If these ideas are new, begin with:
Understand the idea
Vertices are entities; edges encode a relation. Direction, weight, parallel edges, and self-loops affect valid algorithms. Model exactly what one traversal step means before choosing a method.
Work through an example
A friendship edge is usually undirected; a prerequisite A→B is directed. Travel time needs weights, while one move in an equal-cost grid can use unweighted edges.
Time and space complexity
V vertices/E edges determine graph bounds; directed E counts each stored directed edge.
Why it works
The graph contains precisely the transitions permitted by the problem.
Common mistakes
- Treating one-way roads as undirected invents routes.
- Reachability differs from cheapest-path distance.
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}