Chromatic polynomial properties
$begingroup$
I have some properties which I do not understand how to form a proof for.
Most of them are by induction, which is not my strong point
Any help would be fantastic.
Looking at the chromatic polynomial $$P_G(k)=P_{G-e}(k) - P_{G/e}(k)$$
Properties
1.If G has n vertices then $ P_g(k)$ has degree n and the coefficient of $k^n $ is 1
The signs of $P_g(k) $ alternate
If G has q components the smalles non-zero term of $P_g(k)$ is the term in $k^q$
linear-algebra graph-theory
edited Jan 16 at 17:41
Thomas Lesgourgues
1,320220
asked Jan 16 at 17:12
p sp s
428
$endgroup$
add a comment |
$begingroup$
I have some properties which I do not understand how to form a proof for.
Most of them are by induction, which is not my strong point
Any help would be fantastic.
Looking at the chromatic polynomial $$P_G(k)=P_{G-e}(k) - P_{G/e}(k)$$
Properties
1.If G has n vertices then $ P_g(k)$ has degree n and the coefficient of $k^n $ is 1
The signs of $P_g(k) $ alternate
If G has q components the smalles non-zero term of $P_g(k)$ is the term in $k^q$
linear-algebra graph-theory
edited Jan 16 at 17:41
Thomas Lesgourgues
1,320220
asked Jan 16 at 17:12
p sp s
428
$endgroup$
add a comment |
$begingroup$
I have some properties which I do not understand how to form a proof for.
Most of them are by induction, which is not my strong point
Any help would be fantastic.
Looking at the chromatic polynomial $$P_G(k)=P_{G-e}(k) - P_{G/e}(k)$$
Properties
1.If G has n vertices then $ P_g(k)$ has degree n and the coefficient of $k^n $ is 1
The signs of $P_g(k) $ alternate
If G has q components the smalles non-zero term of $P_g(k)$ is the term in $k^q$
linear-algebra graph-theory
edited Jan 16 at 17:41
Thomas Lesgourgues
1,320220
asked Jan 16 at 17:12
p sp s
428
$endgroup$
I have some properties which I do not understand how to form a proof for.
Most of them are by induction, which is not my strong point
Any help would be fantastic.
Looking at the chromatic polynomial $$P_G(k)=P_{G-e}(k) - P_{G/e}(k)$$
Properties
1.If G has n vertices then $ P_g(k)$ has degree n and the coefficient of $k^n $ is 1
The signs of $P_g(k) $ alternate
If G has q components the smalles non-zero term of $P_g(k)$ is the term in $k^q$
linear-algebra graph-theory
linear-algebra graph-theory
edited Jan 16 at 17:41
Thomas Lesgourgues
1,320220
asked Jan 16 at 17:12
p sp s
428
edited Jan 16 at 17:41
Thomas Lesgourgues
1,320220
asked Jan 16 at 17:12
p sp s
428
edited Jan 16 at 17:41
Thomas Lesgourgues
1,320220
edited Jan 16 at 17:41
Thomas Lesgourgues
1,320220
edited Jan 16 at 17:41
Thomas Lesgourgues
1,320220
1,320220
asked Jan 16 at 17:12
p sp s
428
asked Jan 16 at 17:12
p sp s
428
asked Jan 16 at 17:12
p sp s
428
428
add a comment |
add a comment |
1 Answer
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$begingroup$
The proof works by strong induction on $m=|E(G)|$. We also prove that the coefficient of $k^{n-1}$ is $m$ the number of edges.
Base case : $m=0$. Then $G$ is $n$ isolated vertices. And clearly $P_G(k) = k^n$. Satisfying all results
Induction : Suppose the hypothesis are true for all graphs with at most $m$ edges. Let $G$ be a graph on $m+1$ edegs, $n$ vertices, $q$ components, and let $e in E(G)$. Then
$G-e$ has $m$ edges, $n$ vertices, $q$ or $q+1$ components
$G/e$ has at most $m$ edges, with $n-1$ vertices, and $q$ components.
Using the induction hypothesis :
begin{align}
P_{G-e}(k) = k^n - m &k^{n-1} + ldots - ldots pm bk^q \
P_{G/e}(k) = &k^{n-1} - ldots + ldots mp ck^q
end{align}
With $bgeq 0$ and $c>0$. Therefore
$$ P_G(k) = k^n - (m+1) k^n + ldots - ldots + ldots pm (b+c)k^q$$
With $b+c >0$. Provind that $P_G(k)$ is monic, with alternating signs, with second coefficient $m+1$ its number of edges, and smallest coefficient the number $q$ of components.
answered Jan 17 at 12:27
Thomas LesgourguesThomas Lesgourgues
1,320220
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The proof works by strong induction on $m=|E(G)|$. We also prove that the coefficient of $k^{n-1}$ is $m$ the number of edges.
Base case : $m=0$. Then $G$ is $n$ isolated vertices. And clearly $P_G(k) = k^n$. Satisfying all results
Induction : Suppose the hypothesis are true for all graphs with at most $m$ edges. Let $G$ be a graph on $m+1$ edegs, $n$ vertices, $q$ components, and let $e in E(G)$. Then
$G-e$ has $m$ edges, $n$ vertices, $q$ or $q+1$ components
$G/e$ has at most $m$ edges, with $n-1$ vertices, and $q$ components.
Using the induction hypothesis :
begin{align}
P_{G-e}(k) = k^n - m &k^{n-1} + ldots - ldots pm bk^q \
P_{G/e}(k) = &k^{n-1} - ldots + ldots mp ck^q
end{align}
With $bgeq 0$ and $c>0$. Therefore
$$ P_G(k) = k^n - (m+1) k^n + ldots - ldots + ldots pm (b+c)k^q$$
With $b+c >0$. Provind that $P_G(k)$ is monic, with alternating signs, with second coefficient $m+1$ its number of edges, and smallest coefficient the number $q$ of components.
answered Jan 17 at 12:27
Thomas LesgourguesThomas Lesgourgues
1,320220
$endgroup$
add a comment |
$begingroup$
The proof works by strong induction on $m=|E(G)|$. We also prove that the coefficient of $k^{n-1}$ is $m$ the number of edges.
Base case : $m=0$. Then $G$ is $n$ isolated vertices. And clearly $P_G(k) = k^n$. Satisfying all results
Induction : Suppose the hypothesis are true for all graphs with at most $m$ edges. Let $G$ be a graph on $m+1$ edegs, $n$ vertices, $q$ components, and let $e in E(G)$. Then
$G-e$ has $m$ edges, $n$ vertices, $q$ or $q+1$ components
$G/e$ has at most $m$ edges, with $n-1$ vertices, and $q$ components.
Using the induction hypothesis :
begin{align}
P_{G-e}(k) = k^n - m &k^{n-1} + ldots - ldots pm bk^q \
P_{G/e}(k) = &k^{n-1} - ldots + ldots mp ck^q
end{align}
With $bgeq 0$ and $c>0$. Therefore
$$ P_G(k) = k^n - (m+1) k^n + ldots - ldots + ldots pm (b+c)k^q$$
With $b+c >0$. Provind that $P_G(k)$ is monic, with alternating signs, with second coefficient $m+1$ its number of edges, and smallest coefficient the number $q$ of components.
answered Jan 17 at 12:27
Thomas LesgourguesThomas Lesgourgues
1,320220
$endgroup$
add a comment |
$begingroup$
The proof works by strong induction on $m=|E(G)|$. We also prove that the coefficient of $k^{n-1}$ is $m$ the number of edges.
Base case : $m=0$. Then $G$ is $n$ isolated vertices. And clearly $P_G(k) = k^n$. Satisfying all results
Induction : Suppose the hypothesis are true for all graphs with at most $m$ edges. Let $G$ be a graph on $m+1$ edegs, $n$ vertices, $q$ components, and let $e in E(G)$. Then
$G-e$ has $m$ edges, $n$ vertices, $q$ or $q+1$ components
$G/e$ has at most $m$ edges, with $n-1$ vertices, and $q$ components.
Using the induction hypothesis :
begin{align}
P_{G-e}(k) = k^n - m &k^{n-1} + ldots - ldots pm bk^q \
P_{G/e}(k) = &k^{n-1} - ldots + ldots mp ck^q
end{align}
With $bgeq 0$ and $c>0$. Therefore
$$ P_G(k) = k^n - (m+1) k^n + ldots - ldots + ldots pm (b+c)k^q$$
With $b+c >0$. Provind that $P_G(k)$ is monic, with alternating signs, with second coefficient $m+1$ its number of edges, and smallest coefficient the number $q$ of components.
answered Jan 17 at 12:27
Thomas LesgourguesThomas Lesgourgues
1,320220
$endgroup$
The proof works by strong induction on $m=|E(G)|$. We also prove that the coefficient of $k^{n-1}$ is $m$ the number of edges.
Base case : $m=0$. Then $G$ is $n$ isolated vertices. And clearly $P_G(k) = k^n$. Satisfying all results
Induction : Suppose the hypothesis are true for all graphs with at most $m$ edges. Let $G$ be a graph on $m+1$ edegs, $n$ vertices, $q$ components, and let $e in E(G)$. Then
$G-e$ has $m$ edges, $n$ vertices, $q$ or $q+1$ components
$G/e$ has at most $m$ edges, with $n-1$ vertices, and $q$ components.
Using the induction hypothesis :
begin{align}
P_{G-e}(k) = k^n - m &k^{n-1} + ldots - ldots pm bk^q \
P_{G/e}(k) = &k^{n-1} - ldots + ldots mp ck^q
end{align}
With $bgeq 0$ and $c>0$. Therefore
$$ P_G(k) = k^n - (m+1) k^n + ldots - ldots + ldots pm (b+c)k^q$$
With $b+c >0$. Provind that $P_G(k)$ is monic, with alternating signs, with second coefficient $m+1$ its number of edges, and smallest coefficient the number $q$ of components.
answered Jan 17 at 12:27
Thomas LesgourguesThomas Lesgourgues
1,320220
answered Jan 17 at 12:27
Thomas LesgourguesThomas Lesgourgues
1,320220
answered Jan 17 at 12:27
Thomas LesgourguesThomas Lesgourgues
1,320220
answered Jan 17 at 12:27
Thomas LesgourguesThomas Lesgourgues
1,320220
1,320220
add a comment |
add a comment |
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