Proof that $K_{3,3}$ is non planar using Euler's formula.












2












$begingroup$


I'm struggling to understand the proof that $K_{3,3}$ is nonplanar. Using Euler's formula we know that $3f leq 2e$. The proof goes like this:
If we had drawn the graph in the plane, there would be no triangles: this is because in any triangle either two wells or two houses would have to be connected, but that is not possible. So, summing up the sides of every face we get $4f leq 2e$. I don't understand where the 4f comes from.



Why is it that every face has at least four edges?






graph-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    3,3 thank you for the correction
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 5:54
















2












$begingroup$


I'm struggling to understand the proof that $K_{3,3}$ is nonplanar. Using Euler's formula we know that $3f leq 2e$. The proof goes like this:
If we had drawn the graph in the plane, there would be no triangles: this is because in any triangle either two wells or two houses would have to be connected, but that is not possible. So, summing up the sides of every face we get $4f leq 2e$. I don't understand where the 4f comes from.



Why is it that every face has at least four edges?






graph-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    3,3 thank you for the correction
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 5:54














2












2








2





$begingroup$


I'm struggling to understand the proof that $K_{3,3}$ is nonplanar. Using Euler's formula we know that $3f leq 2e$. The proof goes like this:
If we had drawn the graph in the plane, there would be no triangles: this is because in any triangle either two wells or two houses would have to be connected, but that is not possible. So, summing up the sides of every face we get $4f leq 2e$. I don't understand where the 4f comes from.



Why is it that every face has at least four edges?






graph-theory







share|cite|improve this question











$endgroup$




I'm struggling to understand the proof that $K_{3,3}$ is nonplanar. Using Euler's formula we know that $3f leq 2e$. The proof goes like this:
If we had drawn the graph in the plane, there would be no triangles: this is because in any triangle either two wells or two houses would have to be connected, but that is not possible. So, summing up the sides of every face we get $4f leq 2e$. I don't understand where the 4f comes from.



Why is it that every face has at least four edges?






graph-theory




graph-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 5:54







Matteo Ciccozzi

















asked Jan 11 at 5:49









Matteo CiccozziMatteo Ciccozzi

469




469












  • $begingroup$
    3,3 thank you for the correction
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 5:54


















  • $begingroup$
    3,3 thank you for the correction
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 5:54
















$begingroup$
3,3 thank you for the correction
$endgroup$
– Matteo Ciccozzi
Jan 11 at 5:54




$begingroup$
3,3 thank you for the correction
$endgroup$
– Matteo Ciccozzi
Jan 11 at 5:54










1 Answer
1






active

oldest

votes


















2












$begingroup$

As $K_{3,3}$ is a bipartite graph, each face is bounded by an even number of edges, so at least four. If there are $f$ faces, then the total number of edges
in their boundaries is $ge 4f$, but that total number is $2e$ as each edge
is in two faces, so $2ege 4f$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    but I don't understand the triangle all the way to the left has only three edges, yet it is a face
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 5:55












  • $begingroup$
    Ahhh, I think I see it now, the face I'm talking about is not actually a face because the intersection of the two lines does not contain a vertex. Correct?
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 6:08










  • $begingroup$
    @MatteoCiccozzi Which triangle and where is left? There is no triangle in a bipartite graph
    $endgroup$
    – Hagen von Eitzen
    Jan 11 at 6:09










  • $begingroup$
    Lets say we have u1, u2, u3 connecting to v1, v2, v3. We then have the following edges that form a triangle (or so I thought) {u1, v1}, {u1, v2}, and {u2, v1}
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 6:18












  • $begingroup$
    @MatteoCiccozzi Perhaps confusion arose because one should make clearer distinctions: The edges of a graph (as abstract object) do not intersect. In a planar embedding of a graph, the arcs that represent the edges might intersect - but then again, if they do, we do not actually have a planar embedding. - And $K_{3,3}$ can readily be "drawn" in 3D space
    $endgroup$
    – Hagen von Eitzen
    Jan 11 at 7:13











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%2f3069511%2fproof-that-k-3-3-is-non-planar-using-eulers-formula%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









2












$begingroup$

As $K_{3,3}$ is a bipartite graph, each face is bounded by an even number of edges, so at least four. If there are $f$ faces, then the total number of edges
in their boundaries is $ge 4f$, but that total number is $2e$ as each edge
is in two faces, so $2ege 4f$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    but I don't understand the triangle all the way to the left has only three edges, yet it is a face
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 5:55












  • $begingroup$
    Ahhh, I think I see it now, the face I'm talking about is not actually a face because the intersection of the two lines does not contain a vertex. Correct?
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 6:08










  • $begingroup$
    @MatteoCiccozzi Which triangle and where is left? There is no triangle in a bipartite graph
    $endgroup$
    – Hagen von Eitzen
    Jan 11 at 6:09










  • $begingroup$
    Lets say we have u1, u2, u3 connecting to v1, v2, v3. We then have the following edges that form a triangle (or so I thought) {u1, v1}, {u1, v2}, and {u2, v1}
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 6:18












  • $begingroup$
    @MatteoCiccozzi Perhaps confusion arose because one should make clearer distinctions: The edges of a graph (as abstract object) do not intersect. In a planar embedding of a graph, the arcs that represent the edges might intersect - but then again, if they do, we do not actually have a planar embedding. - And $K_{3,3}$ can readily be "drawn" in 3D space
    $endgroup$
    – Hagen von Eitzen
    Jan 11 at 7:13
















2












$begingroup$

As $K_{3,3}$ is a bipartite graph, each face is bounded by an even number of edges, so at least four. If there are $f$ faces, then the total number of edges
in their boundaries is $ge 4f$, but that total number is $2e$ as each edge
is in two faces, so $2ege 4f$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    but I don't understand the triangle all the way to the left has only three edges, yet it is a face
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 5:55












  • $begingroup$
    Ahhh, I think I see it now, the face I'm talking about is not actually a face because the intersection of the two lines does not contain a vertex. Correct?
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 6:08










  • $begingroup$
    @MatteoCiccozzi Which triangle and where is left? There is no triangle in a bipartite graph
    $endgroup$
    – Hagen von Eitzen
    Jan 11 at 6:09










  • $begingroup$
    Lets say we have u1, u2, u3 connecting to v1, v2, v3. We then have the following edges that form a triangle (or so I thought) {u1, v1}, {u1, v2}, and {u2, v1}
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 6:18












  • $begingroup$
    @MatteoCiccozzi Perhaps confusion arose because one should make clearer distinctions: The edges of a graph (as abstract object) do not intersect. In a planar embedding of a graph, the arcs that represent the edges might intersect - but then again, if they do, we do not actually have a planar embedding. - And $K_{3,3}$ can readily be "drawn" in 3D space
    $endgroup$
    – Hagen von Eitzen
    Jan 11 at 7:13














2












2








2





$begingroup$

As $K_{3,3}$ is a bipartite graph, each face is bounded by an even number of edges, so at least four. If there are $f$ faces, then the total number of edges
in their boundaries is $ge 4f$, but that total number is $2e$ as each edge
is in two faces, so $2ege 4f$.






share|cite|improve this answer









$endgroup$



As $K_{3,3}$ is a bipartite graph, each face is bounded by an even number of edges, so at least four. If there are $f$ faces, then the total number of edges
in their boundaries is $ge 4f$, but that total number is $2e$ as each edge
is in two faces, so $2ege 4f$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 11 at 5:53









Lord Shark the UnknownLord Shark the Unknown

103k1160132




103k1160132












  • $begingroup$
    but I don't understand the triangle all the way to the left has only three edges, yet it is a face
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 5:55












  • $begingroup$
    Ahhh, I think I see it now, the face I'm talking about is not actually a face because the intersection of the two lines does not contain a vertex. Correct?
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 6:08










  • $begingroup$
    @MatteoCiccozzi Which triangle and where is left? There is no triangle in a bipartite graph
    $endgroup$
    – Hagen von Eitzen
    Jan 11 at 6:09










  • $begingroup$
    Lets say we have u1, u2, u3 connecting to v1, v2, v3. We then have the following edges that form a triangle (or so I thought) {u1, v1}, {u1, v2}, and {u2, v1}
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 6:18












  • $begingroup$
    @MatteoCiccozzi Perhaps confusion arose because one should make clearer distinctions: The edges of a graph (as abstract object) do not intersect. In a planar embedding of a graph, the arcs that represent the edges might intersect - but then again, if they do, we do not actually have a planar embedding. - And $K_{3,3}$ can readily be "drawn" in 3D space
    $endgroup$
    – Hagen von Eitzen
    Jan 11 at 7:13


















  • $begingroup$
    but I don't understand the triangle all the way to the left has only three edges, yet it is a face
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 5:55












  • $begingroup$
    Ahhh, I think I see it now, the face I'm talking about is not actually a face because the intersection of the two lines does not contain a vertex. Correct?
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 6:08










  • $begingroup$
    @MatteoCiccozzi Which triangle and where is left? There is no triangle in a bipartite graph
    $endgroup$
    – Hagen von Eitzen
    Jan 11 at 6:09










  • $begingroup$
    Lets say we have u1, u2, u3 connecting to v1, v2, v3. We then have the following edges that form a triangle (or so I thought) {u1, v1}, {u1, v2}, and {u2, v1}
    $endgroup$
    – Matteo Ciccozzi
    Jan 11 at 6:18












  • $begingroup$
    @MatteoCiccozzi Perhaps confusion arose because one should make clearer distinctions: The edges of a graph (as abstract object) do not intersect. In a planar embedding of a graph, the arcs that represent the edges might intersect - but then again, if they do, we do not actually have a planar embedding. - And $K_{3,3}$ can readily be "drawn" in 3D space
    $endgroup$
    – Hagen von Eitzen
    Jan 11 at 7:13
















$begingroup$
but I don't understand the triangle all the way to the left has only three edges, yet it is a face
$endgroup$
– Matteo Ciccozzi
Jan 11 at 5:55






$begingroup$
but I don't understand the triangle all the way to the left has only three edges, yet it is a face
$endgroup$
– Matteo Ciccozzi
Jan 11 at 5:55














$begingroup$
Ahhh, I think I see it now, the face I'm talking about is not actually a face because the intersection of the two lines does not contain a vertex. Correct?
$endgroup$
– Matteo Ciccozzi
Jan 11 at 6:08




$begingroup$
Ahhh, I think I see it now, the face I'm talking about is not actually a face because the intersection of the two lines does not contain a vertex. Correct?
$endgroup$
– Matteo Ciccozzi
Jan 11 at 6:08












$begingroup$
@MatteoCiccozzi Which triangle and where is left? There is no triangle in a bipartite graph
$endgroup$
– Hagen von Eitzen
Jan 11 at 6:09




$begingroup$
@MatteoCiccozzi Which triangle and where is left? There is no triangle in a bipartite graph
$endgroup$
– Hagen von Eitzen
Jan 11 at 6:09












$begingroup$
Lets say we have u1, u2, u3 connecting to v1, v2, v3. We then have the following edges that form a triangle (or so I thought) {u1, v1}, {u1, v2}, and {u2, v1}
$endgroup$
– Matteo Ciccozzi
Jan 11 at 6:18






$begingroup$
Lets say we have u1, u2, u3 connecting to v1, v2, v3. We then have the following edges that form a triangle (or so I thought) {u1, v1}, {u1, v2}, and {u2, v1}
$endgroup$
– Matteo Ciccozzi
Jan 11 at 6:18














$begingroup$
@MatteoCiccozzi Perhaps confusion arose because one should make clearer distinctions: The edges of a graph (as abstract object) do not intersect. In a planar embedding of a graph, the arcs that represent the edges might intersect - but then again, if they do, we do not actually have a planar embedding. - And $K_{3,3}$ can readily be "drawn" in 3D space
$endgroup$
– Hagen von Eitzen
Jan 11 at 7:13




$begingroup$
@MatteoCiccozzi Perhaps confusion arose because one should make clearer distinctions: The edges of a graph (as abstract object) do not intersect. In a planar embedding of a graph, the arcs that represent the edges might intersect - but then again, if they do, we do not actually have a planar embedding. - And $K_{3,3}$ can readily be "drawn" in 3D space
$endgroup$
– Hagen von Eitzen
Jan 11 at 7:13


















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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3069511%2fproof-that-k-3-3-is-non-planar-using-eulers-formula%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

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
Read more

Antonio Litta Visconti Arese

Image
Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
Read more