@@ -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.