Number of paths in high-girth graphs
3
$begingroup$
In a graph $G$ with girth $g$, say $g = Omega(log n)$, can we deduce an upper bound on the number of paths between two nodes $u$ and $v$?
combinatorics graph-theory
asked Dec 18 '18 at 21:07
HTVHTV
1067
$endgroup$
add a comment |
3
$begingroup$
In a graph $G$ with girth $g$, say $g = Omega(log n)$, can we deduce an upper bound on the number of paths between two nodes $u$ and $v$?
combinatorics graph-theory
asked Dec 18 '18 at 21:07
HTVHTV
1067
$endgroup$
$begingroup$
This is a great question. If you don't care about regularity, take any bounded-degree graph with lots of paths between two vertices and replace each edge with a path of $log n$ vertices, then the girth will be large. This isn't very interesting, but I do suspect that the girth bound may not limit the number of paths
$endgroup$
– Mike
Dec 19 '18 at 0:35
1
$begingroup$
My suspect is that if the max-degree is $Delta$, then the number of paths is bounded by $Delta^{O(n/g)}$.
$endgroup$
– HTV
Dec 19 '18 at 1:13
1
$begingroup$
I'm not so sure. Let us consider the set of all paths of $G$ nevermind the ending points (as there are only $n^2$ such choices for the starting and ending points this is OK). Let us call a walk legal from a starting vertex $u_0$ if it is a path. I would imagine that for "many" legal walks $W$ of length $ell < epsilon n$ in $G$, there are at least (say) $frac{k}{3}$ ways to extend $W$ to a legal walk of length $ell+1$. This should imply $exp(Omega(n))$ total paths which would imply a $u-v$ pair such that there are $exp(Omega(n))$ paths between $u$ and $v$
$endgroup$
– Mike
Dec 19 '18 at 21:12
1
$begingroup$
We could also assume that $G$ is an expander of course as there are high-girth expanders, and there are even explicit constructions of such graphs
$endgroup$
– Mike
Dec 19 '18 at 21:16
add a comment |
3
3
3
1
$begingroup$
In a graph $G$ with girth $g$, say $g = Omega(log n)$, can we deduce an upper bound on the number of paths between two nodes $u$ and $v$?
combinatorics graph-theory
asked Dec 18 '18 at 21:07
HTVHTV
1067
$endgroup$
In a graph $G$ with girth $g$, say $g = Omega(log n)$, can we deduce an upper bound on the number of paths between two nodes $u$ and $v$?
combinatorics graph-theory
combinatorics graph-theory
asked Dec 18 '18 at 21:07
HTVHTV
1067
asked Dec 18 '18 at 21:07
HTVHTV
1067
asked Dec 18 '18 at 21:07
HTVHTV
1067
asked Dec 18 '18 at 21:07
HTVHTV
1067
asked Dec 18 '18 at 21:07
HTVHTV
1067
1067
$begingroup$
This is a great question. If you don't care about regularity, take any bounded-degree graph with lots of paths between two vertices and replace each edge with a path of $log n$ vertices, then the girth will be large. This isn't very interesting, but I do suspect that the girth bound may not limit the number of paths
$endgroup$
– Mike
Dec 19 '18 at 0:35
1
$begingroup$
My suspect is that if the max-degree is $Delta$, then the number of paths is bounded by $Delta^{O(n/g)}$.
$endgroup$
– HTV
Dec 19 '18 at 1:13
1
$begingroup$
I'm not so sure. Let us consider the set of all paths of $G$ nevermind the ending points (as there are only $n^2$ such choices for the starting and ending points this is OK). Let us call a walk legal from a starting vertex $u_0$ if it is a path. I would imagine that for "many" legal walks $W$ of length $ell < epsilon n$ in $G$, there are at least (say) $frac{k}{3}$ ways to extend $W$ to a legal walk of length $ell+1$. This should imply $exp(Omega(n))$ total paths which would imply a $u-v$ pair such that there are $exp(Omega(n))$ paths between $u$ and $v$
$endgroup$
– Mike
Dec 19 '18 at 21:12
1
$begingroup$
We could also assume that $G$ is an expander of course as there are high-girth expanders, and there are even explicit constructions of such graphs
$endgroup$
– Mike
Dec 19 '18 at 21:16
add a comment |
$begingroup$
This is a great question. If you don't care about regularity, take any bounded-degree graph with lots of paths between two vertices and replace each edge with a path of $log n$ vertices, then the girth will be large. This isn't very interesting, but I do suspect that the girth bound may not limit the number of paths
$endgroup$
– Mike
Dec 19 '18 at 0:35
1
$begingroup$
My suspect is that if the max-degree is $Delta$, then the number of paths is bounded by $Delta^{O(n/g)}$.
$endgroup$
– HTV
Dec 19 '18 at 1:13
1
$begingroup$
I'm not so sure. Let us consider the set of all paths of $G$ nevermind the ending points (as there are only $n^2$ such choices for the starting and ending points this is OK). Let us call a walk legal from a starting vertex $u_0$ if it is a path. I would imagine that for "many" legal walks $W$ of length $ell < epsilon n$ in $G$, there are at least (say) $frac{k}{3}$ ways to extend $W$ to a legal walk of length $ell+1$. This should imply $exp(Omega(n))$ total paths which would imply a $u-v$ pair such that there are $exp(Omega(n))$ paths between $u$ and $v$
$endgroup$
– Mike
Dec 19 '18 at 21:12
1
$begingroup$
We could also assume that $G$ is an expander of course as there are high-girth expanders, and there are even explicit constructions of such graphs
$endgroup$
– Mike
Dec 19 '18 at 21:16
$begingroup$
This is a great question. If you don't care about regularity, take any bounded-degree graph with lots of paths between two vertices and replace each edge with a path of $log n$ vertices, then the girth will be large. This isn't very interesting, but I do suspect that the girth bound may not limit the number of paths
$endgroup$
– Mike
Dec 19 '18 at 0:35
$begingroup$
This is a great question. If you don't care about regularity, take any bounded-degree graph with lots of paths between two vertices and replace each edge with a path of $log n$ vertices, then the girth will be large. This isn't very interesting, but I do suspect that the girth bound may not limit the number of paths
$endgroup$
– Mike
Dec 19 '18 at 0:35
1
1
$begingroup$
My suspect is that if the max-degree is $Delta$, then the number of paths is bounded by $Delta^{O(n/g)}$.
$endgroup$
– HTV
Dec 19 '18 at 1:13
$begingroup$
My suspect is that if the max-degree is $Delta$, then the number of paths is bounded by $Delta^{O(n/g)}$.
$endgroup$
– HTV
Dec 19 '18 at 1:13
1
1
$begingroup$
I'm not so sure. Let us consider the set of all paths of $G$ nevermind the ending points (as there are only $n^2$ such choices for the starting and ending points this is OK). Let us call a walk legal from a starting vertex $u_0$ if it is a path. I would imagine that for "many" legal walks $W$ of length $ell < epsilon n$ in $G$, there are at least (say) $frac{k}{3}$ ways to extend $W$ to a legal walk of length $ell+1$. This should imply $exp(Omega(n))$ total paths which would imply a $u-v$ pair such that there are $exp(Omega(n))$ paths between $u$ and $v$
$endgroup$
– Mike
Dec 19 '18 at 21:12
$begingroup$
I'm not so sure. Let us consider the set of all paths of $G$ nevermind the ending points (as there are only $n^2$ such choices for the starting and ending points this is OK). Let us call a walk legal from a starting vertex $u_0$ if it is a path. I would imagine that for "many" legal walks $W$ of length $ell < epsilon n$ in $G$, there are at least (say) $frac{k}{3}$ ways to extend $W$ to a legal walk of length $ell+1$. This should imply $exp(Omega(n))$ total paths which would imply a $u-v$ pair such that there are $exp(Omega(n))$ paths between $u$ and $v$
$endgroup$
– Mike
Dec 19 '18 at 21:12
1
1
$begingroup$
We could also assume that $G$ is an expander of course as there are high-girth expanders, and there are even explicit constructions of such graphs
$endgroup$
– Mike
Dec 19 '18 at 21:16
$begingroup$
We could also assume that $G$ is an expander of course as there are high-girth expanders, and there are even explicit constructions of such graphs
$endgroup$
– Mike
Dec 19 '18 at 21:16
add a comment |
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$begingroup$
This is a great question. If you don't care about regularity, take any bounded-degree graph with lots of paths between two vertices and replace each edge with a path of $log n$ vertices, then the girth will be large. This isn't very interesting, but I do suspect that the girth bound may not limit the number of paths
$endgroup$
– Mike
Dec 19 '18 at 0:35
1
$begingroup$
My suspect is that if the max-degree is $Delta$, then the number of paths is bounded by $Delta^{O(n/g)}$.
$endgroup$
– HTV
Dec 19 '18 at 1:13
1
$begingroup$
I'm not so sure. Let us consider the set of all paths of $G$ nevermind the ending points (as there are only $n^2$ such choices for the starting and ending points this is OK). Let us call a walk legal from a starting vertex $u_0$ if it is a path. I would imagine that for "many" legal walks $W$ of length $ell < epsilon n$ in $G$, there are at least (say) $frac{k}{3}$ ways to extend $W$ to a legal walk of length $ell+1$. This should imply $exp(Omega(n))$ total paths which would imply a $u-v$ pair such that there are $exp(Omega(n))$ paths between $u$ and $v$
$endgroup$
– Mike
Dec 19 '18 at 21:12
1
$begingroup$
We could also assume that $G$ is an expander of course as there are high-girth expanders, and there are even explicit constructions of such graphs
$endgroup$
– Mike
Dec 19 '18 at 21:16