Mutually disjoint triangles in certain planar graph












1














Let G be a connected, planar graph for which every vertex has degree 3, except that one vertex has degree 2.



Is it possible to construct an example of such G whose only odd faces are triangles, and for which no two such triangles share a common vertex?



(For my purposes I may assume there are no faces of length 1 or 2, in which case there must be an odd face of length at least 3. By the handshake lemma, there are an even number of, hence at least 2, such odd faces.)






combinatorics discrete-mathematics graph-theory







share|cite|improve this question
























  • For my own edification.. what is an odd face ?
    – T. Ford
    2 hours ago










  • Face with odd number of sides.
    – Zachary Hunter
    2 hours ago










  • I think "those bounded by an odd length cycle" is a sensible definition.
    – Finallysignedup
    2 hours ago
















1














Let G be a connected, planar graph for which every vertex has degree 3, except that one vertex has degree 2.



Is it possible to construct an example of such G whose only odd faces are triangles, and for which no two such triangles share a common vertex?



(For my purposes I may assume there are no faces of length 1 or 2, in which case there must be an odd face of length at least 3. By the handshake lemma, there are an even number of, hence at least 2, such odd faces.)






combinatorics discrete-mathematics graph-theory







share|cite|improve this question
























  • For my own edification.. what is an odd face ?
    – T. Ford
    2 hours ago










  • Face with odd number of sides.
    – Zachary Hunter
    2 hours ago










  • I think "those bounded by an odd length cycle" is a sensible definition.
    – Finallysignedup
    2 hours ago














1












1








1







Let G be a connected, planar graph for which every vertex has degree 3, except that one vertex has degree 2.



Is it possible to construct an example of such G whose only odd faces are triangles, and for which no two such triangles share a common vertex?



(For my purposes I may assume there are no faces of length 1 or 2, in which case there must be an odd face of length at least 3. By the handshake lemma, there are an even number of, hence at least 2, such odd faces.)






combinatorics discrete-mathematics graph-theory







share|cite|improve this question















Let G be a connected, planar graph for which every vertex has degree 3, except that one vertex has degree 2.



Is it possible to construct an example of such G whose only odd faces are triangles, and for which no two such triangles share a common vertex?



(For my purposes I may assume there are no faces of length 1 or 2, in which case there must be an odd face of length at least 3. By the handshake lemma, there are an even number of, hence at least 2, such odd faces.)






combinatorics discrete-mathematics graph-theory




combinatorics discrete-mathematics graph-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago

























asked 2 hours ago









Finallysignedup

495




495












  • For my own edification.. what is an odd face ?
    – T. Ford
    2 hours ago










  • Face with odd number of sides.
    – Zachary Hunter
    2 hours ago










  • I think "those bounded by an odd length cycle" is a sensible definition.
    – Finallysignedup
    2 hours ago


















  • For my own edification.. what is an odd face ?
    – T. Ford
    2 hours ago










  • Face with odd number of sides.
    – Zachary Hunter
    2 hours ago










  • I think "those bounded by an odd length cycle" is a sensible definition.
    – Finallysignedup
    2 hours ago
















For my own edification.. what is an odd face ?
– T. Ford
2 hours ago




For my own edification.. what is an odd face ?
– T. Ford
2 hours ago












Face with odd number of sides.
– Zachary Hunter
2 hours ago




Face with odd number of sides.
– Zachary Hunter
2 hours ago












I think "those bounded by an odd length cycle" is a sensible definition.
– Finallysignedup
2 hours ago




I think "those bounded by an odd length cycle" is a sensible definition.
– Finallysignedup
2 hours ago










2 Answers
2






active

oldest

votes


















2














Here is one such graph - or, more precisely, one such plane embedding of a graph, since the lengths of the faces are not properties of the graph itself. It has two faces of length 3 (including the external face), two faces of length 4, and two faces of length 6.



enter image description here



(Motivation: we start with a triangular pyramid, which has all the required properties except for the degree-2 vertex, and modify it a little to make it work.)






share|cite|improve this answer





















  • Thanks. I now wish to only have degree 3 vertices. I've made a new (hopefully final) question.
    – Finallysignedup
    1 hour ago



















2














Take two cycles of the same order 2n+1, $C_a$ and $C_b$ and number each in clockwise fashion.



"Glue together" $v_{1a}$ and $v_{1b}$, and then do the same for $v_{2a}$ and $v_{2b}$, so that there is just one edge between $v_1$ and $v_2$.



Then, for each $i > 2$ create an edge between $v_{ia}$ and $v_{ib}$.



Finally, subdivide the edge between $v_1$ and $v_2$.






share|cite|improve this answer








New contributor




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


















  • Thanks. Sorry that I could not tick both answers.
    – Finallysignedup
    1 hour ago










  • Understandable, it is the fault of my laziness in creating diagrams.
    – Zachary Hunter
    1 hour ago











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056515%2fmutually-disjoint-triangles-in-certain-planar-graph%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









2














Here is one such graph - or, more precisely, one such plane embedding of a graph, since the lengths of the faces are not properties of the graph itself. It has two faces of length 3 (including the external face), two faces of length 4, and two faces of length 6.



enter image description here



(Motivation: we start with a triangular pyramid, which has all the required properties except for the degree-2 vertex, and modify it a little to make it work.)






share|cite|improve this answer





















  • Thanks. I now wish to only have degree 3 vertices. I've made a new (hopefully final) question.
    – Finallysignedup
    1 hour ago
















2














Here is one such graph - or, more precisely, one such plane embedding of a graph, since the lengths of the faces are not properties of the graph itself. It has two faces of length 3 (including the external face), two faces of length 4, and two faces of length 6.



enter image description here



(Motivation: we start with a triangular pyramid, which has all the required properties except for the degree-2 vertex, and modify it a little to make it work.)






share|cite|improve this answer





















  • Thanks. I now wish to only have degree 3 vertices. I've made a new (hopefully final) question.
    – Finallysignedup
    1 hour ago














2












2








2






Here is one such graph - or, more precisely, one such plane embedding of a graph, since the lengths of the faces are not properties of the graph itself. It has two faces of length 3 (including the external face), two faces of length 4, and two faces of length 6.



enter image description here



(Motivation: we start with a triangular pyramid, which has all the required properties except for the degree-2 vertex, and modify it a little to make it work.)






share|cite|improve this answer












Here is one such graph - or, more precisely, one such plane embedding of a graph, since the lengths of the faces are not properties of the graph itself. It has two faces of length 3 (including the external face), two faces of length 4, and two faces of length 6.



enter image description here



(Motivation: we start with a triangular pyramid, which has all the required properties except for the degree-2 vertex, and modify it a little to make it work.)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 hours ago









Misha Lavrov

43.8k555104




43.8k555104












  • Thanks. I now wish to only have degree 3 vertices. I've made a new (hopefully final) question.
    – Finallysignedup
    1 hour ago


















  • Thanks. I now wish to only have degree 3 vertices. I've made a new (hopefully final) question.
    – Finallysignedup
    1 hour ago
















Thanks. I now wish to only have degree 3 vertices. I've made a new (hopefully final) question.
– Finallysignedup
1 hour ago




Thanks. I now wish to only have degree 3 vertices. I've made a new (hopefully final) question.
– Finallysignedup
1 hour ago











2














Take two cycles of the same order 2n+1, $C_a$ and $C_b$ and number each in clockwise fashion.



"Glue together" $v_{1a}$ and $v_{1b}$, and then do the same for $v_{2a}$ and $v_{2b}$, so that there is just one edge between $v_1$ and $v_2$.



Then, for each $i > 2$ create an edge between $v_{ia}$ and $v_{ib}$.



Finally, subdivide the edge between $v_1$ and $v_2$.






share|cite|improve this answer








New contributor




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


















  • Thanks. Sorry that I could not tick both answers.
    – Finallysignedup
    1 hour ago










  • Understandable, it is the fault of my laziness in creating diagrams.
    – Zachary Hunter
    1 hour ago
















2














Take two cycles of the same order 2n+1, $C_a$ and $C_b$ and number each in clockwise fashion.



"Glue together" $v_{1a}$ and $v_{1b}$, and then do the same for $v_{2a}$ and $v_{2b}$, so that there is just one edge between $v_1$ and $v_2$.



Then, for each $i > 2$ create an edge between $v_{ia}$ and $v_{ib}$.



Finally, subdivide the edge between $v_1$ and $v_2$.






share|cite|improve this answer








New contributor




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


















  • Thanks. Sorry that I could not tick both answers.
    – Finallysignedup
    1 hour ago










  • Understandable, it is the fault of my laziness in creating diagrams.
    – Zachary Hunter
    1 hour ago














2












2








2






Take two cycles of the same order 2n+1, $C_a$ and $C_b$ and number each in clockwise fashion.



"Glue together" $v_{1a}$ and $v_{1b}$, and then do the same for $v_{2a}$ and $v_{2b}$, so that there is just one edge between $v_1$ and $v_2$.



Then, for each $i > 2$ create an edge between $v_{ia}$ and $v_{ib}$.



Finally, subdivide the edge between $v_1$ and $v_2$.






share|cite|improve this answer








New contributor




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









Take two cycles of the same order 2n+1, $C_a$ and $C_b$ and number each in clockwise fashion.



"Glue together" $v_{1a}$ and $v_{1b}$, and then do the same for $v_{2a}$ and $v_{2b}$, so that there is just one edge between $v_1$ and $v_2$.



Then, for each $i > 2$ create an edge between $v_{ia}$ and $v_{ib}$.



Finally, subdivide the edge between $v_1$ and $v_2$.







share|cite|improve this answer








New contributor




Zachary Hunter 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 answer



share|cite|improve this answer






New contributor




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









answered 2 hours ago









Zachary Hunter

1245




1245




New contributor




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





New contributor





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






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












  • Thanks. Sorry that I could not tick both answers.
    – Finallysignedup
    1 hour ago










  • Understandable, it is the fault of my laziness in creating diagrams.
    – Zachary Hunter
    1 hour ago


















  • Thanks. Sorry that I could not tick both answers.
    – Finallysignedup
    1 hour ago










  • Understandable, it is the fault of my laziness in creating diagrams.
    – Zachary Hunter
    1 hour ago
















Thanks. Sorry that I could not tick both answers.
– Finallysignedup
1 hour ago




Thanks. Sorry that I could not tick both answers.
– Finallysignedup
1 hour ago












Understandable, it is the fault of my laziness in creating diagrams.
– Zachary Hunter
1 hour ago




Understandable, it is the fault of my laziness in creating diagrams.
– Zachary Hunter
1 hour ago


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056515%2fmutually-disjoint-triangles-in-certain-planar-graph%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

404 Error Contact Form 7 ajax form submitting

Image
0 The contact form does not work, and affects loading after one try. The chrome console shows the error message: POST http://example.com/wp-json/contact-form-7/v1/contact-forms/50/feedback 404 (Not Found) send @ jquery.js?ver=1.12.4:4 ajax @ jquery.js?ver=1.12.4:4 wpcf7.submit @ scripts.js?ver=5.0.3:346 (anonymous) @ scripts.js?ver=5.0.3:53 dispatch @ jquery.js?ver=1.12.4:3 r.handle @ jquery.js?ver=1.12.4:3 How to solve this issue? thanks in advance javascript wordpress contact-form-7 share | improve this question edited Jul 17 '18 at 5:39 Kiran Shahi 4,745 3 16 43
Read more

How to know if a Active Directory user can login interactively

Image
1 I would like to know if and how is it possible to know if a AD user can login interactively (on a server) in a domain. I need to know if I can find it out using a LDAP search. Thanks. windows active-directory login user-permissions user-profile share | improve this question edited 1 hour ago yagmoth555 ♦ 11.6k 3 17 42 asked 1 hour ago Luigi Luigi 6 1 New contributor
Read more

TypeError: fit_transform() missing 1 required positional argument: 'X'

Image
0 I am trying to do Feature Scaling in a dataset, but I get an error and have no idea how to proceed: > Traceback (most recent call last): > > File "<ipython-input-10-71bea414b4d0>", line 22, in <module> > x_train = sc_X.fit_transform(x_train) > > TypeError: fit_transform() missing 1 required positional argument: 'X' and here is my code: import pandas as pd # Importing the dataset dataset = pd.read_csv('Data.csv') X = dataset.iloc[:, :-1].values y = dataset.iloc[:, 3].values # Taking care of missing data from sklearn.preprocessing import Imputer imputer = Imputer(missing_values="NaN", strategy="mean", axis=0) imputer = Imputer.fit(imputer,X[:,1:3]) X[:, 1:3] = Imputer.transform(imputer,X[
Read more