Discrete MathematicsAn Open Introduction

Section4.1Definitions

Investigate!29

Which (if any) of the graphs below are the same?

The graphs above are unlabeled. Usually we think of a graph as having a specific set of vertices. Which (if any) of the graphs below are the same?

Actually, all the graphs we have seen above are just drawings of graphs. A graph is really an abstract mathematical object consisting of two sets \(V\) and \(E\) where \(E\) is a set of 2-element subsets of \(V\text{.}\) Are the graphs below the same or different?

Graph 1:

\(V = \{a, b, c, d, e\}\text{,}\)

\(E = \{\{a,b\}, \{a, c\}, \{a,d\}, \{a,e\}, \{b,c\}, \{d,e\}\}\text{.}\)

Graph 2:

\(V = \{v_1, v_2, v_3, v_4, v_5\}\text{,}\)

\(E = \{\{v_1, v_3\}, \{v_1, v_5\}, \{v_2, v_4\}, \{v_2, v_5\}, \{v_3, v_5\}, \{v_4, v_5\}\}\text{.}\)

Before we start studying graphs, we need to agree upon what a graph is. While we almost always think of graphs as pictures (dots connected by lines) this is fairly ambiguous. Do the lines need to be straight? Does it matter how long the lines are or how large the dots are? Can there be two lines connecting the same pair of dots? Can one line connect three dots?

The way we avoid ambiguities in mathematics is to provide concrete and rigorous definitions. Crafting good definitions is not easy, but it is incredibly important. The definition is the agreed upon starting point from which all truths in mathematics proceed. Is there a graph with no edges? We have to look at the definition to see if this is possible.

We want our definition to be precise and unambiguous, but it also must agree with our intuition for the objects we are studying. It needs to be useful: we could define a graph to be a six legged mammal, but that would not let us solve any problems about bridges. Instead, here is the (now) standard definition of a graph.

Strange. Nowhere in the definition is there talk of dots or lines. From the definition, a graph could be

Here we have a graph with four vertices (the letters \(a, b, c, d\)) and four edges (the pairs \(\{a,b\}, \{a,c\}, \{b,c\}, \{b,d\}, \{c,d\})\)).

Looking at sets and sets of 2-element sets is difficult to process. That is why we often draw a representation of these sets. We put a dot down for each vertex, and connect two dots with a line precisely when those two vertices are one of the 2-element subsets in our set of edges. Thus one way to draw the graph described above is this:

However we could also have drawn the graph differently. For example either of these:

We should be careful about what it means for two graphs to be “the same.” Actually, given our definition, this is easy: Are the vertex sets equal? Are the edge sets equal? We know what it means for sets to be equal, and graphs are nothing but a pair of two special sorts of sets.

Even if two graphs are not equal, they might be basically the same. The graphs in the previous example could be drawn like this:

Graphs that are basically the same (but perhaps not equal) are called isomorphic. We will give a precise definition of this term after a quick example:

Example4.1.2

Consider the graphs:

\(G_1 = \{V_1, E_1\}\) where \(V_1 = \{a, b, c\}\) and \(E_1 = \{\{a,b\}, \{a,c\}, \{b,c\}\}\text{;}\)

\(G_2 = \{V_2, E_2\}\) where \(V_2 = \{u,v,w\}\) and \(E_2 = \{\{u,v\}, \{u,w\}, \{v,w\}\}\text{.}\)

Are these graphs the same?

Solution

The two graphs are NOT equal. It is enough to notice that \(V_1 \ne V_2\) since \(a \in V_1\) but \(a \notin V_2\text{.}\) However, both of these graphs consist of three vertices with edges connecting every pair of vertices. We can draw them as follows:

Clearly we want to say these graphs are basically the same, so while they are not equal, they will be isomorphic. The reason is we can rename the vertices of one graph and get the second graph as the result.

Intuitively, graphs are isomorphic if they are basically the same, or better yet, if they are the same except for the names of the vertices. To make the concept of renaming vertices precise, we give the following definitions:

An isomorphism is simply a function which renames the vertices. It must be a bijection so every vertex gets a new name. These newly named vertices must be connected by edges precisely if they were connected by edges with their old names.

Example4.1.3

Decide whether the graphs \(G_1 = \{V_1, E_1\}\) and \(G_2 = \{V_2, E_2\}\) are equal or isomorphic.

\(V_1 = \{a,b,c,d\}\text{,}\) \(E_1 = \{\{a,b\}, \{a,c\}, \{a,d\}, \{c,d\}\}\)

\(V_2 = \{a,b,c,d\}\text{,}\) \(E_2 = \{\{a,b\}, \{a,c\}, \{b,c\}, \{c,d\}\}\)

Solution

The graphs are NOT equal, since \(\{a,d\} \in E_1\) but \(\{a,d\} \notin E_2\text{.}\) However, since both graphs contain the same number of vertices and same number of edges, they might be isomorphic (this is not enough in most cases, but it is a good start).

We can try to build an isomorphism. How about we say \(f(a) = b\text{,}\) \(f(b) = c\text{,}\) \(f(c) = d\) and \(f(d) = a\text{.}\) This is definitely a bijection, but to make sure that the function is an isomorphism, we must make sure it respects the edge relation. In \(G_1\text{,}\) vertices \(a\) and \(b\) are connected by an edge. In \(G_2\text{,}\) \(f(a) = b\) and \(f(b) = c\) are connected by an edge. So far, so good, but we must check the other three edges. The edge \(\{a,c\}\) in \(G_1\) corresponds to \(\{f(a), f(c)\} = \{b,d\}\text{,}\) but here we have a problem. There is no edge between \(b\) and \(d\) in \(G_2\text{.}\) Thus \(f\) is NOT an isomorphism.

Not all hope is lost, however. Just because \(f\) is not an isomorphism does not mean that there is no isomorphism at all. We can try again. At this point it might be helpful to draw the graphs to see how they should match up.

Alternatively, notice that in \(G_1\text{,}\) the vertex \(a\) is adjacent to every other vertex. In \(G_2\text{,}\) there is also a vertex with this property: \(c\text{.}\) So build the bijection \(g:V_1 \to V_2\) by defining \(g(a) = c\) to start with. Next, where should we send \(b\text{?}\) In \(G_1\text{,}\) the vertex \(b\) is only adjacent to vertex \(a\text{.}\) There is exactly one vertex like this in \(G_2\text{,}\) namely \(d\text{.}\) So let \(g(b) = d\text{.}\) As for the last two, in this example, we have a free choice: let \(g(c) = b\) and \(g(d) = a\) (switching these would be fine as well).

We should check that this really is an isomorphism. It is definitely a bijection. We must make sure that the edges are respected. The four edges in \(G_1\) are

\begin{equation*} \{a,b\}, \{a,c\}, \{a,d\}, \{c,d\} \end{equation*}

Under the proposed isomorphism these become

\begin{equation*} \{g(a), g(b)\}, \{g(a), g(c)\}, \{g(a), g(d)\}, \{g(c), g(d)\} \end{equation*} \begin{equation*} \{c,d\}, \{c,b\}, \{c,a\}, \{b,a\} \end{equation*}

which are precisely the edges in \(G_2\text{.}\) Thus \(g\) is an isomorphism, so \(G_1 \cong G_2\)

Sometimes we will talk about a graph with a special name (like \(K_n\) or the Peterson graph) or perhaps draw a graph without any labels. In this case we are really referring to all graphs isomorphic to any copy of that particular graph. A collection of isomorphic graphs is often called an isomorphism class.¹This is not unlike geometry, where we might have more than one copy of a particular triangle. There instead of isomorphic we say congruent.

There are other relationships between graphs that we care about, other than equality and being isomorphic. For example, compare the following pair of graphs:

These are definitely not isomorphic, but notice that the graph on the right looks like it might be part of the graph on the left, especially if we draw it like this:

We would like to say that the smaller graph is a subgraph of the larger.

We should give a careful definition of this. In fact, there are two reasonable notions for what a subgroup should mean.

Notice that every induced subgraph is also an ordinary subgraph, but not conversely. Think of a subgraph as the result of deleting some vertices and edges from the larger graph. For the subgraph to be an induced subgraph, we can still delete vertices, but now we only delete those edges that included the deleted vertices.

Example4.1.4

Consider the graphs:

Here both \(G_2\) and \(G_3\) are subgraphs of \(G_1\text{.}\) But only \(G_2\) is an induced subgraph. Every edge in \(G_1\) that connects vertices in \(G_2\) is also an edge in \(G_2\text{.}\) In \(G_3\text{,}\) the edge \(\{a,b\}\) is in \(E_1\) but not \(E_3\text{,}\) even though vertices \(a\) and \(b\) are in \(V_3\text{.}\)

The graph \(G_4\) is NOT a subgraph of \(G_1\text{,}\) even though it looks like all we did is remove vertex \(e\text{.}\) The reason is that in \(E_4\) we have the edge \(\{c,f\}\) but this is not an element of \(E_1\text{,}\) so we don't have the required \(E_4 \subseteq E_1\text{.}\)

Back to some basic graph theory definitions. Notice that all the graphs we have drawn above have the property that no pair of vertices is connected more than once, and no vertex is connected to itself. Graphs like these are sometimes called simple, although we will just call them graphs. This is because our definition for a graph says that the edges form a set of 2-element subsets of the vertices. Remember that it doesn't make sense to say a set contains an element more than once. So no pair of vertices can be connected by an edge more than once. Also, since each edge must be a set containing two vertices, we cannot have a single vertex connected to itself by an edge.

That said, there are times we want to consider double (or more) edges and single edge loops. For example, the “graph” we drew for the Bridges of Königsberg problem had double edges because there really are two bridges connecting a particular island to the near shore. We will call these objects multigraphs. This is a good name: a multiset is a set in which we are allowed to include a single element multiple times.

The graphs above are also connected: you can get from any vertex to any other vertex by following some path of edges. A graph that is not connected can be thought of as two separate graphs drawn close together. For example, the following graph is NOT connected because there is no path from \(a\) to \(b\text{:}\)

Most of the time, it makes sense to treat non-connected graphs as separate graphs (think of the above graph as two squares), so unless otherwise stated, we will assume all our graphs are connected.

Vertices in a graph do not always have edges between them. If we add all possible edges, then the resulting graph is called complete. That is, a graph is complete if every pair of vertices is connected by an edge. Since a graph is determined completely by which vertices are adjacent to which other vertices, there is only one complete graph with a given number of vertices. We give these a special name: \(K_n\) is the complete graph on \(n\) vertices.

Each vertex in \(K_n\) is adjacent to \(n-1\) other vertices. We call the number of edges emanating from a given vertex the degree of that vertex. So every vertex in \(K_n\) has degree \(n-1\text{.}\) How many edges does \(K_n\) have? One might think the answer should be \(n(n-1)\text{,}\) since we count \(n-1\) edges \(n\) times (once for each vertex). However, each edge is incident to 2 vertices, so we counted every edge exactly twice. Thus there are \(n(n-1)/2\) edges in \(K_n\text{.}\) Alternatively, we can say there are \({n \choose 2}\) edges, since to draw an edge we must choose 2 of the \(n\) vertices.

In general, if we know the degrees of all the vertices in a graph, we can find the number of edges. The sum of the degrees of all vertices will always be twice the number of edges, since each edge adds to the degree of two vertices. Notice this means that the sum of the degrees of all vertices in any graph must be even!

One final definition: we say a graph is bipartite if the vertices can be divided into two sets, \(A\) and \(B\text{,}\) with no two vertices in \(A\) adjacent and no two vertices in \(B\) adjacent. The vertices in \(A\) can be adjacent to some or all of the vertices in \(B\text{.}\) If each vertex in \(A\) is adjacent to all the vertices in \(B\text{,}\) then the graph is a complete bipartite graph, and gets a special name: \(K_{m,n}\text{,}\) where \(|A| = m\) and \(|B| = n\text{.}\) The graph in the houses and utilities puzzle is \(K_{3,3}\text{.}\)

Named Graphs

Some graphs are used more than others, and get special names.

\(K_n\)

The complete graph on \(n\) vertices.

\(K_{m,n}\)

The complete bipartite graph with sets of \(m\) and \(n\) vertices.

\(C_n\)

The cycle on \(n\) vertices, just one big loop.

\(P_n\)

The path on \(n\) vertices, just one long path.

There are a lot of definitions to keep track of in graph theory. Here is a glossary of the terms we have already used and will soon encounter.

SubsectionExercises

¶

1

If 10 people each shake hands with each other, how many handshakes took place? What does this question have to do with graph theory?

Solution

This is asking for the number of edges in \(K_{10}\text{.}\) Each vertex (person) has degree (shook hands with) 9 (people). So the sum of the degrees is \(90\text{.}\) However, the degrees count each edge (handshake) twice, so there are 45 edges in the graph. That is how many handshakes took place.

2

Among a group of 5 people, is it possible for everyone to be friends with exactly 2 of the people in the group? What about 3 of the people in the group?

Solution

It is possible for everyone to be friends with exactly 2 people. You could arrange the 5 people in a circle and say that everyone is friends with the two people on either side of them (so you get the graph \(C_5\)). However, it is not possible for everyone to be friends with 3 people. That would lead to a graph with an odd number of odd degree vertices which is impossible since the sum of the degrees must be even.

3

Is it possible for two different (non-isomorphic) graphs to have the same number of vertices and the same number of edges? What if the degrees of the vertices in the two graphs are the same (so both graphs have vertices with degrees 1, 2, 2, 3, and 4, for example)? Draw two such graphs or explain why not.

Solution

Yes. For example, both graphs below contain 6 vertices, 7 edges, and have degrees (2,2,2,2,3,3).

4

Are the two graphs below equal? Are they isomorphic? If they are isomorphic, give the isomorphism. If not, explain.

Graph 1: \(V = \{a,b,c,d,e\}\text{,}\) \(E = \{\{a,b\}, \{a,c\}, \{a,e\}, \{b,d\}, \{b,e\}, \{c,d\}\}\text{.}\)

Graph 2:

Solution

The graphs are not equal. For example, graph 1 has an edge \(\{a,b\}\) but graph 2 does not have that edge. They are isomorphic. One possible isomorphism is \(f:G_1 \to G_2\) defined by \(f(a) = d\text{,}\) \(f(b) = c\text{,}\) \(f(c) = e\text{,}\) \(f(d) = b\text{,}\) \(f(e) = a\text{.}\)

5

Consider the following two graphs:

\(G_1\)

\(V_1=\{a,b,c,d,e,f,g\}\)

\(E_1=\{\{a,b\},\{a,d\},\{b,c\},\{b,d\},\{b,e\},\{b,f\},\{c,g\},\{d,e\},\)

\(\{e,f\},\{f,g\}\}\text{.}\)

\(G_2\)

\(V_2=\{v_1,v_2,v_3,v_4,v_5,v_6,v_7\}\text{,}\)

\(E_2=\{\{v_1,v_4\},\{v_1,v_5\},\{v_1,v_7\},\{v_2,v_3\},\{v_2,v_6\},\)

\(\{v_3,v_5\},\{v_3,v_7\},\{v_4,v_5\},\{v_5,v_6\},\{v_5,v_7\}\}\)

1. Let \(f:G_1 \rightarrow G_2\) be a function that takes the vertices of Graph 1 to vertices of Graph 2. The function is given by the following table:

   \(x\)	\(a\)	\(b\)	\(c\)	\(d\)	\(e\)	\(f\)	\(g\)
   \(f(x)\)	\(v_4\)	\(v_5\)	\(v_1\)	\(v_6\)	\(v_2\)	\(v_3\)	\(v_7\)

   Does \(f\) define an isomorphism between Graph 1 and Graph 2? Explain.

2. Define a new function \(g\) (with \(g\not=f\)) that defines an isomorphism between Graph 1 and Graph 2.

3. Is the graph pictured below isomorphic to Graph 1 and Graph 2? Explain.

   

6

Which of the graphs below are bipartite? Justify your answers.

Solution

Three of the graphs are bipartite. The one which is not is \(C_7\) (second from the right). To see that the three graphs are bipartite, we can just give the bipartition into two sets \(A\) and \(B\text{,}\) as labeled below:

The graph \(C_7\) is not bipartite because it is an odd cycle. You would want to put every other vertex into the set \(A\text{,}\) but if you travel clockwise in this fashion, the last vertex will also be put into the set \(A\text{,}\) leaving two \(A\) vertices adjacent (which makes it not a bipartition).

7

For which \(n \ge 3\) is the graph \(C_n\) bipartite?

8

For each of the following, try to give two different unlabeled graphs with the given properties, or explain why doing so is impossible.

1. Two different trees with the same number of vertices and the same number of edges. A tree is a connected graph with no cycles.

2. Two different graphs with 8 vertices all of degree 2.

3. Two different graphs with 5 vertices all of degree 4.

4. Two different graphs with 5 vertices all of degree 3.

Solution

1. For example:

   

   

2. This is not possible if we require the graphs to be connected. If not, we could take \(C_8\) as one graph and two copies of \(C_4\) as the other.

3. Not possible. If you have a graph with 5 vertices all of degree 4, then every vertex must be adjacent to every other vertex. This is the graph \(K_5\text{.}\)

4. This is not possible. In fact, there is not even one graph with this property (such a graph would have \(5\cdot 3/2 = 7.5\) edges).