Fundamental Problems on Bounded-Treewidth Graphs: The Real Source of Hardness
It is known for many algorithmic problems that if a tree decomposition of width $t$ is given in the input, then the problem can be solved with exponential dependence on $t$. A line of research by Lokshtanov, Marx, and Saurabh [SODA 2011] produced lower bounds showing that in many cases known algorit...
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Zusammenfassung: | It is known for many algorithmic problems that if a tree decomposition of
width $t$ is given in the input, then the problem can be solved with
exponential dependence on $t$. A line of research by Lokshtanov, Marx, and
Saurabh [SODA 2011] produced lower bounds showing that in many cases known
algorithms achieve the best possible exponential dependence on $t$, assuming
the SETH. The main message of our paper is showing that the same lower bounds
can be obtained in a more restricted setting: a graph consisting of a block of
$t$ vertices connected to components of constant size already has the same
hardness as a general tree decomposition of width $t$.
Formally, a $(\sigma,\delta)$-hub is a set $Q$ of vertices such that every
component of $Q$ has size at most $\sigma$ and is adjacent to at most $\delta$
vertices of $Q$.
$\bullet$ For every $\epsilon> 0$, there are $\sigma,\delta> 0$ such that
Independent Set/Vertex Cover cannot be solved in time $(2-\epsilon)^p\cdot n$,
even if a $(\sigma,\delta)$-hub of size $p$ is given in the input, assuming the
SETH. This matches the earlier tight lower bounds parameterized by the width of
the tree decomposition. Similar tight bounds are obtained for Odd Cycle
Transversal, Max Cut, $q$-Coloring, and edge/vertex deletions versions of
$q$-Coloring.
$\bullet$ For every $\epsilon>0$, there are $\sigma,\delta> 0$ such that
Triangle-Partition cannot be solved in time $(2-\epsilon)^p\cdot n$, even if a
$(\sigma,\delta)$-hub of size $p$ is given in the input, assuming the Set Cover
Conjecture (SCC). In fact, we prove that this statement is equivalent to the
SCC, thus it is unlikely that this could be proved assuming the SETH.
$\bullet$ For Dominating Set, we can prove a non-tight lower bound ruling out
$(2-\epsilon)^p\cdot n^{O(1)}$ algorithms, assuming either the SETH or the SCC,
but this does not match the $3^p\cdot n^{O(1)}$ upper bound. |
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DOI: | 10.48550/arxiv.2402.07331 |