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

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

Both algorithms are fine. The second one is a direct translation from recursive to stack-based. All mutating state are stored in the stack. G never changes during the execution of the algorithm.

The algorithms will produce a spanning tree for each disconnected region, based on the order the algorithm visited each node. The trees will be represented both with references to the parent-nodes (u.π), and as segment trees (u.d and u.f).

A child will have a reference to its parent node (or NULL if it is a root), as well as having it's range (child.d .. child.f) contained within it's parent's range.

parent.d < child.d < child.f < parent.f

child.π = parent

---

Edit: I found a mistake in the translation. You are not actually pushing the current state into the stack, but a future method argument. In addition, you are not coloring the popped nodes GRAY and setting the f field.

Here is a rewrite of the original first algorithm:

algorithm Stack-DFS(G)
    for each vertex u ∈ G.V
        u.color ← WHITE
        u.π ← NIL
    time ← 0
    S ← empty stack
    for each vertex u ∈ G.V
        if u.color = WHITE
            # Start of DFS-VISIT
            step ← 1
            index ← 0
            loop unconditionally
                if step = 1
                    # Before the loop
                    u.color ← GRAY
                    time ← time + 1
                    u.d ← time
                    step ← 2
                if step = 2
                    # Start/continue looping
                    for each vertex v ∈ G.Adj[u]
                        i ← index of v
                        if i ≥ index and v.color = WHITE
                            v.π ← u
                            # Push current state
                            push((u, 2, i + 1), S)
                            # Update variables for new call
                            u = v
                            step ← 1
                            index ← 0
                            # Make the call
                            jump to start of unconditional loop
                    # No more adjacent white nodes
                    step ← 3
                if step = 3
                    # After the loop
                    u.color ← BLACK
                    time ← time + 1
                    u.right ← time
                    # Return
                    if S is empty
                        break unconditional loop
                    else
                        u, step, index ← pop(S)

There is probably a few places that could be optimized, but should at least work now.

Result:

Name   d    f   π
q      1   16   NULL
s      2    7   q
v      3    6   s
w      4    5   v
t      8   15   q
x      9   12   t
z     10   11   x
y     13   14   t
r     17   20   NULL
u     18   19   r

Answer 2

int stackk[100];
int top=-1;
void graph::dfs(int v){
 stackk[++top]=v;
// visited[v]=1;
 while(top!=-1){
   int x=stackk[top--];
   if(!visited[x])
    {visited[x]=1;
     cout<<x<<endl;
    }
   for(int i=V-1;i>=0;i--)
   {
        if(!visited[i]&&adj[x][i])
        {   //visited[i]=1;
            stackk[++top]=i;
        }
   }
 }
}
void graph::Dfs_Traversal(){
 for(int i=0;i<V;i++)
  visited[i]=0;
 for(int i=0;i<V;i++)
  if(!visited[i])
    g.dfs(i);