preorder: visit each node before its children.
postorder: visit each node after its children.
inorder (for binary trees only): visit left subtree, node, right subtree.
We also saw another kind of traversal, topological ordering, when I talked about shortest paths.
Today, we'll see two other traversals: breadth first search (BFS) and depth first search (DFS). Both of these construct spanning trees with certain properties useful in other graph algorithms. We'll start by describing them in undirected graphs, but they are both also very useful for directed graphs.
breadth first search:
unmark all vertices
choose some starting vertex x
mark x
list L = x
tree T = x
while L nonempty
choose some vertex v from front of list
visit v
for each unmarked neighbor w
mark w
add it to end of list
add edge vw to T
It's very important that you remove vertices from the other end of
the list than the one you add them to, so that the list acts as a
queue (fifo storage) rather than a stack (lifo). The "visit v" step
would be filled out later depending on what you are using BFS for,
just like the tree traversals usually involve doing something at
each vertex that is not specified as part of the basic algorithm.
If a vertex has several unmarked neighbors, it would be equally
correct to visit them in any order. Probably the easiest method to
implement would be simply to visit them in the order the adjacency
list for v is stored in.
Let's prove some basic facts about this algorithm. First, each vertex is clearly marked at most once, added to the list at most once (since that happens only when it's marked), and therefore removed from the list at most once. Since the time to process a vertex is proportional to the length of its adjacency list, the total time for the whole algorithm is O(m).
We also want to know that T is a spanning tree, i.e. that if the graph is connected (every vertex has some path to the root x) then every vertex will occur somewhere in T. We can prove this by induction on the length of the shortest path to x. If v has a path of length k, starting v-w-...-x, then w has a path of length k-1, and by induction would be included in T. But then when we visited w we would have seen edge vw, and if v were not already in the tree it would have been added.
Breadth first traversal of G corresponds to some kind of tree traversal on T. But it isn't preorder, postorder, or even inorder traversal. Instead, the traversal goes a level at a time, left to right within a level (where a level is defined simply in terms of distance from the root of the tree). For instance, the following tree is drawn with vertices numbered in an order that might be followed by breadth first search:
1
/ | \
2 3 4
/ \ |
5 6 7
| / | \
8 9 10 11
The proof that vertices are in this order by breadth first search
goes by induction on the level number. By the induction hypothesis,
BFS lists all vertices at level k-1 before those at level k.
Therefore it will place into L all vertices at level k before all
those of level k+1, and therefore so list those of level k before
those of level k+1. (This really is a proof even though it sounds
like circular reasoning.)
Breadth first search trees have a nice property: Every edge of G can be classified into one of three groups. Some edges are in T themselves. Some connect two vertices at the same level of T. And the remaining ones connect two vertices on two adjacent levels. It is not possible for an edge to skip a level.
Therefore, the breadth first search tree really is a shortest path tree starting from its root. Every vertex has a path to the root, with path length equal to its level (just follow the tree itself), and no path can skip a level so this really is a shortest path.
A second use of breadth first search arises in certain pattern matching problems. For instance, if you're looking for a small subgraph such as a triangle as part of a larger graph, you know that every vertex in the triangle has to be connected by an edge to every other vertex. Since no edge can skip levels in the BFS tree, you can divide the problem into subproblems, in which you look for the triangle in pairs of adjacent levels of the tree. This sort of problem, in which you look for a small graph as part of a larger one, is known as subgraph isomorphism. In a recent paper, I used this idea to solve many similar pattern-matching problems in linear time.
preorder(node v)
{
visit(v);
for each child w of v
preorder(w);
}
To turn this into a graph traversal algorithm, we basically replace
"child" by "neighbor". But to prevent infinite loops, we only want
to visit each vertex once. Just like in BFS we can use marks to
keep track of the vertices that have already been visited, and not
visit them again. Also, just like in BFS, we can use this search to
build a spanning tree with certain useful properties.
dfs(vertex v)
{
visit(v);
for each neighbor w of v
if w is unvisited
{
dfs(w);
add edge vw to tree T
}
}
The overall depth first search algorithm then simply initializes a
set of markers so we can tell which vertices are visited, chooses a
starting vertex x, initializes tree T to x, and calls dfs(x). Just
like in breadth first search, if a vertex has several neighbors it
would be equally correct to go through them in any order. I didn't
simply say "for each unvisited neighbor of v" because it is very
important to delay the test for whether a vertex is visited until
the recursive calls for previous neighbors are finished.
Just like we did for BFS, we can use DFS to classify the edges of G into types. Either an edge vw is in the DFS tree itself, v is an ancestor of w, or w is an ancestor of v. (These last two cases should be thought of as a single type, since they only differ by what order we look at the vertices in.) What this means is that if v and w are in different subtrees of v, we can't have an edge from v to w. This is because if such an edge existed and (say) v were visited first, then the only way we would avoid adding vw to the DFS tree would be if w were visited during one of the recursive calls from v, but then v would be an ancestor of w.
As an example of why this property might be useful, let's prove the following fact: in any graph G, either G has some path of length at least k. or G has O(kn) edges.
Proof: look at the longest path in the DFS tree. If it has length at least k, we're done. Otherwise, since each edge connects an ancestor and a descendant, we can bound the number of edges by counting the total number of ancestors of each descendant, but if the longest path is shorter than k, each descendant has at most k-1 ancestors. So there can be at most (k-1)n edges.
This fact can be used as part of an algorithm for finding long paths in G, another subgraph isomorphism problem closely related to the traveling salesman problem. If k is a small constant (like say 5) you can find paths of length k in linear time (measured as a function of n). But measured as a function of k, the time is exponential, which isn't surprising because this problem is closely related to the traveling salesman problem. For more on this particular problem, see Michael R. Fellows and Michael A. Langston, "On search, decision and the efficiency of polynomial-time algorithms", 21st ACM Symp. Theory of Computing, 1989, pp. 501-512.
With a little care it is possible to make BFS and DFS look almost the same as each other (as similar as, say, Prim's and Dijkstra's algorithms are to each other):
bfs(G)
{
list L = empty
tree T = empty
choose a starting vertex x
search(x)
while(L nonempty)
remove edge (v,w) from start of L
if w not yet visited
{
add (v,w) to T
search(w)
}
}
dfs(G)
{
list L = empty
tree T = empty
choose a starting vertex x
search(x)
while(L nonempty)
remove edge (v,w) from end of L
if w not yet visited
{
add (v,w) to T
search(w)
}
}
search(vertex v)
{
visit(v);
for each edge (v,w)
add edge (v,w) to end of L
}
Both of these search algorithms now keep a list of
edges to explore; the only difference between the two is that,
while both algorithms adds items to the end of L, BFS removes them
from the beginning, which results in maintaining the list as a
queue, while DFS removes them from the end, maintaining the list as
a stack.
For directed graphs, too, we can prove nice properties of the BFS and DFS tree that help to classify the edges of the graph. For BFS in directed graphs, each edge of the graph either connects two vertices at the same level, goes down exactly one level, or goes up any number of levels. For DFS, each edge either connects an ancestor to a descendant, a descendant to an ancestor, or one node to a node in a previously visited subtree. It is not possible to get "forward edges" connecting a node to a subtree visited later than that node. We'll use this property next time to test if a directed graph is strongly connected (every vertex can reach every other one).
ICS 161 -- Dept.
Information & Computer Science -- UC Irvine
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