Commit c2313c73 authored by Nikolaj Tatti's avatar Nikolaj Tatti
Browse files

polishing the text

parent 363b3a78
......@@ -44,20 +44,20 @@ The resulting intervals are indeed forming a timeline with a total span of~$\ell
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 $[2n + 1, 2n + 1]$,
then this yields a timeline $\tl^*$ covering $G$.
The total span intervals $\tl^*$ is $n$.
Thus, $\spn{\tl^*} \leq n$.
then this yields a timeline $\tl$ covering $G$.
The total span of intervals in $\tl$ is $n$.
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 implies that $1 \notin \aint{u_i}$ and so $1 \in \aint{v_i}$.
In summary, if $0 \in \aint{v_i}$, then $\len{\aint{v_i}} = 1$.
This implies that if $\spn{\tl^*} \leq \ell$,
then we have at most $\ell$ vertices covered at $0$.
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}
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}
$k$-\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.
Luckily, we can consider meaningful subproblems.
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 $v\in V$,
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 $m_v$,
......@@ -218,8 +218,8 @@ and we test the median
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
by $1/4$, that is, only $\bigO{\log m}$ iterations are needed,
% We can determine the medians $m(i)$ and the sizes $\abs{W(i)}$
% 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
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
yielding an $\bigO{m \log m}$ algorithm.
In our experiments
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.
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 $p=0.5$.
\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.
\spara{Evaluation metrics.}
......
\section{Algorithms \prbint and {\Large $k$}-\prbint}
\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.
To simplify the notation,
......@@ -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.
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.
\end{proposition}
......@@ -222,7 +222,7 @@ at the beginning of $j$-th round they are equal to
%{e_i \in Q(v) \atop i < j} \alp
%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 $b[v]$.
\begin{lemma}
......
......@@ -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 $\alpha_e \geq 0$
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}
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.
In order to do that, we first bound the counters.
\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}
\label{eq:bounda}
a_{r + 1}[v, i] \leq \theta_{vr}
......@@ -248,7 +248,7 @@ For each vertex $v$, index $i = 1, \ldots, k$, and iteration $r \in S_{vi}$,
and
\begin{equation}
\label{eq:boundb}
b_{r}[v, i] \geq 0.
b_{r + 1}[v, i] \geq 0.
\end{equation}
\end{lemma}
......@@ -263,7 +263,7 @@ we have
\[
b_{r + 1}[v, i] = \min(b_{r}[v, i], \eta_{vr}) - \alpha_r \geq 0.
\]%
% This proves the claim.
This proves the claim.
\end{proof}
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,
making the dual solution feasible.
\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
$\adj{v; r, i} \leq \eta_{vr}$ and $\adj{v; r, i - 1} \leq \theta_{vr}$.
\end{lemma}
\begin{proof}
Eq.~\ref{eq:maxa} and Eq.~\ref{eq:bounda} gives us
Eq.~\ref{eq:maxa} and Eq.~\ref{eq:bounda} give us
\[
\adj{v; r, i - 1} = a_{r + 1}[v, i] \leq \theta_{vr}.
\]
Moreover, $\adj{v; r, i - 1}$ remains constant in the later rounds.
Eq.~\ref{eq:maxb} guarantees that
$b_{r + 1}[v, i] \leq \eta_{vr} - \adj{v; r, i}$. Combining with Eq.~\ref{eq:boundb} gives us
Let $s$ be the last index in $S_{vi}$. Then
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.
\end{proof}
\noindent
......
......@@ -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 $u$.
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,
\[
\diam{\ktl} = \max_{j \in [1,k]}\max_{u \in V} \len{\kaint{u}{j}}.
\]
......@@ -101,7 +101,7 @@ Now the extensions of Problems \prbsum and \prbmax to the case of multiple inter
\begin{problem}{\em (}$k$-\prbmax{\em )}
\label{problem:kmax}
Given a temporal network $G = (V, E)$
Given a temporal network $G = (V, E)$,
find a timeline $\ktl=\{\kaint{v}{j}\}_{v\in V, j\in[1,k]}$
that covers $G$ and minimizes the max-span $\diam{\ktl}$.
\end{problem}
......
......@@ -75,7 +75,7 @@ From an application point-of-view,
our work is loosely related with papers that aim to
process large amounts of data and create maps
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}
and scientific publications~\cite{shahaf2012metro}.
However, their approach is not directly comparable to ours,
......
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