@@ -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.