Number of vertices and edges lies on the boundary of bounded cells in line arrangement

Question

I am learning computational geometry by myself using these lecture notes from ETH Zurich. Here is an exercise (8.9) I have been stuck for a few days:

For a line arrangement $A$ of a set of $n$ lines in $R^2$, let $F$ denote the union of the closure of all bounded cells. Show that the complexity (number of vertices and edges of the arrangement lying on the boundary) of $F$ is $O(n)$.

My intuition is to use induction, proving that when a new line $l$ is added, the number of added edges is a constant. This is similar to how we proved the zone theorem. However, I cannot figure it out. Can someone provide me with some hints?

Observation: Although each cell is convex, the union of all cells is not necessarily. Example: https://imgur.com/ooiYWS6 — j_random_hacker, Nov 12 '20 at 17:33

Vladislav Bezhentsev · Accepted Answer · 2020-11-12T22:03:01.920

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Consider 3 additional lines such that their arrangement consists of a single bounded (triangular) cell $C$ and $F \subset C$. Then every edge of $F$ belongs to a zone (using a terminology from your textbook) of at least one of the 3 added lines. And all 3 zones of 3 added lines contain at most $30n$ edges according to Zone Theorem. So $F$ has no more than $30n$ edges which is $O(n)$.

edited Nov 12 '20 at 22:03

answered Nov 12 '20 at 20:54

Vladislav Bezhentsev

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Or use a bounding triangle for $\le 30n$ edges. – j_random_hacker Nov 12 '20 at 22:00
@j_random_hacker Thank you, I've edited the answer. – Vladislav Bezhentsev Nov 12 '20 at 22:03

Number of vertices and edges lies on the boundary of bounded cells in line arrangement

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