Exists a bound for threewidth in $Delta(G)leq 3$ graphs?
$begingroup$
Is it true that $Delta(G)leq3$ implies $TW(G)leq6$? (where $TW$ is the "treewidth"). I know that for $Delta(G)leq4$ there is no bound for treewidth because the $ntimes n$ grid has $TW=n$. For $Deltaleq2$ it is easy to see that $TWleq 2$, and the case $Deltaleq1$ is trivial. I think this is the right bound only because I cannot find bigger counterexamples; but I dont have any idea to prove this.
graph-theory
$endgroup$
add a comment |
$begingroup$
Is it true that $Delta(G)leq3$ implies $TW(G)leq6$? (where $TW$ is the "treewidth"). I know that for $Delta(G)leq4$ there is no bound for treewidth because the $ntimes n$ grid has $TW=n$. For $Deltaleq2$ it is easy to see that $TWleq 2$, and the case $Deltaleq1$ is trivial. I think this is the right bound only because I cannot find bigger counterexamples; but I dont have any idea to prove this.
graph-theory
$endgroup$
add a comment |
$begingroup$
Is it true that $Delta(G)leq3$ implies $TW(G)leq6$? (where $TW$ is the "treewidth"). I know that for $Delta(G)leq4$ there is no bound for treewidth because the $ntimes n$ grid has $TW=n$. For $Deltaleq2$ it is easy to see that $TWleq 2$, and the case $Deltaleq1$ is trivial. I think this is the right bound only because I cannot find bigger counterexamples; but I dont have any idea to prove this.
graph-theory
$endgroup$
Is it true that $Delta(G)leq3$ implies $TW(G)leq6$? (where $TW$ is the "treewidth"). I know that for $Delta(G)leq4$ there is no bound for treewidth because the $ntimes n$ grid has $TW=n$. For $Deltaleq2$ it is easy to see that $TWleq 2$, and the case $Deltaleq1$ is trivial. I think this is the right bound only because I cannot find bigger counterexamples; but I dont have any idea to prove this.
graph-theory
graph-theory
asked Dec 20 '18 at 2:22
Veridian DynamicsVeridian Dynamics
1,909312
1,909312
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As part of their work on graph minors leading to the Robertson–Seymour theorem and the graph structure theorem, Neil Robertson and Paul Seymour proved that a family $F$ of finite graphs has unbounded treewidth if and only if the minors of graphs in $F$ include arbitrarily large square grid graphs, or equivalently subgraphs of the hexagonal tiling formed by intersecting it with arbitrarily large disks. (Wikipedia)
$endgroup$
add a comment |
$begingroup$
A simple proof goes this way:
Take $K_n$ the complete graph in $n$ vertices. Take $vin G$, replace it with a cycle of length $d(v)=n-1$ and connect the $d(v)$ vertices in a 1-to-1 fashion to a vertex of the cycle. The result is that every vertex in the cycle has degree $3$. Repeat this until no vertex of degree $>3$ left. The graph $G$ obtained is $3$-regular and (clearly) $K_nleq G$. Then, $text{TW}(G)geqtext{TW}(K_n)=n-1$. So, the treewidth is unbounded even in the class of $3$-regular graphs.
$endgroup$
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%2f3047080%2fexists-a-bound-for-threewidth-in-deltag-leq-3-graphs%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As part of their work on graph minors leading to the Robertson–Seymour theorem and the graph structure theorem, Neil Robertson and Paul Seymour proved that a family $F$ of finite graphs has unbounded treewidth if and only if the minors of graphs in $F$ include arbitrarily large square grid graphs, or equivalently subgraphs of the hexagonal tiling formed by intersecting it with arbitrarily large disks. (Wikipedia)
$endgroup$
add a comment |
$begingroup$
As part of their work on graph minors leading to the Robertson–Seymour theorem and the graph structure theorem, Neil Robertson and Paul Seymour proved that a family $F$ of finite graphs has unbounded treewidth if and only if the minors of graphs in $F$ include arbitrarily large square grid graphs, or equivalently subgraphs of the hexagonal tiling formed by intersecting it with arbitrarily large disks. (Wikipedia)
$endgroup$
add a comment |
$begingroup$
As part of their work on graph minors leading to the Robertson–Seymour theorem and the graph structure theorem, Neil Robertson and Paul Seymour proved that a family $F$ of finite graphs has unbounded treewidth if and only if the minors of graphs in $F$ include arbitrarily large square grid graphs, or equivalently subgraphs of the hexagonal tiling formed by intersecting it with arbitrarily large disks. (Wikipedia)
$endgroup$
As part of their work on graph minors leading to the Robertson–Seymour theorem and the graph structure theorem, Neil Robertson and Paul Seymour proved that a family $F$ of finite graphs has unbounded treewidth if and only if the minors of graphs in $F$ include arbitrarily large square grid graphs, or equivalently subgraphs of the hexagonal tiling formed by intersecting it with arbitrarily large disks. (Wikipedia)
answered Dec 20 '18 at 7:54
Alex RavskyAlex Ravsky
40.4k32282
40.4k32282
add a comment |
add a comment |
$begingroup$
A simple proof goes this way:
Take $K_n$ the complete graph in $n$ vertices. Take $vin G$, replace it with a cycle of length $d(v)=n-1$ and connect the $d(v)$ vertices in a 1-to-1 fashion to a vertex of the cycle. The result is that every vertex in the cycle has degree $3$. Repeat this until no vertex of degree $>3$ left. The graph $G$ obtained is $3$-regular and (clearly) $K_nleq G$. Then, $text{TW}(G)geqtext{TW}(K_n)=n-1$. So, the treewidth is unbounded even in the class of $3$-regular graphs.
$endgroup$
add a comment |
$begingroup$
A simple proof goes this way:
Take $K_n$ the complete graph in $n$ vertices. Take $vin G$, replace it with a cycle of length $d(v)=n-1$ and connect the $d(v)$ vertices in a 1-to-1 fashion to a vertex of the cycle. The result is that every vertex in the cycle has degree $3$. Repeat this until no vertex of degree $>3$ left. The graph $G$ obtained is $3$-regular and (clearly) $K_nleq G$. Then, $text{TW}(G)geqtext{TW}(K_n)=n-1$. So, the treewidth is unbounded even in the class of $3$-regular graphs.
$endgroup$
add a comment |
$begingroup$
A simple proof goes this way:
Take $K_n$ the complete graph in $n$ vertices. Take $vin G$, replace it with a cycle of length $d(v)=n-1$ and connect the $d(v)$ vertices in a 1-to-1 fashion to a vertex of the cycle. The result is that every vertex in the cycle has degree $3$. Repeat this until no vertex of degree $>3$ left. The graph $G$ obtained is $3$-regular and (clearly) $K_nleq G$. Then, $text{TW}(G)geqtext{TW}(K_n)=n-1$. So, the treewidth is unbounded even in the class of $3$-regular graphs.
$endgroup$
A simple proof goes this way:
Take $K_n$ the complete graph in $n$ vertices. Take $vin G$, replace it with a cycle of length $d(v)=n-1$ and connect the $d(v)$ vertices in a 1-to-1 fashion to a vertex of the cycle. The result is that every vertex in the cycle has degree $3$. Repeat this until no vertex of degree $>3$ left. The graph $G$ obtained is $3$-regular and (clearly) $K_nleq G$. Then, $text{TW}(G)geqtext{TW}(K_n)=n-1$. So, the treewidth is unbounded even in the class of $3$-regular graphs.
answered Dec 22 '18 at 17:14
Veridian DynamicsVeridian Dynamics
1,909312
1,909312
add a comment |
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.
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%2f3047080%2fexists-a-bound-for-threewidth-in-deltag-leq-3-graphs%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