1
$\begingroup$

This may be a very trivial proof, but I was wondering if anyone could show/prove why you cannot have a PLANAR graph where every node is joined to at least 5 others.

  • 0
    I'm confused. A complete graph is when every node is connected to every other node. So the "at least 5" just means you have 6 or more vertices. Do you mean that only the complete graphs on 4 or fewer vertices can be drawn without edge intersections in $2$-d? – 2017-02-24
  • 0
    Right okay well I'm probably using incorrect terminology then. What I mean is a graph where all possible vertices have been drawn except those which intersect with others. The best illustration I can come up with is a triangle within a larger inverted triangle, then the remaining corners joined. – 2017-02-24
  • 0
    Is this problem from something? Can you post the source, if so? That would help. – 2017-02-24
  • 0
    Sadly it's just something I came up with while doodling, and I'm not allowed to post pictures. I stopped studying maths formally two years ago and don't really have the terminology to explain this properly, but this is how you can phrase it: – 2017-02-24
  • 0
    Ok, no worries. It looks like some people had a go at it. – 2017-02-24
  • 0
    Draw some dots on a page. There is no way that you can join the dots with lines, without any lines crossing, in a way that means every dot is joined to at least 5 others. (You can do it so each is joined to at least 4. But as far as I can tell, not 5). – 2017-02-24
  • 0
    You mean a complete PLANAR graph. – 2017-02-24
  • 0
    Ah brilliant, thanks. Updated the question. Do those proofs look like they're answering my poorly-worded question to you? As far as I can tell they're showing that a complete graph cannot be planar above n=4, not that there couldn't possibly be a planar graph with 5 connections to each node. – 2017-02-24
  • 0
    It is possible, but you need more nodes, an example is the graph of the platonic solid known as the icosahedron. ( If a graph can be embedded in the sphere it can also be embedded in the plane and vice-versa). – 2017-02-24
  • 0
    Ah well there you go, my premise was wrong. Thanks! – 2017-02-24

2 Answers 2

4

Suppose we have a planar graph, let $E,F,V$ be the number of edges faces and vertices.

Let $\mathcal F$ be the set of faces, notice that $\sum\limits_{f\in \mathcal F} |f|=2E$ because every edge is incident to two faces.

This implies that $2E\geq 3F\implies F\leq \frac{2E}{3}$, now use euler's formula:

We have $V-E+F=2\implies V-E+\frac{2E}{3}\geq 2\implies V-2\geq \frac{E}{3}\implies E\leq3V-6$

Notice that $K_5$ has $5$ vertices and more than $3\cdot5-6=9$ edges, so it is not planar.

Reading the comments, you seem to be wondering whether planar $5$-regular graphs exist, they do. For example you can consider the graph of the icosahedron.

enter image description here

1

In fact the result is slightly stronger - every complete graph $K_n$ with $n \ge 5$ is not planar - that is, cannot be drawn in the plane without edge crossings.

You can get this by examining the Euler relationship, $v-e+f=2$ that should hold for a graph that can be drawn in the plane, and considering that every face $f$ should be bounded by at least $3$ edges. So for $K_5$, $v=5, e=10$ and thus we expect $f=7$, which then require more than $10$ edges to delineate - so $K_5$ is not planar.