Hopefully this chapter has given you some sense for the wide variety of graph theory topics as well as why these studies are interesting. There are many more interesting areas to consider and the list is increasing all the time; graph theory is an active area of mathematical research.
One reason graph theory is such a rich area of study is that it deals with such a fundamental concept: any pair of objects can either be related or not related. What the objects are and what βrelatedβ means varies on context, and this leads to many applications of graph theory to science and other areas of math. The objects can be countries, and two countries can be related if they share a border. The objects could be land masses which are related if there is a bridge between them. The objects could be websites which are related if there is a link from one to the other. Or we can be completely abstract: the objects are vertices which are related if their is an edge between them.
What question we ask about the graph depends on the application, but often leads to deeper, general and abstract questions worth studying in their own right. Here is a short summary of the types of questions we have considered:
Can the graph be drawn in the plane without edges crossing? If so, how many regions does this drawing divide the plane into?
Is it possible to color the vertices of the graph so that related vertices have different colors using a small number of colors? How many colors are needed?
Can you find subgraphs with certain properties? For example, when does a (bipartite) graph contain a subgraph in which all vertices are only related to one other vertex?
Not surprisingly, these questions are often related to each other. For example, the chromatic number of a graph cannot be greater than 4 when the graph is planar. Whether the graph has an Euler path depends on how many vertices each vertex is adjacent to (and whether those numbers are always even or not). Even the existence of matchings in bipartite graphs can be proved using paths.
The first and the third graphs are the same (try dragging vertices around to make the pictures match up), but the middle graph is different (which you can see, for example, by noting that the middle graph has only one vertex of degree 2, while the others have two such vertices).
Consider the graph \(G = (V, E)\) with \(V = \{a,b,c,d,e,f,g\}\) and \(E = \{ab, ac, af, bg,
cd, ce\}\) (here we are using the shorthand for edges: \(ab\) really means \(\{a,b\}\text{,}\) for example).
Is the graph \(G\) isomorphic to \(G' = (V', E')\) with \(V' = \{t, u, v, w, x, y, z\}\) and \(E' = \{tz, uv,
uy, uz, vw,
vx\}\text{?}\) If so, give the isomoprhism. If not, explain how you know.
Yes, the graphs are isomorphic, which you can see by drawing them. One isomorphism is:
\begin{equation*}
f = \begin{pmatrix}
a \amp b \amp c \amp d \amp e \amp f \amp g \\
u \amp z \amp v \amp x \amp w \amp y \amp t
\end{pmatrix}\text{.}
\end{equation*}
Any labeling of this graph will be not isomorphic to \(G\text{.}\) For example, we could take \(V'' = \{a,b,c,d,e,f,g\}\) and \(E'' = \{ab,
ac, ad, be, cf, dg\} \text{.}\)
In general this should be possible: the degree sequence does not determine the graphβs isomorphism class. However, in this case, I was almost certain this was not possible. That is, until I stumbled up this:
The chromatic number of \(G\) is 2. It shouldnβt be hard to give a 2-coloring (for example, color \(a, d, e, g\) red and \(b, c, f\) blue), but we know that all bipartite graphs have chromatic number 2.
It is clear from the drawing that there is no Euler path, let alone an Euler circuit. Also, since there are more than 2 vertices of odd degree, we know for sure there is no Euler path.
12. Such a graph would have \(\frac{5n}{2}\) edges. If the graph is planar, then \(n - \frac{5n}{2} + f = 2\) so there would be \(\frac{4+3n}{2}\) faces. Also, we must have \(3f \le 2e\text{,}\) since the graph is simple. So we must have \(3\left(\frac{4 + 3n}{2}\right) \le 5n\text{.}\) Solving for \(n\) gives \(n \ge 12\text{.}\)
Among a group of \(n\) people, is it possible for everyone to be friends with an odd number of people in the group? If so, what can you say about \(n\text{?}\)
Assuming you are successful in building your new 16-faced polyhedron, could every vertex be the joining of the same number of faces? Could each vertex join either 3 or 4 faces? If so, how many of each type of vertex would there be?
No. The 9 triangles each contribute 3 edges, and the 6 pentagons contribute 5 edges. This gives a total of 57, which is exactly twice the number of edges, since each edge borders exactly 2 faces. But 57 is odd, so this is impossible.
Now adding up all the edges of all the 16 polygons gives a total of 64, meaning there would be 32 edges in the polyhedron. We can then use Eulerβs formula \(v - e + f = 2\) to deduce that there must be 18 vertices.
If you add up all the vertices from each polygon separately, we get a total of 64. This is not divisible by 3, so it cannot be that each vertex belongs to exactly 3 faces. Could they all belong to 4 faces? That would mean there were \(64/4 = 16\) vertices, but we know from Eulerβs formula that there must be 18 vertices. We can write \(64 = 3x + 4y\) and solve for \(x\) and \(y\) (as integers). We get that there must be 10 vertices with degree 4 and 8 with degree 3. (Note the number of faces joined at a vertex is equal to its degree in graph theoretic terms.)
How many edges does the graph \(K_{n,n}\) have? For which values of \(n\) does the graph contain an Euler circuit? For which values of \(n\) is the graph planar?
The graph \(G\) has 6 vertices with degrees \(1, 2, 2, 3, 3, 5\text{.}\) How many edges does \(G\) have? If \(G\) was planar how many faces would it have? Does \(G\) have an Euler path?
\(G\) has 8 edges (since the sum of the degrees is 16). If \(G\) is planar, then it will have 4 faces (since \(6 - 8 + 4 = 2\)). \(G\) does not have an Euler path since there are more than 2 vertices of odd degree.
What is the smallest number of colors you need to properly color the vertices of \(K_{7}\text{.}\) Can you say whether \(K_7\) is planar based on your answer?
What is the smallest number of colors you need to properly color the vertices of \(K_{3,4}\text{?}\) Can you say whether \(K_{3,4}\) is planar based on your answer?
The chromatic number of \(K_{3,4}\) is 2, since the graph is bipartite. You cannot say whether the graph is planar based on this coloring (the converse of the Four Color Theorem is not true). In fact, the graph is not planar, since it contains \(K_{3,3}\) as a subgraph.
We have that \(K_{3,4}\) has 7 vertices and 12 edges (each vertex in the group of 3 has degree 4). Then by Eulerβs formula we have that \(7 - 12 + f = 2\) so if the graph were planar, it would have \(f = 7\) faces. However, since the girth of the graph is 4 (there are no cycles of length 3) we get that \(4f \le 2e\text{.}\) But this would mean that \(28 \le 24\text{,}\) a contradiction.
For all these questions, we are really coloring the vertices of a graph. You get the graph by first drawing a planar representation of the polyhedron and then taking its planar dual: put a vertex in the center of each face (including the outside) and connect two vertices if their faces share an edge.
Since the planar dual of a dodecahedron contains a 5-wheel, itβs chromatic number is at least 4. Alternatively, suppose you could color the faces using 3 colors without any two adjacent faces colored the same. Take any face and color it blue. The 5 pentagons bordering this blue pentagon cannot be colored blue. Color the first one red. Its two neighbors (adjacent to the blue pentagon) get colored green. The remaining 2 cannot be blue or green, but also cannot both be red since they are adjacent to each other. Thus a 4th color is needed.
The planar dual of the dodecahedron is itself a planar graph. Thus by the 4-color theorem, it can be colored using only 4 colors without two adjacent vertices (corresponding to the faces of the polyhedron) being colored identically.
The cube can be properly 3-colored. Color the βtopβ and βbottomβ red, the βfrontβ and βbackβ blue, and the βleftβ and βrightβ green.
False. To prove this, we can give an example of a pair of graphs with the same chromatic number that are not isomorphic. For example, \(K_{3,3}\) and \(K_{3,4}\) both have chromatic number 2, but are not isomorphic.
False. The previous example does not work, but you can easily draw two trees that have the same number of vertices and edges but are not isomorphic. Since all trees have chromatic number 2, this is a counterexample.
True. If there is an isomorphism from \(G_1\) to \(G_2\text{,}\) then we have a bijection that tells us how to match up vertices between the graph. Any proper vertex coloring of \(G_1\) will tell us how to properly color \(G_2\text{,}\) simply by coloring \(f(v_i)\) the same color as \(v_i\text{,}\) for each vertex \(v_i \in V\text{.}\) That is, color the vertices in \(G_2\) exactly how you color the corresponding vertices in \(G_1\text{.}\) Similarly, any proper vertex coloring of \(G_2\) corresponds to a proper vertex coloring of \(G_1\text{.}\) Thus the smallest number of colors needed to properly color \(G_1\) cannot be smaller than the smallest number of colors needed to properly color \(G_2\text{,}\) and vice-versa, so the chromatic numbers must be equal.
The graph does have an Euler path, but not an Euler circuit. There are exactly two vertices with odd degree. The path starts at one and ends at the other.
For each part below, say whether the statement is true or false. Explain why the true statements are true, and give counterexamples for the false statements.
True. The graph is bipartite so it is possible to divide the vertices into two groups with no edges between vertices in the same group. Thus we can color all the vertices of one group red and the other group blue.
Let \(G\) be a connected graph with \(v\) vertices and \(e\) edges. Use mathematical induction to prove that if \(G\) contains exactly one cycle (among other edges and vertices), then \(v = e\text{.}\)
You might want to give the proof in two parts. First prove by induction that the cycle \(C_n\) has \(v=e\text{.}\) Then consider what happens if the graph is more than just the cycle.