Jordan Polygon Parity Lemma

Definition
Let $P$ be a polygon embedded in $\R^2$.

Let $U \subseteq \R^2 \setminus \partial P$ be a path-connected subset of $\R^2 \setminus \partial P$.

Let $q \in U$, and let $\mathbf v \in \R^2 \setminus \left\{ {\mathbf 0}\right\}$ be a non-zero vector.

Let $\mathcal L = \left\{ {q + s \mathbf v: s \in \R_{\ge 0} }\right\}$ be a ray with start point $q$.

Then:


 * $(1): \quad$ The parity $\operatorname{par} \left({q}\right)$ is independent of the choice of $v$.
 * $(2): \quad$ All points in $U$ have the same parity.

Proof
Define $g: \R \to \R^2$ by $g \left({\theta}\right) = \left({\cos \theta, \sin \theta}\right)$.

Define $\mathcal L_\theta = \left\{ {q + s g \left({\theta}\right): s \in \R_{\ge 0} }\right\}$ as a ray with start point $q$.

Note that the image of $g$ is equal to the unit circle.

Then $\dfrac{v}{\left\Vert{\mathbf v}\right\Vert} = g\left({\theta'}\right)$ for some $\theta' \in \R$, where $\left\Vert{\mathbf v}\right\Vert$ denotes the Euclidean norm of $v$.

Then $\mathcal L = \mathcal L_{\theta'}$, as $q + s \mathbf v = q + \left({ s \left\Vert{\mathbf v}\right\Vert }\right) g \left({\theta'}\right)$.

$\mathcal L_\theta \cap \partial P$ consists of a finite number of line segments, some of which are crossings.

Let $N_\theta \left({q}\right) \in \N$ be the number of crossings of $\mathcal L_\theta$.

As the value of $\theta$ increases, the value of $N_\theta \left({q}\right) \in \N$ only changes when $\mathcal L_\theta$ crosses a vertex of $P$.

If $\theta_0 \in \R$ is a value for which $\mathcal L_{\theta_0}$ crosses exactly one vertex $A$ or side $S$, there are three possibilities:


 * $\displaystyle N_{\theta_0} \left({q}\right) = \lim_{\theta \to \theta_0} N_\theta \left({q}\right)$, if $A$ or $S$ is not part of a crossing.


 * $\displaystyle N_{\theta_0} \left({q}\right) = \lim_{\theta \to \theta_0^-} N_\theta \left({q}\right) = \lim_{\theta \to \theta_0^+} N_\theta \left({q}\right) - 2$, if $A$ or $S$ is part of a crossing, and $\mathcal L_{\theta_0}$ intersects both lines adjacent to $A$ or $S$ for some $\theta > \theta_0$.


 * $\displaystyle N_{\theta_0} \left({q}\right) = \lim_{\theta \to \theta_0^-} N_\theta \left({q}\right) - 2 = \lim_{\theta \to \theta_0^+} N_\theta \left({q}\right)$, if $A$ or $S$ is part of a crossing, and $\mathcal L_{\theta_0}$ intersects both lines adjacent to $A$ or $S$ for some $\theta < \theta_0$.

Here, $\displaystyle \lim_{\theta \to \theta_0^-} $ denotes a limit from the left, and $\displaystyle \lim_{\theta \to \theta_0^+} $ denotes a limit from the right.


 * [[File:JordanPolygon2.png]]

The figure shows the change of $N_\theta \left({q}\right)$ each time $\mathcal L_\theta$ intersects a vertex.

The change of $N_\theta \left({q}\right)$ is for increasing values of $\theta$, which corresponds to a counterclockwise rotation of $\mathcal L_\theta$ around $q$.

If $\mathcal L_{\theta_0}$ crosses more than one vertex, $\displaystyle N_{\theta_0} \left({q}\right)$ may change be a larger number, but always by a multiple of $2$.

Hence, $N_\theta \left({q}\right) \bmod 2$ is independent of $\theta$, so we define the parity of $q$ as $\operatorname{par} \left({q}\right) := N_\theta \left({q}\right) \bmod 2$.

Since $\mathcal L = \mathcal L_{\theta'}$ for some $\theta' \in \R$, it follows that $\operatorname{par} \left({q}\right)$ can also be defined as the number of crossings modulo $2$ between $\mathcal L$ and $\partial P$.

Let $\sigma: \left[{0\,.\,.\,1}\right] \to \R^2 \setminus \partial P$ be any path in $U$ with initial point $q$.

For fixed $\theta \in \R$ and $t \in \left[{0\,.\,.\,1}\right]$, define $\mathcal L_{\theta, t} = \left\{ {\sigma \left({t}\right) + s g \left({\theta}\right): s \in \R_{\ge 0} }\right\}$ as a ray with start point $\sigma \left({t}\right)$.

A similar argument to the one above shows that for all $t_0 \in \left[{0\,.\,.\,1}\right]$:


 * $\displaystyle \lim_{t \to t_0} N_\theta \left({\sigma \left({t}\right) + s g \left({\theta}\right) }\right) \equiv N_{\theta} \left({\sigma \left({t_0}\right) + s g \left({\theta}\right) }\right) \pmod 2$

So all point in $\operatorname{Im} \left({\sigma}\right)$ have the same parity, which shows that all points in $U$ have the same parity.