Extending the centerpoint theorem to multiple points

Note that condition (1) is equivalent to the condition that every open halfplane containing more than $\frac{4n}{5}$ of the points of $P$ must contain both $p_{1}$ and $p_{2}$. Similarly, condition (2) is equivalent to the condition that every open halfplane containing more than $\frac{3n}{5}$ of the points of $P$ must contain one of $p_{1}$ and $p_{2}$. We will now construct two points $p_{1}$ and $p_{2}$ satisfying both these conditions.

Let $C$ be the intersection of all open halfplanes containing more than $\frac{4n}{5}$ of the points of $P$. Clearly $C$ is convex. Also, note that $C$ is closed. The centerpoint region is a strict subset of $C$ and thus $C$ has a non-empty interior. In order to satisfy condition (1), both $p_{1}$ and $p_{2}$ have to be placed in $C$.

Let now $H$ be the set of all open halfplanes containing more than $\frac{3n}{5}$ of the points of $P$. For any $h_{i}$ in $H$ let $c_{i}$ be the intersection of $h_{i}$ and $C$. In order to also satisfy condition (2), we need to find two points $p_{1}$ and $p_{2}$ such that every $c_{i}$ contains at least one of them. To this end, we partition $H$ into two subsets $L$ and $R$. The set $L$ contains all halfplanes that lie on the left side of their respective boundary lines. Analogously, $R$ contains all halfplanes that lie on the right side of their respective boundary lines. For a halfplane $h_{i}$ that has a horizontal boundary line, we put $h_{i}$ in $L$ if and only if it lies above its boundary line.

Note that any three halfplanes in $L$ have a non-empty intersection: Consider the inclusion-minimal halfplane $h\in L$ with horizontal boundary line and its intersection $r$ with the boundary of the convex hull of $P$. As $h$ is open, $r$ is not in $h$. However, we claim that any point $r^{\prime}$ in $h$ on the convex hull boundary of $P$ in an $\varepsilon$-neighborhood of $r$ is in any halfplane of $L$. Indeed, if there was a halfplane in $L$ not containing $r^{\prime}$, it would contain a strict subset of the intersection of the convex hull of $P$ with $h$; however, this would contradict the minimality of $h$. The analogous holds for $R$.

We will now show that for any two halfplanes $h_{1}$ and $h_{2}$ in $L$, their corresponding regions $c_{1}$ and $c_{2}$ have a non-empty intersection. The same arguments hold for any two halfplanes in $R$. Assume for the sake of contradiction that $c_{1}$ and $c_{2}$ do not intersect. As $C$ and $h_{1}\cap h_{2}$ are convex, this means that there is an open halfplane $g$ containing more than $\frac{4n}{5}$ of the points of $P$ such that the intersection of the boundary lines of $h_{1}$ and $h_{2}$ lies in $\overline{g}$, the complement of $g$ (see Figure 1). In particular, $g\cap h_{1}$ is a strict subset of $\overline{h_{2}}$. As $\overline{g}$ contains strictly fewer than $\frac{n}{5}$ of the points of $P$ and $\overline{h_{1}}$ contains strictly fewer than $\frac{2n}{5}$ of the points of $P$, $g\cap h_{1}$ must contain strictly more than $\frac{2n}{5}$ of the points of $P$. However, being a subset of $\overline{h_{2}}$, which also contains strictly fewer than $\frac{2n}{5}$ of the points of $P$, this is a contradiction. Thus, by contradiction, $c_{1}$ and $c_{2}$ intersect.

As neither three halfplanes in $L$ nor two halfplanes in $L$ and $C$ have an empty intersection, Helly’s Theorem entails that there exists a point in both $C$ and all halfplanes in $L$, i.e., all $c_{i}$s associated to $L$ have a non-empty intersection $D_{L}$. Again, the same holds for $R$, with a non-empty intersection $D_{R}$. Placing $p_{1}$ in $D_{L}$ and $p_{2}$ in $D_{R}$, we have thus constructed two points such that the conditions (1) and (2) hold. ∎

The result is straightforward for $d=1$, so assume $d\geq 2$. Again the condition that for each closed halfspace $h^{\prime}$ containing $j$ of the points $p_{1},\ldots,p_{k}$ we have $\mu(h^{\prime})\geq\alpha_{j}$ is equivalent to the condition that every open halfspace $h$ with $\mu(h)>1-\alpha_{j}$ must contain at least $k-j$ of the points $p_{1},\ldots,p_{k}$. Let $\alpha_{0}=0$. For $1\leq j\leq k$, we call an open halfspace $h$ a $j$-halfspace if $1-\alpha_{k-j+1}<\mu(h)\leq 1-\alpha_{k-j}$. Consider the $x_{1}$-$x_{2}$-plane, denoted by $X$, and for each vector $v=(v_{1},v_{2},\ldots,v_{d})$ in $\mathbb{R}^{d}$ let $\pi(v)=(v_{1},v_{2},0,\ldots,0)$ be the projection of $v$ to $X$. Let $v_{1},\ldots,v_{k}$ be $k$ unit vectors in $X$ with the property that the angle between any $v_{i}$ and $v_{i+1}$ is $\frac{2\pi}{k}$. Note that this implies that also the angle between $v_{k}$ and $v_{1}$ is $\frac{2\pi}{k}$. For each $v_{i}$ we construct a principal set $V_{i}$ of halfspaces as follows: For each $j$, consider all $j$-halfspaces. For any such halfspace $h$, let $n(h)$ be the normal vector to its bounding hyperplane that points into $h$. Let $h$ be in $V_{i}$ if the angle between $\pi(n(h))$ and $v_{i}$ is at most $\frac{j\pi}{k}$. If $\pi(n(h))=0$, place $h$ arbitrarily in $j$ of the $V_{i}$’s. Note that with this construction each $j$-halfspace is contained in exactly $j$ principal sets. Thus, if, for each principal set, we can pick a point in all its halfplanes, then each $j$-halfplane contains $j$ points.

It remains to show that the halfspaces in each principal set have a common intersection. Let $h_{1},\ldots,h_{d+1}$ be $d+1$ halfspaces in $V_{i}$ and assume for the sake of contradiction that they have no common intersection. Then the positive hull (conical hull) of their projected normal vectors must be $X$, and in particular there are three of them, w.l.o.g. $h_{1}$, $h_{2}$ and $h_{3}$, whose projected normal vectors already have $X$ as their positive hull. Further, among those three halfspaces, there are two of them, w.l.o.g. $h_{1}$ and $h_{2}$, such that the angles between their projected normal vectors and $v_{i}$ sum up to more than $\pi$. If $h_{1}$ is a $j_{1}$-halfspace, then by construction of $V_{i}$ we have that the angle between $\pi(n(h_{1}))$ and $v_{i}$ is at most $\frac{j_{1}\pi}{k}$. Analogously, if $h_{2}$ is a $j_{2}$-halfspace, the angle between $\pi(n(h_{2}))$ and $v_{i}$ is at most $\frac{j_{2}\pi}{k}$. By the choice of $h_{1}$ and $h_{2}$ we thus have $\frac{(j_{1}+j_{2})\pi}{k}>\pi$, which is equivalent to $j_{1}+j_{2}>k$, and to $j_{1}+j_{2}\geq k+1$, as $j_{1}$ and $j_{2}$ are integers. By definition of a $j$-halfspace we have

Furthermore we have $\mu(h_{i})>1-\alpha_{k}$ for every $i\in\{1,\ldots,d+1\}$, and thus

which is equivalent to

As $k+1-j_{1}+k+1-j_{2}=2k+2-(j_{1}+j_{2})\leq k+1$, we have that $(d-1)\alpha_{k}+\alpha_{k+1-j_{1}}+\alpha_{k+1-j_{2}}\leq 1$ and thus $\mu(h_{1})+\ldots+\mu(h_{d+1})>d$, which is a contradiction to Observation 4. ∎

Let $C$ be the intersection of all open halfplanes containing more than $\frac{5n}{6}$ of the points of $P$. Just as in the proof of Theorem 3, condition (1) is equivalent to $p_{1}$, $p_{2}$ and $p_{3}$ lying in $C$. Similarly, condition (2) is equivalent to the following statement: for every open halfplane $h$ containing more than $\frac{n}{2}$ of the points of $P$, $h$ contains at least one of $p_{1}$, $p_{2}$ and $p_{3}$. For the latter, let $n_{i}$ be a vector and let $N_{i}$ be the set of all of these halfplanes that have $n_{i}$ as their normal vector. The intersection of all elements of $N_{i}$ is a closed halfspace $h_{i}$. Let $K$ be the set of all of these $h_{i}$, i.e., for every possible direction of $n_{i}$. Then, condition (2) is equivalent to the following statement: for every halfplane $h$ in $K$, $h$ contains at least one of $p_{1}$, $p_{2}$ and $p_{3}$. For each such halfplane $h_{i}$, let $c_{i}$ be the intersection of $h_{i}$ and $C$. Note that $c_{i}$ is compact. We thus want to show the claim that we can find three points $p_{1}$, $p_{2}$ and $p_{3}$ such that each $c_{i}$ contains at least one of them. Let $H$ be the set of $c_{i}$s that are minimal with respect to set inclusion. Clearly, it is enough to show the claim just for the elements of $H$. Let $N$ be the set of normal vectors of the defining halfplanes of the elements of $H$. Note that there is a natural mapping from $N$ to a subset of the unit circle $S^{1}$. We will say that a subset of $N$ is connected, if its image under this mapping to $S^{1}$ is connected.

First we show that among any three elements of $H$, two of them intersect. Let $c_{1}$, $c_{2}$ and $c_{3}$ be elements of $H$, and let $h_{1}$, $h_{2}$ and $h_{3}$ be their associated halfplanes. Assume for the sake of contradiction that $c_{1}$, $c_{2}$ and $c_{3}$ are pairwise disjoint. Let $n_{i}$ be the normal vector of $h_{i}$. Let $A$ be the positive hull of $n_{1}$, $n_{2}$ and $n_{3}$. Then $A$ is either a cone or the whole plane. See Figure 3. If $A$ is the whole plane, then $h_{1}$, $h_{2}$ and $h_{3}$ have no common intersection. Otherwise, if $A$ is a cone, then one of $n_{1}$, $n_{2}$ and $n_{3}$ can be described as a positive linear combination of the other two. In particular, $h_{1}$, $h_{2}$ and $h_{3}$ have a common intersection and one of them is redundant in the description of $h_{1}\cap h_{2}\cap h_{3}$. We thus consider two cases, namely whether $h_{1}$, $h_{2}$ and $h_{3}$ have a common intersection or not.

First, assume that $h_{1}$, $h_{2}$ and $h_{3}$ have no common intersection. Then $h_{1}$, $h_{2}$ and $h_{3}$ partition the plane into seven regions (see Figure 4): $h_{i}\cap h_{j}$ for $i\neq j$, $h_{i}\setminus(h_{j}\cup h_{k})$, for $i,j,k$ all different and $h_{1}^{c}\cap h_{2}^{c}\cap h_{3}^{c}$. Note that each $h_{i}\cap h_{j}$ contains strictly fewer than $\frac{n}{6}$ of the points of $P$, as otherwise the corresponding $c_{i}$ and $c_{j}$ intersect. In particular, $h_{2}\setminus h_{1}$ contains more than $\frac{n}{2}-\frac{n}{6}=\frac{n}{3}$ points of $P$. It follows that $(h_{1}\cup h_{2})^{c}$ and thus also $h_{3}\setminus(h_{1}\cup h_{2})$ contains strictly fewer than $\frac{n}{6}$ of the points of $P$. The number of points in $h_{3}$ is the sum of the number of points in $h_{3}\cap h_{1}$, $h_{3}\cap h_{2}$ and $h_{3}\setminus(h_{1}\cup h_{2})$. All of these sets contain strictly fewer than $\frac{n}{6}$ of the points of $P$, implying that $h_{3}$ contains fewer than $\frac{n}{2}$ of the points of $P$, which is a contradiction. This concludes the first case.

For the second case assume that $h_{1}$, $h_{2}$ and $h_{3}$ have a common intersection and one of the halfplanes is redundant in the description of $h_{1}\cap h_{2}\cap h_{3}$; assume without loss of generality that it is $h_{3}$. Just as in the first case, each $h_{i}\cap h_{j}$ contains strictly fewer than $\frac{n}{6}$ of the points of $P$. Again it follows that $h_{3}\setminus(h_{1}\cup h_{2})$ contains strictly fewer than $\frac{n}{6}$ of the points of $P$. The sets $h_{3}\cap h_{1}$, $h_{3}\cap h_{2}$ and $h_{3}\setminus(h_{1}\cup h_{2})$ cover $h_{3}$, implying that $h_{3}$ contains fewer than $\frac{n}{2}$ of the points of $P$, which is again a contradiction. This concludes the proof that among any three elements of $H$, two intersect.

It remains to show that we can find three points $p_{1}$, $p_{2}$ and $p_{3}$ such that each element of $H$ contains at least one of them. This can be achieved by picking one element of $H$ and placing two points $p_{1}$ and $p_{2}$ at the extreme intersection points with the boundary of $C$; since any three elements of $H$ intersect, any two elements not containing $p_{1}$ and $p_{2}$ must intersect and we may apply Helly’s theorem in dimension one. However, we actually have more flexibility in choosing $p_{1}$.Note that the normal vectors pointing into the halfplanes defining the elements of $H$ define a circular order on $H$. Place $p_{1}$ at a topmost point of the boundary of $C$. Let $h_{1}$ be the first element of $H$ in counterclockwise direction whose defining halfplane does not contain $p_{1}$ in its interior. Place $p_{2}$ at the intersection of the defining line of $h_{1}$ with the boundary of $C$ that is furthest in counterclockwise direction from $p_{1}$. Since $h_{1}$ is minimal, any element of $H$ intersecting $h_{1}$ contains either $p_{1}$ or $p_{2}$. (Note that so far $h_{1}$ does not contain any of $p_{1}$ or $p_{2}$.) Therefore, all elements of $H$ that do not intersect $h_{1}$ have a common intersection, in which we place $p_{3}$. Recall that the defining halfplanes are open and therefore there is no element of $H$ intersecting $h_{1}$ in a single point. We therefore may move $p_{2}$ slightly in clockwise direction, such that it is also contained in $h_{1}$. ∎

Let $\gamma$ be the boundary of the convex hull of all points below the $k$-level, and let $\nu$ be the intersection of $h$ with the $l$-level. Note that $\nu$ might not be connected. Suppose we want our line to be the counterclockwise bitangent of $\gamma$ and $\nu$ (i.e., from left to right, it first intersects $\gamma$, which has no point above it, and then $\nu$). Our algorithm works by obtaining tangents to $\nu$ through points on $\gamma$. Matoušek’s $O(n\log^{2}n)$ algorithm for determining the tangent to a level through a given point that is to the right of that point [11, Lemma 3.2] (our Lemma 7) also directly works for parts of a level such as $\nu$: It requires a sub-algorithm that decides in $O(n\log n)$ time whether a given line $\ell$ intersects the level (or, in our case, the partial level $\nu$). This can be done by sorting the intersection of the lines of the arrangements along $\ell$ (see also [11, Lemma 3.1]) as well as along the line bounding $h$; $\ell$ either intersects the relevant part of $\nu$, or we can compare the intersection of $h$ with $\ell$ to the intersections of $h$ with $\nu$ to determine whether there is a point of $\nu$ below $\ell$.

Suppose first we are given $\gamma$. (It requires $O(n\log^{4}n)$ time though to obtain it, so we eventually get rid of this assumption.) The convex hull of a level is known to have at most $n$ vertices [11, Lemma 2.1]. For a point $p$ on $\gamma$, we can find in $O(n\log^{2}n)$ time the point $q$ on $\nu$ such that the line $pq$ has no point on $\nu$ below it. We can thus find, by binary search on the $O(n)$ vertices of $\gamma$, a vertex $p$ with $q$ on $\nu$ such that $pq$ separates $\gamma$ and $\nu$. This gives an $O(n\log^{4}n)$ time algorithm for obtaining the bitangent. To improve on that bound, we need to get rid of the explicit construction of $\gamma$ to find the tangents to $\nu$.

To this end, we consider Matoušek’s algorithm for constructing the convex hull boundary $\gamma$ of a level and compute only the relevant part (see [11, Section 4]). In particular, the algorithm works by finding, for a constant $c$ and two vertical lines, $(c-1)$ further vertical lines between the given ones such that there are at most $n^{2}/c$ crossings of the arrangement between two of these verticals. This can be done in $O(n)$ time (as described in [12]). The tangents on $\gamma$ at the intersection points with the vertical lines can be computed in $O(n\log^{3}n)$ time [11, Lemma 3.3]. It is shown in [11] that, when choosing $c=64$, there are at most $n/2$ lines of the arrangement relevant for the construction of $\gamma$ between two such vertical lines, and these lines can be found in $O(n)$ time. The original algorithm proceeds recursively within each interval defined by two neighboring vertical lines after removing the non-relevant lines. In our adaption, however, we find the interval that contains the point $p$ on $\gamma$ such that a tangent to $\gamma$ through the vertex $p$ with $q$ on $\nu$ such that $pq$ separates $\gamma$ and $\nu$. (We do this by considering the tangent to $\gamma$ at each of the constant number of intersection of a vertical line with $\gamma$.) When we have found this interval, we can prune $n/2$ of the lines and recurse inside this interval. Note, however, that we cannot prune the set of lines when looking for a tangent to $\nu$. Thus, in each recursive call, we need $O(n\log^{2}n)$ time for computing the tangent. As the recursion depth is $O(\log n)$, this amounts to $O(n\log^{3}n)$ in total. Also, for $n_{i}$ lines during the $i$th recursion, we need $O(n_{i}\log^{3}n_{i})\subseteq O(n_{i}\log^{3}n)$ time for determining the intervals. As $n_{i}$ decreases geometrically, this also amounts to $O(n\log^{3}n)$. This is the total running time for finding the bitangent, as claimed. ∎

To find a point $p_{1}$ in the intersection of $C$ and $D_{U}$, observe first that we can restrict our attention in the dual to the convex hull of the points above the $\left\lfloor(1-\alpha_{1})n\right\rfloor$-level of the dual line arrangement. This is because any primal line with more than $(1-\alpha_{1})n$ points above it (which corresponds to a dual point below the $\left\lceil\alpha_{1}n\right\rceil$-level) also defines a halfplane in $U$. A point in the intersection of $D_{U}$ and $C$ thus corresponds to a line on or above the $\left\lceil\alpha_{2}n\right\rceil$-level and on or below the $\left\lfloor(1-\alpha_{1})n\right\rfloor$-level. We find a bitangent to these two levels in $O(n\log^{3}n)$ time using Lemma 8 (with $h=\mathbb{R}^{2}$). The primal point of this line is $p_{1}$; see the point indicated by the brown box in Figure 5 (right). We obtain $p_{2}$ analogously. ∎

Consider the dual line arrangement of the point set. The points $p_{1},p_{2},p_{3}$ dualize to three lines $p_{1}^{*},p_{2}^{*},p_{3}^{*}$ that are between the $\left\lceil\frac{n}{6}\right\rceil$-level and the $\left\lfloor\frac{5n}{6}\right\rfloor$-level of the arrangement s.t. every point on the middle level has at least one of these lines above it and one of these lines below it. (We assume for simplicity that $n$ is odd and the middle level is the $\left\lfloor\frac{n}{2}\right\rfloor$-level of the arrangement; if $n$ is even, one has to consider the points between the $\frac{n}{2}$-level and the $(\frac{n}{2}+1)$-level.) Theorem 6 asserts that such lines exist, and its proof tells us that we can choose one of these lines to be an arbitrary tangent of one of the levels not intersecting the interior of the other one. We denote by $\gamma_{b}$ and $\gamma_{t}$ the convex hull boundaries of the points on or below the $\left\lceil\frac{n}{6}\right\rceil$-level and of the points on or above the $\left\lfloor\frac{5n}{6}\right\rfloor$-level, respectively.

We let $p^{*}_{1}$ be the clockwise bitangent of $\gamma_{b}$ and $\gamma_{t}$, which we can obtain in $O(n\log^{3}n)$ time using Lemma 8. For simplicity of explanation, we also compute the counterclockwise bitangent $\ell$. (This step may be omitted in an actual implementation, but assuming it to be given facilitates the explanation and does not change the asymptotic running time.)

The line $p^{*}_{1}$ intersects the middle level of the arrangement. Let $\mu_{1}$ be the parts of the middle level below $p^{*}_{1}$, and $\mu_{2}$ be the part above it. Note that each of these parts may be disconnected. Using Lemma 8, we search for the counterclockwise bitangent between $\gamma_{b}$ (or, equivalently, the $\left\lceil\frac{n}{6}\right\rceil$-level) and $\mu_{1}$ (which is the intersection of the middle level with a halfspace defined by $p^{*}_{1}$) in $O(n\log^{3}n)$ time. If it exists, and its intersection point with $\gamma_{b}$ is between the intersections of $\gamma_{b}$ with $p^{*}_{1}$ and $\ell$, we choose this line to be $p^{*}_{2}$. Otherwise, we continue our search on $\gamma_{t})$ in the same way (i.e., we look for the counterclockwise bitangent between $\gamma_{t}$ and $\mu_{1}$). The line $p^{*}_{3}$ can be found in an analogous manner. ∎