Talk:Planar graph

message: 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)
replies: text: 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)
thread_title:  Fourth picture 

message: 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)
replies: text: 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)

text: 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)
thread_title:  Modifications 

message: 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)
replies: text: I think I answered my own question. See Four color theorem#Formal statement in graph theory for the graphic. Chris53516 03:01, 5 July 2006 (UTC)
thread_title:  Newbie 

message: No exact definition was supplied for  plane graph - I hope mine is good enough.  Rp 22:26, 5 July 2006 (UTC)
replies: text: 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)
thread_title:  Plane graph 

message: 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)
thread_title:  Too technical 

message: 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)
thread_title:  Genus 

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

text: 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)
thread_title:  Maximal Planar Graphs 

message: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)
replies: text: It's not that it's confusing, it's wrong. Simple enough fix though, thanks. - Taxman Talk 13:02, 1 June 2008 (UTC)
thread_title:  Kuratowski's theorem 

message: 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)
thread_title:  MacLane's Planarity Criterion 

message: 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)
replies: text: 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)
thread_title:  Average degree - is 2 e >=  3 f ?! 

message: 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)
thread_title:  Is the caracterization of convex graphs correct?