2
$\begingroup$

I have a set $\mathcal{P}$ of $n$ convex $k$-gons (convex $k$-vertex polygons) on the (Euclidean) plane. These define a partition of the plane, or rather the plane sans the pointer on the contour of any of the polygons, into contiguous regions, each defined by the subset of polygons covering it.

For a given $n$ and $k$, and over all sets $\mathcal{P}$ - what is the maximum size of the induced partition?

Note: The polygons are of finite size.


Example: Consider a set $\mathcal{P}$ of two squares, one being the same as the other but rotated by $\pi/4$:

enter image description here

So, $n = \left|\mathcal{P}\right|$ and $k = 4$. The induced partition of the plane has 9 regions - the 8 right-angle triangles, and the octagonal interior - plus the "outside", for 10 overall.

$\endgroup$
3
  • $\begingroup$ @einpoklum Can you please elaborate on how the parameter $k$ plays its role in the set systems? Say you have two triangles ($n=2$, $k=3$). Can you point out how that differs from the above Venn diagrams? And what changes when you have two squares ($n=2$, $k=4$)? $\endgroup$
    – codeR
    Commented Jun 14 at 8:09
  • $\begingroup$ @codeR: The Venn diagrams don't reflect $k$ at all. I'll try to replace them with another diagram soon. $\endgroup$
    – einpoklum
    Commented Jun 14 at 13:39
  • $\begingroup$ @HEKTO: See edit. $\endgroup$
    – einpoklum
    Commented Jun 14 at 13:49

1 Answer 1

2
$\begingroup$

Consider two $k$-gons, $A$ and $B$. Since we are talking about convex polygons, each edge of $A$ can at most cut $B$ at two points, with one point entering the polygon and with the other exiting. Thus, we will have a total of $2k$ intersection points. Now consider the planar graph of the vertices of the two $k$-gons and these $2k$ intersection points. Originally, there were $k$ edges in each of the $k$-gons (total $2k$). Each new intersection point splits two of the participating edges into two parts, thus introducing two new edges. Thus, $|V| = k + k + 2k = 4k$ and $|E| = k + k + 2\times2k = 6k$.

Using Eular's formula, $|V| - |E| + |F| = 2$, we have $|F| = 6k - 4k + 2 = 2k + 2$. Since in your case $k = 4$, you have $2\times4 + 2 = 10$ faces. Here, we are assuming the two polygons maximally intersect each other to maximize the number of faces.

I hope you can now generalize this for $n$ number of $k$-gons.

Use recurrences:
$|V_{n+1}| = |V_{n}| + k + n\times 2k$ with $V_1 = k$
$|E_{n+1}| = |E_{n}| + $_____ with $E_1 = k$
Hence $|F_{n+1}| = |F_{n}| + $____ with $F_1 = 2$

PS: The counting idea is based on this video (made by 3Blue1Brown) on the maximum number of regions that can be formed in a circle by the number of chords of the circle.

$\endgroup$
3
  • $\begingroup$ Yes thus $|V| = 4 + 4 + 8 = 16$ which is same as $4k$. $\endgroup$
    – codeR
    Commented Jun 14 at 16:18
  • $\begingroup$ So, $|F_{n+1}| = |F_n| + (n-1) \times 2k$ then? And $F_n = 2+k \times n \times (n-1)$? $\endgroup$
    – einpoklum
    Commented Jun 15 at 21:38
  • $\begingroup$ In the recurrence, i think it should be $n$ (not $n-1$). Why not verify the final answer with a few examples? $\endgroup$
    – codeR
    Commented Jun 16 at 17:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.