DSA Quiz
A minimum spanning objective and a shortest-path objective can return different answers on the same weighted connections. Explain the question before comparing their numbers.
Your brief
The fixture connects three locations with undirected edges. One algorithm minimizes the total cost of connecting all locations; another minimizes the cost of travelling from gate to pond. The shortest route need not be the route formed by walking through the selected spanning tree. A correct answer to one objective is not evidence that the other objective was solved. Predict both totals before running the source.
Run the code
Select Try in playground to load this complete source into the existing Python editor. It does not run automatically; choose Run when ready. These examples use only the Python standard library and small fictional inputs. They do not need database services, extra packages or personal files.
For local work, save the complete source as quiz.py and run python quiz.py. The local review uses Python 3.12.14; the browser uses the existing Pyodide 0.27.7 runtime. Keep inputs small, respect explicit table/tour budgets and account for recursion limits before increasing a fixture’s size.
Worked solution
Show the worked solution
def validate_edges(vertices, edges):
if len(set(vertices)) != len(vertices) or any(not isinstance(vertex, str) for vertex in vertices):
raise ValueError("Unique string vertices required")
known = set(vertices)
for first, second, weight in edges:
if first not in known or second not in known or type(weight) is not int:
raise ValueError("Invalid undirected edge")
import heapq
def prim_forest(vertices, edges):
validate_edges(vertices, edges)
adjacent = {vertex: [] for vertex in vertices}
for first, second, weight in edges:
adjacent[first].append((second, weight))
adjacent[second].append((first, weight))
seen, selected, total = set(), [], 0
for start in vertices:
if start in seen:
continue
pending = [(0, start, None)]
while pending:
weight, vertex, parent = heapq.heappop(pending)
if vertex in seen:
continue
seen.add(vertex)
if parent is not None:
selected.append((parent, vertex, weight))
total += weight
for target, next_weight in adjacent[vertex]:
if target not in seen:
heapq.heappush(pending, (next_weight, target, vertex))
return selected, total
def validate_graph(graph, weighted=False, nonnegative=False):
if not isinstance(graph, dict):
raise ValueError("Expected an adjacency dictionary")
if any(not isinstance(vertex, str) for vertex in graph):
raise ValueError("String vertex labels required")
for vertex, neighbors in graph.items():
for edge in neighbors:
if weighted:
target, weight = edge
if type(weight) is not int or (nonnegative and weight < 0):
raise ValueError("Invalid integer edge weight")
else:
target = edge
if target not in graph:
raise ValueError("Unknown target vertex")
import heapq
def dijkstra(graph, start):
validate_graph(graph, weighted=True, nonnegative=True)
if start not in graph:
raise ValueError("Unknown start vertex")
distance = {vertex: float("inf") for vertex in graph}
distance[start] = 0
parent, pending = {start: None}, [(0, start)]
while pending:
cost, vertex = heapq.heappop(pending)
if cost != distance[vertex]:
continue
for target, weight in graph[vertex]:
candidate = cost + weight
if candidate < distance[target]:
distance[target] = candidate
parent[target] = vertex
heapq.heappush(pending, (candidate, target))
return distance, parent
vertices = ["gate", "oak", "pond"]
edges = [("gate", "oak", 2), ("oak", "pond", 2), ("gate", "pond", 3)]
_, connection_cost = prim_forest(vertices, edges)
graph = {vertex: [] for vertex in vertices}
for first, second, weight in edges:
graph[first].append((second, weight))
graph[second].append((first, weight))
distance, _ = dijkstra(graph, "gate")
print(connection_cost, distance["pond"])
Expected result
4 3
Connecting every location costs four, while the direct shortest gate-to-pond route costs three. The numerical difference reflects different optimization problems, not a disagreement between two interchangeable solvers. Add constraints or change the meaning of weights only after restating the objective.
Time, space and assumptions
This example pays for both a lazy-heap forest calculation and a lazy-heap shortest-path calculation. The added adjacency construction is O(V+E); their separate costs should not be hidden under one invented universal graph bound.
Extend and test
Construct a graph where a spanning-tree route is strictly longer than the direct shortest route, then show both valid witnesses and explain their distinct objectives.
Consult the Python 3.12 standard-library reference for container and language contracts; an example result is not a proof of every possible input or real-world performance.