If these ideas are new, begin with:
Understand the idea
Define a path's cost as the sum of its edge weights. Clarify direction, unreachable nodes, and whether constraints such as stops or remaining fuel belong in the state. One distance per vertex is insufficient when future choices depend on additional state.
Work through an example
Cheapest route with at most k stops needs (vertex,edges_used), because reaching a vertex cheaply after exhausting the budget can be worse than a pricier early arrival.
Time and space complexity
Model size determines complexity; expanded states multiply vertices and edges.
Why it works
Distances represent the cheapest known cost for the complete chosen state.
Common mistakes
- Minimizing edge count is different from minimizing weight.
- Ignoring resource state can discard useful paths.
Chapter lab · Python implementation
Weighted objectives require precise distinctions. Dijkstra finds cheapest routes from one source with nonnegative weights. Bellman-Ford permits negative edges and detects reachable negative cycles. A minimum spanning tree minimizes the total edges needed to connect all vertices; it does not minimize every source-to-vertex route.
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.
import heapq
def dijkstra(graph,start):
if any(w<0 for edges in graph.values() for _,w in edges):
raise ValueError('nonnegative weights required')
distances={start:0}; heap=[(0,start)]
while heap:
cost,node=heapq.heappop(heap)
if cost!=distances[node]: continue
for neighbor,weight in graph.get(node,[]):
candidate=cost+weight
if candidate<distances.get(neighbor,float('inf')):
distances[neighbor]=candidate
heapq.heappush(heap,(candidate,neighbor))
return distances
def bellman_ford(n,edges,start):
d=[float('inf')]*n; d[start]=0
for _ in range(n-1):
changed=False
for a,b,w in edges:
if d[a]!=float('inf') and d[a]+w<d[b]:
d[b]=d[a]+w; changed=True
if not changed: break
if any(d[a]!=float('inf') and d[a]+w<d[b] for a,b,w in edges):
raise ValueError('reachable negative cycle')
return dRun 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',4),('C',1)],'C':[('B',2)],'B':[]}
assert dijkstra(g,'A')=={'A':0,'B':3,'C':1}
assert bellman_ford(3,[(0,1,4),(0,2,1),(2,1,-2)],0)==[0,-1,1]
try: bellman_ford(2,[(0,1,-1),(1,0,-1)],0)
except ValueError: pass
else: raise AssertionError('negative cycle not detected')What if the destination is unreachable?
Further reading
Original documentation for the implementation details used in this chapter.
Princeton shortest paths Princeton minimum spanning trees