How to prove that the max edge coloring is 2D - 1 (where D is max vertex degree)?
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I have this problem that is giving me fits. This problem seems similar to proving that the max coloring of vertices, where the max vertex degree is D, is D + 1.
We cannot have two edges that share the same endpoint be the same color.
My intuition is that we have to use induction on the number of edges. However, I'm not sure how to piece it together with the endpoint vertex involved.
Does anyone know how to approach this type of problem?
graph-theory coloring
asked Dec 4 '18 at 17:43
Mr.MipsMr.Mips
185
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add a comment |
$begingroup$
I have this problem that is giving me fits. This problem seems similar to proving that the max coloring of vertices, where the max vertex degree is D, is D + 1.
We cannot have two edges that share the same endpoint be the same color.
My intuition is that we have to use induction on the number of edges. However, I'm not sure how to piece it together with the endpoint vertex involved.
Does anyone know how to approach this type of problem?
graph-theory coloring
asked Dec 4 '18 at 17:43
Mr.MipsMr.Mips
185
$endgroup$
$begingroup$
Each edge "touches" 2 vertices, and hence at most 2D-1 edges (since an edge doesn't touch itself). Does this help?
$endgroup$
– Don Thousand
Dec 4 '18 at 17:51
$begingroup$
The "max edge coloring" is the number of edges, because you can give every edge a different color.
$endgroup$
– bof
Dec 4 '18 at 23:04
add a comment |
$begingroup$
I have this problem that is giving me fits. This problem seems similar to proving that the max coloring of vertices, where the max vertex degree is D, is D + 1.
We cannot have two edges that share the same endpoint be the same color.
My intuition is that we have to use induction on the number of edges. However, I'm not sure how to piece it together with the endpoint vertex involved.
Does anyone know how to approach this type of problem?
graph-theory coloring
asked Dec 4 '18 at 17:43
Mr.MipsMr.Mips
185
$endgroup$
I have this problem that is giving me fits. This problem seems similar to proving that the max coloring of vertices, where the max vertex degree is D, is D + 1.
We cannot have two edges that share the same endpoint be the same color.
My intuition is that we have to use induction on the number of edges. However, I'm not sure how to piece it together with the endpoint vertex involved.
Does anyone know how to approach this type of problem?
graph-theory coloring
graph-theory coloring
asked Dec 4 '18 at 17:43
Mr.MipsMr.Mips
185
asked Dec 4 '18 at 17:43
Mr.MipsMr.Mips
185
asked Dec 4 '18 at 17:43
Mr.MipsMr.Mips
185
asked Dec 4 '18 at 17:43
Mr.MipsMr.Mips
185
asked Dec 4 '18 at 17:43
Mr.MipsMr.Mips
185
185
$begingroup$
Each edge "touches" 2 vertices, and hence at most 2D-1 edges (since an edge doesn't touch itself). Does this help?
$endgroup$
– Don Thousand
Dec 4 '18 at 17:51
$begingroup$
The "max edge coloring" is the number of edges, because you can give every edge a different color.
$endgroup$
– bof
Dec 4 '18 at 23:04
add a comment |
$begingroup$
Each edge "touches" 2 vertices, and hence at most 2D-1 edges (since an edge doesn't touch itself). Does this help?
$endgroup$
– Don Thousand
Dec 4 '18 at 17:51
$begingroup$
The "max edge coloring" is the number of edges, because you can give every edge a different color.
$endgroup$
– bof
Dec 4 '18 at 23:04
$begingroup$
Each edge "touches" 2 vertices, and hence at most 2D-1 edges (since an edge doesn't touch itself). Does this help?
$endgroup$
– Don Thousand
Dec 4 '18 at 17:51
$begingroup$
Each edge "touches" 2 vertices, and hence at most 2D-1 edges (since an edge doesn't touch itself). Does this help?
$endgroup$
– Don Thousand
Dec 4 '18 at 17:51
$begingroup$
The "max edge coloring" is the number of edges, because you can give every edge a different color.
$endgroup$
– bof
Dec 4 '18 at 23:04
$begingroup$
The "max edge coloring" is the number of edges, because you can give every edge a different color.
$endgroup$
– bof
Dec 4 '18 at 23:04
add a comment |
1 Answer
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I think the statement is $chi'(G) leq 2triangle(G)-1$. This can be done using induction on the number of edges of a graph.
So
Base case: Let $G$ be a graph such that $|E(G)|=1$, then $triangle(G) =1$,where $triangle(G)$ is the maximum degree of $G$. so then $chi'(G)=1leq2*triangle(G)-1$.
Inductive step: Suppose the claim is true for all graph such that $|E(G)|=n$, then now consider a graph $G$ such that $|E(G)|=n+1$. Pick an arbitrary edge $uv=ein E(G)$, and consider $G'=G-e$. $chi'(G')leq 2triangle(G')-1 leq 2triangle(G)-1$ by induction
hypothesis. Now $deg(u) leq triangle(G)-1$ and $deg(v) leq triangle(G)-1$. So we can color $e$ using one more color different than all color on the edges incident to $u$ and $v$. So we can use one of $2triangle(G)-1$ color to color $e$. so $chi(G) leq 2triangle(G)-1$. I hope this is correct.
answered Dec 4 '18 at 22:59
nafhgoodnafhgood
1,805422
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1 Answer
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$begingroup$
I think the statement is $chi'(G) leq 2triangle(G)-1$. This can be done using induction on the number of edges of a graph.
So
Base case: Let $G$ be a graph such that $|E(G)|=1$, then $triangle(G) =1$,where $triangle(G)$ is the maximum degree of $G$. so then $chi'(G)=1leq2*triangle(G)-1$.
Inductive step: Suppose the claim is true for all graph such that $|E(G)|=n$, then now consider a graph $G$ such that $|E(G)|=n+1$. Pick an arbitrary edge $uv=ein E(G)$, and consider $G'=G-e$. $chi'(G')leq 2triangle(G')-1 leq 2triangle(G)-1$ by induction
hypothesis. Now $deg(u) leq triangle(G)-1$ and $deg(v) leq triangle(G)-1$. So we can color $e$ using one more color different than all color on the edges incident to $u$ and $v$. So we can use one of $2triangle(G)-1$ color to color $e$. so $chi(G) leq 2triangle(G)-1$. I hope this is correct.
answered Dec 4 '18 at 22:59
nafhgoodnafhgood
1,805422
$endgroup$
add a comment |
$begingroup$
I think the statement is $chi'(G) leq 2triangle(G)-1$. This can be done using induction on the number of edges of a graph.
So
Base case: Let $G$ be a graph such that $|E(G)|=1$, then $triangle(G) =1$,where $triangle(G)$ is the maximum degree of $G$. so then $chi'(G)=1leq2*triangle(G)-1$.
Inductive step: Suppose the claim is true for all graph such that $|E(G)|=n$, then now consider a graph $G$ such that $|E(G)|=n+1$. Pick an arbitrary edge $uv=ein E(G)$, and consider $G'=G-e$. $chi'(G')leq 2triangle(G')-1 leq 2triangle(G)-1$ by induction
hypothesis. Now $deg(u) leq triangle(G)-1$ and $deg(v) leq triangle(G)-1$. So we can color $e$ using one more color different than all color on the edges incident to $u$ and $v$. So we can use one of $2triangle(G)-1$ color to color $e$. so $chi(G) leq 2triangle(G)-1$. I hope this is correct.
answered Dec 4 '18 at 22:59
nafhgoodnafhgood
1,805422
$endgroup$
add a comment |
$begingroup$
I think the statement is $chi'(G) leq 2triangle(G)-1$. This can be done using induction on the number of edges of a graph.
So
Base case: Let $G$ be a graph such that $|E(G)|=1$, then $triangle(G) =1$,where $triangle(G)$ is the maximum degree of $G$. so then $chi'(G)=1leq2*triangle(G)-1$.
Inductive step: Suppose the claim is true for all graph such that $|E(G)|=n$, then now consider a graph $G$ such that $|E(G)|=n+1$. Pick an arbitrary edge $uv=ein E(G)$, and consider $G'=G-e$. $chi'(G')leq 2triangle(G')-1 leq 2triangle(G)-1$ by induction
hypothesis. Now $deg(u) leq triangle(G)-1$ and $deg(v) leq triangle(G)-1$. So we can color $e$ using one more color different than all color on the edges incident to $u$ and $v$. So we can use one of $2triangle(G)-1$ color to color $e$. so $chi(G) leq 2triangle(G)-1$. I hope this is correct.
answered Dec 4 '18 at 22:59
nafhgoodnafhgood
1,805422
$endgroup$
I think the statement is $chi'(G) leq 2triangle(G)-1$. This can be done using induction on the number of edges of a graph.
So
Base case: Let $G$ be a graph such that $|E(G)|=1$, then $triangle(G) =1$,where $triangle(G)$ is the maximum degree of $G$. so then $chi'(G)=1leq2*triangle(G)-1$.
Inductive step: Suppose the claim is true for all graph such that $|E(G)|=n$, then now consider a graph $G$ such that $|E(G)|=n+1$. Pick an arbitrary edge $uv=ein E(G)$, and consider $G'=G-e$. $chi'(G')leq 2triangle(G')-1 leq 2triangle(G)-1$ by induction
hypothesis. Now $deg(u) leq triangle(G)-1$ and $deg(v) leq triangle(G)-1$. So we can color $e$ using one more color different than all color on the edges incident to $u$ and $v$. So we can use one of $2triangle(G)-1$ color to color $e$. so $chi(G) leq 2triangle(G)-1$. I hope this is correct.
answered Dec 4 '18 at 22:59
nafhgoodnafhgood
1,805422
answered Dec 4 '18 at 22:59
nafhgoodnafhgood
1,805422
answered Dec 4 '18 at 22:59
nafhgoodnafhgood
1,805422
answered Dec 4 '18 at 22:59
nafhgoodnafhgood
1,805422
1,805422
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$begingroup$
Each edge "touches" 2 vertices, and hence at most 2D-1 edges (since an edge doesn't touch itself). Does this help?
$endgroup$
– Don Thousand
Dec 4 '18 at 17:51
$begingroup$
The "max edge coloring" is the number of edges, because you can give every edge a different color.
$endgroup$
– bof
Dec 4 '18 at 23:04