1. Breadth-First Search L0 L1 L2 C B A E D F Breadth-First Search
4/18/2019 5:55 PM
Breadth-First Search C B A E D L0 L1 F L2
Breadth-First Search

2. Breadth-First Search
Breadth-first search (BFS) is a general technique for traversing a graph
Similar to breadth-first traversal in trees
“level by level”
“children” -> “neighbors”
Level is number of edges from the starting node
Visit each node only once
Breadth-First Search

3. Example unexplored vertex visited vertex unexplored edgeC B A E D L0 L1 F A
unexplored vertex A
visited vertex
unexplored edge
discovery edge
cross edge L0 L0 A A L1 L1 B C D B C D E F E F
Breadth-First Search

4. Example (cont.) L0 L1 L0 L1 L2 L0 L1 L2 L0 L1 L2 C B A E D F C B A E D
Breadth-First Search

5. Example (cont.) L0 L1 L2 L0 L1 L2 L0 L1 L2 C B A E D F A B C D E F C B
Breadth-First Search

6. BFS Algorithm Algorithm BFS(G, s)
Queue  new empty queue
enqueue(Queue, s)
while Queue.isEmpty()
v  dequeue(Queue)
if getLabel(v) != VISITED // **
visit v
setLabel(v, VISITED)
neighbors  getAdjacentVertices(G, v)
for each w  neighbors if getLabel(w) != VISITED // “children” enqueue(Queue, w)
The algorithm uses a mechanism for setting and getting “labels” of vertices and edges
Container C in the book is a queue. **
An unexplored vertex might be might be put on the queue more than once
After the first one is visited, the rest can be skipped.
A queue doesn’t usually have an operation to find an entry on the queue
What is another way to check if an entry is already on the queue? Hint: getLabel()
Breadth-First Search

7. Analysis
Setting/getting a vertex/edge label takes O(1) time
Each vertex is labeled twice
once as UNEXPLORED
once as VISITED
O(n)
getAdjacentVertices(v) is called once for each vertex that is not visited
Each edge in the graph is used at most twice
Each edge has two vertices
O(m)
BFS runs in O(n + m) time provided the graph is represented by the adjacency list structure
[less precisely, since m is O(n2), BFS is also O(n2)]
Breadth-First Search

8. Properties Notation Property 1 Property 2 Property 3 (not in DFS)
Gs: connected component of s
Property 1
BFS(G, s) visits all the vertices and edges of Gs
Property 2
The discovery edges labeled by BFS(G, s) form a spanning tree Ts of Gs
Property 3 (not in DFS)
For each vertex v in Li
The path of Ts from s to v has i edges
Every path from s to v in Gs has at least i edges A B C D E F L0 A L1 B C D L2 E F
Breadth-First Search