In general, the problems polygon
models are used to represent are real-world applications, and thus
it is appropriate to consider motions between configurations in 3 dimensions.
To this end, we introduce the notion of motion in terms of
pivots
, where a polygonal chain of polygon P in R
d is reflected
across a Hyperplane H which supports the convex hull of P and contains
at least two vertices of P. Since we are considering the case where d=3
(i.e. R
3 or 3D), the H is simply a halfplane which contains 2
points on the convex hull of P:
Theorem 2: Any planar
convex polygon P can be reconfigured into any other planar convex
polygon P' using pivots.
Proof:
Locate a quadrangle V
1V
2 V
3
V
4 where the vertices are labelled -, +, -, +, respectively,
in P. Then, step through the following motions:
- Pivot on V1V3, rotating the entire subpolygon
until V2 is over V1V3.
- Pivot on V2V4 to bring the quadrangle into
a planar configuration.
- Pivot on V4V1 to bring the subpolygon defined
by those two vertices into the plane of the quadrangle.
- Pivot on the remaining three edges to bring each of their subpolygons
into the plane of the quadrangle..
Lemma 4: Let V
1
V
2V
3V
4 be a planar convex quadrangle.
After two pivots, suppose the quadrangle is again planar, resulting
in a quadrangle, V
1V
2 V
3V
4
. Then
< V
2V
1 V
4 will be at least
the original value of the expression |
<V
2 V
1 V
3 -
< V
4 V
1
V
3 |.
Proof: If both pivots are on the diagonal
V
2V
4, then the angle at V
1
has not changed, and so we must only consider the remaining cases,
where the first (or both) pivot is on V
1V
3, and
where the first pivot is on V
2V
4 and the second
is on V
1 V
3 .
In the former case, the pivot must occur on a planar
quadrangle, with the vertex of the quadrangle on the moving subpolygon--say
it is V
2-- tracing a circle around V
1V
3 as it rotates, and thus the angle V
2V
1
V
3 remains constant throughout the pivot. Since
V
4 remains fixed during the motion, the angle V
4
V
1 V
3 also remains constant, and so the
angle after the pivot must be at least the difference |
<
V
2 V
1V
3 -
< V
4 V
1 V
3|.
In the latter case, the two pivots must bring the quadrangle
into a planar configuration, and the distance V
2V
4 is increasing during the motion, so by the law of sines,
< V
2V
1V
4 must also have increased,
and so is greater than |
<V
2V
1 V
3 -
< V
4V
1V
3
|.
This leads to the main theorem for this part:
Theorem 3: There exist
polygons which require
arbitrarily many pivots
to achieve a goal configuration.
Proof: Consider
the case of a pair of pivots that result in a planar non-intersecting
quadrangle after the motions depicted in the figure below. The parallelogram
has the property that <V1 can be made arbitrarily close
to zero, and since it is not a rhombus,
Also, since for small angles <x, sin x
~
x, and so as <V
1 decreases, for every two pivots <V
1 is only reduced to an angle <V
1'
where