Reconfiguration with Pivots

    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 Rd 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. R3 or 3D), the H is simply a halfplane which contains 2 points on the convex hull of P:

flipping over a hyperplane

Theorem 2: Any planar convex polygon P can be reconfigured into any other planar convex polygon P' using pivots.

Proof: Locate a quadrangle V1V2 V3 V 4 where the vertices are labelled -, +, -, +, respectively, in P. Then, step through the following motions:

Lemma 4: Let V1 V 2V3V4 be a planar convex quadrangle. After two pivots, suppose the quadrangle is again planar, resulting in a quadrangle, V1V2 V3V4 . Then
V2V1 V4 will be at least the original value of the expression | <V2 V 1 V3 - < V4 V1 V 3 |.

Proof: If both pivots are on the diagonal V2V 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 V1V3, and where the first pivot is on V 2V4 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 V1V 3 as it rotates, and thus the angle V2V1 V 3 remains constant throughout the pivot. Since V 4 remains fixed during the motion, the angle V4 V 1 V3 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 V2V 4 is increasing during the motion, so by the law of sines, < V2V1V4 must also have increased, and so is greater than |<V2V1 V 3 - < V4V1V3 |.
End of proof

2 pivots preserve planarity

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 <V1 decreases, for every two pivots <V 1 is only reduced to an angle <V1' where

more angles

    And so since <V1approaches but does not attain zero (since we disallow the "flat" self-intersecting configuration that would result), a goal configuration with an arbitrarily low <V 1can be chosen, so that an arbitrary number of pivots are needed to reconfigure the polygon. This is easily extended for arbitrary pivots by considering any arbitrary pivot as a pair of pivots on the same diagonal, where the first pivot brings the quadrangle into a planar non-intersecting position and the second produces the original pivot desired.

A parallelogram can be found that requires arbitrarily many pivots

End of proof

    So it is observed that there does indeed exist a motion between any two planar convex polygons using only pivots, but this motion is not bounded by the number of edges of the polygon-- it can be achieved with arbitrarily many pivots if we choose a configuration as in Theorem 3.

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