Commit c2313c73 authored by Nikolaj Tatti's avatar Nikolaj Tatti
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polishing the text

parent 363b3a78
...@@ -44,20 +44,20 @@ The resulting intervals are indeed forming a timeline with a total span of~$\ell ...@@ -44,20 +44,20 @@ The resulting intervals are indeed forming a timeline with a total span of~$\ell
To prove the other direction, To prove the other direction,
first note that if we cover each $v_i$ by an interval $[0, 1]$ and each $u_i$ by an interval $[2n + 1, 2n + 1]$, first note that if we cover each $v_i$ by an interval $[0, 1]$ and each $u_i$ by an interval $[2n + 1, 2n + 1]$,
then this yields a timeline $\tl^*$ covering $G$. then this yields a timeline $\tl$ covering $G$.
The total span intervals $\tl^*$ is $n$. The total span of intervals in $\tl$ is $n$.
Thus, $\spn{\tl^*} \leq n$. Thus, $\spn{\tl^*} \leq \spn{\tl} = n$.
This guarantees that if $0 \in \aint{v_i}$, then $2n + 1 \notin \aint{v_i}$, so $2n + 1 \in \aint{u_i}$. This guarantees that if $0 \in \aint{v_i}$, then $2n + 1 \notin \aint{v_i}$, so $2n + 1 \in \aint{u_i}$.
This implies that $1 \notin \aint{u_i}$ and so $1 \in \aint{v_i}$. This implies that $1 \notin \aint{u_i}$ and so $1 \in \aint{v_i}$.
In summary, if $0 \in \aint{v_i}$, then $\len{\aint{v_i}} = 1$. In summary, if $0 \in \aint{v_i}$, then $\len{\aint{v_i}} = 1$.
This implies that if $\spn{\tl^*} \leq \ell$, This implies that if $\spn{\tl^*} \leq \ell$,
then we have at most $\ell$ vertices covered at $0$. then we have at most $\ell$ vertices covered at $0$.
Let $U$ be the set of those vertices. Let $U$ be the set of those vertices.
Since $\tl^*$ is timeline covering $G$, then $U$ is a vertex cover for $H$. Since $\tl^*$ is a timeline covering $G$, then $U$ is a vertex cover for $H$.
\end{proof} \end{proof}
Our second result states that not only the problems with multiple intervals are Our second result states that not only the problems with multiple intervals are
intractrabel, but we cannot even approximate them. intractrable, but we cannot even approximate them.
\begin{proposition} \begin{proposition}
$k$-\prbsum and $k$-\prbmax are inapproximable, unless $\poly = \np$. $k$-\prbsum and $k$-\prbmax are inapproximable, unless $\poly = \np$.
...@@ -102,7 +102,7 @@ on the fact that we are dealing with edges and not hyper-edges. ...@@ -102,7 +102,7 @@ on the fact that we are dealing with edges and not hyper-edges.
Luckily, we can consider meaningful subproblems. Luckily, we can consider meaningful subproblems.
Assume that we are given a temporal network Assume that we are given a temporal network
$G = (V, E)$ and we also given a set of time point $\set{m_v}_{v \in V}$, $G = (V, E)$ and we also given a set of time points $\set{m_v}_{v \in V}$,
i.e., one time point $m_v$ for each vertex $v\in V$, i.e., one time point $m_v$ for each vertex $v\in V$,
and we are asked whether we can find an optimal activity timeline $\tl=\set{\aint{u}}_{u\in V}$ and we are asked whether we can find an optimal activity timeline $\tl=\set{\aint{u}}_{u\in V}$
so that the interval $\aint{v}$ of vertex $v$ contains the corresponding time point $m_v$, so that the interval $\aint{v}$ of vertex $v$ contains the corresponding time point $m_v$,
...@@ -218,8 +218,8 @@ and we test the median ...@@ -218,8 +218,8 @@ and we test the median
of all $m(i)$ (weighted by $\abs{W(i)}$) as a new budget. of all $m(i)$ (weighted by $\abs{W(i)}$) as a new budget.
We can show that at each iteration $\sum \abs{W(i)}$ is reduced We can show that at each iteration $\sum \abs{W(i)}$ is reduced
by $1/4$, that is, only $\bigO{\log m}$ iterations are needed, by $1/4$, that is, only $\bigO{\log m}$ iterations are needed,
% We can determine the medians $m(i)$ and the sizes $\abs{W(i)}$ We can determine the medians $m(i)$ and the sizes $\abs{W(i)}$
% in linear time since $T$ is sorted, and we can determine the weighted median in linear time by using a modified median-of-medians algorithm. This leads to running time in linear time since $T$ is sorted, and we can determine the weighted median in linear time by using a modified median-of-medians algorithm. This leads to running time
yielding an $\bigO{m \log m}$ algorithm. yielding an $\bigO{m \log m}$ algorithm.
In our experiments In our experiments
we use a straightforward binary search by testing $(U + L) / 2$ as a budget. we use a straightforward binary search by testing $(U + L) / 2$ as a budget.
......
...@@ -25,7 +25,7 @@ even better than the ground-truth solution. ...@@ -25,7 +25,7 @@ even better than the ground-truth solution.
In all cases \synth has $10\,000$ interactions in total. In all cases \synth has $10\,000$ interactions in total.
Unless specified, we report results averaged over $100$ runs and test a fairly complex case of overlap parameter $p=0.5$. Unless specified, we report results averaged over $100$ runs and test a fairly complex case of overlap parameter $p=0.5$.
\spara{$k$ intervals.} \spara{multiple intervals.}
For the problem version with $k>1$ activity intervals per vertex we generate the \synthk dataset. We use the same construction method as for \synth, but plant $k=10$ activity intervals for each vertex with $i=10$ interactions each. Similar to \synth, we control overlaps using a parameter $p\in[0,1]$. \synthk also has $10\,000$ interactions in total. For the problem version with $k>1$ activity intervals per vertex we generate the \synthk dataset. We use the same construction method as for \synth, but plant $k=10$ activity intervals for each vertex with $i=10$ interactions each. Similar to \synth, we control overlaps using a parameter $p\in[0,1]$. \synthk also has $10\,000$ interactions in total.
\spara{Evaluation metrics.} \spara{Evaluation metrics.}
......
\section{Algorithms \prbint and {\Large $k$}-\prbint} \section{Algorithms \prbint and {\Large $k$}-\prbint}
\label{sec:middle} \label{sec:middle}
In this section we present a 2-approximation linear-time algorithm for the In this section we present 2-approximation linear-time algorithms for the
problems \prbint and $k$-\prbint. problems \prbint and $k$-\prbint.
To simplify the notation, To simplify the notation,
...@@ -82,7 +82,7 @@ Our next result shows that a maximal solution can be used to obtain a 2-approxim ...@@ -82,7 +82,7 @@ Our next result shows that a maximal solution can be used to obtain a 2-approxim
Consider a maximal solution $\set{\alpha_e}_{e \in E}$ to the dual program. Consider a maximal solution $\set{\alpha_e}_{e \in E}$ to the dual program.
Define a set of intervals $\tl = \set{\aint{v}}$ by $\aint{v} = [\min \set{\te{i} \mid i \in X_v}, \max \set{\te{i} \mid i \in X_v}]$, where Define a set of intervals $\tl = \set{\aint{v}}$ by $\aint{v} = [\min \set{\te{i} \mid i \in X_v}, \max \set{\te{i} \mid i \in X_v}]$, where
\[ \[
X_v = \set{m_v} \cup \set{i \in \inds{v} \mid \adj{v; i} = |\te{i} - m_v|}. X_v = \set{i \in \inds{v} \mid \adj{v; i} = |\te{i} - m_v|}.
\] \]
Then $\tl$ is a 2-approximation solution for the problem \prbint. Then $\tl$ is a 2-approximation solution for the problem \prbint.
\end{proposition} \end{proposition}
...@@ -222,7 +222,7 @@ at the beginning of $j$-th round they are equal to ...@@ -222,7 +222,7 @@ at the beginning of $j$-th round they are equal to
%{e_i \in Q(v) \atop i < j} \alp %{e_i \in Q(v) \atop i < j} \alp
%a[v] = \sum_{e_i \in Q(v) \atop i < j} \alpha_i %a[v] = \sum_{e_i \in Q(v) \atop i < j} \alpha_i
\] \]
Moreover, weights $\alpha_i = 0$ for $i \geq j$. Moreover, we have weights $\alpha_i = 0$, for $i \geq j$.
The following lemma tells us how to update $\alpha_j$ using $a[v]$ and $b[v]$. The following lemma tells us how to update $\alpha_j$ using $a[v]$ and $b[v]$.
\begin{lemma} \begin{lemma}
......
...@@ -145,7 +145,7 @@ Let~$\tlopt$ be the optimal solution. Then ...@@ -145,7 +145,7 @@ Let~$\tlopt$ be the optimal solution. Then
\] \]
where the second equality follows from the definition of $X_{vi}$ and $Y_{vi}$, the first inequality follows from the fact that $\alpha_e \geq 0$ where the second equality follows from the definition of $X_{vi}$ and $Y_{vi}$, the first inequality follows from the fact that $\alpha_e \geq 0$
and the intervals in $\tl$ do not intersect, and the last inequality follows from primal-dual theory. and the intervals in $\tl$ do not intersect, and the last inequality follows from primal-dual theory.
This proves the claim. %This proves the claim.
\end{proof} \end{proof}
We now move to describe the algorithm for obtaining a left-maximal solution. We now move to describe the algorithm for obtaining a left-maximal solution.
...@@ -240,7 +240,7 @@ Our next step is to prove the feasibility of the output of $k$-\algmaxgreedy. ...@@ -240,7 +240,7 @@ Our next step is to prove the feasibility of the output of $k$-\algmaxgreedy.
In order to do that, we first bound the counters. In order to do that, we first bound the counters.
\begin{lemma} \begin{lemma}
For each vertex $v$, index $i = 1, \ldots, k$, and iteration $r \in S_{vi}$, For each vertex $v$, index $i = 1, \ldots, k$, and $r \in S_{vi}$,
\begin{equation} \begin{equation}
\label{eq:bounda} \label{eq:bounda}
a_{r + 1}[v, i] \leq \theta_{vr} a_{r + 1}[v, i] \leq \theta_{vr}
...@@ -248,7 +248,7 @@ For each vertex $v$, index $i = 1, \ldots, k$, and iteration $r \in S_{vi}$, ...@@ -248,7 +248,7 @@ For each vertex $v$, index $i = 1, \ldots, k$, and iteration $r \in S_{vi}$,
and and
\begin{equation} \begin{equation}
\label{eq:boundb} \label{eq:boundb}
b_{r}[v, i] \geq 0. b_{r + 1}[v, i] \geq 0.
\end{equation} \end{equation}
\end{lemma} \end{lemma}
...@@ -263,7 +263,7 @@ we have ...@@ -263,7 +263,7 @@ we have
\[ \[
b_{r + 1}[v, i] = \min(b_{r}[v, i], \eta_{vr}) - \alpha_r \geq 0. b_{r + 1}[v, i] = \min(b_{r}[v, i], \eta_{vr}) - \alpha_r \geq 0.
\]% \]%
% This proves the claim. This proves the claim.
\end{proof} \end{proof}
To prove the feasibility, we first show that $\alpha_r \geq 0$. To prove the feasibility, we first show that $\alpha_r \geq 0$.
...@@ -293,22 +293,24 @@ Next lemma shows that $\set{\alpha_r}$ satisfies the constraints, ...@@ -293,22 +293,24 @@ Next lemma shows that $\set{\alpha_r}$ satisfies the constraints,
making the dual solution feasible. making the dual solution feasible.
\begin{lemma} \begin{lemma}
Let $v \in V$, $i = 1 \ldots, k + 1$, and $r \in S_{vi}$. Let $v \in V$, $i = 1, \ldots, k + 1$, and $r \in S_{vi}$.
Then Then
$\adj{v; r, i} \leq \eta_{vr}$ and $\adj{v; r, i - 1} \leq \theta_{vr}$. $\adj{v; r, i} \leq \eta_{vr}$ and $\adj{v; r, i - 1} \leq \theta_{vr}$.
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
Eq.~\ref{eq:maxa} and Eq.~\ref{eq:bounda} gives us Eq.~\ref{eq:maxa} and Eq.~\ref{eq:bounda} give us
\[ \[
\adj{v; r, i - 1} = a_{r + 1}[v, i] \leq \theta_{vr}. \adj{v; r, i - 1} = a_{r + 1}[v, i] \leq \theta_{vr}.
\] \]
Moreover, $\adj{v; r, i - 1}$ remains constant in the later rounds.
Eq.~\ref{eq:maxb} guarantees that Let $s$ be the last index in $S_{vi}$. Then
$b_{r + 1}[v, i] \leq \eta_{vr} - \adj{v; r, i}$. Combining with Eq.~\ref{eq:boundb} gives us Eq.~\ref{eq:maxb} and Eq.~\ref{eq:boundb} state that
\[ \[
0 \leq b_{r + 1}[v, i] \leq \eta_{vr} - \adj{v; r, i}. 0 \leq b_{s + 1}[v, i] \leq \eta_{vr} - \adj{v; r, i}.
\] \]
Moreover, the sum $\adj{v; r, i}$ remains constant in the later rounds.
Thus, $\set{\alpha_r}$ is a feasible dual solution. Thus, $\set{\alpha_r}$ is a feasible dual solution.
\end{proof} \end{proof}
\noindent \noindent
......
...@@ -85,7 +85,7 @@ We define the $k$-\emph{sum-span} of a $k$-activity timeline $\ktl$ by ...@@ -85,7 +85,7 @@ We define the $k$-\emph{sum-span} of a $k$-activity timeline $\ktl$ by
\] \]
where $\len{\kaint{u}{j}} = \keint{u}{j}-\ksint{u}{j}$ is the duration of the $j$-th activity interval of vertex $u$. where $\len{\kaint{u}{j}} = \keint{u}{j}-\ksint{u}{j}$ is the duration of the $j$-th activity interval of vertex $u$.
The {\em max-span} of the timeline $\ktl$ is defined similarly as the duration of the longest interval: The {\em max-span} of the timeline $\ktl$ is defined similarly as the duration of the longest interval,
\[ \[
\diam{\ktl} = \max_{j \in [1,k]}\max_{u \in V} \len{\kaint{u}{j}}. \diam{\ktl} = \max_{j \in [1,k]}\max_{u \in V} \len{\kaint{u}{j}}.
\] \]
...@@ -101,7 +101,7 @@ Now the extensions of Problems \prbsum and \prbmax to the case of multiple inter ...@@ -101,7 +101,7 @@ Now the extensions of Problems \prbsum and \prbmax to the case of multiple inter
\begin{problem}{\em (}$k$-\prbmax{\em )} \begin{problem}{\em (}$k$-\prbmax{\em )}
\label{problem:kmax} \label{problem:kmax}
Given a temporal network $G = (V, E)$ Given a temporal network $G = (V, E)$,
find a timeline $\ktl=\{\kaint{v}{j}\}_{v\in V, j\in[1,k]}$ find a timeline $\ktl=\{\kaint{v}{j}\}_{v\in V, j\in[1,k]}$
that covers $G$ and minimizes the max-span $\diam{\ktl}$. that covers $G$ and minimizes the max-span $\diam{\ktl}$.
\end{problem} \end{problem}
......
...@@ -75,7 +75,7 @@ From an application point-of-view, ...@@ -75,7 +75,7 @@ From an application point-of-view,
our work is loosely related with papers that aim to our work is loosely related with papers that aim to
process large amounts of data and create maps process large amounts of data and create maps
that present the available information in a succinct and easy-to-understand manner. that present the available information in a succinct and easy-to-understand manner.
Shahaf and co-authors have considered this problem Shahaf et al.\ have considered this problem
in the context of news articles~\cite{shahaf2012trains,shahaf2013information} in the context of news articles~\cite{shahaf2012trains,shahaf2013information}
and scientific publications~\cite{shahaf2012metro}. and scientific publications~\cite{shahaf2012metro}.
However, their approach is not directly comparable to ours, However, their approach is not directly comparable to ours,
......
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