If these ideas are new, begin with:
Understand the idea
Breadth-first search processes vertices by nondecreasing distance from the source. First discovery gives a shortest unweighted path; parent pointers reconstruct one such path.
Work through an example
The lab discovers B,C at distance 1 and D at distance 2. D's chosen parent depends on neighbor order, but distance does not.
Time and space complexity
O(V+E) time/O(V) state for adjacency lists.
Why it works
Before processing distance d+1, all queued distance d vertices are processed.
Common mistakes
- Weighted edges invalidate the layer-cost argument.
- Unreachable nodes need an explicit infinity or absent representation.
Chapter lab · Python implementation
A graph models entities and connections; edges can be directed or undirected, weighted or unweighted. Our lab uses adjacency lists and marks BFS vertices when enqueuing. The diagram has a cycle A-B-D-C-A, so visited state is essential. Distances count edges because every edge has equal cost.
This shared lab connects the chapter's topics. Follow the current lesson's worked example first, then trace the complete implementation. Additional algorithms described in topic exercises are variations you can implement independently.
from collections import deque
def bfs(graph,start):
distances={start:0}
parent={start:None}
queue=deque([start])
while queue:
node=queue.popleft()
for neighbor in graph.get(node,[]):
if neighbor not in distances:
distances[neighbor]=distances[node]+1
parent[neighbor]=node
queue.append(neighbor)
return distances,parent
def components(graph):
seen=set(); count=0
for start in graph:
if start in seen: continue
count+=1; seen.add(start); stack=[start]
while stack:
node=stack.pop()
for neighbor in graph[node]:
if neighbor not in seen:
seen.add(neighbor); stack.append(neighbor)
return countRun and test the lab
Copy the implementation and these assertions into a Python 3 file and run it. An assertion failure means the actual result differs from the expectation. Add the current topic's exercise and boundary tests to deepen your understanding.
g={'A':['B','C'],'B':['A','D'],'C':['A','D'],'D':['B','C'],'E':[]}
d,p=bfs(g,'A'); assert d=={'A':0,'B':1,'C':1,'D':2}
assert p['D'] in ['B','C']
assert components(g)==2
assert bfs({'X':[]},'X')[0]=={'X':0}