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

5019




5019












  • $begingroup$
    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












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1 Answer
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1 Answer
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active

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active

oldest

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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

108k1162136




108k1162136












  • $begingroup$
    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


















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