Open Problems in Sublinear Algorithms - User contributions [en]

Open Problems:42

2016-04-11T17:40:13Z

89.109.239.35: Typo

{{Header
|source=bertinoro11
|who=Noga Alon
}}
For a graph $G$, a $k$-disc around a vertex $v$ is the subgraph induced by
the vertices that are at distance at most $k$ from $v$.
The frequency vector of $k$-discs of $G$
is a vector indexed by all
isomorphism types of
$k$-discs of vertices in $G$ which counts, for each such isomorphism
type $K$,
the fraction of $k$-discs of vertices of $G$ that are isomorphic to $K$.
The following is a known fact observed in a discussion with
Lovász. It is proved by
a simple compactness argument.

'''Fact:''' For every $\epsilon > 0$, there is an $M=M(\epsilon)$ such that
for every $3$-regular graph $G$, there exists a $3$-regular
graph $H$ on at most $M(\epsilon)$
vertices (independent on $|V(G)|$), such that variation distance
between the frequency vector of the $100$-discs
in $G$ and the frequency vector
of the $100$-discs
in $H$ is at most $\epsilon$.

'''Question:''' Find ''any'' explicit estimate on $M(\epsilon)$. Nothing is
currently known.

'''Update:''' Partial progress has been made in [http://drops.dagstuhl.de/opus/volltexte/2015/5336/ FPS15] where it is proved that if all $k$-discs in $G$ are trees (i.e., $G$ has girth greater than $2k+1$), then $|V(H)| \leq \frac{10^{10^{50}}}{\epsilon}$. The result generalizes to arbitrary degree bound $d$ and $k \geq 0$, and $H$ can be constructed in constant time at the cost of roughly an additional factor $\epsilon^{-1}$. It was sketched by the same authors at the [http://math.ucsd.edu/~sbuss/SPB_Workshops/AlgCPTC_1.html Workshop on Algorithms in Communication Complexity, Property Testing and Combinatorics in Moscow in 2016] that one can also construct $H$ if $G$ is planar by using the planar separator theorem. In this case, $|V(H)| \in O(\epsilon^{-4})$.

Open Problems:42

2016-04-11T17:38:48Z

89.109.239.35: Partial progress

{{Header
|source=bertinoro11
|who=Noga Alon
}}
For a graph $G$, a $k$-disc around a vertex $v$ is the subgraph induced by
the vertices that are at distance at most $k$ from $v$.
The frequency vector of $k$-discs of $G$
is a vector indexed by all
isomorphism types of
$k$-discs of vertices in $G$ which counts, for each such isomorphism
type $K$,
the fraction of $k$-discs of vertices of $G$ that are isomorphic to $K$.
The following is a known fact observed in a discussion with
Lovász. It is proved by
a simple compactness argument.

'''Fact:''' For every $\epsilon > 0$, there is an $M=M(\epsilon)$ such that
for every $3$-regular graph $G$, there exists a $3$-regular
graph $H$ on at most $M(\epsilon)$
vertices (independent on $|V(G)|$), such that variation distance
between the frequency vector of the $100$-discs
in $G$ and the frequency vector
of the $100$-discs
in $H$ is at most $\epsilon$.

'''Question:''' Find ''any'' explicit estimate on $M(\epsilon)$. Nothing is
currently known.

'''Update:''' Partial progress has been made in [http://drops.dagstuhl.de/opus/volltexte/2015/5336/ FPS15] where it is proved that if all $k$-discs in $G$ are trees (i.e., $G$ has girth greater than $2k+1$), then $|H| \leq \frac{10^{10^{50}}}{\epsilon}$. The result generalizes to arbitrary degree bound $d$ and $k \geq 0$, and $H$ can be constructed in constant time at the cost of roughly an additional factor $\epsilon^{-1}$. It was sketched by the same authors at the [http://math.ucsd.edu/~sbuss/SPB_Workshops/AlgCPTC_1.html Workshop on Algorithms in Communication Complexity, Property Testing and Combinatorics in Moscow in 2016] that one can also construct $H$ if $G$ is planar by using the planar separator theorem. In this case, $|H| \in O(\epsilon^{-4})$.