Let \(\bfG\) be a connected planar graph with \(n\) vertices and \(m\) edges. Then every planar drawing of \(\bfG\) has \(f\) faces, where \(f\) satisfies \(n-m+f=2\text{.}\)
Notice that Eulerβs Formula requires that \(\bfG\) be a connected planar graph. Find a counterexample to show that the formula \(n-m+f=2\) does not hold if \(\bfG\) is a planar graph that is not connected.
Since \(\bfG\) must be connected, we start by considering the minimal case in which \(\bfG\) has no cycles (but for the general number \(n\) of vertices). What type of graph does that mean \(\bfG\) is? How many edges does \(\bfG\) have (in terms of \(n\))? (In future questions, Iβll call this value of \(m\) by the name \(m_{0}\text{.}\)) Verify Eulerβs formula in this case.
Draw a planar drawing of a connected planar graph \(\bfG\) with with \(n=7\) or \(n=8\) vertices and more than \(m_{0}\) edges. Make sure your graph has at least one vertex of degree \(1\text{.}\)
Is there an edge \(e\) you can erase from your graph and leave a graph \(\bfG'\) that is still connected? What determines whether or not an edge is safe to remove while still leaving a connected graph?
Erase an edge \(e\) from \(\bfG\) to give you a planar drawing of a connected planar graph \(\bfG'\text{.}\) Let \(n'\text{,}\)\(m'\text{,}\) and \(f'\) be the number of vertices, edges, and faces respectively of \(\bfG'\text{.}\) What are \(n',m',f'\) for your graph?
Now letβs suppose that \(\bfG\) is a connected planar graph with \(n\) vertices and \(m\) edges where \(m\gt m_{0}\text{.}\) Suppose that Eulerβs formula holds for any connected planar graph with fewer than \(m\) edges. (What type of proof are we setting up here?)
Let \(n'\text{,}\)\(m'\text{,}\) and \(f'\) be the number of vertices, edges, and faces respectively of \(\bfG'\text{.}\) What are \(n',m',f'\) in terms of \(n\text{,}\)\(m\text{,}\) and \(f\text{?}\)
In the grand scheme of things, Eulerβs formula doesnβt seem that useful for determining if a graph is planar or not. However, we can deduce something useful from it. In this problem, weβre going to imagine a fixed planar drawing of a connected planar graph \(\bfG\) and let \(n\text{,}\)\(m\text{,}\) and \(f\) be the number of vertices, edges, and faces, respectively, in our planar drawing. Weβll assume that \(n\geq 4\) to avoid some annoying things.
If I pick an edge \(e_{1}\) of \(\bfG\text{,}\) what is the largest possible number of faces that \(e_{1}\) can be part of the boundary of? So what does that tell us about the number of ordered pairs in \(S\) that have \(e_{1}\) as the first coordinate?
If I pick a face \(F_{1}\) of \(\bfG\text{,}\) what is the smallest possible number of edges that can be used to form the boundary of \(F_{1}\text{?}\)
Recall that the complete bipartite graph \(\bfK_{3,3}\) is the bipartite graph with two independent sets each of size three and all the possible edges (nine in total) between vertices in different independent sets.
Notice that since \(\bfK_{3,3}\) does not contain any vertices of degree \(1\text{,}\) every face of a planar drawing would be bounded by a cycle, and thus every edge is part of the boundary of two faces. Since \(\bfK_{3,3}\) is bipartite, what can you say about the length of every cycle in \(\bfK_{3,3}\text{?}\)
Let \(f_{4}\) be the number of faces bounded by a cycle of length \(4\) and let \(f_{6}\) be the number of faces bounded by a cycle of length \(6\text{.}\) Express the total number of faces \(f\) in terms of \(f_{4}\) and \(f_{6}\text{.}\) Then express the number of edge-face pairs in terms of \(f_{4}\) and \(f_{6}\text{.}\)
What does Eulerβs formula tell you \(f\) would have to be for \(\bfK_{3,3}\text{?}\) What lower bound does this give you for the number of edge-face pairs, and how does this show that \(\bfK_{3,3}\) is not planar?