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Discrete Mathematics An Open Introduction, 4th Edition

Oscar Levin

Section 2.3 Planar Graphs

Subsection Section Preview

Investigate!
When a connected graph can be drawn without any edges crossing, it is called planar. When a planar graph is drawn in this way, it divides the plane into regions called faces.
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  1. Draw, if possible, two different planar graphs with the same number of vertices, edges, and faces.
  2. Draw, if possible, two different planar graphs with the same number of vertices and edges, but a different number of faces.
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When is it possible to draw a graph so that none of the edges cross? If this is possible, we say the graph is planar (since you can draw it on the plane).
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Notice that the definition of planar includes the phrase β€œit is possible to.” This means that even if a graph does not look like it is planar, it still might be. Perhaps you can redraw it in a way in which no edges cross. For example, this is a planar graph:
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A drawing of K2,3 with two vertices in a top row, each adjacent to each of the three vertices on the bottom row.
That is because we can redraw it like this:
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Another drawing of K2,3. A single vertex on a top row is adjacent to three vertices in a row below it. Each of these vertices are adjacent to a single vertex below (and to the right of) them.
The graphs are the same, so if one is planar, the other must be, too. However, the original drawing of the graph was not a planar representation of the graph.
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When a planar graph is drawn without edges crossing, the edges and vertices of the graph divide the plane into regions. We will call each region a face. The graph above has 3 faces (yes, we do include the β€œoutside” region as a face). The number of faces does not change no matter how you draw the graph (as long as you do so without the edges crossing), so it makes sense to ascribe the number of faces as a property of the planar graph.
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WARNING: you can only count faces when the graph is drawn in a planar way. For example, consider these two representations of the same graph:
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A drawing of K4 with four vertices in a square and edges forming the sides of the square plus two more crossing through the center.
A drawing of K4 with four vertices arranged in a square and edges forming the sides of the square. Another edge crosses from the bottom left to the top right corners. A curved edge loops outside of the square from the top left to bottom right vertices. No edges intersect.
If you try to count faces using the graph on the left, you might say there are 5 faces (including the outside). But drawing the graph with a planar representation shows that in fact there are only 4 faces.
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Worksheet Preview Activity
A4US

1.
(a)
Draw a connected planar graph with 5 vertices and 5 edges. How many faces (including the β€œoutside” face) does your graph have?
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Number of faces:
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(b)
Now add a single edge to your graph, between two vertices that are not already adjacent. Assuming the resulting graph is still planar, list the number of vertices, edges, and faces it now has.
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Vertices: ; Edges: ; Faces:
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(c)
Now add another edge to the graph, this time to a new vertex. Assuming the resulting graph is still planar, list the number of vertices, edges, and faces it now has.
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Vertices: ; Edges: ; Faces:
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2.
Now draw at least three more connected, planar graphs, each with at least six vertices. Count the number of vertices v, edges e, and faces f for each graph and record your data in the table below.
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v e f
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3.
Do you notice any patterns? What happens to the numbers if you add an edge between two non-adjacent vertices? What happens if you add a new vertex and connect it to an existing vertex?
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Conjecture an expression that involves the number of vertices v, the number of edges e, and the number of faces f that remains constant for all connected planar graphs. What is that constant?
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Conjectured expression: .
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Hint.
You might conjecture an expression like v+ef. But this is not right, because there is a planar graph for which this would be 5+52=5 and another planar graph for which the expression would be 6+73β‰ 5.
What sort of expression will stay constant if v and e both increase by 1? And also stay constant if e and f both increase by 1?
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4.
A cube is made of six squares, each of which shares an edge with each of its neighbors. Vertices of the cube join three of the squares.
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(a)
How many vertices, edges, and faces does a cube have?
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Vertices: ; Edges: ; Faces:
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(b)
Does this match the relationship you conjectured above?
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  • Yes
  • No
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Subsection Euler’s Formula for Planar Graphs

There is a connection between the number of vertices (v), the number of edges (e), and the number of faces (f) in any connected planar graph. This relationship is called Euler’s formula.
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Euler’s Formula for Planar Graphs.

For any connected planar graph with v vertices, e edges, and f faces, we have
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vβˆ’e+f=2.
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Why is Euler’s formula true? One way to convince ourselves of its validity is to draw a planar graph step by step. Start with the graph P2:
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Two vertices connected by an edge.
Any connected graph (besides just a single isolated vertex) must contain this subgraph. Now we build up to our graph by adding edges and vertices. Each step will consist of either adding a new vertex connected by a new edge to part of your graph (so creating a new β€œspike”) or by connecting two vertices already in the graph with a new edge (completing a circuit).
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A graph with four vertices arranged in a square. The top left vertex is adjacent to the other three vertices, and the top right and bottom right vertices are also adjacent. A dashed edge leads from the top right vertex of the square to a fifth vertex below and to its right.
A graph with four vertices arranged in a square. The top left vertex is adjacent to the other three vertices, and the top right and bottom right vertices are also adjacent. A dashed edge connects the bottom two vertices.
What do these β€œmoves” do? When adding the spike, the number of edges increases by 1, the number of vertices increases by 1, and the number of faces remains the same. But this means that vβˆ’e+f does not change. Completing a circuit adds one edge, adds one face, and keeps the number of vertices the same. So again, vβˆ’e+f does not change.
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Since we can build any graph using a combination of these two moves, and doing so never changes the quantity vβˆ’e+f, that quantity will be the same for all graphs. But notice that our starting graph P2 has v=2, e=1, and f=1, so vβˆ’e+f=2.
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The argument we have outlined above is not quite correct, since we made the unjustified assumption that all graphs can be built up from P2 using only the two moves we described. To avoid this issue, we can use a minimal criminal argument. You are asked to do this in the exercises, but the idea is essentially the same as we have here, except that we start with a minimal connected planar graph that does not satisfy the formula, then remove either an edge or a vertex (and its edge) to get a smaller connected planar graph that does satisfy the formula. But just like the adding moves we have described above, removing an edge or a vertex does not change the quantity vβˆ’e+f.
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Subsection Non-planar Graphs

Investigate!
For the complete graphs Kn, we would like to be able to say something about the number of vertices, edges, and (if the graph is planar) faces. Let’s first consider K3:
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  1. How many vertices does K3 have? How many edges?
  2. If K3 is planar, how many faces should it have?
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Repeat parts (1) and (2) for K4, K5, and K23.
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What about complete bipartite graphs? How many vertices, edges, and faces (if it were planar) does K7,4 have? For which values of m and n are Kn and Km,n planar?
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Not all graphs are planar. If there are too many edges and too few vertices, then some of the edges will need to intersect. The smallest graph where this happens is K5.
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A copy of K5: five vertices arranged in a pentagon with edges connecting every vertex to every other vertex.
If you try to redraw this without edges crossing, you quickly get into trouble. There seems to be one edge too many. In fact, we can prove that no matter how you draw it, K5 will always have edges crossing.
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Proof.

The proof is by contradiction. So assume that K5 is planar. Then the graph must satisfy Euler’s formula for planar graphs. K5 has 5 vertices and 10 edges, so we get
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5βˆ’10+f=2,
which says that if the graph is drawn without any edges crossing, there would be f=7 faces.
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Now consider how many edges surround each face. Each face must be surrounded by at least 3 edges. Let B be the total number of boundaries around all the faces in the graph. Thus we have that 3f≀B. But also B=2e, since each edge is used as a boundary exactly twice. Putting this together we get
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3f≀2e.
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But this is impossible, since we have already determined that f=7 and e=10, and 21β‰°20. This is a contradiction, so in fact K5 is not planar.
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The other simplest graph which is not planar is K3,3
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A drawing of K3,3 with a row of three vertices on top, each adjacent to the three vertices in a row directly below.
Proving that K3,3 is not planar answers the classic houses and utilities puzzle: it is not possible to connect each of three houses to each of three utilities without the lines crossing.
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Proof.

Again, we proceed by contradiction. Suppose K3,3 were planar. Then by Euler’s formula, there will be 5 faces, since v=6, e=9, and 6βˆ’9+f=2.
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How many boundaries surround these 5 faces? Let B be this number. Since each edge is used as a boundary twice, we have B=2e. Also, Bβ‰₯4f since each face is surrounded by 4 or more boundaries. We know this is true because K3,3 is bipartite, so does not contain any 3-edge cycles. Thus
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4f≀2e.
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But this would say that 20≀18, which is clearly false. Thus K3,3 is not planar.
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Note the similarities and differences in these proofs. Both are proofs by contradiction, and both start with using Euler’s formula to derive the (supposed) number of faces in the graph. Then we find a relationship between the number of faces and the number of edges based on how many edges surround each face. This is the only difference. In the proof for K5, we got 3f≀2e and for K3,3 we had 4f≀2e. The coefficient of f is the key. It is the smallest number of edges that could surround any face. If some number of edges surround a face, then these edges form a cycle. So that number is the size of the smallest cycle in the graph.
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In general, if we let g be the size of the smallest cycle in a graph (g stands for girth, which is the technical term for this) then for any planar graph we have gf≀2e. When this disagrees with Euler’s formula, we know for sure that the graph cannot be planar.
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Subsection Polyhedra

Investigate!
A cube is an example of a convex polyhedron. It contains 6 identical squares for its faces, 8 vertices, and 12 edges. The cube is a regular polyhedron (also known as a Platonic solid) because each face is an identical regular polygon and each vertex joins an equal number of faces.
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There are exactly four other regular polyhedra: the tetrahedron, octahedron, dodecahedron, and icosahedron, with 4, 8, 12, and 20 faces respectively. How many vertices and edges do each of these have?
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Another area of mathematics where you might have heard the terms β€œvertex,” β€œedge,” and β€œface” is geometry. A polyhedron is a geometric solid made up of flat polygonal faces joined at edges and vertices. We are especially interested in convex polyhedra, which means that any line segment connecting two points on the interior of the polyhedron must be entirely contained inside the polyhedron.
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An alternative definition for convex is that the internal angle formed by any two faces must be less than 180deg⁑.
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Notice that since 8βˆ’12+6=2, the vertices, edges, and faces of a cube satisfy Euler’s formula for planar graphs. This is not a coincidence. We can represent a cube as a planar graph by projecting the vertices and edges onto the plane. One such projection looks like this:
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Eight vertices arranged as a smaller square inside a larger square. Edges from the perimeters of both squares, and edges connect each vertex of the small square to its closest vertex of the larger square.
In fact, every convex polyhedron can be projected onto the plane without edges crossing. Think of placing the polyhedron inside a sphere, with a light at the center of the sphere. The edges and vertices of the polyhedron cast a shadow onto the interior of the sphere. You can then cut a hole in the sphere in the middle of one of the projected faces and β€œstretch” the sphere to lie down flat on the plane. The face that was punctured becomes the β€œoutside” face of the planar graph.
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The point is, we can apply what we know about graphs (in particular planar graphs) to convex polyhedra. Since every convex polyhedron can be represented as a planar graph, we see that Euler’s formula for planar graphs holds for all convex polyhedra as well. We also can apply the same sort of reasoning we use for graphs in other contexts to convex polyhedra. For example, we know that there is no convex polyhedron with 11 vertices all of degree 3, as this would make 33/2 edges.
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Example 2.3.3.

Is there a convex polyhedron consisting of three triangles and six pentagons? What about three triangles, six pentagons, and five heptagons (7-sided polygons)?
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Solution.
How many edges would such polyhedra have? For the first proposed polyhedron, the triangles would contribute a total of 9 edges, and the pentagons would contribute 30. However, this counts each edge twice (as each edge borders exactly two faces), giving 39/2 edges, an impossibility. There is no such polyhedron.
The second polyhedron does not have this obstacle. The extra 35 edges contributed by the heptagons give a total of 74/2 = 37 edges. So far so good. Now how many vertices does this supposed polyhedron have? We can use Euler’s formula. There are 14 faces, so we have vβˆ’37+14=2 or equivalently v=25. But now use the vertices to count the edges again. Each vertex must have degree at least three (that is, each vertex joins at least three faces since the interior angle of all the polygons must be less that 180∘), so the sum of the degrees of vertices is at least 75. Since the sum of the degrees must be exactly twice the number of edges, this says that there are strictly more than 37 edges. Again, there is no such polyhedron.
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To conclude this application of planar graphs, consider the regular polyhedra. We claimed there are only five. How do we know this is true? We can prove it using graph theory.
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Proof.

Recall that all the faces of a regular polyhedron are identical regular polygons and that each vertex has the same degree. Consider four cases, depending on the type of regular polygon.
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Case 1: Each face is a triangle. Let f be the number of faces. There are then 3f/2 edges. Using Euler’s formula, we have vβˆ’3f/2+f=2 so v=2+f/2. Now each vertex has the same degree, say k. So the number of edges is also kv/2. Putting this together gives
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e=3f2=k(2+f/2)2,
which says
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k=6f4+f.
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Both k and f must be positive integers. Note that 6f4+f is an increasing function for positive f, bounded above by a horizontal asymptote at k=6. Thus the only possible values for k are 3, 4, and 5. Each of these is possible. To get k=3, we need f=4 (this is the tetrahedron). For k=4 we take f=8 (the octahedron). For k=5 take f=20 (the icosahedron). Thus there are exactly three regular polyhedra with triangles for faces.
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Case 2: Each face is a square. Now we have e=4f/2=2f. Using Euler’s formula, we get v=2+f, and counting edges using the degree k of each vertex gives us
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e=2f=k(2+f)2.
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Solving for k gives
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k=4f2+f=8f4+2f.
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This is again an increasing function, but this time the horizontal asymptote is at k=4, so the only possible value that k could take is 3. This produces 6 faces, and we have a cube. There is only one regular polyhedron with square faces.
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Case 3: Each face is a pentagon. We perform the same calculation as above, this time getting e=5f/2 so v=2+3f/2. Then
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e=5f2=k(2+3f/2)2,
so
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k=10f4+3f.
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Now the horizontal asymptote is at 103. This is less than 4, so we can only hope to have k=3. We can do so by using 12 pentagons, getting the dodecahedron. This is the only regular polyhedron with pentagons as faces.
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Case 4: Each face is an n-gon with nβ‰₯6. Following the same procedure as above, we deduce that
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k=2nf4+(nβˆ’2)f,
which will be increasing to a horizontal asymptote of 2nnβˆ’2. When n=6, this asymptote is at k=3. Any larger value of n will give an even smaller asymptote. Therefore no regular polyhedra exist with faces larger than pentagons.
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Notice that you can tile the plane with hexagons. This is an infinite planar graph; each vertex has degree 3. These infinitely many hexagons correspond to the limit as fβ†’βˆž to make k=3.
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Reading Questions Reading Questions

2.

Suppose you draw a graph with 10 vertices and 14 edges in such a way that no edges cross. How many faces could your graph have? Explain your answer(s).
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3.

What questions do you have after reading this section? Write at least one question about the content of this section that you are curious about.
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Exercises Practice Problems

1.

Are the following statements true or false?
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  1. K3,6 is planar
  2. K2,5 is not planar
  3. K6 is not planar
  4. K2,4 is planar
  5. K3,10 is not planar
  6. K4,5 is planar
  7. K2,10 is not planar
  8. K8 is not planar
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2.

Suppose G is a planar connected graph. It has 22 edges, and 10 faces. How many vertices does G have?
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3.

Suppose a connected graph has 5 vertices, and every vertex has degree 2.
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  1. How many edges does the graph have?
    e= .
  2. If the graph were planar, how many faces would it have?
    f= .
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4.

Let’s prove that K8 is not planar:
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First, how many vertices and how many edges does K8 have?
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v= and e= .
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If we assume that K8 were planar, then how many faces would it have?
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However, since every face is bounded by at least edges, and every edge borders exactly faces, we can get a bound on the number of faces. What is the largest number of faces possible based on this line of reasoning?
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f≀ .
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This is a contradiction, so K8 is not planar. QED.
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5.

Suppose the graph is planar but not connected, and has 4 components. Draw enough examples to derive a variant of Euler’s formula for this case.
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vβˆ’e+f= .
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Exercises Additional Exercises

1.

Is it possible for a planar graph to have 6 vertices, 10 edges, and 5 faces? Explain.
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2.

The graph G has 6 vertices with degrees 2,2,3,4,4,5. How many edges does G have? Could G be planar? If so, how many faces would it have? If not, explain.
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3.

Is it possible for a connected graph with 7 vertices and 10 edges to be drawn so that no edges cross and create 4 faces? Explain.
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Hint.
What would Euler’s formula tell you?
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4.

Is it possible for a graph with 10 vertices and edges to be a connected planar graph? Explain.
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5.

Is there a connected planar graph with an odd number of faces where every vertex has degree 6? Prove your answer.
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Hint.
You can use the handshake lemma to find the number of edges, in terms of v, the number of vertices.
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6.

I’m thinking of a polyhedron containing 12 faces. Seven are triangles and four are quadrilaterals. The polyhedron has 11 vertices including those around the mystery face. How many sides does the last face have?
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7.

Consider some classic polyhedrons.
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  1. An octahedron is a regular polyhedron made up of 8 equilateral triangles (it sort of looks like two pyramids with their bases glued together). Draw a planar graph representation of an octahedron. How many vertices, edges, and faces does an octahedron (and your graph) have?
  2. The traditional design of a soccer ball is a (spherical projection of a) truncated icosahedron. This consists of 12 regular pentagons and 20 regular hexagons. No two pentagons are adjacent (so the edges of each pentagon are shared only by hexagons). How many vertices, edges, and faces does a truncated icosahedron have? Explain how you arrived at your answers. Bonus: draw the planar graph representation of the truncated icosahedron.
  3. Your β€œfriend” claims that he has constructed a convex polyhedron out of 2 triangles, 2 squares, 6 pentagons, and 5 octagons. Prove that your friend is lying. Hint: each vertex of a convex polyhedron must border at least three faces.
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8.

Prove Euler’s formula using a minimal criminal argument, where minimum means smallest number of edges
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9.

Prove Euler’s formula using a minimal criminal argument, where minimum means smallest number of vertices.
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10.

Euler’s formula (vβˆ’e+f=2) holds for all connected planar graphs. What if a graph is not connected? Suppose a planar graph has two components. What is the value of vβˆ’e+f now? What if it has k components?
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11.

Prove that the Petersen graph (below) is not planar.
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A drawing of the Petersen graph: ten vertices arranged as a larger pentagon around a smaller pentagram (five pointed star). Edges form the outside of the larger pentagon and the crossing lines of the pentagram. Each vertex of the larger pentagon is adjacent to the closest vertex of the inside pentagram.
Hint.
What is the length of the shortest cycle? (This quantity is usually called the girth of the graph.)
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12.

Prove that any planar graph with v vertices and e edges satisfies e≀3vβˆ’6.
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13.

Prove that any planar graph must have a vertex of degree 5 or less.
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14.

Give a careful proof that the graph below is not planar.
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A graph with 11 vertices. A single vertex in the center, then five vertices equally spaced around a ring around it, and five more equally spaced around a ring around those. Edges form the sides of a pentagon for the outer ring of vertices. Each outer vertex is also adjacent to two inner vertices: the two on either side of the vertex closest to it. Finally, every inner vertex is also adjacent to the center vertex.
Hint.
The girth of the graph is 4.
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15.

Explain why we cannot use the same sort of proof we did in Exercise 14 to prove that the graph below is not planar. Then explain how you know the graph is not planar anyway.
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A graph with 11 vertices. A single vertex in the center, then five vertices equally spaced around a ring around it, and five more equally spaced around a ring around those. Edges form the sides of a pentagon for the outer ring of vertices and also the inner ring of vertices. Each outer vertex is also adjacent to two inner vertices: the two on either side of the vertex closest to it. Finally, every inner vertex is also adjacent to the center vertex.
Hint.
What has happened to the girth? Careful: We have a different number of edges as well. Better check Euler’s formula.
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