Biconnnected components

This can be formalized by defining:   a is an articulation point of graph G if and only if there exists vertices v,w such that v, w, a are distinct and every path from v to w contains a.

G is biconnected if and only if there are no articulation points.   That is, it requires deletion of at least 2 vertices to disconnect G.

A maximal subgraph of G that is biconnected is called a biconnected component of G.

Note that all the edges in a cycle will be in the same biconnected component.   We can define an equivalence relation on the edges e₁ is equivalent to e₂ if either e₁ = e₂ or there exists a cycle including both e₁ and e₂.   This equivalence relation partitions the set of edges into equivalence classes, where two edges belong to the same equivalence class if and only if they are equivalent.

If E is an equivalence class (of edges) and V is the set of vertices appearing as endpoints of the edges in E then it will be the case that the graph (V,E) is a biconnected component of G.

Lemma.   For any connected graph G:

1. Any 2 distinct biconnected components have at most one vertex in common.
2. a is an articulation point of G if and only if a is common to more than one of the biconnected components of G.

We will use DFS to find the biconnected components of a graph.   A DFS tree has no cross-edges.   That is, if v is neither an ancestor nor a descendent of w then there is no edge (v,w).

If a is an articulation point of graph G then removing a will split G into two or more parts.   One of these parts will consist of a son of a, s, together with all of s's descendents in the DFS tree.   This is summarized in the next lemma.

Lemma.   a is an articulation point of G if and only if either
1. a is the root and has more than one son, or
2. a is not the root, and there exists a son s of a such that there is no backedge between any descendent of s (including s) and a proper ancestor of a.

Define LOW(v) to be the most remote ancestor of v that can be reached from v if we are allowed to go down the tree from v and then go up at most one backedge.

By the lemma, if the vertices are numbered by DFS numbers then:
if a is not the root then a is an articulation point if and only if a has a son s such that LOW(s) ≥ DFN[a].

We will embed in the DFS algorithm the calculation of LOW, using the following recurrence.

LOW(v) = minimum x such that

either	1. x = DFN[v]	(no backedge taken)
or	2. x = LOW(s), where s is a son of v	(backedge from a proper descendent of v)
or	3. x = DFN[w], where (v,w) is a backedge	(backedge from v)

Depth first search with LOW computation

    Find articulation points and biconnected components
       T := Ø
       COUNT := 0
       STACK := Λ
       forall vertices v do FLAG[v] := "new"
       SEARCHB(1)

    SEARCHB(v)
       FLAG[v] := "old"
       COUNT := COUNT + 1
       DFN[v] := COUNT
1:     LOW[v] := DFN[v]
       Let L[v] be the list of vertices adjacent to v
       while L[v] ≠ Λ do
          w := pop( L[v] )
          if (v,w) is not yet on the stack
                 /** (v,w) is already on the stack if and only if        **/
                 /** (v,w) was a backedge w to v, for w descendent of v. **/
                 /** This can be expressed by:    FATHER[v] = w  or      **/
                 /**        (FLAG[w]="old" and DFN[v] < DFN[w])          **/
             then put (v,w) on STACK    
          if FLAG[w] = "new" then
             T := T union {(v,w)}
             FATHER[w] := v
             SEARCHB(w)
             if LOW[w] ≥ DFN[v] then
                v is an articulation point
                and the next biconnected component is
                the set of edges popped from STACK up to and including (v,w)
2:           LOW[v] := min{ LOW[v], LOW[w] }
          else if w ≠ FATHER[v] then
3:           LOW[v] := min{ LOW[v], DFN[w] }

The time complexity is O(n + e) because the changes from simple DFS are LOW computation and STACK manipulations, each of which takes time O(e).

One can prove by induction on the number of biconnected components of G that, when SEARCHB(w) is completed, if LOW[w] ≥ DFN[v] then the set of edges above (v,w) on the stack constitutes the biconnected component containing edge (v,w).