Peeling a polygonal vegetable

Question

When you peel a vegetable, such as a potato or a cucumber, you usually remove its head, then contiually remove parts of its skin, until you remain with the pulp alone. I would like to formalize this process.

For concreteness, let's assume that the vegetable is a planar polygon $V$, the head is a polygon contained in $V$ (grey in the image below) and the pulp is another polygon contained in $V$ and disjoint from the head (green):

A peeling function is a function $P(t)$, from $[0,1]$ to subsets of $V$, with the following properties:

1. $P(0)$ is the head;
2. $P(1)$ is the complement of the pulp;
3. $P(t)$ is monotonically increasing with $t$, i.e. if $t'>t$ then $P(t')\supset P(t)$;
4. For every $t\in[0,1]$, both $P(t)$ and $V\setminus P(t)$ are connected polygons (possibly with holes).

The following image shows (in grey) six values of a possible peeling function on the example vegetable:

Does a peeling function always exist?

A necessary condition for its existence is that its values in 0 and 1 fulfil the connectivity requirement. This requires that the head, the pulp and their complements are all connected polygons. Equivalently, the head and the pulp are simply-connected.

MY QUESTION IS:

If the head and the pulp are simply-connected, does there always exist a peeling function?

EDIT: I asked a simpler question in Math.SE. Rahul's answer seem to be relevant here.

Answer 1

I'm going to assume the head and pulp are always contained in the interior of $V$, although the other three cases where parts of the boundary may be shared can be treated similarly.

Suppose you have already solved an easy vegetable $V$, e.g., you have a peeling function $P$ roughly specified by the sequence of pictures in your question. For any other vegetable $V'$ satisfying your conditions, there exists a PL homeomorphism $g: V \to V'$ from your original vegetable that takes heads to heads and pulps to pulps (this uses the existence of suitably fine polygonal subdivisions on $V$ and $V'$). Then, it is straightforward to check that the composite $g \circ P$ is a peeling function on $V'$.

Answer 2

This subverts your intent, but in this example, $P(t)$ cannot remain connected as it must jump to other hole, and then outside the pulp.

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So perhaps you should assume that the pulp polygon is simply connected...

Answer 3

Here are a few observations that fall short of a convincing algorithm.

The outermost polygon $V$ might have holes; call them $h_i$. The pulp is just like a hole; call that $h_1$. So now $V$ has holes $h_1,h_2,\ldots,h_k$. Let $P=P(0)$ be the initial polygon.

Find disjoint paths from each $h_i$ to the boundary $\partial V$ of $V$. Connect each $h_i$ to $\partial V$ by a thin corridor. Now we have a simple (no holes) polygon $V'$ containing $P$. If $P$ can be grown inside $V'$ so that $P(t)$ is connected and $\overline{P(t)} = V' \setminus P(t)$ is connected, then it would be easy to fill in the thin corridors methodically to maintain "dual connectivity," connectivity of $P(t)$ and $\overline{P(t)}$. So now the problem has been reduced to growing $P$ within a simple polygon $V'$.

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For the latter, I think this could work. Connect $P$ to $\partial V'$ with a thin corridor so that now we have just a simple polygon $V''$ with some edges originally belonging to $P$. Compute the straight skeleton of $V''$. This partitions $V''$ into faces, each associated with an edge of $\partial V''$; see this impressive CGAL image. I think it is possible to methodically grow from the $P$ edges through the face structure defined by the straight skeleton, to fill $V''$ while maintaining dual connectivity.