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I dislike the proof of Kuratowski's theorem in my textbook, but the book mentions a theorem of William Tutte:

Theorem: A graph $G$ is planar if and only if the conflict graph of each cycle of $G$ is bipartite.

Where the conflict graph of the cycle is, more or less, the graph having the "chords" of the cycle as vertices, and two chords are adjacent in the conflict graph iff they conflict, in the sense that they'll necessarily cross if either both are drawn inside the cycle, or both are drawn outside the cycle.

I would like a proof of the above theorem that doesn't rely on Kuratowski's theorem.

The book references this paper, but I can't find anything in there that resembles the theorem I'm looking for. I don't know anything about matroids at all, so presumably the theorem is in there, but expressed in language that makes it unrecognizable to me.

I'd appreciate either a link to a more graph-theoretical proof, or an explanation as to which proposition in the linked paper is the one I want, and how I can decode it from the language of matroids into something I'll understand.

MJD
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Jack M
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1 Answers1

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AFAIS “chord” of cycle in your note is a path which has no vertices of cycle except the first and the last ones (in contrary to the usual definition that chord is an edge). Otherwise subdivide each edge of any non-planar graph to get counterexample. Also I assume that for two “chords” their common vertex may be considered as virtual “chord” conflicting with both of them. Otherwise another counterexample can be given. However saying that such virual “chord” should be inside the cycle I mean that corresponding vertex should be outside and vice versa.

$\Rightarrow$. If graph $G$ has a cycle $C$ with non-bipartite conflict graph $H$ then $G$ is not planar.

Suppose that $G$ is plane and consider odd length cycle $h_1h_2\ldots h_{2k+1}h_1$ in graph $H$, where $h_i$ are “chords” of $C$. If $h_i$ is inside $C$ then $h_{i+1}$ should be outside $C$ not to intersect with $h_i$ and vice versa. If $h_1$ is inside $C$ then $h_{2k+1}$ is inside $C$, too, and they intersect. The same if $h_1$ is outside $C$. $\square$

$\Leftarrow$. If graph $G$ has no cycle with non-bipartite conflict graph then $G$ is planar.

Let start with empty graph (which is planar) and inductively add edges of $G$. By induction hypothesis $G - e$ is planar. If there is no cycle in $G$ containg $e = uv$ then $G$ is obviously planar (since $e$ just connects two planar connected components $G_u$ and $G_v$ and it is possible to make any face of planar graph to be outer face; so we can make $u$ to belong outer face of $G_u$, $v$ to belong outer face of $G_v$ and add $e = uv$). So let $e$ belong to some cycle of $G$. Now for each such cycle $C$ of $G - e$ such that $\{\,u, v\,\} \subset V(C)$ we know whether $e$ needs to be inside or outside $C$ not to intersect with other chords (for some cycles both inside and outside locations are possible, but such cycles give no restrictions at all). The only possible problem for $G$ to be planar is that there few cycles containing $e$ as chord have conflicting requirements. Then we can produce the outermost cycle $C$ containing $u$ and $v$ using vertices and edges of these cycles. Cycle would have two chords conflicting with $e$ and either they conflict with each other or the inner chord has at least one common vertex with another chord that conflicts with the outer one. In both cases we get contradition with condition that conflict graph of $C$ is bipartite. $\square$

P. S. I understand that the last part of my proof is not strict enough however if it is not clear I can make picture or think about better description.

Smylic
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