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\title{Randomized Incremental Construction of a 3D-Convex Hull}
\author{Wolfgang Mulzer}
\begin{document}
\maketitle
\section{The Algorithm}
Let $P = \{p_1, p_2, \dots, p_n\}$ a set of $n$ point in $\R^3$ in
general position (i.e., no four points from $P$ lie on a common
plane). We would like to compute
$\conv(P)$, the convex hull of $P$.
For $i = 4, \dots, n$, set $P_i = \{p_1, p_2, \dots, p_i\}$.
The incremental algorithm successively computes
$\conv(P_4), \conv(P_5), \dots, \conv(P_n)$. In each step,
we find $\conv(P_i)$ by inserting the point
$p_i$ into $\conv(P_{i-1})$.
Each convex hull is stored as a DCEL, representing the planar
graph $G(\conv(P_i))$.
To facilitate the insertion of the next point, the algorithm stores
\emph{conflict information}:
for each point $p \in P \setminus P_i$, there is a pointer $C_p$ to
a facet $f$ of $\conv(P_i)$, so that $p$ lies to the left of the
oriented plane through $f$.
The facet $f$ muss be deleted when $p$ is inserted. Thus, it is called
a \emph{conflict facet} of $p$.
For each facet $f$ of $\conv(P_i)$, we store a \emph{conflict list} $C_f$
that contains all points $p \in P \setminus P_i$ with $C_p = f$.
Initially, $\conv(P_4)$ and the conflict information can be computed
easily.
Now we describe how to use the conflict information in order to insert
$p_i$ into $\conv(P_{i-1})$.
If $C_{p_i} = \perp$, then $p_i$ lies inside of $\conv(P_{i-1})$,
and nothing has to be done.
Otherwise, we know that the facet $f = C_{p_i}$ must be deleted.
We perform a breadth-first search from $f$ in the dual graph of
$G(\conv(P_{i-1}))$, visiting only facets of $G(\conv(P_{i-1})$ that
are in conflict with $p_i$.
That is, we only insert an unvisited facet $f'$ into the BFS-queue,
if $p_i$ lies to the left of the oriented plane spanned by $f'$.
Let $D_i$ be the set of all facets that are in conflict with $p_i$,
the \emph{conflict region of $p_i$}.
The conflict region is bounded by a simple cycle in
$G(\conv(P{i-1}))$. We call this cycle $Z_i$.
In $\conv(P_i)$, we create new facets between $p_i$ and each
edge of $Z_i$, and we delete all facets in $D_i$.
It remains to update the conflict information.
For this, we consider for each conflict facet $f' \in D_i$
the corresponding conflict list $C_{f'}$.
For each point $q \in C_{f'}$ we do the following: similarly
to $p_i$, we perform a BFS from $f'$ to determine the conflict
region of $q$.
However, we restrict the BFS to the facets in $D_i$.
In this way, we find all edges of $Z_i$ that are incident to
a facet that is in conflict with both
$q$ and $p_i$. For each such edge $e$, we test whether $q$ is
in conflict with the newly created facet of $\conv(P_i)$ incident to $e$.
If so, we update the conflict pointer $C_q$ and the conflict list for
the corresponding facet. If no such facet is found, we set
$C_q = \perp$.
This concludes the description of one insertion.
We repeat this process until we have found
$\conv(P_n) = \conv(P)$.
\section{Running Time Analysis}
\subsection{Structural Change}
Each facet is deleted at most once, so it suffices to bound the
number of facets that are created. In the worst case, this can
be quadratic. Using backwards analysis, however, we can show that
the expected structural change is $O(n)$ (see class).
\subsection{Conflict Change}
While updating the conflict information, each facet $f$ is
traversed at most once for each point that is in conflict with
$f$. Thus, the total time for the conflict change is
proportional to the total number of conflicts that exist
between the created facets and the points in $P$. We can show
that this number is $O(n \log n)$ in expectation. For this,
we need a three-dimensional version of Clarkson's theorem,
which states that the number of oriented planes that are
spanned by three points of $P$ and that have at most $k$
points from $P$ to the left is at most $O(nk^2)$.
\end{document}