Minimizing the Diameter of a Spanning Tree for Imprecise Points

Abstract

We study the diameter of a spanning tree, i.e., the length of its longest simple path, under the imprecise points model, in which each point is assigned its own occurrence region instead of an exact location. We prove that the minimum diameter of a spanning tree of n points each of which is selected from its respective disk (or axis-parallel square) is polynomial-time computable, contrasting with the fact that minimizing the cost of a spanning tree, i.e. the sum of its edge lengths, for imprecise points is known to be NP-hard. This difference is very interesting since for exact points minimizing the cost seems faster than minimizing the diameter [$O(n\log n)$ versus almost cubic running time]. As the first work regarding the minimum diameter of a spanning tree for imprecise points, our main objective is to investigate essential properties that distinguish the problem from NP-hard instead of minimizing the running time [$O(n^9)$ for general disks or $O(n^6)$ for unit ones]. Our algorithm mainly utilizes farthest-site Voronoi diagrams for disks and axis-parallel squares, and these diagrams would have their own merits.

Appendices

Appendix 1: Implementation of Elementary Operations

1.1 Fact 1

We distinguish cases according to the location of $t\in e$:

• 1. $t\in D_p\cap D_q$: Any point $t\in e\cap (D_p\cap D_q)$ satisfies $d(t, D_p)+d(t, D_q)=0$, and $e\cap (D_p\cap D_q)$ can be computed in O(1) time.

• 2. $t\in D_p\setminus D_q$ (resp. $t\in D_q\setminus D_t$): $d(t, D_p)+d(t, D_q)=d(t, D_q)=|{\overline{tc_q}}|-r_q$ (resp. $d(t, D_p)+d(t, D_q)=d(t, D_p)=|{\overline{tc_p}}|-r_p$), and thus the location of t minimizing $d(t, D_p)+d(t, D_q)$ can be trivially computed in O(1) time.

• 3. $t\not \in D_p\cup D_q$: $d(t, D_p)+d(t, D_q)=|{\overline{tc_p}}|+|{\overline{tc_q}}|-r_p-r_q$. We discuss below how to compute t minimizing $d(t, D_p)+d(t, D_q)$ for this case in O(1) time.

For the sake of simplicity, we use p to denote $c_p$, q to denote $c_q$ and, since $r_p$ and $r_q$ are fixed, we attempt to minimize $|{\overline{pt}}|+|{\overline{qt}}|$. Such a point $t\in e$ can be located either in the interior of e or at one of its endpoints. In the latter case we can easily find the endpoint of e resulting in the smallest value by checking both of them. In the former case, if e and the segment ${\overline{pq}}$ intersect, then t is trivially the intersection point. Otherwise we distinguish between the cases where e is a line segment, a circular arc, or a hyperbolic arc.

Fig. 13

Reflection on a segment (left) and on a circular arc (right)

If e is a line segment, it is well-known that the line perpendicular to e and passing through t bisects the angle $\angle ptq$ (Fig. 13a). This point can be computed in O(1) time as the intersection between ${\overline{p'q}}$ and e, where $p'$ is the reflection point of p with respect to the line passing through e.

If e is a circular arc, the line passing through t and the center c of the circle induced by e bisects the angle $\angle ptq$ (Fig. 13b). To compute this point we adopt the Newton–Raphson method. In order to do so, we formulate a function f(z) such that $f(z)=0$ if and only if a ray shot from p in the direction (1, z) first hits e, then reflects on e, and finally passes through q. In other words, if $f(z)=0$, then t is the first intersection between e and the ray shot from p along the direction (1, z).

For simplicity, assume that e induces a unit circle C with center $c=(h, k)$. Let $(x_p+\tau ,y_p+z\tau ), \tau \ge 0$ be the ray shooting from $p=(x_p,y_p)$ in the direction (1, z). This ray hits C at the point $(x_p+f_1(z),y_p+f_1(z)z)$ such that

$$\begin{aligned} a&= x_p-h,\, b =y_p-k,\\ f_1(z)&=\frac{-(2a+2bz)-\sqrt{4(1-a^2)z^2+8abz+(1-b^2)}}{2(z^2+1)}. \end{aligned}$$

After hitting C the ray reflects at $(x_p+f_1(z),y_p+f_1(z)z)$ and continues in the direction $(x_p+f_1(z)+\tau ,y_p+f_1(z)z+f_3(z)\tau ), \tau \ge 0$, where

$$\begin{aligned} f_2(z)&=\frac{2(b-az)f_1(z)}{(b+f_1(z)z)^2+(a+f_1(z))^2},\\ f_3(z)&=\frac{-(a+f_1(z))f_2(z)-f_1(z)z}{(b+f_1(z)z)f_2(z)-f_1(z)}. \end{aligned}$$

Since the ray will pass through $q=(x_q,y_q)$, we have $(y_q-y_p)-f_1(z)(z-f_3(z))-f_3(z)(x_q-x_p) = 0$. Thus, we can set f as

$$\begin{aligned} f(z)&= (y_q-y_p)-f_1(z)(z-f_3(z))-f_3(z)(x_q-x_p). \end{aligned}$$

To apply the Newton–Raphson method we need an interval $[z_l,z_r]$ containing the root $z_0$ of f(z) and such that for every $z\in [z_l,z_0)$ and $z\in (z_0,z_r]$ it holds $f'(z)\ne 0$. We can pick as $z_l$ and $z_r$ two distinct values such that the rays shot from p along the directions $(1, z_l)$ and $(1, z_r)$ are tangent to C.

For the analysis of the running time, since all the sub-functions of f have a second derivative in the interval $[z_l,z_r]$, f also has a second derivative in $[z_l,z_r]$. It is well-known if f has a second derivative in $[z_l,z_r]$, the convergence rate of the Newton–Raphson method is quadratic in $[z_l,z_r]$. According to the time complexity analysis for the Newton–Raphson method by Borwein and Borwein [3], if the convergence rate is quadratic the root $z_0$ can be found in O(M(m)) bit-operations, where M(m) is the number of bit-operations required to multiply two m-bits words. By the standard assumption of using the RAM model of computation, the multiplication between two numbers is considered a constant-time operation. The time required to find a point t on a circular arc e minimizing $|{\overline{pt}}|+|{\overline{qt}}|$ is therefore also in O(1).

In the case where e is a hyperbolic arc, a reflection property similar to the circular arc case holds. The above method can be generalized in a straightforward fashion for hyperbolic arcs.

1.2 Fact 2

We denote the circles induced by $e_1$ and $e_2$ respectively as $C_1$ and $C_2$, their centers respectively as $c_1=(h_1, k_1)$ and $c_2=(h_2, k_2)$, and their radii respectively as $r_1$ and $r_2$. We distinguish four cases depending on the locations of $s\in e_1$ and $t\in e_2$.

• 1. $s\in D_p$ and $t\in D_q$: $d(s, D_p)+|{\overline{st}}|+d(t, D_q)= |{\overline{st}}|$, and the locations of s and t to minimize $|{\overline{st}}|$ can be trivially computed in O(1) time

• 2. $s\in D_p$ but $t\not \in D_q$: $d(s, D_p)+|{\overline{st}}|+d(t, D_q)= |{\overline{st}}|+d(t, D_q)$, and the locations of s and t to minimize $|{\overline{st}}|+d(t, D_q)$ can be computed in O(1) time by Fact 1 where s is replaced with $C_1$.

• 3. $t\in D_q$ but $s\not \in D_p$: It is symmetric to the second case.

• 4. $s\not \in D_p$ and $t\not \in D_q$: $d(s, D_p)+|{\overline{st}}|+d(t, D_q)= |{\overline{c_pt}}|+|{\overline{st}}|+|{\overline{tc_q}}|$. We discuss below how to compute s, t minimizing $d(t, D_p)+|{\overline{st}}|+d(t, D_q)$ for this case in O(1) time.

We use p to denote $c_p$, q to denote $c_q$, and assume that $C_1$ and $C_2$ are unit circles. We distinguish three possibilities for the locations of $s\in \text{ cl }(e_1)$ and $t\in \text{ cl }(e_2)$:

• 1. Both s and t are endpoints of their corresponding arcs: We can then check all 4 possibilities and find the one resulting in the smallest value.

• 2. Either s or t is is an endpoint of its corresponding arc: We can then apply the Newton–Raphson method presented in Sect. 1 for each endpoint and find the one resulting in the smallest value. For example, if s is an endpoint of $e_1$, we find a point $t\in C_2$ minimizing $|{\overline{st}}|+|{\overline{tq}}|$.

• 3. Both s and t lie in the interior of their corresponding arcs: We propose the method below to compute their locations in O(1) time.

We distinguish between the cases where the segment ${\overline{pq}}$ intersects both $C_1$ and $C_2$, where it intersects only one among $C_1$ and $C_2$, and where it intersects neither $C_1$ nor $C_2$.

If ${\overline{pq}}$ intersects both $C_1$ and $C_2$ the points s and t can be computed by setting s as one of the two intersections between ${\overline{pq}}$ and $C_1$ and t as one of the two intersections between ${\overline{pq}}$ and $C_2$.

Fig. 14

Reflection on two circular arcs

If ${\overline{pq}}$ intersects neither $C_1$ nor $C_2$ then the line passing through $c_1$ and s will bisect $\angle qst$ and the line passing through $c_2$ and t will bisect $\angle stq$ (Fig. 14). We design a Newton–Raphson method algorithm similar to the one in Sect. 1 to compute s and t in O(1) time. Again, we formulate a function f(z) such that $f(z)=0$ if and only if a ray shot from $p=(x_p, y_p)$ in the direction (1, z) first reflects on $C_1$, then reflects on $C_2$, and finally passes through $q=(x_q, y_q)$. In other words, if $f(z)=0$, then s is the first intersection between $e_1$ and the ray shot from p along the direction (1, z), and t is the first intersection between $e_2$ and the ray reflecting at s. The function f can be defined in a similar way as in the previous case:

$$\begin{aligned} f(z) \,=\,&((f_3(z)+f_6(z))f_7(z)+f_6(z)f_5(z))x_q\\&+((f_4(z)+f_6(z)f_5(z))f_7(z)-f_6(z))y_q\\&-((f_4(z)+f_6(z)f_5(z))f_7(z)-f_6(z))(y_p+f_1(z)z+f_6(z)f_5(z))\\&-((f_3(z)+f_6(z))f_7(z)+f_6(z)f_5(z))(x_p+f_1(z)+f_6(z)), \end{aligned}$$

where

$$\begin{aligned} a&= x_p-h_1, \, b =y_p-k_1,\\ f_1(z)&= \frac{-(2a+2bz)-\sqrt{4(1-a^2)z^2+8abz+(1-b^2)}}{2(z^2+1)},\\ f_2(z)&=\frac{2(b-az)f_1(z)}{(b+f_1(z)z)^2+(a+f_1(z))^2},\\ f_3(z)&=(x_p-h_2)+f_1(z),\\ f_4(z)&=(y_p-k_2)+f_1(z)z,\\ f_5(z)&=\frac{-(a+f_1(z))f_2(z)-f_1(z)z}{(b+f_1(z)z)f_2(z)-f_1(z)},\\ f_6(z)&= \tfrac{-(2f_3(z)+2f_4(z)f_5(z))- \sqrt{4(1-f_3(z)^2)f_5(z)^2+8f_3(z)f_4(z)f_5(z)+(1-f_4(z)^2)}}{2(f_5(z)^2+1)},\\ f_7(z)&= \frac{2(f_4(z)-f_3(z)f_5(z))f_6(z)}{(f_4(z)+f_6(z)f_5(z))^2+(f_3(z)+f_6(z))^2}. \end{aligned}$$

The Newton–Raphson method requires an interval $[z_l,z_r]$ containing the root $z_0$ of f such that for every $z\in [z_l,z_0)$ and $z\in (z_0,z_r]$, $f'(z)\ne 0$. We can pick two distinct values as $z_l$ and $z_r$ such that the rays shot from p along the directions $(1, z_l)$ and $(1, z_r)$ will first reflect at $C_1$ and then be tangent to $C_2$. Since also in this case all the sub-functions of f have a second derivative in $[z_l,z_r]$, the Newton–Raphson method converges to the root $z_0$ in O(1) time.

For the last case, assume without loss of generality that ${\overline{pq}}$ intersects only $C_1$. If ${\overline{pt}}$ passes through $C_1$, then we can set t as any of the two intersection between ${\overline{pt}}$ and $C_1$, while the point s can be computed using the Newton–Raphson method in Sect. 1. If ${\overline{pt}}$ does not pass through $C_1$, we can instead use the Newton–Raphson method in this section.

Appendix 2: Proofs for Farthest-Disk Voronoi Diagrams

Proof of Theorem 2

Proof

Mehlhorn et al. [25] prove that if the bisecting curve system satisfies several properties, the resulting farthest-site Voronoi diagram is a tree and has O(n) faces. After removing from ${\mathcal {D}}$ all disks that contains another disk, the bisectors among the remaining disks will satisfy these properties. Note that, using the amended bisectors, if a disk D contains another one then $\text{ FV }(D, {\mathcal {D}})$ is empty; thus, the removal of D will not change $\text{ FV }({\mathcal {D}})$, implying that $\text{ FV }({\mathcal {D}})$ has O(n) faces and is a tree.

To analyze the structural complexity of ${\mathcal {FV}}_2({\mathcal {D}})$, since this complexity is proportional to the number of vertices of ${\mathcal {FV}}_2({\mathcal {D}})$ plus the number of unbounded edges (by Euler’s formula), we will bound the two numbers, respectively.

We mainly adopt the well-known Clarkson-and-Shor technique [7, 27]: Let S be a set of n objects, and let A(S) be a set of configurations with respect to S. For example, S is a set of n points, and A(S) is a set of circles spanned by points of S. Each configuration of A(S) is assigned two disjoint subsets of S as its defining set and conflict set, respectively. For instance, the defining set of a circle is the set of points that span it, and the conflict set of a circle is the set of points in its interior. Let $C_i(S)$ be the set of configurations in A(S) whose conflict sets consist of i objects, $C_{\le i}(S)$ be $\bigcup _{0\le j\le i}C_j(S)$, $N_i(S)$ be $|N_i(S)|$, and $N_{\le i}(S)$ be $|C_{\le i}(S)|$. We also use d denote the maximum cardinality of the define set of a configuration over A(S), and use $N_i(n)$ (resp. $N_{\le i}(n)$) denote the maximum of $N_i(S)$ (resp. $N_{\le i}(S)$), over all sets S of n objects of this kind. The Clarkson-and-Shor technique is the following formula:

$$\begin{aligned} N_{\le k}(S)=O(k^d N_0(n/k)). \end{aligned}$$

Now, we analyze the number of vertices of ${\mathcal {FV}}_2({\mathcal {D}})$. For simplicity, we assume that the degree of each Voronoi vertex is exactly three; actually, without this assumption, the number of vertices will not be larger. Each vertex of $\text{ FV }_1({\mathcal {D}})$ is the center of a circle which is tangent to three disks in ${\mathcal {D}}$ and whose exterior does not fully contain any other disk in ${\mathcal {D}}$, and each vertex of ${\mathcal {FV}}_2({\mathcal {D}})$ is the center of a circle which is tangent to three disks in ${\mathcal {D}}$ and whose exterior fully contains at most one other disk in ${\mathcal {D}}$, i.e., zero or one. Therefore, let S be ${\mathcal {D}}$ and A(S) be the set of circles that are tangent to three disks in ${\mathcal {D}}$. For each circle in A(S), let its defining set be the three disks tangent to it, and its conflict set be the disks that fully lies in its exterior. Under these circumstances, $N_0(S)$ is the number of vertices of $\text{ FV }_1({\mathcal {D}})$, and $N_{\le 1}(S)$ is the number of vertices of ${\mathcal {FV}}_2({\mathcal {D}})$. Since d is trivially 3 and $N_0(n)$ is O(n), we have

$$\begin{aligned} N_{\le 1}(S)=N_{\le 1}(n)=O(k^dN_0(n/k))=O(1^3N_0(n))=O(n). \end{aligned}$$

We continue to derive the number of unbounded edges of ${\mathcal {FV}}_2({\mathcal {D}})$. An unbounded endpoint of $\text{ FV }_1({\mathcal {D}})$ corresponds to a hyperplane which is tangent to two disks and does not fully contain any other disk in one of its two half-spaces, and an unbounded endpoint of ${\mathcal {FV}}_2({\mathcal {D}})$ corresponds to a hyperplane which is tangent to two disks and fully contains at most one other disk in one of its two half-spaces. Then, we can define S similar to the case for vertices. Since d is trivially 2 and $N_0(n)$ is O(n), we have

$$\begin{aligned} N_{\le 1}(S)=N_{\le 1}(n)=O(k^dN_0(n/k))=O(1^2N_0(n))=O(n). \end{aligned}$$

$\square$

Proof of Theorem 3

Proof

Cheong et al. [6] proposes a divide-and-conquer algorithm computing farthest-polygon Voronoi diagrams in $O(n\log ^3 n)$ time. If we first remove from ${\mathcal {D}}$ all disks containing another one (for example, with a simple plane-sweep technique running in $O(n\log n)$ time), the algorithm is also applicable to ${\mathcal {D}}$, even under the modified bisectors. Since a disk containing another one does not have regions in $\text{ FV }({\mathcal {D}})$, we can thus compute $\text{ FV }({\mathcal {D}})$ in $O(n\log ^3 n)$ time.

Moreover, one of the $O(\log n)$ factors in the running time comes from the handling of closed merge curves. In the “divide” stage, if we separate the disks according to the x-coordinates of their centers, there will be no closed merge curve at all. Therefore, $\text{ FV }({\mathcal {D}})$ can be computed in $O(n\log ^2 n)$ time. To compute ${\mathcal {FV}}_2({\mathcal {D}})$ we further partition each nonempty $\text{ FV }(D, {\mathcal {D}})$ according to $\text{ FV }({\mathcal {D}}\setminus \{D\})$. Since $\text{ FV }({\mathcal {D}})$ has O(n) faces, the computation of ${\mathcal {FV}}_2({\mathcal {D}})$ takes $O(n)\cdot O(n\log ^2 n)=O(n^2\log ^2 n)$ time. Finally, if ${\mathcal {D}}$ consists of unit disks, then $\text{ FV }({\mathcal {D}})=\text{ FV }(\mathcal {C})$ and ${\mathcal {FV}}_2({\mathcal {D}})={\mathcal {FV}}_2(\mathcal {C})$. Ho et al. [16] showed how to compute these diagrams in $O(n\log n)$ time.$\square$