# Breadth first search and depth first search

*转载自ICS161,学英语,自娱自乐*

# Traversal of graphs and digraphs

To traverse(遍历) means to visit the vertices in some systematic order. You should be familiar with various traversal methods for trees:

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(拓扑排序), 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

This can be throught of as being like Dijkstra's algorithm for shortest paths, but with every edge having the same length. However it is a lot simpler and doesn't need any data structures. We just keep a tree (the breadth first search tree), a list of nodes to be added to the tree, and markings (Boolean variables) on the vertices(顶点) to tell whether they are in the tree or list.

**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
```

C++实现:

```
int BFS()
{
queue<node> q;
node now, next;
now = start
q.push(now);//初始状态将起点放进队列Q
vis(now) = true;//设置节点已访问
while (!q.empty())//队列不为空，继续搜索
{
now = q.front();//取出队列的头Vn
q.pop();//从队列中移除
if (now == end)//找到终点
return ans;//返回结果
for(int i = 0; i < whatever; i++)
{
next = n + dir[i];
if (check(next) && !vis[next])
//Vw是一个合法的节点并且为白色节点
{
q.push(next);//加入队列Q
vis(next) = true;//设置节点颜色
}
}
}
return -1;//无解
}
```

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(先进先出) storage) rather than a stack (lifo(后进先出). 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(顶点) 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(实现) would be simply to visit them in the order the adjacency(毗邻) 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(成正比) to the length of its adjacency list, the total time for the whole algorithm is O(m).

Next, let's look at the tree T constructed by the algorithm. Why is it a tree? If you think of each edge vw as pointing "upward" from w to v, then each edge points from a vertex visited later to one visited earlier. Following successive(连续的) edges upwards can only get stopped at x (which has no edge going upward from it) so every vertex in T has a path to x. This means that T is at least a connected subgraph(子图) of G. Now let's prove that it's a tree. A tree is just a connected and acyclic(非循环的) graph, so we need only to show that T has no cycles. In any cycle, no matter how you orient(向东?) the edges so that one direction is "upward" and the other "downward", there is always a "bottom" vertex having two upward edges out of it. But in T, each vertex has at most one upward edge, so T can have no cycles. Therefore T really is a tree. It is known as a *breadth first search tree*.

We also want to know that T is a *spanning tree(生成树)*, i.e. that if the graph is connected (every vertex has some path to the root x) then every vertex will occur(存在) 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(假说), 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(循环论证).)

Breadth first search trees have a nice property(性质): 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.

Breadth first search has several uses in other graph algorithms, but most are too complicated to explain in detail here. One is as part of an algorithm for *matching*, which is a problem in which you want to pair up the n vertices of a graph by n/2 edges. If you have a partial matching(部分匹配), pairing up only some of the vertices, you can extend it by finding an *alternating path* connecting two unmatched vertices; this is a path in which every other edge is part of the partial matching. If you remove those edges in the path from the matching, and add the other path edges back into the matching, you get a matching with one more edge. Alternating paths can be found using a version of breadth first search.

A second use of breadth first search arises in certain pattern matching(特定模式匹配) 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(子图同构)*. In a recent paper, I used this idea to solve many similar pattern-matching problems in linear(线性) time.

# Depth first search

Depth first search is another way of traversing graphs, which is closely related to preorder traversal of a tree. Recall that preorder traversal simply visits each node before its children. It is most easy to program as a recursive routine(递归程序):

```
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
}
}
```

C++实现:

```
void dfs(node now)
{
vis(now) = true;//设置节点颜色
for (int i = 0; i < whatever; i++)
{
next = now + dir[i];
dfs(dx, dy);//递归调用
}
}
```

注意:上述代码可能不具有一般性!

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(递归) calls for previous neighbors are finished.

The proof that this produces a spanning tree (the *depth first search tree*) is essentially the same as that for BFS, so I won't repeat it. However while the BFS tree is typically "short and bushy"(短粗), the DFS tree is typically "long and stringy"(细长).

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(祖先) 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(后裔), 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.

# Relation between BFS and DFS

It may not be clear from the pseudo-code(伪代码) above, but BFS and DFS are very closely related to each other. (In fact in class I tried to describe a search in which I modified the "add to end of list" line in the BFS pseudocode to "add to start of list" but the resulting traversal algorithm was not the same as DFS.)

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.

# BFS and DFS in directed graphs

Although we have discussed them for undirected graphs, the same search routines work essentially unmodified for directed graphs. The only difference is that when exploring a vertex v, we only want to look at edges (v,w) going out of v; we ignore the other edges coming into v.

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).