Non-recursive Depth-First Search (DFS) Using a Sta

Question

Ok this is my first post on Stack Overflow I have been reading for a little while and really admire the site. I am hoping this is something that will be acceptable to ask. So I have been reading through Intro to Algorithms (Cormen. MIT Press) all the way through and I am up to the graph algorithms. I have been studying the formal algorithms laid out for breadth and depth first search in very fine detail.

Here is the psuedo-code given for depth-first search:

DFS(G)
-----------------------------------------------------------------------------------
1  for each vertex u ∈ G.V
2      u.color ← WHITE       // paint all vertices white; undiscovered
3      u.π ← NIL
4      time ← 0              // global variable, timestamps
5  for each vertex u ∈ G.V
6      if u.color = WHITE
7          DFS-VISIT(G,u)

DFS-VISIT(G, u)
-----------------------------------------------------------------------------------
1  u.color ← GRAY          // grey u; it is discovered
2  time ← time + 1 
3  u.d ← time
4  for each v ∈ G.Adj[u]   // explore edge (u,v)
5      if v.color == WHITE
6          v.π ← u
7          DFS-VISIT(G,v) 
8  u.color ← BLACK         // blacken u; it is finished
9  time ← time + 1
10 u.f ← time

This algorithm is recursive and it proceeds from multiple sources (will discovered every vertex in unconnected graph). It has several properties that most language specific implementations might leave out. It distinguishes between 3 different 'colors' of vertices. It initially paints all of them white, then when they are 'discovered' (visited in DFS-VISIT) they are then painted gray. The DFS-VISIT algorithm runs a loop recursively calling itself on the adjacency list of the current vertex and only paints a vertex black when it has no more edges to any white node.

Also two other attributes of each vertex are maintained u.d and u.f are the time stamps to when the vertex was discovered (painted gray) or when a vertex is finished (painted black). The first time a node is painted it has a time stamp of one and it is incremented to the next integer value for each time another is painted (whether it be grey or black). u.π is also maintained which is just a pointer to the node from which u was discovered.

Algorithm Non-Recursive-DFS(G)
-----------------------------------------------------------------------------------
1   for each vertex u ∈ G.V
2       u.color ← WHITE
3       u.π ← NIL
4   time = 0
5   for each vertex u ∈ G.V
6       if u.color = WHITE
7           u.color ← GRAY
8           time ← time + 1
9           u.d ← time
7           push(u, S)
8           while stack S not empty
9               u ← pop(S)
10              for each vertex v ∈ G.Adj[u]
11                  if v.color = WHITE
12                      v.color = GRAY
13                      time ← time + 1
14                      v.d ← time
15                      v.π ← u
16                      push(v, S)
17              u.color ← BLACK 
18              time ← time + 1
19              u.f ← time

* EDIT 10/31/12 * This is embarrassing that my algorithm has been incorrect for so long, it would work in most cases, but not all. I just got a popular question badge for the question and I saw where Irfy had spotted the problem in his answer below, so that is where the credit goes. I am just fixing it here for anyone that needs this in the future.

Does anyone see a flaw with the above algorithm? I am by far no expert on graph theory, but I think I have a pretty good grasp on recursion and iteration and I believe this should work the same. I am looking to make it functionally equivalent to the recursive algorithm. It should maintain all the attributes of the first algorithm even if they are not needed.

When I began writing it I did not think I would have a triply loops but that's the way it turned out. I have seen other iterative algorithms when I looked around Google that only have a doubly nested loop, however, they do not appear to proceed from multiple sources. (i.e. they will not discover every vertex of unconnected graph)

Answer 1

I believe that there is at least one case where the recursive and stack versions are not functionally equivalent. Consider the case of a Triangle - vertices A, B, and C connected to each other. Now, with the Recursive DFS, the predecessor graph that one would obtain with source A, would be either A->B->C OR A->C->B ( A->B implies A is the parent of B in depth first tree).

However, if you use the stack version of DFS, parents of both B and C, would always be recorded as A. It can never be the case that parent of B is C or vice-versa (which is always the case for recursive DFS). This is because, when exploring the adjacency list of any vertex (here A), we push all the members of adjacency list (here B and C) in one go.

This may become relevant, if you try to use DFS for finding articulation points in a graph[1]. One example would be that the following statement holds true only if we use the recursive version of DFS.

A root vertex is an articulation point if and only if it has at least two children in the depth first tree.

In a triangle, there is obviously no articulation point, but the stack-DFS still gives two children for any source vertex in the depth-first tree (A has children B and C). It's only if we create the depth first tree using recursive DFS that the above statement holds true.

[1] Introduction to Algorithms, CLRS - Problem 22-2 (Second and Third Edition)

Answer 2

In the non-recursive version we need another color that reflects the state in the recursive stack. So, we will add a color=RED to indicate all children of the node were pushed to the stack. I will also assume the stack has a peek() method(which otherwise could be simulated with pop and immediate push)

So, with that addition the updated version of original post should look like:

for each vertex u ∈ G.V
      u.color ← WHITE
      u.π ← NIL
  time = 0
  for each vertex u ∈ G.V
      if u.color = WHITE
          u.color ← GRAY
          time ← time + 1
          u.d ← time
          push(u, S)
          while stack S not empty
              u ← peek(S)
              if u.color = RED
                  //means seeing this again, time to finish
                  u.color ← BLACK
                  time ← time + 1
                  u.f ← time
                  pop(S) //discard the result
              else
                  for each vertex v ∈ G.Adj[u]
                     if v.color = WHITE
                         v.color = GRAY
                         time ← time + 1
                         v.d ← time
                         v.π ← u
                         push(v, S)
                   u.color = RED