15
A PPENDIX
Proof of Lemma 5
Proof: Consider first crossing across two adjacent
sectors. Let the line segment uw intersect the circular
sectors pi vi pi+1 and vi−1 pi vi such that u and w are
the intersection points (see Fig. 11). Consider the convex
quadrilateral vi w pi u . Observe that |vi w | = |vi pi | =
|u pi | = r. Referring to Fig. 11 (left), we calculate the
w as follows:
length of u
| = (g + l)2 + (r − k − h)2
|u w = g 2 + 2gl + l2 + r2 + k 2 + h2 − 2rk − 2rh + 2hk.
Using Pythagoras’ theorem, g can be calculated by
g 2 = r2 − (r − h)2 = r2 − (r2 − 2rh + h2 ) = 2rh − h2 .
Similarly, for l holds l2 = 2rk − k 2 . Substituting g 2 and l2
in the equation above andusing that g, h, l, k √
are positive,
we have |uw| ≥ |u w | = 2gl + r2 + 2hk ≥ r2 = r.
If the two sectors are not adjacent, then uw (that is,
uw) must intersect one of existing line segments vi pi ,
as depicted in Fig. 11 (right). We extend the arcs on
intermediate adjacent sectors such that they intersect
u w in u and w . Applying the arguments above gives
|u w | ≥ |u w | ≥ r. Thus, any line segment uw intersecting S as specified above has length |uw| ≥ r.
Relation of Sweep Circle, Twisting Triangle, and planar subgraph traversals—formal proofs
Lemma 11: An edge vw of a Sweep Circle (SC) traversal path is never intersected by an edge ux of the Gabriel
subgraph.
Proof: Let CSC be the Sweep Circle with v and w on
its perimeter, ux an edge of the Gabriel subgraph, and
CGG the Gabriel circle having ux as diameter. Suppose
the edge ux intersects vw. Since CSC is empty, u and x
must be outside CSC such that ux intersects CSC in two
points which we denote u and x (see Figure 25). Note
that the diameter of CGG is restricted by r and cannot
exceed the diameter of CSC . The sum of the angles ∠wu v
and ∠wx v is π. Then the sum of the angles ∠wuv and
∠wxv is then smaller than π. Consider the quadrangle
wuvx. The Gabriel graph condition requires that w and
v are outside the circle CGG , thus the angles at w and v
(∠uwx and ∠uvx) are smaller than π/2. This means that
the sum of all angles in the quadrangle wuvx is smaller
than 2π, which is a contradiction.
Theorem 2: Let PGG be a sequence of nodes by a
boundary traversal on a Gabriel subgraph. Let PSC be a
Sweep Circle traversal path, and PTT a Twisting Triangle
traversal path. Then it holds that (a) PGG ⊆ PSC and (b)
PGG ⊆ PTT and PSC ⊆ PTT .
Proof: a) Let w be the next node in PSC after node v.
Then w and v are on the perimeter of the empty Sweep
Circle CSC with diameter r. Consider the Gabriel circle
CGG having vw as diameter (i.e. its radius is ≤ r) as
shown in Figure 26. There are no points above vw that
are contained in both circles. Below vw there are two
areas, one is their common intersection, which does not
contain any point. There is another area which is in CGG
but not in CSC . If in that area there is a node x then vw is
not in the Gabriel subgraph but there is a path from v via
x to w in the Gabriel subgraph, and the SC traversal path
containing the edge vw becomes a subsequence of the
path along the Gabriel subgraph. The latter path cannot
intersect vw by Lemma 11. If there is no such node x
then vw is also in the Gabriel subgraph.
b) Fig. 22 and 23 give examples where nodes are part
of PGG or PSC , but not part of PTT and vice versa.
Corollary 3: The Sweep Circle traversal path is of the
same length or shorter than a boundary traversal on a
Gabriel subgraph.
Proof: This follows directly from Theorem 2. An
example is shown in Figure 5.
The Twisting Triangle traversal is not a subsequence
of the Gabriel graph traversal, but a subsequence of a
traversal of the relative neighborhood subgraph (RNG,
[19]), a sparser subgraph structure [26]. However, this
does not imply that worst case RNG constructions are
also worst case constructions for the TT traversal.
Example network
Fig. 25: Illustration
Lemma 11
for
Fig. 26: Illustration for Theorem 2 (1)
Fig. 27: Unit disk graph with density 6 and circular void region in
the center.