How to Draw a Graph Planar

Planar Graphs

• When drawing graphs, we usually endeavour to make them look "prissy".
  • Of class, there's no obvious definition of that.
  • Simply i thing we probably do want if possible: no edges crossing.
  • For case, the drawing on the correct is probably "meliorate"

• Sometimes, it's really important to be able to depict a graph without crossing edges.

  
  From Wikipedia Testpad.JPG.

  • From the text: connecting utilities (electricity, water, natural gas) to houses. If nosotros can keep from crossing those lines, it will be safer and easier to install.
  • Connecting components on a excursion lath: the connections on a circuit lath cannot cross. If we can connect them without resorting to another layer of traces, it will be cheaper to produce.
  • Subway system: if subway lines need to cross, we've got some serious engineering to do.
• A graph that can be drawn on a plane without edges crossing is called planar.
• For instance, nosotros drew \(Q_3\) in a non-planar style originally, but it is really planar:
  • Like being bipartite or isomorphic, nosotros tin can't just describe the graph one way and determine it'southward non planar.
  • In that location might exist some other way to depict it so information technology is planar.

Euler's Formula

• When nosotros draw a planar graph, it divides the plane up into regions.
  • For example, this graph divides the airplane into iv regions: 3 within and the exterior.

  • While we're counting, on this graph \(|V|=half-dozen\) and \(|E|=eight\).
• It's possibly not obvious that the number of regions is the same for any planar representation of this graph.
  • …or whatever representation of a item planar graph.

• Theorem: [Euler's Formula] For a continued planar unproblematic graph \(G=(Five,East)\) with \(e=|E|\) and \(v=|5|\), if we let \(r\) exist the number of regions that are created when drawing a planar representation of the graph, then \(r=e-five+ii\).

  Proof idea: We proceed past induction on the number of edges. For \(east=one\), we can only have this graph, which has \(five=2\) and \(r=1\), and so the theorem holds.

  If we assume the determination for \(e-ane\) edges, and add together an edge to get our planar graph, at that place are only two possibilities to consider.

  Case 1: we add together a pendant vertex and edge. In this example \(e\) and \(five\) both increment past one and the equality still holds.

  Instance 2: we add an edge across ii existing vertices (that doesn't cross any other edges, of course). Here, \(r\) and \(eastward\) both increase by i, preserving the equality.

  So nosotros can build whatever planar graph with \(east\) edges and it volition ever have \(r=due east-v+2\).

• For case:
  • \(Q_3\) has \(r=6\), \(east=12\), and \(v=8\).
  • \(C_n\) has \(r=ii\), \(due east=n\), and \(v=northward\).

Testing for Planarity

• Nosotros know a fashion to decide that a graph is planar: draw information technology without edges crossing.
  • … but that's all we know and then far.
  • It would be nice to have a fashion to determine for sure whether or not a graph is planar, without worrying that we haven't been clever plenty in our drawing.
• Here are two graphs I promise aren't planar: \(K_{3,iii}\) and \(K_5\).
  • We don't accept any theorems for that (however), but whatever way yous try to draw them, you'll eventually be unable to draw some of the edges without crossing.
  • In some way, these are the "smallest" non-planar graphs.
  • We'll see what "smallest" means presently.

• Theorem: For a simple connected planar graph with \(v\ge 3\) vertices and \(e\) edges, \(e\le 3v-6\).

  Proof: Let \(r\) be the number of regions in a planar representation of the graph, and for a region \(R\), allow \(\operatorname{deg}(R)\) be the number of edges that are side by side to the region, so each border is next to 2 regions.

  We know that \(\operatorname{deg}(R)\ge iii\) for each region since the graph doesn't take multiple edges (for interior regions) and has at least three vertices (for the exterior region). Since each edge is adjacent to 2 regions, \[2e=\sum \operatorname{deg}(R) \ge 3r\,.\]

  We tin can use this with Euler'southward formula (\(r=eastward-v+2\)) to get \[\brainstorm{align*} 3r &\le 2e \\ three(east-five+two) &\le 2e \\ e &\le 3v-vi\,.\quad ∎ \\ \end{align*}\]

• This theorem isn't if-and-only-if, then exist careful.
  • It does show that \(K_5\) is not-planar: \(v=5\), \(e=10\).
  • But for \(K_{3,3}\), we have \(v=6\) and \(due east=nine\). It satisfies the inequality, but is non-planar.

• Corallary: A simple connected planar graph with \(v\ge 3\) has a vertex of degree five or less.

  Proof: Suppose every vertex has degree 6 or more. Then the total number of edges is \(2e\ge 6v\). But, considering the graph is planar, \[\sum \operatorname{deg}(v) = 2e\le 6v-12\,. \] We have a contradiction.∎

• Theorem: For a unproblematic continued planar graph with \(v\ge 3\) vertices and \(e\) edges, and no circuits of length three, \(e\le 2v-four\).

  Proof idea: Since we take no loops of length three, we can echo the above proof with \(\operatorname{deg}(R)\ge 4\).

• This theorem should convince y'all that \(K_{three,iii}\) is non-planar: \(5=6\) and \(e=ix\).
  • At that place are still graphs that obey this inequality, but are non-planar.
  • Here's 1 with \(v=xv\) and \(east=eighteen\):

  • … and I know information technology's not planar considering of the next theorem.

Kuratowski's Theorem

• It turns out that \(K_{three,3}\) and \(K_5\) are the "smallest" non-planar graphs in that every not-planar graph contains them.
  • … just not only equally subgraphs: the in a higher place example doesn't have either every bit a subgraph.
• Nosotros will take a more permissive version of "contains" here: contains a graph homeomorphic to.
  • One graph is homeomorphic to another if you lot can turn one into another by adding or removing degree-2 vertices:

• Theorem: [Kuratowski's Theorem] A graph is non-planar if and only if information technology contains a subgraph homeomorphic to \(K_{iii,3}\) or \(K_5\).

• A graph is non-planar iff we can turn it into \(K_{three,3}\) or \(K_5\) by:
  1. Removing edges and vertices. (Making a subgraph.)
  2. Collapsing degree-two vertices into a unmarried border.
  3. Applying an isomorphism to plow it into \(K_{3,iii}\) or \(K_5\).
• This animation explains it better than I ever could: the starting graph here can be turned into \(K_{three,3}\) by these operations.

  
  From Wikipedia Kuratowski.gif.

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Source: https://www.cs.sfu.ca/~ggbaker/zju/math/planar.html

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