*
Reconfiguring Between Two Convex Configurations*

We begin with a sidebar for identifying whether a sequence of edge lengths
admits a convex configuration:

*Fact 1* :
The edge lengths L

_{1} , L

_{2} , ..., L

_{
n} admit a convex configuration precisely if , for all i,

Note that Fact 1 is always a useful piece of knowledge
to have as support when one is delving into the world of convexity.

Now, starting with a pair of distinct convex configurations
C and C', we devise a way of reconfiguring C to reach C'. First, note
that every vertex angle in C will either be equal to the corresponding
vertex angle in C', or it will have to be increased or decreased by reconfigurations
on the way from C to C'. So, labelling each angle of C +, -, or 0 depending
on whether it has to increase, decrease or not change at all to match
the corresponding angle in C', we have a labelling of vertices that can
be used in conjuction with the following lemma:

*Lemma 2**
(Cauchy-Steinitz Lemma)*: In a
+,-,0 labelling that comes from two distinct convex configurations, there
are at least four sign alterations.

Proof: Each configuration is distinct
and circular, and thus not all labels are zero, and the number of sign
alterations must be even. This number also cannot be two, since then we
have one chain of increasing angles, and one chain of decreasing angles,
and this implies the lengthening of one chain and shortening of the other,
which cannot be achieved.

Taking four vertices V

_{i} , V

_{j} , V

_{
k}, V

_{ l} around the polygon with respective alternating
signs of +,-,+,-, the quadrangle defined by the vertices can be

*flexed*
to achieve a label of 0 for one of the vertex angles-- this can be
seen more clearly through the use of the lemma 3.

*Lemma 3*: Given a convex quadrangle
V

_{1}, V

_{2}, V

_{ 3}, V

_{4}, there
is a motion that decreases the angles of V

_{ 1} and V

_{3}
, and increases the angles of V

_{2} and V

_{ 4}.
The motion can continue until one of the angles reaches 0 or PI.

Proof: Consider a motion where V

_{1}
is held fixed and V

_{3} moves along the directed line from V

_{
1} to V

_{3}, and the distances between each vertex on the
quadrangle and the vertices adjacent to it (V

_{1} and V

_{2}
, V

_{2} and V

_{3}, etc) remain constant. Then we can apply

Euclid's Proposition I.25
to triangle V

_{1}, V

_{2}, V

_{3}, and note that
the angles at V

_{2} and V

_{4} must be increasing since
|V

_{1} - V

_{3}| is increasing. Since the angles are restricted
to lie between 0 and PI, the convex quadrangle is maintained throughout
the motion, and by Lemma 2 we have four sign alterations in the vertex
angle labels, we see that the angles at V

_{1} and V

_{3}
are decreasing throughout the motion.

Thus we can now define the main theorem for this part:

*Theorem 1***
:** Given two convex configurations C and C' of the same
sequence of edge lengths L1, ..., Ln, there is an angle-monotone motion
from C to C' that involves O(n) moves each of which changes only four vertex
angles at once.

**Proof:** By Lemma 2, we can find
vertices Vi, Vj, Vk, V

_{l} in cyclic order around the polygon in
configuration C, and whose angles are labelled +, - , + , -, in that order.
We obtain a convex quadrangle by specifying the rigidity of the subchains
between each of the vertices, and apply motions to this quadrangle as
per Lemma 3 until the label on one angle becomes 0. Repeating this process
until all vertex angle labels are 0 yields a sequence of motions from
C to C' with at most n moves, since each motion changes at least one non-zero
label to zero, and each zero label remains unchanged during subsequent motions.

*Proposition 1:* Computing
the motion in Theorem 1 can be done in O(n) time on a pointer machine
with real numbers.

**Proof:** The first step in computing
the motion in Theorem 1 is to maintain vertices V

_{i} , V

_{
j}, V

_{k}, V

_{ l} throughout the motion.
In order to do this, we maintain four consecutive blocks I, J, K, L
of the same sign--so I is a block of consecutive vertices labelled "+",
J, is a block of consecutive vertices labelled "-", and so on. Each block
can be created in linear time with an initial scan of the polygon's vertices
in order, and each block contains pointers to its first and last vertices.
The quadrangle vertices V

_{ i} , V

_{j} , V

_{ k}
, V

_{l} are then identified as the first vertex in the corresponding
block (so V

_{i} is the first vertex in block I, V

_{j}
is the first vertex in J, ...), and when a motion changes a vertex angle's
label to 0, the pointer for that vertex moves to the next one in the block.
If block I becomes empty (i.e. the pointer for I is incremented after the
last vertex in the block obtains a zero label), move block I to the set
of "+"-labelled vertices after L, and if L becomes empty, move it to the
block of -'s before I. If block K becomes empty, it advances to the block
after L, J and L merge to form a new J block, and L advances to the block
after the new K. Likewise, if J becomes empty, J moves to the block of
-'s before I, blocks I and K merge to form a new K block, and I moves to
the block of +'s before J (see below).

Now we apply the quadrangle motions of Lemma 3, including
computing the finishing times for each of the vertex angles to reach a
0 label, and taking the vertex with the lowest of these computed times
as the focal vertex in the Lemma. Since the algorithm effects a constant
number of steps for each vertex in the polygon, and the number of iterations
is a multiple of the number of vertices in the polygon, we have an O(n)
algorithm for angle-monotone motions between two convex polygons with a
common sequence of edge lengths.

** Thus we see that a reconfiguration between any two convex polygons
is indeed possible, and it can be computed in linear time with the most
direct moves specified above.**