Talk:Planar graph

old comments
I'm guessing that this sentence "The number of unlabeled (non-isomorphic) planar graphs on n vertices is between 27.2^n and 30.06^n" is dealing with asymptotics, since it clearly doesn't hold for small n.

— Preceding unsigned comment added by 98.232.136.74 (talk) 16:41, 23 April 2014 (UTC) In Kazimierz Kuratowski, the following version of Kuratowski's theorem was given:
 * a graph with no vertices of order 2 is nonplanar if and only if it contains a copy of K 5 or K 3,3.

This statement is incorrect. Take K5, pick one of its edges, say A --- B, and introduce two new vertices X and Y to change the edge to

Y            | |     A --- X --- B

The resulting graph is non-planar, has no degree-two vertex, and has no subgraph of form K 5 or K 3,3. AxelBoldt 00:58 13 Jun 2003 (UTC)

I removed the paragraph
 * For a connected planar graph G we may construct a graph whose vertices are the regions into which G divides the plane (including a single external region). The edges represent adjacency of regions: there is one for each edge of G, and can be shown as crossing it. The resulting graph G* is naturally also planar: it is called the planar dual graph, or just dual graph, with respect to the given plane embedding of G. We have G** = G, justifying the name dual.

The case of a tree shows that G** need not equal G, and so this operation needs to be defined differently to deserve the name "dual". Maybe it should be restricted to graphs arising from simple polyhedra? AxelBoldt 19:57, 26 Oct 2003 (UTC)

OK, the double dual clearly doesn't work unless crossing an edge takes you into a different region. That looks like the only condition?

Charles Matthews 06:43, 27 Oct 2003 (UTC)

I think so, if we allow our graphs to have multiple edges. AxelBoldt 14:24, 27 Oct 2003 (UTC)

The dual graph is always taken to be a multigraph and the imbedding in the plane is important. The dual of a tree is a single vertex with a whole lot of loops on it (one loop for each edge of the tree), but the way that the loops are imbedded in the plane (which loops are inside which other loops) encodes the tree structure so it is still true that G** = G. --Zero 14:59, 27 Oct 2003 (UTC)

Aha! Now we're getting somewhere. We only need a clear description of how to get from one embedded connected multigraph to the (or a?) dual embedded multigraph.

If G is the graph consisting of a single edge connecting two vertices, would G* be the one-vertex graph with two separate loops, or with one loop inside the other? Or do we work on the sphere where the two are the same? AxelBoldt 09:40, 28 Oct 2003 (UTC)

It's a vertex with one loop; you put a vertex in the distinct regions (only 1) and where there's an edge (only 1) you cross it, so you only have a loop. Dysprosia 10:03, 28 Oct 2003 (UTC)

There is always one edge in G** for each edge of G. The dual of the graph you mention has one vertex with one loop on it. To answer the last part, the imbedding is regarded as being on the sphere for most purposes. That's one of the reasons it's a bit hard to define the concept of a dual precisely without more background theory on the nature of imbeddings. --Zero 10:10, 28 Oct 2003 (UTC)

Oops, above I meant the graph G with three vertices, connected by two edges. It seems there are the two possibilities for the dual I mentioned above; but if we do work on the sphere they are the same and all is well. AxelBoldt 10:25, 28 Oct 2003 (UTC)

If you mean something like * \   *  / *

the dual will be  _ / \ | * | \  /   o  * / \ | * |  \_/

(the *s are to show the relative position of the previous vertices) Like Zero mentioned, we have an edge crossing for an edge. We can't have just one loop around that apex vertex. Dysprosia 10:28, 28 Oct 2003 (UTC)

__________    /   _      \   |   /  \     |    \   | * |    |    \  \  /     |     -- o  *   / \   /         *

would be another possibility for the dual; on a sphere, the two are equivalent, but not in the plane. AxelBoldt 11:41, 28 Oct 2003 (UTC)

Yepyep :) These graphs are isomorphic, I think, too... Dysprosia 11:41, 28 Oct 2003 (UTC)

If we allow infinite planar graphs: does Kuratowski remain true for countably infinite graphs? How about the four-color theorem? AxelBoldt 13:10, 30 Oct 2003 (UTC)

A graph is k-colorable if all its finite subgraphs are k-colorable, so 4CT is true even without the "countable". I think Kuratowski is more complicated. --Zero 13:41, 30 Oct 2003 (UTC)

Kuratowski showed:
 * that the only non-planar graphs are those that contain a subdivision of K 5 or K 3,3 obtained by replacing edges with paths.

The G** = G equality holds even for simple connected duals iff the edge-connectivity of G is strictly greater than 2. --- John Fremlin  Sun Jan 18 03:00:37 GMT 2004

Fourth picture
I believe that it is possible to redraw the fourth picture without any intersections by moving the top-right point to a point a little to the right of the lower-left, and moving the top-left to down below the bottom. Is this an error, or am I missing something?

--Miko 10:45, July 23, 2005 (UTC)


 * If you referring to the K3,3 picture, you are mistaken. It is impossible.  --Zero 11:11, 23 July 2005 (UTC)


 * It is possible to draw this graph with only one crossing. I can't understand your description, but if you take a closer look at your redrawing you should be able to find your crossing. Deco 21:31, 23 July 2005 (UTC)

infinite graphs
Does Kuratowski's theorem apply to infinite graphs? Infinite trees are mentioned - what other types of infinite graphs are there? It's clear that there are $$2^{\aleph_0}$$ infinite graphs - how many infinite planar graphs are there?
 * Define "infinite graph" first. mikka (t) 19:41, 16 August 2005 (UTC)


 * The definition is the same. A graph is a set V of vertices, and a set E of edges, each of which is a set of two vertices.  If V is infinite, the graph is infinite.  The graph, whether finite or infinite, is planar if it can be embedded in the plane so that no edges intersect.


 * With this definition, the reunion of all infinite graphs is not a set (under ZFC) and thus has no cardinal --81.202.212.241 17:28, 3 October 2006 (UTC)


 * Graph theorists often talk about the graph K&omega;, for example; this is the graph which has a countably infinite set of vertices, and an edge between each distinct pair of vertices. -- Dominus 14:26, 8 November 2005 (UTC)


 * More-or-less, yes. See "Planarity and Duality of Finite and Infinite Graphs", Carsten Thomassen, J. Comb. Theory B, 29 244-271 (1980).


 * The corresponding theorem is as follows. A graph is planar iff 1. it has at most continuum many vertices, 2. it has at most countably many vertices with degree 3 or more, 3. it does not have a topological K5 or K3,3. Kope (talk) 20:17, 16 December 2008 (UTC)

Modifications
The characterization by forbidden minors is due to Wagner, not to Kuratowksi (both characterization have been published in 1937). I have added the reference to Wagner's paper.

I have added several links to further characterizations, but many are missing.


 * What exactly is the difference between the two statements? A subgraph homeomorphic to $$K_{3,3}$$ or $$K_5$$ IS a minor of the graph isomorphic to $$K_{3,3}$$ or $$K_5$$. I have always heard both statements referred to interchangeably as Kuratowski's Theorem. I guess I'll leave it this way for now. ---SpaceMoose 12:15, 22 February 2006 (UTC)

The difference is that a minor of $$K_5$$ is not necessarily a subdivision of $$K_5$$. It happens that a minor of $$K_5$$ is either a subdivision of $$K_5$$ or a subdivision of $$K_{3,3}$$ and thus Wagner and Kuratowski statements are equivalent. If you consider Hadwiger conjecture (a graph with chromatic number k has a $$K_k$$ minor) and Hajos conjecture (a graph with chromatic number k includes a subgraph homeomorphic to $$K_k$$) then the difference is essential: the first conjecture is still open while the second is false. It has also to be noticed that Wagner's characterization was motivation of Wagner's conjecture, later proved by Robertson and Seymour in a long series of papers: any minor closed familly of finite graphs is characterized by a finite set of forbidden minors.

--pom 00:32, 2 March 2006 (UTC)


 * Seems to me that a minor of $$K_5$$ has at most 5 vertices, so it can't be a subdivision of $$K_{3,3}$$. Maybe it is the other way around?  --Zero 12:41, 2 March 2006 (UTC)


 * Sorry, I "took my feet in the carpet" as we say in french. The good formulation is that if a graph has a $$K_5$$ minor, it does not need to include a subdivision of $$K_5$$. However, graphs with a $$K_5$$ minor include a subdivision of $$K_5$$ or $$K_{3,3}$$ pom 16:13, 2 March 2006 (UTC)

Newbie
I'm new to this, and I just want someone to clarify what is meant by "redrawing" the second picture. If you redraw it so that it is planar, could it look like this?: .___________. . Well, something like that? Sorry, I'm not very good at ASCII art. Chris53516 02:55, 5 July 2006 (UTC)
 * I think I answered my own question. See Four color theorem for the graphic. Chris53516 03:01, 5 July 2006 (UTC)

Plane graph
No exact definition was supplied for  plane graph - I hope mine is good enough. Rp 22:26, 5 July 2006 (UTC)
 * Actually, the usual definition of a plane graph is a graph embedded in the plane (up to topological equivalence). A plane embedding is defined by the circular order of the edges around the vertices (which define an embedding in the sphere) and the description of the unbounded face. pom 19:08, 3 October 2006 (UTC)

Too technical
I remember simplifying the 1st paragraph of the article in March. But already the second and third paragraphs are way too technical. This article should be readable by youngsters curious of mathematics. I suggest we try to present elementary notions and results (e.g., proof that K3,3 is not planar based on Euler equation for planer graphs) without using cryptic language. PhS 15:02, 20 November 2006 (UTC)

Genus
The article should probably explain somewhere the equivalence between "embeddable in a plane" and "embeddable in a sphere", mention that planar graphs are graphs of genus zero and hint to the fact that the genus gives a finer and more useful distinction than planar / non-planar. I would add this myself if I knew how to best present it. Ideas welcome. Morana (talk) 21:08, 7 April 2008 (UTC)

Maximal Planar Graphs
This is just a little something for everyone. It's a derivation I saw, but can't find a RS to confirm it. If we could it would be a good contribution to the article. So, the derivation is this:

Take any maximal planar graph G. It follows $$$$ $$V(G),E(G),F(G):\rightarrow V,3(V-2),2(V-2)$$

Basically, the number of faces and edges can be derived from the number of vertices and visa versa. Has anyone seen anything on this before? InfoNation101 |  talk  | 18:35, 22 April 2008 (UTC)


 * Nevermind. Didn't catch that part on the page until after a couple times through.  InfoNation101  |  talk  | 18:38, 22 April 2008 (UTC)

Standard terminology: "Plane triangulation" = maximal planar graph. "Triangular graph" = line graph of complete graph. "Triangulated graph" = chordal graph. I've noted these errors in the article. (Weisstein is not a reliable authority.) Zaslav (talk) 03:53, 4 August 2012 (UTC)
 * Calling them "errors" is too much editorialization. If those terms more commonly mean something else, we should say that, but we should not be trying to judge what is erroneous. —David Eppstein (talk) 05:01, 4 August 2012 (UTC)

Kuratowski's theorem
Either we need to clarify that a graph can be a subdivision of itself (subdividing zero times) or we need to add to the statement of Kuratowski's theorem that a graph can't also contain a subgraph that is isomorphic to K5, etc. The article's definition of subdivision mentions repeating subdividing zero or more times which doesn't mean subdividing zero times. The concept of subdividing links to Homeomorphism (graph theory), which may make this all clear if I had studied it and thus been able to understand the article, but I haven't. The second option above is the way I've seen it in textbooks, which is what made me assume a graph is not a subdivision of itself. - Taxman Talk 14:43, 31 May 2008 (UTC)


 * Your assumption is mistaken. A graph is allowed to be a subdivision of itself. See the word "zero" in the phrase you quoted, "subdividing zero or more times". —David Eppstein (talk) 18:35, 31 May 2008 (UTC)
 * But you missed the emphasis above making the sentence mean something different. Here's the whole phrase from the article: "A subdivision of a graph results from inserting vertices into edges (for example, changing an edge •——• to •—•—•) and repeating this zero or more times.". That zero refers to repeating the subdivision process, not subdividing zero times. That is what I was referring to above. So, if you're correct, the sentence needs to be adjusted, but I don't yet see how to do this without being unwieldy. - Taxman Talk 03:01, 1 June 2008 (UTC)
 * Take out the word "repeating" that you find so confusing, and just say that the edge is subdivided zero or more times? —David Eppstein (talk) 03:44, 1 June 2008 (UTC)
 * It's not that it's confusing, it's wrong. Simple enough fix though, thanks. - Taxman Talk 13:02, 1 June 2008 (UTC)

MacLane's Planarity Criterion
This page links to http://en.wikipedia.org/wiki/Mac_Lane%27s_planarity_criterion and I can't follow the proof given:

The proof states that the cycle space for the complete graph K5 is 7-dimensional, however for a connected graph isn't the dimension of the cycle space |E|-|V|+1 = 10-5+1=6? This formula is given in the page, Cycle_space whether over GF(2) or the rationals. That is the rank of the |V|×|E| incidence matrix, M, of a connected graph (V,E) is |V|-1 so the right null space (the cycle space) has dimension |E| - rank(M) =|E|-|V|+1. Likewise the proof states that the cycle space of K3,3 is 5-dimensional, however 9-6+1=4 dimensions. If C(G) for K5 is 6-dimensional, then there would be only at least 18 non-zeros rather than 21 on which the proof depends. Likewise for K33 there would be only at least 16 non-zeros instead of 20 on which the proof also depends. — Preceding unsigned comment added by StephenK51 (talk • contribs) 16:14, 7 March 2012 (UTC)

Average degree - is 2 e >= 3 f ?!
In the subsection "average degree" contains the claim:


 * $$2e \ge 3f$$ (one face has minimum 3 edges and each edge has maximum two faces)...

this claim is certainly false for the graph with two vertices and a single edge, since there 2e=2 while 3f=3. Also, it does not true that one face has minimum 3 edges. What is the correct statement? --Erel Segal (talk) 07:47, 24 August 2017 (UTC)
 * If we count face-edge incidences it is true for every connected planar graph larger than a single edge. —David Eppstein (talk) 14:36, 24 August 2017 (UTC)

Is the caracterization of convex graphs correct?
In this page it says: A plane graph is said to be convex if all of its faces (including the outer face) are convex polygons. A planar graph may be drawn convexly if and only if it is a subdivision of a 3-vertex-connected planar graph.

I think this is wrong: Can somebody please have a look at my counter-example? Wikipedia does not let me upload pictures (I do not know why, I am new to Wikipedia...) but you can see the example there: https://matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=257382&post_id=1869130 — Preceding unsigned comment added by Noskar (talk • contribs) 15:01, 3 February 2022 (UTC)

I have come to the conclusion that the caracterization is wrong. I will change it. Noskar (talk) 13:02, 4 February 2022 (UTC)