Section 2.3 Euler Formula
Problem 2.3.1.
For each of the regular polyhedra count the number of vertices (points), edges (line segments), and faces (2D shapes). 3D models are in FigureĀ 2.3.3 to FigureĀ 2.3.7. You may also use the ZomeTool models in the instructor's office.
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Tetrahedron
4 Vertices
6 Edges
4 Faces (Triangles)
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Cube
8 Vertices
6 Edges
4 Faces (Triangles)
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Octahedron
6 Vertices
12 Edges
8 Faces (Triangles)
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Dodecahedron
20 Vertices
30 Edges
12 Faces (Pentagons)
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Icosahedron
12 Vertices
30 Edges
20 Faces (Triangles)
Problem 2.3.2.
For each of the regular polyhedra draw a graph whose vertices are the vertices of the polyhedra and whose edges are the edges of the polyhedra. Try to draw each graph with no edges crossing.
It is possible to draw all these shapes as a graph with no edges crossing.
Tetrahedron
Cube
Octahedron
Dodecahedron
Icosahedron
Drag to change viewpoint.
Drag to change viewpoint.
Drag to change viewpoint.
Drag to change viewpoint.
Drag to change viewpoint.
Definition 2.3.8.
A figure 8 graph, denoted \(F8_{m,n}\text{,}\) consists of two cycles \(C_m\) and \(C_n\) that share exactly one vertex.
Example 2.3.9. Figure 8 Graph Examples.
\(F8_{4,3}\)
\(F8_{5,3}\)
Definition 2.3.10.
A graph is triangular if and only if every cycle of size at least four has an additional edge.
Example 2.3.11. Triangular Graphs.
Lemma 2.3.12.
Conjecture and prove a formula for the number of edges in terms of the number of vertices for each of the following types of graphs.
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Trees
The number of edges in a tree equals to: \(f(n)=n-1\) where \(n\) is the amount of vertices in the tree. This formula is proved in TheoremĀ 2.2.2.
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Cycles (\(C_n\) for arbitrary \(n\))
Recall DefinitionĀ 1.3.3. Thus for a given set of vertices, a cycles's amount of edges must equal the amount to vertices, i.e. E = V.
Figure 8's (\(F_{m,n}\) for arbitrary \(m\) and \(n\))
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Triangular graphs
Conjecture: The minimum number of edges in a triangular graph with \(n\) vertices is \(n-1\)
This is a direct proof. It may seem surprising that the conjecture is so uninspired, but what becomes clear from exploring this idea is that a statement about the edges in terms of the vertices alone is insufficient to adequately describe the number of edges contained in a triangular graph.
The definition of a triangular graph provided in this text sets no requirement that the graph be connected, so let's address that first. Suppose G is a triangular graph and is disconnected. It is possible to separate G into individual, connected components (even if they are just a single point) that can be analyzed individually. The results of each connected component can then be summed at the end. Thus, an analysis is possible.
Next, the definition provided in this text makes no requirement that a graph have any cycles. Suppose G is a connected component of a triangular graph that has no cycles. Then G is a tree and has \(n-1\) edges.
Now suppose that G is a connected component of a graph that contains one or more cycles. By the definition provided, each of these cycles must be a three-cycle, or a combination of three-cycles (think of a honeycomb structure). Recall that the question is framed in such a way that it requests a formula for the number of edges in terms of the number of vertices alone. The best we can do is state that the number of edges in a triangular graph containing cycles must be greater than \(n-1\text{.}\) It is tempting to invoke that each cycle contains a removable edge, but the number of cycles present in a triangular graph can vary without varying the number of vertices. This is best illustrated with an example:
Consider a triangular graph with 6 vertices.
By the definition that was given, this is a connected triangular graph with six vertices and seven edges.
But it is possible to add more edges without violating the definition or adding more vertices.
This is also a connected triangular graph with six vertices. But this one has nine edges.
And it is possible to add even more edges without violating the definition or adding vertices.
This is also a connected triangular graph with six vertices. But this one has eleven edges.
So clearly the number of edges present in a connected triangular graph (or a connected component of a triangular graph) does not depend solely on the number of vertices in that graph. Thus the best we can do is to say that a connected component or a triangular graph has a minimum of \(n-1\) edges. It is tempting to start discussing a notion of counting all the cycles, but that is the motivation behind part 4 of LemmaĀ 2.3.13 so we will leave the discussion here.
Lemma 2.3.13.
Conjecture and prove a formula for the number edges in terms of the number faces and the number vertices for each of the following types of graphs.
Trees
Cycles (\(C_n\) for arbitrary \(n\))
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Figure 8's (\(F_{m,n}\) for arbitrary \(m\) and \(n\))
Suppose that \(G\) is the figure-eight graph \(F8_{m,n}\text{.}\) Then \(G\) is formed by joining the cycles \(C_m\) and \(C_n\) at a single vertex. It follows that \(G\) has \(m+n\) edges and \(m+n-1\) vertices. So \(V - E = (m+n-1)-(m+n) = -1\) .
Triangular graphs
Proof.
Let\(G\)be a connected graph on\(n\)vertices. We prove each case in turn.
Suppose \(G\) is a tree. Then it has no faces. So \(V - E + F = n - (n-1) + 0 = 1\)
Suppose \(G\) is a cycle. Then it has one face. Thus \(V - E + F = n - n + 1 = 1\)
Suppose \(G\) is a figure eight. Then \(G\) has two faces. Hence \(V - E + F = n - (n+1) + 2 = 1\)
The last case is more complicated. Suppose \(G\) is triangular and \(n\geq 4\) (Smaller \(n\) reduce to other cases.) Then every face of \(G\) is a 3-cycle. Such a graph can be constructed by successively adding triangular faces to a single 3-cycle. When adding a face, the new face shares either 0, 1, or 2 edges with an existing face. In each of these cases \(V-E+F\) is unchanged. Since we began with a 3-cycle, \(V-E+F=1\)
Suppose that we add a new face that shares no edges with an existing face. Then we must add three edges and two vertices. So the new sum is \((V+2) - (E+3) + (F+1) = V-E+F\text{,}\) so the sum is unchanged.
Suppose instead that we add a new face that shares exactly one edge with an existing face. Then we must add one new vertex and two new edges. The new sum is \((V+1)-(E+2)+(F+1) = V-E+F\text{,}\) so again the sum is unchanged.
Finally, suppose that we add a face that shares two edges with existing faces. Then we add no new vertices and one new edge. So the sum is \(V - (E+1) + (F+1) = V - E + F,\) so the new graph has the same sum as the old graph.
It is possible that the original graph had some number of arbitrary trees which cannot be recreated in this manner. However, by the tree case of this theorem, we can add any number of arbitrary trees to our graph without changing \(V-E+F\text{.}\)
This shows inductively that for any triangular graph\(V-E+F=1\)
Theorem 2.3.14.
Generalize your conjecture above to all graphs meeting the following conditions. The graph is connected. The graph can be drawn in the plane without edges crossing; thus āfacesā makes sense.