Update 22-2.md

1. Insert code from 22-3 that should have been here.
2. Add a special case to 22-2.f

docs/Chap22/Problems/22-2.md

@@ -48,7 +48,7 @@ Thus, we need only check $v.low$ versus $v.d$ to decide in constant time whether
 
 **e.** An edge $(u, v)$ lies on a simple cycle if and only if there exists at least one path from $u$ to $v$ which doesn't contain the edge $(u, v)$, if and only if removing $(u, v)$ doesn't disconnect the graph, if and only if $(u, v)$ is not a bridge.
 
-**f.** A edge $(u, v)$ lies on a simple cycle in an undirected graph if and only if either both of its endpoints are articulation points, or one of its endpoints is an articulation point and the other is a vertex of degree $1$. Since we can compute all articulation points in $O(E)$ and we can decide whether or not a vertex has degree $1$ in constant time, we can run the algorithm in part (d) and then decide whether each edge is a bridge in constant time, so we can find all bridges in $O(E)$ time.
+**f.** A edge $(u, v)$ lies on a simple cycle in an undirected graph if and only if either both of its endpoints are articulation points, or one of its endpoints is an articulation point and the other is a vertex of degree $1$. There's also a special case where there's only one edge whose incident vertices are both degree $1$. We can check this case in constant time. Since we can compute all articulation points in $O(E)$ and we can decide whether or not a vertex has degree $1$ in constant time, we can run the algorithm in part (d) and then decide whether each edge is a bridge in constant time, so we can find all bridges in $O(E)$ time.
 
 **g.** It is clear that every nonbridge edge is in some biconnected component, so we need to show that if $C_1$ and $C_2$ are distinct biconnected components, then they contain no common edges. Suppose to the contrary that $(u, v)$ is in both $C_1$ and $C_2$.
 
@@ -69,3 +69,12 @@ $$a, b, p_1, \ldots, p_k, u, q_m, \ldots, q_1, d, c, q_l , \ldots, q_{m + 1}, v,
 is a simple cycle containing $(a, b)$ and $(c, d)$, a contradiction. Thus, the biconnected components form a partition.
 
 **h.** Locate all bridge edges in $O(E)$ time using the algorithm described in part (f). Remove each bridge from $E$. The biconnected components are now simply the edges in the connected components. Assuming this has been done, run the following algorithm, which clearly runs in $O(|E|)$ where $|E|$ is the number of edges originally in $G$.
+
+```cpp
+VISIT-BCC(G, u, k)
+    u.color = GRAY
+    for each v ∈ G.Adj[u]
+        (u, v).bcc = k
+        if v.color == WHITE
+            VISIT-BCC(G, v, k)
+```