Commit c2313c73 by Nikolaj Tatti

### 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$r \in S_{vi}$, \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 \label{eq:boundb} b_{r}[v, i] \geq 0. b_{r + 1}[v, i] \geq 0. \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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