If these ideas are new, begin with:
Understand the idea
A grid is an implicit graph. Cells are vertices; valid neighbor moves are edges. Coordinate bounds and blocked-cell rules define the relation without building adjacency lists explicitly.
Work through an example
Four-way moves from (1,1) are (0,1),(2,1),(1,0),(1,2). Diagonal moves are excluded unless specified. Multi-source BFS begins with all sources at distance zero.
Time and space complexity
O(RC) for constant-degree grid traversal; O(RC) visited/queue worst case.
Why it works
Only valid, permitted cells are scheduled; each is discovered once.
Common mistakes
- Diagonal versus orthogonal connectivity changes components.
- Check bounds before indexing.
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}