Approximating (k,ℓ)-Median Clustering for Polygonal Curves

Define \(\rho :\mathcal {H} \rightarrow \mathbb {R}_{\ge 0}, h \mapsto \max \limits _{t\in [0,1]} \Vert \sigma (t) - \tau (h(t)) \Vert\) with image \(R = \lbrace \rho (h) \mid h \in \mathcal {H} \rbrace\). Per definition, we have \(\text{d}_\text{F}(\sigma , \tau) = \inf R = r\).

For any non-empty subset \(X\) of \(\mathbb {R}\) that is bounded from below and for every \(\varepsilon \gt 0,\) it holds that there exists an \(x \in X\) with \(\inf X \le x \lt \inf X + \varepsilon\), by definition of the infimum. Since \(R \subseteq \mathbb {R}\) and \(\inf R\) exists, for every \(\varepsilon \gt 0\) there exists an \(r^\prime \in R\) with \(\inf R \le r^\prime \lt \inf R + \varepsilon\).

Now, let \(a_i = 1/i\) be a zero sequence. For every \(i \in \mathbb {N,}\) there exists an \(r_i \in R\) with \(r \le r_i \lt r + a_i\), thus \(\lim \limits _{i \rightarrow \infty } r_i = r\).

Let \(\rho ^{-1}(r^\prime) = \lbrace h \in \mathcal {H} \mid \rho (h) = r^\prime \rbrace\) be the preimage of \(\rho\). Since \(\rho\) is a function, \(\vert \rho ^{-1}(r^\prime) \vert \ge 1\) for each \(r^\prime \in R\). Now, for \(i \in \mathbb {N}\), let \(h_i\) be an arbitrary element from \(\rho ^{-1}(r_i)\). By definition, it holds that \(\begin{equation*} \lim \limits _{i \rightarrow \infty } \max \limits _{t \in [0,1]} \Vert \sigma (t) - \tau (h_i(t)) \Vert = \lim _{i \rightarrow \infty } \rho (h_i) = \lim _{i \rightarrow \infty } r_i = r = \inf R, \end{equation*}\) which proves the claim.□

First, we know that \(\text{d}_\text{F}(\tau , \text{simpl}( \alpha ,\tau )) \le \alpha \cdot \text{d}_\text{F}(\tau , c^\ast)\), for each \(\tau \in T\), by Definition 2.6.

Now, there are at least \(\frac{n}{2}\) curves in \(T\) that are within distance at most \(\frac{2\text{cost}( T,c^\ast ) }{n}\) to \(c^\ast\). Otherwise, the cost of the remaining curves would exceed \(\text{cost}( T,c^\ast )\), which is a contradiction. Hence, each \(s \in S\) has probability at least \(\frac{1}{2}\) to be within distance \(\frac{2\text{cost}( T,c^\ast ) }{n}\) to \(c^\ast\).

Since the elements of \(S\) are sampled independently, we conclude that the probability that every \(s \in S\) has distance to \(c^\ast\) greater than \(\frac{2\text{cost}( T,c^\ast ) }{n}\) is at most \((1-\frac{1}{2})^{\vert S \vert } \le \exp (-\frac{2(\ln (2)+\ln (1/\delta))}{2}) = \frac{\delta }{2}\).

Now, assume there is an \(s \in S\) with \(\text{d}_\text{F}(s, c^\ast) \le \frac{2 \text{cost}( T,c^\ast ) }{n}\). We do not want any \(\tau \in S \setminus \lbrace s\rbrace\) with \(\text{cost}( T,\tau ) \gt 2 \text{cost}( T,s)\) to have \(\text{cost}( W,\tau ) \le \text{cost}( W,s)\). Using Theorem 2.8, we conclude that this happens with probability at most \( \begin{equation*} \exp \left(-\frac{64 (\ln (1/\delta) + \ln (\lceil 4(\ln (2)+\ln (1/\delta))\rceil))}{64}\right) \le \frac{\delta }{\lceil 4 (\ln (2) + \ln (1/\delta)) \rceil } \le \frac{\delta }{2 \vert S \vert }, \end{equation*} \) for each \(\tau \in S \setminus \lbrace s\rbrace\).

Using a union bound over all bad events, we conclude that with probability at least \(1-\delta\), Algorithm 1 samples a curve \(s \in S\), with \(\text{d}_\text{F}(s, c^\ast) \le 2 \text{cost}( T,c^\ast ) /n\) and returns the simplification \(c = \text{simpl}( \alpha ,t)\) of a curve \(t \in S\), with \(\text{cost}( T,t) \le 2 \text{cost}( T,s)\). The triangle inequality yields \(\begin{equation*} \sum _{\tau \in T} (\text{d}_\text{F}(t, c^\ast) - \text{d}_\text{F}(\tau , c^\ast)) \le \sum _{\tau \in T} \text{d}_\text{F}(t, \tau) \le 2 \sum _{\tau \in T} \text{d}_\text{F}(s, \tau) \le 2 \sum _{\tau \in T} (\text{d}_\text{F}(\tau , c^\ast) + \text{d}_\text{F}(c^\ast , s)), \end{equation*}\) which is equivalent to \(\begin{equation*} n \cdot \text{d}_\text{F}(t, c^\ast) \le 2 \text{cost}( T,c^\ast ) + \text{cost}( T,c^\ast ) + 2 n \frac{2\text{cost}( T,c^\ast ) }{n} \iff {} \text{d}_\text{F}(t, c^\ast) \le \frac{7\text{cost}( T,c^\ast ) }{n}. \end{equation*}\)

Hence, we have \(\begin{align*} \text{cost}( T,c) & ={} \sum _{\tau \in T} \text{d}_\text{F}(\tau , \text{simpl}( \alpha ,t)) \le \sum _{\tau \in T} (\text{d}_\text{F}(\tau , t) + \text{d}_\text{F}(t, \text{simpl}( \alpha ,t))) \\ & \le {} 2 \text{cost}( T,s) + \sum _{\tau \in T} \alpha \cdot \text{d}_\text{F}(t, c^\ast) \le {} 2 \sum _{\tau \in T} (\text{d}_\text{F}(\tau , c^\ast) + \text{d}_\text{F}(c^\ast , s)) + 7\alpha \cdot \text{cost}( T,c^\ast ) \\ & \le {} 2 \text{cost}( T,c^\ast ) + 4 \text{cost}( T,c^\ast ) + 7\alpha \cdot \text{cost}( T,c^\ast ) ={} (6 + 7\alpha) \text{cost}( T,c^\ast ) . \end{align*}\)

The lower bound \(\text{cost}( T,c^\ast ) \le \text{cost}( T,c)\) follows from the fact that the returned curve has \(\ell\) vertices and that \(c^\ast\) has minimum cost among all curves with \(\ell\) vertices.□

We use Lemma 3.2 together with Theorem 3.1, which yields an approximation factor of 34.

Now, drawing the samples takes time \(O(\ln (1/\delta))\) each. Evaluating the samples against each other takes time \(O(m^2 \log (m) \ln ^2(1/\delta))\) and simplifying one of the curves that evaluates best takes time \(O(m^3 \log m)\). We conclude that Algorithm 1 has running time \(O(m^2 \log (m) \ln ^2(1/\delta) + m^3 \log m)\).□

Let \(v^\sigma _1, \ldots , v^\sigma _{\vert \sigma \vert }\) be the vertices of \(\sigma\). Further, let \(t^\sigma _1, \ldots , t^\sigma _{\vert \sigma \vert }\) and \(t^\tau _1, \ldots , t^\tau _{\vert \tau \vert }\) be the instants of \(\sigma\) and \(\tau\), respectively. In addition, for \(h \in \mathcal {H}\) (recall that \(\mathcal {H}\) is the set of all continuous bijections \(h:[0,1] \rightarrow [0,1]\) with \(h(0) = 0\) and \(h(1) = 1\)), let \(r_h = \max \nolimits _{t \in [0,1]} \Vert \sigma (t) - \tau (h(t)) \Vert\) be the distance realized by \(h\). We know from Lemma 2.4 that there exists a sequence \((h_x)_{x=1}^\infty\) in \(\mathcal {H}\), such that \(\lim \nolimits _{x \rightarrow \infty } r_{h_x} = \text{d}_\text{F}(\sigma , \tau) = r\).

Now, fix an arbitrary \(h \in \mathcal {H}\) and assume that there is a vertex \(v^\sigma _i\) of \(\sigma\), with instant \(t^\sigma _i\), which is not contained in any of \(B(v^\tau _1, r_h), \ldots , B(v^\tau _{\vert \tau \vert }, r_h)\). Let \(j\) be the maximum of \([\vert \tau \vert - 1]\), such that \(t^\tau _j \le h(t^\sigma _i) \le t^\tau _{j+1}\). So \(v^\sigma\) is matched to \(\overline{\tau (t^\tau _j) \tau (t^\tau _{j+1})}\) by \(h\). We modify \(\sigma\) in such a way that \(v^\sigma _i\) is replaced by two new vertices that are elements of \(B(v^\tau _j, r_h)\) and \(B(v^\tau _{j+1}, r_h)\), respectively.

Namely, let \(t^-\) be the maximum of \([0, t^\sigma _i)\), such that \(\sigma (t^-) \in B(v^\tau _j, r_h),\) and let \(t^+\) be the minimum of \((t^\sigma _i, 1]\), such that \(\sigma (t^+) \in B(v^\tau _{j+1}, r_h)\). These are the instants when \(\sigma\) leaves \(B(v^\tau _j, r_h)\) before visiting \(v^\sigma _i\) and \(\sigma\) enters \(B(v^\tau _{j+1}, r_h)\) after visiting \(v^\sigma _i\), respectively. Let \(\sigma ^\prime _h\) be the piecewise defined curve, defined just like \(\sigma\) on \([0,t^-]\) and \([t^+,1]\), but on \((t^-, t^+)\) it connects \(\sigma (t^-)\) and \(\sigma (t^+)\) with the line segment \(s(t) =(1-\frac{t-t^-}{t^+-t^-}) \tau (t^-) + \frac{t-t^-}{t^+-t^-} \tau (t^+)\).

We know that \(\Vert \sigma (t^-) - \tau (h(t^-)) \Vert \le r_h\) and \(\Vert \sigma (t^+) - \tau (h(t^+)) \Vert \le r_h\). Note that \(t^\tau _j \le h(t^-)\) and \(h(t^+) \le t^\tau _{j+1}\) since \(\sigma (t^-)\) and \(\sigma (t^+)\) are the closest points to \(v^\sigma _i\) on \(\sigma\) that have distance \(r_h\) to \(v^\tau _j\) and \(v^\tau _{j+1}\), respectively, by definition. Therefore, \(\tau\) has no vertices between the instants \(h(t^-)\) and \(h(t^+)\). Now, \(h\) can be used to match \(\sigma ^\prime _h\vert _{[0,t^-)}\) to \(\tau \vert _{[0,h(t^-))}\) and \(\sigma ^\prime _h\vert _{(t^+,1]}\) to \(\tau \vert _{(t^+,1]}\) with distance at most \(r_h\). Since \(\sigma ^\prime _h\vert _{[t^-, t^+]}\) and \(\tau \vert _{[h(t^-), h(t^+)]}\) are just line segments, they can be linearly matched to each other with distance at most \(\max \lbrace \Vert \sigma ^\prime _h(t^-) - \tau (h(t^-)) \Vert , \Vert \sigma ^\prime _h(t^+) - \tau (h(t^+)) \Vert \rbrace \le r_h\). We conclude that \(\text{d}_\text{F}(\sigma ^\prime _h, \tau) \le r_h\).

Because this modification works for every \(h \in \mathcal {H}\), we have \(\text{d}_\text{F}(\sigma ^\prime _{h}, \tau) \le r_h\) for every \(h \in \mathcal {H}\). Thus, \(\lim \nolimits _{x \rightarrow \infty } \text{d}_\text{F}(\sigma ^\prime _{h_x}, \tau) \le \text{d}_\text{F}(\sigma , \tau) = r\).

Now, to prove the claim, for every \(h \in \mathcal {H}\) we apply this modification to \(v^\sigma _i\) and successively to every other vertex \(v^{\sigma ^\prime _h}_i\) of the resulting curve \(\sigma ^\prime _h\), not contained in one of the balls, until every vertex of \(\sigma ^\prime _h\) is contained in a ball. Note that the modification is repeated at most \(\vert \sigma \vert - 2\) times for every \(h \in \mathcal {H}\), since the start and end vertex of \(\sigma\) must be contained in \(B(v^\tau _1, r_h)\) and \(B(v^\tau _{\vert \tau \vert }, r_h)\), respectively. Therefore, the number of vertices of every \(\sigma ^\prime _h\) can be bounded by \(2 \cdot (\vert \sigma \vert - 2) + 2\) since every other vertex must not lie in a ball and for each such vertex one new vertex is created. Thus, \(\vert \sigma ^\prime _h \vert \le 2 \vert \sigma \vert - 2\).□

We assume that \(\vert T^\prime \vert \ge \frac{1}{\beta } \vert T \vert\). Let \(n^\prime\) be the number of sampled curves in \(S\) that are elements of \(T^\prime\). Clearly, \(E\left[n^\prime \right] \ge \sum _{i=1}^{\vert S \vert } \frac{1}{\beta } = \frac{\vert S \vert }{\beta }\). In addition, \(n^\prime\) is the sum of independent Bernoulli trials. A Chernoff bound (see Lemma 2.11) yields the following: \(\begin{align*} \Pr \left[n^\prime \lt \frac{\vert S \vert }{2\beta }\right] \le \Pr \left[n^\prime \lt \frac{1}{2}E\left[n^\prime \right]\right] \le \exp \left(-\frac{1}{4}\frac{\vert S \vert }{2\beta }\right) \le \exp \left(\frac{\ln (\delta)-\ln (4)}{\varepsilon } \right) = \left(\frac{\delta }{4}\right)^{\frac{1}{\varepsilon }} \le \frac{\delta }{4}. \end{align*}\)

In other words, with probability at most \(\delta /4,\) no subset \(S^\prime \subseteq S\), of cardinality at least \(\frac{\vert S \vert }{2\beta }\), is a subset of \(T^\prime\). We condition the rest of the proof on the contrary event, denoted by \(\mathcal {E}_{T^\prime }\), namely that there is a subset \(S^\prime \subseteq S\) with \(S^\prime \subseteq T^\prime\) and \(\vert S^\prime \vert \ge \frac{\vert S \vert }{2\beta }\). Note that \(S^\prime\) is then a uniform and independent sample of \(T^\prime\) (see Section 2.1.1).

Now, let \(c^\ast \in \text{arg min}_{c \in \mathbb {R}^d_\ell }\; \text{cost}\;({T^\prime }{c})\) be an optimal \(\ell\)-median for \(T^\prime\). The expected distance between \(s \in S^\prime\) and \(c^\ast\) is \(\begin{equation*} E\left[\text{d}_\text{F}(s, c^\ast)\ \vert \ \mathcal {E}_{T^\prime }\right] = \sum _{\tau \in T^\prime } \text{d}_\text{F}(c^\ast , \tau) \cdot \frac{1}{\vert T^\prime \vert } = \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }. \end{equation*}\) By linearity, we have \(E\left[\text{cost}( S^\prime ,c^\ast ) \ \vert \ \mathcal {E}_{T^\prime }\right] = \frac{\vert S^\prime \vert }{\vert T^\prime \vert } \text{cost}( T^\prime ,c^\ast )\). Markov’s inequality yields the following: \(\begin{align*} \Pr \left[ \frac{\delta \vert T^\prime \vert }{4\vert S^\prime \vert }\text{cost}( S^\prime ,c^\ast ) \gt \text{cost}( T^\prime ,c^\ast ) \ \Big \vert \ \mathcal {E}_{T^\prime }\right] \le \frac{\delta }{4}. \end{align*}\) We conclude that with probability at most \(\delta /4,\) we have \(\frac{\delta \vert T^\prime \vert }{4\vert S^\prime \vert } \text{cost}( S^\prime ,c^\ast ) \gt \text{cost}( T^\prime ,c^\ast )\).

Using Markov’s inequality again, for every \(s \in S^\prime ,\) we have \(\begin{equation*} \Pr \left[\text{d}_\text{F}(s, c^\ast) \gt (1+\varepsilon) \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }\ \Big \vert \ \mathcal {E}_{T^\prime }\right] \le \frac{1}{1+\varepsilon }, \end{equation*}\) and therefore by independence, \(\begin{equation*} \Pr \left[\min _{s \in S^\prime } \text{d}_\text{F}(s, c^\ast) \gt (1+\varepsilon)\frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }\ \Big \vert \ \mathcal {E}_{T^\prime }\right] \le \frac{1}{(1+\varepsilon)^{\vert S^\prime \vert }} \le \exp \left(-\frac{\varepsilon }{2}\frac{\vert S \vert }{2\beta }\right). \end{equation*}\) Hence, with probability at most \(\exp (-\frac{\varepsilon \left\lceil \frac{8\beta (\ln (1/\delta) + \ln (4))}{\varepsilon } \right\rceil }{4\beta }) \le \delta ^2/16 \le \delta /4,\) there is no \(s \in S^\prime\) with \(\text{d}_\text{F}(s, c^\ast) \le (1+\varepsilon) \cdot \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }\). In addition, with probability at most \(\delta /4,\) Algorithm 1 fails to compute a 34-approximate \(\ell\)-median \(c \in \mathbb {R}^d_\ell\) for \(S^\prime\) (see Corollary 3.3).

Using Lemma 2.10, we conclude that with probability at least \(1-\delta ,\) all of the following events occur simultaneously:

Since \(c^\ast _{S^\prime }\) is an optimal \(\ell\)-median for \(S^\prime ,\) we get the following from the last two items: \(\begin{align*} \text{cost}( T^\prime ,c^\ast ) \ge \frac{\delta \vert T^\prime \vert }{4 \vert S^\prime \vert } \text{cost}( S^\prime ,c^\ast ) \ge \frac{\delta \vert T^\prime \vert }{4 \vert S^\prime \vert } \text{cost}( S^\prime ,c^\ast _{S^\prime }) \ge \frac{\delta \vert T^\prime \vert }{4 \vert S^\prime \vert } \frac{\text{cost}( S^\prime ,c) }{34}. \end{align*}\)

We now distinguish between two cases.

Case 1. \(\text{d}_\text{F}(c,c^\ast) \ge (1+2\varepsilon) \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }.\)

The triangle inequality yields \(\begin{align*} \text{d}_\text{F}(c,s) & \ge {} \text{d}_\text{F}(c,c^\ast) - \text{d}_\text{F}(c^\ast ,s) \ge \text{d}_\text{F}(c,c^\ast) - (1+\varepsilon) \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert } \\ & \ge {} (1+2\varepsilon) \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert } - (1+\varepsilon) \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert } = \varepsilon \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }. \end{align*}\)

As a consequence, \(\text{cost}( S^\prime ,c) \ge \varepsilon \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert } \iff \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert } \le \frac{1}{\varepsilon } \text{cost}( S^\prime ,c)\).

Now, let \(v^{s}_1, \ldots , v^{s}_{\vert s \vert }\) be the vertices of \(s\). By Lemma 4.1, there exists a polygonal curve \(c^\prime\) with up to \(2\ell - 2\) vertices, every vertex contained in one of \(B(v^{s}_1, \text{d}_\text{F}(c^\ast , s)), \ldots , B(v^{s}_{\vert s \vert }, \text{d}_\text{F}(c^\ast , s))\) and \(\text{d}_\text{F}(s, c^\prime) \le \text{d}_\text{F}(s, c^\ast) \le (1+\varepsilon) \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert } \le (1+\varepsilon) \frac{\text{cost}( S^\prime ,c) }{\varepsilon }\).

In the set of candidates that Algorithm 2 returns, a curve \(c^{\prime \prime }\) with up to \(2\ell -2\) vertices from the union of the grid covers and distance at most \(\frac{\varepsilon \frac{\delta n}{2\vert S \vert }\text{cost}( S^\prime ,c) }{n} \le \frac{\varepsilon \frac{\delta \vert T^\prime \vert }{4\vert S^\prime \vert }\text{cost}( S^\prime ,c) }{\vert T^\prime \vert } \le \varepsilon \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }\) between every corresponding pair of vertices of \(c^\prime\) and \(c^{\prime \prime }\) is contained. We conclude that \(\text{d}_\text{F}(c^\prime , c^{\prime \prime }) \le \frac{\varepsilon \text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }\).

We can now bound the cost of \(c^{\prime \prime }\) as follows: \(\begin{align*} \text{cost}( T^\prime ,c^{\prime \prime }) & ={} \sum _{\tau \in T^\prime } \text{d}_\text{F}(\tau , c^{\prime \prime }) \le \sum _{\tau \in T^\prime } \left(\text{d}_\text{F}(\tau , c^\prime) + \frac{\varepsilon \text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }\right) \\ & \le {} \sum _{\tau \in T^\prime } (\text{d}_\text{F}(\tau , c^\ast) + \text{d}_\text{F}(c^\ast , c^\prime)) + \varepsilon \text{cost}( T,c^\ast ) \\ & \le {} \sum _{\tau \in T^\prime } (\text{d}_\text{F}(\tau , c^\ast) + \text{d}_\text{F}(c^\ast , s) + \text{d}_\text{F}(s, c^\prime)) + \varepsilon \text{cost}( T^\prime ,c^\ast ) \le {} (3+3\varepsilon) \text{cost}( T^\prime ,c^\ast ) . \end{align*}\)

Case 2. \(\text{d}_\text{F}(c,c^\ast) \lt (1+2\varepsilon) \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }.\)

The cost of \(c\) can easily be bounded: \(\begin{align*} \text{cost}( T^\prime ,c) \le \sum _{\tau \in T^\prime } (\text{d}_\text{F}(\tau , c^\ast) + \text{d}_\text{F}(c^\ast , c)) \lt \text{cost}( T^\prime ,c^\ast ) + (1+2\varepsilon) \text{cost}( T^\prime ,c^\ast ) = (2+2\varepsilon) \text{cost}( T^\prime ,c^\ast ) . \end{align*}\)

The claim follows by rescaling \(\varepsilon\) by \(\frac{1}{3}\).□

Let \(c^\ast \in \text{arg min} _{c \in \mathbb {R}^d_\ell }\; \text{cost}\;({T}{c})\) be an optimal \(\ell\)-median for \(T\). The expected distance between \(s \in S\) and \(c^\ast\) is \(\begin{equation*} \text{E}\left[\text{d}_\text{F}(s, c^\ast)\right] = \sum _{i=1}^n \text{d}_\text{F}(c^\ast , \tau _i) \cdot \frac{1}{n} = \frac{\text{cost}( T,c^\ast ) }{n}. \end{equation*}\)

Now using Markov’s inequality, for every \(s \in S\) we have \(\begin{equation*} \Pr [\text{d}_\text{F}(s, c^\ast) \gt (1+\varepsilon)\text{cost}( T,c^\ast ) /n] \le \frac{\text{cost}( T,c^\ast ) n^{-1}}{(1+\varepsilon)\text{cost}( T,c^\ast ) n^{-1}} = \frac{1}{1+\varepsilon }, \end{equation*}\) and therefore by independence, \(\begin{equation*} \Pr \left[\min _{s \in S} \text{d}_\text{F}(s, c^\ast) \gt (1+\varepsilon)\text{cost}( T,c^\ast ) /n\right] \le \frac{1}{(1+\varepsilon)^{\vert S \vert }} \le \exp (-\frac{\varepsilon \vert S \vert }{2}). \end{equation*}\) Hence, with probability at most \(\exp (-\frac{\varepsilon \left\lceil \frac{2(\ln (1/\delta) + \ln (4))}{\varepsilon } \right\rceil }{2}) \le \delta /4,\) there is no \(s \in S\) with \(\text{d}_\text{F}(s, c^\ast) \le (1+\varepsilon) \frac{\text{cost}( T,c^\ast ) }{n}\). Now, assume there is an \(s \in S\) with \(\text{d}_\text{F}(s, c^\ast) \le (1+\varepsilon) \text{cost}( T,c^\ast ) /n\). We do not want any \(\tau \in S \setminus \lbrace s\rbrace\) with \(\text{cost}( T,\tau ) \gt (1+\varepsilon)\text{cost}( T,s)\) to have \(\text{cost}( W,\tau ) \le \text{cost}( W,s)\). Using Theorem 2.8, we conclude that this happens with probability at most \(\begin{equation*} \exp \left(-\frac{\varepsilon ^2 \lceil 64\varepsilon ^{-2}(\ln (1/\delta) + \ln (\lceil 8 (\varepsilon ^\prime)^{-1} (\ln (1/\delta)+\ln (4))\rceil)) \rceil }{64}\right) \le \frac{\delta }{\lceil -8 (\varepsilon ^\prime)^{-1} (\ln (\delta)-\ln (4))\rceil } \le \frac{\delta }{4 \vert S \vert }, \end{equation*}\) for each \(\tau \in S \setminus \lbrace s\rbrace\). In addition, with probability at most \(\delta /2,\) Algorithm 1 fails to compute a 34-approximate \(\ell\)-median \(\widehat{c} \in \mathbb {R}^d_\ell\) for \(T\) (see Corollary 3.3).

Using a union bound over these bad events, we conclude that with probability at least \(1-\delta\),

Let \(v^{t}_1, \ldots , v^{t}_{\vert t \vert }\) be the vertices of \(t\). By Lemma 4.1, there exists a polygonal curve \(c^\prime\) with up to \(2\ell - 2\) vertices, every vertex contained in one of \(B(v^{t}_1, \text{d}_\text{F}(c^\ast , t)), \ldots , B(v^{t}_{\vert t \vert }, \text{d}_\text{F}(c^\ast , t))\) and \(\text{d}_\text{F}(t, c^\prime) \le \text{d}_\text{F}(t, c^\ast)\). Using the triangle inequality yields \(\begin{align*} \sum _{\tau \in T} (\text{d}_\text{F}(t, c^\ast) - \text{d}_\text{F}(\tau , c^\ast)) \le \sum _{\tau \in T} \text{d}_\text{F}(t, \tau) \le (1+\varepsilon) \sum _{\tau \in T} \text{d}_\text{F}(s, \tau) \le (1+\varepsilon) \sum _{\tau \in T} (\text{d}_\text{F}(\tau , c^\ast) + \text{d}_\text{F}(c^\ast , s)), \end{align*}\) which is equivalent to \(\begin{align*} n \cdot \text{d}_\text{F}(t, c^\ast) \le (2+\varepsilon) \text{cost}( T,c^\ast ) + (1+\varepsilon) n (1+\varepsilon) \text{cost}( T,c^\ast ) /n \iff & \text{d}_\text{F}(t, c^\ast) \le (3 + 4\varepsilon) \text{cost}( T,c^\ast ) /n. \end{align*}\) Hence, we have \(\text{d}_\text{F}(t, c^\prime) \le \text{d}_\text{F}(t, c^\ast) \le (3+4\varepsilon) \text{cost}( T,c^\ast ) /n \le (3+ 4\varepsilon) \Delta /n\).

In the last step, Algorithm 3 returns a curve \(c^{\prime \prime }\) from the set \(C\) of all curves with up to \(2\ell -2\) vertices from \(P\), the union of the grid covers, that evaluates best. We can assume that \(c^{\prime \prime }\) has distance at most \(\frac{\varepsilon \Delta _l}{n} \le \varepsilon \frac{\text{cost}( T,c^\ast ) }{n}\) between every corresponding pair of vertices of \(c^\prime\) and \(c^{\prime \prime }\). We conclude that \(\text{d}_\text{F}(c^\prime , c^{\prime \prime }) \le \frac{\varepsilon \Delta _l}{n} \le \varepsilon \frac{\text{cost}( T,c^\ast ) }{n}\).

We can now bound the cost of \(c^{\prime \prime }\) as follows: \(\begin{align*} \text{cost}( T,c^{\prime \prime }) & ={} \sum _{\tau \in T} \text{d}_\text{F}(\tau , c^{\prime \prime }) \le \sum _{\tau \in T} \left(\text{d}_\text{F}(\tau , c^\prime) + \frac{\varepsilon \Delta _l}{n}\right) \le \sum _{\tau \in T} (\text{d}_\text{F}(\tau , t) + \text{d}_\text{F}(t, c^\prime)) + \varepsilon \text{cost}( T,c^\ast ) \\ & \le {} (1+\varepsilon) \text{cost}( T,s) + (3+5\varepsilon) \text{cost}( T,c^\ast ) \\ & \le {} (1+\varepsilon) \sum _{\tau \in T}(\text{d}_\text{F}(\tau , c^\ast) + \text{d}_\text{F}(c^\ast , s)) + (3+5\varepsilon) \text{cost}( T,c^\ast ) \\ & \le {} (1+\varepsilon)\text{cost}( T,c^\ast ) + (1+\varepsilon)^2 \text{cost}( T,c^\ast ) + (3+5\varepsilon) \text{cost}( T,c^\ast ) \\ & \le {} (5 + 9\varepsilon) \text{cost}( T,c^\ast ) . \end{align*}\)

The claim follows by rescaling \(\varepsilon\) by \(\frac{1}{9}\).□

Algorithm 1 has running time \(O(m^2 \log (m) \ln ^2(1/\delta) + m^3 \log m)\). The sample \(S\) has size \(O(\frac{\ln (1/\delta)}{\varepsilon })\), and the sample \(W\) has size \(O(\frac{\ln (1/\delta)}{\varepsilon ^2})\). Evaluating each curve of \(S\) against \(W\) takes time \(O(\frac{m^2 \log (m) \ln ^2(1/\delta)}{\varepsilon ^3})\), using the algorithm of Alt and Godau [4] to compute the distances.

Now, \(c\) has up to \(m\) vertices, and every grid consists of \((\frac{\frac{(3+\varepsilon)\Delta }{n}}{\frac{2\varepsilon ^\prime \Delta }{n\sqrt {d}}})^d = (\frac{(3+\varepsilon)\sqrt {d}}{2 \varepsilon ^\prime })^d \in O(\frac{1}{\varepsilon ^d})\) points. Therefore, we have \(O(\frac{m}{\varepsilon ^{d}})\) points in \(P,\) and Algorithm 3 enumerates all combinations of \(2\ell -2\) points from \(P\) taking time \(O(\frac{m^{2\ell -2}}{\varepsilon ^{(2\ell -2)d}})\). Afterward, these candidates are evaluated, which takes time \(O(n m \log (m))\) per candidate using the algorithm of Alt and Godau [4] to compute the distances. All in all, we then have running time \(O(\frac{nm^{2\ell -1} \log (m)}{\varepsilon ^{(2\ell -2)d}} + \frac{m^3 \log (m) \ln ^2(1/\delta)}{\varepsilon ^3})\).□

Let \(\ell = \vert \sigma \vert\). For the sake of simplicity, we assume that \(\frac{\varepsilon n}{2\ell }\) is integral. For \(i \in [n]\), let \(v^{\tau _i}_1, \ldots , v^{\tau _i}_{\vert \tau _i \vert }\) be the vertices of \(\tau _i\) with instants \(t^{\tau _i}_{1}, \ldots , t^{\tau _i}_{\vert \tau _i \vert }\), and let \(v^\sigma _1, \ldots , v^\sigma _{\ell }\) be the vertices of \(\sigma\) with instants \(t^\sigma _1, \ldots , t^\sigma _{\ell }\). In addition, for \(h \in \mathcal {H}\) (recall that \(\mathcal {H}\) is the set of all continuous bijections \(h:[0,1] \rightarrow [0,1]\) with \(h(0) = 0\) and \(h(1) = 1\)) and \(i \in [n]\), let \(r_{i,h} = \max \limits _{t \in [0,1]} \Vert \sigma (t) - \tau _i(h(t)) \Vert\) be the distance realized by \(h\) with respect to \(\tau _i\). We know from Lemma 2.4 that for each \(i \in [n],\) there exists a sequence \((h_{i,x})_{x=1}^\infty\) in \(\mathcal {H}\), such that \(\lim \limits _{x \rightarrow \infty } r_{i,h_{i,x}} = \text{d}_\text{F}(\sigma , \tau _i) = r_i\).

In the following, given arbitrary \(h_1, \ldots , h_n \in \mathcal {H}\), we describe how to modify \(\sigma\), such that its vertices can be found in the balls around the vertices of the \(\tau \in T\), of radii determined by \(h_1, \ldots , h_n\). Later we will argue that this modification can be applied using the \(h_{1,x}, \ldots , h_{n,x}\), for each \(x \in \mathbb {N}\), in particular.

Now, fix arbitrary \(h_1, \ldots , h_n \in \mathcal {H,}\) and for an arbitrary \(k \in \lbrace 2, \ldots , \vert \sigma \vert -1\rbrace\), fix the vertex \(v^\sigma _k\) of \(\sigma\) with instant \(t^\sigma _k\). For \(i \in [n]\), let \(s_i\) be the maximum of \([\vert \tau _i \vert -1]\), such that \(t^{\tau _i}_{s_i} \le h_i(t^\sigma _k) \le t^{\tau _i}_{s_{i}+1}\). Namely, \(v^\sigma _k\) is matched to a point on the line segment \(\overline{v^{\tau _1}_{s_1}v^{\tau _1}_{s_1+1}}, \ldots , \overline{v^{\tau _n}_{s_n}v^{\tau _n}_{s_n+1}}\), respectively, by \(h_1, \ldots , h_n\).

For \(i \in [n]\), let \(t^{-}_i\) be the maximum of \([0, t^\sigma _k]\), such that \(\sigma (t^{-}_i) \in B(v^{\tau _i}_{s_i}, r_{i,h_i}),\) and let \(t^{+}_i\) be the minimum of \([t^\sigma _k, 1]\), such that \(\sigma (t^+_i) \in B(v^{\tau _i}_{s_i+1}, r_{i,h_i})\). These are the instants when \(\sigma\) visits \(B(v^{\tau _i}_{s_i}, r_{i,h_i})\) before or when it visits \(v^\sigma _k\) and \(\sigma\) visits \(B(v^{\tau _i}_{s_i+1}, r_{i,h_i})\) when or after it visits \(v^\sigma _k\), respectively. Furthermore, there is a permutation \(\alpha \in \mathcal {S}_n\) of the index set \([n]\), such that \(\begin{equation*} t^{-}_{\alpha ^{-1}(1)} \le \dots \le t^{-}_{\alpha ^{-1}(n)}. \qquad \qquad \qquad \qquad {\rm (I)} \end{equation*}\) In addition, there is a permutation \(\zeta \in \mathcal {S}_n\) of the index set \([n]\), such that \(\begin{equation*} t^{+}_{\zeta ^{-1}(1)} \le \dots \le t^{+}_{\zeta ^{-1}(n)}. \qquad \qquad \qquad \qquad {\rm (II)} \end{equation*}\) As well, for each \(i \in [n],\) we have \(\begin{equation*} t^{\tau _i}_{s_i} \le h_i(t^{-}_i) \qquad \qquad \qquad \qquad {\rm (III)} \end{equation*}\) and \(\begin{equation*} h_i(t^{+}_i) \le t^{\tau _i}_{s_i+1}, \qquad \qquad \qquad \qquad {\rm (IV)} \end{equation*}\) because \(\sigma (t^-_i)\) and \(\sigma (t^+_i)\) are the closest points to \(v^\sigma\) on \(\sigma\) that have distance at most \(r_{i,h_i}\) to \(v^{\tau _i}_{s_i}\) and \(v^{\tau _i}_{s_i+1}\), respectively, by definition. We will now use Equations (I) to (IV) to prove that an advanced shortcut only affects matchings among line segments, and hence we can easily bound the resulting distances for at least \((1-\varepsilon)n\) of the curves.

Let \(\begin{equation*} I_{v^\sigma _k}(h_1, \ldots , h_n) = \lbrace \tau _{\alpha ^{-1}((1-\frac{\varepsilon }{2\ell })n + 1)}, \ldots , \tau _{\alpha ^{-1}(n)}\rbrace ,\ O_{v^\sigma _k}(h_1, \ldots , h_n) = \lbrace \tau _{\zeta ^{-1}(1)}, \ldots , \tau _{\zeta ^{-1}(\frac{\varepsilon n}{2\ell })} \rbrace . \end{equation*}\) \(I_{v^\sigma _k}(h_1, \ldots , h_n)\) is the set of the last \(\frac{\varepsilon n}{2\ell }\) curves whose balls are visited by \(\sigma\), before or when \(\sigma\) visits \(v^\sigma _k\). Similarly, \(O_{v^\sigma _k}(h_1, \ldots , h_n)\) is the set of the first \(\frac{\varepsilon n}{2\ell }\) curves whose balls are visited by \(\sigma\), when or immediately after \(\sigma\) visited \(v^\sigma _k\). We now modify \(\sigma\), such that \(v^\sigma _k\) is replaced by two new vertices that are elements of at least one \(B(v^{\tau _i}_{j}, r_{i,h_i})\), for a \(\tau _i \in I_{v^\sigma _k}(h_1, \ldots , h_n)\), respectively for a \(\tau _i \in O_{v^\sigma _k}(h_1, \ldots , h_n)\), and \(j \in [\vert \tau _i \vert ]\), each.

Let \(\sigma ^\prime _{h_1, \ldots , h_n}\) be the piecewise defined curve, defined just like \(\sigma\) on \([0,t^-_{\alpha ^{-1}(k_1)}]\) and \([t^+_{\zeta ^{-1}(k_2)},1]\) for arbitrary \(k_1 \in \lbrace (1-\frac{\varepsilon }{2\ell })n + 1, \ldots , n\rbrace\) and \(k_2 \in [\frac{\varepsilon n}{2\ell }]\), but on \((t^-_{\alpha ^{-1}(k_1)}, t^+_{\zeta ^{-1}(k_2)})\) it connects \(\sigma (t^-_{\alpha ^{-1}(k_1)})\) and \(\sigma (t^+_{\zeta ^{-1}(k_2)})\) with the line segment \(\begin{equation*} \gamma (t) = \left(1-\frac{t-t^-_{\alpha ^{-1}(k_1)}}{t^+_{\zeta ^{-1}(k_2)}-t^-_{\alpha ^{-1}(k_1)}}\right) \sigma \left(t^-_{\alpha ^{-1}(k_1)}\right) + \frac{t-t^-_{\alpha ^{-1}(k_1)}}{t^+_{\zeta ^{-1}(k_2)}-t^-_{\alpha ^{-1}(k_1)}} \sigma \left(t^+_{\zeta ^{-1}(k_2)}\right). \end{equation*}\) We now argue that for all \(\tau _i \in T \setminus (I_{v^\sigma _k}(h_1, \ldots , h_n) \cup O_{v^\sigma _k}(h_1, \ldots , h_n)),\) the Fréchet distance between \(\sigma ^\prime _{h_1, \ldots , h_n}\) and \(\tau _i\) is upper bounded by \(r_{i,h_i}\). First, note that by definition, \(h_1, \ldots , h_n\) are strictly increasing functions, since they are continuous bijections that map 0 to 0 and 1 to 1. As immediate consequence, we have that \(\begin{equation*} t^{\tau _i}_{s_i} \le h_i(t^-_i) \le h_i\left(t^-_{\alpha ^{-1}(k_1)}\right)\qquad \qquad \qquad \qquad {\rm (V)} \end{equation*}\) for each \(\tau _i \in T \setminus I_{v^\sigma _k}(h_1, \ldots , h_n)\) and \(\begin{equation*} h_i\left(t^+_{\zeta ^{-1}(k_2)}\right) \le h_i(t^+_i) \le t^{\tau _i}_{s_i+1} \qquad \qquad \qquad \qquad {\rm (VI)} \end{equation*}\) for each \(\tau _i \in T \setminus O_{v^\sigma _k}(h_1, \ldots , h_n)\), using Equations (I) to (IV).. Therefore, each \(\tau _i \in T \setminus (I_{v^\sigma _k}(h_1, \ldots , h_n) \cup O_{v^\sigma _k}(h_1, \ldots , h_n))\) has no vertex between the instants \(h_i(t^-_{\alpha ^{-1}(k_1)})\) and \(h_i(t^+_{\zeta ^{-1}(k_2)})\). We also know that for each \(\tau _i \in T,\) \(\begin{equation*} \left\Vert \sigma \left(t^-_{\alpha ^{-1}(k_1)}\right) - \tau _i\left(h_i\left(t^-_{\alpha ^{-1}(k_1)}\right)\right) \right\Vert \le r_{i,h_i} \qquad \qquad \qquad \qquad {\rm (VII)} \end{equation*}\) and \(\begin{equation*} \left\Vert \sigma \left(t^+_{\zeta ^{-1}(k_2)}\right) - \tau _i\left(h_i\left(t^+_{\zeta ^{-1}(k_2)}\right)\right) \right\Vert \le r_{i,h_i}. \qquad \qquad \qquad \qquad {\rm (VIII)} \end{equation*}\)

Let \(D_{s,\sigma } = [0,t^-_{\alpha ^{-1}(k_1)})\), \(D_{m,\sigma } = [t^-_{\alpha ^{-1}(k_1)}, t^+_{\zeta ^{-1}(k_2)}]\) and \(D_{e,\sigma } = (t^+_{\zeta ^{-1}(k_2)}, 1]\). In addition, for \(i \in [n]\), let \(D_{s,\tau _i} = [0,h_i(t^-_{\alpha ^{-1}(k_1)}))\), \(D_{m,\tau _i} = [h_i(t^-_{\alpha ^{-1}(k_1)}), h_i(t^+_{\zeta ^{-1}(k_2)})]\), and \(D_{e,\tau _i} = (h_i(t^+_{\zeta ^{-1}(k_2)}),1]\). Now, for each \(\tau _i \in T \setminus (I_{v^\sigma _k}(h_1, \ldots , h_n) \cup O_{v^\sigma _k}(h_1, \ldots , h_n)),\) we use \(h_i\) to match \(\sigma ^\prime _{h_1, \ldots , h_n}\vert _{D_{s,\sigma }}\) to \(\tau _i\vert _{D_{s,\tau _i}}\) and \(\sigma ^\prime _{h_1, \ldots , h_n}\vert _{D_{e,\sigma }}\) to \(\tau _i\vert _{D_{e,\tau _i}}\) with distance at most \(r_{i,h_i}\). Since \(\sigma ^\prime _{h_1, \ldots , h_n}\vert _{D_{m,\sigma }}\) and \(\tau _i\vert _{D_{m,\tau _i}}\) are just line segments by Equations (V) and (VI), they can be linearly matched to each other with distance at most \(\begin{equation*} \max \left\lbrace \left\Vert \sigma \left(t^-_{\alpha ^{-1}(k_1)}\right) - \tau _i\left(h_i\left(t^-_{\alpha ^{-1}(k_1)}\right)\right) \right\Vert , \left\Vert \sigma \left(t^+_{\zeta ^{-1}(k_2)}\right) - \tau _i\left(h_i\left(t^+_{\zeta ^{-1}(k_2)}\right)\right) \right\Vert \right\rbrace , \end{equation*}\) which is at most \(r_{i,h_i}\) by Equations (VII) to (VIII). We conclude that \(\text{d}_\text{F}(\sigma ^\prime _{h_1, \ldots , h_n}, \tau _i) \le r_{i,h_i}\).

Because this modification works for every \(h_1, \ldots , h_n \in \mathcal {H}\), we conclude that \(\text{d}_\text{F}(\sigma ^\prime _{h_1, \ldots , h_n}, \tau _i) \le r_{i,h_i}\) for every \(h_1, \ldots , h_n \in \mathcal {H}\) and \(\tau _i \in T \setminus (I_{v^\sigma _k}(h_1, \ldots , h_n) \cup O_{v^\sigma _k}(h_1, \ldots , h_n))\). Thus, \(\lim \limits _{x \rightarrow \infty } \text{d}_\text{F}(\sigma ^\prime _{h_{1,x}, \ldots , h_{n,x}}, \tau _i) \le \text{d}_\text{F}(\sigma , \tau _i) = r_i\) for each \(\tau _i \in T \setminus (I_{v^\sigma _k}(h_{1,x}, \ldots , h_{n,x}) \cup O_{v^\sigma _k}(h_{1,x}, \ldots , h_{n,x}))\).

Now, to prove the claim, for each combination \(h_1, \ldots , h_n \in \mathcal {H}\), we apply this modification to \(v^\sigma _k\) and successively to every other vertex \(v^{\sigma ^\prime _{h_1, \ldots , h_n}}_l\) of the resulting curve \(\sigma ^\prime _{h_1, \ldots , h_n}\), except \(v^{\sigma ^\prime _{h_1, \ldots , h_n}}_1\) and \(v^{\sigma ^\prime _{h_1, \ldots , h_n}}_{\vert \sigma ^\prime _{h_1, \ldots , h_n}\vert }\), since these must be elements of \(B(v^{\tau _i}_1, r_{i,h_i})\) and \(B(v^{\tau _i}_{\vert \tau _i \vert }, r_{i,h_i})\), respectively, for each \(i \in [n]\), by definition of the Fréchet distance.

Since the modification is repeated at most \(\vert \sigma \vert - 2\) times for each combination \(h_1, \dots h_n \in \mathcal {H}\), we conclude that the number of vertices of each \(\sigma ^\prime _{h_1, \ldots , h_n}\) can be bounded by \(2 \cdot (\vert \sigma \vert - 2) + 2\).

\(T_1, \ldots , T_{2\ell -4}\) are therefore all the \(I_{v^\sigma _k}(h_{1,x}, \ldots , h_{n,x})\) and \(O_{v^\sigma _k}(h_{1,x}, \ldots , h_{n,x})\) for \(k \in \lbrace 2, \ldots , 2\vert \sigma \vert - 3\rbrace\), when \(x \rightarrow \infty\). Note that every \(I_{v^\sigma _k}(h_{1,x}, \ldots , h_{n,x})\) and \(O_{v^\sigma _k}(h_{1,x}, \ldots , h_{n,x})\) is determined by the visiting order of the balls, and since their radii converge, these sets do as well.□

We assume that \(\vert T^\prime \vert \ge \frac{1}{\beta } \vert T \vert\) and \(\ell \gt 2\). Let \(n^\prime\) be the number of sampled curves in \(S\) that are elements of \(T^\prime\). Clearly, \(E\left[n^\prime \right] \ge \sum _{i=1}^{\vert S \vert } \frac{1}{\beta } = \frac{\vert S \vert }{\beta }\). In addition, \(n^\prime\) is the sum of independent Bernoulli trials. A Chernoff bound (see Lemma 2.11) yields the following: \(\begin{align*} \Pr \left[n^\prime \lt \frac{\vert S \vert }{2\beta }\right] \le \Pr \left[n^\prime \lt \frac{E\left[n^\prime \right]}{2}\right] \le \exp \left(-\frac{1}{4}\frac{\vert S \vert }{2\beta }\right) \le \exp \left(\frac{\ell \ln \left(\frac{\delta }{4(2\ell -4)}\right)}{\varepsilon } \right) \le \left(\frac{\delta ^\ell }{8^\ell }\right)^{\frac{1}{\varepsilon }} \le \frac{\delta }{8}. \end{align*}\)

In other words, with probability at most \(\delta /8,\) no subset \(S^\prime \subseteq S\), of cardinality at least \(\frac{\vert S \vert }{2\beta }\), is a subset of \(T^\prime\). We condition the rest of the proof on the contrary event, denoted by \(\mathcal {E}_{T^\prime }\), namely that there is a subset \(S^\prime \subseteq S\) with \(S^\prime \subseteq T^\prime\) and \(\vert S^\prime \vert \ge \frac{\vert S \vert }{2\beta }\). Note that \(S^\prime\) is then a uniform and independent sample of \(T^\prime\) (see Section 2.1.1).

Now, let \(c^\ast \in \mathbb {R}^d_\ell\) be an optimal \(\ell\)-median for \(T^\prime\). The expected distance between \(s \in S^\prime\) and \(c^\ast\) is \(\begin{equation*} E\left[\text{d}_\text{F}(s, c^\ast)\ \vert \ \mathcal {E}_{T^\prime }\right] = \sum _{\tau \in T^\prime } \text{d}_\text{F}(c^\ast , \tau) \cdot \frac{1}{\vert T^\prime \vert } = \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }. \end{equation*}\) By linearity, we have \(E\left[\text{cost}( S^\prime ,c^\ast ) \ \vert \ \mathcal {E}_{T^\prime }\right] = \frac{\vert S^\prime \vert }{\vert T^\prime \vert } \text{cost}( T^\prime ,c^\ast )\). Markov’s inequality yields the following: \(\begin{align*} \Pr \left[ \frac{\delta \vert T^\prime \vert }{4\vert S^\prime \vert }\text{cost}( S^\prime ,c^\ast ) \gt \text{cost}( T^\prime ,c^\ast ) \ \Big \vert \ \mathcal {E}_{T^\prime }\right] \le \frac{\delta }{4}. \end{align*}\)

We conclude that with probability at most \(\delta /4,\) we have \(\frac{\delta \vert T^\prime \vert }{4\vert S^\prime \vert } \text{cost}( S^\prime ,c^\ast ) \gt \text{cost}( T^\prime ,c^\ast )\).

Now, from Lemma 6.1, we know that there are \(2\ell -4\) subsets \(T^\prime _1, \ldots , T^\prime _{2\ell -4} \subseteq T^\prime\), of cardinality \(\frac{\varepsilon \vert T^\prime \vert }{2\ell }\) each and which are not necessarily disjoint, such that for every set \(W \subseteq T^\prime\) that contains at least one curve \(\tau \in T^\prime _i\) for each \(i \in [2\ell -4]\), there exists a curve \(c^\prime \in \mathbb {R}^d_{2\ell -2}\) that has all of its vertices contained in \(\begin{equation*} \bigcup \limits _{\tau \in W} \bigcup \limits _{j \in [\vert \tau \vert ]} B\left(v^{\tau }_j, \text{d}_\text{F}(\tau , c^\ast)\right) \end{equation*}\) and for at least \((1-\varepsilon) \vert T^\prime \vert\) curves \(\tau \in T^\prime \setminus (T^\prime _1 \cup \dots \cup T^\prime _{2\ell -4})\) it holds that \(\text{d}_\text{F}(\tau , c^\prime) \le \text{d}_\text{F}(\tau , c^\ast)\).

There are up to \(\frac{\varepsilon \vert T^\prime \vert }{4\ell }\) curves with distance to \(c^\ast\) at least \(\frac{4\ell \text{cost}( T^\prime ,c^\ast ) }{\varepsilon \vert T^\prime \vert }\). Otherwise, the cost of these curves would exceed \(\text{cost}( T^\prime ,c^\ast )\), which is a contradiction. Later we will prove that each ball we cover has radius at most \(\frac{4\ell \text{cost}( T^\prime ,c^\ast ) }{\varepsilon \vert T^\prime \vert }\). Therefore, for each \(i \in [2\ell -4],\) we have to ignore up to half of the curves \(\tau \in T^\prime _i\), since we do not cover the balls of radius \(\text{d}_\text{F}(\tau , c^\ast)\) centered at their vertices. For each \(i \in [2\ell -4]\) and \(s \in S^\prime ,\) we now have \(\begin{equation*} \Pr \left[s \in T^\prime _i \wedge \text{d}_\text{F}(s, c^\ast) \le \frac{4\ell \text{cost}( T^\prime ,c^\ast ) }{\varepsilon \vert T^\prime \vert } \ \Big \vert \ \mathcal {E}_{T^\prime } \right] \ge \frac{\varepsilon }{4\ell }. \end{equation*}\) Therefore, by independence, for each \(i \in [2\ell -4],\) the probability that no \(s \in S^\prime\) is an element of \(T^\prime _i\) and has distance to \(c^\ast\) at most \(\frac{4\ell \text{cost}( T^\prime ,c^\ast ) }{\varepsilon \vert T^\prime \vert }\) is at most \((1-\frac{\varepsilon }{4\ell })^{\vert S^\prime \vert } \le \exp (-\frac{\varepsilon }{4\ell } \frac{4\ell (\ln (1/\delta)+\ln (4(2\ell -4)))}{\varepsilon }) = \exp (\ln (\frac{\delta }{4(2\ell -4)})) = \frac{\delta }{4(2\ell -4)}\). In addition, with probability at most \(\delta /4,\) Algorithm 1 fails to compute a 34-approximate \(\ell\)-median \(c \in \mathbb {R}^d_\ell\) for \(S^\prime\) (see Corollary 3.3).

Using Lemma 2.10, we conclude that with probability at least \(1-7/8 \delta ,\) all of the following events occur simultaneously:

Let \(B_{c^\ast } = \lbrace \tau \in T^\prime \mid \text{d}_\text{F}(\tau , c^\ast) \le \frac{\text{cost}( T^\prime ,c^\ast ) }{\varepsilon ^2 \vert T^\prime \vert } \rbrace\) and \(B_{c} = \lbrace \tau \in T^\prime \mid \text{d}_\text{F}(\tau , c) \le \varepsilon \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert } \rbrace\). First, note that \(\vert T^{\prime } \setminus B_{c^\ast } \vert \le \varepsilon ^2 \vert T^\prime \vert\); otherwise, \(\text{cost}( T^\prime \setminus B_{c^\ast },c^\ast ) \gt \text{cost}( T^\prime ,c^\ast )\), which is a contradiction, and therefore \(\vert B_{c^\ast } \vert \ge (1-\varepsilon ^2) \vert T^\prime \vert\). We now distinguish two cases.

Case 1. \(\vert B_{c^\ast } \setminus B_{c} \vert \gt 2 \varepsilon \vert B_{c^\ast } \vert .\)

We have \(2 \varepsilon \vert B_{c^\ast } \vert \ge (1-\varepsilon ^2)2\varepsilon \vert T^\prime \vert \ge \varepsilon \vert T^\prime \vert\), and hence \(\Pr [\text{d}_\text{F}(s,c) \gt \varepsilon \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }\ \vert \ \mathcal {E}_{T^\prime } ] \ge \varepsilon\) for each \(s \in S^\prime\). Using independence, we conclude that with probability at most \(\begin{equation*} (1-\varepsilon)^{\vert S^\prime \vert } \le \exp \left(-\varepsilon \frac{4\ell (\ln (1/\delta)+\ln (4(2\ell -4)))}{\varepsilon }\right) \le \frac{\delta ^{4\ell }}{4^{4\ell }} \le \frac{\delta }{8} \end{equation*}\) no \(s \in S^\prime\) has distance to \(c\) greater than \(\varepsilon \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }\). Including this bad event, by Lemma 2.10 we conclude that with probability at least \(1-\delta ,\) Items 1 to 4 occur simultaneously and at least one \(s \in S^\prime\) has distance to \(c\) greater than \(\varepsilon \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }\), and hence \(\text{cost}( S^\prime ,c) \gt \varepsilon \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert } \iff \frac{\text{cost}( S^\prime ,c) }{\varepsilon } \gt \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }\) and thus we indeed cover the balls of radius at most \(\frac{4\ell \text{cost}( T^\prime ,c^\ast ) }{\varepsilon \vert T^\prime \vert } \lt \frac{4\ell }{\varepsilon } \frac{\text{cost}( S^\prime ,c^\ast ) }{\varepsilon }\).

In the last step, Algorithm 4 returns a set \(C\) of all curves with up to \(2\ell -2\) vertices from the grids that contains one curve, denoted by \(c^{\prime \prime }\) with same number of vertices as \(c^\prime\) (recall that this is the curve guaranteed from Lemma 6.1) and distance at most \(\frac{\varepsilon }{n} \Delta _l \le \frac{\varepsilon }{\vert T^\prime \vert } \text{cost}( T^\prime ,c^\ast )\) between every corresponding pair of vertices of \(c^\prime\) and \(c^{\prime \prime }\). We conclude that \(\text{d}_\text{F}(c^\prime , c^{\prime \prime }) \le \frac{\varepsilon }{\vert T^\prime \vert } \text{cost}( T^\prime ,c^\ast )\). In addition, recall that \(\text{d}_\text{F}(\tau , c^\prime) \le \text{d}_\text{F}(\tau , c^\ast)\) for \(\tau \in T^\prime \setminus (T^\prime _1 \cup \dots \cup T^\prime _{2\ell -4})\). Further, \(T^\prime\) contains at least \(\frac{\vert T^\prime \vert }{2}\) curves with distance at most \(\frac{2\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }\) to \(c^\ast\); otherwise, the cost of the remaining curves would exceed \(\text{cost}( T^\prime ,c^\ast )\), which is a contradiction, and since \(\varepsilon \lt \frac{1}{2}\), there is at least one curve \(\sigma \in T^\prime \setminus (T^\prime _1 \cup \dots \cup T^\prime _{2\ell -4})\) with \(\text{d}_\text{F}(\sigma , c^\prime) \le \text{d}_\text{F}(\sigma , c^\ast) \le \frac{2\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }\) by the pigeonhole principle. We can now bound the cost of \(c^{\prime \prime }\) as follows: \(\begin{align*} \text{cost}( T^\prime ,c^{\prime \prime }) & ={} \sum _{\tau \in T^\prime } \text{d}_\text{F}(\tau , c^{\prime \prime }) \le \sum _{\tau \in T^\prime \setminus (T^\prime _1 \cup \dots \cup T^\prime _{2\ell -4})} \left(\text{d}_\text{F}(\tau , c^\prime) + \frac{\varepsilon }{\vert T^\prime \vert } \text{cost}( T^\prime ,c^\ast ) \right)\ + \\ & \ \ \ \ \sum _{\tau \in (T^\prime _1 \cup \dots \cup T^\prime _{2\ell -4})} \left(\text{d}_\text{F}(\tau , c^\ast) + \text{d}_\text{F}(c^\ast , \sigma) + \text{d}_\text{F}(\sigma , c^{\prime }) + \text{d}_\text{F}(c^\prime , c^{\prime \prime }) \right) \\ & \le {} (1+\varepsilon) \text{cost}( T^\prime ,c^\ast ) + \sum _{\tau \in (T^\prime _1 \cup \dots \cup T^\prime _{2\ell -4})} \left((2+2+\varepsilon)\frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert } \right) \\ & \le {} \text{cost}( T^\prime ,c^\ast ) + \varepsilon \text{cost}( T^\prime ,c^\ast ) + 5\varepsilon \text{cost}( T^\prime ,c^\ast ) = (1 + 6 \varepsilon) \text{cost}( T^\prime ,c^\ast ) . \end{align*}\)

Case 2. \(\vert B_{c^\ast } \setminus B_{c} \vert \le 2 \varepsilon \vert B_{c^\ast } \vert .\)

Again, we distinguish two cases.

Case 2.1. \(\text{d}_\text{F}(c,c^\ast) \le 4\varepsilon \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }.\)

We can easily bound the cost of \(c\): \(\begin{align*} \text{cost}( T^\prime ,c) \le \sum _{\tau \in T^\prime } (\text{d}_\text{F}(\tau , c^\ast) + \text{d}_\text{F}(c^\ast , c)) \le (1+4\varepsilon) \text{cost}( T^\prime ,c^\ast ) . \end{align*}\)

Case 2.2. \(\text{d}_\text{F}(c,c^\ast) \gt 4\varepsilon \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert }.\)

Recall that \(\vert B_{c^\ast } \vert \ge (1-\varepsilon ^2) \vert T^\prime \vert\). We have \(\begin{align*} \vert T^\prime \setminus B_{c} \vert & \le {} \vert T^\prime \setminus B_{c^\ast } \vert + 2\varepsilon \vert B_{c^\ast } \vert = \vert T^\prime \vert - (1-2\varepsilon) \vert B_{c^\ast } \vert \le \vert T^\prime \vert - (1-2\varepsilon)(1-\varepsilon ^2) \vert T^\prime \vert \\ & = (2\varepsilon + \varepsilon ^2 - 2\varepsilon ^3) \vert T^\prime \vert \lt \frac{1}{3} \vert T^\prime \vert . \end{align*}\)

Hence, \(\vert B_{c} \vert \ge (1-2\varepsilon -\varepsilon ^2 + 2 \varepsilon ^3) \vert T^\prime \vert \gt \frac{2}{3} \vert T^\prime \vert\). Assume that we assign all curves to \(c\) instead of to \(c^\ast\). For \(\tau \in B_{c}\), we now have decrease in cost \(\text{d}_\text{F}(\tau , c^\ast) - \text{d}_\text{F}(\tau , c)\), which can be bounded as follows: \(\begin{align*} \text{d}_\text{F}(\tau , c^\ast) - \text{d}_\text{F}(\tau , c) & \ge {} \text{d}_\text{F}(\tau , c^\ast) - \varepsilon \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert } \ge \text{d}_\text{F}(c,c^\ast) - \text{d}_\text{F}(\tau , c) - \varepsilon \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert } \\ & \ge \text{d}_\text{F}(c, c^\ast) - 2\varepsilon \frac{\text{cost}( T^\prime ,c^\ast ) }{\vert T^\prime \vert } \gt \frac{1}{2} \text{d}_\text{F}(c,c^\ast). \end{align*}\)

For \(\tau \in T^\prime \setminus B_{c}\), we have an increase in cost \(\text{d}_\text{F}(\tau , c) - \text{d}_\text{F}(\tau , c^\ast) \le \text{d}_\text{F}(c, c^\ast)\). Let the overall increase in cost be denoted by \(\alpha\), which can be bounded as follows: \(\begin{align*} \alpha \lt \vert T^\prime \setminus B_{c} \vert \cdot \text{d}_\text{F}(c, c^\ast) - \vert B_{c} \vert \cdot \frac{\text{d}_\text{F}(c,c^\ast)}{2}. \end{align*}\)

By the fact that \(\vert T^\prime \setminus B_{c} \vert \lt \frac{1}{2} \vert B_{c} \vert\) for our choice of \(\varepsilon\), we conclude that \(\alpha \lt 0\), which is a contradiction because \(c^\ast\) is an optimal \(\ell\)-median for \(T^\prime\). Therefore, Case 2.2 cannot occur. Rescaling \(\varepsilon\) by \(\frac{1}{6}\) proves the claim.□

For \(k=1\), the claim trivially holds. We now distinguish two cases. In the first case, the principle of the proof is presented in all of its detail. In the second case, we only show how to generalize the first case to \(k \gt 2\).

Case 1. \(k=2.\)

Let \(C^\ast = \lbrace c^\ast _1, c^\ast _2 \rbrace\) be an optimal set of \(k\) medians for \(T\) with clusters \(T^\ast _1\) and \(T^\ast _2\), respectively, that form a partition of \(T\). For the sake of simplicity, assume that \(n\) is a power of 2 and w.l.o.g. assume that \(\vert T^\ast _1 \vert \ge \frac{1}{2} \vert T \vert \gt \frac{1}{\beta } \vert T \vert\). Let \(C_1\) be the set of candidates returned by Median-Candidates in the initial call. With probability at least \(1-\delta /k\), there is a \(c_1 \in C_1\) with \(\text{cost}({T^\ast _1}{c_1}) \le (\alpha +\varepsilon)\ \text{cost}({T^\ast _1}{c^\ast _1})\). We distinguish two cases.

Case 1.1. There exists a recursive call with parameters \((T^\prime ,\lbrace c_1\rbrace ,1,\beta ,\delta ,\varepsilon)\) and \(\vert T^\ast _2 \cap T^\prime \vert \ge \frac{1}{\beta } \vert T^\prime \vert\).

First, we assume that \(T^\prime\) is the maximum cardinality input with \(\vert T^\ast _2 \cap T^\prime \vert \ge \frac{1}{\beta } \vert T^\prime \vert\), occurring in a recursive call of the algorithm. Let \(C_2\) be the set of candidates returned by Median-Candidates in this call. With probability at least \(1-\delta /k\), there is a \(c_2 \in C_2\) with \({\rm cost}({T^\ast _2 \cap T^\prime },{c_2}) \le (\alpha +\varepsilon) {\rm cost}({T^\ast _2 \cap T^\prime },{\widetilde{c}_2})\), where \(\widetilde{c}_2\) is an optimal median for \(T^\ast _2 \cap T^\prime\).

Let \(P\) be the set of elements of \(T\) removed in the \(m \in \mathbb {N}\), \(m \le \log (n)\), pruning phases between obtaining \(c_1\) and \(c_2\). Without loss of generality, we assume that \(P \ne \emptyset\). For \(i \in [m]\), let \(P_i\) be the elements removed in the \(i\)^th (in the order of the recursive calls occurring) pruning phase. Note that the \(P_i\) are pairwise disjoint, and we have that \(P = \cup _{i=1}^t P_i\) and \(\vert P_i \vert = \frac{n}{2^{i}}\). Since \(T = T^\ast _1 \uplus (T^\ast _2 \cap T^\prime) \uplus (T^\ast _2 \cap P)\), we have \(\begin{align*} \text{cost}( T,\lbrace c_1, c_2) \le \text{cost}( T^\ast _1,c_1) + \text{cost}( T^\ast _2 \cap T^\prime ,c_2) + \text{cost}( T^\ast _2 \cap P,c_1) \qquad \qquad \qquad \qquad {\rm (I)} \end{align*}\) Our aim is now to prove that the number of elements wrongly assigned to \(c_1\) (i.e., \(T^\ast _2 \cap P\)) is small and further, that their cost is a fraction of the cost of the elements correctly assigned to \(c_1\) (i.e., \(T^\ast _1\)).

We define \(R_0 = T,\) and for \(i \in [m],\) we define \(R_i = R_{i-1} \setminus P_i\). The \(R_i\) are the elements remaining after the \(i\)^th pruning phase. Note that by definition, \(\vert R_i \vert = \frac{n}{2^i} = \vert P_i \vert\). Since \(R_{m} = T^\prime\) is the maximum cardinality input, with \(\vert T^\ast _2 \cap T^\prime \vert \ge \frac{1}{\beta } \vert T^\prime \vert\), we have that \(\vert T^\ast _2 \cap R_i \vert \lt \frac{1}{\beta } \vert R_i \vert\) for all \(i \in [m-1]\). In addition, for each \(i \in [m],\) we have \(P_i \subseteq R_{i-1}\), and therefore \(\begin{align*} \vert T^\ast _2 \cap P_i \vert \le \vert T^\ast _2 \cap R_{i-1} \vert \lt \frac{1}{\beta } \vert R_{i-1} \vert = \frac{2}{\beta } \frac{n}{2^i} \qquad \qquad \qquad \qquad {\rm (II)} \end{align*}\) and as an immediate consequence \(\begin{align*} \vert T^\ast _1 \cap P_i \vert = \vert P_i \vert - \vert T^\ast _2 \cap P_i \vert \gt \vert P_i \vert - \frac{1}{\beta } \vert R_{i-1} \vert = \left(1-\frac{2}{\beta }\right) \frac{n}{2^i}. \qquad \qquad \qquad \qquad {\rm (III)} \end{align*}\) This tells us that mainly the elements of \(T^\ast _1\) are removed in the pruning phase and only very few elements of \(T^\ast _2\). By definition, we have for all \(i \in [m-1]\), \(\sigma \in P_i\) and \(\tau \in P_{i+1}\) that \(\rho (\sigma , c_1) \le \rho (\tau , c_1)\), hence \(\begin{equation*} \frac{1}{\vert T^\ast _2 \cap P_i \vert } \text{cost}( T^\ast _2 \cap P_i,c_1) \le \frac{1}{\vert T^\ast _1 \cap P_{i+1} \vert } \text{cost}( T^\ast _1 \cap P_{i+1},c_1) . \end{equation*}\) Combining this inequality with Equations (II) and (III), we obtain for \(i \in [m-1]\): \(\begin{align*} & \frac{\beta 2^i}{2n} \text{cost}( T^\ast _2 \cap P_i,c_1) \lt \frac{2^{i+1}}{(1-2/\beta)n} \text{cost}( T^\ast _1 \cap P_{i+1},c_1) \\ \iff & \text{cost}( T^\ast _2 \cap P_i,c_1) \lt \frac{2^{i+1} 2n}{(1-2/\beta)n\beta 2^i} \text{cost}( T^\ast _1 \cap P_{i+1},c_1) = \frac{4}{(\beta -2)} \text{cost}( T^\ast _1 \cap P_{i+1},c_1) . \qquad \qquad \qquad \qquad {\rm (IV)} \end{align*}\) We still need such a bound for \(i=m\). Since \(\vert R_m \vert = \vert P_m \vert\) and also \(R_m \subseteq R_{m-1}\), we can use Equation (II) to obtain \(\begin{align*} \vert T^\ast _1 \cap R_m \vert = \vert R_m \vert - \vert T^\ast _2 \cap R_m \vert \ge \vert R_m \vert - \vert T^\ast _2 \cap R_{m-1} \vert \gt \left(1-\frac{2}{\beta }\right)\frac{n}{2^m}. \qquad \qquad \qquad \qquad {\rm (V)} \end{align*}\) In addition, we have for all \(\sigma \in P_m\) and \(\tau \in R_m\) that \(\rho (\sigma , c_1) \le \rho (\tau , c_1)\) by definition, thus \(\begin{equation*} \frac{1}{\vert T^\ast _2 \cap P_m \vert } \text{cost}( T^\ast _2 \cap P_m,c_1) \le \frac{1}{\vert T^\ast _1 \cap R_{m} \vert } \text{cost}( T^\ast _1 \cap R_m,c_1) . \end{equation*}\) We combine this inequality with Equations (II) and (V) and obtain \(\begin{align*} & \frac{\beta 2^m}{2n} \text{cost}( T^\ast _2 \cap P_m,c_1) \lt \frac{2^m2n}{(1-2/\beta)n\beta 2^m} \text{cost}( T^\ast _1 \cap R_m,c_1) \\ \iff & \text{cost}( T^\ast _2 \cap P_m,c_1) \lt \frac{2}{(\beta -2)} \text{cost}( T^\ast _1 \cap R_m,c_1) .\qquad \qquad \qquad \qquad {\rm (VI)} \end{align*}\) We are now ready to bound the cost of the elements of \(T^\ast _2\) wrongly assigned to \(c_1\). Combining Equations (IV) and (VI) yields \(\begin{align*} \text{cost}( T^\ast _2 \cap P,c_1) & ={} \sum _{i=1}^m \text{cost}( T^\ast _2 \cap P_i,c_1) \lt \frac{4}{\beta -2} \sum _{i=1}^{m-1} \text{cost}( T^\ast _1 \cap P_{i+1},c_1) + \frac{2}{\beta -2} \text{cost}( T^\ast _1 \cap R_m,c_1) \\ & \lt \frac{4}{\beta -2} \text{cost}( T^\ast _1,c_1) . \end{align*}\) Here, the last inequality holds, because \(P_2, \ldots , P_m\) and \(R_m\) are pairwise disjoint. In addition, we have \(\begin{align*} \text{cost}( T^\ast _2 \cap T^\prime ,c_2) \le (\alpha +\varepsilon) \text{cost}( T^\ast _2 \cap T^\prime ,\widetilde{c_2}) \le (\alpha +\varepsilon) \text{cost}( T^\ast _2 \cap T^\prime ,c^\ast _2) \le (\alpha +\varepsilon) \text{cost}( T^\ast _2,c^\ast _2) . \end{align*}\) Finally, using Equation (I) and a union bound, with probability at least \(1-\delta ,\) the following holds: \(\begin{align*} \text{cost}( T,\lbrace c_1,c_2) & \lt (\alpha +\varepsilon) \text{cost}( T^\ast _1,c^\ast _1) + (\alpha +\varepsilon) \text{cost}( T^\ast _2,c^\ast _2) + \frac{4}{\beta -2} (\alpha +\varepsilon) \text{cost}( T^\ast _1,c^\ast _1) \\ & \lt \left(1+\frac{4}{\beta -2}\right) (\alpha + \varepsilon) \text{cost}( T,C^\ast ) = \left(1+\frac{4k}{k\beta -2k}\right) (\alpha + \varepsilon) \text{cost}( T,C^\ast ) \\ & \le {} \left(1+\frac{4k^2}{\beta -2k}\right) (\alpha + \varepsilon) \text{cost}( T,C^\ast ) . \end{align*}\)

Case 1.2. For all recursive calls with parameters \((T^\prime ,\lbrace c_1\rbrace ,1,\beta ,\delta ,\varepsilon),\) it holds that \(\vert T^\ast _2 \cap T^\prime \vert \lt \frac{1}{\beta } \vert T^\prime \vert\).

After \(\log (n)\) pruning phases, we end up with a singleton \(\lbrace \sigma \rbrace = T^\prime\) as the input set. Since \(\vert T^\ast _2 \cap T^\prime \vert \lt \frac{1}{\beta } \vert T^\prime \vert\), it must be that \(0 = \vert T^\ast _2 \cap T^\prime \vert \lt \frac{1}{\beta } \vert T^\prime \vert = \frac{1}{\beta } \lt 1\) and thus \(\sigma \in T^\ast _1\).

Let \(C_2\) be the set of candidates returned by Median-Candidates in this call. With probability at least \(1-\delta /k,\) there is a \(c_2 \in C_2\) with \(\text{cost}( \lbrace \sigma \rbrace ,c_2) \le (\alpha + \varepsilon) \text{cost}( \lbrace \sigma \rbrace ,\widetilde{c}_2) \le (\alpha + \varepsilon) {\rm cost}({\lbrace \sigma \rbrace },{c^\ast _1})\), where \(\widetilde{c}_2\) is an optimal median for \(\lbrace \sigma \rbrace\). Since \({\rm cost}({T^\ast _2 \cap P},{c_1})\) is bounded as in Case 1.1, by a union bound we have with probability at least \(1- \delta\): \(\begin{align*} \text{cost}( T,\lbrace c_1, c_2) & \le {} \text{cost}( T^\ast _1 \setminus \lbrace \sigma \rbrace ,c_1) + \text{cost}( T^\ast _2 \cap P,c_1) + \text{cost}( \lbrace \sigma \rbrace ,c_2) \\ & \le (\alpha + \varepsilon) \text{cost}( T^\ast _1,c^\ast _1) + \text{cost}( T^\ast _2 \cap P,c_1) \\ & \le \left(1+\frac{4}{\beta -2}\right) (\alpha + \varepsilon) \text{cost}( T,C^\ast ) \\ & \le \left(1+\frac{4k^2}{\beta -2k}\right) (\alpha + \varepsilon) \text{cost}( T,C^\ast ) . \end{align*}\)

Case 2. \(k \gt 2.\)

We only prove the generalization of Case 1.1 to \(k \gt 2\), and the remainder of the proof is analogous to Case 1. Let \(C^\ast = \lbrace c^\ast _1, \ldots , c^\ast _k \rbrace\) be an optimal set of \(k\) medians for \(T\) with clusters \(T^\ast _1, \ldots , T^\ast _k\), respectively, that form a partition of \(T\). For the sake of simplicity, assume that \(n\) is a power of 2 and w.l.o.g. assume that \(\vert T^\ast _1 \vert \ge \dots \ge \vert T^\ast _k \vert\). For \(i \in [k]\) and \(j \in [k] \setminus [i],\) we define \(T^\ast _{i,j} = \uplus _{t=i}^j T^\ast _t\).

Let \(\mathcal {T}_0 = T,\) and let \((\mathcal {T}_j = \mathcal {T}_{j-1} \setminus \mathcal {P}_j)_{j=1}^m\) be the sequence of input sets in the recursive calls of the \(m \in \mathbb {N}\), \(m \le \log (n)\), pruning phases, where \(\mathcal {P}_j\) is the set of elements removed in the \(j\)^th (in the order of the recursive calls occurring) pruning phase. Let \(\mathcal {T} = \lbrace \mathcal {T}_0\rbrace \cup \lbrace \mathcal {T}_j \mid j \in [m] \rbrace\). For \(i \in [k]\), let \(T_i\) be the maximum cardinality set in \(\mathcal {T}\), with \(\vert T^\ast _i \cap T_i \vert \ge \frac{1}{\beta } \vert T_i \vert\). Note that by assumption and since \(\beta \gt 2k\), \(T_1 = T\) must hold and also \(T_j \subset T_i\) for \(j \in [k] \setminus [i]\).

Using a union bound, with probability at least \(1-\delta\), for each \(i \in [k]\) the call of Median-Candidates with input \(T_i\) yields a candidate \(c_i\) with \(\begin{align*} \text{cost}( T^\ast _i \cap T_i,c_i) \le (\alpha +\varepsilon) \text{cost}( T^\ast _i \cap T_i,\widetilde{c}_i) \le (\alpha +\varepsilon) \text{cost}( T^\ast _i \cap T_i,c^\ast _i) \le (\alpha +\varepsilon) \text{cost}( T^\ast _i,c^\ast _i) , \qquad \qquad \qquad \qquad {\rm (I)} \end{align*}\) where \(\widetilde{c}_i\) is an optimal 1-median for \(T^\ast _i \cap T_i\). Let \(C = \lbrace c_1, \ldots , c_k\rbrace\) be the set of these candidates, and for \(i \in [k-1]\), let \(P_i = T_{i} \setminus T_{i+1}\) denote the set of elements of \(T\) removed by the pruning phases between obtaining \(c_{i}\) and \(c_{i+1}\). Note that the \(P_i\) are pairwise disjoint.

By definition, the sets \(\begin{equation*} T^\ast _1 \cap T_1, \ldots , T^\ast _k \cap T_k, T^\ast _{2,k} \cap P_1, \ldots , T^\ast _{k,k} \cap P_{k-1} \end{equation*}\) form a partition of \(T\), and therefore \(\begin{align*} \text{cost}( T,\lbrace c_1, \ldots , c_k ) & \le {} \sum _{i=1}^k {\rm cost}\left({T^\ast _i \cap T_i},{c_i}\right) + \sum _{i=1}^{k-1} \text{cost}( T^\ast _{i+1,k} \cap P_{i},\lbrace c_1, \ldots , c_{i} ) \\ & \le {} (\alpha + \varepsilon) \sum _{i=1}^k {\rm cost}\left({T^\ast _i},{c^\ast _i}\right) + \sum _{i=1}^{k-1} \text{cost}( T^\ast _{i+1,k} \cap P_{i},\lbrace c_1, \ldots , c_{i} ) . \qquad \qquad \qquad \qquad {\rm (II)} \end{align*}\) Now, it only remains to bound the cost of the wrongly assigned elements of \(T^\ast _{i+1,k}\). For \(i \in [k]\), let \(n_i = \vert T_i \vert ,\) and w.l.o.g. assume that \(P_i \ne \emptyset\) for each \(i \in [k-1]\). Each \(P_i\) is the disjoint union \(\uplus _{j=1}^{m_i} P_{i,j}\) of \(m_i \in \mathbb {N}\) sets of elements of \(T\) removed in the interim pruning phases, and it holds that \(\vert P_{i,j} \vert = \frac{n_{i}}{2^j}\). We now prove for each \(i \in [k-1]\) and \(j \in [m_i]\) that \(P_i\) contains many elements from \(T^\ast _{1,i}\) and only a few elements from \(T^\ast _{i+1,k}\).

For \(i \in [k-1]\), we define \(R_{i,0} = T_i\), and for \(j \in [m_i],\) we define \(R_{i,j} = R_{i,j-1} \setminus P_{i,j}\). By definition, \(\vert R_{i,j} \vert = \frac{n_i}{2^j} = \vert P_{i,j} \vert\), \(R_{i,j_1} \supset R_{i,j_2}\) for each \(j_1 \in [m_i]\) and \(j_2 \in [m_i] \setminus [j_1]\), also \(R_{i,m_i} = T_{i+1}\). Thus, \(\vert T^\ast _t \cap R_{i,j} \vert \lt \frac{1}{\beta } \vert R_{i,j} \vert\) for all \(i \in [k-1], j \in [m_i],\) and \(t \in [k] \setminus [i]\). As an immediate consequence, we obtain \(\vert T^\ast _{i+1,k} \cap R_{i,j} \vert \le \frac{k}{\beta } \vert R_{i,j} \vert\). Since \(P_{i,j} \subseteq R_{i,j-1}\) for all \(i \in [k-1]\) and \(j \in [m_i]\), we have \(\begin{align*} \vert T_{i+1,k} \cap P_{i,j} \vert \le \vert T_{i+1,k} \cap R_{i,j-1} \vert \le \frac{k}{\beta } \vert R_{i,j-1} \vert = \frac{2k}{\beta } \frac{n_i}{2^j}, \qquad \qquad \qquad \qquad {\rm (III)} \end{align*}\) which immediately yields \(\begin{align*} \vert T_{1,i} \cap P_{i,j} \vert = \vert P_{i,j} \vert - \vert T_{i+1,k} \cap P_{i,j} \vert \ge \left(1-\frac{2k}{\beta }\right) \frac{n_i}{2^j}. \qquad \qquad \qquad \qquad {\rm (IV)} \end{align*}\) Now, by definition, we know that for all \(i \in [k-1]\), \(j \in [m_i] \setminus \lbrace m_i\rbrace\), \(\sigma \in P_{i,j}\), and \(\tau \in P_{i,j+1}\) that \(\min \nolimits _{c \in \lbrace c_1, \ldots , c_{i}\rbrace }\rho (\sigma , c) \le \min \nolimits _{c \in \lbrace c_1, \ldots , c_{i}\rbrace }\rho (\tau , c)\). Thus, \(\begin{align*} \frac{\text{cost}( T^\ast _{i+1,k} \cap P_{i,j},\lbrace c_1, \ldots , c_i) }{\vert T^\ast _{i+1,k} \cap P_{i,j} \vert } \le \frac{\text{cost}( T^\ast _{1,i} \cap P_{i,j+1},\lbrace c_1, \ldots , c_i) }{\vert T^\ast _{1,i} \cap P_{i,j+1} \vert }. \end{align*}\) Combining this inequality with Equations (III) and (IV) yields for \(i \in [k-1]\) and \(j \in [m_i] \setminus \lbrace m_i\rbrace\): \(\begin{align*} & \frac{\beta 2^j}{2kn_i} \text{cost}( T^\ast _{i+1,k} \cap P_{i,j},\lbrace c_1, \ldots , c_i) \le \frac{2^{j+1}}{(1-\frac{2k}{\beta })n_i} \text{cost}( T^\ast _{1,i} \cap P_{i,j+1},\lbrace c_1, \ldots , c_i) \\ \iff & \text{cost}( T^\ast _{i+1,k} \cap P_{i,j},\lbrace c_1, \ldots , c_i) \le \frac{4k}{\beta -2k} \text{cost}( T^\ast _{1,i} \cap P_{i,j+1},\lbrace c_1, \ldots , c_i) . \qquad \qquad \qquad \qquad {\rm (V)} \end{align*}\) For each \(i \in [k-1],\) we still need an upper bound on \({\rm cost}({T^\ast _{i+1,k} \cap P_{i,m_i}},{\lbrace c_1, \ldots , c_i\rbrace })\). Since \(\vert R_{i,m_i} \vert = \vert P_{i,m_i} \vert\) and also \(R_{i,m_i} \subseteq R_{i,m_i-1}\), we can use Equation (III) to obtain \(\begin{align*} \vert T^\ast _{1,i} \cap R_{i,m_i} \vert = \vert R_{i,m_i} \vert - \vert T^\ast _{i+1,k} \cap R_{i,m_i} \vert \ge \vert R_{i,m_i} \vert - \vert T^\ast _{i+1,k} \cap R_{i,m_i-1} \vert \gt \left(1-\frac{2k}{\beta }\right)\frac{n_i}{2^{m_i}}. \qquad \qquad \qquad \qquad {\rm (VI)} \end{align*}\) By definition, we also know that for all \(i \in [k-1]\), \(\sigma \in P_{i,m_i}\) and \(\tau \in R_{i,m_i}\) that \(\min \limits _{c \in \lbrace c_1, \ldots , c_{i}\rbrace }\rho (\sigma , c) \le \min \limits _{c \in \lbrace c_1, \ldots , c_{i}\rbrace }\rho (\tau , c)\). Thus, \(\begin{equation*} \frac{\text{cost}( T^\ast _{i+1,k} \cap P_{i,m_i},\lbrace c_1, \ldots , c_i ) }{\vert T^\ast _{i+1,k} \cap P_{i,m_i} \vert } \le \frac{\text{cost}( T^\ast _{1,i} \cap R_{i,m_i},\lbrace c_1, \ldots , c_i ) }{\vert T^\ast _{1,i} \cap R_{i,m_i} \vert }. \end{equation*}\) Combining this inequality with Equations (III) and (VI) yields \(\begin{align*} & \frac{\beta 2^{m_i}}{2kn_i} \text{cost}( T^\ast _{i+1,k} \cap P_{i,m_i},\lbrace c_1, \ldots , c_i ) \lt \frac{2^{m_i}}{(1-\frac{2k}{\beta })n_i} \text{cost}( T^\ast _{1,i} \cap R_{i,m_i},\lbrace c_1, \ldots , c_i ) \\ \iff & \text{cost}( T^\ast _{i+1,k} \cap P_{i,m_i},\lbrace c_1, \ldots , c_i ) \lt \frac{2k}{\beta -2k} \text{cost}( T^\ast _{1,i} \cap R_{i,m_i},\lbrace c_1, \ldots , c_i ) . \qquad \qquad \qquad \qquad {\rm (VII)} \end{align*}\) We can now give the following bound, combining Equations (V) and (VII), for each \(i \in [k-1]\): \(\begin{align*} \text{cost}( T^\ast _{i+1,k} \cap P_i,\lbrace c_1, \ldots , c_i) & ={} \sum _{j=1}^{m_i} \text{cost}( T^\ast _{i+1,k} \cap P_{i,j},\lbrace c_1, \ldots , c_i) \\ &\quad \lt \sum _{j=1}^{m_i-1} \frac{4k}{\beta -2k} \text{cost}( T^\ast _{1,i} \cap P_{i,j+1},\lbrace c_1, \ldots , c_i) \\ &\quad + \frac{2k}{\beta -2k} \text{cost}( T^\ast _{1,i} \cap R_{i,m_i},\lbrace c_1, \ldots , c_i ) \\ &\quad \lt \frac{4k}{\beta -2k} \text{cost}( T^\ast _{1,i} \cap T_i,\lbrace c_1, \ldots , c_i ) . \qquad \qquad \qquad \qquad {\rm (VIII)} \end{align*}\) Here, the last inequality holds, because \(P_{i,2}, \ldots , P_{i,m_i}\) and \(R_{i,m_i}\) are pairwise disjoint subsets of \(T_i\).

Now, we plug this bound into Equation (II). Note that \(T^\ast _j \cap T_i \subseteq T^\ast _j \cap T_j\) for each \(i \in [k]\) and \(j \in [i]\) by definition. We obtain \(\begin{align*} \text{cost}( T,\lbrace c_1, \ldots , c_k ) & \le {} (\alpha + \varepsilon) \sum _{i=1}^k \text{cost}( T^\ast _i,c^\ast _i) + \sum _{i=1}^{k-1} \text{cost}( T^\ast _{i+1,k} \cap P_{i},\lbrace c_1, \ldots , c_{i} ) \\ &\quad \lt (\alpha + \varepsilon) \sum _{i=1}^k \text{cost}( T^\ast _i,c^\ast _i) + \frac{4k}{\beta -2k} \sum _{i=1}^{k-1} \text{cost}( T^\ast _{1,i} \cap T_i,\lbrace c_1, \ldots , c_i ) \\ & \le {} (\alpha + \varepsilon) \sum _{i=1}^k \text{cost}( T^\ast _i,c^\ast _i) + \frac{4k}{\beta -2k} \sum _{i=1}^{k-1} \sum _{t=1}^i \text{cost}( T^\ast _{t} \cap T_i,c_t) \\ & \le {} (\alpha + \varepsilon) \sum _{i=1}^k \text{cost}( T^\ast _i,c^\ast _i) + \frac{4k}{\beta -2k} \sum _{i=1}^{k-1} \sum _{t=1}^i \text{cost}( T^\ast _{t} \cap T_t,c_t) \\ & \le (\alpha + \varepsilon) \sum _{i=1}^k \text{cost}( T^\ast _i,c^\ast _i) + \frac{4k^2}{\beta -2k} \sum _{i=1}^{k-1} \text{cost}( T^\ast _{i} \cap T_i,c_i) \\ & \le {} \left(1+\frac{4k^2}{\beta -2k}\right) (\alpha + \varepsilon) \sum _{i=1}^k \text{cost}( T^\ast _i,c^\ast _i) = \left(1+\frac{4k^2}{\beta -2k}\right) (\alpha + \varepsilon) \text{cost}( T,C^\ast ) . \end{align*}\) The last inequality follows from Equation (I).□

Let \(T(n, \kappa , \beta , \delta , \varepsilon)\) denote the worst-case running time of Algorithm 5 for input set \(T\) with \(\vert T \vert = n\). For the sake of simplicity, we assume that \(n\) is a power of 2. Note that we always have \(\kappa \le n\).

If \(\kappa = 0\), Algorithm 5 has running time \(c_1 \in O(1)\). If \(n \ge \kappa \ge 1\), Algorithm 5 has running time at most \(c_2 \cdot (n \cdot T_\rho + n) \in O(n \cdot T_\rho)\) to obtain \(P\), \(T(n/2, \kappa , \beta , \delta , \varepsilon)\) for the recursive call in the pruning phase; \(T_1(n, \beta , \delta , \varepsilon)\) to obtain the candidates \(C(n, \beta , \delta , \varepsilon) \cdot T(n, \kappa - 1, \beta , \delta , \varepsilon)\) for the recursive calls in the candidate phase, one for each candidate; and \(c_3 \cdot n \cdot T_\rho \cdot C(n, \beta , \delta , \varepsilon) \in O(n \cdot T_\rho \cdot C(n, \beta , \delta , \varepsilon))\) to eventually evaluate the candidate sets. Let \(c = c_1 + c_2 + c_3 + 1\). We obtain the following recurrence relation: \(\begin{align*} T(n, \kappa , \beta , \delta , \varepsilon) \le {\left\lbrace \begin{array}{ll} c & \text{if } \kappa = 0 \\ C(n, \beta , \delta , \varepsilon) \cdot T(n, \kappa -1, \beta , \delta , \varepsilon) + T(n/2, \kappa , \beta , \delta , \varepsilon) \\ + T_1(n, \beta , \delta , \varepsilon) + c n \cdot T_\rho \cdot C(n, \beta , \delta , \varepsilon)) & \text{else} \end{array}\right.}. \end{align*}\)

Let \(f(n, \beta , \delta , \varepsilon) = \frac{1}{cn} \cdot T_1(n, \beta , \delta , \varepsilon) + T_\rho \cdot C(n, \beta , \delta , \varepsilon)\).

We prove that \(T(n, \kappa , \beta , \delta , \varepsilon) \le c \cdot 4^{\kappa } \cdot C(n,\beta ,\delta ,\varepsilon)^{\kappa +1} \cdot n \cdot f(n, \beta , \delta , \varepsilon)\), by induction on \(n, \kappa\).

For \(\kappa = 0,\) we have \(T(n,\kappa , \beta , \delta , \varepsilon) \le c \le cn \le c \cdot 4^0 \cdot C(n,\beta ,\delta ,\varepsilon) \cdot n \cdot f(n, \beta , \delta , \varepsilon)\).

Now, let \(n \ge \kappa \ge 1,\) and assume the claim holds for \(T(n^\prime ,\kappa ^\prime , \beta , \delta , \varepsilon)\), for each \(\kappa ^\prime \in \lbrace 0, \ldots , \kappa -1\rbrace\) and \(n^\prime \in [n-1]\). We have \(\begin{align*} T(n, \kappa , \beta , \delta , \varepsilon) & \le {} C(n, \beta , \delta , \varepsilon) \cdot T(n, \kappa -1, \beta , \delta , \varepsilon) + T(n/2, \kappa , \beta , \delta , \varepsilon) \\ & \ \ \ + T_1(n, \beta , \delta , \varepsilon) + cn \cdot T_\rho \cdot C(n, \beta , \delta , \varepsilon) \\ & \le {} C(n, \beta , \delta , \varepsilon) \cdot c \cdot 4^{\kappa -1} \cdot C(n,\beta ,\delta ,\varepsilon)^{\kappa } \cdot n \cdot f(n, \beta , \delta , \varepsilon) \\ & \ \ \ + c \cdot 4^{\kappa } \cdot C(n/2,\beta ,\delta ,\varepsilon)^{\kappa +1} \cdot \frac{n}{2} \cdot f(n/2, \beta , \delta , \varepsilon) \\ & \ \ \ + cn \cdot f(n, \beta , \delta , \varepsilon) \\ & \le {} \left(\frac{1}{4} + \frac{1}{2} + \frac{1}{4^\kappa C(n,\beta ,\delta ,\varepsilon)^{\kappa +1}}\right) c \cdot 4^{\kappa } \cdot C(n,\beta ,\delta ,\varepsilon)^{\kappa +1} \cdot n \cdot f(n, \beta , \delta , \varepsilon) \\ & \le {} c \cdot 4^{\kappa } \cdot C(n,\beta ,\delta ,\varepsilon)^{\kappa +1} \cdot n \cdot f(n, \beta , \delta , \varepsilon). \end{align*}\)

The last inequality holds, because \(\frac{1}{4^\kappa C(n,\beta ,\delta ,\varepsilon)^{\kappa +1}} \le \frac{1}{4}\), and the claim follows by induction.□