Behavior of the slope number

2.3. Behavior of the slope number

slopes. First we consider the number of slopes ofP_G^∗ for a fixed Hamiltonian path under different embeddings of G.

Letw_iw_i+1, w_i+1w_i+2, . . . , wi+k−1w_i+k be bad edges of the same type (upwards/down-wards). We call the vertices a chain of lengthk. If all the edges of the chain are of the same front or back type the chain is called monotone.

Lemma 2.6. Let p be a Hamiltonian path in G and C be a chain of bad edges. Any segment representation ofP_G^∗, whose vertex order along e^∗ is given byp, requires at least 2(d^|C|₂ e+ 1)slopes.

Proof. Consider a segment representation ofP_G^∗. Without loss of generality, letC⁰ be a chain of front upward bad edges. For a bad front edge we know that the segment with the upper part has a larger slope then the segment with the lower part. By Observation 2.5, the bad edges connecting vertices in the chain have slopes that lie between the slopes of their incident vertices. Thus the segment representation uses at least2(|C⁰|+ 1)slopes;

|C⁰|+ 1slopes for the vertices incident to the bad edges,|C⁰|for the bad edges, and one for e^∗.

e^∗

Figure 2.8.: No good edge separates the ends ofe^∗from the unbounded face.

Therefore, we only have to show that a chain of bad edges of length |C| leads to a monotone chain of bad edgesC⁰of lengthd|C|/2e. We observe that the unbounded face of P_G^∗−B, where B is the set of bad edges, is the face that is incident to the two ends of the Hamiltonian path: Otherwise we have good edge that separates both ends of e^∗ from the unbounded face, which is not possible as shown in Figure 2.8.

On the other hand, we can obtain a pseudosegment representation of each such em-bedding by flipping one front bad edge over to a back bad as shown in Figure 2.9. This flipping splits the chain into two chains, one of which has the claimed length.

We show that there is a family of planar triangulations that have only Hamiltonian paths including long chains of bad edges. This way, we prove that there are graphs G_n, such that P_G^∗ requires a large number of slopes in any segment representation.

Therefore, we need the following tools.

2.3. Behavior of the slope number

e^∗ e^∗

Figure 2.9.: Flipping a front bad edge to the back.

Proposition 2.7. By P andR we denote the graphs from Figure 2.10.

1. Let G be a planar triangulation that contains P as proper subgraph, such that the vertices of the outer triangle are the only vertices of P that are adjacent to vertices of G−P. Then each Hamiltonian path of G traverses all vertices of P consecutively, or has an endpoint in P.

2. LetG be a planar triangulation that containsR as proper subgraph, such that the three outer vertices of Rare a separating triangle. Then each Hamiltonian path of Ghas one endpoint insideR. In addition, the vertices of the connected components of the inner vertices are traversed consecutively, unless both end vertices of the Hamiltonian path lie inside of R.

P R

Figure 2.10.: Gadgets for fixing a Hamiltonian path according to Proposition 2.7.

Proof. The proof of this proposition is based on the observation that removing a sep-arator S from G results in a subgraph with at most |S|+ 1 connected components if G has a Hamiltonian path p (and at most |S| components if G has a Hamiltonian cycle [Chv73]). If the removal of S leaves exactly |S|+ 1connected components, then the two ends ofh lie in two different components of G−S. These observations follow from the fact that a Hamiltonian path p traverses at least one vertex of S whenever it goes from one connected componentG−S to another. If the number of connected components ofG−S is exactly |S|+ 1, this implies that all vertices of one component ofG−S are traversed consecutively byp.

To show part 1 we assume that G is a planar graph with proper subgraph P, such that the three outer vertices are the only vertices ofP connected toG−P. We observe that the three outer vertices along with the central blue vertex of P in Figure 2.10 are a separator. Removing these four vertices leads to a graph with at least four connected components, three inside of the triangle and a fourth component outside, because we assumed that P is a proper subgraph ofG. Assume Ghas a Hamiltonian path p with both endpoints in G−P, then H traverses the three connected components inside of P consecutively: Since one vertex of the separator is traversed between the connected components by the Hamiltonian path p between two vertices of different connected components we have shown part 1.

To show part 2 we consider a planar graphGwithR as a proper subgraph, such that the outer vertices ofRare a separator ofG. The six green vertices ofRare a separator of G. The removal of these vertices leads to (at least) seven connected components. Thus G has no Hamiltonian path with the ends in the same component. This implies that also one of the components of the red vertices ofRcontains one end of the Hamiltonian path. If one end vertex of the Hamiltonian path lies in the connected component that is separated from the inner vertices of R by the its outer triangle, then the vertices of R are traversed consecutively by the path.

We use this proposition to construct triangulations with almost fixed Hamiltonian path.

Lemma 2.8. Let G be a planar graph with Hamiltonian path p = v1, . . . , vn. There exists a planar graphG⁰ with Θ(|V(G)|) vertices, such thatG is an induced subgraph of G⁰ and each Hamiltonian path in each triangulation of G⁰ visits all vertices ofG in the same order (up to the reverse) as p.

Proof. Given a planar graphGwith a Hamiltonian pathp=v1, . . . , vn, we fix the order of the vertices of pathpin any Hamiltonian path of a graphG⁰ as shown in Figure 2.11.

We add a copy ofRfrom Proposition 2.7 tov1 and one copy tovnby identifyingv1 and v_n with one of the points of the outer triangle. For each edgev_iv_i+1 ofpwe glue a copy of P from Proposition 2.7 onto the edge by identifying v_i and v_i+1 with two vertices from the outer triangle ofP. We call this new graph G⁰.

By Proposition 2.7.2 the two green triangles contain the end vertices of each Hamil-tonian path p⁰ of G⁰. Since the vertices of those triangles are traversed consecutively by p⁰ the verticesv₁ andv_n are the first and last vertex of p⁰ restricted on the vertices of G. When the path p⁰ traverses v₁ it enters the copy ofP that is glued on the edge v₁v₂. Since the vertices of a copy of P are traversed consecutively by Proposition 2.7.1, the vertexv₂ is the second vertex inp⁰ restricted on the vertices ofG. Since two copies