# 12.3: Paths and Cycles

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Recall the definition of a walk. As we saw in Example 12.2.1, the vertices and edges in a walk do not need to be distinct.

There are many circumstances under which we might not want to allow edges or vertices to be re-visited. Efficiency is one possible reason for this. We have a special name for a walk that does not allow vertices to be re-visited.

Definition: Path

A walk in which no vertex appears more than once is called a **path**.

Notation

For \(n ≥ 0\), a graph on \(n + 1\) vertices whose only edges are those used in a path of length \(n\) (which is a walk of length \(n\) that is also a path) is denoted by \(P_n\). (Notice that \(P_0 \cong K_1\) and \(P_1 \cong K_2\).)

Notice that if an edge were to appear more than once in a walk, then both of its endvertices would also have to appear more than once, so a path does not allow vertices or edges to be re-visited.

Example \(\PageIndex{1}\)

In the graph

\((a, f, c, h)\) is a path of length \(3\). However, \((a, f, c, h, d, f)\) is not a path, even though no edges are repeated, since the vertex \(f\) appears twice. Both are walks.

Proposition \(\PageIndex{1}\)

Suppose that \(u\) and \(v\) are in the same connected component of a graph. Then any \(u − v\) walk of minimum length is a path. In particular, if there is a \(u − v\) walk, then there is a \(u − v\) path.

**Proof**-
Since \(u\) and \(v\) are in the same connected component of a graph, there is a \(u − v\) walk.

Towards a contradiction, suppose that we have a \(u − v\) walk of minimum length that is not a path. By the definition of a path, this means that some vertex \(x\) appears more than once in the walk, so the walk looks like:

\[(u = u_1, . . . , u_i = x, . . . , u_j = x, . . . , u_k = v),\]

and \(j > i\). Observe that the following is also a \(u − v\) walk:

\[(u = u_1, . . . , u_i = x, u_{j+1}, u_{j+2}, . . . , u_k = v).\]

Since consecutive vertices were adjacent in the first sequence, they are also adjacent in the second sequence, so the second sequence is a walk. The length of the first walk is \(k − 1\), and the length of the second walk is \(k − 1 − (j − i)\). Since \(j > i\), the second walk is strictly shorter than the first walk. In particular, the first walk was not a \(u − v\) walk of minimum length. This contradiction serves to prove that every \(u − v\) walk of minimum length is a path.

This allows us to prove another interesting fact that will be useful later.

Proposition \(\PageIndex{2}\)

Deleting an edge from a connected graph can never result in a graph that has more than two connected components.

**Proof**-
Let \(G\) be a connected graph, and let \(u_v\) be an arbitrary edge of \(G\). If \(G \setminus \{u_v\}\) is connected, then it has only one connected component, so it satisfies our desired conclusion. Thus, we assume in the remainder of the proof that \(G \setminus \{u_v\}\) is not connected.

Let \(G_u\) denote the connected component of \(G \setminus \{u_v\}\) that contains the vertex \(u\), and let \(G_v\) denote the connected component of \(G \setminus \{u_v\}\) that contains the vertex \(v\). We aim to show that \(G_u\) and \(G_v\) are the only connected components of \(G \setminus \{u_v\}\).

Let \(x\) be an arbitrary vertex of \(G\), and suppose that \(x\) is a vertex that is not in \(G_u\). Since \(G\) is connected, there is a \(u − x\) walk in \(G\), and therefore by Proposition 12.3.1 there is a \(u − x\) path in \(G\). Since \(x\) is not in \(G_u\), this \(u−x\) path must use the edge \(u−v\), so must start with this edge since \(u\) only occurs at the start of the path. Therefore, by removing the vertex \(u\) from the start of this path, we obtain a \(v − x\) path that does not use the vertex \(u\). This path cannot use the edge \(u_v\), so must still be a path in \(G \setminus \{u_v\}\). Therefore \(x\) is a vertex in \(G_v\).

Since \(x\) was arbitrary, this shows that every vertex of \(G\) must be in one or the other of the connected components \(G_u\) and \(G_v\), so there are at most two connected components of \(G \setminus \{u_v\}\). Since \(u_v\) was an arbitrary edge of \(G\) and \(G\) was an arbitrary connected graph, this shows that deleting any edge of a connected graph can never result in a graph with more than two connected components.

A cycle is like a path, except that it starts and ends at the same vertex. The structures that we will call cycles in this course, are sometimes referred to as **circuits**.

Definition: Cycle

A walk of length at least \(1\) in which no vertex appears more than once, except that the first vertex is the same as the last, is called a **cycle**.

Notation

For \(n ≥ 3\), a graph on \(n\) vertices whose only edges are those used in a cycle of length \(n\) (which is a walk of length \(n\) that is also a cycle) is denoted by \(C_n\).

The requirement that the walk have length at least \(1\) only serves to make it clear that a walk of just one vertex is not considered a cycle. In fact, a cycle in a simple graph must have length at least \(3\).

Example \(\PageIndex{2}\)

In the graph from Example 12.3.1, \((a, e, f, a)\) is a cycle of length \(3\), and \((b, g, d, h, c, f, b)\) is a cycle of length \(6\).

Here are drawings of some small paths and cycles:

We end this section with a proposition whose proof will be left as an exercise.

Proposition \(\PageIndex{3}\)

Suppose that \(G\) is a connected graph. If \(G\) has a cycle in which \(u\) and \(v\) appear as consecutive vertices (so \(u_v\) is an edge of \(G\)) then \(G \setminus \{u_v\}\) is connected.

Exercise \(\PageIndex{1}\)

1) In the graph

(a) Find a path of length \(3\).

(b) Find a cycle of length \(3\).

(c) Find a walk of length \(3\) that is neither a path nor a cycle. Explain why your answer is correct.

2) Prove that in a graph, any walk that starts and ends with the same vertex and has the smallest possible non-zero length, must be a cycle.

3) Prove Proposition 12.3.3.

4) Prove by induction that if every vertex of a connected graph on \(n ≥ 2\) vertices has valency \(1\) or \(2\), then the graph is isomorphic to \(P_n\) or \(C_n\).

5) Let \(G\) be a (simple) graph on \(n\) vertices. Suppose that \(G\) has the following property: whenever \(u \nsim v\), \(dG(u) + dG(v) ≥ n − 1\). Prove that \(G\) is connected.