1 Walks, Paths, and Cycles
A walk in a graph G is a sequence of vertices v0,v1,…,vk such that {vi−1,vi}∈E(G) for each i. The integer k is the length of the walk.
A walk in which all edges are distinct is called a trail.
A walk in which all vertices are distinct is called a path.
A trail with v0=vk and k≥1 is called a closed trail.
A walk with v0=vk, v0,v1,…,vk−1 all distinct, and k≥3 is called a cycle.
A path of length k and a cycle of length k are also denoted Pk and Ck, respectively.
A path P4 (left) and a cycle C4 (right):
If there is a walk from u to v in a graph G, then there is a path from u to v.
Among all walks from u to v, choose one of minimum length, say W=v0,v1,…,vk with v0=u and vk=v. If W has a repeated vertex vi=vj with i<j, then the walk v0,…,vi,vj+1,…,vk is shorter, contradicting the minimality of W. Therefore W is a path. □
2 Connectivity and Connected Components
A graph G is connected if for every pair of vertices u,v∈V(G), there exists a path from u to v.
The relation "there is a path between u and v" is an equivalence relation on V(G). The subgraphs induced by the equivalence classes are called the connected components of G.
In a connected graph G, the distanced(u,v) between two vertices u and v is the length of a shortest path from u to v. The diameter of G is diam(G)=maxu,v∈Vd(u,v).
3 Cut Vertices, Bridges, and Connectivity
A vertex v of a connected graph G is a cut vertex if the graph G−v (obtained by removing v and all edges incident to v) is disconnected. Similarly, an edge e is a bridge if G−e is disconnected.
In the following graph, v is a cut vertex and the edge {v,d} is a bridge:
An edge e is a bridge of G if and only if e does not lie on any cycle.
(⇒) We prove the contrapositive. Suppose the edge e={u,v} lies on a cycle C. In G−e, the vertices u and v are still connected by the portion of C that avoids e. Any path in G that uses e can be rerouted through this portion, so G−e remains connected. Hence e is not a bridge.(⇐) Suppose e={u,v} lies on no cycle. If G−e contained a path P from u to v, then P together with e would form a cycle, a contradiction. Therefore u and v are in different components of G−e, and e is a bridge. □
The vertex connectivityκ(G) of a connected graph G is the minimum number of vertices whose removal disconnects G (with the convention κ(Kn)=n−1). The edge connectivityλ(G) is the minimum number of edges whose removal disconnects G.
For any connected graph G, κ(G)≤λ(G)≤δ(G),
where δ(G) denotes the minimum degree of G.
λ(G)≤δ(G): Removing all edges incident to a vertex v of minimum degree isolates v, so λ(G)≤deg(v)=δ(G).κ(G)≤λ(G): Let F={e1,…,ek} be a minimum edge cut with λ(G)=k, and let G−F have components G1 and G2. Write ei={ui,vi} with ui∈V(G1) and vi∈V(G2). If ∣V(G2)∣>k, then removing the (at most k) distinct vertices among v1,…,vk disconnects G1 from the remainder of G2, so κ(G)≤k. If ∣V(G2)∣≤k, removing all of V(G2) uses at most k vertices and disconnects G, so again κ(G)≤k. □
4 Characterization of Bipartite Graphs
A graph G is bipartite if and only if it contains no cycle of odd length.
(⇒) Suppose G is bipartite with partition V=X⊔Y. In any cycle v0,v1,…,vk=v0, the vertices alternate between X and Y. If v0∈X, then vi∈X if and only if i is even. Since vk=v0∈X, the length k must be even.(⇐) We may assume G is connected (apply the argument to each component). Fix a vertex s and set X={v∣d(s,v) is even} and Y={v∣d(s,v) is odd}. Suppose an edge {u,v}∈E has both endpoints in the same part. Then a shortest path from s to u and a shortest path from s to v, together with the edge {u,v}, yield a closed walk of odd length, which contains an odd cycle — a contradiction. □
5 Menger's Theorem
A collection of s-t paths is internally vertex-disjoint if no two paths share an internal vertex (i.e., a vertex other than s or t).
Let s and t be non-adjacent vertices of a graph G. Then the minimum size of a vertex cut separating s from t equals the maximum number of internally vertex-disjoint s-t paths.
If k internally vertex-disjoint paths exist, then at least one internal vertex must be removed from each path to separate s from t, so the minimum cut has size at least k. The reverse inequality — that the existence of a minimum cut of size k guarantees k disjoint paths — can be established by induction on the number of edges. The essential idea mirrors the max-flow min-cut theorem, and the most elegant proof proceeds via the theory of network flows (which we treat in a later chapter). □