The number of vertices, edges, and faces of a map or a graph are related by Euler’s formula, provided that we consider the unbounded surrounding area as one of the faces. We define the symbols V, E, and F as follows:
• V is the number of vertices.
• E is the number of edges.
• F is the number of faces. Remember to count the unbounded surrounding area.
According to Euler’s formula [Ref. 4], for a map or a graph (or even a polyhedron):
V + F = E + 2
The number of vertices plus the number of faces is two more than the number of edges.
For the map shown above:
• V = 20. On a map, vertices are where the edges intersect. We marked the vertices with small dots (•) on the diagram above to help you count them.
There are 5 for the inner pentagon, 10 for the decagon, and another 5 for the outer pentagon.
• E = 30. On a map, an edge is any line segment (or part of one) or curve that separates two regions (or which separates a region from the unbounded surrounding area). The inner pentagon has 5, there are 5 connecting the inner pentagon to the decagon, there are 10 along the decagon, there are 5
connecting the decagon to the outer pentagon, and there are 5 along the outer pentagon.
• F = 12. On a map, the faces are regions. The twelve regions are A, B, C, D, E, F, G, H, I, J, K, and L. This includes the unbounded surrounding area as a face.
Check the formula for the map on the left: V + F = 20 + 12 = 32 and E + 2 = 30 + 2 = 32. Since V + F and E + 2 both equal 32 for this map, we see that Euler’s formula agrees with it.
For the graph shown above:
• V = 12. On a graph, the vertices are regions where lines intersect. These are regions A, B, C, D, E, F, G, H, I, J, K, and L.
• E = 30. On a graph, each edge is a line or curve that connects a pair of regions. The edges are AB, AE, AF, AG, AK, BC, BF, BG, BH (but don’t count BA since that’s the same as AB, which was already counted), CD, CF, CH, CI, DE, DF, DI, DJ, EF, EJ, EK, GH, GK, GL, HI, HL, IJ, IL, JK, JL, and KL.
• F = 20. On a graph, the faces are areas formed between edges. The faces are ABF, ABG, AEF, AEK, AGK, BCF, BCH, BGH (but don’t count BAF or BAG since they are the same as ABF and ABG, which were already counted), CDF, CDI, CHI, DEF, DEJ, DIJ, EJK, GKL, HIL, IJL, JKL, and the infinite area lying outside of the graph.
Note that the graph shown above corresponds to the map on the previous page, where we included region L explicitly on the graph and still counted the infinite area lying outside of the graph as its own face. On the map, L represented the unbounded surrounding area and was counted only as a face.
If you are wondering if this may be inconsistent, there is a reason for drawing
the map and its corresponding graph as we have done here. The values of V and F are swapped for the map compared to the graph when we draw them this way. Doing so emphasizes the fact that the graph is considered to be a dual representation of the map. (We’ll elaborate on what we mean by a dual representation in the solution to Problem 1.)
For the graph, V + F = 12 + 20 = 32 and E + 2 = 30 + 2 = 32, agreeing with Euler’s formula. Note that the map and graph give the same values of V, F, and E as a dodecahedron (a 12-sided polyhedron) and its dual polyhedron, which is an icosahedron (a 20-sided polyhedron). If you imagine cutting the map along the five outer radial edges and then folding the map up, you might be able to visualize how it can fold into the shape of a dodecahedron.
Note that if you remove region L from the previous graph, Euler’s formula will still work. In that case, you would have one less vertex (region) so V = 11, five fewer edges (since L connects to G, H, I, J, and K) so E = 25, and four fewer faces (you wouldn’t have GKL, HIL, IJL, or JKL) so F = 16 (note that we still include the infinite area outside the graph as a face). In this case V + F = 11 + 16 = 27 and E + 2 = 25 + 2 = 27, still satisfying Euler’s formula.
When we include region L in the graph, note that 2E = 3F. That is, two times the number of edges is equal to three times the number of faces. When we include region L, 2E = 2(30) = 60 and 3F = 3(20) = 60. The formula 2E = 3F applies to any MPG. (Recall that MPG stands for “maximal planar graph”
and that for a MPG every face in the graph has three “sides;” we’ll call each edge a side whether it appears straight or curved.) To get the number of faces F from the number of edges for a MPG, divide E by 3 because each face has 3 edges and then multiply this by 2 because each edge is shared by 2 faces: F
= 2E/3, from which it follows that F/E = 2/3. In the previous graph, for example, note that edge AE is part of both triangles AEF and AEK. The ratio of F to E is a maximum for a MPG. In this case, the ratio is 2/3, which is approximately 0.67 when rounded to two decimal places. (Note that if region L isn’t included in the graph, it would be a PG, not a MPG. In that case, the ratio is slightly smaller: 16/25 = 0.64.) In contrast, if every face were a pentagon, the ratio would be 2/5 = 0.40.
Now we will do a little arithmetic to obtain a classic result related to the four-color theorem. This is generally done in a more formal, technical manner, but
we’ll try to keep it simple. Note that the formulas from this chapter apply if V is at least 3.
Consider the special case where every vertex of the graph has the same degree, meaning that the same number of edges intersect at every vertex. If every vertex has the same degree and if we define D to be the degree of each vertex (equal to the number of edges intersecting at each vertex), then DV/2
= E, which is equivalent to DV = 2E. This well-known result is referred to as the handshaking lemma. If V people each shake hands with D other people, DV/2 equals the number of handshakes that will occur. We divide by two in order to avoid double counting. For example, if there are 8 people at a gathering and each person shakes hands with 7 other people, there are 8(7)/2
= 56/2 = 28 handshakes. If their names are Mr. A, Mr. B, Mr. C, etc., thru Mr. H, then we divide by 2 because 56 counts Mr. A shaking hands with Mr.
B and Mr. B shaking hands with Mr. A separately, and similarly for all other pairs (like Mr. A and Mr. C which is the same as Mr. C with Mr. A).
What if the vertices don’t all have the same degree? This will be the case with most graphs. We can still use the formula DV = 2E provided that we interpret D as the average degree of the vertices.
If we multiply both sides of Euler’s formula by 2, we get 2V + 2F = 2E + 4.
Substitute the equation DV = 2E into this equation to obtain 2V + 2F = DV + 4. Multiply both sides by 3 to get 6V + 6F = 3DV + 12.
For a MPG, we noted that 2E = 3F. Combine DV = 2E with 2E = 3F to see that DV = 3F. Multiply both sides by 2 to get 2DV = 6F. Substitute this into the last equation from the previous paragraph to obtain 6V + 2DV = 3DV + 12. Subtract 3DV from both sides to get 6V – DV = 12. Factor out the V to obtain V(6 – D) = 12. We finally obtain the formula V = 12/(6 – D) for a MPG (recall that every face of a MPG has three sides). Recall that V is the number of vertices (which are regions) and that D is the average number of edges intersecting at each vertex on the graph.
When each vertex has degree D = 2, we get V = 12/(6 – 2) = 12/4 = 3. Every vertex on the MPG will connect to 2 edges if there are exactly 3 vertices.
When each vertex has degree D = 3, we get V = 12/(6 – 3) = 12/3 = 4. Every vertex on the MPG will connect to 3 edges if there are exactly 4 vertices.
When each vertex has degree D = 4, we get V = 12/(6 – 4) = 12/2 = 6. Every vertex on the MPG can connect to 4 edges if there are exactly 6 vertices.
When each vertex has degree D = 5, we get V = 12/(6 – 5) = 12/1 = 12.
Every vertex on the MPG can connect to 5 edges if there are exactly 12 vertices.
According to the formula V = 12/(6 – D), there doesn’t exist a solution where every vertex can connect to 6 edges. It isn’t possible to draw a MPG (or any type of PG, as we’ll see) where every vertex connects to 6 edges.
In the formula V = 12/(6 – D), the only way to get a positive number for V is if D is less than 6. This means that the average degree of the vertices must be less than 6. On average, there need to be fewer than 6 edges connecting to each vertex. There can be vertices on a graph that connect to more than 6 edges, provided that the average is under 6. This shows that there must be at least one vertex on any MPG with a degree less than 6.
What if the PG isn’t triangulated? In that case, the ratio of F to E will be smaller, as we noted earlier in this chapter. Let’s look at a specific case with numbers. Suppose every face is a quadrilateral (a polygon with 4 sides) instead of a triangle. For quadrilateral faces, 2E = 4F (instead of 2E = 3F for triangular faces). This time, we combine DV = 2E with 2E = 4F to get DV = 4F. Return to the equation 2V + 2F = DV + 4, multiply both sides by 2 to get 4V + 4F = 2DV + 8, and plug in DV = 4F to obtain 4V + DV = 2DV + 8.
Subtract 2DV to get 4V – DV = 8, which we can write as V(4 – D) = 8. If every face is a quadrilateral, then we find that the average degree has to be less than 4. This restriction is even more severe than it was for a MPG. Since the ratio of F to E is maximum for a MPG, the case of triangulation results in more edges per vertex. Thus, if there must be at least one vertex with a degree of less than 6 for a MPG, we can conclude that there must be at least one vertex with a degree of less than 6 for any PG (triangulated or not) [Ref.
2].
Returning to MPG’s, it will be useful in later chapters to relate the number of edges (E) to the number of vertices (V):
• 2E = 3F for a MPG. which is equivalent to E = 3V – 6. This formula applies if V is at least 3.
(For very small values of V, it shouldn’t be necessary to bother with the formula.)
This shows that a MPG with V vertices has E = 3V – 6 edges. Following are some examples.
Since every PG has an average degree value less than 6 (and thus has at least one vertex with a degree of 5 or less), it follows that:
• If every vertex of a MPG has an even degree number, there will be at least one vertex with degree 4. (See Challenge Problem 4 in Chapter 11.)
• If a MPG doesn’t have any vertices with a degree less than 5, it must have at least one edge that joins together two vertices with degree 5 or at least one edge that joins together vertices with degree 5 and 6.
CHAPTER 4 EXERCISES
1. For the map below on the left, determine the number of vertices, edges, and faces. For the graph below on the right, determine the number of vertices, edges, and faces. Are this map and graph duals to one another? Explain.
Verify Euler’s formula for the map. Also verify Euler’s formula for the graph. Determine the ratio of F to E for the graph. What can you conclude from the ratio of F to E for this graph?
2. For the graph below, determine the number of vertices, edges, and faces.
Verify Euler’s formula for the graph. Determine the ratio of F to E for the graph. What can you conclude from the ratio of F to E for this graph? If the graph is not a MPG, add edges to triangulate the graph, and determine V, E, F, and F/E for the MPG.
3. For the graph below determine the number of vertices, edges, and faces.
Verify Euler’s formula for the graph. Determine the ratio of F to E for the graph. What can you conclude from the ratio of F to E for this graph? If the graph is not a MPG, add edges to triangulate the graph, and determine V, E, F, and F/E for the MPG.
4. If every face of a PG is a quadrilateral, what is the ratio of F to E? Is this ratio larger or smaller than the ratio of F to E for a MPG?
5. If a MPG has 15 vertices, how many edges and how many faces does it
11. If every vertex of a MPG has degree five, how many vertices, edges, and faces does it have?
12. If one-half of the vertices of a MPG have degree five and the other half have degree six, how many vertices, edges, and faces does it have?
13. If a MPG has 2 vertices with degree 8, 2 vertices with degree 3, and each
of the remaining vertices have degree 4, how many vertices, edges, and faces does it have?