Prove or disprove: If graph $G$ has at most $3n-1$ edges, then $chi(G) leq 6$
0
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The question is to prove or disprove that if graph $G$ has at most $3n-1$ edges, then the chromatic number is at most $6$. I tried to work it out using the fact that $chi(G) leq triangle(G) + 1$, but I can't think of a relation between the max degree and the number of edges.
graph-theory
asked Dec 9 '18 at 14:27
Mera InsanMera Insan
176
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show 1 more comment
0
$begingroup$
The question is to prove or disprove that if graph $G$ has at most $3n-1$ edges, then the chromatic number is at most $6$. I tried to work it out using the fact that $chi(G) leq triangle(G) + 1$, but I can't think of a relation between the max degree and the number of edges.
graph-theory
asked Dec 9 '18 at 14:27
Mera InsanMera Insan
176
$endgroup$
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
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– José Carlos Santos
Dec 9 '18 at 14:35
1
$begingroup$
This seems impossible: for every fixed $k$, consider the complete graph on $k$ vertices and append to it a long linear portion, then, if the number of vertices is $n$, when $ntoinfty$, the number of edges is $n+O(1)$ and the chromatic number is (at least) $k$. For your homework, take $k=7$.
$endgroup$
– Did
Dec 9 '18 at 14:37
$begingroup$
But maybe $n$ is not the number of vertices of $G$... What is $n$?
$endgroup$
– Did
Dec 9 '18 at 14:42
$begingroup$
n is the number of vertices of G. I don't get what you meant that the number of edges increases by 1 when n goes to infinity. Thanks for your help
$endgroup$
– Mera Insan
Dec 9 '18 at 14:47
$begingroup$
Or what did you mean by n+O(1)
$endgroup$
– Mera Insan
Dec 9 '18 at 14:49
|
show 1 more comment
0
0
0
$begingroup$
The question is to prove or disprove that if graph $G$ has at most $3n-1$ edges, then the chromatic number is at most $6$. I tried to work it out using the fact that $chi(G) leq triangle(G) + 1$, but I can't think of a relation between the max degree and the number of edges.
graph-theory
asked Dec 9 '18 at 14:27
Mera InsanMera Insan
176
$endgroup$
The question is to prove or disprove that if graph $G$ has at most $3n-1$ edges, then the chromatic number is at most $6$. I tried to work it out using the fact that $chi(G) leq triangle(G) + 1$, but I can't think of a relation between the max degree and the number of edges.
graph-theory
graph-theory
asked Dec 9 '18 at 14:27
Mera InsanMera Insan
176
asked Dec 9 '18 at 14:27
Mera InsanMera Insan
176
edited Dec 9 '18 at 14:46
Mera Insan
asked Dec 9 '18 at 14:27
Mera InsanMera Insan
176
asked Dec 9 '18 at 14:27
Mera InsanMera Insan
176
asked Dec 9 '18 at 14:27
Mera InsanMera Insan
176
176
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 14:35
1
$begingroup$
This seems impossible: for every fixed $k$, consider the complete graph on $k$ vertices and append to it a long linear portion, then, if the number of vertices is $n$, when $ntoinfty$, the number of edges is $n+O(1)$ and the chromatic number is (at least) $k$. For your homework, take $k=7$.
$endgroup$
– Did
Dec 9 '18 at 14:37
$begingroup$
But maybe $n$ is not the number of vertices of $G$... What is $n$?
$endgroup$
– Did
Dec 9 '18 at 14:42
$begingroup$
n is the number of vertices of G. I don't get what you meant that the number of edges increases by 1 when n goes to infinity. Thanks for your help
$endgroup$
– Mera Insan
Dec 9 '18 at 14:47
$begingroup$
Or what did you mean by n+O(1)
$endgroup$
– Mera Insan
Dec 9 '18 at 14:49
|
show 1 more comment
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 14:35
1
$begingroup$
This seems impossible: for every fixed $k$, consider the complete graph on $k$ vertices and append to it a long linear portion, then, if the number of vertices is $n$, when $ntoinfty$, the number of edges is $n+O(1)$ and the chromatic number is (at least) $k$. For your homework, take $k=7$.
$endgroup$
– Did
Dec 9 '18 at 14:37
$begingroup$
But maybe $n$ is not the number of vertices of $G$... What is $n$?
$endgroup$
– Did
Dec 9 '18 at 14:42
$begingroup$
n is the number of vertices of G. I don't get what you meant that the number of edges increases by 1 when n goes to infinity. Thanks for your help
$endgroup$
– Mera Insan
Dec 9 '18 at 14:47
$begingroup$
Or what did you mean by n+O(1)
$endgroup$
– Mera Insan
Dec 9 '18 at 14:49
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 14:35
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 14:35
1
1
$begingroup$
This seems impossible: for every fixed $k$, consider the complete graph on $k$ vertices and append to it a long linear portion, then, if the number of vertices is $n$, when $ntoinfty$, the number of edges is $n+O(1)$ and the chromatic number is (at least) $k$. For your homework, take $k=7$.
$endgroup$
– Did
Dec 9 '18 at 14:37
$begingroup$
This seems impossible: for every fixed $k$, consider the complete graph on $k$ vertices and append to it a long linear portion, then, if the number of vertices is $n$, when $ntoinfty$, the number of edges is $n+O(1)$ and the chromatic number is (at least) $k$. For your homework, take $k=7$.
$endgroup$
– Did
Dec 9 '18 at 14:37
$begingroup$
But maybe $n$ is not the number of vertices of $G$... What is $n$?
$endgroup$
– Did
Dec 9 '18 at 14:42
$begingroup$
But maybe $n$ is not the number of vertices of $G$... What is $n$?
$endgroup$
– Did
Dec 9 '18 at 14:42
$begingroup$
n is the number of vertices of G. I don't get what you meant that the number of edges increases by 1 when n goes to infinity. Thanks for your help
$endgroup$
– Mera Insan
Dec 9 '18 at 14:47
$begingroup$
n is the number of vertices of G. I don't get what you meant that the number of edges increases by 1 when n goes to infinity. Thanks for your help
$endgroup$
– Mera Insan
Dec 9 '18 at 14:47
$begingroup$
Or what did you mean by n+O(1)
$endgroup$
– Mera Insan
Dec 9 '18 at 14:49
$begingroup$
Or what did you mean by n+O(1)
$endgroup$
– Mera Insan
Dec 9 '18 at 14:49
|
show 1 more comment
1 Answer
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Ok so, consider $K_7$, It has $21$ edges which $3*7$. But then add a path of length $2$ to it to get the graph $G$ like in the following picture:
Then $chi(G)=7$, $n=9$and $E(G)=23 <3*n-1=26$. So dosent't work!
answered Dec 9 '18 at 14:58
nafhgoodnafhgood
1,803422
$endgroup$
$begingroup$
Thanks! I get it now
$endgroup$
– Mera Insan
Dec 9 '18 at 15:05
add a comment |
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1 Answer
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$begingroup$
Ok so, consider $K_7$, It has $21$ edges which $3*7$. But then add a path of length $2$ to it to get the graph $G$ like in the following picture:
Then $chi(G)=7$, $n=9$and $E(G)=23 <3*n-1=26$. So dosent't work!
answered Dec 9 '18 at 14:58
nafhgoodnafhgood
1,803422
$endgroup$
$begingroup$
Thanks! I get it now
$endgroup$
– Mera Insan
Dec 9 '18 at 15:05
add a comment |
1
$begingroup$
Ok so, consider $K_7$, It has $21$ edges which $3*7$. But then add a path of length $2$ to it to get the graph $G$ like in the following picture:
Then $chi(G)=7$, $n=9$and $E(G)=23 <3*n-1=26$. So dosent't work!
answered Dec 9 '18 at 14:58
nafhgoodnafhgood
1,803422
$endgroup$
$begingroup$
Thanks! I get it now
$endgroup$
– Mera Insan
Dec 9 '18 at 15:05
add a comment |
1
1
1
$begingroup$
Ok so, consider $K_7$, It has $21$ edges which $3*7$. But then add a path of length $2$ to it to get the graph $G$ like in the following picture:
Then $chi(G)=7$, $n=9$and $E(G)=23 <3*n-1=26$. So dosent't work!
answered Dec 9 '18 at 14:58
nafhgoodnafhgood
1,803422
$endgroup$
Ok so, consider $K_7$, It has $21$ edges which $3*7$. But then add a path of length $2$ to it to get the graph $G$ like in the following picture:
Then $chi(G)=7$, $n=9$and $E(G)=23 <3*n-1=26$. So dosent't work!
answered Dec 9 '18 at 14:58
nafhgoodnafhgood
1,803422
answered Dec 9 '18 at 14:58
nafhgoodnafhgood
1,803422
answered Dec 9 '18 at 14:58
nafhgoodnafhgood
1,803422
answered Dec 9 '18 at 14:58
nafhgoodnafhgood
1,803422
1,803422
$begingroup$
Thanks! I get it now
$endgroup$
– Mera Insan
Dec 9 '18 at 15:05
add a comment |
$begingroup$
Thanks! I get it now
$endgroup$
– Mera Insan
Dec 9 '18 at 15:05
$begingroup$
Thanks! I get it now
$endgroup$
– Mera Insan
Dec 9 '18 at 15:05
$begingroup$
Thanks! I get it now
$endgroup$
– Mera Insan
Dec 9 '18 at 15:05
add a comment |
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$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 14:35
1
$begingroup$
This seems impossible: for every fixed $k$, consider the complete graph on $k$ vertices and append to it a long linear portion, then, if the number of vertices is $n$, when $ntoinfty$, the number of edges is $n+O(1)$ and the chromatic number is (at least) $k$. For your homework, take $k=7$.
$endgroup$
– Did
Dec 9 '18 at 14:37
$begingroup$
But maybe $n$ is not the number of vertices of $G$... What is $n$?
$endgroup$
– Did
Dec 9 '18 at 14:42
$begingroup$
n is the number of vertices of G. I don't get what you meant that the number of edges increases by 1 when n goes to infinity. Thanks for your help
$endgroup$
– Mera Insan
Dec 9 '18 at 14:47
$begingroup$
Or what did you mean by n+O(1)
$endgroup$
– Mera Insan
Dec 9 '18 at 14:49