Confusion about a proof on Mycielski construction and chromatic number
Theorem 10.10 of the textbook "A First Course in Graph Theory (2012)" by Gary Chartrand and Ping Zhang is as follows:
For every integer $k ge 3$, there exists a triangle-free graph with chromatic number $k$.
This is proved by induction on $k$ and in the inductive step it actually shows that
From a $(k-1)$-chromatic triangle-free graph $F$, Mycielski's construction produces a $k$-chromatic triangle-free graph $G$.
The Mycielski's construction (wiki) of $F$ is obtained from $F$ by first adding, for each vertex of $F$, a new vertex $v'$, called the shadow vertex of $v$, and joining $v'$ to the neighbors of $v$ in $F$ and then adding a new vertex $z$ and joining $z$ to all the shadow vertices.
For the part the correctness of $chi(G) = k$, it proceeds as follows:
Assume, to the contrary, that $chi(G) = k-1$. Let there be a $(k-1)$-coloring $c$ of $G$, say with colors $1, 2, ldots, k-1$. We may assume that $c(z) = k-1$. Since $z$ is adjacent to every shadow vertex in $G$, it follows that the shadow vertices are colored with the colors $1, 2, ldots, k-2$. For every shadow vertex $x'$ of $G$, the color $c(x')$ is different from the colors assigned to the neighbors of $x$. Therefore, if for each vetex $y$ of $G$ belonging to $F$, the color $c(y)$ is replaced by $c(y')$, we have a $(k-2)$-coloring of $F$. This is impossible, however, since $chi(F) = k - 1$.
My Problem: The argument in bold does not seem clear to me. How to make sure that the resulting coloring is a proper coloring? In other words, how to show that for each pair of adjacent vertices $u,v$ in $F$, $c(u') neq c(v')$?
graph-theory proof-explanation coloring
asked Dec 26 at 8:31
hengxin
1,5431428
add a comment |
Theorem 10.10 of the textbook "A First Course in Graph Theory (2012)" by Gary Chartrand and Ping Zhang is as follows:
For every integer $k ge 3$, there exists a triangle-free graph with chromatic number $k$.
This is proved by induction on $k$ and in the inductive step it actually shows that
From a $(k-1)$-chromatic triangle-free graph $F$, Mycielski's construction produces a $k$-chromatic triangle-free graph $G$.
The Mycielski's construction (wiki) of $F$ is obtained from $F$ by first adding, for each vertex of $F$, a new vertex $v'$, called the shadow vertex of $v$, and joining $v'$ to the neighbors of $v$ in $F$ and then adding a new vertex $z$ and joining $z$ to all the shadow vertices.
For the part the correctness of $chi(G) = k$, it proceeds as follows:
Assume, to the contrary, that $chi(G) = k-1$. Let there be a $(k-1)$-coloring $c$ of $G$, say with colors $1, 2, ldots, k-1$. We may assume that $c(z) = k-1$. Since $z$ is adjacent to every shadow vertex in $G$, it follows that the shadow vertices are colored with the colors $1, 2, ldots, k-2$. For every shadow vertex $x'$ of $G$, the color $c(x')$ is different from the colors assigned to the neighbors of $x$. Therefore, if for each vetex $y$ of $G$ belonging to $F$, the color $c(y)$ is replaced by $c(y')$, we have a $(k-2)$-coloring of $F$. This is impossible, however, since $chi(F) = k - 1$.
My Problem: The argument in bold does not seem clear to me. How to make sure that the resulting coloring is a proper coloring? In other words, how to show that for each pair of adjacent vertices $u,v$ in $F$, $c(u') neq c(v')$?
graph-theory proof-explanation coloring
asked Dec 26 at 8:31
hengxin
1,5431428
add a comment |
Theorem 10.10 of the textbook "A First Course in Graph Theory (2012)" by Gary Chartrand and Ping Zhang is as follows:
For every integer $k ge 3$, there exists a triangle-free graph with chromatic number $k$.
This is proved by induction on $k$ and in the inductive step it actually shows that
From a $(k-1)$-chromatic triangle-free graph $F$, Mycielski's construction produces a $k$-chromatic triangle-free graph $G$.
The Mycielski's construction (wiki) of $F$ is obtained from $F$ by first adding, for each vertex of $F$, a new vertex $v'$, called the shadow vertex of $v$, and joining $v'$ to the neighbors of $v$ in $F$ and then adding a new vertex $z$ and joining $z$ to all the shadow vertices.
For the part the correctness of $chi(G) = k$, it proceeds as follows:
Assume, to the contrary, that $chi(G) = k-1$. Let there be a $(k-1)$-coloring $c$ of $G$, say with colors $1, 2, ldots, k-1$. We may assume that $c(z) = k-1$. Since $z$ is adjacent to every shadow vertex in $G$, it follows that the shadow vertices are colored with the colors $1, 2, ldots, k-2$. For every shadow vertex $x'$ of $G$, the color $c(x')$ is different from the colors assigned to the neighbors of $x$. Therefore, if for each vetex $y$ of $G$ belonging to $F$, the color $c(y)$ is replaced by $c(y')$, we have a $(k-2)$-coloring of $F$. This is impossible, however, since $chi(F) = k - 1$.
My Problem: The argument in bold does not seem clear to me. How to make sure that the resulting coloring is a proper coloring? In other words, how to show that for each pair of adjacent vertices $u,v$ in $F$, $c(u') neq c(v')$?
graph-theory proof-explanation coloring
asked Dec 26 at 8:31
hengxin
1,5431428
Theorem 10.10 of the textbook "A First Course in Graph Theory (2012)" by Gary Chartrand and Ping Zhang is as follows:
For every integer $k ge 3$, there exists a triangle-free graph with chromatic number $k$.
This is proved by induction on $k$ and in the inductive step it actually shows that
From a $(k-1)$-chromatic triangle-free graph $F$, Mycielski's construction produces a $k$-chromatic triangle-free graph $G$.
The Mycielski's construction (wiki) of $F$ is obtained from $F$ by first adding, for each vertex of $F$, a new vertex $v'$, called the shadow vertex of $v$, and joining $v'$ to the neighbors of $v$ in $F$ and then adding a new vertex $z$ and joining $z$ to all the shadow vertices.
For the part the correctness of $chi(G) = k$, it proceeds as follows:
Assume, to the contrary, that $chi(G) = k-1$. Let there be a $(k-1)$-coloring $c$ of $G$, say with colors $1, 2, ldots, k-1$. We may assume that $c(z) = k-1$. Since $z$ is adjacent to every shadow vertex in $G$, it follows that the shadow vertices are colored with the colors $1, 2, ldots, k-2$. For every shadow vertex $x'$ of $G$, the color $c(x')$ is different from the colors assigned to the neighbors of $x$. Therefore, if for each vetex $y$ of $G$ belonging to $F$, the color $c(y)$ is replaced by $c(y')$, we have a $(k-2)$-coloring of $F$. This is impossible, however, since $chi(F) = k - 1$.
My Problem: The argument in bold does not seem clear to me. How to make sure that the resulting coloring is a proper coloring? In other words, how to show that for each pair of adjacent vertices $u,v$ in $F$, $c(u') neq c(v')$?
graph-theory proof-explanation coloring
graph-theory proof-explanation coloring
asked Dec 26 at 8:31
hengxin
1,5431428
asked Dec 26 at 8:31
hengxin
1,5431428
edited Dec 26 at 9:17
asked Dec 26 at 8:31
hengxin
1,5431428
asked Dec 26 at 8:31
hengxin
1,5431428
asked Dec 26 at 8:31
hengxin
1,5431428
1,5431428
add a comment |
add a comment |
1 Answer
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Based on the assumption that $c(z)=k-1$, since $z$ is only adjacent to the set of shadow vertices of $F$, we know that the set of shadow vertices is $k-2$ colorable.
Now to address your question; When applying a proper coloring to a Mycielski's Constructed graph, it is possible to assign $c(y)=c(y')$. That is, a shadow vertex $y'$ can have the same color as $y$. This is because each shadow is not adjacent it's origin in the constructed graph.
Hence, because the set of shadow vertices is $k-2$ colorable, without any colors already imposed on $F$, we color $F$ so that $c(y)=c(y')$ for each $yin V(F)$. Thus, providing a $k-2$ coloring on $F$ which contradicts how $F$ was defined.
answered Dec 26 at 14:45
Steve Schroeder
1759
add a comment |
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Based on the assumption that $c(z)=k-1$, since $z$ is only adjacent to the set of shadow vertices of $F$, we know that the set of shadow vertices is $k-2$ colorable.
Now to address your question; When applying a proper coloring to a Mycielski's Constructed graph, it is possible to assign $c(y)=c(y')$. That is, a shadow vertex $y'$ can have the same color as $y$. This is because each shadow is not adjacent it's origin in the constructed graph.
Hence, because the set of shadow vertices is $k-2$ colorable, without any colors already imposed on $F$, we color $F$ so that $c(y)=c(y')$ for each $yin V(F)$. Thus, providing a $k-2$ coloring on $F$ which contradicts how $F$ was defined.
answered Dec 26 at 14:45
Steve Schroeder
1759
add a comment |
Based on the assumption that $c(z)=k-1$, since $z$ is only adjacent to the set of shadow vertices of $F$, we know that the set of shadow vertices is $k-2$ colorable.
Now to address your question; When applying a proper coloring to a Mycielski's Constructed graph, it is possible to assign $c(y)=c(y')$. That is, a shadow vertex $y'$ can have the same color as $y$. This is because each shadow is not adjacent it's origin in the constructed graph.
Hence, because the set of shadow vertices is $k-2$ colorable, without any colors already imposed on $F$, we color $F$ so that $c(y)=c(y')$ for each $yin V(F)$. Thus, providing a $k-2$ coloring on $F$ which contradicts how $F$ was defined.
answered Dec 26 at 14:45
Steve Schroeder
1759
add a comment |
Based on the assumption that $c(z)=k-1$, since $z$ is only adjacent to the set of shadow vertices of $F$, we know that the set of shadow vertices is $k-2$ colorable.
Now to address your question; When applying a proper coloring to a Mycielski's Constructed graph, it is possible to assign $c(y)=c(y')$. That is, a shadow vertex $y'$ can have the same color as $y$. This is because each shadow is not adjacent it's origin in the constructed graph.
Hence, because the set of shadow vertices is $k-2$ colorable, without any colors already imposed on $F$, we color $F$ so that $c(y)=c(y')$ for each $yin V(F)$. Thus, providing a $k-2$ coloring on $F$ which contradicts how $F$ was defined.
answered Dec 26 at 14:45
Steve Schroeder
1759
Based on the assumption that $c(z)=k-1$, since $z$ is only adjacent to the set of shadow vertices of $F$, we know that the set of shadow vertices is $k-2$ colorable.
Now to address your question; When applying a proper coloring to a Mycielski's Constructed graph, it is possible to assign $c(y)=c(y')$. That is, a shadow vertex $y'$ can have the same color as $y$. This is because each shadow is not adjacent it's origin in the constructed graph.
Hence, because the set of shadow vertices is $k-2$ colorable, without any colors already imposed on $F$, we color $F$ so that $c(y)=c(y')$ for each $yin V(F)$. Thus, providing a $k-2$ coloring on $F$ which contradicts how $F$ was defined.
answered Dec 26 at 14:45
Steve Schroeder
1759
answered Dec 26 at 14:45
Steve Schroeder
1759
answered Dec 26 at 14:45
Steve Schroeder
1759
answered Dec 26 at 14:45
Steve Schroeder
1759
1759
add a comment |
add a comment |
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