Tirgul 8

1
Tirgul 8

• Graph algorithms
• Strongly connected components

2
Reminder white path theorem

• In a depth forest of a graph G, a vertex v is a
  descendant of a vertex u iff in time d.u, when
  the DFS algorithm discovers u there is a path
  from u to v which consists only of white vertices
  (vertices which were not discovered yet)

3
Reminder the parenthesis theorem

• The parenthesis theorem
• Let G be a graph (directed or undirected) then
  after DFS on the graph
• For each two vertices u, v exactly one of the
  following is true
• u.d, u.f and v.d, v.f are disjoint.
• u.d, u.f ? v.d, v.f and u is a descendant of
  v.
• v.d, v.f ? u.d, u.f and v is a descendant of
  u.
• Immediate conclusion a vertex v is a descendant
  of a vertex u in the first-depth forest iff v.d,
  v.f ? u.d, u.f.

4
Reminder after DFS After DFS on directed
graph. Each vertex has a time stamp. The edges
in gray are the edges in the depth-first
forest. The first-depth forest of the above
graph. There is a correspondence between the
discover and finish times of each vertex and the
parenthesis structure below.
5
Strongly connected components

• Definition
• First we define the following binary relation
  between two vertices
• R(u,v) there is a path from u to v and a path
  from v to u.
• Note that this is an equivalence relation and the
  graph can be divided into equivalence classes
• Those classes are the strongly connected
  components of the graph.

6
SCC algorithm

• STRONGLY-CONNECTED-COMPONENTS(G)
• Call DFS(G) to compute finishing times u.f for
  each vertex u.
• Compute GT, by reversing the direction of the
  edges in G.
• Call DFS(GT), but in the main loop of DFS,
  consider the vertices in order of decreasing u.f
• The vertices of each tree in the last search form
  a SCC

7
Lemma 1

• Claim
• Let u and v be vertices of the same SCC, then
  each path between u and v is contained in this
  SCC.
• Proof
• Let u and v be vertices of the same SCC
• According to the SCC definition there is a path
  from u to v and a path from v to u.

8
Lemma 1 cont.

• Let w be some vertex on some path between u and
  v, u?w?v.
• According to the selection of w, there is a path
  from u to w.
• u and v belong to the same connected component.
  Thus, there is a path from v to u, v?u.
• Again from the selection of w, there is a path
  from w to v, w?v.
• By concatenating the paths w?v and v?u we get a
  path from w to u.
• That implies that u and w are in the same
  connected component.

9
Lemma 2

• Claim
• After DFS, all the vertices that belongs to the
  same SCC, belongs to a single tree in the depth
  forest.
• Proof
• Let r be the first vertex of the SCC that DFS
  discovers. In the time that DFS discovers r, all
  the other vertices in the same connected
  component are white. There is a path form r to
  each one of those vertices. According to the
  previous lemma, the path is contained in the SCC
  and thus white. According to the white path
  theorem, each vertex in the SCC becomes a
  descendant of r in the depth forest

10
Forefather - definition

• Definition
• Let G be a graph with DFS finish times.
• Let ?(u)argmaxww.f s.t. u?w
• We will say that ?(u) is the forefather of u.
• In other words, the forefather of u is the vertex
  that is reachable from u that finished last in
  the DFS.
• Note ?(u)u is possible.

11
Forefather - properties

• We will show that ?(?(u))?(u).
• for two vertices u, v it is obvious that
• u?v ? ?(v).f ? ?(u).f
• u??(u) by definition. Thus, ?(?(u)).f ? ?(u).f
• And of course ?(u).f ? ?(?(u)).f (by definition
  of forefather)
• Therefore, ?(u).f ?(?(u)).f which implies that
  ?(?(u))?(u).

12
SCC representative

• In the following two theorems we will prove that
  in each SCC there is one vertex which is the
  forefather of all the vertices in this SCC. This
  forefather is the representative of the SCC.
  It is the first vertex in the SCC which is
  discovered and the last that finishes. In the
  DFS on GT it is the root of the depth forest.
• Theorem 1
• ?(u) is an ancestor of u in the depth forest.
• Theorem 2
• Two vertices are in the same SCC iff they have
  the same forefather in DFS on G.

13
Theorem 1

• Theorem
• ?(u) is an ancestor of u in the depth forest.
• Conclusion
• u and ?(u) belongs to the same SCC. (immediate)
• Proof
• If ?(u)u we are done.
• Otherwise, let us consider the color of ?(u) in
  time u.d
• If ?(u) is black than ?(u).f ? u.f -
  contradiction!
• If ?(u) is gray than it is an ancestor of u and
  we are done.
• We only need to prove that ?(u) can not be white
  in time u.d

14
Theorem 1 cont.

• Assume that ?(u) is white. Let us consider the
  vertices on the path from u to ?(u).
• Case 1 all the vertices on the path are white
• Then, according to the white path theorem ?(u) is
  a descendant of u in the depth forest ? ?(u).f ?
  u.f, again contradiction.
• Case 2 there is a non-white vertex on the path
• Let t be the last vertex which is not white. t
  must be gray, since there is no edge from a black
  vertex to a white vertex. But than there is a
  white path from t to ?(u). Thus, ?(u).f ? t.f.
  Contradiction to the definition of ?(u).

15
Theorem 2

• Theorem
• Two vertices u,v are in the same SCC iff
  ?(u) ?(v)
• Proof
• Assume that ?(u)?(v). According to the
  conclusion of the previous theorem
• u and ?(u) are in the same SCC.
• v and ?(v) are in the same SCC.
• From transitivity, u and v are in the same SCC

16
Theorem 2 cont.

• ?Assume that u and v are in the same SCC.
• There is a path from u to v and vice versa.
  Thus, the set of the vertices that are reachable
  from u is equal to the set of vertices that are
  reachable from v. By definition of forefather,
  ?(u) ?(v).
• Now we have enough tools to understand the SCC
  algorithm.

17
DFS on GT, why ?

• Let us look on the vertex r, with the largest
  finishing time according to the times that were
  computed by the first DFS.
• By the definition of a forefather r must be one,
  since it is the forefather of itself. The SCC of
  r, as we proved, is all the vertices that r is
  their forefather. Since r has the largest
  finishing time, those are simply all the vertices
  that r can be reached from, or, in other words,
  all the vertices that can be reached from r in
  GT.
• Afterwards, the algorithm, continues by choosing
  the vertex r, which is the vertex with the
  largest finishing time that is not in the same
  SCC as r. Again r is the forefather of itself
  and the same process is repeated.

18
SCC correctness

• Theorem
• The procedure SCC(G) correctly computes the
  strongly connected components of a graph G.
• Proof
• By induction on the number of depth trees in DFS
  on GT.
• Each step proves that a depth tree that is
  constructed in the DFS is a SCC, assuming that
  all the previous trees are SCC.
• Induction basis Trivial, The first tree has no
  previous trees, so the assumption is true.

19
SCC correctness cont.

• Let us look at a depth tree T, with a root r
  which is created in DFS on GT. We will denote
  C(r)v ?(v)r
• Now we prove that u?T ? u?C(r).
• Assume u?C(r).
• From theorem 2 we know that u is in the same SCC
  as r. From lemma 2, u will be in the same depth
  forest as r.
• ?Assume u?T
• We will show that for each vertex w such that
  ?(w)?r, w is not in T and since u?T, ?(u) will be
  r.

20
SCC correctness cont.

• If ?(w).f gt r.f then by induction w is already in
  the tree which is rooted by ?(w) and cant be in
  the current tree.
• If ?(w).f lt r.f. If w is in T that implies w?r,
  and by definition of forefather, r would be its
  forefather, since its finish time is larger.
  Contradiction !