Chromatic polynomial with constant sign on $x > n -1$ and $0<x<1$
Let $G$ be a connected graph with $n$ vertices and chromatic polynomial $p_G(x)$. Prove that:
a) $(-1)^{n-1}p_G(x) > 0$ for $0<x<1$.
b) $p_G(x)>0$ for $x>n-1$.
For a), the case of a tree is clear, having the explicit formula $p_G(x) = x(x-1)^{n-1}$. It might be possible to nicely use that any connected graph has a spanning tree, but unfortunately there is no nice inequality such as $p_H leq p_G$ (or $geq$) if $H$ is a subgraph of $G$. Similarly, for b) the complete graph case is nice and $G$ is necessarily a subgraph but this doesn't look enough.
Any help appreciated!
graph-theory coloring
asked yesterday
DesmondMiles
669
add a comment |
Let $G$ be a connected graph with $n$ vertices and chromatic polynomial $p_G(x)$. Prove that:
a) $(-1)^{n-1}p_G(x) > 0$ for $0<x<1$.
b) $p_G(x)>0$ for $x>n-1$.
For a), the case of a tree is clear, having the explicit formula $p_G(x) = x(x-1)^{n-1}$. It might be possible to nicely use that any connected graph has a spanning tree, but unfortunately there is no nice inequality such as $p_H leq p_G$ (or $geq$) if $H$ is a subgraph of $G$. Similarly, for b) the complete graph case is nice and $G$ is necessarily a subgraph but this doesn't look enough.
Any help appreciated!
graph-theory coloring
asked yesterday
DesmondMiles
669
Actually, the first part can use the tree for an induction argument on the number of edges. Indeed, $p_G = p_{G-e} - p_{G/e}$ and $G-e$, $G/e$ satisfy the induction hypothesis and have $|G|$ and $|G|-1$ vertices. Then we are fine. The second part?
– DesmondMiles
yesterday
add a comment |
Let $G$ be a connected graph with $n$ vertices and chromatic polynomial $p_G(x)$. Prove that:
a) $(-1)^{n-1}p_G(x) > 0$ for $0<x<1$.
b) $p_G(x)>0$ for $x>n-1$.
For a), the case of a tree is clear, having the explicit formula $p_G(x) = x(x-1)^{n-1}$. It might be possible to nicely use that any connected graph has a spanning tree, but unfortunately there is no nice inequality such as $p_H leq p_G$ (or $geq$) if $H$ is a subgraph of $G$. Similarly, for b) the complete graph case is nice and $G$ is necessarily a subgraph but this doesn't look enough.
Any help appreciated!
graph-theory coloring
asked yesterday
DesmondMiles
669
Let $G$ be a connected graph with $n$ vertices and chromatic polynomial $p_G(x)$. Prove that:
a) $(-1)^{n-1}p_G(x) > 0$ for $0<x<1$.
b) $p_G(x)>0$ for $x>n-1$.
For a), the case of a tree is clear, having the explicit formula $p_G(x) = x(x-1)^{n-1}$. It might be possible to nicely use that any connected graph has a spanning tree, but unfortunately there is no nice inequality such as $p_H leq p_G$ (or $geq$) if $H$ is a subgraph of $G$. Similarly, for b) the complete graph case is nice and $G$ is necessarily a subgraph but this doesn't look enough.
Any help appreciated!
graph-theory coloring
graph-theory coloring
asked yesterday
DesmondMiles
669
asked yesterday
DesmondMiles
669
asked yesterday
DesmondMiles
669
asked yesterday
DesmondMiles
669
asked yesterday
DesmondMiles
669
669
Actually, the first part can use the tree for an induction argument on the number of edges. Indeed, $p_G = p_{G-e} - p_{G/e}$ and $G-e$, $G/e$ satisfy the induction hypothesis and have $|G|$ and $|G|-1$ vertices. Then we are fine. The second part?
– DesmondMiles
yesterday
add a comment |
Actually, the first part can use the tree for an induction argument on the number of edges. Indeed, $p_G = p_{G-e} - p_{G/e}$ and $G-e$, $G/e$ satisfy the induction hypothesis and have $|G|$ and $|G|-1$ vertices. Then we are fine. The second part?
– DesmondMiles
yesterday
Actually, the first part can use the tree for an induction argument on the number of edges. Indeed, $p_G = p_{G-e} - p_{G/e}$ and $G-e$, $G/e$ satisfy the induction hypothesis and have $|G|$ and $|G|-1$ vertices. Then we are fine. The second part?
– DesmondMiles
yesterday
Actually, the first part can use the tree for an induction argument on the number of edges. Indeed, $p_G = p_{G-e} - p_{G/e}$ and $G-e$, $G/e$ satisfy the induction hypothesis and have $|G|$ and $|G|-1$ vertices. Then we are fine. The second part?
– DesmondMiles
yesterday
add a comment |
1 Answer
1
active
oldest
votes
For the second part:
We can express the chromatic polynomial as a sum over partitions into independent sets. For every such partition ${I_1, I_2, dots, I_k}$, there are $x^{|I_1|} (x-1)^{|I_2|} dotsm (x-k+1)^{|I_k|}$ ways to color the graph by $x$ colors if we want to make $I_1, I_2, dots, I_k$ be the color classes, divided by some constant term to account for symmetry. (Conversely, for every proper coloring of the graph by $x$ colors, the color classes form such a partition.)
For $x>n-1$, each factor in each $x^{|I_1|} (x-1)^{|I_2|} dotsm (x-k+1)^{|I_k|}$ term is positive, and therefore the chromatic polynomial is positive.
answered yesterday
Misha Lavrov
43.6k555103
Wow, amazing! Thanks a lot.
– DesmondMiles
yesterday
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052054%2fchromatic-polynomial-with-constant-sign-on-x-n-1-and-0x1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
For the second part:
We can express the chromatic polynomial as a sum over partitions into independent sets. For every such partition ${I_1, I_2, dots, I_k}$, there are $x^{|I_1|} (x-1)^{|I_2|} dotsm (x-k+1)^{|I_k|}$ ways to color the graph by $x$ colors if we want to make $I_1, I_2, dots, I_k$ be the color classes, divided by some constant term to account for symmetry. (Conversely, for every proper coloring of the graph by $x$ colors, the color classes form such a partition.)
For $x>n-1$, each factor in each $x^{|I_1|} (x-1)^{|I_2|} dotsm (x-k+1)^{|I_k|}$ term is positive, and therefore the chromatic polynomial is positive.
answered yesterday
Misha Lavrov
43.6k555103
Wow, amazing! Thanks a lot.
– DesmondMiles
yesterday
add a comment |
For the second part:
We can express the chromatic polynomial as a sum over partitions into independent sets. For every such partition ${I_1, I_2, dots, I_k}$, there are $x^{|I_1|} (x-1)^{|I_2|} dotsm (x-k+1)^{|I_k|}$ ways to color the graph by $x$ colors if we want to make $I_1, I_2, dots, I_k$ be the color classes, divided by some constant term to account for symmetry. (Conversely, for every proper coloring of the graph by $x$ colors, the color classes form such a partition.)
For $x>n-1$, each factor in each $x^{|I_1|} (x-1)^{|I_2|} dotsm (x-k+1)^{|I_k|}$ term is positive, and therefore the chromatic polynomial is positive.
answered yesterday
Misha Lavrov
43.6k555103
Wow, amazing! Thanks a lot.
– DesmondMiles
yesterday
add a comment |
For the second part:
We can express the chromatic polynomial as a sum over partitions into independent sets. For every such partition ${I_1, I_2, dots, I_k}$, there are $x^{|I_1|} (x-1)^{|I_2|} dotsm (x-k+1)^{|I_k|}$ ways to color the graph by $x$ colors if we want to make $I_1, I_2, dots, I_k$ be the color classes, divided by some constant term to account for symmetry. (Conversely, for every proper coloring of the graph by $x$ colors, the color classes form such a partition.)
For $x>n-1$, each factor in each $x^{|I_1|} (x-1)^{|I_2|} dotsm (x-k+1)^{|I_k|}$ term is positive, and therefore the chromatic polynomial is positive.
answered yesterday
Misha Lavrov
43.6k555103
For the second part:
We can express the chromatic polynomial as a sum over partitions into independent sets. For every such partition ${I_1, I_2, dots, I_k}$, there are $x^{|I_1|} (x-1)^{|I_2|} dotsm (x-k+1)^{|I_k|}$ ways to color the graph by $x$ colors if we want to make $I_1, I_2, dots, I_k$ be the color classes, divided by some constant term to account for symmetry. (Conversely, for every proper coloring of the graph by $x$ colors, the color classes form such a partition.)
For $x>n-1$, each factor in each $x^{|I_1|} (x-1)^{|I_2|} dotsm (x-k+1)^{|I_k|}$ term is positive, and therefore the chromatic polynomial is positive.
answered yesterday
Misha Lavrov
43.6k555103
edited yesterday
answered yesterday
Misha Lavrov
43.6k555103
answered yesterday
Misha Lavrov
43.6k555103
answered yesterday
Misha Lavrov
43.6k555103
43.6k555103
Wow, amazing! Thanks a lot.
– DesmondMiles
yesterday
add a comment |
Wow, amazing! Thanks a lot.
– DesmondMiles
yesterday
Wow, amazing! Thanks a lot.
– DesmondMiles
yesterday
Wow, amazing! Thanks a lot.
– DesmondMiles
yesterday
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052054%2fchromatic-polynomial-with-constant-sign-on-x-n-1-and-0x1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Actually, the first part can use the tree for an induction argument on the number of edges. Indeed, $p_G = p_{G-e} - p_{G/e}$ and $G-e$, $G/e$ satisfy the induction hypothesis and have $|G|$ and $|G|-1$ vertices. Then we are fine. The second part?
– DesmondMiles
yesterday