Prove Vizing Class 1 if maximum degree vertices induce forest by induction











up vote
1
down vote

favorite












I tried to prove that if for a simple graph $G$, vertices of maximum degree induce a forest, then $chi'(G) leq Delta(G)$, in other words, the graph is in Vizing Class 1.



When I applied induction on $|E(G)|$, I encountered the problem, that if there is only one vertex $v in V(G)$ having maximum degree, I cannot ensure that I can apply the induction hypothesis on $Gsetminus e$ for an edge $e$ incident with $v$.



If I remove $e$, there might be vertices of degree $Delta(G) - 1$ inducing a cycle.



How can I solve this problem? Is there a way to fix this proof?






graph-theory







share|cite|improve this question







New contributor




Lydia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Check this math.stackexchange.com/questions/1009600/…
    – hbm
    14 hours ago










  • i tried to do the same, but i think there is a problem there. I can't apply the induction hypothesis, because $v$ might be the only vertex having maximum degree
    – Lydia
    12 hours ago










  • the proof still applies. In the case of only one vertex, just remove any edge incident with it.
    – hbm
    12 hours ago















up vote
1
down vote

favorite












I tried to prove that if for a simple graph $G$, vertices of maximum degree induce a forest, then $chi'(G) leq Delta(G)$, in other words, the graph is in Vizing Class 1.



When I applied induction on $|E(G)|$, I encountered the problem, that if there is only one vertex $v in V(G)$ having maximum degree, I cannot ensure that I can apply the induction hypothesis on $Gsetminus e$ for an edge $e$ incident with $v$.



If I remove $e$, there might be vertices of degree $Delta(G) - 1$ inducing a cycle.



How can I solve this problem? Is there a way to fix this proof?






graph-theory







share|cite|improve this question







New contributor




Lydia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Check this math.stackexchange.com/questions/1009600/…
    – hbm
    14 hours ago










  • i tried to do the same, but i think there is a problem there. I can't apply the induction hypothesis, because $v$ might be the only vertex having maximum degree
    – Lydia
    12 hours ago










  • the proof still applies. In the case of only one vertex, just remove any edge incident with it.
    – hbm
    12 hours ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I tried to prove that if for a simple graph $G$, vertices of maximum degree induce a forest, then $chi'(G) leq Delta(G)$, in other words, the graph is in Vizing Class 1.



When I applied induction on $|E(G)|$, I encountered the problem, that if there is only one vertex $v in V(G)$ having maximum degree, I cannot ensure that I can apply the induction hypothesis on $Gsetminus e$ for an edge $e$ incident with $v$.



If I remove $e$, there might be vertices of degree $Delta(G) - 1$ inducing a cycle.



How can I solve this problem? Is there a way to fix this proof?






graph-theory







share|cite|improve this question







New contributor




Lydia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I tried to prove that if for a simple graph $G$, vertices of maximum degree induce a forest, then $chi'(G) leq Delta(G)$, in other words, the graph is in Vizing Class 1.



When I applied induction on $|E(G)|$, I encountered the problem, that if there is only one vertex $v in V(G)$ having maximum degree, I cannot ensure that I can apply the induction hypothesis on $Gsetminus e$ for an edge $e$ incident with $v$.



If I remove $e$, there might be vertices of degree $Delta(G) - 1$ inducing a cycle.



How can I solve this problem? Is there a way to fix this proof?






graph-theory




graph-theory






share|cite|improve this question







New contributor




Lydia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Lydia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Lydia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 23 hours ago









Lydia

62




62




New contributor




Lydia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Lydia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Lydia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • Check this math.stackexchange.com/questions/1009600/…
    – hbm
    14 hours ago










  • i tried to do the same, but i think there is a problem there. I can't apply the induction hypothesis, because $v$ might be the only vertex having maximum degree
    – Lydia
    12 hours ago










  • the proof still applies. In the case of only one vertex, just remove any edge incident with it.
    – hbm
    12 hours ago


















  • Check this math.stackexchange.com/questions/1009600/…
    – hbm
    14 hours ago










  • i tried to do the same, but i think there is a problem there. I can't apply the induction hypothesis, because $v$ might be the only vertex having maximum degree
    – Lydia
    12 hours ago










  • the proof still applies. In the case of only one vertex, just remove any edge incident with it.
    – hbm
    12 hours ago
















Check this math.stackexchange.com/questions/1009600/…
– hbm
14 hours ago




Check this math.stackexchange.com/questions/1009600/…
– hbm
14 hours ago












i tried to do the same, but i think there is a problem there. I can't apply the induction hypothesis, because $v$ might be the only vertex having maximum degree
– Lydia
12 hours ago




i tried to do the same, but i think there is a problem there. I can't apply the induction hypothesis, because $v$ might be the only vertex having maximum degree
– Lydia
12 hours ago












the proof still applies. In the case of only one vertex, just remove any edge incident with it.
– hbm
12 hours ago




the proof still applies. In the case of only one vertex, just remove any edge incident with it.
– hbm
12 hours ago















active

oldest

votes











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',
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
});


}
});






Lydia is a new contributor. Be nice, and check out our Code of Conduct.










 

draft saved


draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2995381%2fprove-vizing-class-1-if-maximum-degree-vertices-induce-forest-by-induction%23new-answer', 'question_page');
}
);

Post as a guest























active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes








Lydia is a new contributor. Be nice, and check out our Code of Conduct.










 

draft saved


draft discarded


















Lydia is a new contributor. Be nice, and check out our Code of Conduct.













Lydia is a new contributor. Be nice, and check out our Code of Conduct.












Lydia is a new contributor. Be nice, and check out our Code of Conduct.















 


draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2995381%2fprove-vizing-class-1-if-maximum-degree-vertices-induce-forest-by-induction%23new-answer', 'question_page');
}
);

Post as a guest






































































-

Popular posts from this blog

Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

Image
2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
Read more

ComboBox Display Member on multiple fields

Image
2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
Read more

Is it possible to collect Nectar points via Trainline?

Image
2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
Read more