Show that the number of connected components are less or equal of the degree of vertex in a cut set. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Proving if G is a 3-regular graph, then the size of edge cut equals size of min size of vertex cutGiven a simple connected bipartite graph $G$ with degree of vertices equal to $k$, where $kge 2$. Prove that there is no cut vertex exist in $G$.Number of connected components for a graph of binary n-tuples.Is a connected graph in which every vertex with degree at least two being a cut vertex necessarily a tree?In a 2-connected graph find minimum no of edgesShow that in a graph with $n$ vertices, if $delta + Delta ge n - 1 $, then the graph must be connected.Number of connected components in a even graphWhat is the minimum number of edges a graph can have with strongly connected components of at least $3$ vertices each?Proving $k(G) leq k'(G) $ where $k(G)$ is the minumum vertex cut and $k'(G)$ is minimum edge cut.Show that if $kappa(G) leq 1$ then $lambda(G) leq fracDelta(G)2$
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Show that the number of connected components are less or equal of the degree of vertex in a cut set.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Proving if G is a 3-regular graph, then the size of edge cut equals size of min size of vertex cutGiven a simple connected bipartite graph $G$ with degree of vertices equal to $k$, where $kge 2$. Prove that there is no cut vertex exist in $G$.Number of connected components for a graph of binary n-tuples.Is a connected graph in which every vertex with degree at least two being a cut vertex necessarily a tree?In a 2-connected graph find minimum no of edgesShow that in a graph with $n$ vertices, if $delta + Delta ge n - 1 $, then the graph must be connected.Number of connected components in a even graphWhat is the minimum number of edges a graph can have with strongly connected components of at least $3$ vertices each?Proving $k(G) leq k'(G) $ where $k(G)$ is the minumum vertex cut and $k'(G)$ is minimum edge cut.Show that if $kappa(G) leq 1$ then $lambda(G) leq fracDelta(G)2$
$begingroup$
The exercise is defined as it follows:
Let $G$ connected and $W subseteq V(G)$ a cut set with minimum cardinality. Show that if $x in W$ then $omega(G-W) leq delta(x)$.
Where $omega$ is the number of connected components in $G$.
Proof: Since $W$ is a cut set with minimum cardinality we've got that $|W| = kappa(G)$ thus, each cut point is contained in at most one connected component, therefore we have: $$ omega(G - W) leq kappa(G) \ kappa(G) leq delta(G) \ delta(G) leq delta(x) \ therefore omega(G - W) leq delta(x) $$.
I'm not sure about the first argument, where I argue that each cut point is in at most one connected component, also I don't see where I use the hypothesis that if $ x in W$, why should I use it? Is the proof wrong? If it is, there are any hints?
graph-theory graph-connectivity
asked Apr 9 at 3:01
Eric ToporekEric Toporek
10910
$endgroup$
add a comment |
$begingroup$
The exercise is defined as it follows:
Let $G$ connected and $W subseteq V(G)$ a cut set with minimum cardinality. Show that if $x in W$ then $omega(G-W) leq delta(x)$.
Where $omega$ is the number of connected components in $G$.
Proof: Since $W$ is a cut set with minimum cardinality we've got that $|W| = kappa(G)$ thus, each cut point is contained in at most one connected component, therefore we have: $$ omega(G - W) leq kappa(G) \ kappa(G) leq delta(G) \ delta(G) leq delta(x) \ therefore omega(G - W) leq delta(x) $$.
I'm not sure about the first argument, where I argue that each cut point is in at most one connected component, also I don't see where I use the hypothesis that if $ x in W$, why should I use it? Is the proof wrong? If it is, there are any hints?
graph-theory graph-connectivity
asked Apr 9 at 3:01
Eric ToporekEric Toporek
10910
$endgroup$
add a comment |
$begingroup$
The exercise is defined as it follows:
Let $G$ connected and $W subseteq V(G)$ a cut set with minimum cardinality. Show that if $x in W$ then $omega(G-W) leq delta(x)$.
Where $omega$ is the number of connected components in $G$.
Proof: Since $W$ is a cut set with minimum cardinality we've got that $|W| = kappa(G)$ thus, each cut point is contained in at most one connected component, therefore we have: $$ omega(G - W) leq kappa(G) \ kappa(G) leq delta(G) \ delta(G) leq delta(x) \ therefore omega(G - W) leq delta(x) $$.
I'm not sure about the first argument, where I argue that each cut point is in at most one connected component, also I don't see where I use the hypothesis that if $ x in W$, why should I use it? Is the proof wrong? If it is, there are any hints?
graph-theory graph-connectivity
asked Apr 9 at 3:01
Eric ToporekEric Toporek
10910
$endgroup$
The exercise is defined as it follows:
Let $G$ connected and $W subseteq V(G)$ a cut set with minimum cardinality. Show that if $x in W$ then $omega(G-W) leq delta(x)$.
Where $omega$ is the number of connected components in $G$.
Proof: Since $W$ is a cut set with minimum cardinality we've got that $|W| = kappa(G)$ thus, each cut point is contained in at most one connected component, therefore we have: $$ omega(G - W) leq kappa(G) \ kappa(G) leq delta(G) \ delta(G) leq delta(x) \ therefore omega(G - W) leq delta(x) $$.
I'm not sure about the first argument, where I argue that each cut point is in at most one connected component, also I don't see where I use the hypothesis that if $ x in W$, why should I use it? Is the proof wrong? If it is, there are any hints?
graph-theory graph-connectivity
graph-theory graph-connectivity
asked Apr 9 at 3:01
Eric ToporekEric Toporek
10910
asked Apr 9 at 3:01
Eric ToporekEric Toporek
10910
edited Apr 11 at 7:09
Eric Toporek
asked Apr 9 at 3:01
Eric ToporekEric Toporek
10910
asked Apr 9 at 3:01
Eric ToporekEric Toporek
10910
asked Apr 9 at 3:01
Eric ToporekEric Toporek
10910
10910
add a comment |
add a comment |
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