Confusion about a proof on Mycielski construction and chromatic number
Multi tool use
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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