2.2 Applications
2.2.3 Planar Generic Circuits
Proof. Let H ̸=C3 be an almost 4-regular plane graph and let R be a SLTR of H. The three suspensions inRare the three degree-2 vertices. Since all the vertices inH have even degree, the dual is a bipartite graph. We abuse notation and denote the bounded faces inR that contain the suspension vertices, by suspension of the dual. Since they are all adjacent to the outer face ofR, the suspensions are all in the same color class of the bipartition, say in the white class.
a3
a1
a2
G H
Figure 2.39: An SLTR of an almost 4-regular graphHwith a 2-coloring of its faces and its underlying graphG.
LetG△be the graph whose vertices correspond to the white triangles ofRtogether with an extra vertex v∞. The edges of G△ are the contacts between white triangles together with an edge between each of the suspensions andv∞. The degree ofv∞ is 3 and each corner of a white triangle is responsible for a contact, hence, every vertex ofG△ has degree at least 3.
Claim. G△is 3-connected.
Suppose there is a separating setU of size at most 2. LetC be be a component ofG△\U such thatv∞̸∈C. The convex hullHCof the corners of triangles inChas at least 3 corners.
Covering all the corners ofHCwith only two triangles results in a cornerpofHCthat has a contact to a triangleT ∈U such thatphas an angle larger thanπin the skeleton ofC+T. Sincepis a vertex ofH and angles larger than πdo not occur at vertices of degree 4 of an
SLTR, this is a contradiction. △
Therefore, ifH ̸=C3has an SLTR thenG△is 3-connected. On the other hand, if there is an internally 3-connected planar graphGsuch thatH is its medial graph, then by Lemma 2.29 it follows thatH has an SLTR.
Recall that a Henneberg type 2 step consists of subdividing an edge and connecting the new vertex to a third vertex (see page 5). To show that every generic circuit admits an SLTR, we will show that an SLTR can be extended along a plane Henneberg type 2 step. A plane Henneberg type 2 step is a Henneberg type 2 step that takes place ‘within one face’, i.e., an edge to subdivide is selected and the new vertex is connected to a vertex in one of the two faces bounded by this edge. For a planar generic circuit, we fix an embedding. The construction that we know to exist due to Theorem 2.33 gives us a reverse order. Reverse steps never violate planarity, hence, a 3-connected, plane generic circuit can be constructed fromK4 with plane Henneberg type 2 steps. We start by showing how to get an assignment of the extended graph. Then we will show that this assignment is indeed a GFAA. Since we only consider planar graphs and thus plane Henneberg type 2 steps, we omit the word
‘plane’ in the sequel.
The assignment. Given a graph G and a GFAA ψ of G. Let uv be the edge that is subdivided,xthe new vertex andwthe third vertex to whichxis connected (see Figure 2.40).
We denote the face incident to uv and w by f. After the Henneberg step, the new face incident touis denoted byfu and the other new face by fv. The third face incident touv is denoted byfx. The resulting graph is denoted byG+. We will construct an assignment ψ+ forG+ and prove thatψ+ is a GFAA.
There are three vertices not assigned to f under ψ: we will call them corners of f. We consider two cases, firstly fu is incident to all corners of f, secondly, fu is incident to precisely two corners off. The cases are depicted in Figure 2.40. Note that, ifwis a corner of f, it will be a corner for bothfu and fv. The vertices different from u, v, w, xthat are assigned to f under ψ will be assigned in the trivial way under ψ+, i.e., such a vertex is assigned tofu(or fv) if inG+ it is incident tofu (orfv).
Case 1: fu is incident to all corners off. Ifuorwis assigned tof underψ, it is assigned to fu underψ+. The vertexv is assigned tofxandxtofu underψ+.
Case 2: fu is incident to precisely two corners off. Ifuorw is assigned tof under ψ, it is assigned tofu underψ+, ifvwas assigned tof it is assigned tofv underψ+ andx is assigned to fx.
This yields an assignmentψ+ forG+.
Theorem 2.34. Given a 3-connected, plane graphGwith a GFAAψ. LetG+ be the result of a Henneberg type 2 step applied toG and letψ+ be the updated assignment. Thenψ+ is a GFAA andG+ admits an SLTR.
Proof. It is trivial thatψ+ satisfies Cv and Cf and hence, is an FAA.
We consider the induced families of pseudosegments, Σ andΣ+ ofψ and ψ+ respectively.
Since ψ is a Good FAA, we know that every subset of Σ has at least three free points or cardinality at most one. We will show that every subset of pseudosegments ofΣ+ of cardi-nality at least two has at least three free points. LetS⊆Σ+be a subset of pseudosegments of cardinality at least two. An endpoint of a pseudosegment inS that is not a free point of S is said to becovered.
Case 1: Let sx and sv be the pseudosegment that has x respectively v as interior point and letswthe pseudosegment containing the edge vw. IfS does not containsx, sv or
Henneberg Step
sv sw sx
sx
sw
sc Stretching
u
fu x v
fv f
u v w
fu fu
w x u
v
u
w v x
fv
w
v x
u
fv fu
w w
f
u v
c fv
Case 2:
Case 1:
Figure 2.40: A stretched representation of the original face, and of the results of a Henneberg type 2 step in Case 1 and Case 2.
sw thenS is also a subset of Σ, hence, it must have at least three free points. Sup-pose S ⊆ {sx, sv, sw}, then S has three free points, since, no two pseudosegments of {sx, sv, sw}touch twice and ifS={sx, sv, sw}there are precisely three of the six end-points covered. So supposeS contains at least one pseudosegment not of{sx, sv, sw}.
Consider the comparable set S′ ofΣ, that is
• If sx ∈ S, then, replacesx by the pseudosegment s′x of Σthat has u and v as interior points.
• If sv ∈ S, then, replace sv by the pseudosegment s′v of Σ that ends in v and contains all the edges ofsv but the edge vx.
• Ifsw∈S, then, delete sw.
Now we have S′∈Σ, thusS′ has three free points unless|S′|= 1.
• Ifsx∈S, then,sx contributes the same free points toS ass′xto S′.
• If sv ∈S, then, if v was a free point for S′, then x is forS, since, in this case sx̸∈S. Hence,sv contributes the same number of free points toS ass′v toS′.
• If sw ∈S and|S′|= 1, then, sw contributes at least one free point to S and it covers no other points, thus S has three free points. If|S′|>1 then S′ has at least three free points, adding sw does not cover any of them, and therefore, S has at least three free points.
We conclude thatS has at least three free points in this case.
Case 2: Ifwis a corner off, then, in this case it must be opposite to the subdivided edge, and, it follows immediately thatψ+ is a GFAA.
So supposew is not a corner off. Letsxandsw be the pseudosegments that havex respectivelywas an interior point and letscbe the pseudosegment containing the edge from w to the corner of f which is incident to fv. If S does not containsx, sw or sc thenSis also a subset ofΣ, hence, it must have at least three free points. SupposeS⊆ {sx, sw, sc}, thenShas three free points, since, no two pseudosegments of{sx, sw, sc} touch twice and ifS ={sx, sw, sc}there are precisely three of the six endpoints covered.
So suppose S contains at least one pseudosegment not of {sx, sw, sc}. Consider the comparable setS′ ofΣ, that is:
• If sx ∈S, then, replace sx by the pseudosegment s′x of Σthat has u and v as interior points,
• If sw ∈ S, then, replace sw by the pseudosegment s′w of Σ that has w as an interior point,
• Ifsc∈S, then, ifsw̸∈S, replacesc by the pseudosegment ofΣthat haswas an interior point, otherwise, deletesc.
Now we have S′ ∈ Σ, thus S′ has three free points unless |S′| = 1. If |S′| = 1 then S ⊆ {sw, sc} which contradicts the assumption that S contains at least one pseudosegment not of {sx, sw, sc}, thus|S′|>1.
• Ifsx∈S, then,sx contributes the same free points toS ass′xto S′.
• Ifsw∈S, then, supposexis covered inS. Note thatc, the corner off that is now infv, is a vertex of the pseudosegment ofΣthat containswas an interior point.
Since xis covered, sx ∈S, and either c is not free in S′ or it is an endpoint of two pseudosegments inS′. In the latter case,cis also a free point ofS assx∈S.
The other endpoint of sw is an endpoint of s′w. We conclude that replacing s′w bysw leaves the number of free points intact.
• If sc ∈S and sw ̸∈S, then, sc contributes at least as many free points to S as s′w toS′, so assume alsosw ∈S. The free points that s′w contributes to S′ are also free points of S as the endpoints of s′w are included in the endpoints of the set {sw, sc}. Hence,S has at least three free points.
We conclude thatS has at least three free points, hence,ψ+ is a GFAA.
Theorem 2.35. Every 3-connected, plane generic circuit admits an SLTR.
Proof. A 3-connected, generic circuit can be constructed with Henneberg type 2 steps from K4 (Berg and Jordan [BJ03]) and K4 admits an SLTR. Every plane 3-connected generic circuit can be constructed with Henneberg type 2 steps fromK4 such that all intermediate graphs are plane. Now it follows from Theorem 2.34 that every 3-connected, plane generic circuit admits an SLTR.