\section{Conclusion}\label{sec:concl}
In this paper, we initiated the study of computational complexity of several variants of explainable clustering. Concluding, we would like to outline some further directions of research and state a number of open problems. 

We showed that \probClustExpl admits a polynomial-time approximation with a factor of $(k-1)$. Can this factor be improved in polynomial-time, say, to $\log k$? We proved that \probEMeanClust and  \ProblemName{Explainable $k$-median} can be solved in $n^{2d}\cdot (dn)^{\mathcal{O}(1)}$ time. Is this result tight? Or is it possible to obtain an $f(d)\cdot (dn)^{\mathcal{O}(1)}$ time algorithm for some function $f$? Also, is it possible to obtain approximation schemes parameterized by $k$, i.e., $(1+\varepsilon)$-approximation in $g(k,\varepsilon) (nd)^{\mathcal{O}(1)}$ time for some function $g$. Regular $k$-means/median admits such approximation schemes \cite{KumarSS10}. 


\section{Hardness of Explainable Clustering} 
\subsection{Hardness}
\label{sec:hardexplncl}


In this section, we show our hardness results for \probEMeanClust and \ProblemName{Explainable $k$-median}: the problems are  \ensuremath{\operatorClassNP}-complete, \classW{2}-hard when parameterized by $k$, and cannot be solved in $f(k)\cdot n^{o(k)}$ time for a computable function $f(\cdot)$ unless ETH fails.
Moreover, the hardness holds even if the input points are binary. More precisely, we prove the following theorem.


\begin{theorem}
\label{thm:exp-ethhard} 
Given a collection of $n$ points $\bfX\subseteq \{0,1\}^d$, a positive integer $k\leq n$, and a nonnengative integer $B$, it is \classW{2}-hard to decide whether $\bfX$ admits an explainable $k$-clustering of  mean (median, respectively) cost at most $B$, when the problem is parameterized by $k$. Moreover, the problems are  \ensuremath{\operatorClassNP}-complete and cannot be solved in $f(k)\cdot n^{o(k)}$ time for a computable function $f(\cdot)$ unless ETH fails.
\end{theorem}

\begin{proof}
We show the theorem for the means costs and then briefly explain how the proof is modified for medians. The case of median cost is easier due to the fact that for a collection of binary points, its median is also a binary vector. 

We reduce from the \probHS problem. The task of the problem is, given a family of sets $\mathcal{W}=\{W_1,\ldots,W_m\}$ over  universe $U=\{u_1,\ldots,u_n\}$, and a positive integer $k$, decide whether there  is  $S\subseteq U$ of size at most $k$ that is a  \emph{hitting} set for $\mathcal{W}$, i.e.  such that  $S\cap W_i\neq\emptyset$ for every $i\in\{1,\ldots,m\}$. This problem is well-known to be \classW{2}-complete  when parameterized by $k$ even if the input is restricted to families of sets of the same size (see~\cite{DowneyF13}). 

Let $\mathcal{W}=\{W_1,\ldots,W_m\}$ be a family of sets over $U=\{u_1,\ldots,u_n\}$ with $|W_1|=\cdots=|W_m|=r$, and let $k$ be a positive integer. We construct the following points.
\begin{itemize}
\item Let $s=8nm^2$. For each $i\in\{1,\ldots,n\}$, we construct a vector $\bfu_i\in\{0,1\}^n$ by setting 
$$\bfu_i[j]=
\begin{cases}
1&\mbox{if }j=i,\\
0&\mbox{otherwise}.
\end{cases}
$$
We also define $\mathbf{U}_i$ to be the collection of $s$ points identical to $\bfu_i$.
\item For each $i\in\{1,\ldots,m\}$, let $\bfw_i$ be the characteristic vector of $W_i$.
That is, 
$$\bfw_i[j]=
\begin{cases}
1&\mbox{if }u_j\in W_i,\\
0&\mbox{otherwise},
\end{cases}
$$
 We define $\bfW=\{\bfw_1,\ldots,\bfw_m\}$.
\item For $t=16ns^2$ we construct a collection $\bfZ$ of $t$ zero points.
\end{itemize}
Finally,  we define
 \begin{itemize}
 \item $\bfX=\big(\bigcup_{i=1}^n\mathbf{U}_i\big)\cup\bfW\cup\bfZ$,
 \item $k'=k+1$, and
 \item $B=m(r-1)+s(n-k)$.
 \end{itemize}
 
 Clearly, $\bfX$, $k'$ and $B$ can be constructed form the considered instance of \probHS in polynomial time.
 We claim that $\mathcal{W}$ has a hitting set $S$ of size at most $k$ if and only if $\bfX$ has an explainable $k'$-clustering of means cost at most $B$.
 
 For the forward direction, assume that $S$ is a hitting set of size at most $k$. Without loss of generality we assume that $|S|=k$. Let $S=\{u_{i_1},\ldots,u_{i_k}\}$. We define the $k'$-clustering $\{\bfC_1,\ldots,\bfC_{k+1}\}$ as follows. We set $\bfC_1=\{\bfx\in\bfX\mid \bfx[i_1]=1\}$ and for $j=2,\ldots,k+1$, 
 $$\bfC_j=\{\bfx\in\bfX\mid \bfx[i_j]=1\}\setminus\bigcup_{h=1}^{j-1}\bfC_h.$$ 
 Observe that the constructed clustering is explainable. To see this, let $X_0=\bfX$ and set $X_j=\bfX\setminus \bigcup_{h=1}^{j-1}\bfC_h$. Then $(X_j,\bfC_j)={\sf Cut}_{i_j,0}(X_{j-1})$ for $j\in \{1,\ldots,k+1\}$. Notice that $\mathbf{U}_{i_j}\subseteq \bfC_j$ for every $j\in\{1,\ldots,k\}$, and we have that $\bfZ\subseteq \bfC_{k+1}$ and $\mathbf{U}_h\subseteq \bfC_{k+1}$ for all $h\in\{1,\ldots,m\}\setminus \{i_1,\ldots,i_k\}$. Moreover, because $S$ is a hitting set, each $\bfw_j\in\bfW$ is included in some cluster $\bfC_h$ for some $h\in\{1,\ldots,k\}$, that is, $\bfC_{k+1}=\bfZ\cup \bfR$ , where $\bfR= \bigcup_{h\in\{1,\ldots,m\}\setminus \{i_1,\ldots,i_k\}}\mathbf{U}_h$. 
 
 For every $j\in\{1,\ldots,k\}$, we define $\bfc_j=\bfu_{i_j}$, and we set $\bfc_{k+1}$ be the zero vector. Clearly, for every $j\in\{1,\ldots,k+1\}$, ${\sf cost}_2(\bfC_j)\leq \sum_{\bfx\in \bfC_j}\|\bfx-\bfc_j||_2^2$. Let $\bfW_j=\bfW\cap\bfC_j$ for $j\in\{1,\ldots,k\}$. Note that $\bfC_j=\bfW_j\cup\mathbf{U}_{i_j}$ for $j\in\{1,\ldots,k\}$.
 Because $\bfx[i_j]=1$ for every $\bfx\in\bfC_j$ for  $j\in \{1,\ldots,k\}$ and $\bfx$ has exactly $r$ nonzero elements, we have that $\|\bfx-\bfc_j\|_2^2=r-1$ for every $\bfx\in\bfW_j$ for $j\in\{1,\ldots,k\}$. 
 If $\bfx\in \bfR$, then $\|\bfx-\bfc_{k+1}\|_2^2=1$. We obtain that 
 \begin{multline*}
 {\sf cost}_2(\bfC_1,\ldots,\bfC_{k+1})\leq \sum_{j=1}^{k+1}\sum_{\bfx\in \bfC_j}\|\bfx-\bfc_j\|_2^2
 =\sum_{j=1}^k\sum_{\bfx\in\bfW_j}\|\bfx-\bfc_j\|_2^2+\sum_{\bfx\in\bfR}\|\bfx-\bfc_{k+1}\|_2^2\\
 =m(r-1)+s(n-k)=B.
 \end{multline*}
 Thus, $\{\bfC_1,\ldots,\bfC_{k+1}\}$ is an explainable $k'$-clustering of means cost at most $B$. 
 
 For the opposite direction, let $\{\bfC_1,\ldots,\bfC_{k+1}\}$ be an explainable $k'$-clustering with \linebreak ${\sf cost}_2(\bfC_1,\ldots,\bfC_{k+1})\leq B$. Let also $(T,k',\varphi)$ be a canonical threshold tree for this clustering.  Notice that because every point of $\bfX$ is binary, for every nonleaf $v\in V(T)$, $\varphi(v)=(i,0)$ for some $i\in\{1,\ldots,d\}$.

We show the following property of  $(T,k',\varphi)$: for every nonleaf node $v\in V(T)$, its right child is a leaf. Suppose that this is not the case. Denote by $x$ the root of $T$ and let $P=x_1\cdots x_p$ be the root-leaf path, where $x_1=x$ and $x_i$ is the left child of $x_{i-1}$ for $i\in\{2,\ldots,p\}$, that is, $P$ is constructed starting from the root and following the left children until we achieve the leftmost leaf.  Because  $T$ has $k'$ leaves and at least one right  child is not a leaf, we have that $p\leq k'-1=k$. Denote by $\bfC_q$ the cluster corresponding to the leaf $x_p$ of $T$. Notice that $\bfZ\subseteq \bfC_q$ and $n-(p-1)$ collections $\mathbf{U}_i$ are included in $\bfC_q$ for some $i\in\{1,\ldots,n\}$.
 Let $i_1,\ldots,i_{n-p+1}\in\{1,\ldots,n\}$ be the distinct indices such that $\mathbf{U}_{i_j}\subseteq \bfC_q$.   Denote by $\bfc$ the mean of $\bfC_q$. Observe that for each $j\in\{1,\ldots,n\}$, the multiset of the $j$-th coordinates of the points of $\bfC_q$ contains at most $s+m$ ones and at least $t$ zeros.  Then for every $j\in \{1,\ldots,n\}$,
 $$\bfc[j]\leq \frac{m+s}{|\bfC_q|}\leq \frac{m+s}{t}\leq\frac{2s}{t}=\frac{1}{8ns}\leq \frac{1}{2s},$$
 and we obtain that 
 \begin{align*}
 {\sf cost}_2(\bfC_q)\geq& \sum_{j=1}^{n-p+1}\sum_{\bfx\in \mathbf{U}_{i_j}}\|\bfx-\bfc\|_2^2=s\sum_{j=1}^{n-p+1}\|\bfu_{j_i}-\bfc\|_2^2 \\
 \geq&s(n-k+1)\big(1-\frac{1}{2s}\big)^2> s(n-k+1)\big(1-\frac{1}{s}\big)\\ 
 \geq &s(n-k)+s-n \geq s(n-k)+2nm-n\\
 >&m(n-1)+s(n-k)\geq B,
 \end{align*}
 contradicting that  ${\sf cost}_2(\bfC_1,\ldots,\bfC_{k+1})\leq B$. 
 
 Because for every nonleaf node $v\in V(T)$, its right child is a leaf,  we have that the clustering is obtained by consecutive cutting each cluster from the set of points.  
 Then  we can assume without loss of generality that there a $k$-tuple of distinct indices $(i_1,\ldots,i_k)$ from $\{1,\ldots,n\}$ such that 
  $\bfC_1=\{\bfx\in\bfX\mid \bfx[i_1]=1\}$ and  
 $$\bfC_j=\{\bfx\in\bfX\mid \bfx[i_j]=1\}\setminus\bigcup_{h=1}^{j-1}\bfC_h$$ 
 for $j=2,\ldots,k+1$. We claim that $S=\{u_{i_1},\ldots,u_{i_k}\}$ is a hitting set for $\mathcal{W}$.
 
The proof is by contradiction. Suppose that $S$ is not a hitting set. 
For every $i\in\{1,\ldots,k+1\}$, let $\bfW_i=\bfC_i\cap \bfW$. Notice that because $S$ is not a hitting set, $\bfW_{k+1}\neq\emptyset$. 
We analyse the structure of clusters and upper bound their means costs. For this, denote by $\bfc_1,\ldots,\bfc_{k+1}$ the means of $\bfC_1,\ldots,\bfC_{k+1}$. 

Let $j\in \{1,\ldots,k\}$ and consider $\bfC_j$ with the mean $\bfc_j$. We have that $\bfC_j=\mathbf{U}_{i_j}\cup W_j$. Then $\bfc_j[i_j]=1$. Let $h\in \{1,\ldots,n\}$ be distinct from $i_j$. If $\bfx\in \mathbf{U}_{i_j}$, then $\bfx[h]=0$.
Then the multiset of the $h$-th coordinates of the points of $\bfC_j$ contains at most $|\bfW_j|\leq m$
ones and at least  $s$ zeros, and we have that  
 $$\bfc_j[h]\leq \frac{m}{|\bfC_j|}\leq \frac{m}{s}\leq\frac{1}{8mn}.$$ 
 Recall that $\bfx\in \bfW$ has exactly $r$ elements that are equal to one. This implies, that for every $\bfx\in \bfW_j$,
 \begin{equation*}
 \|\bfx-\bfc_j\|_2^2\geq (r-1)\big(1-\frac{1}{8mn}\big)^2\geq (r-1)\big(1-\frac{1}{4mn}\big)
 \end{equation*}
 and, therefore,
  \begin{equation}\label{eq:first-clusters}
 {\sf cost}_2(\bfC_j)\geq (r-1)\big(1-\frac{1}{4mn}\big)|\bfW_j|
 \geq (r-1)|\bfW_j|-\frac{1}{4m}|\bfW_j|,
 \end{equation} 
 because $r\leq n$.
 
 
 Now we consider $\bfC_{k+1}$ and the corresponding mean $\bfc_{k+1}$. Notice that $\bfC_{k+1}=\bfZ\cup \bfW_{k+1}\cup \bfR$, where $\bfR= \bigcup_{h\in\{1,\ldots,m\}\setminus \{i_1,\ldots,i_k\}}\mathbf{U}_h$.
 Let $h\in \{1,\ldots,n\}$. We have that $\bfx[h]=0$ for $\bfx\in \bfZ$. Hence, the multiset of of $h$-th coordinates of the points of $\bfC_{k+1}$ contains at most $|\bfW_{k+1}|+s\leq m+s$
ones and at least $t$ zeros. Therefore,   
 $$\bfc_{k+1}[h]\leq \frac{m+s}{|\bfC_{k+1}|}\leq \frac{2s}{t}\leq\frac{1}{8sn}.$$ 
Since $\bfx\in \bfW$ has exactly $r\leq n$ elements that are equal to one,  for every $\bfx\in \bfW_{k+1}$,
\begin{equation*}
\|\bfx-\bfc_{k+1}\|_2^2\geq  r\big(1-\frac{1}{8sn}\big)^2 \geq r\big(1-\frac{1}{8mn}\big)^2
\geq r\big(1-\frac{1}{4mn}\big)\geq r-\frac{1}{4m}.
\end{equation*}
For every $\bfx\in \bfR$,  $\bfx$ contains a unique nonzero element and, therefore, 
\begin{equation*}
\|\bfx-\bfc_{k+1}\|_2^2\geq \big(1-\frac{1}{8sn}\big)^2\geq 1-\frac{1}{4sn}.
\end{equation*}
Note that $\bfR$ contains exactly $s(n-k)$ points.
Then 
\begin{multline}\label{eq:last-cluster}
{\sf cost}_2(\bfC_{k+1})\geq\sum_{\bfx\in \bfW_{k+1}}\|\bfx-\bfc_{k+1}\|_2^2+\sum_{\bfx\in \bfR}\|\bfx-\bfc_{k+1}\|_2^2\\
\geq \big(r-\frac{1}{4m}\big)|\bfW_{k+1}|+s(n-k)\big(1-\frac{1}{4sn}\big)\\
\geq r|\bfW_{k+1}|+s(n-k)-\frac{1}{4m}|\bfW_{k+1}|-\frac{1}{4}.
\end{multline}

Recall that $|\bfW_1|+\cdots+|\bfW_{k+1}|=m$. Then combining (\ref{eq:first-clusters}) and (\ref{eq:last-cluster}), we obtain that 
\begin{equation*}
 {\sf cost}_2(\bfC_1,\ldots,\bfC_{k+1})=\sum_{j=1}^{k+1}{\sf cost}_2(\bfC_j)
 \geq m(r-1)+s(n-k)+|\bfW_{k+1}|-\frac{1}{2}
 \geq B+\frac{1}{2},
\end{equation*}
because $\bfW_{k+1}\neq\emptyset$.
However, this contradicts that the means cost of $\{\bfC_1,\ldots,\bfC_{k'}\}$ is at most $B$. This means that $S$ is a hitting set for $\mathcal{W}$. 

This concludes the hardness proof for \probEMeanClust. For the median cost, we use exactly the same reduction from \probHS. Then we show that $\mathcal{W}$ has a hitting set $S$ of size at most $k$ if and only if $\bfX$ has an explainable $k'$-clustering of median cost at most $B$. The proof for the forward direction is identical to the proof for the means up to the replacement of ${\sf cost}_2$ by ${\sf cost}_1$ and of $\|\cdot\|_2^2$ by $\|\cdot\|_1$. For the opposite direction, 
the proof follows the same lines as the above proof for means but gets simplified, because we can assume that medians are  binary. In particular, if the multiset of $h$-th coordinates of the points of a cluster $\bfC_{j}$ contains 
more zeros that ones, then $\bfc_j[h]=0$ for the median $\bfc_j$. Notice that the crucial part of the proof for the means is obtaining upper bounds for the values $\bfc_j[h]$. Now we can immediately assume that $\bfc_j[h]=0$ in all considered cases whenever we upper bound $\bfc_j[h]$, and the lower bounds for the costs in the proof become straightforward. 

To see the second part of the theorem, note that our reduction from  \probHS is polynomial. This immediately implies \ensuremath{\operatorClassNP}-hardness of the considered problems for both measures. For the lower bound up to ETH, we use the well-known fact (see, e.g.~\cite[Chapter~14]{CyganFKLMPPS15}) that \probHS does not admit an algorithm with running time $f(k)\cdot (n+m)^{o(k)}$ for any computable function $f(\cdot)$ up ETH. Because our reduction is polynomial and, moreover, the value of the parameter in the constructed instance is $k'=k+1$, we obtain that an algorithm with running time $f(k)\cdot (n+m)^{o(k)}$ for our problems would contradict ETH.
\end{proof}

\section{Explainable Clustering}
\label{sec:explainable}
\paragraph{Explainable $k$-means/median{} clustering.} We consider the \probEMeanClust (resp. \ProblemName{Explainable $k$-median}) problem where given a collection $\bfX\subseteq \mathbb{R}^d$ of $n\geq k$ points, the task is to find an explainable $k$-clustering $\{\bfC_1,\ldots,\bfC_k\}$ of $\bfX$ of minimum $k$-means (resp. $k$-median) cost.



\subsection{Exact algorithms} 

Our $(nd)^{k+\mathcal{O}(1)}$ time algorithm is indeed very simple and based on branching technique. At each non-leaf node of threshold tree, we would like to find an optimal cut. As we focus on canonical threshold trees, the number of distinct choices for branching is at most $nd$. Also as the number of non-leaf nodes in the threshold binary tree is $k-1$, we have the following theorem. 

\begin{theorem}\label{obs:naive}
 \probEMeanClust and  \ProblemName{Explainable $k$-median} can be solved in $(nd)^{k+\mathcal{O}(1)}$ time.
 \end{theorem}
 
 
 Our $n^{2d}\cdot (dn)^{\mathcal{O}(1)}$ time algorithm is based on dynamic programming, which we describe in the following.  For two vectors $\bfx,\bfy\in\mathbb{R}^d$, we write $\bfx\leq \bfy$  ($\bfx< \bfy$, respectively) to denote that $\bfx[i]\leq \bfy[i]$ ($\bfx[i]< \bfy[i]$, respectively) for every $i\in\{1,\ldots,d\}$. We highlight that when we write $\bfx<\bfy$, we require the strict  inequality for \emph{every} coordinate.  
 
 \begin{theorem}\label{prop:DP}
  \probEMeanClust and  \ProblemName{Explainable $k$-median} can be solved in $n^{2d}\cdot (dn)^{\mathcal{O}(1)}$ time.
  \end{theorem}
   
 \begin{proof} 
 The algorithms for both problems are almost the same. Hence, we demonstrate it for \probEMeanClust. For simplicity, we only show how to find the minimum cost of clustering but the algorithm can be easily modified to produce an optimal clustering as well by standard arguments.
 
 Let $(\bfX,k)$ be an instance of the problem with $\bfX=\{\bfx_1,\ldots,\bfx_n\}$ and $\bfX\subseteq\mathbb{R}^d$. We say that a vector $\bfz\in(\mathbb{R}\cup\{\pm\infty\})^d$ is \emph{canonical}  if $\bfz[i]\in {\sf coord}_i(\bfX)\cup\{\pm\infty\}$ for every $i\in\{1,\ldots,d\}$. For every pair of canonical vectors $(\bfa,\bfb)$ such that $\bfa\leq \bfb$ and every positive integer $s\leq k$, we compute the minimum means cost of an explainable $s$-clustering of $\bfX_{\bfa,\bfb}=\{\bfx_i\in \bfX\mid \bfa< \bfx_i\leq\bfb\}$ and denote this value $\omega(\bfa,\bfb,s)$.  We assume that $\omega(\bfa,\bfb,s)=+\infty$ if $\bfX_{\bfa,\bfb}$ does not admit an explainable $s$-clustering.  It is also convenient to assume that  $\omega(\bfa,\bfb,s)=+\infty$ if $\bfX_{\bfa,\bfb}=\emptyset$, because we are not interested in empty clusters.
  Notice that the minimum means cost of an explainable $k$-clustering of $\bfX$ is $\omega(\bfa^*,\bfb^*,k)$, where $\bfa^*[i]=-\infty$ and $\bfb^*[i]=+\infty$ for $i\in\{1,\ldots,d\}$. We compute the table of values of $\omega(\bfa,\bfb,s)$ consecutively for $s=1,2,\ldots,k$.
 
 If $s=1$, then by definition, 
 $$\omega(\bfa,\bfb,s)=
 \begin{cases}
 {\sf cost}_2(\bfX_{\bfa,\bfb})&\mbox{if }\bfX_{\bfa,\bfb}\neq\emptyset,\\
 +\infty&\mbox{if }\bfX_{\bfa,\bfb}=\emptyset, 
 \end{cases}
 $$ 
 and this value can be computed in polynomial time.  Let $s\geq 2$ and assume that the tables are already constructed for the lesser values of $s$. Consider a pair $(\bfa,\bfb)$ of canonical vectors of 
 $(\mathbb{R}
 \cup\{\pm\infty\})^d$ such that $\bfa\leq \bfb$. For $i\in \{1,\ldots,d\}$ and $\theta\in {\sf coord}_i(\bfX)$ such that $\bfa[i]< \theta<\bfb[i]$, we define the vectors $\bfa^{i,\theta}$ and $\bfb^{i,\theta}$ by setting
 $$\bfa^{i,\theta}[j]=
 \begin{cases}
 \theta&\mbox{if }j=i,\\
 \bfa[j]&\mbox{if }j\neq i,
 \end{cases}
 \text{ and }
\bfb^{i,\theta}[j]=
 \begin{cases}
 \theta&\mbox{if }j=i,\\
 \bfb[j]&\mbox{if }j\neq i.
 \end{cases}
$$
Then we compute $\omega(\bfa,\bfb,s)$ using the following recurrence
\begin{align}\label{eq:rec-dp}
\omega(\bfa,\bfb,s)=\min\{\omega(\bfa,&\bfb^{i,\theta},s_1)+\omega(\bfa^{i,\theta},\bfb,s_2)\nonumber\\
&\text{for }  1\leq i\leq d, \theta\in{\sf coord}_i(\bfX),\nonumber\\
& \bfa[i]< \theta< \bfb[i],\nonumber\\
& s_1,s_2\geq 1,\text{ and }s_1+s_2=s\}.
\end{align}
 The correctness of (\ref{eq:rec-dp}) follows form the definition of the explainable clustering. It is sufficient to observe that to compute the optimum means cost of an explainable $s$-clustering of $\bfX_{\bfa,\bfb}$, we have to take minimum 
 over the sums of  optimum costs of explainable  $s_1$-clusterings and $s_2$-clusterings of $X_1$ and $X_2$, respectively, where $(X_1,X_2)={\sf Cut}_{i,\theta}(\bfX_{\bfa,\bfb})$ for some $i\in \{1,\ldots,d\}$, $\theta\in{\sf coord}_i(\bfX_{\bfa,\bfb})$ and $s_1+s_2=s$, and this is exactly what is done in (\ref{eq:rec-dp}).  
 
To evaluate the running time, observe that to compute $\omega(\bfa,\bfb,s)$ using (\ref{eq:rec-dp}), we consider $d$ values of $i$, at most $n$ values of $\theta$ and at most $k\leq n$ values of $s_1$ and $s_2$, that is, we go over at most  $dn^2$ choices. Thus, computing $\omega(\bfa,\bfb,s)$ for $s\geq 2$ and fixed $\bfa$ and $\bfb$  can be done in $\mathcal{O}(dn^2)$ time. Since there are at most $(n+2)^{2d}$ pairs of canonical vectors $\bfa$ and  $\bfb$, we obtain that the  time to compute the table of values of $\bfX_{\bfa,\bfb}$ for all pairs of vectors is $n^{2d}\cdot (dn)^{\mathcal{O}(1)}$. Since the table for $s=1$ can be constructed in $n^{2d}\cdot (dn)^{\mathcal{O}(1)}$ and we iterate using (\ref{eq:rec-dp}) $k-1\leq n$ times, the total running time is $n^{2d}\cdot (dn)^{\mathcal{O}(1)}$.
 \end{proof} 



\input{Explainable_hardness}


\section{Clustering Explanation}\label{sec:explanation}



\paragraph{Clustering explanation.}  In  the \probClustExpl problem, the input contains a $k$-clustering $\{\bfC_1,\ldots,\bfC_k\}$ of $\bfX\subseteq \mathbb{R}^d$ and a nonnegative integer $s$, and the task is to decide whether there is a collection of points $W\subseteq \bfX$ with $|W|\leq s$ such that $\{\bfC_1\setminus W,\ldots,\bfC_k\setminus W\}$ is an explainable $k$-clustering. Note that some $\bfC_i\setminus W$ may be empty here.  

\input{greedy.tex}


\subsection{Exact algorithm}




 
 Our $2^{2\min\{s,k\}}\cdot n^{2d}\cdot (dn)^{\mathcal{O}(1)}$  time algorithm is based on a novel dynamic programming scheme. Here, we briefly describe the algorithm. 
Our first observation is that each subproblem can be defined w.r.t. a bounding box in $\mathbb{R}^d$, as each cut used to split a point set in any threshold tree is an axis-parallel hyperplane. The number of such distinct bounding boxes is at most $n^{2d}$, as in each dimension a box is specified by two bounding values. This explains the $n^{2d}$ factor in the running time. Now, consider a fixed bounding box corresponding to a subproblem containing a number of given clusters, may be partially.    
If a new canonical cut splits a cluster, then one of the two resulting parts has to be removed, and this choice has to be passed on along the dynamic programming. As we remove at most $s$ points and the number of clusters is at most $k$, the number of such distinct choices can be bounded by $2^{2\min\{s,k\}}$. This roughly gives us the following theorem. 

\begin{theorem}\label{thm:DP-expl}
\probClustExpl  can be solved in $2^{2\min\{s,k\}}\cdot n^{2d}\cdot (dn)^{\mathcal{O}(1)}$  time.
\end{theorem}

Before we move to the formal proof of the theorem, let us introduce some specific notations.
Let $\bfa,\bfb\in \mathbb{R}^d$ be such that $\bfa< \bfb$. We denote $(\bfa,\bfb]=\{\bfx\in\mathbb{R}^d\mid \bfa<\bfx\leq \bfb\}$ and call $(\bfa,\bfb]$ an \emph{interval}. For a collection of points $X\subseteq \mathbb{R}^d$, we say that $X$ is \emph{in} $(\bfa,\bfb]$ if $X\subseteq (\bfa,\bfb]$, $X$ is \emph{outside} $(\bfa,\bfb]$ if $X\cap (\bfa,\bfb]=\emptyset$,   
  and we say that $(\bfa,\bfb]$ \emph{splits} $X$ if $X\cap(\bfa,\bfb]\neq\emptyset$ and $X\setminus(\bfa,\bfb]\neq\emptyset$.  

Let $\mathcal{X}$ be a family of disjoint collections of points of $\mathbb{R}^d$.
A subfamily $\mathcal{Y}\subseteq\mathcal{X}$ is said to be \emph{$(\bfa,\bfb]$-proper} if
(i) every $X\in\mathcal{X}$ that is in $(\bfa,\bfb]$ is in $\mathcal{Y}$, and (ii) 
every $X\in\mathcal{X}$ that is outside $(\bfa,\bfb]$ is not included in $\mathcal{Y}$. Note that $X\subseteq\mathcal{X}$ that are split by $(\bfa,\bfb]$ may be either in $\mathcal{Y}$ or not in $\mathcal{Y}$.

 The  \emph{truncation} of $\mathcal{X}$ with respect to $(\bfa,\bfb]$,
is the family  
$${\sf tr}_{(\bfa,\bfb]}(\mathcal{X})=\{X\cap(\bfa,\bfb]\mid X\in\mathcal{X}\text{ s.t. }X\cap(\bfa,\bfb]\neq\emptyset\}.$$


 %
 
  For an integer $s\geq 0$, 
  $\mathcal{X}$ is \emph{$s$-feasible}  with respect to $(\bfa,\bfb]$ if $(\bfa,\bfb]$ splits at most $s$ collections in $\mathcal{X}$.  
 
   
  \begin{proof}[Proof of Theorem~\ref{thm:DP-expl}]
 Let $(\mathcal{C},s)$ be an instance of \probClustExpl, where $\mathcal{C}=\{\bfC_1,\ldots,\bfC_k\}$ for disjoint collections of points $\bfC_i$ of $\mathbf{R}^d$. Let $\bfX=\bigcup_{i=1}^k\bfC_i$.
 Following the proof of Proposition~\ref{prop:DP}, we say that a vector $\bfz\in(\mathbb{R}\cup\{\pm\infty\})^d$ is \emph{canonical}  if $\bfz[i]\in {\sf coord}_i(\bfX)\cup\{\pm\infty\}$ for every $i\in\{1,\ldots,d\}$. 
 
 For every pair of canonical vectors $(\bfa,\bfb)$ such that $\bfa<\bfb$ and $\mathcal{C}$ is $s$-feasible with respect to $(\bfa,\bfb]$, and every
 $(\bfa,\bfb]$-proper $\mathcal{S}=\{\bfS_1,\ldots,\bfS_\ell\}\subseteq\mathcal{C}$, 
 we denote by $\omega(\bfa,\bfb,\mathcal{S})$ the minimum size of a collection of points $\bfW\subseteq \bfX\cap(\bfa,\bfb]$ such that $\{\bfS_1'\setminus \bfW,\ldots,\bfS_\ell'\setminus \bfW\}$, where $\{\bfS_1',\ldots,\bfS_\ell'\}={\sf tr}_{(\bfa,\bfb]}(\mathcal{S})$,  is an explainable  $\ell$-clustering. We assume that $\omega(\bfa,\bfb,\mathcal{S})=0$ if $\mathcal{S}$ is empty.



 We compute 
 \begin{equation}\label{eq:def-w}
 w(\bfa,\bfb,\mathcal{S})=\omega(\bfa,\bfb,\mathcal{S}) +\sum_{\bfC_i\in \mathcal{S}}|\bfC_i\setminus (\bfa,\bfb]|+\sum_{\bfC_i\in\mathcal{C}\setminus\mathcal{S}}|\bfC_i\cap(\bfa,\bfb]|.
 \end{equation}
 Since we are interested only in clustering that can be obtained by deleting at most $s$ points, we assume that  $\omega(\bfa,\bfb,\mathcal{S})=+\infty$ if this value is bigger than $s$. This slightly informal agreement simplifies arguments.  In particular, observe the two sums in (\ref{eq:def-w}) give the value that is bigger than $s$ if $\mathcal{S}$ is not $s$-feasible with respect to $(\bfa,\bfb]$. In fact, this is the reason why these sums are included in (\ref{eq:def-w}).  
 
  Notice that $(\mathcal{C},s)$ is a yes-instance of \probClustExpl if and only if $w(\bfa^*,\bfb^*,\mathcal{C})\leq s$, where $\bfa^*[i]=-\infty$ and $\bfb^*[i]=+\infty$ for $i\in\{1,\ldots,d\}$. 
 
 The values $w(\bfa,\bfb,\mathcal{S})$ are computed depending on $\ell=|\mathcal{S}|$.  If $\ell=0$, that is, $\mathcal{S}=\emptyset$, then $\omega(\bfa,\bfb,\mathcal{S})=0$ and
 $w(\bfa,\bfb,\mathcal{S})=\sum\limits_{\bfC_j\in\mathcal{C}}|\bfC_j\cap(\bfa,\bfb]|$. 
 If $\ell =1$, then $\omega(\bfa,\bfb,\mathcal{S})=0$ by definition. Then  $\mathcal{S}=\{\bfC_i\}$ for some $i\in\{1,\ldots,k\}$ such that $\bfC_i\cap(\bfa,\bfb]\neq\emptyset$ and 
 $$w(\bfa,\bfb,\mathcal{S})=|\bfC_i\setminus (\bfa,\bfb]|+\sum_{\bfC_j\in\mathcal{C}\setminus\{\bfC_i\}}|\bfC_j\cap(\bfa,\bfb]|.$$
 
 Assume that $\ell\geq 2$, and the  values of   $\omega(\bfa',\bfb',\mathcal{S}')$ 
 are computed for $|\mathcal{S}'|<\ell$. 
 
 For $i\in \{1,\ldots,d\}$ and $\theta\in {\sf coord}_i(\bfX)$ such that $\bfa[i]< \theta<\bfb[i]$, we define the vectors $\bfa^{i,\theta}$ and $\bfb^{i,\theta}$ by setting
 $$\bfa^{i,\theta}[j]=
 \begin{cases}
 \theta&\mbox{if }j=i,\\
 \bfa[j]&\mbox{if }j\neq i,
 \end{cases}
 \text{ and }
\bfb^{i,\theta}[j]=
 \begin{cases}
 \theta&\mbox{if }j=i,\\
 \bfb[j]&\mbox{if }j\neq i.
 \end{cases}
$$
 We also say that $(i,\theta)$ is \emph{$s$-feasible} if $\mathcal{C}$ is $s$-feasible with respect to $(\bfa,\bfb^{i,\theta}]$ and $(\bfa^{i,\theta},
 \bfb]$.
For an $s$-feasible $(i,\theta)$, a partition  $(\mathcal{S}_1,\mathcal{S}_2)$ of $\mathcal{S}$ is \emph{$(i,\theta)$-proper} if $\mathcal{S}_1$ and $\mathcal{S}_2$ are  $(\bfa,\bfb^{i,\theta}]$ and $(\bfa^{i,\theta},
 \bfb]$-proper, respectively. We define 
 \begin{equation*}  
 \delta_{i,\theta}(\mathcal{S})=\sum_{\substack{\bfC_i\in\mathcal{S}\colon \bfC_i\cap(\bfa,\bfb^{i,\theta}]\neq\emptyset\\\text{and }\bfC_i\cap(\bfa^{i,\theta},\bfb]\neq\emptyset}} |\bfC_i\cap(\bfa,\bfb]|. 
 \end{equation*}
 

We compute $\omega(\bfa,\bfb,\mathcal{S})$ by the following recurrence.
\begin{align}\label{eq:rec-dp-expl}
w(\bfa&,\bfb,\mathcal{S})= \min\{(**) + (***)\},

\end{align}
where the right part is denoted by $(*)$, and
\begin{equation*}
(**)=\min\{|\bfC_i\setminus(\bfa,\bfb]|+\sum_{\bfC_j\in\mathcal{S}\setminus\{C_i\}}|\bfC_j|+\sum_{\bfC_j\in \mathcal{C}\setminus\mathcal{S} }|\bfC_j\cap(\bfa,\bfb]|\mid\bfC_i\in\mathcal{S} \},
\end{equation*}
 \begin{multline*
(***)=\min \{w(\bfa,\bfb^{i,\theta},\mathcal{S}_1)+w(\bfa^{i,\theta},\bfb,\mathcal{S}_2)-\delta_{i,\theta}(\mathcal{S}) \text{ for } 1\leq i\leq d,~\theta\in {\sf coord}_i(\bfX), \nonumber\\
(\mathcal{S}_1,\mathcal{S}_2)\text{ is  partition of }\mathcal{S}\text{ s.t.}, \bfa[i]< \theta<\bfb[i],(i,\theta)\text{ is }s\text{-feasible}, (\mathcal{S}_1,\mathcal{S}_2)\text{ is  }(i,\theta)\text{-proper}\}.
\end{multline*} 
We assume that $(***)=+\infty$ if there is no triple $(i,\theta,(\mathcal{S}_1,\mathcal{S}_2))$ satisfying the conditions in the definition of the set.
We also assume that $(*)=+\infty$
if its value proves to be bigger than $s$. 



 The correctness of (\ref{eq:rec-dp-expl}) is proved by  showing the inequalities between the left and rights parts in both directions.  


First, we show that $w(\bfa,\bfb,\mathcal{S})\geq (*)$. This is trivial  if $w(\bfa,\bfb,\mathcal{S})=+\infty$. Assume that this is not the case. Then by our assumption, $w(\bfa,\bfb,\mathcal{S})\leq s$. 
Recall that 
$w(\bfa,\bfb,\mathcal{S})=\omega(\bfa,\bfb,\mathcal{S})+\sum_{\bfC_i\in \mathcal{S}}|\bfC_i\setminus (\bfa,\bfb]|+\sum_{\bfC_i\in\mathcal{C}\setminus\mathcal{S}}|\bfC_i\cap(\bfa,\bfb]|$. 
Let $r=\omega(\bfa,\bfb,\mathcal{S})$ and let $\bfW\subseteq \bfX\cap(\bfa,\bfb]$ be a collection of  $r$ points such that $\mathcal{S}'=\{\bfS_1'\setminus \bfW,\ldots,\bfS_\ell'\setminus \bfW\}$, where $\{\bfS_1',\ldots,\bfS_\ell'\}={\sf tr}_{(\bfa,\bfb]}(\mathcal{S})$,  is an explainable  $\ell$-clustering. Assume that $\mathcal{S}=\{\bfC_{i_1},\ldots,\bfC_{i_\ell}\}$ and $\bfS_j'=\bfC_{i_j}\cap (\bfa,\bfb]$.
 Let $\bfW_i=\bfW\cap \bfS_i'$ for $i\in\{1,\ldots,\ell\}$. 

Notice that it may happen that $\bfW_j=\bfS_j'$ for some $j\in\{1,\ldots,\ell\}$. Then $\bfC_{i_j}\cap(\bfa,\bfb]=\bfW_j$.  
Observe, however, that $\bfC_{i_j}\cap(\bfa,\bfb]=\bfW_j$ for at most $\ell-1$ values of $j$. Suppose that there is $h\in \{1,\ldots,\ell\}$
such that $\bfC_{i_h}\cap(\bfa,\bfb]\neq \bfW_h$ and $\bfC_{i_j}\cap(\bfa,\bfb]=\bfW_j$ for every $h\in\{1,\ldots,\ell\}$ such that $j\neq h$.
In this case, we obtain that 
\begin{equation*}
\omega(\bfa,\bfb,\mathcal{S})=\sum_{\bfC_j\in\mathcal{S}\setminus\{C_{i_h}\}}|\bfC_j\cap(\bfa,\bfb]|
\end{equation*}
and 
\begin{equation*}
w(\bfa,\bfb,\mathcal{S})=|\bfC_{i_h}\setminus(\bfa,\bfb]| + \sum_{\bfC_j\in\mathcal{S}\setminus\bfC_{i_h}}|\bfC_j|+\sum_{\bfC_j\in \mathcal{C}\setminus\mathcal{S} }|\bfC_j\cap(\bfa,\bfb]|.
\end{equation*}
Then 
$w(\bfa,\bfb,\mathcal{S})\geq (**)\geq (*)$.
 Assume from now that this is not the case and $\bfS_j\setminus \bfW_j\neq\emptyset$ for at least two distinct indices $j\in\{1,\ldots,\ell\}$.
Then we show that $w(\bfa,\bfb,\mathcal{S})\geq (***)$.

Because we separate at least two nonempty collections of points, the definition of explainable clustering implies that 
there are $i\in\{1,\ldots,d\}$ and $\theta\in{\sf coord}_i(\bfX)$, such that $\bfa[i]<\theta<\bfb[i]$ and 
there is a partition $(I_1,I_2)$ of $\{1,\ldots,\ell\}$ with the property that (i) $\hat{\mathcal{S}}_1=\{\bfS_i'\setminus \bfW_j\mid j\in I_1 \}$ is an explainable $\ell_1=|I_1|$-clustering with the clusters in $(\bfa,\bfb^{i,\theta}]$, and 
(ii) $\hat{\mathcal{S}}_2=\{\bfS_j'\setminus \bfW_j\mid j\in I_2 \}$ is an explainable $\ell_2=|I_2|$-clustering with the clusters in $(\bfa^{i,\theta},\bfb]$, where both $\hat{\mathcal{S}}_1$ and $\hat{\mathcal{S}}_2$ contain nonempty collections of points. 
Moreover, we assume that if  $\bfS_j\setminus \bfW_j=\emptyset$ for some $j\in\{1,\ldots,\ell\}$, then $\bfS_j\setminus \bfW_j$ is placed in $\hat{\mathcal{S}}_1$ if $\bfS_j$ has a point in $(\bfa,\bfb^{i,\theta}]$ and, otherwise, i.e. if  $\bfS_j$ has only points in $(\bfa^{i,\theta},\bfb]$, it is placed in  $\hat{\mathcal{S}}_2$.

We define $\mathcal{S}_1=\{\bfC_{i_j}\mid j\in I_1\}$ and $\mathcal{S}_2=\{\bfC_{i_j}\mid j\in I_2\}$. For $j\in I_1$, let $\bfW_j^1=(\bfa,\bfb^{i,\theta}]\cap \bfW_j$, and let   $\bfW_j^2=(\bfa^{i,\theta},\bfb]\cap \bfW_j$. 
We set $\bfW^1=\bigcup_{j\in I_1}\bfW_j^1$ and  $\bfW^2= \bigcup_{j\in I_2}\bfW_j^2$. Let also 
$\bfR_1=(\bfa^{i,\theta},\bfb]\cap\big( \bigcup_{j\in I_1}\bfW_j\big)$ and  $\bfR_2=(\bfa,\bfb^{i,\theta}]\cap\big( \bigcup_{j\in I_2}\bfW_j\big)$. Observe that $(\bfW^1,\bfR_1,\bfW^2,\bfR_2)$ is a partition of $\bfW$ where some sets may be empty. 
Denote by $w_1=|\bfW^1|$ and $w_2=|\bfW^2|$, and let $r_1=|\bfR_1|$ and $r_2=|\bfR_2|$. Clearly, $w_1+w_2+r_1+r_2=r$.

Notice that $\mathcal{S}_1'=\{\bfS_j'\setminus \bfR_1\mid j\in I_1\}={\sf tr}_{(\bfa,\bfb^{i,\theta}]}(\mathcal{S}_1)$ and $\mathcal{S}_2'=\{\bfS_j'\setminus \bfR_2\mid j\in I_2\}={\sf tr}_{(\bfa^{i,\theta},\bfb]}(\mathcal{S}_2)$. 
Also $\mathcal{S}_1$ and $\mathcal{S}_2$ are $(\bfa,\bfb^{i,\theta}]$ and $(\bfa^{i,\theta},\bfb]$-proper, respectively. Furthermore, for $h\in\{1,2\}$,
$\hat{\mathcal{S}}_h$ is obtained from $\mathcal{S}_h'$ by deleting the points of $\bfW^1$ from the clusters. 
Also we have that 
\begin{equation}\label{eq:in-one}
\sum_{\bfC_j\in \mathcal{S}_1}|\bfC_j\setminus(\bfa,\bfb^{i,\theta}]|=\sum_{\bfC_j\in \mathcal{S}_1}|\bfC_j\setminus(\bfa,\bfb]|+|\bfR_1|,
\end{equation}
\begin{equation}\label{eq:in-two}
\sum_{\bfC_j\in \mathcal{S}_2}|\bfC_j\setminus(\bfa^{i,\theta},\bfb]|=\sum_{\bfC_j\in \mathcal{S}_2}|\bfC_j\setminus(\bfa,\bfb]|+|\bfR_2|,
\end{equation}
and
\begin{equation}\label{eq:out}
\sum_{\bfC_j\in \mathcal{C}\setminus \mathcal{S}_1}|\bfC_j\cap(\bfa,\bfb^{i,\theta}]|+\sum_{\bfC_j\in \mathcal{C}\setminus \mathcal{S}_2}|\bfC_j\cap(\bfa^{i,\theta},\bfb]|=\sum_{\bfC_j\in \mathcal{C}\setminus\mathcal{S}}|\bfC_j\cap(\bfa,\bfb]|+|\bfR_1|+|\bfR_2|.
\end{equation}
Note also that 
\begin{equation}\label{eq:delta}
\delta_{i,\theta}(\mathcal{S})=|\bfR_1|+|\bfR_2|.
\end{equation}
Then by (\ref{eq:in-one})--(\ref{eq:delta}),
\begin{align}\label{eq:forward}
w(\bfa,\bfb,\mathcal{S})=&(w_1+w_2+r_1+r_2) +\sum_{\bfC_j\in \mathcal{S}}|\bfC_j\setminus (\bfa,\bfb]|+\sum_{\bfC_j\in\mathcal{C}\setminus\mathcal{S}}|\bfC_j\cap(\bfa,\bfb]|    \nonumber\\
=&(w_1+w_2+r_1+r_2)+\sum_{\bfC_j\in \mathcal{S}_1}|\bfC_j\setminus (\bfa,\bfb]|+\sum_{\bfC_j\in \mathcal{S}_2}|\bfC_j\setminus (\bfa,\bfb]|\nonumber\\
 & +\sum_{\bfC_j\in \mathcal{C}\setminus \mathcal{S}_1}|\bfC_j\cap(\bfa,\bfb^{i,\theta}]|  +\sum_{\bfC_j\in \mathcal{C}\setminus \mathcal{S}_2}|\bfC_j\cap(\bfa^{i,\theta},\bfb]|-2r_1-2r_2 \nonumber\\
=& w_1+\sum_{\bfC_j\in \mathcal{S}_1}|\bfC_j\setminus (\bfa,\bfb^{i,\theta}]| +\sum_{\bfC_j\in \mathcal{C}\setminus \mathcal{S}_1}|\bfC_j\cap(\bfa,\bfb^{i,\theta}]|\nonumber\\
&+w_2+\sum_{\bfC_j\in \mathcal{S}_2}|\bfC_j\setminus (\bfa^{i,\theta},\bfb]| +\sum_{\bfC_j\in \mathcal{C}\setminus \mathcal{S}_2}|\bfC_j\cap(\bfa^{i,\theta},\bfb]|-\delta_{i,\theta}(\mathcal{S}).
\end{align}
Recall that $w(\bfa,\bfb,\mathcal{S})\leq s$. 
Note that 
\begin{equation*}\label{eq:feas}
\sum_{\bfC_j\in \mathcal{C}\setminus \mathcal{S}_1}|\bfC_j\cap(\bfa,\bfb^{i,\theta}]|\leq \sum_{\bfC_j\in \mathcal{C}\setminus\mathcal{S}}|\bfC_j\cap(\bfa,\bfb]|+|\bfR_2|.
\end{equation*}
Using (\ref{eq:in-one}), we obtain that 
\begin{equation*} 
\sum_{\bfC_j\in \mathcal{S}_1}|\bfC_j\setminus (\bfa,\bfb^{i,\theta}]|+\sum_{\bfC_j\in \mathcal{C}\setminus \mathcal{S}_1}|\bfC_j\cap(\bfa,\bfb^{i,\theta}]|\leq  w(\bfa,\bfb,\mathcal{S}),
\end{equation*}
which is at most $s$.
This means that $\mathcal{S}_1$ is $(\bfa,\bfb^{i,\theta}]$-proper. Similarly, we have that 
$\mathcal{S}_2$ is  $(\bfa^{i,\theta},\bfb]$-proper.
Therefore, 
\begin{equation*}
w_1+\sum_{\bfC_j\in \mathcal{S}_1}|\bfC_j\setminus (\bfa,\bfb^{i,\theta}]|+\sum_{\bfC_j\in \mathcal{C}\setminus \mathcal{S}_1}|\bfC_j\cap(\bfa,\bfb^{i,\theta}]|\geq w(\bfa,\bfb^{i,\theta},\mathcal{S}_1)
\end{equation*}
and 
\begin{equation*}
w_2+\sum_{\bfC_j\in \mathcal{S}_2}|\bfC_j\setminus (\bfa^{i,\theta},\bfb]|+\sum_{\bfC_j\in \mathcal{C}\setminus \mathcal{S}_2}|\bfC_j\cap(\bfa^{i,\theta},\bfb]|\geq w(\bfa^{i,\theta},\bfb,\mathcal{S}_2).
\end{equation*}


This allows us to extend (\ref{eq:forward}) and conclude that 
\begin{equation*}
w(\bfa,\bfb,\mathcal{S})\geq w(\bfa,\bfb^{i,\theta},\mathcal{S}_1)+w(\bfa^{i,\theta},\bfb,\mathcal{S}_2)\geq (***).
\end{equation*}
This shows that $w(\bfa,\bfb,\mathcal{S})\geq(***)\geq(*)$ and
concludes the proof of the first inequality.

Now we show that $w(\bfa,\bfb,\mathcal{S})\leq (*)$. The inequality is trivial if $(*)=+\infty$.  Suppose that this is not the case. Then, by our assumption about assigning the value to $(*)$, $(*)\leq s$. 

Suppose that the minimum in $(*)$ is achieved in the first part, that is, $(**)\leq (***)$.
Then 
\begin{equation*}
(*)=|\bfC_i\setminus(\bfa,\bfb]|+\sum_{\bfC_j\in\mathcal{S}\setminus\{C_i\}}|\bfC_j|\nonumber+\sum_{\bfC_j\in \mathcal{C}\setminus\mathcal{S} }|\bfC_j\cap(\bfa,\bfb]|
\end{equation*} 
for some $\bfC_i\in\mathcal{S}$.
We define $\bfW=\bigcup\limits_{\bfC_j\in \mathcal{S}\setminus \{\bfC_\}}|\bfC_j\cap (\bfa,\bfb]|$.
We have that the family obtained from $tr_{(\bfa,\bfb]}(\mathcal{S})$ by the deletion of the points of $\bfW$ is an explainable $\ell$-clustering, because we deleted the points in $(\bfa,\bfb]$ of every $\bfC_j\in\mathcal{S}$ excepts $\bfC_i$.
This implies that $\omega(\bfa,\bfb,\mathcal{S})\leq|\bfW|$ and we obtain that 
\begin{multline*} 
w(\bfa,\bfb,\mathcal{S})\leq|\bfW|+\sum_{\bfC_i\in \mathcal{S}}|\bfC_i\setminus (\bfa,\bfb]|+\sum_{\bfC_i\in\mathcal{C}\setminus\mathcal{S}}|\bfC_i\cap(\bfa,\bfb]|\\
=|\bfC_i\setminus(\bfa,\bfb]|+\sum_{\bfC_j\in\mathcal{S}\setminus\{C_i\}}|\bfC_j|\nonumber+\sum_{\bfC_j\in \mathcal{C}\setminus\mathcal{S} }|\bfC_j\cap(\bfa,\bfb]|,
\end{multline*}
which is exactly $(**)$, so $\omega(\bfa,\bfb,\mathcal{S})\leq (*)$.    


Assume from now that the minimum in $(*)$ is achieved for the second part, that is, $(*)=(***)<(**)$.
Suppose that 
$i\in\{1,\ldots,d\}$, $\theta\in {\sf coord}_i(\bfX)$ where $\bfa[i]\leq \theta\leq \bfb[i]$ and $(i,\theta)$ is $s$-feasible, and a partition $(\mathcal{S}_1,\mathcal{S}_2)$ of $\mathcal{S}$ that is $(i,\theta)$-proper are chosen in such a way that $(**)$ achieves the minimum value for them, that is, $(***)=w(\bfa,\bfb^{i,\theta},\mathcal{S}_1)+w(\bfa^{i,\theta},\bfb,\mathcal{S}_2)-\delta_{i,\theta}(\mathcal{S})$.  
  
Let $\bfR_1=(\bfa^{i,\theta},\bfb]\cap\big(\bigcup_{\bfC_j\in\mathcal{S}_1}\bfC_j)$ and  $\bfR_2=(\bfa,\bfb^{i,\theta}]\cap\big(\bigcup_{\bfC_j\in\mathcal{S}_2}\bfC_j)$. Then we obtain that 
\begin{equation}\label{eq:in-one-b}
\sum_{\bfC_j\in \mathcal{S}_1}|\bfC_j\setminus(\bfa,\bfb^{i,\theta}]|=\sum_{\bfC_j\in \mathcal{S}_1}|\bfC_j\setminus(\bfa,\bfb]|+|\bfR_1|,
\end{equation}
\begin{equation}\label{eq:in-two-b}
\sum_{\bfC_j\in \mathcal{S}_2}|\bfC_j\setminus(\bfa^{i,\theta},\bfb]|=\sum_{\bfC_j\in \mathcal{S}_2}|\bfC_j\setminus(\bfa,\bfb]|+|\bfR_2|,
\end{equation}
\begin{equation}\label{eq:out-b}
\sum_{\bfC_j\in \mathcal{C}\setminus \mathcal{S}_1}|\bfC_j\cap(\bfa,\bfb^{i,\theta}]|+\sum_{\bfC_j\in \mathcal{C}\setminus \mathcal{S}_2}|\bfC_j\cap(\bfa^{i,\theta},\bfb]|
=\sum_{\bfC_j\in \mathcal{C}\setminus\mathcal{S}}|\bfC_j\cap(\bfa,\bfb]|+|\bfR_1|+|\bfR_2|,
\end{equation}
and 
\begin{equation}\label{eq:delta-b}
\delta_{i,\theta}(\mathcal{S})=|\bfR_1|+|\bfR_2|.
\end{equation}

Let $w_1=\omega(\bfa,\bfb^{i,\theta},\mathcal{S}_1)$ and $w_2=\omega(\bfa^{i,\theta},\bfb,\mathcal{S}_2)$. Then there is a collection $\bfW_1\subseteq (\bfa,\bfb^{i,\theta}]\cap\bfX$ such that  the collection of sets obtained from the collections of ${\sf tr}_{(\bfa,\bfb^{i,\theta}]}(\mathcal{S}_1)$ by the deletions of the points of $\bfW_1$ is an explainable $|\mathcal{S}_1|$-clustering. Similarly, there is a collection $\bfW_2\subseteq (\bfa^{i,\theta},\bfb]\cap\bfX$ such that  the family of points collections obtained from the collections of ${\sf tr}_{(\bfa^{i,\theta},\bfb]}(\mathcal{S}_2)$ by the deletions the points of $\bfW_2$ is an explainable $|\mathcal{S}_2|$-clustering. Consider $\bfW=\bfW_1\cup \bfW_2\cup \bfR_1\cup \bfR_2$. The crucial observation is that the  family  obtained from the collections of ${\sf tr}_{(\bfa,\bfb]}(\mathcal{S})$ by the deletions of the points of $\bfW$ is an explainable $\ell$-clustering, where the first cut is ${\sf Cut}_{i,\theta}$. Using (\ref{eq:in-one-b})--(\ref{eq:delta-b}), we obtain that
\begin{multline*}
(***)=w(\bfa,\bfb^{i,\theta},\mathcal{S}_1)+w(\bfa^{i,\theta},\bfb,\mathcal{S}_2)-\delta_{i,\theta}(\mathcal{S})
=|\bfW|+\sum_{\bfC_i\in \mathcal{S}}|\bfC_i\setminus (\bfa,\bfb]|+\sum_{\bfC_i\in\mathcal{C}\setminus\mathcal{S}}|\bfC_i\cap(\bfa,\bfb]|\\
\geq \omega(\bfa,\bfb,\mathcal{S})+\sum_{\bfC_i\in \mathcal{S}}|\bfC_i\setminus (\bfa,\bfb]|+\sum_{\bfC_i\in\mathcal{C}\setminus\mathcal{S}}|\bfC_i\cap(\bfa,\bfb]|
=w(\bfa,\bfb,\mathcal{S}).
\end{multline*} 
 Since $(*)=(***)$, this completes the correctness proof of the recurrence (\ref{eq:rec-dp-expl}).
 
 In the final stage of the proof, we evaluate the running time. We construct the table of values of $w(\bfa,\bfb,\mathcal{S})$ for pairs $(\bfa,\bfb)$ of canonical vectors such that $\bfa< \bfb$. The total number of such pairs is at most $(n+2)^{2d}$ and they can be constructed in $n^{2d}\cdot (dn)^{\mathcal{O}(1)}$ time. We are interested only in $\bfa$ and $\bfb$ such that $\mathcal{C}$ is $s$-feasible with respect to $(\bfa,\bfb]$. Clearly, for given $\bfa$ and $\bfb$, the  $s$-feasibility can be checked in $(dn)^{\mathcal{O}(1)}$ time. If $\mathcal{C}$ is $s$-feasible with respect to $(\bfa,\bfb]$, then all $(\bfa,\bfb]$-proper subfamilies $\mathcal{S}$ of $\mathcal{C}$ can be listed by brute force as follows. Observe that 
 $\mathcal{X}=\{\bfC_i\in\mathcal{C}\mid \bfC_i\text{ is in }(\bfa,\bfb]\}$ that can be constructed in polynomial time is a subfamily of every $(\bfa,\bfb]$-proper $\mathcal{S}$. Let $\mathcal{Y}=\{\bfC_i\in \mathcal{C}\mid \bfC_i\text{ is split by }(\bfa,\bfb]\}$.
 Since  $\mathcal{C}$ is $s$-feasible, $|\mathcal{Y}|\leq\min\{s,k\}$. We can construct $\mathcal{Y}$ in polynomial time and then generate at most $2^{\min\{s,k\}}$ subfamilies $\mathcal{Z}$ of $\mathcal{Y}$ in total 
 $2^{\min\{s,k\}}\cdot (dn)^{\mathcal{O}(1)}$ time. Then the $(\bfa,\bfb]$-proper subfamilies $\mathcal{S}$ are exactly the families of the form $\mathcal{Z}\cup\mathcal{Y}$.  We obtain that there are at most $2^{\min\{s,k\}}$ the 
 $(\bfa,\bfb]$-proper subfamilies that can be generated in time $2^{\min\{s,k\}}\cdot (dn)^{\mathcal{O}(1)}$ time. Then we conclude that the dynamic programming algorithm computes at most $2^{\min\{s,k\}}\cdot n^{2d}$ values of $w(\bfa,\bfb,\mathcal{S})$. 
  
The  value of $w(\bfa,\bfb,\mathcal{S})$ for $|\mathcal{S}|$ is constructed in $(dn)^{\mathcal{O}(1)}$ time  if $\ell\leq1$. If $\ell\geq 2$, we are using the recurrence (\ref{eq:rec-dp-expl}). 
Computing  $(**)$
 can be done in polynomial time.
To compute $(***)$, we  go through $i\in\{1,\ldots,d\}$, $\theta\in{\sf coord}_i(\bfX)$, and partitions $(\mathcal{S}_1,\mathcal{S})$ of $\mathcal{S}$, where $(\mathcal{S}_1,\mathcal{S}_2)$ is required to be $(i,\theta)$-proper. This implies that   $(***)$ can be computed in $2^{\min\{s,k\}}\cdot (dn)^{\mathcal{O}(1)}$ time. Summarizing, we have that the total running time of the dynamic programming algorithm is    $2^{2\min\{s,k\}}\cdot n^{2d}\cdot (dn)^{\mathcal{O}(1)}$. This concludes the proof.    
 \end{proof} 
 
\subsection{Data reduction}\label{sec:kernel}

  
  
\begin{restatable}{theorem}{kernel}\label{thm:kern}
    Let $r = 2 (s + 1) dk$. There is a polynomial-time algorithm that, given an instance of \probClustExpl, produces an equivalent one with at most $r$ points in $\{1,\ldots, r\}^d$. 
   
\end{restatable}
  
\begin{proof}
Let $(\mathcal{C},s)$ be an instance of \probClustExpl, where $\mathcal{C}=\{\bfC_1,\ldots,\bfC_k\}$ for disjoint collections of points $\bfC_i$ of $\mathbf{R}^d$. Let $\bfX=\bigcup_{i=1}^k\bfC_i$.
  
Our first aim is to reduce the number of points. For this, we use a procedure that marks essential points. 
  
For every $i\in\{1,\ldots,k\}$ and every $j\in\{1,\ldots,d\}$, do the following:
\begin{itemize}
\item Order the points of $\bfC_i$ by the increase of their $j$-th coordinate; the  ties are broken arbitrarily.
\item \emph{Mark} the first $\min\{s+1,|\bfC_i|\}$ points and the last   $\min\{s+1,|\bfC_i|\}$ points in the ordering.
\end{itemize}
  
The procedure marks at most $2(s+1)dk$ points. Then we delete the remaining unmarked points. Formally, we denote by $\bfY$ the collection of marked points and set $\bfS_i=\bfC_i\cap\bfY$ for all $i\in\{1,\ldots,k\}$.   
Then we consider the instance $(\mathcal{S},s)$ of  \probClustExpl, where $\mathcal{S}=\{\bfS_1,\ldots,\bfS_k\}$. We show the following claim.

\begin{claim}\label{cl:expl}
$(\mathcal{C},s)$ is a yes-instance of \probClustExpl if and only if $(\mathcal{S},s)$ is a yes-instance.
\end{claim}

\begin{proof}[Proof of Claim~\ref{cl:expl}]
Trivially, if   $(\mathcal{C},s)$ is a yes-instance, then $(\mathcal{S},s)$ is a yes-instance, because we just deleted some point to construct $(\mathcal{S},s)$. We show that  
if $(\mathcal{S},s)$ is a yes-instance, then $(\mathcal{C},s)$ is a yes-instance. 

Because $(\mathcal{S},s)$ is a yes-instance, there is a collection of at most $s$ points $\bfW\subseteq \bfY$ such that $\{\bfS_1\setminus \bfW,\ldots,\bfS_k\setminus \bfW\}$ is an explainable $k$-clustering. In other words, there is an explainable clustering of $\bfY\setminus \bfW$ with a canonical threshold tree $(T,k,\varphi)$ such that the clusters $\bfS_1\setminus \bfW,\ldots,\bfS_k\setminus \bfW$ correspond to the leaves of the threshold tree. We claim that if we use the same threshold tree for $\bfX\setminus \bfW$, then $\bfC_1\setminus \bfW,\ldots,\bfC_k\setminus \bfW$ correspond to the leaves. 

The proof is by contradiction. Assume that at least one collections of points corresponding to a leaf is distinct from every $\bfC_1\setminus \bfW,\ldots,\bfC_k\setminus \bfW$. Then there is a node $v\in V(T)$ such that for some $j\in\{1,\ldots,k\}$, $\bfC_j\setminus \bfW$ is split by the cut ${\sf Cut}_{i,\theta}$ for $(i,\theta)=\varphi(v)$, that is, for $(\bfA,\bfB)={\sf Cut}_{i,\theta}(\bfX)$, $\bfA\cap (\bfC_j\setminus \bfW)\neq\emptyset$ and $\bfB\cap (\bfC_j\setminus \bfW)\neq\emptyset$. Observe that either $\bfA\cap (\bfS_j\setminus \bfW)=\emptyset$ or $\bfB\cap (\bfS_j\setminus \bfW)=\emptyset$. We assume without loss of generality that $\bfA\cap (\bfS_j\setminus \bfW)=\emptyset$ (the other case is symmetric). This means that there is an unmarked point $\bfx\in \bfC_j\setminus \bfW$ in $\bfA$ and all the marked points of $\bfC_j\setminus \bfW$ are in $\bfB$. Because $\bfC_j$ has an unmarked point, $|\bfC_j|\geq 2(s+1)+1$.
Following the marking procedure, we order the points of  $\bfC_j$ by the increase of the $i$-th coordinate breaking ties exactly as in the marking procedure. Let $L$ be the collection of the first $s+1$ points that are marked. Since $|\bfW|\leq s$, there is $\bfy\in L\setminus \bfW$. Because $L\setminus \bfW\subseteq \bfS_j\setminus \bfW\subseteq \bfB$, we have that $\bfy[i]>\theta$. Then $\bfx[i]\geq\bfy[i]>\theta$ and $\bfx\in \bfB$; a contradiction.  

We conclude that  if we use $(T,k,\varphi)$ to cluster $\bfX\setminus \bfW$, then $\bfC_1\setminus \bfW,\ldots,\bfC_k\setminus \bfW$ correspond to the leaves. This proves that $(\mathcal{C},s)$ is a yes-instance of 
 \probClustExpl.
 \end{proof}   

We obtained the instance $(\mathcal{S},s)$, where $\bfY=\bigcup_{i=1}^k\bfS_i$ has $\ell\leq 2(s+1)dk$ points, that is equivalent to the original instance.  Now we modify  the points to ensure that they are in $\{1,\ldots,\ell\}^d$. For this, we observe that for each $i\in\{1,\ldots,d\}$, the values of the $i$-th coordinates can be changed if we maintain their order. Formally, we do the following. For every $i\in \{1,
\ldots,d\}$, let ${\sf coord}_i(\bfY)=\{\theta_1^i,\ldots,\theta_{r_i}^i\}$, where   $\theta_1^i<\cdots<\theta_{r_i}^i$. For every $\bfy\in\bfY$, we construct a point $\bfz$, by setting
$\bfz[i]=j$, where $\theta_j^i=\bfy[i]$, for each $i\in\{1,\ldots,d\}$. Then for  $\bfS_i$ containing $\bfy$, we replace $\bfy$ by $\bfz$. 
Denote by $\bfZ$ the constructed collection of points, and let $\mathcal{R}=\{\bfR_1,
\ldots,\bfR_k\}$ be the family of the collections of points constructed from $\bfS_1,\ldots,\bfS_k$.

We have that $(\mathcal{R},s)$ is a yes-instance of \probClustExpl if and only if $(\mathcal{S},s)$ is a yes-instance, and $\bfZ\subseteq \{1,\ldots,\ell\}^d$. Then the data reduction algorithm returns $(\mathcal{R},s)$. To complete the proof, it remains to observe that the marking procedure is polynomial, and the coordinates replacement also can be done in polynomial time.
\end{proof}

\input{inapproximability}

\section{Approximate Explainable Clustering}
\label{sec:fractional}

\paragraph{Approximate explainable $k$-means/median{} clustering.} In \probFMeanExpl, we are given a collection of $n$ points $\bfX\subseteq \mathbb{R}^d$, a positive integer $k\leq n$, and a positive real constant $\varepsilon<1$. Then the task is to find a collection of points $\bfY\subseteq \bfX$ with $|\bfY|\geq (1-\varepsilon)|\bfX|$ and an explainable $k$-clustering of $\bfY$ whose $k$-median cost does not exceed the optimum $k$-median cost of an explainable $k$-clustering for the original collection of points $\bfX$. Note that we ask about the construction of $\bfY$ and the corresponding clustering as the decision variant is trivial. Observe also that the optimum cost is unknown a priori. 
 \probFMedExpl differs only by the clustering measure.


\begin{theorem}\label{thm:frac}
    \probFMeanExpl and \probFMedExpl are solvable in $(\frac{8dk}{\epsilon})^k\cdot n^{\mathcal{O}(1)}$ time. 
\end{theorem}

As the proofs for
both problems are identical, we describe only the algorithm for \probFMeanExpl. 

\begin{proof}
    

Let $\bfX\subseteq \mathbb{R}^d$ be an instance of \probFMeanExpl, $(T,k,\varphi)$ be the optimal (canonical) threshold tree for explainable $k$-means clustering and $(C_1, \dots, C_k)$ the clustering induced by $(T,k,\varphi)$. The goal of the algorithm will be to guess an approximation of $(T,k,\varphi)$. Since $T$ is a binary tree with $k$ leaves, guessing $T$ only requires $4^{k}$ tries. Guessing $\varphi$ is more complicated however, as there is $d \cdot n$ choice at each node of $T$, which gives potentially $(dn)^k$ possibilities, where $n = |\bfX|$. The idea here will be to guess for every nonleaf node $u$ of $T$ the second element of $\varphi(u)$ up to a precision of $\mathcal{O}(\frac{\epsilon n}{k})$, which gives only $\mathcal{O}(d \cdot \frac{k}{\epsilon})$ choices at each nonleaf node. 

More formally, let $n' = \lfloor \frac{\epsilon n}{k} \rfloor$ and note first that if $n' = 0$, then $\frac{\epsilon n}{k} < 1$ and thus $n \leq \frac{k}{\epsilon}$. This means that if $n' = 0$, then the algorithm  trying all the possible values of $T$ and $\varphi$, and computing the value of the obtained clustering, runs in time $4^k(\frac{dk}{\epsilon}) \cdot n^{\mathcal{O}(1)}$, which ends the proof. From now on, let us assume that $n' \not = 0$. 

Let $U$ denote the set of nonleaf nodes of $T$. Let  $\varphi'\colon U\rightarrow \{1,\ldots,d\}\times \mathbb{R}$ be the function obtained from $\varphi$ by rounding, for every $u \in U$, the value of the first element of $\varphi(u)$ to the closest multiple of $n'$. In other words, if $\varphi(u) = (j,r)$, then $\varphi'(u) = (j,i \cdot n')$ where $i$ is the largest integer such that $i \cdot n' \leq r$. 

Consider now the clustering obtained from the threshold tree $(T,k,\varphi')$. At each node $v \in T$ such that $\varphi(v) = (j,r)$ and $\varphi'(v) = (j,i \cdot n')$, the points $x$ of $\bfX$ that can be misplaced by the ${\sf Cut}_{\varphi'(u)}(\bfX)$ are exactly the points such that $i \cdot n' < \bfx[i] \leq r$. This means that, if $\bfZ_u$ denotes the set of all the points $x$ such that $i \cdot n' \leq \bfx[j] \leq (i+1)n'$, then $|\bfZ_u| \leq n'$ and the partitions ${\sf Cut}_{\varphi'(u)}(\bfX \setminus \bfZ_u)$ and ${\sf Cut}_{\varphi(u)}(\bfX \setminus \bfZ_u)$ are identical. Therefore, if $\bfZ = \bigcup_{u \in U} \bfZ_u$, then $(T,k,\varphi')$ and $(T,k,\varphi)$ induce the exact same clustering on $\bfX \setminus \bfZ$. Note that $|\bfZ| \leq kn' \leq \epsilon n$. 

Therefore the algorithm will try all possible choices for $T$, and for every nonleaf node $u$, it tries all possible values for $\varphi'(u)$ of the form $(j, i \cdot n')$, where $j \in [d]$ and $i \in [\frac{2k}{\epsilon}]$. For each such try, the algorithm also removes the set $\bf Z$ consisting  of all the points $x$ such that $  i \cdot n' \leq \bfx[j] \leq (i+1) \cdot n'$ whenever there exists $u \in U$ such that $\varphi'(u) = (j, i \cdot n')$ and computes the value of the clustering induces by $(T,k, \varphi')$ on $\bfX \setminus \bfZ$. Finally it outputs the set $\bf X \setminus \bfZ$ as well as the threshold tree $(T,k, \varphi')$ which minimises the value of the clustering. 

Note that for every set of choices of $T$ and $\varphi'(u)$ of the form $(j, i \cdot n')$, the set $\bfZ$ has size at most $k \cdot n' \leq \epsilon n$, which implies that the algorithm indeed outputs the desired set and threshold tree. Moreover, since there are at most $4^{k}$ possible trees $T$ and $ d \cdot \frac{2 k}{\epsilon}$ possible choices of $\varphi'(u)$ for every node of the tree, we conclude that the  running time of the algorithm is $(\frac{8dk}{\epsilon})^kn^{\mathcal{O}(1)}$.
\end{proof}
\section{Introduction }\label{sec:intro}
Interpretation or explanation of decisions produced by learning models, including clustering, is a significant direction in machine learning (ML) and artificial intelligence (AI), and has given rise to the subfield of Explainable AI. Explainable AI has attracted a lot of attention from the researchers in recent years (see the surveys by Carvalho et al.~\cite{carvalho2019machine} and Marcinkevi{\v{c}}s and Vogt~\cite{marcinkevivcs2020interpretability}). All these works can be divided into two main categories: \emph{pre-modelling} \cite{wang2015falling,ustun2016supersparse,hastie1986generalized,feng2017sparse,lu2018deeppink} and \emph{post-modelling} \cite{ribeiro2016should,shrikumar2017learning,breiman2001random,sundararajan2017axiomatic,lundberg2017unified} explainability. While post-modeling explainability focuses on giving reasoning behind decisions made by black box models, pre-modeling explainability deals with ML systems that are inherently understandable or perceivable by humans. 
 One of the canonical approaches to pre-modelling explainability builds on decision trees~\cite{molnar2020interpretable,murdoch2019interpretable}. In fact, a significant amount of work on explainable clustering is based on unsupervised decision trees \cite{BertsimasOW21,fraiman2013interpretable,geurts2007inferring,ghattas2017clustering,lipton2018mythos,MoshkovitzDRF20}. In each node of the decision tree, the data is partitioned according to some features' threshold value. 
While such a \textit{threshold tree} provides a clear interpretation of the resulting clustering, its cost measured by the standard $k$-means/median{} objective can be significantly worse than the cost of the optimal clustering. 
Thus,  on the one hand, the efficient algorithms developed for  $k$-means/median{} clustering~\cite{Aggarwalbook13}  are often challenging to explain. On the other hand, the easily explainable models could output very costly clusterings. Subsequently, Moshkovitz et al. \cite{MoshkovitzDRF20}, in a fundamental work, posed the natural algorithmic question of whether it is possible to kill two birds with one stone? To be precise, is it possible to design an efficient procedure for clustering that 
\begin{itemize}
\item Is explainable by a small decision tree; and
\item  Does not cost significantly more than  the cost of an optimal $k$-means/median{} clustering?
\end{itemize}
 
 
 To address this question,  Moshkovitz et al. \cite{MoshkovitzDRF20} introduced   explainable $k$-means/median{} clustering. In this scheme, a clustering is represented by a binary (\textit{threshold}) tree whose leaves correspond to clusters, and each internal node corresponds to partitioning a collection of points by a threshold on a fixed coordinate. Thus, the number of leaves in such a tree is $k$, the number of clusters sought. Also, any cluster assignment can be explained by the thresholds along the corresponding root-leaf path.  
 
 
  For example, consider Fig.~\ref{fig:fig}: Fig.~\ref{fig:sfig1}  shows an optimal $5$-means clustering of a 2D data set; Fig.~\ref{fig:sfig2} shows an explainable $5$-means clustering of the same data set; The threshold tree inducing the explainable clustering is shown in Fig.~\ref{fig:sfig3}.   The tree has five leaves, corresponding to 5 clusters. Note that in this model of explainability, any clustering has a clear geometric interpretation, where each cluster is formed by a set of axis-aligned cuts defined by the tree. 
 As Moshkovitz et al. argue, the classical $k$-means clustering algorithm leads to more complicated clusters while the threshold tree leads to an easy explanation. The advantage of the explainable approach becomes even more evident in higher dimensions when many feature values in $k$-means contribute to the formation of the clusters. 
 
\begin{figure}[h]
    \centering
  \begin{subfigure}[h]{.3\columnwidth}
    \centering
    \includegraphics[width=.95\linewidth]{figure_1.pdf}
    \subcaption{}
    \label{fig:sfig1}
  \end{subfigure}\hfill
  \begin{subfigure}[h]{.3\columnwidth}
    \centering
    \includegraphics[width=.95\linewidth]{figure_optimal_explainable.pdf}
    \subcaption{}
    \label{fig:sfig2}
  \end{subfigure}\hfill
  \begin{subfigure}[h]{.3\columnwidth}
    \centering
    \includegraphics[width=.95\linewidth]{tree_explainable.pdf}
    \subcaption{}
  \label{fig:sfig3}
  \end{subfigure}
  \caption{(a) An example  of an optimal solution to 5-means. (b) An explainable $5$-means clustering and (c) the corresponding threshold tree.}
  \label{fig:fig}
  \end{figure}


 Moshkovitz et al. \cite{MoshkovitzDRF20} define the quality of any explainable clustering as the ``cost of explainability'', that is the ratio of the cost of the explainable clustering to the cost of an optimal clustering.  Subsequently, they obtain efficient algorithms for computing explainable clusterings whose ``cost of explainability'' is $\mathcal{O}(k)$
for $k$-median and $\mathcal{O}(k^2)$ for $k$-means. They also show that the this ratio is at least $\Omega(\log k)$ in both cases. Recently, a series of works has been dedicated to improving these bounds.
In the low-dimensional setting, Laber and Murtinho~\cite{LaberM21} showed an upper bound of $\mathcal{O}(d \log k)$ and $\mathcal{O}(dk \log k)$ for $k$-median and $k$-means respectively.
In general, Makarychev and Shan~\cite{MakarychevS21}, Gamlath, Jia, Polak, and Svensson~\cite{GamlathJPS21} Esfandiari, Mirrokni, and Narayanan~\cite{EsfandiariMN21} showed independently a $k (\log k)^{\mathcal{O}(1)}$ upper bound for $k$-means and a $(\log k)^{\mathcal{O}(1)}$ upper bound for $k$-median, while also improving the lower bound for $k$-means to $\Omega(k)$. For low dimensions this was improved by Charikar and Hu~\cite{CharikarH21}, who showed an upper bound of $k^{1 - 2/d} (\log k)^{\mathcal{O}(1)}$ for $k$-means.





\medskip\noindent\textbf{Our contributions.} In this work, we propose a new model for explaining a clustering, called \probClustExpl. 
Our approach to explainability is inspired by the research on robustness in statistics and machine learning, especially the vast field of outlier detection and removal in the context of clustering \cite{chen2008constant,10.1145/3301446,feng2019improvedFPT,charikar2001algorithms,ChakrabartyGK16,harris2019lottery,krishnaswamy2018constant}. 
 In this model, we are given a $k$-means/median{} clustering and we would like to explain the clustering by a threshold tree \emph{after removing a subset of points}. To be precise, we are interested in finding a subset of points $S$ (which are to be removed) and a threshold tree $T$ such that the explainable clustering induced by the leaves of $T$ is exactly the same as the given clustering after removing the points in $S$. For the given clustering, we define an optimal (or {best}) explainable clustering to be the one that minimizes the size of $S$, i.e. for which the given clustering can be explained by removing the minimum number of points. Thus in  
  \probClustExpl  we measure the ``explainability'' as the number of outlying points whose removal turns the given clustering into  an explainable clustering. The reasoning behind the new measure of cluster explainability is the following.  In certain situations, we would be satisfied with a small decision tree explaining  clustering of all but a few outlying data points.  We note that for a given clustering that is already an explainable clustering, i.e. can be explained by a threshold tree, the size of $S$ is 0.  

 In  Fig.~\ref{fig:fig1}, we provide an example of an optimal  $5$-means clustering of exactly the same data set as  in Fig.~\ref{fig:fig}. However, the new explainable clustering is obtained in a different way. If we remove a small number of points (in  Fig.~\ref{fig:outliers} these are the 9 red larger points), then the explainable clustering is same as the optimal clustering after removing those 9 points.  
 
\begin{figure}[h]
  \label{fig:1}
  \centering

  \begin{subfigure}[t]{.47\columnwidth}
    \centering
    \includegraphics[width=.85\linewidth]{figure_1.pdf}
    \subcaption{}
    \label{fig:opt1}
  \end{subfigure}\hfill
  \begin{subfigure}[t]{.47\columnwidth}
    \centering
    \includegraphics[width=.85\linewidth]{figure_remove.pdf}
    \subcaption{}
    \label{fig:outliers}
  \end{subfigure}%
  \caption{(a) An optimal $5$-clustering and (b) an explainable clustering that fits this clustering after removing the larger (red) points. }
  \label{fig:fig1}
  \end{figure}

We note that \probClustExpl corresponds to the classical machine learning setting of interpreting a black-box model, i.e. it lies within the scope of post-modeling explainability. Surprisingly, this area is widely unexplored when it comes to rigorous algorithmic analysis of clustering explanation. Consequently, we study \probClustExpl from the perspective of computational complexity. Our new model naturally raises the following algorithmic questions: (i) \emph{Given a clustering, how efficiently can one decide whether  the clustering can be explained by a threshold tree (without removing any points)?} and (ii) \emph{Given a clustering and an integer $s$, how efficiently can one decide whether the clustering can be explained by removing $s$ points?} 

In our work, we design a polynomial time algorithm that resolves the first question. 
Regarding the second question, we give an algorithm that in time 
$2^{2\min\{s,k\}}\cdot n^{2d}\cdot (dn)^{\mathcal{O}(1)}$ decides whether a given  clustering of $n$ points in $\mathbb{R}^d$ could be 
 explained by removing $s$ points. 

 We also give an $n^{\mathcal{O}(1)}$ time  $(k-1)$-approximation algorithm for \probClustExpl. That is, 
 we give a polynomial time algorithm that returns a solution set of at most $s(k-1)$ points that are to be removed, whereas any  best explainable clustering removes $s$ points. Moreover, we provide an efficient data reduction procedure that reduces an instance of   \probClustExpl  to an equivalent instance with at most $r=2(s+1)dk$ points in $\mathbb{R}^d$ with integer coordinates within the range  $\{1,\ldots, r\}$. The procedure can be used to speed up \emph{any} algorithm for \probClustExpl, as long as $n > 2(s + 1) dk$.
 We complement our algorithms by showing a hardness lower bound. In particular, we show that \probClustExpl cannot be approximated within a factor of $F(s)$ in time $f(s) (nd)^{o(s)}$, for any functions $F$ and $f$, unless Exponential Time Hypothesis (ETH)~\cite{ImpagliazzoPZ01} fails. All these results appear in Section \ref{sec:explanation}. 



 



 
We also provide new insight into the computational complexity of the model of Moshkovitz et al. \cite{MoshkovitzDRF20}.  While   the vanilla $k$-median and $k$-means problems are NP-hard  
for $k=2$ \cite{AloiseDHP09,DrineasFKVV04,dasgupta2008hardness} or $d=2$ \cite{MahajanNV12},  this is not the case for explainable clustering! We design two simple algorithms  computing optimal (best) explainable clustering with $k$-means/median{} objective that run in time $(4nd)^{k+\mathcal{O}(1)}$ and $n^{2d}\cdot n^{\mathcal{O}(1)}$, respectively. Hence  for constant $k$ or constant $d$, an optimal explainable clustering can be computed in polynomial time. 
The research on approximation algorithms on the ``cost of explainability'' in~\cite{MoshkovitzDRF20,CharikarH21,EsfandiariMN21,GamlathJPS21,LaberM21,MakarychevS21} implicitly assumes that solving the problem exactly is NP-hard. However,   we did not find a  proof of this fact in the literature.  To fill this gap, we obtain the following  hardness lower bound: An optimal explainable clustering cannot be found in $f(k)\cdot n^{o(k)}$ time for any computable function $f(\cdot)$, unless Exponential Time Hypothesis (ETH) fails. This lower bound demonstrates that asymptotically the running times of our simple algorithms are unlikely to be improved. 
Our reduction also yields that the problem is NP-hard.
These results are described in Section \ref{sec:explainable}.

Finally, we combine the above two explainability models 
 to obtain the Approximate Explainable Clustering model: For a collection of $n$ points in $\mathbb{R}^d$ and a positive real constant $\varepsilon<1$, we seek whether we can
   identify at most $\varepsilon n$ outliers, such that the cost of explainable $k$-means/median{} of the remaining points does not exceed the optimal cost of an explainable $k$-means/median{} clustering of the original data set. Thus, if we are allowed to remove a small number of points, can we do as good as any original optimal solution? 
   While our hardness result of Section~\ref{sec:explainable} holds for explaining the whole dataset, by ``sacrificing'' a small fraction of points it might be possible to solve the problem more efficiently.
  
 And indeed, for this model, we obtain an algorithm whose running time $(\frac{8dk}{\epsilon})^k\cdot n^{\mathcal{O}(1)}$ has a significantly better dependence on $d$ and $k$. For example, compare this with the above time bounds of $(4nd)^{k+\mathcal{O}(1)}$ and $n^{2d}\cdot (dn)^{\mathcal{O}(1)}$. This algorithm appears in Section \ref{sec:fractional}. See Table \ref{table:1} for a summary of all our results. 

\renewcommand{\arraystretch}{1.5}
\begin{center}
\begin{table}[t]
\centering
\begin{tabular}{ |c|c|c| }
 \hline

 \centering \textbf{Model} & \textbf{Algorithms/Upper bounds} & \textbf{Hardness/Lower bounds} \\ 
 \hline
 \centering Clustering Explanation  & \makecell{$2^{2\min\{s,k\}}n^{2d}n^{\mathcal{O}(1)}$\\$(k-1)$-approximation\\Reduction to $\mathcal{O}(sdk)$ points }& No $F(s)$-approx. in $f(s) (nd)^{o(s)}$\\ 
\hline
\centering Explainable Clustering  & \centering $(4nd)^{k+\mathcal{O}(1)}$ $n^{2d}\cdot n^{\mathcal{O}(1)}$ & $f(k)\cdot n^{o(k)}$ \\
\hline
\centering \makecell{Approximate\\Explainable Clustering} & \centering $(\frac{8dk}{\epsilon})^k\cdot n^{\mathcal{O}(1)}$ & \\
\hline
\end{tabular}
\caption{A summary of our results.}
\label{table:1}
\end{table}
\end{center}





 
 

 

 
 


\section{Preliminaries}\label{sec:prelim} 


\paragraph{$k$-means/median.}
Given a collection $\bfX=\{\bfx_1,\ldots,\bfx_n\}$ of $n$ points in $\mathbb{R}^d$ and a positive integer $k$, the task of \emph{$k$-clustering} is to partition $\bfX$ into $k$ parts $\bfC_1,\ldots,\bfC_k$, called \emph{clusters}, such that the \emph{cost} of clustering is minimized. We follow the convention in the previous work  \cite{MoshkovitzDRF20} for defining the cost. 
In particular, for $k$-means, we consider the Euclidean distance and for $k$-median, the Manhattan distance.
 For a collection of points $\bfX'$ of $\mathbb{R}^d$, we define 
\begin{equation}\label{eq:mean}
{\sf cost}_2(\bfX')=\min_{\bfc\in\mathbb{R}^d}\sum_{\bfx\in \bfX'}\|\bfc-\bfx\|_2^2,
\end{equation}
and call the point $\bfc^*\in \mathbb{R}^d$ minimizing the sum in  (\ref{eq:mean}) the \emph{mean} of  $\bfX'$.
For a clustering $\{\bfC_1,\ldots,\bfC_k\}$ of $\bfX\subseteq \mathbb{R}^d$, its \emph{$k$-means} (or simply means) cost is 
${\sf cost}_2(\bfC_1,\ldots,\bfC_k)=\sum_{i=1}^k  {\sf cost}_2(\bfC_i)$.
With respect to the Manhattan distance, we define analogously ${\sf cost}_1(\bfX')=\min_{\bfc\in\mathbb{R}^d}\sum_{\bfx\in \bfX'}\|\bfc-\bfx\|_1$, which is minimized at the \emph{median} of $\bfX'$,  and ${\sf cost}_1(\bfC_1,\ldots,\bfC_k) = \sum_{i=1}^k  {\sf cost}_1(\bfC_i)$, which we call the \emph{$k$-median} (or simply median) cost of the clustering.







 
\paragraph{Explainable clustering.} For a vector $\bfx\in\mathbb{R}^d$, we use $\bfx[i]$ to denote the $i$-th element (coordinate) of the vector for $i\in\{1,\ldots,d\}$. 
Let $\bfX$ be a collection of points of $\mathbb{R}^d$. For $i
\in\{1,\ldots,d\}$ and $\theta\in\mathbb{R}$, we define ${\sf Cut}_{i,\theta}(\bfX)=(\bfX_1,\bfX_2)$, where $\{\bfX_1,\bfX_2\}$ is a partition of $\bfX$ with 
\begin{equation*} 
\bfX_1=\{\bfx\in \bfX\mid \bfx[i]\leq \theta\} \text{ and }\bfX_2=\{\bfx\in \bfX\mid \bfx[i]>\theta\}.
\end{equation*}
Then, given a collection $\bfX\subseteq \mathbb{R}^d$ and a positive integer $k$, we cluster $\bfX$ as follows. 
If $k=1$, then $\bfX$ is the unique cluster.  If $k=2$, then we choose $i\in \{1,
\ldots,d\}$ and $\theta\in\mathbb{R}$ and construct two clusters $\bfC_1$ and $
\bfC_2$, where $(\bfC_1,\bfC_2)={\sf Cut}_{i,\theta}(\bfX)$. For $k>2$, we select $i\in \{1,
\ldots,d\}$ and $\theta\in\mathbb{R}$, and construct a partition $(\bfX_1,\bfX_2)={\sf Cut}_{i,\theta}(\bfX)$ of $\bfX$. 
Then clustering of $\bfX$ is defined recursively as the union of a $k_1$-clustering of $\bfX_1$ and a $k_2$-clustering of $\bfX_2$ for some 
  integers $k_1$ and $k_2$ such that $k_1+k_2=k$.
We say that a clustering $\{\bfC_1,\ldots,\bfC_k\}$ is an \emph{explainable $k$-clustering} of a collection of points $\bfX\subseteq \mathbb{R}^d$ if  $\bfC_1,\ldots,\bfC_k$ can be constructed by the described procedure. 

 




\paragraph{Threshold tree.} It is useful to represent an explainable $k$-clustering as 
a triple $(T,k,\varphi)$, called a \emph{threshold tree}, where $T$ is a rooted binary tree with $k$ leaves, where each nonleaf node has two children called \emph{left} and \emph{right}, respectively, and $\varphi\colon U\rightarrow \{1,
\ldots,d\}\times \mathbb{R}$, where $U$ is the set of nonleaf nodes of $T$. For each node $v$ of $T$, we compute a collection of points $\bfX_v\subseteq \bfX$. For the root $r$, $\bfX_r=\bfX$.   
Let $v$ be a nonleaf node of $T$ and let $u$ and $w$ be its left and right children, respectively,
and assume that $\bfX_v$ is constructed.  We compute  $(\bfX_u,\bfX_w)={\sf Cut}_{\varphi(v)}(\bfX)$. If $v$ is a leaf, then $\bfX_v$ is a cluster. 
A clustering $\{\bfC_1,\ldots,\bfC_k\}$ is an explainable $k$-clustering of a collection of points $\bfX\subseteq \mathbb{R}^d$ if there is a threshold tree $(T,k,\varphi)$ such that $\bfC_1,\ldots,\bfC_k$ are the clusters corresponding to the leaves of $T$. Note that $T$ is a full binary tree with $k$ leaves and the total number of such trees is the $(k-1)$-th Catalan number, which is upper bounded by $4^k$.  


For a collection $\bfX=\{\bfx_1,\ldots,\bfx_n\}$ of $n$ points and $i\in\{1,\ldots,d\}$, we denote by ${\sf coord}_i(\bfX)$ the set of distinct  values of $i$-th coordinates $\bfx_j[i]$ for $j\in\{1,\ldots,n\}$.  
It is easy to observe that in the construction of a threshold tree for a set of points $\bfX\subseteq \mathbb{R}^d$, it is sufficient to consider cuts ${\sf Cut}_{i,\theta}$ with $\theta\in{\sf coord}_i(\bfX)$; we call such values of $\theta$ and cuts  \emph{canonical}. We say that a threshold tree $(T,k,\varphi)$ for a collection of points $\bfX\subseteq \mathbb{R}^d$ is \emph{canonical}, if for every nonleaf node $u\in V(T)$, $\varphi(u)=(i,\theta)$ where $\theta\in {\sf coord}_i(\bfX)$. 
Throughout the paper we consider only canonical threshold trees. 

\paragraph{Parameterized complexity and ETH.} 
A \emph{parameterized problem} $\Pi$ is a subset of $\Sigma^*\times \mathbb{N}$, where $\Sigma$ is a finite alphabet. Thus, an instance of $\Pi$ is a pair $(I,k)$, where $I\subseteq\Sigma^*$ and $k$ is a nonnegative integer called a \emph{parameter}. It is said that a parameterized problem $\Pi$ is \emph{fixed-parameter tractable} (\ensuremath{\operatorClassFPT}\xspace) if it can be solved in $f(k)\cdot |I|^{\mathcal{O}(1)}$ time for some computable function $f(\cdot)$. 
The parameterized complexity theory also provides tools to refute the existence of an \ensuremath{\operatorClassFPT}\xspace algorithm for a parameterized problem. The standard way is to show that the considered problem is hard in the parameterized complexity classes \classW{1} or \classW{2}. We refer to the book~\cite{CyganFKLMPPS15} for the formal definitions of the parameterized complexity classes.  The basic complexity assumption  of the theory is that for the class  \ensuremath{\operatorClassFPT}\xspace, formed by all parameterized fixed-parameter tractable problems, $\ensuremath{\operatorClassFPT}\xspace\subset \classW{1}\subset \classW{2}$. The hardness is proved by demonstrating a parameterized reduction from a problem known to be hard in  the considered complexity class.  
A \emph{parameterized reduction} is a many-one reduction that takes an input $(I,k)$ of the first problem, and in $f(k)|I|^{\mathcal{O}(1)}$ time outputs an equivalent instance $(I',k')$ of the second problem with $k'\leq g(k)$, where $f(\cdot)$ and $g(\cdot)$ are computable functions. Another way to obtain lower bounds is to use the \emph{Exponential Time Hypothesis (ETH)} formulated by Impagliazzo, Paturi and Zane~\cite{ImpagliazzoP99,ImpagliazzoPZ01}. For an integer $k\geq 3$, let $q_k$ be the infimum of the real numbers $c$ such that the \textsc{$k$-Satisfiability} problem can be solved in time $\mathcal{O}(2^{c n})$, where $n$ is the number of variables. Exponential Time Hypothesis states that $\delta_3>3$. In particular, ETH implies that  \textsc{$k$-Satisfiability} cannot be solved in time $2^{o(n)}n^{\mathcal{O}(1)}$.


\subsection{A Polynomial-time $(k-1)$-Approximation}

In the optimization version of \probClustExpl, we are given a $k$-clustering $\mathcal{C}=\{\bfC_1,\ldots,\bfC_k\}$ of $\bfX$ in $\mathbb{R}^d$, and the goal is to find a minimum-sized subset $W\subseteq \bfX$ such that $\{\bfC_1\setminus W,\ldots,\bfC_k\setminus W\}$ is an explainable clustering. In the following, we design an approximation algorithm for this problem based on a greedy scheme. 

For any subset $W\subseteq \bfX$, let $\mathcal{C}-W=\{\bfC_1\setminus W,\ldots,\bfC_k\setminus W\}$. Also, for any subset $Y \subseteq \bfX$, define the clustering induced by $Y$ as $\mathcal{C}(Y)= \{\bfC_1\cap Y,\ldots,\bfC_k\cap Y\}$. Denote by OPT$(Y)$ the size of the minimum-sized subset $W$ such that the clustering $\mathcal{C}(Y)-W$ is explainable. First, we have the following simple observation which follows trivially from the definition of OPT$(.)$.  

\begin{observation}\label{obs:subsetcheaper}
 For any subset $Y \subseteq \bfX$, \emph{OPT}$(Y)\le $ \emph{OPT}$(\bfX)$. 
\end{observation}


\iffalse
\begin{lemma}\label{lem:hereditary}
 Consider any subset $\mathcal{C}'\subseteq \mathcal{C}$ of clusters such that $|\mathcal{C}'|\ge 2$ and let $X$ be the union of the points in the clusters of $\mathcal{C}'$. If $\mathcal{C}$ is explainable, then there exists $i\in \{1,
\ldots,d\}$ and $\theta\in\mathbb{R}$ such that ${\sf Cut}_{i,\theta}(X)=(X_1,X_2)$ has the property that for any cluster $\bfC_j \in \mathcal{C}'$, either $\bfC_j \subseteq X_1$ or $\bfC_j \subseteq X_2$. 
\end{lemma}
 
 \begin{proof}
  Let $B$ be a $d$-dimensional bounding hyperbox that contains all the points in $X$, and suppose $\mathcal{C}$ is explainable. As $\mathcal{C}$ is explainable there must be a set $S=\{(i,\theta)\}$ of cuts or hyperplanes that recursively partition the points in $P$ into the clusters in $\mathcal{C}$. Let $\mathcal{T}$ be the corresponding recursion tree -- each node $u$ is corresponding to a point set $X^u$ and has either no child or exactly two children. If a node has two children, the union of the children's point sets is the point set of the node, which is divided into two by some hyperplane in $S$. The number of leaves is $k$ which are corresponding to the $k$ clusters. The root is corresponding to $P$ and is at level 0 of $\mathcal{T}$. Level of a non-root node is one more than the level of its parent. 
  
 
  Now, consider the hyperplane $H=(i,\theta)$, corresponding to a node $u$ of $\mathcal{T}$, such that ${\sf Cut}_{i,\theta}(X^u)=(X_1^u,X_2^u)$ has the property that $X_1^u$ contains all the points of a cluster in $\mathcal{C}'$ and also $X_2^u$ contains all the points of a (different) cluster in $\mathcal{C}'$. Such a hyperplane must exist, as $B$ contains points from at least two clusters in $\mathcal{C}'$ and the leaves of $\mathcal{T}$ are clusters in $\mathcal{C}$. Note that $H$ must intersect $B$. Moreover, assume that $H$ is such a hyperplane corresponding to a node $u$   
 
  which is at the minimum level in $\mathcal{T}$. As $u$ is at the minimum level, $H$ also pierces $B$ and, hence, induces a partition of $B$ into two smaller hyperboxes $B_1$ and $B_2$. Note that by definition of $\mathcal{T}$, each cluster in $\mathcal{C}'$ must lie completely either in $B_1$ or in $B_2$. Then $X_1 \subseteq B_1$ and $X_2 \subseteq B_2$, where $(X_1,X_2)={\sf Cut}_{i,\theta}(X)$. It folllows that $(X_1,X_2)$ satisfies the property that for any cluster $\bfC_j \in \mathcal{C}'$, either $\bfC_j \subseteq X_1$ or $\bfC_j \subseteq X_2$.        
 \end{proof}

For any  $i\in \{1,
\ldots,d\}$, let $N_i$ be the set of values corresponding to the $i$-th coordinate of the points in $P$. Now, we have the following simple observation. 

\begin{observation}\label{obs:discretize}
 Consider any set of points $X$ and let ${\sf Cut}_{i,\theta}(X)=(X_1,X_2)$ for $i\in \{1,
\ldots,d\}$ and $\theta\in\mathbb{R}$. Also, suppose $x < \theta < y$ where $x,y \in N_i$ and there is no $z\in N_i$ such that $x < z < y$. Then ${\sf Cut}_{i,x}(X)=(X_1,X_2)$.   
\end{observation}

\begin{proof}
 The proof follows from the fact that $X_1=\{\bfX\in X\mid \bfX[i]\leq \theta\}=\{\bfX\in X\mid \bfX[i]\leq x\}$ and $X_2=X\setminus X_1$. 
\end{proof}
\fi

For any cut $(i,\theta)$ where $i\in \{1,
\ldots,d\}$ and $\theta\in{\sf coord}_i(\bfX)$, let $L(i,\theta)=\{\bfx\in \mathbb{R}^d\mid \bfx[i]\le \theta\}$ and  $R(i,\theta)=\{\bfx\in \mathbb{R}^d\mid \bfx[i]> \theta\}$. 

\begin{lemma}\label{lem:cheapcut}
 Consider any subset $Y\subseteq \bfX$ such that $\mathcal{C}(Y)$ contains at least two non-empty clusters. It is possible to select a cut $(i,\theta)$ for $i\in \{1,
\ldots,d\}$ and $\theta\in {\sf coord}_i(Y)$, and a subset $W\subseteq Y$, in polynomial time, such that (i) each cluster in $\mathcal{C}(Y)-W$ is fully contained in either $L(i,\theta)$ or in  $R(i,\theta)$, (ii) at least one cluster in $\mathcal{C}(Y)-W$ is in $L(i,\theta)$, (iii) at least one cluster in $\mathcal{C}(Y)-W$ is in $R(i,\theta)$ and (iv) size of $W$ is at most \emph{OPT}$(Y)$. 
\end{lemma}

Before we prove this lemma, we show how to use it to design the desired approximation algorithm. 

\paragraph{The Algorithm.} We start with the set of all points $\bfX$. We apply the algorithm in Lemma \ref{lem:cheapcut} with $Y=\bfX$ to find a cut $(i,\theta)$  and a subset $W_1\subseteq \bfX$ such that each cluster in $\mathcal{C}(\bfX)-W_1$ is fully contained in either $L(i,\theta)$ or in  $R(i,\theta)$. Let $\bfX_1=(\bfX\setminus W_1)\cap L(i,\theta)$ and $\bfX_2=(\bfX\setminus W_1)\cap R(i,\theta)$. We recursively apply the above step on both $\bfX_1$ and $\bfX_2$ separately. If at some level the point set is a subset of a single cluster, we simply return.  

The correctness of the above algorithm trivially follows from Lemma \ref{lem:cheapcut}. In particular, the recursion tree of the algorithm gives rise to the desired threshold tree. Also, the algorithm runs in polynomial time, as each successful cut $(i,\theta)$ can be found in polynomial time and the algorithm  finds only $k-1$ such cuts that separate the clusters. The last claim follows due to the properties (ii) and (iii) in Lemma \ref{lem:cheapcut}. 

Consider the threshold tree generated by the algorithm. For each internal node $u$, let $X_u$ be the corresponding points and $W_u$ be the points removed from $X_u$ for finding an explainable clustering of the points in $X_u\setminus W_u$. Note that we have at most $k-1$ such nodes. The total number of points removed from $\bfX$ for finding the explainable clustering is $\sum_u |W_u|$. By Lemma \ref{lem:cheapcut}, \[|W_u|\le  \text{OPT}(X_u).\] Now, as $X_u \subseteq \bfX$, by Observation \ref{obs:subsetcheaper}, OPT$(X_u)\le $ OPT$(\bfX)$. It follows that \[\sum_u |W_u|\le (k-1)\cdot \text{OPT}(\bfX). \]

\begin{theorem}\label{thm:k-appr}
 There is a polynomial-time $(k-1)$-approximation algorithm for the optimization version of \probClustExpl. 
\end{theorem}

By noting that OPT$(\bfX)=0$ if $\mathcal{C}$ is an explainable clustering, we obtain the following corollary.  

\begin{corollary}\label{cor:expl}
 Explainability of any given $k$-clustering in $\mathbb{R}^d$ can be tested in polynomial time. 
\end{corollary}

\begin{proof}[Proof of Lemma \ref{lem:cheapcut}]
 We probe all possible choices for cuts $(i,\theta)$ with  $i\in \{1,
\ldots,d\}$ and $\theta\in {\sf coord}_i(Y)$, and select one which incurs the minimum cost. We also select a subset $W$ of points to be removed w.r.t. each cut. The cost of such a cut is exactly the size of $W$. 

Fix a cut $(i,\theta)$. We have the following three cases. In the first case, for all clusters in $\mathcal{C}(Y)$, strictly more than half of the points are contained in $L(i,\theta)$. In this case select a cluster $\bfC$ which has the minimum intersection with $L(i,\theta)$. Put all the points in $\bfC\cap L(i,\theta)$ into $W$. Also, for any other cluster $\bfC' \in \mathcal{C}(Y)$, put the points in $\bfC'\cap R(i,\theta)$ into $W$. The second case is symmetric to the first one -- for all clusters in $\mathcal{C}(Y)$, strictly more than half of the points are contained in $R(i,\theta)$. In this case we again select a cluster $\bfC$ which has the minimum intersection with $R(i,\theta)$. Put all the points in $\bfC\cap R(i,\theta)$ into $W$. Also, for any other cluster $\bfC' \in \mathcal{C}(Y)$, put the points in $\bfC'\cap L(i,\theta)$ into $W$. In both of the above cases, the first three desired properties are satisfied for $\mathcal{C}(Y)-W$. In the third case, for each cluster $\bfC \in \mathcal{C}(Y)$, add the smaller part among $\bfC\cap L(i,\theta)$ and $\bfC\cap R(i,\theta)$ to $W$. In case $|\bfC\cap L(i,\theta)|=|\bfC\cap R(i,\theta)|$, we break the tie in a way so that properties (ii) and (iii) are satisfied. As $\mathcal{C}(Y)$ contains at least two clusters this can always be done. Moreover, property (i) is trivially satisfied.  

In the above we showed that for all the choices of the cuts, it is possible to select $W$ so that the first three properties are satisfied. Let $w_m$ be the minimum size of the set $W$ over all cuts. As we select a cut for which the size of $W$ is minimized, it is sufficient to show that $w_m \le $ OPT$(Y)$. 

Let $k'$ be the number of clusters in $\mathcal{C}(Y)$. Consider any optimal set $W^*$ for $Y$ such that $\mathcal{C}(Y)-W^*$ is explainable. Let $(i^*,\theta^*)$ be the canonical cut corresponding to the root of the threshold tree corresponding to the explainable clustering $\mathcal{C}(Y)-W^*$. Such a cut exists, as $\mathcal{C}(Y)$ contains at least two clusters. Let $\widehat{W}$ be the set selected in our algorithm corresponding to the cut $(i^*,\theta^*)$. In the first of the above mentioned three cases, suppose $W^*$ does not contain the part $\bfC\cap L(i^*,\theta^*)$ fully for any of the $k'$ clusters $\bfC\in \mathcal{C}(Y)$. In other words, $\mathcal{C}(Y)-W^*$ contains points from each such part $\bfC\cap L(i^*,\theta^*)$. But, then even after choosing the root cut $(i^*,\theta^*)$ we still need $k'$ more cuts to separate the points in $(Y\setminus W^*)\cap L(i^*,\theta^*)$, which contains points from all the $k'$ clusters. However, by definition, the threshold tree must use only $k'$ cuts and hence we reach to a contradiction. Hence, $\bfC^*\cap L(i^*,\theta^*)$  must be fully contained in $W^*$ for some $\bfC^*\in \mathcal{C}(Y)$. In this case,  our algorithm adds the points in $\bfC\cap L(i^*,\theta^*)$ to $\widehat{W}$ such that the size $|\bfC\cap L(i^*,\theta^*)|$ is minimized over all $\bfC\in \mathcal{C}(Y)$ and for any other cluster $\bfC' \in \mathcal{C}(Y)$, we put the points in $\bfC'\cap R(i^*,\theta^*)$ into $\widehat{W}$. Thus, $|\widehat{W}| \le |W^*|=$ OPT$(Y)$. The proof for the second case is the same as the one for the first case. We discuss the proof for the third case. Consider the  clusters $\bfC\in \mathcal{C}(Y)$ such that both $\bfC\cap L(i^*,\theta^*)$ and $\bfC\cap R(i^*,\theta^*)$ are non-empty. Note that these are the only clusters whose points are put into $\widehat{W}$. But, then $W^*$ must contain all the points from at least one of the parts $\bfC\cap L(i^*,\theta^*)$ and $\bfC\cap R(i^*,\theta^*)$. For each such cluster $\bfC$, we add the smaller part among $\bfC\cap L(i,\theta)$ and $\bfC\cap R(i,\theta)$ to $\widehat{W}$. Hence, in this case also $|\widehat{W}| \le |W^*|=$ OPT$(Y)$. The lemma follows by noting that $w_m\le |\widehat{W}|$.        
\end{proof}


\subsection{Hardness of approximation}
\label{sec:hardclexpl}

We show that the \probClustExpl problem remains hard when the number of points to delete $s$ is small. Specifically, we provide a parameter-preserving reduction from \textsc{Hitting Set} to \probClustExpl that transfers known results about hardness of approximation for the \textsc{Hitting Set} problem
to \probClustExpl.
Recall that in the \textsc{Hitting Set} problem, the input is a family of sets $\mathcal{A}$ over a universe $U$ together with a parameter $\ell$, and the goal is to decide whether there is a set $H \subset U$ of size at most $\ell$ such that $H$ has non-empty intersection with each set in $\mathcal{A}$. Whenever we consider \ensuremath{\operatorClassFPT}\xspace algorithms for \textsc{Hitting Set}, we assume that the problem is parameterized by $\ell$.

The starting point of our reduction is the following result for \textsc{Hitting Set} by Karthik, Laekhanukit, and Manurangsi~\cite{KarthikLM19}.

\begin{theorem}[Theorem 1.4 in~\cite{KarthikLM19}]
    Assuming ETH, no $f(\ell)(|U||\mathcal{A}|)^{o(\ell)}$-time algorithm can approximate \textsc{Hitting Set} to within a factor of $F(\ell)$, for any functions $f$ and $F$ of $\ell$.
    \label{thm:hitting_set_hardness}
\end{theorem}

Note that Theorem 1.4 in~\cite{KarthikLM19} is stated for the \textsc{Dominating Set} problem, however by the standard parameter-preserving reduction from \textsc{Dominating Set} to \textsc{Hitting Set} (see e.g. Theorem 13.28 in \cite{CyganFKLMPPS15}), the statement above immediately follows.

Now, intuitively, given an instance $(U, \mathcal{A}, \ell)$ of \probHS, our reduction constructs clusters $\bfC_0$, \ldots, $\bfC_m$ in $\mathbb{R}^{\sum_{j \in [m]} |A_j|}$. The clusters $\bfC_1$, \ldots, $\bfC_m$ represent the sets in the family $\mathcal{A}=\{A_1,\ldots,A_m\}$, and $\bfC_0$ is a special cluster that needs to be separated from each of $\bfC_1$, \ldots, $\bfC_m$ so that the clustering is explainable. The separation can only be performed by removing special points from $\bfC_0$ each of which corresponds to an element of the universe $U$. Removing such a point allows for separation between $\bfC_0$ and each $\bfC_j$ such that the corresponding set $A_j$ contains the corresponding universe element. The two clusters can be separated along a special coordinate where only that special point ``blocks'' the separation. This is the crux of the reduction, which results in the following theorem.

\begin{theorem}
    For any functions $f$ and $F$, there is no algorithm that approximates \probClustExpl within a factor of $F(s)$ in time $f(s) (nd)^{o(s)}$, unless ETH fails. 
    \label{thm:explanation_hardness}
\end{theorem}



\begin{proof}
    We show a reduction from \textsc{Hitting Set}. Consider an instance of \textsc{Hitting Set} over a universe $U$ with a family of sets $\mathcal{A}$ and the size of the target hitting set $\ell$. We construct the following instance of \probClustExpl, denote $|\mathcal{A}| = m$. The target dimension $d$ is equal to the sum of set sizes in the family $\mathcal{A}$, $d = \sum_{S \in \mathcal{A}} |S|$. The number of clusters in the constructed instance is $m + 1$, and for clarity we denote them by $\bfC_0$, \ldots, $\bfC_m$. The target parameter $s$, that is, the number of points to remove from the cluster, is set exactly to $\ell$.

    Now we describe how the clusters are composed. Intuitively, the clusters $\bfC_1$, \ldots, $\bfC_m$ represent the sets in the family $\mathcal{A}$, and $\bfC_0$ is a special cluster that needs to be separated from each of $\bfC_1$, \ldots, $\bfC_m$ so that the clustering is explainable. The separation can only be performed by removing special points from $\bfC_0$ that each correspond to an element of the universe $U$. Removing such a point allows for separation between $\bfC_0$ and each $\bfC_j$ such that the corresponding set $S_j$ contains the corresponding universe element. The two clusters can be separated along a special coordinate where only that special point ``blocks'' the separation. This is the crux of the reduction.

    Formally, we define point sets $\bfC_0$, \ldots, $\bfC_m$ in terms of coordinates in $\mathbb{R}^d$. The constructed instance is binary, that is, only values zero and one are used in the vectors.
    Order arbitrarily the sets in the family $\mathcal{A}$, i.e. $\mathcal{A} = \{S_1, \ldots, S_m\}$, for each $j \in [m]$, the cluster $\bfC_j$ corresponds to the set $S_j$. The $d$ coordinates are partitioned between the $m$ sets in the following way. For each $j \in [m]$, denote $I_j = \left[\sum_{j' < j} |S_{j'}| + 1, \sum_{j' \le j} |S_{j'}|\right] \subset [d]$. That is, the first range of coordinates $I_1$ are the first $|S_1|$ coordinates, $I_2$ are the following $|S_2|$ coordinates, and so on. Now, for $j \in [m]$, the set $\bfC_j$ consists of $F(\ell) \cdot \ell + 1$ identical points $\bfw_j$, where $\bfw_j[i] = 1$ for $j \in I_j$ and $\bfw_j[i] = 0$ for $j \notin I_j$.
    The set $\bfC_0$ consists of two parts, $\bfC_0 = \bfO \cup \bfV$. First, there are $F(\ell) \cdot \ell + 1$ identical zero vectors in $\bfC_0$, we denote this set by $\bfO$. Second, for each element $u$ in the universe $U$, there is a point $\bfv_u$ in $\bfV$. The coordinates of $\bfv_u$ are set so that for every $j \in [m]$ such that $u \in S_j$, there is exactly one coordinate in $I_j$ where $\bfv_u$ is set to one, and this coordinate is unique for each $u \in S_j$. If for $j \in [m]$, $u \notin S_j$, then $\bfv_u[i] = 0$ for all $i \in I_j$. More specifically, for each $S_j \in \mathcal{A}$, order arbitrarily the elements of $S_j$, $S_j = \{u_1, \ldots, u_{|S_j|}\}$. For $i \in [|S_j|]$, $\bfv_{u_i}[\sum_{j' < j} |S_j| + i] = 1$. After performing the above for each $j \in [m]$, set the remaining coordinates of each vector $\bfv_u$ to zero. This concludes the construction.
    
    Now we show that an $F(s)$-approximate solution to the constructed \probClustExpl instance would imply an $F(\ell)$-approximate solution to the original \textsc{Hitting Set} instance.
    Specifically, we prove the following.
    \begin{claim}
    Whenever there exists a set $\bfW \subset \bfX$ of size at most $F(s) \cdot s$ such that $\{\bfC_0 \setminus \bfW, \ldots, \bfC_m \setminus \bfW\}$ is an explainable clustering, there also exists a set $H \subset U$ that is a hitting set of $\mathcal{A}$ and $|H| \le |\bfW|$.
    On the other hand, for any hitting set $H \subset U$ there exists a solution $\bfW$ to the \probClustExpl instance such that $|\bfW| = |H|$.
    \label{claim:approximation}
    \end{claim}

    \begin{proof}
    In the forward direction, consider such a set $\bfW \subset \bfX$. We may assume that $\bfW \cap \bfC_j = \emptyset$ for each $j \in [m]$ and $\bfW \cap \bfO = \emptyset$, since these sets consist of $F(s) \cdot s + 1$ identical points, and replacing $\bfW$ by a smaller set not intersecing the sets above is still a solution. Thus, $\bfW \subset \bfV$. Recall that each of the points in $\bfV$ corresponds to an element of the universe $U$ in the \textsc{Hitting Set} instance. Denote by $H$ the subset of $U$ corresponding to $\bfW$, that is, $H = \{u \in U : \bfv_u \in \bfW\}$. Clearly $|H| \le |\bfW|$, we claim that $H$ is a solution to the \textsc{Hitting Set} instance.
   
    Consider a threshold tree that provides explanation for $\{\bfC_0 \setminus \bfW, \ldots, \bfC_m \setminus \bfW\}$, by the above all these clusters are non-empty. For $j \in \{0\}\cup [m]$, denote $\bfC_j' = \bfC_j \setminus \bfW$, recall that we assume $\bfC_j = \bfC_j'$ for all $j \in [m]$. We now show that for any $j \in [m]$, the set $H$ has a non-empty intersection with the set $S_j$.
    Since the tree represents the clustering, for each $j \in [m]$, there exists a cut in the tree that separates $\bfC_j'$ and $\bfC_0'$. That is, there exists a cut ${\sf Cut}_{i, \theta} (\bfX') = (\bfX'_1, \bfX'_2)$ in the tree for $\bfX'  \subset \bfX$ such that $\bfC'_j \cup \bfC'_0 \subset \bfX'$, and $\bfC'_j \subset \bfX'_1$, $\bfC'_0 \subset \bfX'_2$ (w.l.o.g). The dimension of the cut $i$ necessarily belongs to $I_j$, since in all other coordinates $\bfC_j'$ is indistinguishable from $\bfO$. For $i \in I_j$, there exists a point $\bfv_u \in \bfV$ such that $\bfv_u[i] = 1$ and $u \in S_j$. Since the points in $\bfC_j'$ are set to one in this coordinate and the points in $\bfO$ to zero, $\bfv_u$ has to be in $\bfW$. Thus, by construction, $u \in H$, and the set $S_j$ is hit by $H$.

    In the other direction, assume that there exists a hitting set $H$ for $\mathcal{A}$.
    We set $\bfW$ to be the corresponding to $H$ subset of $\bfV$, $|\bfW| = |H|$. For every $j \in [m]$, there exists an element $u \in H$ such 
    that $u \in S_j$. Consider the corresponding element $\bfv_u$ of $\bfW$, and the coordinate $i \in I_j$ where $\bfv_u[i] = 1$.
    By construction,  $\bfv_u$ is the only element of $\bfC_0$ that has one in the $i$-th coordinate, and for any $j' \ne j$, the elements of $\bfC_{j'}$ have zero in thes coordinate. Thus, cutting over the dimension $i$ with the threshold $\theta$ set to zero separates $\bfC_j$ from all the other clusters. Finally, the threshold tree for $\{\bfC_0 \setminus \bfW, \ldots, \bfC_m \setminus \bfW\}$ is constructed by separating out each $\bfC_j$ one by one via the corresponding cut.
    \end{proof}
    
    The theorem follows easily from Claim~\ref{claim:approximation}. Namely, assume there exists an $F(s)$-approximate algorithm for \probClustExpl with running time $f(s) (nd)^{o(s)}$, that is, an algorithm that correctly decides either that the input instance has a solution of size at most $F(s) \cdot s$, or that it has no solution of size at most $s$. We construct an $F(\ell)$-approximate algorithm for \textsc{Hitting Set} as follows. First, construct a \probClustExpl instance via the reduction above. Second, run the \probClustExpl algorithm
    on that instance and output its answer. If it returns that there is a solution $\bfW$ of size at most $F(s) \cdot s$, by Claim~\ref{claim:approximation}
    there is a solution $H$ to the original \textsc{Hitting Set} instance of size at most $F(s) \cdot s = F(\ell) \cdot \ell$.
    If there is no solution to the \probClustExpl instance of size at most $s$, by Claim~\ref{claim:approximation} there also cannot be a solution to the \textsc{Hitting Set} instance of size at most $s = \ell$. This shows that the constructed algorithm provides indeed an $F(\ell)$-approximation for the \textsc{Hitting Set} problem. Finally, its running time can be bounded as $f(s) (nd)^{o(s)} = f(\ell)(F(\ell)|U| m)^{o(\ell)}$,
    which contradicts Theorem~\ref{thm:hitting_set_hardness} unless ETH fails.
\end{proof}

\section{Acknowledgments}



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