Convex Polygonal Duals
Corollary 5.1. Deciding whether a plane graph admits a t-sided convex polygonal dual is solvable in polynomial time for graphs of max degree 3, partial 3-trees, and
5.7 Further Applications of Our Technique
In addition to the fixed-parameter tractability results derived in the previous sec-tions, in this section we give alternate proofs for some interesting existing results using the technique we have developed. First, we give a simple proof for a result of [17]:
Theorem 5.6 ( [17]). Each maximal plane graph admits a 6-sided convex polygonal
dual.
Proof. Our alternative proof relies on Theorem 4.1, which guarantees that maximal plane graphs admit rectilinear duals using only upside-down T-shapes and their degen-erations. Fig. 5.8(1.1) lists the set of allowed shapes, while shapes listed in Fig. 5.8(1.2) are not allowed.
Given a maximal plane graph G (see Fig. 5.8(2.1)), an FAA of G∗ is constructed naturally according to a rectilinear dual as shown in Fig. 5.8(2.2). To be precise, we make an assignment at each 180◦corner not in the boundary of the drawing. Note that the concave corners (bends) of any rectilinear polygon are not vertices in G∗. As such an FAA may not lead to a stretchable drawing, we do some adjustments by unassigning
some vertices according to the rules specified in Fig. 5.8(3). See Fig. 5.8(2.3) for an example of an FAA after the adjustment.
It is easy to see that the resulting FAA is a FAA. Prior to the adjustment, it is a 6-FAA since only a convex corner of a polygon can be a combinatorial convex corner, and since each polygon has at most 6 convex corners. Though each adjustment increases the number of combinatorial convex corner of a face by one, we can apply it only when we have a nearby convex corner that is not a vertex in G∗. Therefore, the FAA after the adjustment is still a 6-FAA.
What is left to be done is to show that each cycle has at least 3 strongly-free vertices.
Let C be any cycle in G∗. Consider the sub-drawing, which is a rectilinear polygon, of C in the rectilinear dual. Let ab and cd be its highest and lowest horizontal segments, respectively, as shown in Fig. 5.8(4). It is immediate that a and b are strongly-free vertices of C. Suppose that there is no strongly-free vertex on cd. Then, c and d must be bends in the drawing (i.e. not a vertex in G∗), and no adjustment is applied on cd.
This implies that there is no line segment touching cd from out(C), meaning that there is a non-convex polygon F in out(C) incident to cd, which is a contradiction to the allowed set of shapes (i.e., upside-down T-shapes and their degeneracies). Therefore, we conclude that there is a free vertex in cd, and hence C has at least 3 strongly-free vertices. See Fig. 5.8(2.4) for the resulting convex polygonal dual.
As each Hamiltonian maximal plane graph admits a rectilinear dual using only L-shape and rectangles [43], following a similar approach, our technique can also be uti-lized to give a simple proof for the following:
Theorem 5.7 ( [43]). Each Hamiltonian maximal plane graph admits a 5-sided convex polygonal dual.
Next, we showcase a quick proof for the main result of [28]:
Theorem 5.8 ( [28]). Each triconnected cubic plane graph admits a proper touching triangle representation.
a b
c d F
(1.1) (1.2)
(2.1) (2.2) (2.3) (2.4)
(3) (4)
Figure 5.8: Illustration of the proof of Theorem 5.6.
Proof. A proper touching triangle representation is just a 3-sided convex polygonal dual whose boundary is a triangle.
Let G be a triconnected cubic plane graph, and we construct its associated G∗ as described in Section 5.3. We let FO(G) = (v1, v2, . . . , vs) be the outer face of G. Note that we must have s ≥ 3 since G is simple. It is easy to see that the subgraph H of G∗ induced by the faces corresponding to vertices in V (G)\ V (FO(G)) (the shaded area in Fig. 5.9) is biconnected, since otherwise G is not triconnected.
We contract most of the boundary edges, only leaving a boundary edge for each of F1, F2, and F3, where Fiis the face in G∗corresponding to vi. We let the 3-FAA contain only ui → Fi, i∈ {1, 2, 3}, where ui ∈ V (G∗) is the shared non-boundary vertex of Fi and Fi+1. See Fig. 5.9 for an illustration. We claim that our edge contraction and FAA work.
It is immediate that the assignment is a 3-FAA such that the boundary in the resulting drawing is a triangle whose three corners are c1, c2 and c3 in Fig. 5.9. What remains to be done is to verify that each cycle C has 3 free vertices:
• If C contains none of c1, c2 and c3, it belongs entirely to H (the shaded area).
v1
v2
v3
v4
v5
F1
F2
F3
F4
F5
u1
u2
u3
c1
c2
c3
G G* D(G)
Figure 5.9: Illustration of the proof for Theorem 5.8.
Then, certainly all its vertices are free, as they are not assigned to in(C).
• If C contains exactly one of c1, c2and c3, the one it contains must be c3(since c1, c2 have only one adjacent non-boundary vertex). Let x and y be the two neighboring vertices of c3 in C. It is clear that x, c3and y are 3 free vertices in C, Since x and y are either unassigned or assigned to out(C), and since c3 is unassigned.
• If C contains exactly two of c1, c2 and c3, as these two corners already contribute two free vertices to C, the only situation that makes C to have less than 3 free ver-tices is that all verver-tices in C\ {c1, c2, c3} are assigning to in(C). However, since only u1, u2 and u3 are involved in our FAA (i.e. V (C)⊆ {c1, c2, c3, u1, u2, u3}), we can assure that it never happen by examining a small bounded amount of pos-sibilities.
• If C contains all of c1, c2and c3, these three corners form 3 free vertices of C.
Adapting our approach, the laborious process of explicitly assigning positions for all junction points and all segments to construct a drawing, which inevitably appears in many works on contact graph representations in non-rectilinear situation, can be pre-vented.