by the shape of $\mathcal{R}_2$. Then we shorten two segments to avoid the intersections corresponding to $a'b_j$ and $b_jc$. The segment $b_{j+1}$ is moved away to avoid the interior of the hexagon containing $\mathcal{R}_2$. Then $\mathcal{R}_1$ is drawn inside the triangle bordered by $b_{j+1}$, $a'$ and $c$. This is possible because the hexagons containing $\mathcal{R}_1$ and $\mathcal{R}_2$ do not overlap.

If $b_j c$ and $b_{j+1} c'$ are chords, we proceed similarly (see the right part of Figure 16). The difference is that, here, $T_1$ is the triangulation inside the cycle $(b_{j+1}, a', c')$ and let $T_2$ be the one obtained from the interior of the 5-cycle $(c', b_j, a, c, b_{j+1})$ by adding the edges $ab_{j+1}$ and $ac'$ outside.

If $ab_{j+1}$ and $a'b_j$ are chords, we proceed similarly (see Figure 17). One difference is that, $T_1$ is the triangulation inside the cycle $(a, c, b_{j+1})$ and that $T_2$ is the one obtained from the interior of the 5-cycle $(b_{j+1}, a', c', b_j, a)$ by adding the edges $ac'$ and $b_{j+1}c'$. Another difference is that we proceed differently according to the shape of $\mathcal{R}_2$. We distinguish the case where the intersection point of $a'$ and $b_j$ lies inside the triangle $ab_{j+1}c'$ or outside it.

Finally, note that the moves of $b_j$ and $b_{j+1}$ are of proportional magnitudes (up to some constant depending on the shapes of $\mathcal{R}'$ or $\mathcal{R}_2$). This property propagates from $b_1$, to $b_t$, and it is thus possible to assume that all these moves are arbitrarily small. The procedure hence leads to the sought 3-slope segment representation, and
<figure><img src="image_18.png" /><figcaption>Figure 17: The case where $ab_{j+1}$ and $a'b_j$ are chords. (left) If in $\mathbb{R}_2$, the intersection point of $a'$ and $b_j$ is inside the triangle $ab_{j+1}c'$. The two subcases whether $b_j$ is moved to the left or to the right of $p$. (right) If in $\mathbb{R}_2$, the intersection point of $a'$ and $b_j$ is outside the triangle $ab_{j+1}c'$.</figcaption></figure>

this concludes the proof of the proposition.

□

# 5 Conclusion

Our result implies that for $k \le 3$, planar graphs that are $k$-colorable admit a $k$-slopes segment representation, where parallel segments induce an independent set. These graphs have a so-called *PURE-$k$-DIR$*$ representation. Unfortunately, this does not extend to the final case $k = 4$ as conjectured by D. West [18]. The author [9] built a counter-example based on a construction of Kardoš and Narboni [12]. Their construction is an example of a signed planar graph that is not 4-colorable, in the sense of signed graphs. However, it remains open to know whether (4-colorable) planar graphs admit a PURE-$k$-DIR representation (resp. a non-necessarily pure one) for some $k > 4$ (resp. $k > 2$).

**Acknowledgements.** The author is thankful to the anonymous reviewers, especially for a simplified proof of Lemma 8, and for pointing out a gap in a preliminary version of the proof. The author is also thankful to Pascal Ochem for bringing [12] to his attention, and to Marc de Visme for fruitful discussions on this topic.

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