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Find a sub-graph where each node in the sub-graph has minimum degree d.
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Given an undirected graph $G$ and integer $d$, find a sub-graph $H$ of $G$ such that every node in $H$ has minimum degree $d$.
I wonder if there is any condition to decide whether $H$ exists or not. If such $H$ exists, is there any polynomial algorithm to find such $H$.
graph-theory
asked Feb 2 at 23:14
ParadoxParadox
909
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add a comment |
$begingroup$
Given an undirected graph $G$ and integer $d$, find a sub-graph $H$ of $G$ such that every node in $H$ has minimum degree $d$.
I wonder if there is any condition to decide whether $H$ exists or not. If such $H$ exists, is there any polynomial algorithm to find such $H$.
graph-theory
asked Feb 2 at 23:14
ParadoxParadox
909
$endgroup$
add a comment |
$begingroup$
Given an undirected graph $G$ and integer $d$, find a sub-graph $H$ of $G$ such that every node in $H$ has minimum degree $d$.
I wonder if there is any condition to decide whether $H$ exists or not. If such $H$ exists, is there any polynomial algorithm to find such $H$.
graph-theory
asked Feb 2 at 23:14
ParadoxParadox
909
$endgroup$
Given an undirected graph $G$ and integer $d$, find a sub-graph $H$ of $G$ such that every node in $H$ has minimum degree $d$.
I wonder if there is any condition to decide whether $H$ exists or not. If such $H$ exists, is there any polynomial algorithm to find such $H$.
graph-theory
graph-theory
asked Feb 2 at 23:14
ParadoxParadox
909
asked Feb 2 at 23:14
ParadoxParadox
909
asked Feb 2 at 23:14
ParadoxParadox
909
asked Feb 2 at 23:14
ParadoxParadox
909
asked Feb 2 at 23:14
ParadoxParadox
909
909
add a comment |
add a comment |
1 Answer
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Remove all vertices from $G$ of degree less than $d$ and call the resulting graph $G_1$; if there is such a subgraph $H$ of $G$, then $H$ contains no vertex in $G setminus G_1$. Then remove from $G_1$ all vertices of degree(in the smaller graph $G_1$) of degree less than $d$ and call the resulting graph $G_2$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_1 setminus G_2$. Continue on in a similar fashion: For general $ell$,
remove from $G_{ell}$ all vertices of degree(in $G_{ell}$) of degree less than $d$ and call the resulting graph $G_{ell+1}$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_{ell} setminus G_{ell+1}$. Stop this process when either the resulting graph is empty or when every vertex in the resulting graph $H'$ has degree at least $d$.
If there is indeed a subgraph of $G$ where every vertex has degree $d$, then the resulting graph $H'$ will be nonempty (as in each step above you didn't remove any vertices from that graph) and (as every vertex in $H'$ has degree at least $d$) is the graph you are seeking.
answered Feb 2 at 23:29
MikeMike
4,641512
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1 Answer
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$begingroup$
Remove all vertices from $G$ of degree less than $d$ and call the resulting graph $G_1$; if there is such a subgraph $H$ of $G$, then $H$ contains no vertex in $G setminus G_1$. Then remove from $G_1$ all vertices of degree(in the smaller graph $G_1$) of degree less than $d$ and call the resulting graph $G_2$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_1 setminus G_2$. Continue on in a similar fashion: For general $ell$,
remove from $G_{ell}$ all vertices of degree(in $G_{ell}$) of degree less than $d$ and call the resulting graph $G_{ell+1}$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_{ell} setminus G_{ell+1}$. Stop this process when either the resulting graph is empty or when every vertex in the resulting graph $H'$ has degree at least $d$.
If there is indeed a subgraph of $G$ where every vertex has degree $d$, then the resulting graph $H'$ will be nonempty (as in each step above you didn't remove any vertices from that graph) and (as every vertex in $H'$ has degree at least $d$) is the graph you are seeking.
answered Feb 2 at 23:29
MikeMike
4,641512
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add a comment |
$begingroup$
Remove all vertices from $G$ of degree less than $d$ and call the resulting graph $G_1$; if there is such a subgraph $H$ of $G$, then $H$ contains no vertex in $G setminus G_1$. Then remove from $G_1$ all vertices of degree(in the smaller graph $G_1$) of degree less than $d$ and call the resulting graph $G_2$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_1 setminus G_2$. Continue on in a similar fashion: For general $ell$,
remove from $G_{ell}$ all vertices of degree(in $G_{ell}$) of degree less than $d$ and call the resulting graph $G_{ell+1}$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_{ell} setminus G_{ell+1}$. Stop this process when either the resulting graph is empty or when every vertex in the resulting graph $H'$ has degree at least $d$.
If there is indeed a subgraph of $G$ where every vertex has degree $d$, then the resulting graph $H'$ will be nonempty (as in each step above you didn't remove any vertices from that graph) and (as every vertex in $H'$ has degree at least $d$) is the graph you are seeking.
answered Feb 2 at 23:29
MikeMike
4,641512
$endgroup$
add a comment |
$begingroup$
Remove all vertices from $G$ of degree less than $d$ and call the resulting graph $G_1$; if there is such a subgraph $H$ of $G$, then $H$ contains no vertex in $G setminus G_1$. Then remove from $G_1$ all vertices of degree(in the smaller graph $G_1$) of degree less than $d$ and call the resulting graph $G_2$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_1 setminus G_2$. Continue on in a similar fashion: For general $ell$,
remove from $G_{ell}$ all vertices of degree(in $G_{ell}$) of degree less than $d$ and call the resulting graph $G_{ell+1}$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_{ell} setminus G_{ell+1}$. Stop this process when either the resulting graph is empty or when every vertex in the resulting graph $H'$ has degree at least $d$.
If there is indeed a subgraph of $G$ where every vertex has degree $d$, then the resulting graph $H'$ will be nonempty (as in each step above you didn't remove any vertices from that graph) and (as every vertex in $H'$ has degree at least $d$) is the graph you are seeking.
answered Feb 2 at 23:29
MikeMike
4,641512
$endgroup$
Remove all vertices from $G$ of degree less than $d$ and call the resulting graph $G_1$; if there is such a subgraph $H$ of $G$, then $H$ contains no vertex in $G setminus G_1$. Then remove from $G_1$ all vertices of degree(in the smaller graph $G_1$) of degree less than $d$ and call the resulting graph $G_2$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_1 setminus G_2$. Continue on in a similar fashion: For general $ell$,
remove from $G_{ell}$ all vertices of degree(in $G_{ell}$) of degree less than $d$ and call the resulting graph $G_{ell+1}$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_{ell} setminus G_{ell+1}$. Stop this process when either the resulting graph is empty or when every vertex in the resulting graph $H'$ has degree at least $d$.
If there is indeed a subgraph of $G$ where every vertex has degree $d$, then the resulting graph $H'$ will be nonempty (as in each step above you didn't remove any vertices from that graph) and (as every vertex in $H'$ has degree at least $d$) is the graph you are seeking.
answered Feb 2 at 23:29
MikeMike
4,641512
edited Feb 2 at 23:40
answered Feb 2 at 23:29
MikeMike
4,641512
answered Feb 2 at 23:29
MikeMike
4,641512
answered Feb 2 at 23:29
MikeMike
4,641512
4,641512
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