Suppose a graph $G$ is connected with $n$ vertices and $e$ edges. If $n geq 3$ and $G$ has exactly one cycle,...
$begingroup$
Suppose a graph $G$ is connected with $n$ vertices and $e$ edges. If $n ge 3$ and $G$ has exactly one cycle, prove that $e = n$.
graph-theory
edited Dec 8 '18 at 12:51
amWhy
192k28225439
asked Dec 8 '18 at 12:31
Siddiqa AlhaykiSiddiqa Alhayki
72
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add a comment |
$begingroup$
Suppose a graph $G$ is connected with $n$ vertices and $e$ edges. If $n ge 3$ and $G$ has exactly one cycle, prove that $e = n$.
graph-theory
edited Dec 8 '18 at 12:51
amWhy
192k28225439
asked Dec 8 '18 at 12:31
Siddiqa AlhaykiSiddiqa Alhayki
72
$endgroup$
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
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– Shaun
Dec 8 '18 at 12:43
add a comment |
$begingroup$
Suppose a graph $G$ is connected with $n$ vertices and $e$ edges. If $n ge 3$ and $G$ has exactly one cycle, prove that $e = n$.
graph-theory
edited Dec 8 '18 at 12:51
amWhy
192k28225439
asked Dec 8 '18 at 12:31
Siddiqa AlhaykiSiddiqa Alhayki
72
$endgroup$
Suppose a graph $G$ is connected with $n$ vertices and $e$ edges. If $n ge 3$ and $G$ has exactly one cycle, prove that $e = n$.
graph-theory
graph-theory
edited Dec 8 '18 at 12:51
amWhy
192k28225439
asked Dec 8 '18 at 12:31
Siddiqa AlhaykiSiddiqa Alhayki
72
edited Dec 8 '18 at 12:51
amWhy
192k28225439
asked Dec 8 '18 at 12:31
Siddiqa AlhaykiSiddiqa Alhayki
72
edited Dec 8 '18 at 12:51
amWhy
192k28225439
edited Dec 8 '18 at 12:51
amWhy
192k28225439
edited Dec 8 '18 at 12:51
amWhy
192k28225439
192k28225439
asked Dec 8 '18 at 12:31
Siddiqa AlhaykiSiddiqa Alhayki
72
asked Dec 8 '18 at 12:31
Siddiqa AlhaykiSiddiqa Alhayki
72
asked Dec 8 '18 at 12:31
Siddiqa AlhaykiSiddiqa Alhayki
72
72
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 8 '18 at 12:43
add a comment |
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 8 '18 at 12:43
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 8 '18 at 12:43
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 8 '18 at 12:43
add a comment |
2 Answers
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Let $C$ be the only cycle in $G$. Consider any two adjacent vertices $u$ & $v$ that belong to $C$. Even if I remove the edge between $u$ & $v$, $G$ will remain connected. Let $G'$ be the new graph after removing the edge $uv$. $G'$ is connected, and has no cycle (bcoz we removed the $uv$ edge), hence $G'$ is a tree. $G'$ has $n-1$ edges $implies$ $G$ has $n$ edges.
answered Dec 8 '18 at 12:47
Ankit KumarAnkit Kumar
1,352219
$endgroup$
add a comment |
$begingroup$
Consider the spanning tree of the graph $G$, The number of edges in the spanning tree is $n-1$. As $G$ has only one cycle, we can add only one more edge to the spanning tree to get the cycle(as any edges added to the spanning tree create a cycle), so $G$ has $n$ edges.
answered Dec 8 '18 at 19:39
nafhgoodnafhgood
1,801422
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $C$ be the only cycle in $G$. Consider any two adjacent vertices $u$ & $v$ that belong to $C$. Even if I remove the edge between $u$ & $v$, $G$ will remain connected. Let $G'$ be the new graph after removing the edge $uv$. $G'$ is connected, and has no cycle (bcoz we removed the $uv$ edge), hence $G'$ is a tree. $G'$ has $n-1$ edges $implies$ $G$ has $n$ edges.
answered Dec 8 '18 at 12:47
Ankit KumarAnkit Kumar
1,352219
$endgroup$
add a comment |
$begingroup$
Let $C$ be the only cycle in $G$. Consider any two adjacent vertices $u$ & $v$ that belong to $C$. Even if I remove the edge between $u$ & $v$, $G$ will remain connected. Let $G'$ be the new graph after removing the edge $uv$. $G'$ is connected, and has no cycle (bcoz we removed the $uv$ edge), hence $G'$ is a tree. $G'$ has $n-1$ edges $implies$ $G$ has $n$ edges.
answered Dec 8 '18 at 12:47
Ankit KumarAnkit Kumar
1,352219
$endgroup$
add a comment |
$begingroup$
Let $C$ be the only cycle in $G$. Consider any two adjacent vertices $u$ & $v$ that belong to $C$. Even if I remove the edge between $u$ & $v$, $G$ will remain connected. Let $G'$ be the new graph after removing the edge $uv$. $G'$ is connected, and has no cycle (bcoz we removed the $uv$ edge), hence $G'$ is a tree. $G'$ has $n-1$ edges $implies$ $G$ has $n$ edges.
answered Dec 8 '18 at 12:47
Ankit KumarAnkit Kumar
1,352219
$endgroup$
Let $C$ be the only cycle in $G$. Consider any two adjacent vertices $u$ & $v$ that belong to $C$. Even if I remove the edge between $u$ & $v$, $G$ will remain connected. Let $G'$ be the new graph after removing the edge $uv$. $G'$ is connected, and has no cycle (bcoz we removed the $uv$ edge), hence $G'$ is a tree. $G'$ has $n-1$ edges $implies$ $G$ has $n$ edges.
answered Dec 8 '18 at 12:47
Ankit KumarAnkit Kumar
1,352219
answered Dec 8 '18 at 12:47
Ankit KumarAnkit Kumar
1,352219
answered Dec 8 '18 at 12:47
Ankit KumarAnkit Kumar
1,352219
answered Dec 8 '18 at 12:47
Ankit KumarAnkit Kumar
1,352219
1,352219
add a comment |
add a comment |
$begingroup$
Consider the spanning tree of the graph $G$, The number of edges in the spanning tree is $n-1$. As $G$ has only one cycle, we can add only one more edge to the spanning tree to get the cycle(as any edges added to the spanning tree create a cycle), so $G$ has $n$ edges.
answered Dec 8 '18 at 19:39
nafhgoodnafhgood
1,801422
$endgroup$
add a comment |
$begingroup$
Consider the spanning tree of the graph $G$, The number of edges in the spanning tree is $n-1$. As $G$ has only one cycle, we can add only one more edge to the spanning tree to get the cycle(as any edges added to the spanning tree create a cycle), so $G$ has $n$ edges.
answered Dec 8 '18 at 19:39
nafhgoodnafhgood
1,801422
$endgroup$
add a comment |
$begingroup$
Consider the spanning tree of the graph $G$, The number of edges in the spanning tree is $n-1$. As $G$ has only one cycle, we can add only one more edge to the spanning tree to get the cycle(as any edges added to the spanning tree create a cycle), so $G$ has $n$ edges.
answered Dec 8 '18 at 19:39
nafhgoodnafhgood
1,801422
$endgroup$
Consider the spanning tree of the graph $G$, The number of edges in the spanning tree is $n-1$. As $G$ has only one cycle, we can add only one more edge to the spanning tree to get the cycle(as any edges added to the spanning tree create a cycle), so $G$ has $n$ edges.
answered Dec 8 '18 at 19:39
nafhgoodnafhgood
1,801422
answered Dec 8 '18 at 19:39
nafhgoodnafhgood
1,801422
answered Dec 8 '18 at 19:39
nafhgoodnafhgood
1,801422
answered Dec 8 '18 at 19:39
nafhgoodnafhgood
1,801422
1,801422
add a comment |
add a comment |
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$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 8 '18 at 12:43